├── .gitignore ├── Final report ├── 于峥期末报告.pdf ├── 傅凌玥+期末报告.pdf ├── 刘成锴+期末报告.pdf ├── 卢禹杰+期末报告.pdf ├── 吕优+期末报告.pdf ├── 吴怀瑾+期末报告.pdf ├── 吴润哲期末报告.pdf ├── 唐泽-期末报告.pdf ├── 姚远+期末报告.pdf ├── 寻之扬期末报告.pdf ├── 庄永昊 期末报告.pdf ├── 张宇恒 期末报告.pdf ├── 张志成+期末报告.pdf ├── 张扬天+期末报告.pdf ├── 李云卿+期末报告.pdf ├── 李孜睿期末报告.pdf ├── 杨宗翰+期末报告+fix2.pdf ├── 柳志轩 期末报告.pdf ├── 毛昕渝-期末报告.pdf ├── 王一+期末报告.pdf ├── 董海辰+期末报告.pdf ├── 计家宝+期末报告.pdf ├── 贾雨祺-期末报告.pdf ├── 赖睿航-期末报告.pdf ├── 郭林松+期末报告.pdf ├── 郭睿涵+期末报告.pdf ├── 金弘义+期末报告.pdf ├── 陆嘉馨 期末报告.pdf ├── 陈彤+期末报告.pdf ├── 顾逸+期末报告.pdf └── 马浩博+期末报告.pdf ├── README.md ├── Week1 ├── (quiz)boom_balls_childrenSelect_董海辰 │ ├── quizzes.pdf │ └── quizzes.tex ├── A proof of A^A = 2^A and A = AA-毛昕渝 │ ├── discussion.pdf │ ├── discussion.tex │ └── ref.bib ├── AC&WellOrder&Zorn_庄永昊 │ ├── 10.pdf │ └── 10.tex ├── A_proof_of_A^A=2^A_吴润哲 │ ├── A proof of A^A=2^A.pdf │ └── A proof of A^A=2^A.tex ├── An Introduction for Axiom of Choice_姚远 │ ├── An Introduction for Axiom of Choice.pdf │ └── An Introduction for Axiom of Choice.tex ├── BijectionBetween[0,1]andR_计家宝 │ ├── BijectionBetween[0,1]andR.pdf │ ├── BijectionBetween[0,1]andR.tex │ ├── WasCantorSurprised_AfterReading.md │ └── cantor.png ├── BijectionFrom01toR 刘成锴 │ ├── HW1.pdf │ └── HW1.tex ├── Bijection_between_[0, 1]_and_R_姚远 │ ├── Bijection_between_[0, 1]_and_R.pdf │ └── Bijection_between_[0, 1]_and_R.tex ├── Construction_of_4_sets_to_satisfies_3_rules_马浩博 │ ├── discussion.pdf │ └── discussion.tex ├── Expansion of uniqueness theorem_郭睿涵 │ ├── Expansion of uniqueness theorem_郭睿涵.pdf │ ├── Expansion of uniqueness theorem_郭睿涵.tex │ └── tmp ├── Further discussion of Exercise 4_傅凌玥 │ ├── Further discussion of Exercise 4.pdf │ └── Further discussion of Exercise 4.tex ├── Parallel_sets_don't_exist_张志成 │ ├── parallel_sets_don't_exist.pdf │ └── parallel_sets_don't_exist.tex ├── Poset_R^3_Cannot_Be_Stuffed_into_R^2_寻之扬 │ ├── Poset R^3 Cannot Be Stuffed into R^2.pdf │ └── Poset R^3 Cannot Be Stuffed into R^2.tex ├── dual_Cantor-Bernstein_theorem_李孜睿 │ ├── dual_Cantor-Bernstein_Theorem_李孜睿.pdf │ ├── dual_Cantor-Bernstein_Theorem_李孜睿.tex │ └── 参考资料：The_Dual_Cantor-Bernstein_Theorem_and_the_Partition_Principle.pdf ├── exercise3&4_刘成锴 │ ├── HW1.pdf │ └── HW1.tex ├── guessing_discontinuity_chain_liar_庄永昊_金弘义_赖睿航_董海辰 │ ├── discuss.pdf │ └── discuss.tex ├── quiz_balls_吕优 │ ├── quiz.pdf │ └── quiz.tex ├── quiz_balls_吴怀瑾 │ ├── quiz_balls_吴怀瑾.pdf │ └── quiz_balls_吴怀瑾.tex ├── quiz_fair_dice_张志成 │ ├── factor.png │ ├── fair_dice.pdf │ ├── fair_dice.tex │ ├── gen.py │ └── solve.png ├── quiz_shuffle_cards_陈彤 │ ├── poker.pdf │ └── poker.tex └── quiz_柳志轩 │ ├── quiz_柳志轩.pdf │ └── quiz_柳志轩.tex ├── Week10 ├── Incomplete subspace of L^2 │ ├── proof.pdf │ └── proof.tex ├── Kullback–Leibler divergence greater or equal to 0-马浩博 │ ├── discussion.pdf │ └── discussion.tex ├── Notes of Expectation 刘成锴 │ ├── Expection.md │ └── Expection.pdf └── Rogers and Hoelder_董海辰 │ ├── An extension of a certain theorem in inequalities.pdf │ ├── Hoelder.jpg │ ├── Rogers and Hoelder.pdf │ ├── Rogers and Hoelder.tex │ ├── Rogers.jpg │ └── Ueber einen Mittelwertsatz.pdf ├── Week11 ├── E9.2 Symmetry Conditional Probability 郭睿涵 │ ├── E9.2 Symmetry Conditional Probability 郭睿涵.pdf │ ├── E9.2 Symmetry Conditional Probability 郭睿涵.tex │ └── tmp ├── E9_Conditional_Expectation_张志成 │ ├── Exercise_E9_Conditional_Expectation.pdf │ └── Exercise_E9_Conditional_Expectation.tex ├── Proof of a Simple Lemma 马浩博 │ ├── discussion.pdf │ └── discussion.tex ├── Symmetry Conditional Expectation │ ├── Symmetry Conditional Expectation.pdf │ └── Symmetry Conditional Expectation.tex └── The Distribution Function of X1 Divided by X2_吴润哲 │ ├── The Distribution Function of X1 Divided by X2.pdf │ └── The Distribution Function of X1 Divided by X2.tex ├── Week12 ├── 5.22note_Conditional_Expectation_and_Martingale_郭睿涵 │ ├── 5.22note_Conditional_Expectation_and_Martingale_郭睿涵.pdf │ ├── 5.22note_Conditional_Expectation_and_Martingale_郭睿涵.tex │ └── tmp ├── E10.1 Polya Urn 郭林松 │ ├── Ploya Urn.pdf │ └── Polya Urn.tex ├── Ex10.2 Martingale formulation of Bellman's Optimality Principle_王一 │ ├── Ex10.2 Martingale formulation of Bellman's Optimality Principle.pdf │ ├── Ex10.2 Martingale formulation of Bellman's Optimality Principle.synctex.gz │ ├── Ex10.2 Martingale formulation of Bellman's Optimality Principle.tex │ └── hw.sty ├── Ex10.6 ABRACADABRA_顾逸 │ ├── Doob's Optional Stopping Theorem and the ABRACADABRA Problem.pdf │ └── Doob's Optional Stopping Theorem and the ABRACADABRA Problem.tex ├── Gamblers Ruin_姚远 │ ├── Gamblers Ruin.pdf │ └── Gamblers Ruin.tex └── optional time operation │ ├── optional time operation.pdf │ └── optional time operation.tex ├── Week14 ├── An Extended Super-Martingale Convergence Theorem with its Proof │ └── An_Extended_Super_Martingale_Convergence_Theorem_with_its_Proof_张扬天.pdf └── σ-algebra Generated By Stopping Time - 郭林松 │ ├── main.tex │ └── σ-algebra Generated By Stopping Time.pdf ├── Week2 ├── A_phi is negligible-寻之扬 │ ├── A_phi is negligible.pdf │ └── A_phi is negligible.tex ├── Cantor-meager-金弘义 │ ├── cantor-meager.pdf │ └── cantor-meager.tex ├── Exercise2 Cantor Set-郭林松 │ ├── discussion.pdf │ └── discussion.tex ├── Lecture Notes-董海辰 │ ├── notes.pdf │ ├── notes.tex │ └── w1fig.pdf ├── Liouville Number-张志成 │ ├── Liouville_Number.pdf │ └── Liouville_Number.tex ├── Principle of inclusion-exclusion-陈彤 │ ├── ex12.pdf │ └── ex12.tex ├── R is uncountable-寻之扬 │ ├── R is uncountable.pdf │ └── R is uncountable.tex ├── SimplyNormalToBase10-计家宝 │ ├── ProveSimplyNormalToBase10.pdf │ └── ProveSimplyNormalToBase10.tex ├── Union-of-countable-intervals_and_baby-vitali-lemma-赖睿航 │ ├── solution.pdf │ └── solution.tex └── WLLN and SLLN-毛昕渝 │ ├── WLLN and SLLN.pdf │ └── WLLN and SLLN.tex ├── Week3 ├── B([a,b]) equals sigma(pi[a,b]) │ ├── Borel&sigma(pi).pdf │ └── Borel&sigma(pi).tex ├── B([a,b])_equals_σ(π[a,b])-赖睿航 │ ├── B([a,b]).md │ ├── B([a,b]).pdf │ ├── B([a,b])_equals_σ(π[a,b]).pdf │ └── B([a,b])_equals_σ(π[a,b]).tex ├── Be careful about interchanging order of set operations-毛昕渝 │ ├── the order matters.pdf │ └── the order matters.tex ├── BorelSetOn[a,b]-is-generated-by-pi-system-of-[a,b]-计家宝 │ ├── BorelSetOn[a,b]-is-generated-by-pi-system-of-[a,b]-计家宝.pdf │ └── BorelSetOn[a,b]-is-generated-by-pi-system-of-[a,b]-计家宝.tex ├── Lecture Notes of Measure Space_傅凌玥 │ ├── Measure Space.pdf │ └── Measure Space.tex ├── Non-Lebesgue-measurable set in R^n_吴润哲 │ ├── non-Lebesgue-measurable-sets-in-R^n.pdf │ └── non-Lebesgue-measurable-sets-in-R^n.tex ├── Notes of Measure Space 刘成锴 │ ├── Measure Space.md │ └── Measure Space.pdf ├── Notes of Random Variable 刘成锴 │ ├── Random Variables.md │ └── Random Variables.pdf ├── Trivial proofs of Inclusion-exclusion Principle_傅凌玥 │ ├── Trival Proof of Inclusion-exclusion Principle.pdf │ └── Trival Proof of Inclusion-exclusion Principle.tex └── Usual length function is countably additive_董海辰 │ ├── additivity.pdf │ └── additivity.tex ├── Week4 ├── A problem of indicator function and lim - 陆嘉馨 │ ├── A problem of indicator function and lim - 陆嘉馨.pdf │ └── A problem of indicator function and lim - 陆嘉馨.tex ├── Can_event_space_countably_infinite-郭睿涵 │ ├── Can_event_space_countably_infinite.pdf │ └── Can_event_space_countably_infinite.tex ├── Distrubution_func_man_not_be_left_continuous-&-h∈mΣ-iff-h+h-∈mΣ-赖睿航 │ ├── Distrubution_func_may_not_be_left_continuous-&-h∈mΣ-iff-h+h-∈mΣ-赖睿航.pdf │ └── Distrubution_func_may_not_be_left_continuous-&-h∈mΣ-iff-h+h-∈mΣ-赖睿航.tex ├── How_An_is_constructed_in_PR_THM_卢禹杰 │ ├── How_An_is_constructed_in_PR_THM.pdf │ └── How_An_is_constructed_in_PR_THM.tex ├── Indicator function and Infinite events-计家宝 │ ├── Indicator function and Infinite events-计家宝.pdf │ └── Indicator function and Infinite events-计家宝.tex ├── Indicator function for lim sup and lim inf-郭林松 │ ├── Indicator function for lim sup and lim inf.pdf │ └── Indicator function for lim sup and lim inf.tex ├── Intersection of Uncountable a.s. Events May Be Empty-马浩博 │ ├── discussion.pdf │ └── discussion.tex ├── Notes of Events 刘成锴 │ ├── Events.md │ └── Events.pdf ├── Notes of Events_傅凌玥 │ ├── Event.pdf │ └── Event.tex ├── PR_thm_for_incompressible_trans_卢禹杰 │ ├── PR_thm_for_incompressible_trans.pdf │ └── PR_thm_for_incompressible_trans.tex ├── Product_of_divergent_series - 张志成_寻之扬 │ ├── Product_of_divergent_series.pdf │ └── Product_of_divergent_series.tex ├── Proof of Poincare's Recurrence Theorem - 赖睿航 │ ├── poincare_recurrence_thm.pdf │ └── poincare_recurrence_thm.tex ├── Second Borel–Cantelli lemma - 陆嘉馨 │ ├── Second Borel–Cantelli lemma.pdf │ └── Second Borel–Cantelli lemma.tex ├── Sigma-measurable-functions-and-simple-functions-赖睿航 │ ├── sigma-measurable-functions-and-simple-functions.pdf │ └── sigma-measurable-functions-and-simple-functions.tex ├── Simple exercise of phi(1-y_n) = 0_吴怀瑾 │ ├── simple exercise of phi(1-y_n) = 0.pdf │ └── simple exercise of phi(1-y_n) = 0.tex ├── The Convergence of Summation and Product_吴润哲 │ ├── The Convergence of Summation and Product.pdf │ └── The Convergence of Summation and Product.tex ├── The application of positive and negative part in simple functions_金弘义 │ ├── discussion.pdf │ └── discussion.tex ├── The continuity of distribution function-郭林松 │ ├── The continuity of distribution function.pdf │ └── The continuity of distribution function.tex ├── Truth-set-of-the-Strong-Law-for- the-subsequence-α-于峥 │ ├── thinking.pdf │ └── thinking.tex ├── Unique_Extension_Lemma-寻之扬 │ ├── Unique_Extension_Lemma.pdf │ └── Unique_Extension_Lemma.tex └── distribution_function_may_not_be_left_continuous_金弘义 │ ├── discussion.pdf │ └── discussion.tex ├── Week5 ├── A proof of Doob Dynkin Lemma_吴润哲 │ ├── A proof of Doob Dynkin Lemma.pdf │ └── A proof of Doob Dynkin Lemma.tex ├── A_history_of_Independence_张扬天 │ ├── A_history_of_Independence.pdf │ └── A_history_of_Independence.tex ├── Independence of coin tossing events_董海辰 │ ├── Independence of coin tossing events.pdf │ └── Independence of coin tossing events.tex ├── Independent of Tail Sigma Algebra │ ├── Independent of Tail Sigma Algebra.pdf │ └── Independent of Tail Sigma Algebra.tex ├── Measurable_function_is_the_limit_of_a_sequence_of_simple_function_李孜睿 │ ├── Measurable_function_is_the_limit_of_a_sequence_of_simple_function_李孜睿.pdf │ └── Measurable_function_is_the_limit_of_a_sequence_of_simple_function_李孜睿.tex ├── Notes of Independence 刘成锴 │ ├── Independence.md │ └── Independence.pdf ├── Sigma-algebra-generated-by-r.v-赖睿航 │ ├── sigma-algebra-generated-by-r.v.pdf │ └── sigma-algebra-generated-by-r.v.tex ├── Simple functions_converging_to_measurable_function_张志成 │ ├── Simple functions_converging_to_measurable_function.pdf │ └── Simple functions_converging_to_measurable_function.tex ├── Surprise Test Paradox_杨宗翰 │ └── A naive research of the Surprise Test Paradox_杨宗翰.pdf ├── independence in infty_庄永昊_董海辰 │ ├── independenceInInfty.pdf │ └── independenceInInfty.tex └── independence_by_distrib_func_卢禹杰 │ ├── depict_independence_by_distrib_func.pdf │ └── depict_independence_by_distrib_func.tex ├── Week6 ├── An application of LLL-hypergraph 2-coloring-毛昕渝 │ ├── An application of LLL-hypergraph 2-coloring-毛昕渝.pdf │ ├── An application of LLL-hypergraph 2-coloring-毛昕渝.tex │ ├── ex.png │ └── ref.bib ├── Any graph can be a minimum dependency graph_杨宗翰.pdf ├── Consecutive_heads_in_coin_tossing_董海辰 │ ├── Consecutive_heads_in_coin_tossing.pdf │ └── Consecutive_heads_in_coin_tossing.tex ├── Converse to SLLN │ ├── Converse to SLLN.pdf │ └── Converse to SLLN.tex ├── E4.2-A disitribution on N related to zeta function-毛昕渝 │ ├── E.4.2.pdf │ ├── E.4.2.tex │ └── ref.bib ├── Events_in_tail_sigma_algebra_李孜睿 │ ├── Events_in_tail_sigma_algebra_李孜睿.pdf │ └── Events_in_tail_sigma_algebra_李孜睿.tex ├── Existence of iid random variables_金弘义 │ ├── Existence of iid variables.pdf │ └── Existence of iid variables.tex ├── Hewitt-Savage Zero-One Law and Random Walk-吴润哲 │ ├── Hewitt-Savage Zero-One Law and Random Walk.pdf │ └── Hewitt-Savage Zero-One Law and Random Walk.tex ├── Independent_normal_distribution_variable_sequence_张志成 │ ├── Independent_normal_distribution_variable_sequence.pdf │ └── Independent_normal_distribution_variable_sequence.tex ├── LLL and k-SAT_傅凌玥 │ ├── Lecure Notes of LLL.pdf │ └── Lecure Notes of LLL.tex ├── Lovasz Local Lemma and k-SAT problem - 陆嘉馨 │ ├── Lovasz Local Lemma and k-SAT problem - Lu Jiaxin.tex │ ├── Lovasz Local Lemma and k-SAT problem- Lu Jiaxin.pdf │ └── ref.bib ├── Lovász Local Lemma and its Application-郭林松 │ ├── Lovász Local Lemma and its Application-郭林松.pdf │ └── main.tex ├── Proof for Lovasz Local Lemma_郭睿涵_陈彤 │ ├── Proof for Lovasz Local Lemma_郭睿涵_陈彤.pdf │ └── Proof for Lovasz Local Lemma_郭睿涵_陈彤.tex ├── Proof of Extended BC1 马浩博 │ ├── discussion.pdf │ └── discussion.tex ├── What's_fair_about_a_fair_game │ ├── whats_fair_about_a_fair_game.pdf │ └── whats_fair_about_a_fair_game.tex ├── Zero-One Law Of First Order Logic 金弘义 │ ├── zero_one_law_of_fo.pdf │ └── zero_one_law_of_fo.tex ├── an extended BC1 庄永昊 │ ├── proof.pdf │ └── proof.tex └── bc1-extension-赖睿航 │ ├── bc1-extension.bib │ ├── bc1-extension.pdf │ └── bc1-extension.tex ├── Week7 ├── Exercise4.1_郭睿涵 │ ├── Exercise4.1_郭睿涵.pdf │ ├── Exercise4.1_郭睿涵.tex │ └── tmp ├── Independence of multiple pi-systems and their conditions - 陆嘉馨 │ ├── Independence of multiple pi-systems and their conditions - 陆嘉馨.pdf │ └── Independence of multiple pi-systems and their conditions - 陆嘉馨.tex ├── Independence of pi-system(E4.1)_傅凌玥，唐泽 │ ├── Independence of pi-system(E4.1).pdf │ └── Independence of pi-system(E4.1).tex ├── Integral-of-positive-simple-functions-赖睿航 │ ├── integral-of-positive-simple-functions.pdf │ └── integral-of-positive-simple-functions.tex ├── Note for Lebesgue integration-陈彤 │ ├── Note for Lebesgue integration.pdf │ └── Note for Lebesgue integration.tex ├── Notes of Integration 刘成锴 │ ├── Integration.md │ └── Integration.pdf ├── P(A_k,i.o.) of throw a coin(E4.4)_傅凌玥 │ ├── P(Ak,i.o.) of throw a coin(E4.4).pdf │ ├── P(Ak,i.o.) of throw a coin(E4.4).tex │ └── P(Ak,i.o.) of throw a coin(E4.4)_傅凌玥.pdf └── Proof of MON 马浩博 │ ├── discussion.pdf │ └── discussion.tex ├── Week8 ├── A proof of a μ-integrable lemma and h(fμ)=(hf)μ_吴润哲 │ ├── A proof of a mu-integrable lemma and h(f mu)=(hf)mu using the standard machine.pdf │ └── A proof of a mu-integrable lemma and h(f mu)=(hf)mu using the standard machine.tex ├── DOM's dominating function is necessary 李孜睿 │ ├── DOM's dominating function is necessary 李孜睿.pdf │ └── DOM's dominating function is necessary 李孜睿.tex ├── Integrals over subsets_董海辰 │ ├── Integrals over subsets_董海辰.pdf │ └── Integrals over subsets_董海辰.tex ├── More to say on Scheffe's Lemma-毛昕渝 │ ├── More to say on Scheffe's Lemma.pdf │ ├── More to say on Scheffe's Lemma.tex │ └── ref.bib ├── Notes on the proof of Radon-Nikodym theorem_金弘义 │ ├── proof of Radon-Nikodym theorem.pdf │ └── proof of Radon-Nikodym theorem.tex ├── Relationship_between_convergences_张志成 │ ├── Relationship_between_convergences.pdf │ └── Relationship_between_convergences.tex ├── The Necessity of Continuity Constraints on Convergence in Distribution_寻之扬 │ ├── The Necessity of Continuity Constraints on Convergence in Distribution.pdf │ └── The Necessity of Continuity Constraints on Convergence in Distribution.tex └── pointwise convergence and convergence in probability_董海辰 │ ├── pointwise convergence and convergence in probability.pdf │ └── pointwise convergence and convergence in probability.tex ├── Week9 └── Lebesgue Covering Dimension-寻之扬 │ ├── Lebesgue Covering Dimension of One-dimensional Euclidean Space Equals to One.pdf │ └── Lebesgue Covering Dimension of One-dimensional Euclidean Space Equals to One.tex ├── example ├── example.pdf └── example.tex └── sheep ├── 吴润哲 ├── dp.cpp ├── sheep.pdf ├── sheep.tex ├── sheet.xlsx └── simulator.cpp ├── 唐泽 ├── 魔法羊的一条死路和两个猜想.pdf └── 魔法羊的一条死路和两个猜想.tex ├── 杨宗翰 ├── sheep-yzh.pdf ├── sheep.tex ├── sheep2-yzh.pdf └── sheep2.tex ├── 毛昕渝 ├── The Mabinogion sheep problem-mxy.pdf └── The Mabinogion sheep problem-mxy.tex ├── 董海辰 ├── sheep-董海辰.pdf └── sheep.tex └── 钱程阳 ├── 数据.xlsx ├── 施了魔法的羊.html └── 施了魔法的羊.ipynb /.gitignore: -------------------------------------------------------------------------------- 1 | .DS_Store 2 | 3 | ## Core latex/pdflatex auxiliary files: 4 | *.aux 5 | *.lof 6 | *.log 7 | *.lot 8 | *.fls 9 | *.out 10 | *.toc 11 | *.fmt 12 | *.fot 13 | *.cb 14 | *.cb2 15 | .*.lb 16 | 17 | ## Intermediate documents: 18 | *.dvi 19 | *.xdv 20 | *-converted-to.* 21 | # these rules might exclude image files for figures etc. 22 | # *.ps 23 | # *.eps 24 | # *.pdf 25 | 26 | ## Generated if empty string is given at "Please type another file name for output:" 27 | .pdf 28 | 29 | ## Bibliography auxiliary files (bibtex/biblatex/biber): 30 | *.bbl 31 | *.bcf 32 | *.blg 33 | *-blx.aux 34 | *-blx.bib 35 | *.run.xml 36 | 37 | ## Build tool auxiliary files: 38 | *.fdb_latexmk 39 | *.synctex 40 | *.synctex(busy) 41 | *.synctex.gz 42 | *.synctex.gz(busy) 43 | *.pdfsync 44 | 45 | ## Auxiliary and intermediate files from other packages: 46 | # algorithms 47 | *.alg 48 | *.loa 49 | 50 | # achemso 51 | acs-*.bib 52 | 53 | # amsthm 54 | *.thm 55 | 56 | # beamer 57 | *.nav 58 | *.pre 59 | *.snm 60 | *.vrb 61 | 62 | # changes 63 | *.soc 64 | 65 | # cprotect 66 | *.cpt 67 | 68 | # elsarticle (documentclass of Elsevier journals) 69 | *.spl 70 | 71 | # endnotes 72 | *.ent 73 | 74 | # fixme 75 | *.lox 76 | 77 | # feynmf/feynmp 78 | *.mf 79 | *.mp 80 | *.t[1-9] 81 | *.t[1-9][0-9] 82 | *.tfm 83 | 84 | #(r)(e)ledmac/(r)(e)ledpar 85 | *.end 86 | *.?end 87 | *.[1-9] 88 | *.[1-9][0-9] 89 | *.[1-9][0-9][0-9] 90 | *.[1-9]R 91 | *.[1-9][0-9]R 92 | *.[1-9][0-9][0-9]R 93 | *.eledsec[1-9] 94 | *.eledsec[1-9]R 95 | *.eledsec[1-9][0-9] 96 | *.eledsec[1-9][0-9]R 97 | *.eledsec[1-9][0-9][0-9] 98 | *.eledsec[1-9][0-9][0-9]R 99 | 100 | # glossaries 101 | *.acn 102 | *.acr 103 | *.glg 104 | *.glo 105 | *.gls 106 | *.glsdefs 107 | 108 | # gnuplottex 109 | *-gnuplottex-* 110 | 111 | # gregoriotex 112 | *.gaux 113 | *.gtex 114 | 115 | # htlatex 116 | *.4ct 117 | *.4tc 118 | *.idv 119 | *.lg 120 | *.trc 121 | *.xref 122 | 123 | # hyperref 124 | *.brf 125 | 126 | # knitr 127 | *-concordance.tex 128 | # TODO Comment the next line if you want to keep your tikz graphics files 129 | *.tikz 130 | *-tikzDictionary 131 | 132 | # listings 133 | *.lol 134 | 135 | # makeidx 136 | *.idx 137 | *.ilg 138 | *.ind 139 | *.ist 140 | 141 | # minitoc 142 | *.maf 143 | *.mlf 144 | *.mlt 145 | *.mtc[0-9]* 146 | *.slf[0-9]* 147 | *.slt[0-9]* 148 | *.stc[0-9]* 149 | 150 | # minted 151 | _minted* 152 | *.pyg 153 | 154 | # morewrites 155 | *.mw 156 | 157 | # nomencl 158 | *.nlg 159 | *.nlo 160 | *.nls 161 | 162 | # pax 163 | *.pax 164 | 165 | # pdfpcnotes 166 | *.pdfpc 167 | 168 | # sagetex 169 | *.sagetex.sage 170 | *.sagetex.py 171 | *.sagetex.scmd 172 | 173 | # scrwfile 174 | *.wrt 175 | 176 | # sympy 177 | *.sout 178 | *.sympy 179 | sympy-plots-for-*.tex/ 180 | 181 | # pdfcomment 182 | *.upa 183 | *.upb 184 | 185 | # pythontex 186 | *.pytxcode 187 | pythontex-files-*/ 188 | 189 | # thmtools 190 | *.loe 191 | 192 | # TikZ & PGF 193 | *.dpth 194 | *.md5 195 | *.auxlock 196 | 197 | # todonotes 198 | *.tdo 199 | 200 | # easy-todo 201 | *.lod 202 | 203 | # xmpincl 204 | *.xmpi 205 | 206 | # xindy 207 | *.xdy 208 | 209 | # xypic precompiled matrices 210 | *.xyc 211 | 212 | # endfloat 213 | *.ttt 214 | *.fff 215 | 216 | # Latexian 217 | TSWLatexianTemp* 218 | 219 | ## Editors: 220 | # WinEdt 221 | *.bak 222 | *.sav 223 | 224 | # Texpad 225 | .texpadtmp 226 | 227 | # Kile 228 | *.backup 229 | 230 | # KBibTeX 231 | *~[0-9]* 232 | 233 | # auto folder when using emacs and auctex 234 | ./auto/* 235 | *.el 236 | 237 | # expex forward references with \gathertags 238 | *-tags.tex 239 | 240 | # standalone packages 241 | *.sta 242 | 243 | # generated if using elsarticle.cls 244 | *.spl 245 | 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\maketitle 10 | 11 | \begin{thm}{}{} 12 | 中原大战，冯玉祥问士兵:空中飞机多还是乌鸦多?众人答:乌鸦多。冯再问:然则 乌鸦拉屎掉到你们头上没有?众人异口同声:没有。冯说:所以嘛，随着飞机投下的 炸弹的命中机会就更少了，大家莫怕! 13 | \end{thm} 14 | 15 | 我认为这里「飞机数量」，「飞机出现」和「飞机投炸弹」这三者的关系与「乌鸦数量」，「乌鸦出现」和「乌鸦拉屎」三者的关系并不一致————飞机的数量分布并不如乌鸦一样是均匀的，而是在这一片区域的出现概率远大于其他区域。且由于飞机可能有瞄准，「飞机投弹」事件也并不与「飞机出现」的事件独立，而「乌鸦拉屎」与「乌鸦出现」却是独立的。 16 | 17 | 设飞机总数为$N$，飞机经过为事件$X$，飞机投炸弹为$Y$，被炸弹命中的事件为$Y \cup X$，有$P(Y \cup X) \neq P(Y) P(X)$。反观乌鸦，乌鸦在任何区域内拉屎的概率是一定的，设乌鸦总数为$N'$，乌鸦经过为事件$X'$，乌鸦拉屎为$Y'$，有$P(Y' \cup X') = P(Y') P(X')$。 18 | 19 | 因此即使在这里冯玉祥说：$N' > N$，这样并不能得出$P(X') > P(X)$。即使是有$P(X') > P(X)$，也不能决定$P(Y \cup X)$和$P(Y' \cup X')$的大小关系。 20 | 21 | \begin{thm}{}{} 22 | 罐子里有$70$个黑球和$30$个白球。每次从中取一个球直到罐子中只含单色球为止。最后罐子中剩的都是白球的概率为多少? 23 | \end{thm} 24 | 25 | 更一般地，设共有$N$个黑球和$M$个白球，$P_m$为最后恰好剩$m$个白球的概率，由于当共有$n$个黑球和$m$个白球时，选出一个黑球的概率为$\frac{n}{n+m}$，有 26 | \begin{align*} 27 | P_m &= \frac{N! \cdot (M \cdot (M-1) \cdots (m+1))}{(N+M) \cdot (N+M-1) \cdots (m+1)} \cdot \binom {N-1+M-m} {M-m} \\ 28 | &= N! \cdot \frac{m!}{(N+M)!} \cdot \frac{M!}{m!} \cdot \frac{(N-1+M-m)!}{(M-m)!(N-1)!}\\ 29 | &= \frac{N \cdot M!}{(N+M)!}\cdot \frac{(N+M-1-m)!}{(M-m)!} 30 | .\end{align*} 31 | 32 | 最后剩下白球的概率为 33 | \begin{align*} 34 | p = \sum_{i=1}^{M} p_i &= \frac{N \cdot M!}{(N+M)!} \sum_{i=1}^{M} \frac{(N+M-1-m)!}{(M-m)!} \\ 35 | &= \frac{N \cdot M!}{(N+M)!} \cdot \frac{(M+N-1)!}{N (M-1)!} \\ 36 | &= \frac{M}{N + M} 37 | .\end{align*} 38 | 39 | 可以发现，最后剩下白球的概率恰为白球在所有球里面的比例。 40 | 41 | \begin{thm}{}{} 42 | 对于生男生女，比如生男生女概率各 $50\%$，每个家庭都生到第一个男孩就不再生，那么产生的男女比例是多少? 43 | \end{thm} 44 | 45 | 设$X$为每个家庭的孩子数量，有$$P(X=k) = \frac{1}{2^{k-1}} \cdot \frac{1}{2} = 2^{-k}.$$ 46 | 47 | 即前$k-1$次均为女孩，最后一次为男孩。 48 | 49 | 则期望为$$E(X) = \sum_{k=1}^{\infty} k P(X=k) = 2.$$ 50 | 51 | 而每个家庭恰好有$1$个男孩，因此男女比为$1:1$。 52 | 53 | \end{document} 54 | -------------------------------------------------------------------------------- /Week1/A proof of A^A = 2^A and A = AA-毛昕渝/discussion.pdf: 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Choice.pdf -------------------------------------------------------------------------------- /Week1/BijectionBetween[0,1]andR_计家宝/BijectionBetween[0,1]andR.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/BijectionBetween[0,1]andR_计家宝/BijectionBetween[0,1]andR.pdf -------------------------------------------------------------------------------- /Week1/BijectionBetween[0,1]andR_计家宝/BijectionBetween[0,1]andR.tex: -------------------------------------------------------------------------------- 1 | % This is a template for lecture notes. 2 | \documentclass{article} 3 | \usepackage[UTF8]{ctex} 4 | \usepackage{amssymb} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{geometry} 8 | \usepackage{booktabs} 9 | \usepackage{bm} 10 | \usepackage{tcolorbox} 11 | \CTEXoptions[today=old] 12 | %Some commonly used notations 13 | %\geometry{a4paper,bottom = 3cm,left = 3cm, right = 3cm} 14 | 15 | %for reference 16 | \usepackage{hyperref} 17 | \usepackage[capitalise]{cleveref} 18 | \crefname{enumi}{}{} 19 | 20 | \newtheorem{theorem}{Theorem} 21 | \newtheorem{lemma}[theorem]{Lemma} 22 | \newtheorem{proposition}[theorem]{Proposition} 23 | \newtheorem{corollary}[theorem]{Corollary} 24 | \newtheorem{fact}[theorem]{Fact} 25 | \newtheorem{definition}[theorem]{Definition} 26 | \newtheorem{remark}[theorem]{Remark} 27 | \newtheorem{question}[theorem]{Question} 28 | \newtheorem{answer}[theorem]{Answer} 29 | \newtheorem{exercise}[theorem]{Exercise} 30 | \newtheorem{example}[theorem]{Example} 31 | %\newenvironment{proof}{\noindent \textbf{Proof:}}{$\Box$} 32 | \newtheorem{observation}[theorem]{Observation} 33 | 34 | %to use newcommand for convenience 35 | \newcommand\field{\mathbb{F}} 36 | \newcommand\Real{\mathbb{R}} 37 | \newcommand\Q{\mathbb{Q}} 38 | \newcommand\Z{\mathbb{Z}} 39 | \newcommand\complex{\mathbb{C}} 40 | 41 | %this is how we define operators. 42 | \DeclareMathOperator{\rank}{rank} % rank 43 | 44 | \title{Bijection between [0, 1] and (0, 1) and $\mathbb{R}$} 45 | \author{Ji Jiabao} 46 | \date{\today} 47 | 48 | \begin{document} 49 | \maketitle 50 | The problem i encountered is that what the image for the endpoint 0,1 should be, 51 | since [0,1] is a closed interval while $\mathbb{R}$ is an open interval. 52 | But we know there is a simple bijection between (0, 1) and $\mathbb{R}$ with the help of $tan$. 53 | And from the reading of Was Cantor Surprised, we can take advantage of the injection 54 | Cantor constructed between [0,1] and [0, 1).\\ 55 | 56 | First we give the simple bijection $f = tan(\frac{\pi}{2} x - \frac{\pi}{2}) (x \in (0, 1))$ between (0, 1) and $\mathbb{R}$.\\ 57 | Next construct the bijection between $(0, 1)$ and [0, 1]. 58 | 59 | \begin{figure}[h] 60 | \centering 61 | \includegraphics[scale = 0.5]{cantor.png} 62 | \caption{Cantor's bijection} 63 | \end{figure} 64 | \begin{quote} 65 | The domain has been divided by a geometric progression, so $b = 1/2$,$ b_1 = 3/4$, and so on; 66 | $a = (0, 1/2)$, $a' = (1/2, 3/4)$, etc. The point $C$ is (1, 1). The points $d' = (1/2, 1/2)$, $d^{''} = (3/4, 3/4)$, etc. 67 | give the corresponding subdivision of the main diagonal. 68 | \end{quote} 69 | I can't explain the curve in english clearly, so i quote the 70 | discription in Was Cantor Surprised \footnote{I also add my notes on this article, see \url{WasCantorSurprised_AfterReading.md} } here. 71 | We can slightly change this bijection by ignoring the endpoint C, by doing so, we get 72 | the desired bijection $g$.\\ 73 | \hspace*{2em} Also, in a $'modern'$ way(as Cantor would say), we can construct the bijection like this. 74 | \begin{equation} 75 | g(x) = \left\{ 76 | \begin{aligned} 77 | \frac{1}{2} &, & x = 0\\ 78 | \frac{1}{n + 2} &, &(x \in \{\frac{1}{n}, n \in \mathbb{N}\}\\ 79 | x &, &(x \notin \{\frac{1}{n}, n \in \mathbb{N}\})\\ 80 | \end{aligned} 81 | \right. 82 | \end{equation} 83 | By $g$, we map $0, 1, \frac{1}{2}...$ to $\frac{1}{2}, \frac{1}{3}...$, since we have infinite $\frac{1}{n}$ in (0, 1), 84 | this can be done. 85 | 86 | To sum up, the bijection between [0, 1] and $\mathbb{R}$ is : $f \circ g$ 87 | 88 | 89 | 90 | 91 | \end{document} -------------------------------------------------------------------------------- /Week1/BijectionBetween[0,1]andR_计家宝/WasCantorSurprised_AfterReading.md: -------------------------------------------------------------------------------- 1 | ## Was Cantor Surprised读后感 2 | 3 | ### 内容简述： 4 | 5 | ​ 主要讲了Cantor和Dedekind的一系列关于“Cantor was the fact that sets of different dimension could have the same cardinality” 的通信。我了解过少，但是这个问题的结果好像是现代数学的相当重要的结果。 6 | 7 | ​ 这个过程中，年轻的Cantor（27、8岁），有更多创造性的想法和思维，而Dedekind（41、2）岁，功成名就有很高的威望。一开始Cantor是主要是询问Dedekind自己的一些想法是否对，自己不确定的结论Dedekind能否给出一个证明或者反驳。而随着Dedekind不断地肯定，Cantor逐渐建立自信并且开始向Dedekind提出自己的与当时的数学结论有悖的关于不同维度的集合完全可以构建一一映射，也就是一个正方形和一条线 "has same power"。并且我们可以看到思考的不断深入，以及"thinking in terms of what later became known as 'Cardinality'"。 8 | 9 | ​ 我感觉其中表现出了几个阶段： 10 | 11 | ​ 1、Cantor对于Cardinality的问题("was there a bijection between the interval $I = [0, 1]$ and the n-fold product $I^n = I × I × · · · × I$)，有一个模糊的疑惑，这个命题是否正确 12 | 13 | ​ 2、Cantor改变了自己基于当时的数学对于这个问题的“definitely no”的结论，转而意识到这个问题结论是“yes”，并且逐渐坚信这一点 14 | 15 | ​ 3、Cantor开始试图证明并向Dedekind求证证明的正确性 16 | 17 | ​ 4、5、6、... 证明有漏洞，补漏洞，有漏洞，补漏洞 18 | 19 | ​ n、完成证明，Dedekind找不出问题 20 | 21 | ​ n+1、著名的"I see it, but i don't believe it" 22 | 23 | 24 | 25 | ### 感想： 26 | 27 | ​ 结合之前看的"Correspondance between Fermat and Pascal"，从这两个书信记录中，我看到了知名的数学家（膜大佬）是如何思考，如何与别人讨论数学问题的。令我惊讶的是，从中发现了一些自己和别人讨论时可能遇到的问题。 28 | 29 | ​ 在Fermat和Pascal的通信中，我先看了对于他们讨论的分配问题的现代解决，枚举剩下的胜负情况，按照胜负情况分配赌资，我的第一感觉是这么简单的算法他们会没想到吗？然后在看他们的书信的过程中(太多了没看完)，我发现他们讨论了许多关于如果对手不承认我再掷硬币，我不能投了如何直接分配赌资这样的问题（这一部分我看不懂，很多数字我不知道是怎么想的）。从已知正确结果的情况下会觉得这种讨论偏离了原命题，但是这像极了我对一道题没有思路的时候会纠结在一些明白后会觉得莫名其妙的地方的情况。 30 | 31 | ​ 在Cantor和Dedekind的通信中，Cantor给的第一个[0, 1]x[0,1] 和[0,1]的双射，他自信正确，但Dedekind发现有漏洞，而这个漏洞不好补。像极了我有一个假结论自信可以A，然后暴毙的惨淡情况（但是Cantor大佬能换条思路补上）。 32 | 33 | ​ 简单的总结： 这两个阅读都告诉我再厉害的结论，或者说越厉害的结论，在做出来之前都经历了一堆大佬的反复思考，试错，找漏洞，补漏洞，补不上换思路（或者问别人）。看到Cantor这样的大佬被指出证明有问题内心甚至偷笑。佩服大佬们的试错，但是大佬们试着试着就能想出结论，而我不行，这大概就是区别吧。 -------------------------------------------------------------------------------- /Week1/BijectionBetween[0,1]andR_计家宝/cantor.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/BijectionBetween[0,1]andR_计家宝/cantor.png -------------------------------------------------------------------------------- /Week1/BijectionFrom01toR 刘成锴/HW1.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/BijectionFrom01toR 刘成锴/HW1.pdf -------------------------------------------------------------------------------- /Week1/BijectionFrom01toR 刘成锴/HW1.tex: -------------------------------------------------------------------------------- 1 | \documentclass{ximera} 2 | \usepackage[UTF8]{ctex} 3 | 4 | \title{Bijection from [0,1] to R} 5 | \author{刘成锴} 6 | 7 | \begin{document} 8 | 9 | \begin{abstract} 10 | 11 | 刘成锴 12 | 13 | 学号：518030910425 14 | \end{abstract} 15 | \maketitle 16 | 17 | 18 | \begin{solution} 19 | $$ 20 | f(x) = \tan({x - \frac{1}{2}})\pi 21 | $$ 22 | 23 | $f:[0, 1] \rightarrow \mathbb{R}$ is a simple injection. 24 | 25 | $$ 26 | g(x)=\left\{ 27 | \begin{array}{lc} 28 | 0 & x = -\infty \\ 29 | \frac{\arctan(x)}{\pi} + \frac{1}{2} & -\infty < x < \infty \\ 30 | 1 & x = \infty 31 | \end{array}\right. 32 | $$ 33 | 34 | $g:\mathbb{R} \rightarrow [0, 1]$ is a simple injection. 35 | 36 | Hence, we find an explicit bijection from [0, 1] to $\mathbb{R}$ 37 | \end{solution} 38 | 39 | 40 | \end{document} -------------------------------------------------------------------------------- /Week1/Bijection_between_[0, 1]_and_R_姚远/Bijection_between_[0, 1]_and_R.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/Bijection_between_[0, 1]_and_R_姚远/Bijection_between_[0, 1]_and_R.pdf -------------------------------------------------------------------------------- /Week1/Bijection_between_[0, 1]_and_R_姚远/Bijection_between_[0, 1]_and_R.tex: -------------------------------------------------------------------------------- 1 | % This is a template for lecture notes. 2 | \documentclass{article} 3 | \usepackage[UTF8]{ctex} 4 | \usepackage{amssymb} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{geometry} 8 | \usepackage{booktabs} 9 | \usepackage{bm} 10 | \usepackage{tcolorbox} 11 | \CTEXoptions[today=old] 12 | %Some commonly used notations 13 | %\geometry{a4paper,bottom = 3cm,left = 3cm, right = 3cm} 14 | 15 | %for reference 16 | \usepackage{hyperref} 17 | \usepackage[capitalise]{cleveref} 18 | \crefname{enumi}{}{} 19 | 20 | \newtheorem{theorem}{Theorem} 21 | \newtheorem{lemma}[theorem]{Lemma} 22 | \newtheorem{proposition}[theorem]{Proposition} 23 | \newtheorem{corollary}[theorem]{Corollary} 24 | \newtheorem{fact}[theorem]{Fact} 25 | \newtheorem{definition}[theorem]{Definition} 26 | \newtheorem{remark}[theorem]{Remark} 27 | \newtheorem{question}[theorem]{Question} 28 | \newtheorem{answer}[theorem]{Answer} 29 | \newtheorem{exercise}[theorem]{Exercise} 30 | \newtheorem{example}[theorem]{Example} 31 | %\newenvironment{proof}{\noindent \textbf{Proof:}}{$\Box$} 32 | \newtheorem{observation}[theorem]{Observation} 33 | 34 | %to use newcommand for convenience 35 | \newcommand\field{\mathbb{F}} 36 | \newcommand\Real{\mathbb{R}} 37 | \newcommand\Q{\mathbb{Q}} 38 | \newcommand\Z{\mathbb{Z}} 39 | \newcommand\complex{\mathbb{C}} 40 | 41 | %this is how we define operators. 42 | \DeclareMathOperator{\rank}{rank} % rank 43 | 44 | \title{Bijection between $[0, 1]$ and $\Real$} 45 | \author{Yao Yuan} 46 | \date{\today} 47 | 48 | \begin{document} 49 | \maketitle 50 | 51 | \begin{tcolorbox} 52 | \begin{theorem} 53 | Find an explicit bijection from $[0, 1]$ to $\Real$. 54 | \end{theorem} 55 | \end{tcolorbox} 56 | 57 | \begin{fact}\label{fact:real} 58 | There exists a bijection between $(0, 1)$ and $\Real$. 59 | \end{fact} 60 | 61 | \begin{proof} 62 | Function $f: y = \tan(\pi x - \frac\pi2)$ is obviously a bijection between $(0, 1)$ and $\Real$. 63 | \end{proof} 64 | 65 | \begin{fact}\label{fact:extension} 66 | There exists a bijection between $(0, 1)$ and $[0, 1]$. 67 | \end{fact} 68 | 69 | \begin{proof} 70 | Choose an infinite sequence $(x_n)_{n \geq 1}$ of distinct elements of $(0,1)$. 71 | Let $X=\{x_{n} \mid n \geq 1\}$, hence $X \subset (0,1) .$ Let $x_0=1 .$ Define $f(x_n) = x_{n+1} $ for every $n \geq 0$ and $ f(x) = x $ for every $ x $ in $(0,1) \backslash X$. Then $f$ is defined on $(0,1]$ and the map $f:(0,1] \rightarrow(0,1)$ is bijective. 72 | 73 | Similarly, we can find a bijection between $(0, 1]$ and $[0, 1]$. Thus there exists a bijection between $(0, 1)$ and $[0, 1]$. 74 | \end{proof} 75 | 76 | %use \cref instead of \ref 77 | Combine \cref{fact:real} and \cref{fact:extension} 78 | and we get a bijection between $[0, 1]$ and $\Real$. 79 | 80 | \end{document} -------------------------------------------------------------------------------- /Week1/Construction_of_4_sets_to_satisfies_3_rules_马浩博/discussion.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/Construction_of_4_sets_to_satisfies_3_rules_马浩博/discussion.pdf -------------------------------------------------------------------------------- /Week1/Construction_of_4_sets_to_satisfies_3_rules_马浩博/discussion.tex: -------------------------------------------------------------------------------- 1 | % This is a template for lecture notes. 2 | \documentclass[12pt]{article} 3 | \usepackage{amssymb} 4 | \usepackage[UTF8]{ctex} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{geometry} 8 | \usepackage{booktabs} 9 | \usepackage{bm} 10 | \usepackage{cite} 11 | %\usepackage{CJK} 12 | \usepackage[many]{tcolorbox} 13 | %\tcbuselibrary{listingsutf8} 14 | %\tcbuselibrary{skins, breakable, theorems, most} 15 | %\geometry{a4paper,bottom = 3cm,left = 3cm, right = 3cm} 16 | \CTEXoptions[today=old] 17 | %for reference 18 | \usepackage{hyperref} 19 | \usepackage[capitalise]{cleveref} 20 | \crefname{enumi}{}{} 21 | 22 | 23 | \newtheoremstyle{mythm}{1.5ex plus 1ex minus .2ex}{1.5ex plus 1ex minus .2ex} 24 | {}{\parindent}{\bfseries}{}{1em}{} 25 | \theoremstyle{mythm} 26 | \newtheorem{theorem}{Theorem} 27 | \newtheorem{lemma}[theorem]{Lemma} 28 | \newtheorem{corollary}[theorem]{Corollary} 29 | \newtheorem{fact}[theorem]{Fact} 30 | \newtheorem{definition}[theorem]{Definition} 31 | \newtheorem*{remark}{Remark} 32 | 33 | %\newenvironment{proof}{\noindent \textbf{Proof:}}{$\Box$} 34 | 35 | %to use newcommand for convenience 36 | \newcommand\field{\mathbb{F}} 37 | \newcommand\Real{\mathbb{R}} 38 | \newcommand\Q{\mathbb{Q}} 39 | \newcommand\Z{\mathbb{Z}} 40 | \newcommand\complex{\mathbb{C}} 41 | \newcommand\cc{\mathcal{C}} 42 | \newcommand\uu{\mathcal{U}} 43 | \newcommand\pp{\mathcal{P}} 44 | \newcommand\ff{\mathcal{F}} 45 | \renewcommand\refname{Reference} 46 | \renewcommand{\proofname}{Proof} 47 | \DeclareMathOperator{\range}{range} 48 | 49 | \title{Construction of 4 sets to satisfy 3 rules} 50 | \author{马浩博 518030910428} 51 | \date{\today} 52 | \begin{document} 53 | \maketitle 54 | 55 | \section*{Exercise 8} 56 | \begin{tcolorbox} 57 | Construct three sets A, B, C and a set E $\subseteq$ $A \times B \times C$ 58 | such that the following hold: 1) For every (a, b, c) $\in$ $A \times B \times C$, E has a matching which saturates {a, b, c}; 59 | 2) E has an (A∪B)-saturated matching, a (B∪C)-saturated matching, and an (A∪C)-saturated matching; 3) E does not have any (A∪B∪C)-saturated matching. 60 | 61 | \end{tcolorbox} 62 | 63 | \begin{proof}[Answer] 64 | 65 | Let me first give out the construction: 66 | $$A = \{1, 0, 4, 7 \}$$ 67 | $$B = \{2, 0, 4, 7 \}$$ 68 | $$C = \{3, 0, 4, 7 \}$$ 69 | $$E = \bigcup\limits_{i=0}^5 \{e_i\}$$ 70 | $$e_0 = (1,7,4)$$ 71 | $$e_1 = (4,2,7)$$ 72 | $$e_2 = (7,4,3)$$ 73 | $$e_3 = (0,7,4)$$ 74 | $$e_4 = (4,0,7)$$ 75 | $$e_5 = (7,4,0)$$ 76 | 77 | 78 | 79 | First, we shall prove it satisfies 3). Assume there is such matching $E'$. 80 | Then $e_0$, $e_1$, $e_2$ must belong to $E'$ because only they have the elements 1,2,3. 81 | Then we know $e_3 \notin E'$ because it's not disjoint with $e_0$, and same as $e_4, e_5$. 82 | Thus there it no element which can cover 0. So such $E'$ doesn't exist.\\ 83 | 84 | Next, we shall prove it satisfies 2). For $A \cup B$, we just need to let $\{e_0,e_1,e_5\}$ be the matching. 85 | Beause of symmetry, we can know there are also $(B \cup C)$-saturated matching and $(C \cup A)$-saturated matching.\\ 86 | 87 | Finally, we need to prove 1). If the $a$ in the choosen $(a,b,c)$ doesn't equal to 1, then we can just use the $(B \cup C)$-saturated matching. 88 | Same way if $b \neq 2$ or $c \neq 3$. So we just need to concern $(1,2,3)$. 89 | To saturate this, we can use the macthing $\{e_0,e_1,e_2\}$.\\ 90 | 91 | Hence, the construction is right. I thought for a long time and tried many constructions. 92 | This should be the most concise one. 93 | 94 | \end{proof} 95 | 96 | \end{document} 97 | -------------------------------------------------------------------------------- /Week1/Expansion of uniqueness theorem_郭睿涵/Expansion of uniqueness theorem_郭睿涵.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/Expansion of uniqueness theorem_郭睿涵/Expansion of uniqueness theorem_郭睿涵.pdf -------------------------------------------------------------------------------- /Week1/Expansion of uniqueness theorem_郭睿涵/tmp: -------------------------------------------------------------------------------- 1 | 2 | -------------------------------------------------------------------------------- /Week1/Further discussion of Exercise 4_傅凌玥/Further discussion of Exercise 4.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/Further discussion of Exercise 4_傅凌玥/Further discussion of Exercise 4.pdf -------------------------------------------------------------------------------- /Week1/Parallel_sets_don't_exist_张志成/parallel_sets_don't_exist.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/Parallel_sets_don't_exist_张志成/parallel_sets_don't_exist.pdf -------------------------------------------------------------------------------- /Week1/Poset_R^3_Cannot_Be_Stuffed_into_R^2_寻之扬/Poset R^3 Cannot Be Stuffed into R^2.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/Poset_R^3_Cannot_Be_Stuffed_into_R^2_寻之扬/Poset R^3 Cannot Be Stuffed into R^2.pdf -------------------------------------------------------------------------------- /Week1/dual_Cantor-Bernstein_theorem_李孜睿/dual_Cantor-Bernstein_Theorem_李孜睿.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/dual_Cantor-Bernstein_theorem_李孜睿/dual_Cantor-Bernstein_Theorem_李孜睿.pdf -------------------------------------------------------------------------------- /Week1/dual_Cantor-Bernstein_theorem_李孜睿/dual_Cantor-Bernstein_Theorem_李孜睿.tex: -------------------------------------------------------------------------------- 1 | % This is a template for lecture notes. 2 | \documentclass{article} 3 | \usepackage[UTF8]{ctex} 4 | \usepackage{amssymb} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{geometry} 8 | \usepackage{booktabs} 9 | \usepackage{bm} 10 | \usepackage{tcolorbox} 11 | \CTEXoptions[today=old] 12 | %Some commonly used notations 13 | %\geometry{a4paper,bottom = 3cm,left = 3cm, right = 3cm} 14 | 15 | %for reference 16 | \usepackage{hyperref} 17 | \usepackage[capitalise]{cleveref} 18 | \crefname{enumi}{}{} 19 | 20 | \newtheorem{theorem}{Theorem} 21 | \newtheorem{lemma}[theorem]{Lemma} 22 | \newtheorem{proposition}[theorem]{Proposition} 23 | \newtheorem{corollary}[theorem]{Corollary} 24 | \newtheorem{fact}[theorem]{Fact} 25 | \newtheorem{definition}[theorem]{Definition} 26 | \newtheorem{remark}[theorem]{Remark} 27 | \newtheorem{question}[theorem]{Question} 28 | \newtheorem{answer}[theorem]{Answer} 29 | \newtheorem{exercise}[theorem]{Exercise} 30 | \newtheorem{example}[theorem]{Example} 31 | %\newenvironment{proof}{\noindent \textbf{Proof:}}{$\Box$} 32 | \newtheorem{observation}[theorem]{Observation} 33 | 34 | %to use newcommand for convenience 35 | \newcommand\field{\mathbb{F}} 36 | \newcommand\Real{\mathbb{R}} 37 | \newcommand\Q{\mathbb{Q}} 38 | \newcommand\Z{\mathbb{Z}} 39 | \newcommand\complex{\mathbb{C}} 40 | \newenvironment{myproof}{\ignorespaces\paragraph{Proof:}}{\hfill $\square$\par\noindent} 41 | %this is how we define operators. 42 | \DeclareMathOperator{\rank}{rank} % rank 43 | 44 | \title{dual Cantor-Bernstein theorem} 45 | \author{李孜睿 518030910424} 46 | \date{\today} 47 | 48 | 49 | \begin{document} 50 | \maketitle 51 | 1. (Cantor-Bernstein theorem) Let $f \in Y^X$ and $g \in X^Y$ be two injective maps. Then there is a bijection $h \in Y^X$ such that $h\subseteq f\cup g^{-1}$. 52 | \begin{myproof} 53 | Let $C_0 = X \backslash g(Y)$, $C_{n+1} = g(f(C_n))$. And $$C = \bigcup_{n=0}^{\infty}C_n$$ 54 | 55 | For every $x\in X$, define 56 | $$h(x) = \left\{ 57 | \begin{array}{lr} 58 | f(x),&x\in C\\ 59 | g^{-1}(x),&x\notin C 60 | \end{array}\right.$$ 61 | 62 | We can easily see $h\subseteq f\cup g^{-1}$ from the definition. Next, prove $h$ is bijective. 63 | 64 | $h$ is injective: Assume $a\neq b\wedge h(a)=h(b)$(h is not injective). If $a\in C \wedge b\in C$, $h(a) = f(a) \neq f(b) = h(b)$. If $a\notin C \wedge b\notin C$, $h(a) = g^{-1}(a) \neq g^{-1}(b) = h(b)$. If $a\in C \wedge b\notin C$, $g^{-1}(b) = f(a)\Rightarrow b = g(f(a))\Rightarrow b\in C$, contradicting to $b\notin C$. Otherwise is the same. All cases contradict to the premise. So $h$ is injective. 65 | 66 | $h$ is surjective: For any $y\in Y$, If $g(y)\in C$, there is a certion $n\geq 1$ such that $g(y)\in C_n$. Also, there is a $x\in C_{n-1}\subseteq X$ that $h(x) = f(x) = y$. If $g(y)\notin C$, there is $g(y)\in X$ that $h(g(y))=g^{-1}(g(y))=y$. Thus $h$ is surjective. 67 | 68 | Above all, $h\in Y^X$ is a bijection and $h\subseteq f\cup g^{-1}$. 69 | \end{myproof} 70 | 71 | 2. (dual Cantor-Bernstein theorem) Let $f \in Y^X$ and $g \in X^Y$ be two surjectve maps. Assuming AC, show that there is a bijection $h \in Y^X$ such that $h\subseteq f\cup g^{-1}$. 72 | \begin{myproof} 73 | Given any surjections $f\in Y^X$ and $g\in X^Y$, by AC (we can get a Right inverse of a surjective map) there are injections $u\subseteq f^{-1}$ and $v\subseteq g^{-1}$. 74 | 75 | By Cantor-Bernstein Theorem, there exits a bijection $h\in Y^X$ such that $h\subseteq u^{-1}\cup v$. 76 | 77 | As $u^{-1}\subseteq f$ and $v\subseteq g^{-1}$, $h\subseteq f\cup g^{-1}$. 78 | \end{myproof} 79 | \end{document} -------------------------------------------------------------------------------- /Week1/dual_Cantor-Bernstein_theorem_李孜睿/参考资料：The_Dual_Cantor-Bernstein_Theorem_and_the_Partition_Principle.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/dual_Cantor-Bernstein_theorem_李孜睿/参考资料：The_Dual_Cantor-Bernstein_Theorem_and_the_Partition_Principle.pdf -------------------------------------------------------------------------------- /Week1/exercise3&4_刘成锴/HW1.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/exercise3&4_刘成锴/HW1.pdf -------------------------------------------------------------------------------- /Week1/exercise3&4_刘成锴/HW1.tex: -------------------------------------------------------------------------------- 1 | \documentclass{ximera} 2 | \usepackage[UTF8]{ctex} 3 | 4 | \title{Basic Probability Theory} 5 | \author{刘成锴} 6 | 7 | \begin{document} 8 | 9 | \begin{abstract} 10 | Week 1 11 | 12 | 刘成锴 13 | 14 | 学号：518030910425 15 | \end{abstract} 16 | \maketitle 17 | 18 | \section{Exercises} 19 | 20 | \subsection{Exercise 3} 21 | 22 | \begin{solution} 23 | $$ 24 | f(x) = \tan({x - \frac{1}{2}})\pi 25 | $$ 26 | 27 | $f:[0, 1] \rightarrow \mathbb{R}$ is a simple injection. 28 | 29 | $$ 30 | g(x)=\left\{ 31 | \begin{array}{lc} 32 | 0 & x = -\infty \\ 33 | \frac{\arctan(x)}{\pi} + \frac{1}{2} & -\infty < x < \infty \\ 34 | 1 & x = \infty 35 | \end{array}\right. 36 | $$ 37 | 38 | $g:\mathbb{R} \rightarrow [0, 1]$ is a simple injection. 39 | 40 | Hence, we find an explicit bijection from [0, 1] to $\mathbb{R}$ 41 | \end{solution} 42 | 43 | \subsection{Exercise 4} 44 | \begin{solution} 45 | I would like to prove $ |\mathbb{R}| = |\mathbb{R}^2| $. 46 | $$ 47 | |\mathbb{R}| \rightarrow |\mathbb{R}^2| 48 | $$ 49 | For any $a \in \mathbb{R}$, there is $(a, a) \in \mathbb{R}^2$. It's an injection. 50 | 51 | $$ 52 | |\mathbb{R}^2| \rightarrow |\mathbb{R}| 53 | $$ 54 | 利用Exercise 3 中的$g$，我们可以将一个属于$\mathbb{R}$中的数映射到区间[0, 1]上。 55 | 56 | Then, for any $(a, b) \in \mathbb{R}^2$, 我们可以通过映射，得到$(g(a), g(b))$。 57 | 58 | 将$g(a), g(b)$分别表示为：$0.a_1a_2a_3...$ 和 $0.b_1b_2b_3...$。 59 | 60 | 然后我们再找一个映射，得到一个实数$c$,且$c \in [0, 1]$。$c = 0.a_1b_1a_2b_2a_3b_3...$。 61 | 62 | 综合以上映射，we get an injection from $\mathbb{R}^2$ to $\mathbb{R}$。 63 | 64 | Hence, there is an bijection from $\mathbb{R}^2$ to $\mathbb{R}$. 65 | 66 | That is, $ |\mathbb{R}| = |\mathbb{R}^2|$. 67 | 68 | \end{solution} 69 | 70 | 71 | \end{document} -------------------------------------------------------------------------------- /Week1/guessing_discontinuity_chain_liar_庄永昊_金弘义_赖睿航_董海辰/discuss.pdf: -------------------------------------------------------------------------------- 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\providecommand{\norm}[1]{\lVert#1\rVert} 11 | \providecommand{\ud}[1]{\underline{#1}} 12 | 13 | \newmdtheoremenv{thm}{Theorem} 14 | \newmdtheoremenv{lemma}[thm]{Lemma} 15 | \newmdtheoremenv{fact}[thm]{Fact} 16 | \newmdtheoremenv{cor}[thm]{Corollary} 17 | \newtheorem{eg}{Example} 18 | \newtheorem{ex}{Exercise} 19 | \newmdtheoremenv{defi}{Definition} 20 | \newenvironment{sol} 21 | {\par\vspace{3mm}\noindent{\it Solution}.} 22 | {\qed \\ \medskip} 23 | 24 | \newcommand{\ov}{\overline} 25 | \newcommand{\ca}{{\cal A}} 26 | \newcommand{\cb}{{\cal B}} 27 | \newcommand{\cc}{{\cal C}} 28 | \newcommand{\cd}{{\cal D}} 29 | \newcommand{\ce}{{\cal E}} 30 | \newcommand{\cf}{{\cal F}} 31 | \newcommand{\ch}{{\cal H}} 32 | \newcommand{\cl}{{\cal L}} 33 | \newcommand{\cm}{{\cal M}} 34 | \newcommand{\cp}{{\cal P}} 35 | \newcommand{\cs}{{\cal S}} 36 | \newcommand{\cz}{{\cal Z}} 37 | \newcommand{\eps}{\varepsilon} 38 | \newcommand{\ra}{\rightarrow} 39 | \newcommand{\la}{\leftarrow} 40 | \newcommand{\Ra}{\Rightarrow} 41 | \newcommand{\dist}{\mbox{\rm dist}} 42 | \newcommand{\bn}{{\mathbb N}} 43 | \newcommand{\bz}{{\mathbb Z}} 44 | 45 | \newcommand{\expe}{{\mathsf E}} 46 | \newcommand{\pr}{{\mathsf{Pr}}} 47 | 48 | 49 | \setlength{\parindent}{0pt} 50 | %\setlength{\parskip}{2ex} 51 | \newenvironment{proofof}[1]{\bigskip\noindent{\itshape #1. }}{\hfill$\Box$\medskip} 52 | 53 | \theoremstyle{definition} 54 | \newtheorem{problem}{Problem} 55 | \newtheorem*{problem*}{Problem} 56 | \newtheorem{solution}{Solution} 57 | \newtheorem*{solution*}{Solution} 58 | 59 | \begin{document} 60 | 61 | \title{Discussions on some quizzes} 62 | \author{吕优 518030910427} 63 | \date{2020 3 9} 64 | 65 | \maketitle 66 | 67 | \begin{problem} 68 | 罐子里有 70 个黑球和 30 个白球。每次从中取一个球直到罐子中只含单色球为止。最后罐子中剩的都是白球的概率为多少？ 69 | \end{problem} 70 | 71 | \begin{solution} 72 | 我们假设罐子里有$n$个黑球和$m$个白球。 73 | 我们可以先考虑这样一个操作，每次从罐子中取一个球直到罐子取空为止，那这样操作得到的序列一共有$\frac{n+m!}{n!m!}$种 74 | 然后为了满足题目的条件，我们删除序列最后一段连续的同色球，假如最后一个球是黑色，我们就从后一直删到球变成白色为止 75 | 我们会发现得到的新序列满足题目的条件，并且两两序列互不相同。这样，每一个序列都代表着一种合法的取球方案，我们考虑 76 | 最后的一段序列是白球的情况个数，这个数目就等于之前序列中最后一个球是白球的数目，一共有$\frac{(n+m-1)!}{n!(m-1)!}$种 77 | 所以最后罐子中剩的都是白球的概率为$\frac{\frac{(n+m-1)!}{n!(m-1)!}}{\frac{n+m!}{n!m!}}=\frac{m}{n+m}$,即为 78 | 白色球占所有球的比例，而且通过上述讨论不难发现，如果罐子中有若干种颜色的球，按题目的要求操作，最后罐子种所剩的球为 79 | 某一特定颜色球的概率就是这种颜色的球所占的比例。 80 | \end{solution} 81 | 82 | \end{document} 83 | 84 | -------------------------------------------------------------------------------- /Week1/quiz_balls_吴怀瑾/quiz_balls_吴怀瑾.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/quiz_balls_吴怀瑾/quiz_balls_吴怀瑾.pdf -------------------------------------------------------------------------------- /Week1/quiz_balls_吴怀瑾/quiz_balls_吴怀瑾.tex: -------------------------------------------------------------------------------- 1 | % This is a template for lecture notes. 2 | \documentclass{article} 3 | \usepackage[UTF8]{ctex} 4 | \usepackage{amssymb} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{geometry} 8 | \usepackage{booktabs} 9 | \usepackage{bm} 10 | \usepackage{tcolorbox} 11 | \CTEXoptions[today=old] 12 | %Some commonly used notations 13 | %\geometry{a4paper,bottom = 3cm,left = 3cm, right = 3cm} 14 | 15 | %for reference 16 | \usepackage{hyperref} 17 | \usepackage[capitalise]{cleveref} 18 | \crefname{enumi}{}{} 19 | 20 | \newtheorem{theorem}{Theorem} 21 | \newtheorem{lemma}[theorem]{Lemma} 22 | \newtheorem{proposition}[theorem]{Proposition} 23 | \newtheorem{corollary}[theorem]{Corollary} 24 | \newtheorem{fact}[theorem]{Fact} 25 | \newtheorem{definition}[theorem]{Definition} 26 | \newtheorem{remark}[theorem]{Remark} 27 | \newtheorem{question}[theorem]{Question} 28 | \newtheorem{answer}[theorem]{Answer} 29 | \newtheorem{exercise}[theorem]{Exercise} 30 | \newtheorem{example}[theorem]{Example} 31 | %\newenvironment{proof}{\noindent \textbf{Proof:}}{$\Box$} 32 | \newtheorem{observation}[theorem]{Observation} 33 | 34 | %to use newcommand for convenience 35 | \newcommand\field{\mathbb{F}} 36 | \newcommand\Real{\mathbb{R}} 37 | \newcommand\Q{\mathbb{Q}} 38 | \newcommand\Z{\mathbb{Z}} 39 | \newcommand\complex{\mathbb{C}} 40 | 41 | %this is how we define operators. 42 | \DeclareMathOperator{\rank}{rank} % rank 43 | 44 | \title{Quiz Balls} 45 | \author{吴怀瑾 518030910414} 46 | \date{\today} 47 | 48 | \begin{document} 49 | \maketitle 50 | \section{} 51 | \begin{question}{}{} 52 | 罐子里有 70 个黑球和 30 个白球。每次从中取一个球直到罐子中只含单色球为止。最后罐子中剩的都是白球的概率为多少? 53 | \end{question} 54 | 55 | \paragraph{solution} 56 | My intuition of the answer is $\frac {3}{10}$. We can discuss this question in a more general way. 57 | Suppose there are $X$ black balls and $Y$ white balls in the pot initially. First, We have $P = \binom {X+Y} Y = \frac {(X+Y)!}{X!Y!}$ different ways to 58 | take all the balls out of the pot. Next we consider how many different ways when last $k$ balls are exactly white which means the last $(k+1)^{th}$ ball must be 59 | black, suppose the number is $w_i$, and the sum is $W$. 60 | \begin{align*} 61 | W = \sum_{i=1}^{Y} w_i &= \sum_{i=1}^{Y} \binom {X+Y-1-i} {X-1}\\ 62 | &= \sum_{i=1}^{Y} ( \binom {X+Y-i} {X} - \binom {X+Y-i-1} {X}) \\ 63 | &= \binom {X+Y-1} {X}\\ 64 | &= \frac{(X+Y-1)!}{X!(Y-1)!} 65 | .\end{align*} 66 | Every way to take out the whole balls has equal probability. So the answer of the question is $\frac W P = \frac {Y} {X+Y}$. 67 | \end{document} -------------------------------------------------------------------------------- /Week1/quiz_fair_dice_张志成/factor.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/quiz_fair_dice_张志成/factor.png -------------------------------------------------------------------------------- /Week1/quiz_fair_dice_张志成/fair_dice.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/quiz_fair_dice_张志成/fair_dice.pdf -------------------------------------------------------------------------------- /Week1/quiz_fair_dice_张志成/fair_dice.tex: -------------------------------------------------------------------------------- 1 | \documentclass[UTF8, 12pt]{ctexart} 2 | \usepackage{enumitem} 3 | \usepackage{amsmath} 4 | \usepackage{amssymb} 5 | \usepackage{amsfonts} 6 | \usepackage{mathrsfs} 7 | \usepackage{XCharter} 8 | \usepackage{fancyhdr} 9 | \setCJKmainfont{DENGL.TTF} 10 | \usepackage{eulervm} 11 | \usepackage{graphicx} 12 | 13 | \topmargin -.5in 14 | \textheight 9in 15 | \oddsidemargin -.25in 16 | \evensidemargin -.25in 17 | \textwidth 7in 18 | \pagestyle{fancy} 19 | 20 | \newenvironment{proof}{\\\ignorespaces\textbf{Proof:}}{\hfill $\square$\par\noindent} 21 | \newenvironment{solution}{\ignorespaces\textbf{Solution:}}{\hfill $\square$\par\noindent} 22 | 23 | \newenvironment{rcases}{\left.\begin{aligned}}{\end{aligned}\right\rbrace} 24 | 25 | \title{Fair Dice} 26 | \author{张志成 518030910439} 27 | \date{\today} 28 | 29 | \begin{document} 30 | \maketitle 31 | 32 | \section{Problem} 33 | 骰子的六个面上分别有数字1,...,6。假设你有办法对骰子内部添东西使得抛掷后各个面出现的概率分布为给定分布。是否可以造出两个骰子，同时抛出后得到的两个数 34 | 之和在2,...,12之间等概率分布？ 35 | 36 | \section{Solution} 37 | I utilize the tool of Generative Function to represent the distribution of each roll. 38 | $$ P(x) = p_1x + p_2x^2 + p_3x^3 + p_4x^4 + p_5x^5 + p_6x^6 $$ 39 | Here, $p_i$ represents the probability of the dice roll being $i$. 40 | 41 | With 2 dices, we can simply multiply two generative functions. 42 | \begin{align*} 43 | P(x) &= p_1x + p_2x^2 + p_3x^3 + p_4x^4 + p_5x^5 + p_6x^6 \\ 44 | Q(x) &= q_1x + q_2x^2 + q_3x^3 + q_4x^4 + q_5x^5 + q_6x^6 \\ 45 | \therefore T(x) &= P(x)Q(x) \\ &= (p_1x + p_2x^2 + p_3x^3 + p_4x^4 + p_5x^5 + p_6x^6)(q_1x + q_2x^2 + q_3x^3 + q_4x^4 + q_5x^5 + q_6x^6) \\ 46 | &= t_2x^2 + ... + t_{12}x^{12} 47 | \end{align*} 48 | The problem asks for a equal distribution between 2 and 12. 49 | Thus, we have the following constraints: 50 | \begin{align} 51 | t_2 = t_3 = ... = t_{12} &= \frac{1}{11} \\ 52 | \sum_{i = 1}^{6} p_i &= 1 \\ 53 | \sum_{i = 1}^{6} q_i &= 1 54 | \end{align} 55 | The is a set of equations. So I tried using mathematica to solve it but it only yields complex number answers, which is not accpetable. Below is the code I use: 56 | \begin{figure}[htbp] 57 | \includegraphics{solve.png} 58 | \caption{Mathematica sucks} 59 | \end{figure} 60 | 61 | So we have to further consider these constrains. 62 | \begin{align*} 63 | P(x)Q(x) &= \frac{1}{11} * (x^2 + ... + x^{12}) \\ 64 | &= \frac{1}{11} * x^2 * \frac{x^{11}-1}{x-1} 65 | \end{align*} 66 | 67 | The problem is really asking if there are any factorization of $x^2 * \frac{x^{11}-1}{x-1} \in \mathbb{R}[x]$. 68 | 69 | But there is none since the only root $x \in \mathbb{R}$ is 1. This result is confirmed using Mathematica, see \textit{factor.png} in the folder. 70 | \section{Conclusion} 71 | I used the above method to show that no solution exists. Futher explorations still interest me whether there is a 72 | pattern of possible distribution feasible under this rule. 73 | \end{document} -------------------------------------------------------------------------------- /Week1/quiz_fair_dice_张志成/gen.py: -------------------------------------------------------------------------------- 1 | ans = [[], [], [], [], [], [], [], [], [], [], [], [], []] 2 | for i in range(1,7): 3 | for j in range(1, 7): 4 | ans[i + j].append("p%d * q%d" % (i,j)) 5 | for i in range(1, 12): 6 | print(" + ".join(ans[i]) + " == 1/11 && ") -------------------------------------------------------------------------------- /Week1/quiz_fair_dice_张志成/solve.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week1/quiz_fair_dice_张志成/solve.png 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4 | \usepackage{amssymb} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{geometry} 8 | \usepackage{booktabs} 9 | \usepackage{bm} 10 | \usepackage{tcolorbox} 11 | \CTEXoptions[today=old] 12 | %Some commonly used notations 13 | %\geometry{a4paper,bottom = 3cm,left = 3cm, right = 3cm} 14 | 15 | %for reference 16 | \usepackage{hyperref} 17 | \usepackage[capitalise]{cleveref} 18 | \crefname{enumi}{}{} 19 | 20 | \newtheorem{theorem}{Theorem} 21 | \newtheorem{lemma}[theorem]{Lemma} 22 | \newtheorem{proposition}[theorem]{Proposition} 23 | \newtheorem{corollary}[theorem]{Corollary} 24 | \newtheorem{fact}[theorem]{Fact} 25 | \newtheorem{definition}[theorem]{Definition} 26 | \newtheorem{remark}[theorem]{Remark} 27 | \newtheorem{question}[theorem]{Question} 28 | \newtheorem{answer}[theorem]{Answer} 29 | \newtheorem{exercise}[theorem]{Exercise} 30 | \newtheorem{example}[theorem]{Example} 31 | %\newenvironment{proof}{\noindent \textbf{Proof:}}{$\Box$} 32 | \newtheorem{observation}[theorem]{Observation} 33 | 34 | 35 | \newtheorem{problem}{Problem} 36 | \newtheorem*{problem*}{Problem} 37 | \newtheorem{solution}{Solution} 38 | \newtheorem*{solution*}{Solution} 39 | 40 | %to use newcommand for convenience 41 | \newcommand\field{\mathbb{F}} 42 | \newcommand\Real{\mathbb{R}} 43 | \newcommand\Q{\mathbb{Q}} 44 | \newcommand\Z{\mathbb{Z}} 45 | \newcommand\complex{\mathbb{C}} 46 | 47 | %this is how we define operators. 48 | \DeclareMathOperator{\rank}{rank} % rank 49 | 50 | \title{Discussions on some quizzes} 51 | \author{柳志轩 518030910426} 52 | \date{\today} 53 | 54 | \begin{document} 55 | \maketitle 56 | \begin{problem} 57 | 中原大战，冯玉祥问士兵：空中飞机多还是乌鸦多？众人答：乌鸦多。冯再问：然则 58 | 乌鸦拉屎掉到你们头上没有？众人异口同声：没有。冯说：所以嘛，随着飞机投下的 59 | 炸弹的命中机会就更少了，大家莫怕！ 60 | \end{problem} 61 | 62 | \begin{solution} 63 | 考虑被炸弹炸到的概率和被乌鸦拉屎砸到的概率，是一件很有意思的事情。如果不考虑乌鸦聚集地(如树林)被砸的概率较高和飞机扔炸弹一般往目标密集的地方扔，炸弹(屎)的覆盖范围不会重叠等因素，假设乌鸦和飞机均在该固定区域随机拉屎或扔炸弹，则p(飞机)=num(炸弹数)*S(炸弹覆盖面积)/S(区域面积)，p(乌鸦)=num(拉屎数)*S(拉屎的覆盖面积)/S(区域面积),考虑到飞机的数量与乌鸦的数量，炸弹覆盖面积与屎的覆盖面积等，实际上被炸弹砸到的概率确实低于被乌鸦拉屎砸到的概率。但是飞机扔炸弹不可能随机扔，而鸟类一般与人类都会保持距离，所以冯玉祥的话是缺乏根据的。 64 | \end{solution} 65 | 66 | \begin{problem} 67 | 太阳连续升起了 100 天。明天照常升起的概率有多大？ 68 | \end{problem} 69 | 70 | \begin{solution} 71 | 拉普拉斯提出该类问题的正确答案是$\frac{n+1}{n+2}$，推导过程使用了拉普拉斯概率计算的第六第七原则。有兴趣的同学可以看看$https://blog.csdn.net/three_body/article/details/14498889$该博客。 72 | \end{solution} 73 | 74 | \begin{problem} 75 | 对于生男生女，比如生男生女概率各 50％，每个家庭都生到第一个男孩就不再生，那 76 | 么产生的男女比例是多少？ 77 | \end{problem} 78 | 79 | \begin{solution} 80 | 这的确是一个反直觉的问题。按照直觉我们会认为男孩的概率会大一些，但实际上男女孩的概率是一样的。此题想要想明白可以这样考虑：在父母们生第一个孩子时，男女孩的概率相同，在第一个孩子中男女数量相同。生了女孩的父母继续生第二个孩子的时候，在第二个孩子中男女数量也相同，不断重复，可以发现最终总的男女孩数量还是相同的。也即男女比例仍是1:1的关系。 81 | \end{solution} 82 | 83 | 84 | 85 | \end{document} 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Conditional Probability 郭睿涵/E9.2 Symmetry Conditional Probability 郭睿涵.tex: -------------------------------------------------------------------------------- 1 | \documentclass[UTF8]{ctexart} 2 | \usepackage{amsmath} 3 | \usepackage{amssymb} 4 | \usepackage{amsthm} 5 | \usepackage{graphicx} 6 | \usepackage{bm} 7 | \usepackage{CJK} 8 | \usepackage{float} 9 | \usepackage{mdframed} 10 | 11 | \usepackage{indentfirst} 12 | \setlength{\parindent}{2em} 13 | 14 | \providecommand{\abs}[1]{\lvert#1\rvert} 15 | \providecommand{\norm}[1]{\lVert#1\rVert} 16 | \providecommand{\ud}[1]{\underline{#1}} 17 | 18 | \newmdtheoremenv{thm}{Theorem} 19 | \newmdtheoremenv{lemma}[thm]{Lemma} 20 | \newmdtheoremenv{fact}[thm]{Fact} 21 | \newmdtheoremenv{cor}[thm]{Corollary} 22 | \newtheorem{eg}{Example} 23 | \newtheorem{ex}{Exercise} 24 | \newmdtheoremenv{defi}{Definition} 25 | \newenvironment{sol} 26 | {\par\vspace{3mm}\noindent{\it Solution}.} 27 | {\qed \\ \medskip} 28 | 29 | \newcommand{\ov}{\overline} 30 | \newcommand{\ca}{{\cal A}} 31 | \newcommand{\cb}{{\cal B}} 32 | \newcommand{\cc}{{\cal C}} 33 | \newcommand{\cd}{{\cal D}} 34 | \newcommand{\ce}{{\cal E}} 35 | \newcommand{\cf}{{\cal F}} 36 | \newcommand{\ch}{{\cal H}} 37 | \newcommand{\cl}{{\cal L}} 38 | \newcommand{\cm}{{\cal M}} 39 | \newcommand{\cp}{{\cal P}} 40 | \newcommand{\cs}{{\cal S}} 41 | \newcommand{\cz}{{\cal Z}} 42 | \newcommand{\eps}{\varepsilon} 43 | \newcommand{\ra}{\rightarrow} 44 | \newcommand{\la}{\leftarrow} 45 | \newcommand{\Ra}{\Rightarrow} 46 | \newcommand{\dist}{\mbox{\rm dist}} 47 | \newcommand{\bn}{{\mathbb N}} 48 | \newcommand{\bz}{{\mathbb Z}} 49 | 50 | \newcommand{\expe}{{\mathsf E}} 51 | \newcommand{\pr}{{\mathsf{Pr}}} 52 | \usepackage{amsthm,amsmath,amssymb} 53 | 54 | \usepackage{mathrsfs} 55 | 56 | \setlength{\parindent}{0pt} 57 | %\setlength{\parskip}{2ex} 58 | \newenvironment{proofof}[1]{\bigskip\noindent{\itshape #1. }}{\hfill$\Box$\medskip} 59 | \usepackage{amsthm,amsmath,amssymb} 60 | 61 | \theoremstyle{definition} 62 | \newtheorem{problem}{Problem} 63 | \newtheorem*{problem*}{Problem} 64 | 65 | \pagenumbering{gobble} 66 | 67 | \begin{document} 68 | 69 | \title{E9.2 Symmetry Conditional Probability} 70 | \date{May. 22, 2020} 71 | 72 | \maketitle 73 | \paragraph{E9.2 } Suppose $X,Y \in \mathcal{L}^1(\Omega,\mathcal{F},P)$, and $E(X|Y) = Y,a.s.,$ $E(Y|X) = X,a.x..$ To proof $P(X=Y)=1$ 74 | \paragraph{Solution} 75 | As the hint said, 76 | \begin{align*} 77 | E((X-Y)I_{Y\leq c}) = E((X-Y)I_{Y\leq c}I_{X\leq c}) + E((X-Y)I_{Y\leq c}I_{X>c})\\ 78 | E((X-Y)I_{X\leq c}) = E((X-Y)I_{Y\leq c}I_{X\leq c}) + E((X-Y)I_{X\leq c}I_{Y>c}) 79 | \end{align*} 80 | At the same time, 81 | \begin{align*} 82 | E((X-Y)I_{X\leq c}) &= E((X-E(X|Y))I_{X\leq c})\\&= E(XI_{X\leq c}) - E((X|Y)I_{X\leq c})\\&= E(XI_{X\leq c}) - E(XI_{X\leq c}) = 0 83 | \end{align*} 84 | So that 85 | 86 | \begin{align*} 87 | E((X-Y)I_{Y\leq c}I_{X>c}) = E((X-Y)I_{X\leq c}I_{Y>c}) 88 | \end{align*} 89 | Because the left side is non-negative and the right side is non-positive, both of them equal to 0. Then we can present the following events: 90 | \begin{align*} 91 | (X c)\\ 92 | (X>Y) = \bigcup_{c\in \mathbb{Q}}(X > c)\cap (Y \leq c) 93 | \end{align*} 94 | They are all null events, which leads to the event$(X\neq Y)$ is a null event. Then we got $P(X = Y) = 1$. 95 | \subparagraph{Question} Is the union of infinite null-set is also a null-set? 96 | \paragraph{Another way to proof} 97 | \begin{align*} 98 | E(X^2) = E(X E(Y|X)) = E(E(XY|X)) = E(XY)\\ 99 | E(Y^2) = E(Y E(X|Y)) = E(E(XY|Y)) = E(XY)\\ 100 | \end{align*} 101 | So that 102 | \begin{align*} 103 | E((X-Y)^2) = E(X^2 + Y^2 - 2XY) = 0 104 | \end{align*} 105 | Which means X and Y is different on a null-set. 106 | \end{document} 107 | 108 | -------------------------------------------------------------------------------- /Week11/E9.2 Symmetry Conditional Probability 郭睿涵/tmp: -------------------------------------------------------------------------------- 1 | 2 | -------------------------------------------------------------------------------- /Week11/E9_Conditional_Expectation_张志成/Exercise_E9_Conditional_Expectation.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week11/E9_Conditional_Expectation_张志成/Exercise_E9_Conditional_Expectation.pdf -------------------------------------------------------------------------------- /Week11/Proof of a Simple Lemma 马浩博/discussion.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week11/Proof of a Simple Lemma 马浩博/discussion.pdf -------------------------------------------------------------------------------- /Week11/Proof of a Simple Lemma 马浩博/discussion.tex: -------------------------------------------------------------------------------- 1 | % This is a template for lecture notes. 2 | \documentclass[12pt]{article} 3 | \usepackage{amssymb} 4 | \usepackage[UTF8]{ctex} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{geometry} 8 | \usepackage{booktabs} 9 | \usepackage{bm} 10 | \usepackage{cite} 11 | %\usepackage{CJK} 12 | \usepackage[many]{tcolorbox} 13 | %\tcbuselibrary{listingsutf8} 14 | %\tcbuselibrary{skins, breakable, theorems, most} 15 | %\geometry{a4paper,bottom = 3cm,left = 3cm, right = 3cm} 16 | \CTEXoptions[today=old] 17 | %for reference 18 | \usepackage{hyperref} 19 | \usepackage[capitalise]{cleveref} 20 | \crefname{enumi}{}{} 21 | 22 | \newtheoremstyle{mythm}{1.5ex plus 1ex minus .2ex}{1.5ex plus 1ex minus .2ex} 23 | {}{\parindent}{\bfseries}{}{1em}{} 24 | \theoremstyle{mythm} 25 | \newtheorem{theorem}{Theorem} 26 | \newtheorem{lemma}[theorem]{Lemma} 27 | \newtheorem{corollary}[theorem]{Corollary} 28 | \newtheorem{fact}[theorem]{Fact} 29 | \newtheorem{definition}[theorem]{Definition} 30 | \newtheorem*{remark}{Remark} 31 | 32 | %\newenvironment{proof}{\noindent \textbf{Proof:}}{$\Box$} 33 | 34 | %to use newcommand for convenience 35 | \newcommand\field{\mathbb{F}} 36 | \newcommand\Real{\mathbb{R}} 37 | \newcommand\Q{\mathbb{Q}} 38 | \newcommand\Z{\mathbb{Z}} 39 | \newcommand\complex{\mathbb{C}} 40 | \newcommand\cc{\mathcal{C}} 41 | \newcommand\uu{\mathcal{U}} 42 | \newcommand\pp{\mathcal{P}} 43 | \newcommand\ff{\mathcal{F}} 44 | \renewcommand\refname{Reference} 45 | \renewcommand{\proofname}{Proof} 46 | \DeclareMathOperator{\range}{range} 47 | 48 | \title{Proof of a Simple Lemma} 49 | \author{马浩博 518030910428} 50 | \date{\today} 51 | \begin{document} 52 | \maketitle 53 | 54 | This is an easy exercise arranged in class. Nobody did it last week, so I give out a simple proof. 55 | 56 | \begin{lemma} 57 | Suppose that $h$ is a Borel measurable function from $R$ to $R$. Then 58 | $$h(X) \in \mathcal{L}^1(\Omega,\mathcal{F},P) \ if \ and \ only \ if \ h \in \mathcal{L}^1(R,\mathcal{B},\Lambda_X )$$ 59 | and then 60 | $$Eh(X) = \Lambda_X(h) = \int_{R} h(x) \Lambda_X\, (dx).$$ 61 | \end{lemma} 62 | 63 | \begin{proof} 64 | Proof is simple cause we can just feed everything into the standard machine. 65 | First if $h = I_B (B \in \mathcal{B})$, by the definition of $\Lambda_X$ we know that $$Eh(X) = \Lambda_X(h) = \int_{R} h(x) \Lambda_X\, (dx).$$ 66 | Then use linearity to tell that the conclusion is still true when $h$ is a simple function on $(R,\mathcal{B})$. 67 | Next, when $h$ is a non-negative function, by MON we can also get that 68 | $$Eh(X) = \Lambda_X(h) = \int_{R} h(x) \Lambda_X\, (dx).$$ 69 | Use linearity again and we finish the proof. 70 | \end{proof} 71 | 72 | \end{document} -------------------------------------------------------------------------------- /Week11/Symmetry Conditional Expectation/Symmetry Conditional Expectation.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week11/Symmetry Conditional Expectation/Symmetry Conditional Expectation.pdf -------------------------------------------------------------------------------- /Week11/Symmetry Conditional Expectation/Symmetry Conditional Expectation.tex: -------------------------------------------------------------------------------- 1 | \documentclass[UTF8, 12pt]{article} 2 | \usepackage{xeCJK} 3 | \usepackage{enumitem} 4 | \usepackage{amsmath} 5 | \usepackage{amssymb} 6 | \usepackage{amsfonts} 7 | \usepackage{mathrsfs} 8 | \usepackage{XCharter} 9 | \usepackage{fancyhdr} 10 | \usepackage{eulervm} 11 | \usepackage{graphicx} 12 | \usepackage{mdframed} 13 | \usepackage{ntheorem} 14 | \usepackage{lipsum} 15 | 16 | \topmargin -.5in 17 | \textheight 9in 18 | \oddsidemargin -.25in 19 | \evensidemargin -.25in 20 | \textwidth 7in 21 | \pagestyle{fancy} 22 | 23 | \newenvironment{proof}{\noindent\ignorespaces\textbf{Proof:}}{\hfill $\square$\par\noindent} 24 | \newenvironment{solution}{\noindent\ignorespaces\textbf{Solution:}}{\hfill $\square$\par\noindent} 25 | 26 | \theoremstyle{break} 27 | \newtheorem{claim}{Claim} 28 | \newtheorem{problem}{Problem} 29 | \newtheorem{lemma}{Lemma} 30 | \newtheorem*{theorem*}{Theorem} 31 | 32 | \newenvironment{rcases}{\left.\begin{aligned}}{\end{aligned}\right\rbrace} 33 | 34 | \title{Symmetry Conditional Expectation} 35 | \author{于峥 518030910437} 36 | \date{\today} 37 | 38 | \begin{document} 39 | \maketitle 40 | 41 | \begin{problem}[\textbf{Exercise 9.2} of \textit{Chapter E}] 42 | Suppose that $X, Y \in \mathcal{L}^1(\Omega, \mathcal{F}, P)$ and that 43 | $$ 44 | E(X|Y) =Y ,a.s., ~~~ E(Y|X) =X ,a.s. 45 | $$ 46 | 47 | Prove that $P(X=Y)=1$. 48 | 49 | Hint, consider $E(X-Y; X > c, Y \leq c) + E(X - Y; X \leq c, Y \leq c)$ 50 | \end{problem} 51 | 52 | \begin{proof} 53 | Because of $E(X|Y)=Y$, we have $E(X ; Y \leq c) = E(Y; Y \leq c)$ for every $c$ 54 | by the definition of conditional expectation. Because $X, Y \in \mathcal{L}^1(\Omega, \mathcal{F}, P)$, so 55 | 56 | \begin{align*} 57 | 0 58 | &= E(X - Y; Y \leq c) \\ 59 | &=E(X-Y; X > c, Y \leq c) + E(X - Y; X \leq c, Y \leq c) 60 | \end{align*} 61 | 62 | Consider $P(X < Y)$, note that event $\{ X < Y \} = \bigcup_{c} \{ X < c < Y\}$, and 63 | $$ 64 | \{ X < c - \frac 1 n < c + \frac 1 n < Y\} \uparrow \{ X < c < Y \} 65 | $$ 66 | 67 | If $P(X < Y) > 0$, $\exists c, n$, s.t. 68 | $P(X < c - \frac 1 n < c + \frac 1 n < Y) = p > 0$, hence, 69 | $$ 70 | E(X-Y; X > c, Y \leq c) \geq \frac 2 n P(X < c - \frac 1 n < c + \frac 1 n < Y) = \frac {2p} {n} > 0 71 | $$ 72 | 73 | And the $E(X - Y; X \leq c, Y \leq c)$ must be zero because the symmetric in $X, Y$, 74 | we get $E(X - Y; Y \leq c) > 0$, it is impossible. Therefore $P(X < Y) = 0$. 75 | 76 | We can prove $P(X > Y) = 0$ from $E(Y|X)=X$ use the same way. Hence 77 | $P(X=Y)=1$. 78 | 79 | 80 | 81 | 82 | 83 | \end{proof} 84 | \end{document} -------------------------------------------------------------------------------- /Week11/The Distribution Function of X1 Divided by X2_吴润哲/The Distribution Function of X1 Divided by X2.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week11/The Distribution Function of X1 Divided by X2_吴润哲/The Distribution Function of X1 Divided by X2.pdf -------------------------------------------------------------------------------- /Week12/5.22note_Conditional_Expectation_and_Martingale_郭睿涵/5.22note_Conditional_Expectation_and_Martingale_郭睿涵.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week12/5.22note_Conditional_Expectation_and_Martingale_郭睿涵/5.22note_Conditional_Expectation_and_Martingale_郭睿涵.pdf -------------------------------------------------------------------------------- /Week12/5.22note_Conditional_Expectation_and_Martingale_郭睿涵/tmp: -------------------------------------------------------------------------------- 1 | 2 | -------------------------------------------------------------------------------- /Week12/E10.1 Polya Urn 郭林松/Ploya Urn.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week12/E10.1 Polya Urn 郭林松/Ploya Urn.pdf -------------------------------------------------------------------------------- /Week12/Ex10.2 Martingale formulation of Bellman's Optimality Principle_王一/Ex10.2 Martingale formulation of Bellman's Optimality Principle.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week12/Ex10.2 Martingale formulation of Bellman's Optimality Principle_王一/Ex10.2 Martingale formulation of Bellman's Optimality Principle.pdf -------------------------------------------------------------------------------- /Week12/Ex10.2 Martingale formulation of Bellman's Optimality Principle_王一/Ex10.2 Martingale formulation of Bellman's Optimality Principle.synctex.gz: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week12/Ex10.2 Martingale formulation of Bellman's Optimality Principle_王一/Ex10.2 Martingale formulation of Bellman's Optimality Principle.synctex.gz -------------------------------------------------------------------------------- /Week12/Ex10.2 Martingale formulation of Bellman's Optimality Principle_王一/Ex10.2 Martingale formulation of Bellman's Optimality Principle.tex: -------------------------------------------------------------------------------- 1 | \documentclass[11pt,a4paper,oneside]{article} 2 | 3 | \usepackage{euler,amsthm,amsmath,amsfonts,graphicx,epigraph,indentfirst,enumerate,comment,listings,fontspec,color,subcaption,listings} 4 | \usepackage{xeCJK} 5 | \usepackage{hw} 6 | \usepackage{pythonhighlight} 7 | 8 | \renewcommand{\hwtitle} {Martingale Formulation of Bellman's Optimality Principle} 9 | \renewcommand{\hwauthor}{Wang Yi - 518030910413} 10 | \renewcommand{\hwdate}{\today} 11 | 12 | \begin{document} 13 | \title{\hwtitle} 14 | \author{\hwauthor} 15 | \date{\hwdate} 16 | \maketitle 17 | 18 | \section*{Martingale Formulation of Bellman's Optimality Principle(Ex10.2)} 19 | \begin{problem}{1} 20 | \statement 21 | Your winnings per unit stake on game $n$ are $\varepsilon_n$, where the $\varepsilon_n$ are IID RVs with 22 | \[ 23 | P(\epsilon_n = +1) = p,P(\epsilon_n = -1) = q, \textrm{where} \; \frac{1}{2} < p = 1 - q < 1 24 | \] 25 | Your stake $C_n$ on game $n$ must lie between $0$ and $Z_{n-1}$, where $Z_{n-1}$ is your fortune at time $n - 1$. Your object is to maximize the expected 'interest rate' $E\log (Z_N/Z_0)$, where $N$ is a given integer representing the length of the game, and $Z_0$, your fortune at time $0$ is a given constant. Let $\mathcal{F}_n = \sigma(\epsilon_1, \dots, \epsilon_n)$ be your 'history' up to time $n$. Show that if $C$ is any previsible strategy, the $\log Z_n - n\alpha$ is a supermartingale, where $\alpha$ denotes the 'entropy' 26 | \[ 27 | \alpha = p \log p + q \log q + \log 2, 28 | \] 29 | so that $E\log (Z_n/Z_0) \leq N\alpha$, but that, for a certain strategy, $\log Z_n - n\alpha$ is a martingale. What is the best strategy? 30 | \solution 31 | \begin{proof} 32 | We can get $Z_n = Z_{n-1} + C_n \cdot \epsilon_n$ by definition of this game. Then we will get 33 | \[ 34 | \begin{split} 35 | E[(\log Z_n - n\alpha) - (\log Z_{n-1} - (n - 1)\alpha)| \mathcal{F}_{n-1}] &= E[\log Z_n - \log Z_{n-1}| \mathcal{F}_{n-1}] - \alpha\\ 36 | &= E[\log (1 + \frac{C_n \cdot \epsilon_n}{Z_{n-1}})| \mathcal{F}_{n-1}] - \alpha \\ 37 | &= p\log (1 + \frac{C_n }{Z_{n-1}}) + q\log (1 - \frac{C_n }{Z_{n-1}}) - \alpha 38 | \end{split} 39 | \] 40 | $p\log (1 + \frac{C_n }{Z_{n-1}}) + q\log (1 - \frac{C_n }{Z_{n-1}}) - \alpha$ reaches the maximum $\alpha$, when $\frac{C_n }{Z_{n-1}} = \frac{p-q}{p + q} = p - q$. The maximium is $p\log (1 + p - q) + q\log (1 - (p - q)) - \alpha = \alpha - \alpha = 0$. Thus 41 | \[ 42 | E[(\log Z_n - n\alpha) - (\log Z_{n-1} - (n - 1)\alpha)| \mathcal{F}_{n-1}] \leq 0. 43 | \] 44 | Therefore $\log Z_n - n\alpha$ is a supermartingale.\\ 45 | When $C_n = (p - q)Z_{n - 1}$, it is a martingale, so the best strategy is $C_n = (p - q)Z_{n - 1}$. 46 | \end{proof} 47 | \end{problem} 48 | \end{document} -------------------------------------------------------------------------------- /Week12/Ex10.6 ABRACADABRA_顾逸/Doob's Optional Stopping Theorem and the ABRACADABRA Problem.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week12/Ex10.6 ABRACADABRA_顾逸/Doob's Optional Stopping Theorem and the ABRACADABRA Problem.pdf -------------------------------------------------------------------------------- /Week12/Gamblers Ruin_姚远/Gamblers Ruin.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week12/Gamblers Ruin_姚远/Gamblers Ruin.pdf -------------------------------------------------------------------------------- /Week12/optional time operation/optional time operation.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week12/optional time operation/optional time operation.pdf -------------------------------------------------------------------------------- /Week12/optional time operation/optional time operation.tex: -------------------------------------------------------------------------------- 1 | \documentclass[UTF8, 12pt]{article} 2 | \usepackage{xeCJK} 3 | \usepackage{enumitem} 4 | \usepackage{amsmath} 5 | \usepackage{amssymb} 6 | \usepackage{amsfonts} 7 | \usepackage{mathrsfs} 8 | \usepackage{XCharter} 9 | \usepackage{fancyhdr} 10 | \usepackage{eulervm} 11 | \usepackage{graphicx} 12 | \usepackage{mdframed} 13 | \usepackage{ntheorem} 14 | \usepackage{lipsum} 15 | 16 | \topmargin -.5in 17 | \textheight 9in 18 | \oddsidemargin -.25in 19 | \evensidemargin -.25in 20 | \textwidth 7in 21 | \pagestyle{fancy} 22 | 23 | \newenvironment{proof}{\noindent\ignorespaces\textbf{Proof:}}{\hfill $\square$\par\noindent} 24 | \newenvironment{solution}{\noindent\ignorespaces\textbf{Solution:}}{\hfill $\square$\par\noindent} 25 | 26 | \theoremstyle{break} 27 | \newtheorem{claim}{Claim} 28 | \newtheorem{problem}{Problem} 29 | \newtheorem{lemma}{Lemma} 30 | \newtheorem*{theorem*}{Theorem} 31 | 32 | \newenvironment{rcases}{\left.\begin{aligned}}{\end{aligned}\right\rbrace} 33 | 34 | \title{Optional Time Operation} 35 | \author{于峥 518030910437} 36 | \date{\today} 37 | 38 | \begin{document} 39 | \maketitle 40 | 41 | \begin{problem}[\textbf{Exercise 10.3} of \textit{Chapter E}] 42 | Suppose that S and T are stopping times 43 | (relative to ($\Omega, \mathcal{F}, \{ \mathcal{F}_n \}$)). 44 | Prove that $S \land T , S \lor T, S + T$ 45 | are stopping time. 46 | \end{problem} 47 | 48 | \begin{proof} 49 | Proof it by definition, $\{S \leq n \}, \{ T \leq n \}, \{S = n \}, \{ T = n \}$ all in $\mathcal{F}_n$, 50 | \begin{align*} 51 | \{ S \land T \leq n \} &= \{ S \leq n \} \cup \{ T \leq n \} \\ 52 | \{ S \lor T \leq n \} &= \{ S \leq n \} \cap \{ T \leq n \} \\ 53 | \{ S + T \leq n \} &= \bigcup_{k=0}^n \{ S = k \} \cap \{T = n-k \} \\ 54 | \end{align*} 55 | 56 | Hence the r.h.s of these equations are $\mathcal{F}_n$-measurable. 57 | \end{proof} 58 | \end{document} -------------------------------------------------------------------------------- /Week14/An Extended Super-Martingale Convergence Theorem with its Proof/An_Extended_Super_Martingale_Convergence_Theorem_with_its_Proof_张扬天.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week14/An Extended Super-Martingale Convergence Theorem with its Proof/An_Extended_Super_Martingale_Convergence_Theorem_with_its_Proof_张扬天.pdf -------------------------------------------------------------------------------- /Week14/σ-algebra Generated By Stopping Time - 郭林松/σ-algebra Generated By Stopping Time.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week14/σ-algebra Generated By Stopping Time - 郭林松/σ-algebra Generated By Stopping Time.pdf -------------------------------------------------------------------------------- /Week2/A_phi is negligible-寻之扬/A_phi is negligible.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week2/A_phi is negligible-寻之扬/A_phi is negligible.pdf -------------------------------------------------------------------------------- /Week2/A_phi is negligible-寻之扬/A_phi is negligible.tex: -------------------------------------------------------------------------------- 1 | % !TEX program = xelatex 2 | \input{D:/template} 3 | 4 | \author{Zhiyang Xun} 5 | \title{Reals Allowing Rational Approximation of Too High Order Are Negligible} 6 | 7 | \begin{document} 8 | \maketitle 9 | 10 | \begin{definition} 11 | $A_\phi$ is the set of reals $x$ in $(0, 1]$ such that 12 | \[ \abs{x - \frac{p}{q}} < \frac{1}{q^2\phi(q)} \] 13 | has infinitely many irreduciple rational solutions $(p, q)$, that is, $(p, q) \in \Z^2$, $q > 0$ and $\gcd(p, q) = 1$. 14 | \end{definition} 15 | 16 | \begin{theorem} 17 | Suppose that $\phi$ is positive. If $\sum_q \frac{1}{q\phi(q)} < \infty$, then $P (A_\phi) = 0$. 18 | \end{theorem} 19 | 20 | \begin{proof} 21 | For each $(p, q) \in \Z^+ \times \Z^+$ satisfying $p \leq q$, we define interval 22 | $I_{(p, q)} = [\frac{p}{q} - \frac{1}{q^2\phi(q)}, \frac{p}{q} + \frac{1}{q^2\phi(q)}]$. 23 | Clearly, $A_\phi \subseteq \bigcup I_{(p, q)}$. 24 | Because each element $x \in A_\phi$ has infinitely many solutions to $\abs{x - \frac{p}{q}} < \frac{1}{q^2\phi(q)}$, and for each $k \geq 1$, only finite $(p, q)$ satisfying $p \leq q < k$, we can see that \[ 25 | A_\phi \subseteq \bigcup_{q=k}^\infty \bigcup_{p=1}^{q} I_{(p, q)} 26 | \] 27 | 28 | Now we only need to prove for each $\epsilon > 0$, there is a positive integer $k$ satisfying $\sum_{q=k}^\infty \sum_{p=1}^{q} \abs{I_{(p, q)}} < \epsilon$ to illustrate $P (A_\phi) = 0$. 29 | 30 | We can see 31 | \begin{align*} 32 | &\sum_{q=k}^\infty \sum_{p=1}^{q} \abs{I_{(p, q)}} \\ 33 | =&\sum_{q=k}^\infty \sum_{p=1}^{q} \frac{2}{q^2\phi(q)} \\ 34 | =&\sum_{q=k}^\infty \frac{2}{q\phi(q)} \\ 35 | =&2\sum_{q=k}^\infty \frac{1}{q\phi(q)} \\ 36 | \end{align*} 37 | 38 | Since $\sum_q \frac{1}{q\phi(q)} < \infty$, for each $\epsilon > 0$, there is a $k \geq 1$ satisfying $\sum_{q=k}^\infty \frac{1}{q\phi(q)} < \frac{\epsilon}{2}$. It implies $\sum_{q=k}^\infty \sum_{p=1}^{q} \abs{I_{(p, q)}} = 2\sum_{q=k}^\infty \frac{1}{q\phi(q)} < \epsilon$. 39 | 40 | Hence, $P(A_\phi) = 0$. 41 | 42 | \end{proof} 43 | 44 | \end{document} -------------------------------------------------------------------------------- /Week2/Cantor-meager-金弘义/cantor-meager.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week2/Cantor-meager-金弘义/cantor-meager.pdf -------------------------------------------------------------------------------- /Week2/Exercise2 Cantor Set-郭林松/discussion.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week2/Exercise2 Cantor Set-郭林松/discussion.pdf -------------------------------------------------------------------------------- /Week2/Lecture Notes-董海辰/notes.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week2/Lecture Notes-董海辰/notes.pdf -------------------------------------------------------------------------------- /Week2/Lecture Notes-董海辰/w1fig.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week2/Lecture Notes-董海辰/w1fig.pdf -------------------------------------------------------------------------------- /Week2/Liouville Number-张志成/Liouville_Number.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week2/Liouville Number-张志成/Liouville_Number.pdf -------------------------------------------------------------------------------- /Week2/Principle of inclusion-exclusion-陈彤/ex12.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week2/Principle of inclusion-exclusion-陈彤/ex12.pdf -------------------------------------------------------------------------------- /Week2/Principle of inclusion-exclusion-陈彤/ex12.tex: -------------------------------------------------------------------------------- 1 | % This is a template for lecture notes. 2 | \documentclass{article} 3 | \usepackage[UTF8]{ctex} 4 | \usepackage{amssymb} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{geometry} 8 | \usepackage{booktabs} 9 | \usepackage{bm} 10 | \usepackage{tcolorbox} 11 | \CTEXoptions[today=old] 12 | %Some commonly used notations 13 | %\geometry{a4paper,bottom = 3cm,left = 3cm, right = 3cm} 14 | 15 | %for reference 16 | \usepackage{hyperref} 17 | \usepackage[capitalise]{cleveref} 18 | \crefname{enumi}{}{} 19 | 20 | \newtheorem{theorem}{Theorem} 21 | \newtheorem{lemma}[theorem]{Lemma} 22 | \newtheorem{proposition}[theorem]{Proposition} 23 | \newtheorem{corollary}[theorem]{Corollary} 24 | \newtheorem{fact}[theorem]{Fact} 25 | \newtheorem{definition}[theorem]{Definition} 26 | \newtheorem{remark}[theorem]{Remark} 27 | \newtheorem{question}[theorem]{Question} 28 | \newtheorem{answer}[theorem]{Answer} 29 | \newtheorem{exercise}[theorem]{Exercise} 30 | \newtheorem{example}[theorem]{Example} 31 | %\newenvironment{proof}{\noindent \textbf{Proof:}}{$\Box$} 32 | \newtheorem{observation}[theorem]{Observation} 33 | 34 | %to use newcommand for convenience 35 | \newcommand\field{\mathbb{F}} 36 | \newcommand\Real{\mathbb{R}} 37 | \newcommand\Q{\mathbb{Q}} 38 | \newcommand\Z{\mathbb{Z}} 39 | \newcommand\complex{\mathbb{C}} 40 | 41 | %this is how we define operators. 42 | \DeclareMathOperator{\rank}{rank} % rank 43 | 44 | \title{Principle of inclusion-exclusion} 45 | \author{Tong CHEN} 46 | \date{\today} 47 | 48 | \begin{document} 49 | \maketitle 50 | \begin{theorem} 51 | (Principle of inclusion-exclusion) Let $A_1, \dots, A_k$ be k finite sets. We have $$|\cup_{i=1}^k A_i |=\sum_i|A_i|-\sum_{in,x+2^{-i} \frac{1}{b - y}, n \in \mathbb{N}}[a, y + \frac{1}{n}]$ and $[a, y + \frac{1}{n}] \in \pi([a,b])$, so $s \in \sigma(\pi([a,b]))$ 68 | \item $s = (x, b]$, $s = \cap_{n > \frac{1}{x - a}, n \in \mathbb{N}}[x - \frac{1}{n}, b]$ and $[x - \frac{1}{n}, b] \in \pi([a,b])$, so $s \in \sigma(\pi([a,b]))$ 69 | \item $s = [a, b]$, $[a, b] \in \pi([a,b])$ so $s \in \sigma(\pi([a,b]))$ 70 | \item $s = (x, y)$, $(x, y) = (x, b] \cap [a, y) \in \sigma(\pi([a,b]))$ 71 | \end{enumerate} 72 | \hspace*{1em}Therefore, $\mathcal{B}([a,b]) \subseteq \sigma(\pi[a,b])$ 73 | \end{proof} 74 | 75 | \end{document} -------------------------------------------------------------------------------- /Week3/Lecture Notes of Measure Space_傅凌玥/Measure Space.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week3/Lecture Notes of Measure Space_傅凌玥/Measure Space.pdf -------------------------------------------------------------------------------- /Week3/Non-Lebesgue-measurable set in R^n_吴润哲/non-Lebesgue-measurable-sets-in-R^n.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week3/Non-Lebesgue-measurable set in R^n_吴润哲/non-Lebesgue-measurable-sets-in-R^n.pdf -------------------------------------------------------------------------------- /Week3/Notes of Measure Space 刘成锴/Measure Space.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week3/Notes of Measure Space 刘成锴/Measure Space.pdf -------------------------------------------------------------------------------- /Week3/Notes of Random Variable 刘成锴/Random Variables.md: -------------------------------------------------------------------------------- 1 | # Random Variables 2 | 3 | ## by 刘成锴 4 | 5 | Let $(S, \Sigma)$ be a measurable space, so that $\Sigma$ is a $\sigma$-algebra on $S$. 6 | 7 | ## Definitions. $\Sigma$-measurable function, $m\Sigma$, $(m\Sigma)^+$, $b\Sigma$ 8 | 9 | Suppose that $h : S \rightarrow \mathbf R$. For $A \in \mathbf R$, define 10 | $$ 11 | h^{-1}(A) := {s \in S : h(s) \in A} 12 | $$ 13 | Then $h$ is called *$\Sigma$-measurable* if $h^{-1} : \mathcal B \rightarrow \Sigma$, that is, $h^{-1}(A) \in \Sigma, \forall A \in \mathcal B$. 14 | 15 | Here is a picture of a $\Sigma$-measurable function $h$: 16 | $$ 17 | S \stackrel{\mathrm{h}}{\longrightarrow} \mathbf{R} 18 | \\ 19 | \Sigma \stackrel{\mathrm{h^{-1}}}{\longleftarrow} \mathcal{B} 20 | $$ 21 | 22 | 23 | $m\Sigma$ : the class of $\Sigma$-measurable functions on $S$ 24 | 25 | $(m\Sigma)^+$ : the class of non-negative elements in $m\Sigma$ 26 | 27 | $b\Sigma$ : the class of bounded $\Sigma$-measurable functions on $S$ 28 | 29 | ### Borel function 30 | 31 | A function $h$ from a topological space $S$ to $\mathbf R$ is called **Borel** if $h$ is $\mathcal B(S)$-measurable. 32 | 33 | The most important case is when $S$ itself is $\mathbf R$. 34 | 35 | 36 | 37 | ## Elementary Propositions on measurability 38 | 39 | (a) The map $h^{-1}$ preserves all set operations: 40 | 41 | ​ $h^{-1}\left(\cup_{\alpha} A_{\alpha}\right)=U_{\alpha} h^{-1}\left(A_{\alpha}\right), \quad h^{-1}\left(A^{c}\right)=\left(h^{-1}(A)\right)^{c}$, etc. 42 | 43 | (b) $\text {If } \mathcal{C} \subseteq \mathcal{B} \text { and } \sigma(\mathcal{C})=\mathcal{B}, \text { then } h^{-1}: \mathcal{C} \rightarrow \Sigma \quad \Rightarrow \quad h \in \mathrm{m} \Sigma$ 44 | 45 | (c) If $S$ is *topological* and $h : S \rightarrow \mathbf R$ is *continuous*, then $h$ is ***Borel***. 46 | 47 | (d) For any measurable space $(S, \Sigma)$, a function $h : S \rightarrow \mathbf R$ is $\Sigma$-measurable if 48 | 49 | ​ $\{h \leq c\}:=\{s \in S: h(s) \leq c\} \in \Sigma \quad(\forall c \in \mathbf{R})$ 50 | 51 | 52 | 53 | ## Lemma. Sums and products of measurable functions are measurable 54 | 55 | $m\Sigma$ is an algebra over $\mathbf R$, that is, 56 | 57 | if $\lambda \in \mathbf R$ and $h, h_1, h_2 \in m\Sigma$, then 58 | $$ 59 | h1 + h2 \in m\Sigma, \quad h_1h_2 \in m\Sigma, \quad \lambda h \in m \Sigma 60 | $$ 61 | 62 | 63 | ## Composition Lemma 复合函数可测性引理 64 | 65 | If $h \in m\Sigma$ and $f \in m \mathcal B$, then $f \ \circ \ h \in m\Sigma$. 66 | 67 | Proof. 68 | $$ 69 | S \stackrel{\mathrm{h}}{\longrightarrow} \mathbf{R} \stackrel{\mathrm{f}}{\longrightarrow} \mathbf{R} 70 | \\ 71 | \Sigma \stackrel{\mathrm{h^{-1}}}{\longleftarrow} \mathcal{B} \stackrel{\mathrm{f^{-1}}}{\longleftarrow} \mathcal{B} 72 | $$ 73 | 74 | 75 | ## 76 | 77 | ## Definition. Random Variable 78 | 79 | Let $(\Omega, \mathcal F)$ be our (sample space, family of events). A *random variable* is an element of $m\mathcal F$. Thus, 80 | $$ 81 | X : \Omega \rightarrow \mathbf R, \quad X ^ {-1} : \mathcal B \rightarrow \mathcal F 82 | $$ 83 | 84 | 85 | -------------------------------------------------------------------------------- /Week3/Notes of Random Variable 刘成锴/Random Variables.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week3/Notes of Random Variable 刘成锴/Random Variables.pdf -------------------------------------------------------------------------------- /Week3/Trivial proofs of Inclusion-exclusion Principle_傅凌玥/Trival Proof of Inclusion-exclusion Principle.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week3/Trivial proofs of Inclusion-exclusion Principle_傅凌玥/Trival Proof of Inclusion-exclusion Principle.pdf -------------------------------------------------------------------------------- /Week3/Trivial proofs of Inclusion-exclusion Principle_傅凌玥/Trival Proof of Inclusion-exclusion Principle.tex: -------------------------------------------------------------------------------- 1 | \input{/Users/fulingyue/Desktop/def.tex} 2 | \newtheorem{theorem}{Theorem} 3 | \title{Trival Proofs of Inclusion-exclusion Principle} 4 | \author{Fu Lingyue} 5 | \date{\today} 6 | \begin{document} 7 | \maketitle 8 | \begin{theorem} 9 | Let $A_1, \dots, A_k$ be k finite sets. We have 10 | $$|\cup_{i=1}^k A_i |=\sum_i|A_i|-\sum_{i 0$. That is, 40 | \begin{align*} 41 | \forall \varepsilon > 0, \exists I_i' := (a_i, b_i + \frac{\varepsilon}{2} ) \in \Sigma_0, I_i \subseteq I_i', \mu_0(I_i') - \mu_0(I_i) = \frac{\varepsilon}{2} < \varepsilon 42 | .\end{align*} 43 | 44 | Also, we have $A \in \Sigma_0$, which means $A = \bigcup_{t=1}^{r} A_t$. Similarly, for a fixed $\varepsilon >0$, for each $A_t$, we can have $[l_t, r_t] \subseteq A_t$ with $\mu_0(A_t) - \mu_0([l_t, r_t]) < \frac{\varepsilon}{r} $. 45 | 46 | From this we can get a infinite open cover of $[l_t, r_t]$. By the property of $\mathbb{R} $, there exists a finite open cover $I_{tk_1}', I_{tk_2}',\cdots I_{tk_M}'$ such that $A_t \subseteq \bigcup_{i=1}^M I_{tk_i}'$. Then by the elementary inequality, 47 | \begin{align*} 48 | \mu_0(A) = \sum_{t=1}^r \mu_0(A_t) < \varepsilon + \sum_{t=1}^r \mu_0([l_t, r_t]) &= \varepsilon + \sum_{t=1}^r \mu_0(\bigcup_{i=1}^M I_{tk_i}) \\ 49 | &\le \varepsilon + \sum_{t=1}^r \sum_{i=1}^M \mu_0(I_{tk_i}') \\ 50 | &\le \varepsilon + \sum_{t=1}^r \sum_{i=1}^M (\mu_0(I_{tk_i}) + \varepsilon_i) \\ 51 | &\le \varepsilon + \sum_{i=1}^\infty \mu_0(I_i) + r\sum_{i=1}^\infty \varepsilon_i 52 | .\end{align*} 53 | 54 | Let $\varepsilon _i = \frac{\varepsilon}{r2^i}$, finally We get $\mu_0(A) \le \sum _{i=1}^\infty \mu_0(I_i) + 2\varepsilon $ for all $\varepsilon > 0$. 55 | 56 | Combine the two sides, resulting in 57 | \begin{align*} 58 | \mu_0(A) = \sum _{i\in\mathbb{N} _+} \mu_0(A_i) 59 | .\end{align*} 60 | 61 | That finishes the proof for the countable additivity of $\mu_0$. 62 | 63 | \end{proof} 64 | 65 | 66 | \end{document} 67 | -------------------------------------------------------------------------------- /Week4/A problem of indicator function and lim - 陆嘉馨/A problem of indicator function and lim - 陆嘉馨.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/A problem of indicator function and lim - 陆嘉馨/A problem of indicator function and lim - 陆嘉馨.pdf -------------------------------------------------------------------------------- /Week4/Can_event_space_countably_infinite-郭睿涵/Can_event_space_countably_infinite.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/Can_event_space_countably_infinite-郭睿涵/Can_event_space_countably_infinite.pdf -------------------------------------------------------------------------------- /Week4/Can_event_space_countably_infinite-郭睿涵/Can_event_space_countably_infinite.tex: -------------------------------------------------------------------------------- 1 | \documentclass[UTF8]{ctexart} 2 | 3 | \usepackage{graphicx,amsmath,amssymb,amsthm, boxedminipage} 4 | 5 | \usepackage{algorithm} 6 | \usepackage{algpseudocode} 7 | 8 | 9 | \newtheorem{theorem}{Theorem}%[section] 10 | \newtheorem{proposition}[theorem]{Proposition} 11 | \newtheorem{lemma}[theorem]{Lemma} 12 | \newtheorem{corollary}[theorem]{Corollary} 13 | \newtheorem{definition}[theorem]{Definition} 14 | 15 | \newcommand{\scalar}[2]{\ensuremath{\langle #1, #2\rangle}} 16 | \newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor} 17 | \newcommand{\ceil}[1]{\left\lceil #1 \right\rceil} 18 | \newcommand{\norm}[1]{\|#1\|} 19 | \newcommand{\pfrac}[2]{\left(\frac{#1}{#2}\right)} 20 | \newcommand{\nth}[1]{#1\textsuperscript{th}} 21 | 22 | % \newcommand{\nth}[1]{#1\textsuperscript{th}} 23 | \newcommand{\E}{\mathop{\mathbb{E\/}}} 24 | \newcommand{\N}{\mathbb{N}} 25 | 26 | \newcommand{\R}{\mathbb{R}} 27 | 28 | \newtheorem{exercise}[theorem]{Exercise} 29 | \newtheorem{exerciseD}[theorem]{*Exercise} 30 | \newtheorem{exerciseDD}[theorem]{**Exercise} 31 | 32 | \let\oldexercise\exercise 33 | \renewcommand{\exercise}{\oldexercise\normalfont} 34 | 35 | \let\oldexerciseD\exerciseD 36 | \renewcommand{\exerciseD}{\oldexerciseD\normalfont} 37 | 38 | \let\oldexerciseDD\exerciseDD 39 | \renewcommand{\exerciseDD}{\oldexerciseDD\normalfont} 40 | 41 | 42 | 43 | \newtheorem{problem}{Problem} 44 | \newtheorem*{problem*}{Problem} 45 | \begin{document} 46 | 47 | \author{郭睿涵} 48 | \date{\today} 49 | 50 | \title{Can an event space be a countably infinite set?\\ 51 | \vspace{3mm} 52 | {\large Shanghai Jiaotong University, Spring 2020\\ 53 | } 54 | } 55 | \maketitle 56 | 57 | \paragraph{Exercise 2.}Can an event space be a countably infinite set? 58 | \paragraph{Solution}First of all, event space in measure theory based probability, it is required to be a sigma-algebra. Then we need to proof a sigma-algebra that contains infinitely many sets must be uncountable.\\ 59 | Assume we have a set $\mathbb{X}$, and an infinite sigma-algebra $\mathbb{S}$ on it. I want to proof that $\mathbb{S}$ is uncountable by contradiction.\\ 60 | \paragraph{Assumption} $\mathbb{S}=\{A_{i}\}_{i=1}^{\infty}$,$B_{x} =\cap_{x\in A_{i}}A_{i}$. Due to $\mathbb{S}$ is countable, $B_{x}$ is made of countable intersection, which means it belongs to $\mathbb{S}$.\\ 61 | \paragraph{Lemma} $c \in B_x \cap B_y \rightarrow B_x = B_y$ 62 | \paragraph{Proof} If $x \notin B_c$, $B_x \backslash B_c \subset S$ with $x \in B_x \backslash B_c$. But $B_x$ is the intersection of all the intervals containing x. Therefore $B_x \backslash B_c = B_x$, which means,$B_x = B_c$.\\ 63 | Analogouly, we have $B_y = B_c$. So we've got $B_x = B_y$ when $c \in B_x \cap B_y$. 64 | \\\\ 65 | If there are finite sets of the form $B_x$, then: $\mathbb{S}$ is a union of a finite number of disjoint sets, which leads to $\mathbb{S}$ is finite. 66 | \\\\ 67 | If there are countable-infinite sets of the form $B_x$, then suppose $G=\{B_x\}_{x\in X}$.By taking all the possible disjoint unions from G you can form $\|P(G)\|$ new different sets(P means power set), hense an uncountable number of different sets.(G is countably infinite, then P(G) is uncountable) 68 | \\\\ 69 | Notice that every possible union of sets in G is a set that belongs to S, since $B_x \in S$ and S is a sigma-algebra. This means that S should be uncountable in order to contain this uncountable number of all possible different unions of the sets in the family G. 70 | \\\\ 71 | In conclusion, an event space can't be a countably infinite set. 72 | \end{document} 73 | -------------------------------------------------------------------------------- /Week4/Distrubution_func_man_not_be_left_continuous-&-h∈mΣ-iff-h+h-∈mΣ-赖睿航/Distrubution_func_may_not_be_left_continuous-&-h∈mΣ-iff-h+h-∈mΣ-赖睿航.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/Distrubution_func_man_not_be_left_continuous-&-h∈mΣ-iff-h+h-∈mΣ-赖睿航/Distrubution_func_may_not_be_left_continuous-&-h∈mΣ-iff-h+h-∈mΣ-赖睿航.pdf -------------------------------------------------------------------------------- /Week4/How_An_is_constructed_in_PR_THM_卢禹杰/How_An_is_constructed_in_PR_THM.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/How_An_is_constructed_in_PR_THM_卢禹杰/How_An_is_constructed_in_PR_THM.pdf -------------------------------------------------------------------------------- /Week4/How_An_is_constructed_in_PR_THM_卢禹杰/How_An_is_constructed_in_PR_THM.tex: -------------------------------------------------------------------------------- 1 | \input{../preamble.tex} 2 | \title{How $A_{n}$ is constructed in the proof of Poincare' Recurrence Theorem} 3 | \begin{document} 4 | \maketitle 5 | The construction of $A_{n}$ in the proof given by Mr. Wu is really tricky when 6 | I take the first look. 7 | \[ 8 | A_{n} = \{x\in E \mid x \not \in T^{-kn}\left( E \right) , \forall k\} 9 | .\] 10 | And now I try to give some intuition behind the construction. 11 | First let us take a look at another version of Poincare' Recurrence Theorem: 12 | \begin{pro} 13 | Let $\left( \Omega,\mathcal{F},P \right) $ be a probability space, and 14 | let $T$ be a map from $\Omega$ to itself with the property 15 | that $T\left( F \right) \in \mathcal{F}, \forall F\in \mathcal{ F}$. Plus 16 | $P\left( T\left( F \right) \right) = P\left( F \right)$. 17 | 18 | For $\forall E \in \mathcal{F}$, prove that $P\left( E \setminus 19 | \limsup_{n \to \infty}T^{n}\left( E \right) \right) =0$ 20 | \end{pro} 21 | \begin{proof} 22 | First we use a simple trick in set operation 23 | \[ 24 | E \setminus \bigcap_{n\ge 1} R_{n} = \bigcup_{n\ge 1} \left( E\setminus 25 | R_{n} \right) 26 | .\] 27 | to simpify the problem. Applying the equation above, our goal as 28 | \[ 29 | P\left( \bigcup_{k\ge 1} \left( 30 | E \setminus \bigcup_{n\ge k} T^{n}(E) \right) \right) = 0 31 | .\] 32 | Otherwise we assume 33 | \[ 34 | \exists k\ge 1, s.t. \ P\left( E \setminus \bigcup_{n\ge k} 35 | T^{n}\left( E \right) \right) > 0 36 | .\] 37 | Therefore this problem is equivalent to prove 38 | \[ 39 | P\left( A_k \right) =0,\text{ in which } A_k = \{ x \in E| x \not\in \bigcup_{n\ge k} 40 | T^{n}\left( E \right) \} 41 | .\] 42 | Notice that \textbf{this is exactly what the original form has proved}. 43 | Here we give a direct proof for it. Since $P\left( X \right) \le \infty$, we 44 | have 45 | \[ 46 | P\left( E \cup \bigcup_{n\ge k} T^{n}\left( E \right) \right) > 47 | P\left( \bigcup_{n\ge k} T^{n}\left( E \right) \right) 48 | .\] 49 | Applying $T$ to the set above $k$ times, there is 50 | \begin{align*} 51 | P\left( T^{k}\left( E \right) \cup 52 | \bigcup_{n\ge 2k} T^{n}\left( E \right) \right) &= 53 | P \left( T^{k}\left( E \cup \bigcup_{n\ge k}T^{n}\left( E \right) \right) \right) \\ 54 | &= P\left( 55 | E \cup \bigcup_{n\ge k} T^{n}\left( E \right) \right) \\ 56 | & > P\left( 57 | \bigcup_{n\ge k} T^{n}\left( E \right) \right) 58 | .\end{align*} 59 | This leads to a contradiction since 60 | \[ 61 | T^{k}\left( E \right) \cup \bigcup_{n\ge 2k} T^{n}(E) 62 | \subset \bigcup_{n\ge k} T^{n}(E) 63 | .\] 64 | Which completes our proof. 65 | \end{proof} 66 | 67 | \noindent \textbf{Remark} : what this proof tells us is that the construction 68 | of $A_{n}$ in the original proof \textbf{does not come from nowhere}. 69 | It is exactly something derived from the objective equation, which reveals 70 | that the original proof is somewhat natural. 71 | 72 | \end{document} 73 | -------------------------------------------------------------------------------- /Week4/Indicator function and Infinite events-计家宝/Indicator function and Infinite events-计家宝.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/Indicator function and Infinite events-计家宝/Indicator function and Infinite events-计家宝.pdf -------------------------------------------------------------------------------- /Week4/Indicator function for lim sup and lim inf-郭林松/Indicator function for lim sup and lim inf.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/Indicator function for lim sup and lim inf-郭林松/Indicator function for lim sup and lim inf.pdf -------------------------------------------------------------------------------- /Week4/Intersection of Uncountable a.s. Events May Be Empty-马浩博/discussion.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/Intersection of Uncountable a.s. Events May Be Empty-马浩博/discussion.pdf -------------------------------------------------------------------------------- /Week4/Intersection of Uncountable a.s. Events May Be Empty-马浩博/discussion.tex: -------------------------------------------------------------------------------- 1 | % This is a template for lecture notes. 2 | \documentclass[12pt]{article} 3 | \usepackage{amssymb} 4 | \usepackage[UTF8]{ctex} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{geometry} 8 | \usepackage{booktabs} 9 | \usepackage{bm} 10 | \usepackage{cite} 11 | %\usepackage{CJK} 12 | \usepackage[many]{tcolorbox} 13 | %\tcbuselibrary{listingsutf8} 14 | %\tcbuselibrary{skins, breakable, theorems, most} 15 | %\geometry{a4paper,bottom = 3cm,left = 3cm, right = 3cm} 16 | \CTEXoptions[today=old] 17 | %for reference 18 | \usepackage{hyperref} 19 | \usepackage[capitalise]{cleveref} 20 | \crefname{enumi}{}{} 21 | 22 | 23 | \newtheoremstyle{mythm}{1.5ex plus 1ex minus .2ex}{1.5ex plus 1ex minus .2ex} 24 | {}{\parindent}{\bfseries}{}{1em}{} 25 | \theoremstyle{mythm} 26 | \newtheorem{theorem}{Theorem} 27 | \newtheorem{lemma}[theorem]{Lemma} 28 | \newtheorem{corollary}[theorem]{Corollary} 29 | \newtheorem{fact}[theorem]{Fact} 30 | \newtheorem{definition}[theorem]{Definition} 31 | \newtheorem*{remark}{Remark} 32 | 33 | %\newenvironment{proof}{\noindent \textbf{Proof:}}{$\Box$} 34 | 35 | %to use newcommand for convenience 36 | \newcommand\field{\mathbb{F}} 37 | \newcommand\Real{\mathbb{R}} 38 | \newcommand\Q{\mathbb{Q}} 39 | \newcommand\Z{\mathbb{Z}} 40 | \newcommand\complex{\mathbb{C}} 41 | \newcommand\cc{\mathcal{C}} 42 | \newcommand\uu{\mathcal{U}} 43 | \newcommand\pp{\mathcal{P}} 44 | \newcommand\ff{\mathcal{F}} 45 | \renewcommand\refname{Reference} 46 | \renewcommand{\proofname}{Proof} 47 | \DeclareMathOperator{\range}{range} 48 | 49 | \title{Intersection of Uncountable a.s. Events May Be Empty} 50 | \author{马浩博 518030910428} 51 | \date{\today} 52 | \begin{document} 53 | \maketitle 54 | 55 | \section*{Exercise 4} 56 | 57 | This exercise is going to say that intersection of uncountable events may be empty. 58 | We already know that for the countable situation, things are different: 59 | \begin{theorem} 60 | $$If \ F_n \in \mathcal{F} \ (n \in N) \ and \ P(F_n) = 1,\ \forall n, \ then $$ 61 | $$P(\bigcap\limits_{n} F_n) = 1 $$ 62 | \end{theorem} 63 | Now the number of $\alpha$ is obviously uncountable, so the theorem above is not working anymore. 64 | And we can get an opposite conclusion that 65 | $$\bigcap\limits_{\alpha} F_\alpha = \emptyset $$ 66 | \begin{proof} 67 | 68 | For any given $\omega$, we just need to find an $\alpha$ with $\omega \notin F_\alpha$. 69 | Then we can easily get the result that $\bigcap\limits_{\alpha} F_\alpha = \emptyset $. 70 | Thus the only thing to concern about is to find the $\alpha$.\\ 71 | 72 | For $\omega = (\omega_1,\omega_2,\dots)$, we can assume without losing generality that 73 | $$\forall m,\ \exists n > m \ \omega_n = H \ (n \in N)$$ 74 | If $\omega$ doesn't satisify this proposition, then because $\omega = \{H,T\}$, we have $\forall m,\ \exists n > m \ \omega_n = T$. 75 | These two cases are just the same, so we only consider the first one. 76 | 77 | In this case, we can generate $\alpha$ in this way: For $m=0$, we can find $a_1>m$ with $\omega_{a_1} = H$. 78 | So we let $\alpha(1)=a_1$. Next, let $m=a_1$, find $a_2$ in the same way and also let $\alpha(2)=a_2$, and so on $\dots$ 79 | Eventually, we get a map $\alpha$ with $\alpha(1)<\alpha(2)< \dots $ Obviously, 80 | $$\frac{\#\{ k \le n: \ \omega_{\alpha(k)} = H \}}{n} \to 1$$ 81 | Therefore we know that $\omega \notin F_\alpha$ and finish the proof. 82 | \end{proof} 83 | 84 | \end{document} -------------------------------------------------------------------------------- /Week4/Notes of Events 刘成锴/Events.md: -------------------------------------------------------------------------------- 1 | # Events 2 | 3 | ## by 刘成锴 4 | 5 | ## Model for experiment: $(\Omega, \mathcal F, P)$ 6 | 7 | A probability triple $(\Omega, \mathcal F, P)$. 8 | 9 | ### Sample space $\Omega$ 10 | 11 | $\Omega$ is a set called the *sample space*. 12 | 13 | ### Sample point $\omega$ 14 | 15 | A point $\omega$ of $\Omega$ is called *a sample point*. 16 | 17 | ### Event 18 | 19 | The $\sigma$-algebra $\mathcal F$ on $\Omega$ is called the family of events（事件类）, so that an *event* is an element of $\mathcal F$, that is, an $\mathcal F$-measurable（$\mathcal F$-可测） subset of $\Omega$. 20 | 21 | $P$ is a probability measure on $(\Omega, \mathcal F)$ 22 | 23 | For $F \in \mathcal F$, $P(F)$ is the probability that $\omega$ is in $F$. 24 | 25 | 26 | 27 | ## Almost surely (a.s.) 28 | 29 | A statement $S$ about outcomes is said to be true *almost surely (a.s.)*, or *with probability* 1 *(w.p.1)*, if 30 | $$ 31 | F:=\{\omega: S(\omega) \text { is true }\} \in \mathcal{F} \text { and } \mathbf{P}(F)=1 32 | $$ 33 | 34 | 35 | ## Some definitions 36 | 37 | Let $(x_n : n \in \mathbf N)$ be a sequence of real numbers. 38 | 39 | (a) 40 | $$ 41 | \limsup x_{n}:=\inf _{m}\left\{\sup _{n \geq m} x_{n}\right\}=\downarrow \lim _{m}\left\{\sup _{n \geq m} x_{n}\right\} \in[-\infty, \infty] 42 | $$ 43 | 44 | 45 | (b) 46 | $$ 47 | \liminf x_{n}:=\sup _{m}\left\{\inf _{n \geq m} x_{n}\right\}=\uparrow \lim _{m}\left\{\inf _{n \geq m} x_{n}\right\} \in[-\infty, \infty] 48 | $$ 49 | 50 | (c) 51 | $$ 52 | x_{n} \text { converges in }[-\infty, \infty] \Longleftrightarrow \limsup x_{n}=\lim \inf x_{n} 53 | $$ 54 | and then $\lim x_{n}=\lim\sup x_{n}=\lim\inf x_{n}$ 55 | 56 | 57 | 58 | ## Definition. $\limsup {E_{n},\left(E_{n}, \text { i.o. }\right)}$ 59 | 60 | If $E$ is an event, then 61 | $$ 62 | E = {\omega : \omega \in E} 63 | $$ 64 | Suppose now that $(E_n : n \in \mathbf N)$ is a sequence of events. 65 | 66 | (a) We define 67 | $$ 68 | \begin{aligned}\left(E_{n}, \text { i.o. }\right): &=\left(E_{n} \text { infinitely often }\right) \\ : &=\lim \sup E_{n}:=\bigcap_{m} \bigcup_{n \geq m} E_{n} \\ &=\left\{\omega: \text { for every } m, \quad \exists n(\omega) \geq m \text { such that } \omega \in E_{n(\omega)}\right\} \\ &=\left\{\omega: \omega \in E_{n} \text { for infinitely many } n\right\} \end{aligned} 69 | $$ 70 | 71 | 72 | (b) (**Reverse Fatou Lemma** - needs *FINITENESS* of $P$) 73 | $$ 74 | P(\lim \sup E_n) \geq \lim \sup P(E_n) 75 | $$ 76 | 77 | 78 | ## First Borel-Cantelli lemma (BC1) 79 | 80 | Let $(E_n : n \in \mathbf N)$ be a sequence of events such that $\sum_{n} P\left(E_{n}\right)<\infty$. Then 81 | $$ 82 | P(\lim \sup E_n) = P(E_n, i.o.) = 0 83 | $$ 84 | 85 | 86 | ## Definitions. $\lim \inf E_n$, $(E_n, \text{ev})$ 87 | 88 | Suppose that $(E_n : n \in \mathbf N)$ is a sequence of events. 89 | 90 | (a) We define 91 | $$ 92 | \begin{aligned}\left(E_{n}, \mathrm{ev}\right) &:=\left(E_{n} \text { eventually }\right) \\ &=\lim \inf E_{n}:=\bigcup_{m} \bigcap_{n \geq m} E_{n} \\ &=\left\{\omega: \text { for some } m(\omega), \quad \omega \in E_{n}, \forall n \geq m(\omega)\right\} \\ &=\left\{\omega: \omega \in E_{n} \text { for all large } n\right\} \end{aligned} 93 | $$ 94 | (b) Note that $\left(E_{n}, \mathrm{ev}\right)^{c}=\left(E_{n}^{c}, \text { i.o. }\right)$. 95 | 96 | (c) (**Fatou's Lemma for sets** - true for *ALL* measure spaces) 97 | $$ 98 | P(\lim \inf E_n) \leq \lim \inf P(E_n) 99 | $$ 100 | -------------------------------------------------------------------------------- /Week4/Notes of Events 刘成锴/Events.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/Notes of Events 刘成锴/Events.pdf -------------------------------------------------------------------------------- /Week4/Notes of Events_傅凌玥/Event.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/Notes of Events_傅凌玥/Event.pdf -------------------------------------------------------------------------------- /Week4/PR_thm_for_incompressible_trans_卢禹杰/PR_thm_for_incompressible_trans.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/PR_thm_for_incompressible_trans_卢禹杰/PR_thm_for_incompressible_trans.pdf -------------------------------------------------------------------------------- /Week4/PR_thm_for_incompressible_trans_卢禹杰/PR_thm_for_incompressible_trans.tex: -------------------------------------------------------------------------------- 1 | \input{../preamble.tex} 2 | \title{\textsc{Poincaré’s Recurrence Theorem for Incompressible 3 | Transformations}} 4 | \begin{document} 5 | \maketitle 6 | 7 | \begin{pro} 8 | Let $\left( \Omega ,\mathcal{F},P \right) $ be a probability space and let $T$ be 9 | an incompressible transformation on it, namely it is measurable and 10 | there is no $F \in \mathcal{F}$ such that $F\subset T^{-1}(F)$ and $P\left( F \right) < P\left( T^{-1}(F) \right) $. 11 | Prove that $\forall E\in \mathcal{F}$ with $P\left( E \right) >0$, almost all points of $E$ return 12 | to $E$ infinitely often under positive iterations by $T$. 13 | \end{pro} 14 | \begin{proof} 15 | This problem is almost the same as the last one with some slight changes. 16 | Our goal is to prove 17 | \[ 18 | P\left( E \setminus \limsup_{n \to \infty} T^{-n}(E) \right) = 0 19 | .\] 20 | The same as the proof of another version of PR theorem, the above is equivalent to prove 21 | \[ 22 | P\left( A_{k} \right) =0, \text{ in which } A_{k} = E \setminus \bigcup_{n\ge k} 23 | T^{-n}(E) 24 | .\] 25 | We will prove by making contradiction. Assume $P\left( A_{k} \right) > 0$, we have 26 | \[ 27 | P\left( B_{k} \right) = P\left( E \cup \bigcup_{n\ge k} T^{-n}(E) \right) > 28 | P\left( \bigcup_{n\ge k} T^{-n}(E)\right) = P\left( C_{k} \right) 29 | .\] 30 | Here $B_{k},C_{k}$ are notations for simplicity. Notice that the definition of incompressible 31 | transformation is exactly 32 | \[ 33 | F\subset T^{-1}\left( F \right) \implies P(F) = P( T^{-1}(F)) 34 | .\] 35 | Hence by $T^{-1}\left( C_{k-1} \right) = C_{k} $ we have 36 | \[ 37 | P\left( C_{k} \right) = P\left( C_{k-1} \right) 38 | .\] 39 | Similarly, as $T^{-1}\left( B_{k} \right) \subset C_{1}$ we derive that 40 | \[ 41 | P\left( B_{k} \right) = P\left( C_{1} \right) = P\left( C_{k} \right) 42 | .\] 43 | Which leads to a contradiction. 44 | \end{proof} 45 | \end{document} 46 | -------------------------------------------------------------------------------- /Week4/Product_of_divergent_series - 张志成_寻之扬/Product_of_divergent_series.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/Product_of_divergent_series - 张志成_寻之扬/Product_of_divergent_series.pdf -------------------------------------------------------------------------------- /Week4/Proof of Poincare's Recurrence Theorem - 赖睿航/poincare_recurrence_thm.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/Proof of Poincare's Recurrence Theorem - 赖睿航/poincare_recurrence_thm.pdf -------------------------------------------------------------------------------- /Week4/Second Borel–Cantelli lemma - 陆嘉馨/Second Borel–Cantelli lemma.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/Second Borel–Cantelli lemma - 陆嘉馨/Second Borel–Cantelli lemma.pdf -------------------------------------------------------------------------------- /Week4/Sigma-measurable-functions-and-simple-functions-赖睿航/sigma-measurable-functions-and-simple-functions.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/Sigma-measurable-functions-and-simple-functions-赖睿航/sigma-measurable-functions-and-simple-functions.pdf -------------------------------------------------------------------------------- /Week4/Simple exercise of phi(1-y_n) = 0_吴怀瑾/simple exercise of phi(1-y_n) = 0.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/Simple exercise of phi(1-y_n) = 0_吴怀瑾/simple exercise of phi(1-y_n) = 0.pdf -------------------------------------------------------------------------------- /Week4/Simple exercise of phi(1-y_n) = 0_吴怀瑾/simple exercise of phi(1-y_n) = 0.tex: -------------------------------------------------------------------------------- 1 | % This is a template for lecture notes. 2 | \documentclass{article} 3 | \usepackage[UTF8]{ctex} 4 | \usepackage{amssymb} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{geometry} 8 | \usepackage{booktabs} 9 | \usepackage{bm} 10 | \usepackage{tcolorbox} 11 | \CTEXoptions[today=old] 12 | %Some commonly used notations 13 | %\geometry{a4paper,bottom = 3cm,left = 3cm, right = 3cm} 14 | 15 | %for reference 16 | \usepackage{hyperref} 17 | \usepackage[capitalise]{cleveref} 18 | \crefname{enumi}{}{} 19 | 20 | \newtheorem{theorem}{Theorem} 21 | \newtheorem{lemma}[theorem]{Lemma} 22 | \newtheorem{proposition}[theorem]{Proposition} 23 | \newtheorem{corollary}[theorem]{Corollary} 24 | \newtheorem{fact}[theorem]{Fact} 25 | \newtheorem{definition}[theorem]{Definition} 26 | \newtheorem{remark}[theorem]{Remark} 27 | \newtheorem{question}[theorem]{Question} 28 | \newtheorem{answer}[theorem]{Answer} 29 | \newtheorem{exercise}[theorem]{Exercise} 30 | \newtheorem{example}[theorem]{Example} 31 | %\newenvironment{proof}{\noindent \textbf{Proof:}}{$\Box$} 32 | \newtheorem{observation}[theorem]{Observation} 33 | 34 | %to use newcommand for convenience 35 | \newcommand\field{\mathbb{F}} 36 | \newcommand\Real{\mathbb{R}} 37 | \newcommand\Q{\mathbb{Q}} 38 | \newcommand\Z{\mathbb{Z}} 39 | \newcommand\complex{\mathbb{C}} 40 | 41 | %this is how we define operators. 42 | \DeclareMathOperator{\rank}{rank} % rank 43 | 44 | \title{20200324 Exercise 4} 45 | \author{吴怀瑾 518030910414} 46 | \date{\today} 47 | 48 | \begin{document} 49 | \maketitle 50 | \section{} 51 | \begin{question}{}{} 52 | Let $(y_n)_{n \in N}$ be a sequence of reals from $[0, 1]$ such that $\sum_{n \in N} y_n = \infty$. Show that $\prod (1−y_n)=0$. 53 | \end{question} 54 | 55 | \paragraph{solution} 56 | Because $\sum_{n \in N} y_n = \infty$, so $\exists \epsilon >0 \exists N $ $\forall n > N$, $y_n > \epsilon$. Then we consider when $n > N$, $\prod_N^M (1-y_n) <= (1-\epsilon) ^{M-N}$. 57 | So $\lim_{M \to \infty} \prod_N^M (1-y_n) <= (1-\epsilon) ^{M-N} = 0$, which means $\prod (1−y_n)=0$. 58 | 59 | \end{document} -------------------------------------------------------------------------------- /Week4/The Convergence of Summation and Product_吴润哲/The Convergence of Summation and Product.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/The Convergence of Summation and Product_吴润哲/The Convergence of Summation and Product.pdf -------------------------------------------------------------------------------- /Week4/The Convergence of Summation and Product_吴润哲/The Convergence of Summation and Product.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt]{article} 2 | 3 | %\usepackage[UTF8]{ctex} 4 | \usepackage{geometry} 5 | \usepackage{amsthm} 6 | \usepackage{amsmath} 7 | \usepackage{amssymb} 8 | \usepackage{mathtools} 9 | \usepackage{enumerate} 10 | \usepackage{hyperref} 11 | \usepackage{tcolorbox} 12 | 13 | \geometry{a4paper, left = 2cm, right = 2cm, top = 2cm} 14 | 15 | \newcommand\problem[1]{\section*{Problem #1}} 16 | 17 | \newcommand\bE{\mathbb{E}} 18 | \newcommand\bF{\mathbb{F}} 19 | \newcommand\bN{\mathbb{N}} 20 | \newcommand\bZ{\mathbb{Z}} 21 | \newcommand\bQ{\mathbb{Q}} 22 | \newcommand\bR{\mathbb{R}} 23 | \newcommand\fC{\mathbf{C}} 24 | \newcommand\fF{\mathbf{F}} 25 | \newcommand\fN{\mathbf{N}} 26 | \newcommand\fQ{\mathbf{Q}} 27 | \newcommand\fR{\mathbf{R}} 28 | \newcommand\fZ{\mathbf{Z}} 29 | \newcommand\cU{\mathcal{U}} 30 | 31 | \newcommand\ce{\coloneqq} 32 | \newcommand\lproof{\item ``$\Leftarrow$'' :} 33 | \newcommand\rproof{\item ``$\Rightarrow$'' :} 34 | 35 | \newcommand{\leb}{\text{Leb}} 36 | \newcommand*{\dif}{\mathop{}\!\mathrm{d}} 37 | \newcommand\ord{\text{ord}} 38 | \newcommand{\floor}[1]{\lfloor {#1}\rfloor} 39 | 40 | \newtheorem{claim}{Claim} 41 | \newtheorem{lemma}{Lemma} 42 | \newtheorem{theorem}{Theorem} 43 | \newtheorem{corollary}{Corollary} 44 | 45 | 46 | \title{The Convergence of Summation and Product} 47 | \author{WU Runzhe\\ 48 | Student ID : 518030910432\\ 49 | \textsc{Shanghai Jiao Tong University}} 50 | \date{\today} 51 | 52 | \begin{document} 53 | \maketitle 54 | 55 | \begin{tcolorbox} 56 | \begin{theorem}\label{t1} 57 | Let $(y_n)_{n\in \bN}$ be a sequence of reals from [0,1] such that $$\sum_{n\in\bN}y_n=\infty.$$ Show that $$\prod_{n\in\bN}(1-y_{n})=0.$$ 58 | \end{theorem} 59 | \end{tcolorbox} 60 | 61 | \begin{lemma}\label{l1} 62 | $e^{-x}\ge1-x$ for every $x\in \bR$. 63 | \end{lemma} 64 | 65 | \begin{proof}[Proof of \ref{l1}] 66 | Let $f(x)=e^{-x}-(1-x)$. Then, $f'(x)=-e^{-x}+1$, and $f'(x)=0$ iff $x=0$. It is easy to check that $f(x)\ge f(0)=0$. 67 | \end{proof} 68 | 69 | \begin{proof}[Proof of theorem \ref{t1}] 70 | 71 | Using lemma \ref{l1}, we have 72 | \begin{align*} 73 | &\prod_{n\in\bN}(1-y_n))\\ 74 | \le& \prod_{n\in\bN} e^{-y_n}\\ 75 | \le& e^{-\sum_{n\in\bN}y_n}\\ 76 | =&0 77 | \end{align*}\ 78 | 79 | \end{proof} 80 | 81 | Similarly, we can also prove the following complementary theorem. 82 | 83 | \begin{tcolorbox} 84 | \begin{theorem}\label{t2} 85 | Let $(y_n)_{n\in \bN}$ be a sequence of reals from [0,1) such that $$\sum_{n\in\bN}y_n<\infty.$$ Show that $$\prod_{n\in\bN}(1-y_{n})>0.$$ 86 | \end{theorem} 87 | \end{tcolorbox} 88 | 89 | \begin{lemma}\label{l2} 90 | $x\ge -\ln(1-x)$ for each $x\in [0,1)$. 91 | \end{lemma} 92 | 93 | It is easy to check the above lemma. Just do what we did in the proof of lemma \ref{l1}. 94 | 95 | \begin{proof}[Proof of theorem \ref{t2}] 96 | 97 | Using lemma \ref{l2}, we have 98 | $$\infty>\sum_{n\in\bN}y_n \ge -\sum_{n\in\bN}\ln(1-y_n).$$ 99 | It means $\sum_{n\in\bN}\ln(1-y_n)$ converges. Let $\sum_{n\in\bN}\ln(1-y_n)=C$. 100 | 101 | Then, 102 | $$\prod_{n\in\bN}(1-y_n)=e^{\sum_{n\in\bN}\ln(1-y_n)}=e^C>0.$$ 103 | 104 | \end{proof} 105 | 106 | 107 | \end{document} 108 | -------------------------------------------------------------------------------- /Week4/The application of positive and negative part in simple functions_金弘义/discussion.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/The application of positive and negative part in simple functions_金弘义/discussion.pdf -------------------------------------------------------------------------------- /Week4/The continuity of distribution function-郭林松/The continuity of distribution function.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/The continuity of distribution function-郭林松/The continuity of distribution function.pdf -------------------------------------------------------------------------------- /Week4/The continuity of distribution function-郭林松/The continuity of distribution function.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | % \usepackage[UTF8]{ctex} 3 | \usepackage{amssymb} 4 | \usepackage{amsmath} 5 | \usepackage{amsthm} 6 | \usepackage{geometry} 7 | \usepackage{booktabs} 8 | \usepackage{bm} 9 | \usepackage{tcolorbox} 10 | \usepackage{setspace} 11 | \renewcommand{\baselinestretch}{1.3} 12 | 13 | %for reference 14 | \usepackage{hyperref} 15 | \usepackage[capitalise]{cleveref} 16 | \crefname{enumi}{}{} 17 | 18 | \newtheorem{theorem}{Theorem} 19 | \newtheorem{lemma}[theorem]{Lemma} 20 | \newtheorem{proposition}[theorem]{Proposition} 21 | \newtheorem{corollary}[theorem]{Corollary} 22 | \newtheorem{fact}[theorem]{Fact} 23 | \newtheorem{definition}[theorem]{Definition} 24 | \newtheorem{remark}[theorem]{Remark} 25 | \newtheorem{question}[theorem]{Question} 26 | \newtheorem{answer}[theorem]{Answer} 27 | \newtheorem{exercise}[theorem]{Exercise} 28 | \newtheorem{example}[theorem]{Example} 29 | \newtheorem{observation}[theorem]{Observation} 30 | 31 | 32 | %this is how we define operators. 33 | \DeclareMathOperator{\rank}{rank} % rank 34 | 35 | \title{The continuity of distribution function} 36 | \author{Guo Linsong~~518030910419} 37 | \date{\today} 38 | 39 | \begin{document} 40 | \maketitle 41 | 42 | \begin{tcolorbox} 43 | \begin{question}\label{question} 44 | Construct an example to show that the distribution function of a random variable may not be left-continuous. 45 | \end{question} 46 | \end{tcolorbox} 47 | 48 | 49 | 50 | \begin{theorem} 51 | Let $X$ be a random variable.The distribution function $F_X(x)=\mathcal{L}_X(-\infty,x]$ is right-continuous. 52 | \end{theorem} 53 | 54 | \begin{proof} 55 | Let $\{a_n=\frac{1}{n},n\in\mathbb{N}\}$ be a countable sequence of numbers.Let $A_n=(-\infty,x+a_n]$ and $A=(-\infty,x]$.Then $A_n,A\in \mathcal{B}$ and $A_n\downarrow A$.$\mathcal{L}_x$ is a probability measure on $(R,\mathcal{B})$.According to lemma $1.10(b)$ in textbook, we have $$\mathcal{L}_X(A_n)\downarrow\mathcal{L}_X(A)$$ 56 | Thus $$\mathcal{L}_X(-\infty, x]=\lim\limits_{n\rightarrow\infty}\mathcal{L}_X(-\infty, x+\frac{1}{n}]$$ 57 | Together with the monotonicity of $\mathcal{L}_X$, we have $\mathcal{L}_X(-\infty, x]=\mathcal{L}_X(-\infty, x^{+}]$.Thus $F_X(x)=F_X(x^{+})$,the distribution function $F_X(x)=\mathcal{L}_X(-\infty,x]$ is right-continuous. 58 | \end{proof} 59 | 60 | However,The distribution function may not left-continuous.Considering two consecutive coin tosses,let $X$ be the number of heads. 61 | 62 | $$F_{X}(x)= \mathcal{L}_X(-\infty, x]=\left\{\begin{array}{ll} 63 | 0, & \text { if } x\in(-\infty,0) \\ 64 | 1 / 4, & \text { if } x \in [0,1) \\ 65 | 3 / 4, & \text { if } x \in [1,2) \\ 66 | 1, & \text { if } x \in [2,+\infty) 67 | \end{array}\right.$$ 68 | 69 | We have $F_X(0^-)\neq F_X(0)=F_X(0^+),F_X(1^-)\neq F_X(1)=F_X(1^+)$ and $F_X(2^-)\neq F_X(2)=F_X(2^+)$.Therefore, $F_X(x)$ is not left-continuous. 70 | 71 | \end{document} 72 | -------------------------------------------------------------------------------- /Week4/Truth-set-of-the-Strong-Law-for- the-subsequence-α-于峥/thinking.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week4/Truth-set-of-the-Strong-Law-for- the-subsequence-α-于峥/thinking.pdf -------------------------------------------------------------------------------- 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1 | \documentclass[a4paper, linespread=1.5]{article} 2 | %\usepackage[UTF8]{ctex} 3 | \usepackage{xeCJK} 4 | \usepackage{geometry} 5 | \usepackage{amsmath} 6 | \usepackage{amssymb} 7 | \usepackage{amsthm} 8 | \usepackage{graphicx} 9 | \usepackage{keyval} 10 | \usepackage[dvipsnames,svgnames,x11names]{xcolor} 11 | \usepackage{float} 12 | \usepackage{ifthen} 13 | \usepackage{calc} 14 | \usepackage{ifplatform} 15 | \usepackage{fancyvrb} 16 | \usepackage{minted} 17 | \usepackage{hyperref} 18 | \usepackage{enumerate} 19 | \usepackage{multicol} 20 | \usepackage[all]{xy} 21 | \usepackage{ulem} 22 | \usepackage{epstopdf} 23 | \usepackage{mathrsfs} 24 | \usepackage{cancel} 25 | \usepackage{algorithm} 26 | \usepackage{algorithmic} 27 | \setlength{\parskip}{0.2\baselineskip} 28 | \setlength{\parindent}{2em} 29 | %\geometry{left=2.7cm,right=2.7cm,top=2.7cm,bottom=2.7cm} 30 | 31 | 32 | \newtheorem{theorem}{Theorem} 33 | \newtheorem{proposition}[theorem]{Proposition} 34 | \newtheorem{lemma}[theorem]{Lemma} 35 | \newtheorem{corollary}[theorem]{Corollary} 36 | \newtheorem{definition}[theorem]{Definition} 37 | \newtheorem{exercise}[theorem]{Exercise} 38 | 39 | \newtheorem{innercustom}{\customname} 40 | \providecommand{\customname}{} 41 | \newcommand{\newcustomtheorem}[2]{ 42 | \newenvironment{#1}[1] 43 | { 44 | \renewcommand\customname{#2} 45 | \renewcommand\theinnercustom{##1} 46 | \innercustom 47 | } 48 | {\endinnercustom} 49 | } 50 | \newcustomtheorem{customthm}{Theorem} 51 | \newcustomtheorem{customprop}{Proposition} 52 | \newcustomtheorem{customlemma}{Lemma} 53 | \newcustomtheorem{customcorollary}{Corollary} 54 | \newcustomtheorem{customdef}{Definition} 55 | \newcustomtheorem{customex}{Exercise} 56 | \newcustomtheorem{customremark}{Remark} 57 | 58 | \newcommand{\Natural}{\mathbb{N}} 59 | \newcommand{\addbigcup}{\bigcup{\kern-1.12em{+}}\kern0.3em} 60 | \newcommand{\nth}[1]{#1\textsuperscript{th}} 61 | 62 | \begin{document} 63 | \title{Distribution function may not be left-continuous} 64 | \author{金弘义\ 518030910333} 65 | \date{\today} 66 | \maketitle 67 | 68 | \begin{customex}{8} 69 | 70 | Construct an example to show that the distribution function of a random variable may not be left-continuous. 71 | \end{customex} 72 | \begin{proof}[Solution] 73 | Let X be a random variable on $([0,1],\mathcal{B}[0,1],Leb)$. 74 | \begin{equation} 75 | X(a):= 76 | \begin{cases} 77 | 0& \mbox{if $a>\frac{1}{2}$}\\ 78 | 1& \mbox{if $a\le\frac{1}{2}$} 79 | \end{cases} 80 | \end{equation} 81 | Then $F_X(1)=P(X\le 1)=P([0,1])=1$, while $\lim\limits_{a\to1^{-}}F_X(a)=P(X<1)=P([0,\frac{1}{2}))=\frac{1}{2}\ne F_X(1)$. So $F_X$ is not left-continuous. 82 | \end{proof} 83 | 84 | 85 | 86 | 87 | \end{document} 88 | -------------------------------------------------------------------------------- /Week5/A proof of Doob Dynkin Lemma_吴润哲/A proof of Doob Dynkin Lemma.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week5/A proof of Doob Dynkin Lemma_吴润哲/A proof of Doob Dynkin Lemma.pdf -------------------------------------------------------------------------------- /Week5/A_history_of_Independence_张扬天/A_history_of_Independence.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week5/A_history_of_Independence_张扬天/A_history_of_Independence.pdf -------------------------------------------------------------------------------- /Week5/Independence of coin tossing events_董海辰/Independence of coin tossing events.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week5/Independence of coin tossing events_董海辰/Independence of coin tossing events.pdf -------------------------------------------------------------------------------- /Week5/Independence of coin tossing events_董海辰/Independence of coin tossing events.tex: -------------------------------------------------------------------------------- 1 | \input{/Users/oscar/Documents/LaTeX_Templates/HW.tex} 2 | 3 | \title{Independence of coin tossing events} 4 | \date{\today} 5 | \author{董海辰 518030910417} 6 | 7 | \begin{document} 8 | \maketitle 9 | 10 | \begin{thm}{}{} 11 | In probability triple $(\Omega , \mathcal{F}, P) = ([0,1),\mathcal{B}[0,1), \mathrm{Leb})$. 12 | Let event 13 | \begin{math} 14 | A_n = [0, \frac{1}{2^n}) \cup [\frac{2}{2^n}, \frac{3}{2^n}) \cup \cdots \cup [\frac{2^n-2}{2^n}, \frac{2^n-1}{2^n}) 15 | .\end{math} 16 | 17 | Then $(A_n: n \in \mathbb{N}_+)$ are independent. 18 | \end{thm} 19 | 20 | \begin{proof}[Proof] 21 | By definition, we should show that 22 | \begin{math} 23 | \mathcal{A}_i := \{\varnothing, A_n, \Omega \backslash A_n, \Omega \}, i \in \mathbb{N} _+ 24 | .\end{math} 25 | are independent $\sigma $-algebras. 26 | 27 | Consider any $B_i \in \mathcal{A}_i$ and distinct $i_1, i_2, \cdots ,i_n$. For some $l \in [n]$, 28 | \begin{itemize} 29 | \item If $B_{i_l} = \varnothing$, then $\bigcap_{k=1}^n B_{i_k} = \varnothing$ and $P(B_i) = 0$ thus $\prod_{k=1}^nP(B_{i_k}) = 0$. The equation holds for any other $B_{i_k}$. 30 | \item If $B_{i_l} = \Omega $, then $\bigcap_{k \in [n]}B_{i_k} = \bigcap_{k \in [n] \backslash \{l\} }B_{i_k}$ and $P(B_{i_l}) = 1$. Thus 31 | \begin{math} 32 | P(\bigcap_{k \in [n]}B_{i_k}) = \prod_{k \in [n]}P(B_{i_k}) \iff 33 | P(\bigcap_{k \in [n] \backslash \{l\} }B_{i_k}) = \prod_{k \in [n] \backslash \{l\} }P(B_{i_k}) 34 | .\end{math} 35 | \end{itemize} 36 | 37 | So we can assume that either $B_{i_k} = A_{i_k}$ or $B_{i_k} = \Omega \backslash A_{i_k}$. 38 | 39 | Let $N = \max_{k \in [n]} i_k $ and $L_i = [\frac{i}{2^N}, \frac{i+1}{2^N}) $. $(L_i)$ are disjoint. Then 40 | \begin{math} 41 | A_{i_k} &= \bigcup_{s=0}^{2^{i_k} - 1} [\frac{2s}{2^{i_k}}, \frac{2s+1}{2^{i_k}}) \\ 42 | &= \bigcup_{s=0}^{2^{i_k} - 1} \bigcup_{i=2^{N-i_k}\cdot 2s}^{2^{N-i_k}(2s+1)-1} L_i \\ 43 | &=\bigcup_{\{i\in \{0,1,\cdots ,2^N-1\} : \left\lfloor \frac{i}{2^{i_k}} \right\rfloor\} \equiv 0(\bmod 2)\}} L_i 44 | .\end{math} 45 | 46 | Similarly, $\Omega \setminus A_{i_k} =\bigcup_{\{i \in \{0,1,\cdots ,2^N-1\} : \left\lfloor \frac{i}{2^{i_k}} \right\rfloor\} \equiv 1(\bmod 2)\}} L_i$. 47 | 48 | Thus 49 | \begin{math} 50 | P(\bigcap_{k \in [n]} B_{i_k}) = P(\bigcup_{\{i \in \{0,1,\cdots ,2^N-1\}: \forall k \in [n], \left\lfloor \frac{i}{2^{i_k}} \right\rfloor \equiv a_k(\bmod 2) \} }L_i) = 2^{-k} 51 | \end{math} 52 | ,where $a_k = \begin{cases} 53 | 0, B_{i_k} = A_{i_k} \\ 54 | 1, B_{i_k} = \Omega \setminus A_{i_k} 55 | \end{cases}$. 56 | 57 | We also have $\forall k, P(\Omega \setminus A_{i_k}) = P(A_{i_k}) = 2^{i_k - 1} \cdot 2^{N-i_k} \cdot 2^{-N} = 2^{-1}$. Therefore 58 | \begin{math} 59 | P(\bigcap_{k \in [n]}B_{i_k}) = \prod_{k \in [n]}P(B_{i_k}) = 2^{-k} 60 | .\end{math} 61 | 62 | This implies that $(\mathcal{A}_i)$ are independent $\sigma $-algebras, and then $(A_n)$ are independent events. 63 | 64 | \end{proof} 65 | 66 | \end{document} 67 | -------------------------------------------------------------------------------- /Week5/Independent of Tail Sigma Algebra/Independent of Tail Sigma Algebra.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week5/Independent of Tail Sigma Algebra/Independent of Tail Sigma Algebra.pdf -------------------------------------------------------------------------------- /Week5/Independent of Tail Sigma Algebra/Independent of Tail Sigma Algebra.tex: -------------------------------------------------------------------------------- 1 | \documentclass[UTF8, 12pt]{article} 2 | \usepackage{xeCJK} 3 | \usepackage{enumitem} 4 | \usepackage{amsmath} 5 | \usepackage{amssymb} 6 | \usepackage{amsfonts} 7 | \usepackage{mathrsfs} 8 | \usepackage{XCharter} 9 | \usepackage{fancyhdr} 10 | \usepackage{eulervm} 11 | \usepackage{graphicx} 12 | \usepackage{mdframed} 13 | \usepackage{ntheorem} 14 | \usepackage{lipsum} 15 | 16 | \topmargin -.5in 17 | \textheight 9in 18 | \oddsidemargin -.25in 19 | \evensidemargin -.25in 20 | \textwidth 7in 21 | \pagestyle{fancy} 22 | 23 | \newenvironment{proof}{\noindent\ignorespaces\textbf{Proof:}}{\hfill $\square$\par\noindent} 24 | \newenvironment{solution}{\noindent\ignorespaces\textbf{Solution:}}{\hfill $\square$\par\noindent} 25 | 26 | \theoremstyle{break} 27 | \newtheorem{claim}{Claim} 28 | \newtheorem{problem}{Problem} 29 | \newtheorem{lemma}{Lemma} 30 | \newtheorem*{theorem*}{Theorem} 31 | 32 | \newenvironment{rcases}{\left.\begin{aligned}}{\end{aligned}\right\rbrace} 33 | 34 | \title{Independent of Tail Sigma Algebra} 35 | \author{于峥 518030910437} 36 | \date{\today} 37 | 38 | \begin{document} 39 | \maketitle 40 | 41 | \begin{problem}[\textbf{Exercise 4.9} of \textit{Chapter E}] 42 | Let $Y_0, Y_1, Y_2, \dots$ be independent random variables with 43 | $$ 44 | P(Y_n = +1) = P(Y_n = -1) = \frac 1 2, ~~ \forall n 45 | $$ 46 | 47 | For n $n \in \mathbb{N}$, define 48 | $$ 49 | X_n := Y_0Y_1\dots Y_n 50 | $$ 51 | 52 | Prove that the variables $X_1, X_2, \dots$ are independent. define 53 | $$ 54 | \mathcal{Y} := \sigma(Y_1,Y_2,\dots), ~~ \mathcal{T}_n := \sigma(X_r : r > n) 55 | $$ 56 | 57 | Prove that 58 | $$ 59 | \mathcal{L} := \bigcup _n \sigma(\mathcal{Y}, \mathcal{T}_n) \not= \sigma \left( \mathcal{Y}, \bigcup_n \mathcal{T}_n \right) =: \mathcal{R} 60 | $$ 61 | \end{problem} 62 | 63 | \begin{proof} 64 | (a) 65 | 66 | For $i < j$, let $Y_{ij} = Y_{i+1}Y_{i+2}\dots Y_j$, then we can know 67 | $P(Y_{ij} = +1) = P(Y_{ij} = -1) = \frac 1 2$ by symmetry. 68 | 69 | And we can easily get for $j > i$, $X_i, Y_{ij}$ be independent variable. 70 | Hence 71 | 72 | \begin{align*} 73 | P(X_i = 1, X_j = 1) &= P(X_i = 1, Y_{in}=1) = P(X_i=1)P(Y_{in}=1) = \frac 1 4 \\ 74 | P(X_i = 1, X_j = -1) &= P(X_i = 1, Y_{in}=-1) = P(X_i=1)P(Y_{in}=-1) = \frac 1 4 \\ 75 | P(X_i = -1, X_j = 1) &= P(X_i = -1, Y_{in}=-1) = P(X_i=-1)P(Y_{in}=-1) = \frac 1 4 \\ 76 | P(X_i = -1, X_j = -1) &= P(X_i = -1, Y_{in}=1) = P(X_i=-1)P(Y_{in}=1) = \frac 1 4 \\ 77 | \end{align*} 78 | 79 | Therefore $X_i$ and $X_j$ are independent for all $i \not= j$. 80 | 81 | \newpage 82 | (b) Obviously, $Y_0$ is independent of $\mathcal{Y}$. $X_0 = Y_0$, so 83 | $Y_0$ is independent of $X_n$ for $n > 0$. Therefore $Y_0$ is independent of 84 | $\mathcal{T}_n$ for all $n$. So $Y_0$ is independent of $\mathcal{R}$. 85 | 86 | Then we can proof $Y_0 \in m\mathcal{L}$, it means $Y_0$ is measurable in 87 | $\mathcal{L}$. And we know that $Y_0 = X_{n+1}/Y_{1,n+1}$, $X_{n+1}$ is measurable 88 | in $\mathcal{T}_n$, and $Y_{1,n+1}$ is measurable in $\mathcal{Y}$. It means that 89 | $X_{n+1}$ and $Y_1, Y_2, \dots, Y_{n+1}$ enables us to solve $Y_0$. Hence $Y_0$ ignorespaces 90 | measurable in $\sigma(\mathcal{F}, \mathcal{T}_n)$ for all $n$. It implies that 91 | $Y_0$ is measurable in $\mathcal{L}$. 92 | 93 | Because $Y_0$ is independent of $\mathcal{R}$ and $Y_0$ is measurable in $\mathcal{L}$, 94 | we conlude that $\mathcal{L} \not= \mathcal{R}$. 95 | \end{proof} 96 | \end{document} -------------------------------------------------------------------------------- /Week5/Measurable_function_is_the_limit_of_a_sequence_of_simple_function_李孜睿/Measurable_function_is_the_limit_of_a_sequence_of_simple_function_李孜睿.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week5/Measurable_function_is_the_limit_of_a_sequence_of_simple_function_李孜睿/Measurable_function_is_the_limit_of_a_sequence_of_simple_function_李孜睿.pdf -------------------------------------------------------------------------------- /Week5/Notes of Independence 刘成锴/Independence.md: -------------------------------------------------------------------------------- 1 | # Independence 2 | 3 | ## by 刘成锴 4 | 5 | Let $(\Omega, \mathcal F, \mathbf P)$ be a probability triple. 6 | 7 | ## Definitions of independence 8 | 9 | ### Independent $\sigma$-algebras 10 | 11 | Sub-$\sigma$-algebras $\mathcal {G_1, G_2, \dots}$ of $\mathcal F$ are called *independent* if, whenever $G_i \in \mathcal G_i (i \in \mathbf N)$ and $i_1, \dots, i_n$ are distinct, then 12 | $$ 13 | P(G_{i_1} \cap \dots \cap G_{i_n} = \prod_{k=1}^n P(G_{i_k}) ) 14 | $$ 15 | 16 | ### Indenpendent random variables 17 | 18 | Random variables $X_1, X_2, \dots$ are called *independent* if the $\sigma$-algebras 19 | $$ 20 | \sigma(X_1), \sigma(X_2), \dots 21 | $$ 22 | are independent. 23 | 24 | ### Independent events 25 | 26 | Events $E_1, E_2, \dots$ are called *independent* if the $\sigma$-algebras $\mathcal{E_1, E_2, \dots}$ are independent, where 27 | $$ 28 | \mathcal E_n \text{ is the $\sigma$-algebra } \{\empty, E_n, \Omega{\setminus{E_n}}, \Omega \} 29 | $$ 30 | Since $\mathcal{E_n} = \sigma(I_{E_n})$, it follows that 31 | 32 | event $E_1, E_2, \dots$ are independent if and only if the random variables $I_{E_1}, I_{E_2}, \dots$ are independent. 33 | 34 | 35 | 36 | ## The $\pi$-system Lemma; and the more familiar definitions 37 | 38 | We study independence via $\pi$-systems rather than $\sigma$-algebras. 39 | 40 | (a) **LEMMA.** Suppose that $\mathcal G$ and $\mathcal H$ are sub-$\sigma$-algebras of $\mathcal F$, and that $\mathcal I$ and $\mathcal J$ are $\pi$-systems with 41 | $$ 42 | \sigma(\mathcal I) = \mathcal G, \quad \sigma(\mathcal J) = \mathcal H 43 | $$ 44 | Then $\mathcal G$ and $\mathcal H$ are *independent* if and only if $\mathcal I$ and $\mathcal J$ are independent in that 45 | $$ 46 | P(I \cap J) = P(I)P(J), \qquad I \in \mathcal I, \quad J \in \mathcal J 47 | $$ 48 | (b) 49 | 50 | Suppose that $X$ and $Y$ are two random variables on $(\Omega, \mathcal F, \mathbf P)$ such that, whenever $x, y \in \mathbf R$, 51 | $$ 52 | P(X \leq x; Y \leq y) = P(X \leq x) P(Y \leq y) 53 | $$ 54 | The $\pi$-systems $\pi(X)$ and $\pi(Y)$ are independent. Hence $\sigma(X)$ and $\sigma(Y)$ are independent. 55 | 56 | 57 | 58 | ## Second Borel-Cantelli Lemma (BC2) 59 | 60 | If $E_n : n \in \mathbf N$ is a sequence of **independent** events, then 61 | $$ 62 | \sum P(E_n) = \infty \Rightarrow P(E_n, \text{ i.o.}) = P(\lim \sup E_n) = 1 63 | $$ 64 | 65 | 66 | ## Definitions. Tail $\sigma$-algebras 67 | 68 | Let $X_1, X_2, \dots$ be random variables. Define 69 | $$ 70 | \mathcal T_n := \sigma(X_{n+1}, X_{n+2}, \dots), \quad \mathcal T := \bigcap_n \mathcal T_n 71 | $$ 72 | The $\sigma$-algebra $\mathcal T$ is called the *tail $\sigma$-algebra* of the sequence $(X_n : n \in \mathbf N)$. 73 | 74 | 75 | 76 | ## Theorem. Kolmogorov's 0-1 Law 77 | 78 | Let $(X_n : n \in \mathbf N)$ be a sequence of **independent** random variables, and let $\mathcal T$ be the tail $\sigma$-algebra of $(X_n : n \in \mathbf N)$. Then $\mathcal T$ is $P$-trivial, that is 79 | 80 | (i) $F \in \mathcal T \Rightarrow P(F) = 0 \text{ or } P(F) = 1$ 81 | 82 | (ii) if $\xi$ is a $\mathcal T$-measurable random variable, then, $\xi$ is almost deterministic in that for some constant $c$ in $[-\infty, +\infty]$, 83 | $$ 84 | P(\xi = c) = 1 85 | $$ 86 | We allow $\xi = \pm \infty$ at (ii) for obvious reasons. 87 | 88 | -------------------------------------------------------------------------------- /Week5/Notes of Independence 刘成锴/Independence.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week5/Notes of Independence 刘成锴/Independence.pdf -------------------------------------------------------------------------------- /Week5/Sigma-algebra-generated-by-r.v-赖睿航/sigma-algebra-generated-by-r.v.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week5/Sigma-algebra-generated-by-r.v-赖睿航/sigma-algebra-generated-by-r.v.pdf -------------------------------------------------------------------------------- /Week5/Simple functions_converging_to_measurable_function_张志成/Simple functions_converging_to_measurable_function.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week5/Simple functions_converging_to_measurable_function_张志成/Simple functions_converging_to_measurable_function.pdf -------------------------------------------------------------------------------- /Week5/Surprise Test Paradox_杨宗翰/A naive research of the Surprise Test Paradox_杨宗翰.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week5/Surprise Test Paradox_杨宗翰/A naive research of the Surprise Test Paradox_杨宗翰.pdf -------------------------------------------------------------------------------- /Week5/independence in infty_庄永昊_董海辰/independenceInInfty.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week5/independence in infty_庄永昊_董海辰/independenceInInfty.pdf -------------------------------------------------------------------------------- /Week5/independence_by_distrib_func_卢禹杰/depict_independence_by_distrib_func.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week5/independence_by_distrib_func_卢禹杰/depict_independence_by_distrib_func.pdf -------------------------------------------------------------------------------- /Week5/independence_by_distrib_func_卢禹杰/depict_independence_by_distrib_func.tex: -------------------------------------------------------------------------------- 1 | \input{../preamble.tex} 2 | \linespread{1.5} \selectfont 3 | \title{\textsc{Depicting Independence of Random Variables by Their Distribution Functions} } 4 | \begin{document} 5 | \maketitle 6 | First we give a definition for distribution for multiple random variables 7 | \begin{defi}[Joint distribution function] 8 | For $n$ random variables $X_1,\ldots,X_{n}$, the joint cumulative distribution function $F_{X_1,X_2\ldots,X_{n}}$ is given by 9 | \[ 10 | F_{X_1,\ldots,X_{n}} = P\left( X_1\le x_1,\ldots, 11 | X_{n} \le x_{n}\right) 12 | .\] 13 | Interpreting the $n$ random variables as a random vector 14 | $\bm{X} = \left( X_1,\ldots,X_{n} \right) ^{\top }$ yields a shorter 15 | notation 16 | \[ 17 | F_{\bm{X}}\left( \bm{x} \right) = 18 | P\left( X_1\le x_1,\ldots,X_{n}\le x_{n} \right) 19 | .\] 20 | \end{defi} 21 | And we also need the lemma for independence of sigma algebra, which is proved 22 | in class. Here I give a generalization form of it. 23 | \begin{lemma} 24 | Suppose $\mathcal{A}_{1}, \ldots, \mathcal{A}_{n}$ are independent 25 | and each $\mathcal{A}_{i}$ is a $\pi$-system, then 26 | $\sigma\left( \mathcal{A}_{1} \right) , \ldots, 27 | \sigma\left( \mathcal{A}_{n} \right) $ are independent. 28 | \end{lemma} 29 | The essence is the same with the case of 2 sigma algebras. 30 | 31 | With the preparation done, we can depict the independence of 32 | random variables by their distribution functions as follows 33 | \begin{thm} 34 | Random variables $X_1,X_2,\ldots,X_{n}$ are independent, if and only if 35 | $\forall \bm{x}=\left( x_1,\ldots,x_{n} \right) \in \R^{n}$ 36 | \[ 37 | P\left( X_1\le x_1,\ldots,X_{n}\le x_{n} \right) 38 | = \prod_{i=1}^{n} F_{i}\left( x_{i} \right) 39 | .\] 40 | \end{thm} 41 | \begin{proof} 42 | Using the definition of joint distribution function, the 43 | formula above is equivalent with 44 | \[ 45 | F_{\bm{X}}\left( \bm{x} \right) = \prod_{i=1}^{n} F_{i}\left( x_{i} \right) 46 | .\] 47 | Which is clean and beautiful. First we prove $\implies$, the result 48 | is obvious by definition. Since 49 | \[ 50 | \forall i \le n, B_{i} :=\{X_i | X_i \le x_i\} \subset \mathcal{B} 51 | .\] 52 | By definition, we obtain that 53 | \[ 54 | P\left( \bigcap_{i\in [n] } B_{i} \right) = 55 | \prod_{i=1}^{n} P\left( B_{i} \right) 56 | .\] 57 | 58 | Now we prove $\impliedby$. Consider $\mathcal{A}_{i} $ be the set of form 59 | $\{X_{i} \le x_{i}\} $. Hence by 60 | \[ 61 | \{ X_{i} \le r \} \cap \{X_{i} \le s\} = \{ X_{i} \le \min\left( r,s \right) \} 62 | .\] 63 | We know that $\mathcal{A}_{i}$ is a $\pi$-system plus we 64 | can write it as $X^{-1}\left( \pi\left( \R \right) \right) $, and they are 65 | independent (from the definition of independence). 66 | Moreover by the conclusion (the second half) proved by Ruihang Lai, 67 | we have 68 | \[ 69 | \sigma\left( \mathcal{A}_{i} \right)= \sigma\left( 70 | X_{i}^{-1}\left( \pi\left( \R \right) \right)\right) =X_{i}^{-1}\left( \mathcal{B} \right) = \sigma\left( X_{i} \right) 71 | .\] 72 | It's obvious that the independence of random variables 73 | is equivalent with the independence of the sigma algebra they generated. 74 | So by \textbf{Lemma 1} we know that $\sigma\left( \mathcal{A}_1 \right) , 75 | \ldots, \sigma\left( \mathcal{A}_{n} \right) $ are independent, which 76 | means $\sigma\left( X_1 \right) ,\ldots,\sigma\left( X_{n} \right) $ are 77 | independent, 78 | hence $X_1,\ldots,X_{n}$ are independent. 79 | \end{proof} 80 | 81 | \end{document} 82 | -------------------------------------------------------------------------------- /Week6/An application of LLL-hypergraph 2-coloring-毛昕渝/An application of LLL-hypergraph 2-coloring-毛昕渝.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/An application of LLL-hypergraph 2-coloring-毛昕渝/An application of LLL-hypergraph 2-coloring-毛昕渝.pdf -------------------------------------------------------------------------------- /Week6/An application of LLL-hypergraph 2-coloring-毛昕渝/ex.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/An application of LLL-hypergraph 2-coloring-毛昕渝/ex.png -------------------------------------------------------------------------------- /Week6/An application of LLL-hypergraph 2-coloring-毛昕渝/ref.bib: -------------------------------------------------------------------------------- 1 | @article{leighton1994packet, 2 | title={Packet routing and job-shop scheduling ino (congestion+ dilation) steps}, 3 | author={Leighton, Frank Thomson and Maggs, Bruce M and Rao, Satish B}, 4 | journal={Combinatorica}, 5 | volume={14}, 6 | number={2}, 7 | pages={167--186}, 8 | year={1994}, 9 | publisher={Springer} 10 | } -------------------------------------------------------------------------------- /Week6/Any graph can be a minimum dependency graph_杨宗翰.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/Any graph can be a minimum dependency graph_杨宗翰.pdf -------------------------------------------------------------------------------- /Week6/Consecutive_heads_in_coin_tossing_董海辰/Consecutive_heads_in_coin_tossing.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/Consecutive_heads_in_coin_tossing_董海辰/Consecutive_heads_in_coin_tossing.pdf -------------------------------------------------------------------------------- /Week6/Consecutive_heads_in_coin_tossing_董海辰/Consecutive_heads_in_coin_tossing.tex: -------------------------------------------------------------------------------- 1 | \input{/Users/oscar/Documents/LaTeX_Templates/HW.tex} 2 | 3 | \title{Consecutive heads in coin tossing} 4 | \date{\today} 5 | \author{董海辰 518030910417} 6 | 7 | \begin{document} 8 | \maketitle 9 | 10 | \begin{thm}{}{} 11 | Suppose that a coin with probability $p$ of heads is tossed repeatedly. 12 | 13 | Let $A_k$ be the event that a sequence of $k$ (or more) consecutive heads occurs amongst tosses numbered $2^k, 2^k + 1,2^k+ 2,..., 2^{k+1} - 1$. Prove that 14 | \begin{math} 15 | P(A_k, \io) = \begin{cases} 16 | 1, \text{ if } p \ge \frac{1}{2} \\ 17 | 0, \text{ if } p < \frac{1}{2} 18 | \end{cases} 19 | .\end{math} 20 | \end{thm} 21 | 22 | \begin{proof}[Proof] 23 | Let $E_{ki}$ be the event that there are $k$ consecutive heads beginning at toss numbered $2^k+i$. There is $P(E_{ki}) = p^k$ for all $i$. We have 24 | \begin{math} 25 | A_k = \bigcup_{i=0}^{2^k-k} E_{ki} 26 | .\end{math} 27 | 28 | By inclusion-exclusion formula, 29 | \begin{math} 30 | P(A_k) \le \sum _{i=0}^{2^k-k} P(E_{ki}) = (2^k-k+1) p^k 31 | .\end{math} 32 | 33 | And if we consider the disjoint blocks from $2^k + ik$ to $2^k + (i+1)k -1$for $i = 0,1,\cdots ,\left\lfloor \frac{2^k}{k} \right\rfloor$, namely $E_{k0}, E_{kk}, \cdots $. 34 | 35 | We have $\bigcup_{j=0}^{\left\lfloor \frac{2^k}{k} \right\rfloor} E_{k(jk)} \subseteq A_k$, thus 36 | \begin{math} 37 | P(A_k^c) \le P((\bigcup _{j=0}^{\left\lfloor \frac{2^k}{k} \right\rfloor} E_{k(jk)})^c) = (1-p^k)^{\left\lfloor \frac{2^k}{k} \right\rfloor} 38 | .\end{math} 39 | 40 | Therefore, 41 | \begin{math} 42 | 1 - (1-p^k)^{\left\lfloor \frac{2^k}{k} \right\rfloor} \le 43 | P(A_k) \le 44 | (2^k-k+1)p^k 45 | .\end{math} 46 | 47 | If $p < \frac{1}{2}$, $\sum _k P(A_k) \le \sum _k(2p)^k < \infty$. By BC1, we know that $P(A_k, \io) = 0$. 48 | 49 | If $p \ge \frac{1}{2}$, 50 | \begin{math} 51 | \sum _k P(A_k) &\ge \sum _k (1-(1-p^k)^{\frac{2^k}{k}}) \\ 52 | &= \sum _k (1 -\exp (\frac{2^k}{k} \ln (1-p^k))) \\ 53 | &= \sum _k (1 -\exp(- \frac{2^k}{k} p^k)) \\ 54 | &\ge \sum _k (1-\exp (-\frac{1}{k})) \\ 55 | &\ge \sum _k \frac{1}{k} \to \infty 56 | .\end{math} 57 | And apparently all events $A_k, k \in \mathbb{N} $ are independent. By BC2, $P(A_k, \io) = 1$. 58 | 59 | As a conclusion, 60 | \begin{math} 61 | P(A_k, \io) = \begin{cases} 62 | 1, \text{ if } p \ge \frac{1}{2} \\ 63 | 0, \text{ if } p < \frac{1}{2} 64 | \end{cases} 65 | .\end{math} 66 | 67 | \end{proof} 68 | 69 | \end{document} 70 | -------------------------------------------------------------------------------- /Week6/Converse to SLLN/Converse to SLLN.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/Converse to SLLN/Converse to SLLN.pdf -------------------------------------------------------------------------------- /Week6/E4.2-A disitribution on N related to zeta function-毛昕渝/E.4.2.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/E4.2-A disitribution on N related to zeta function-毛昕渝/E.4.2.pdf -------------------------------------------------------------------------------- /Week6/E4.2-A disitribution on N related to zeta function-毛昕渝/ref.bib: -------------------------------------------------------------------------------- 1 | @book{tenenbaum2015introduction, 2 | title={Introduction to analytic and probabilistic number theory}, 3 | author={Tenenbaum, G{\'e}rald}, 4 | volume={163}, 5 | year={2015}, 6 | publisher={American Mathematical Soc.} 7 | } -------------------------------------------------------------------------------- /Week6/Events_in_tail_sigma_algebra_李孜睿/Events_in_tail_sigma_algebra_李孜睿.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/Events_in_tail_sigma_algebra_李孜睿/Events_in_tail_sigma_algebra_李孜睿.pdf -------------------------------------------------------------------------------- /Week6/Existence of iid random variables_金弘义/Existence of iid variables.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/Existence of iid random variables_金弘义/Existence of iid variables.pdf -------------------------------------------------------------------------------- /Week6/Hewitt-Savage Zero-One Law and Random Walk-吴润哲/Hewitt-Savage Zero-One Law and Random Walk.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/Hewitt-Savage Zero-One Law and Random Walk-吴润哲/Hewitt-Savage Zero-One Law and Random Walk.pdf -------------------------------------------------------------------------------- /Week6/Independent_normal_distribution_variable_sequence_张志成/Independent_normal_distribution_variable_sequence.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/Independent_normal_distribution_variable_sequence_张志成/Independent_normal_distribution_variable_sequence.pdf -------------------------------------------------------------------------------- /Week6/LLL and k-SAT_傅凌玥/Lecure Notes of LLL.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/LLL and k-SAT_傅凌玥/Lecure Notes of LLL.pdf -------------------------------------------------------------------------------- /Week6/Lovasz Local Lemma and k-SAT problem - 陆嘉馨/Lovasz Local Lemma and k-SAT problem- Lu Jiaxin.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/Lovasz Local Lemma and k-SAT problem - 陆嘉馨/Lovasz Local Lemma and k-SAT problem- Lu Jiaxin.pdf -------------------------------------------------------------------------------- /Week6/Lovasz Local Lemma and k-SAT problem - 陆嘉馨/ref.bib: -------------------------------------------------------------------------------- 1 | @incollection {MR0382050, AUTHOR = {Erd\H{o}s, P. and Lov\'{a}sz, L.}, TITLE = {Problems and results on {$3$}-chromatic hypergraphs and some related questions}, BOOKTITLE = {Infinite and finite sets ({C}olloq., {K}eszthely, 1973; dedicated to {P}. {E}rd\H{o}s on his 60th birthday), {V}ol. {II}}, PAGES = {609--627. Colloq. Math. Soc. J\'{a}nos Bolyai, Vol. 10}, YEAR = {1975}, MRCLASS = {05C15}, MRNUMBER = {0382050}, MRREVIEWER = {A. C.-F. Liu}, } -------------------------------------------------------------------------------- /Week6/Lovász Local Lemma and its Application-郭林松/Lovász Local Lemma and its Application-郭林松.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/Lovász Local Lemma and its Application-郭林松/Lovász Local Lemma and its Application-郭林松.pdf -------------------------------------------------------------------------------- /Week6/Proof for Lovasz Local Lemma_郭睿涵_陈彤/Proof for Lovasz Local Lemma_郭睿涵_陈彤.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/Proof for Lovasz Local Lemma_郭睿涵_陈彤/Proof for Lovasz Local Lemma_郭睿涵_陈彤.pdf -------------------------------------------------------------------------------- /Week6/Proof of Extended BC1 马浩博/discussion.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/Proof of Extended BC1 马浩博/discussion.pdf -------------------------------------------------------------------------------- /Week6/Proof of Extended BC1 马浩博/discussion.tex: -------------------------------------------------------------------------------- 1 | % This is a template for lecture notes. 2 | \documentclass[12pt]{article} 3 | \usepackage{amssymb} 4 | \usepackage[UTF8]{ctex} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{geometry} 8 | \usepackage{booktabs} 9 | \usepackage{bm} 10 | \usepackage{cite} 11 | %\usepackage{CJK} 12 | \usepackage[many]{tcolorbox} 13 | %\tcbuselibrary{listingsutf8} 14 | %\tcbuselibrary{skins, breakable, theorems, most} 15 | %\geometry{a4paper,bottom = 3cm,left = 3cm, right = 3cm} 16 | \CTEXoptions[today=old] 17 | %for reference 18 | \usepackage{hyperref} 19 | \usepackage[capitalise]{cleveref} 20 | \crefname{enumi}{}{} 21 | 22 | 23 | \newtheoremstyle{mythm}{1.5ex plus 1ex minus .2ex}{1.5ex plus 1ex minus .2ex} 24 | {}{\parindent}{\bfseries}{}{1em}{} 25 | \theoremstyle{mythm} 26 | \newtheorem{theorem}{Theorem} 27 | \newtheorem{lemma}[theorem]{Lemma} 28 | \newtheorem{corollary}[theorem]{Corollary} 29 | \newtheorem{fact}[theorem]{Fact} 30 | \newtheorem{definition}[theorem]{Definition} 31 | \newtheorem*{remark}{Remark} 32 | 33 | %\newenvironment{proof}{\noindent \textbf{Proof:}}{$\Box$} 34 | 35 | %to use newcommand for convenience 36 | \newcommand\field{\mathbb{F}} 37 | \newcommand\Real{\mathbb{R}} 38 | \newcommand\Q{\mathbb{Q}} 39 | \newcommand\Z{\mathbb{Z}} 40 | \newcommand\complex{\mathbb{C}} 41 | \newcommand\cc{\mathcal{C}} 42 | \newcommand\uu{\mathcal{U}} 43 | \newcommand\pp{\mathcal{P}} 44 | \newcommand\ff{\mathcal{F}} 45 | \renewcommand\refname{Reference} 46 | \renewcommand{\proofname}{Proof} 47 | \DeclareMathOperator{\range}{range} 48 | 49 | \title{Proof of Extended BC1} 50 | \author{马浩博 518030910428} 51 | \date{\today} 52 | \begin{document} 53 | \maketitle 54 | 55 | \begin{theorem} 56 | If $\lim_{n\to \infty}A_n=0$ and $\sum_{n = 1}^{\infty}P(A_n^c\cap A_{n+1})<\infty$ , then $P(A_n\ i.o.)=0$ 57 | \end{theorem} 58 | 59 | \begin{proof} 60 | Let $G_m=\cup_{n>m} A_n$ and $G_m \downarrow G$, where $G := lim sup A_n$ 61 | By the proof of BC1, we know that: 62 | $$P(A_n i.o.)=P(G) \leq \lim_{m\to \infty}P(G_m)$$ 63 | Because $\sum_{n = 1}^{\infty}P(A_n^c\cap A_{n+1})<\infty$, we know that $\lim_{m\to \infty}\sum_{m}^{\infty}P(A_n^c\cap A_{n+1})=0$. 64 | And by $\cup_{n\geq m} (A_n^c\cap A_{n+1})=G_m/(A_m\cap A_{m+1})$, we can get that: 65 | $$\lim_{m\to \infty}P(G_m) \leq \lim_{m\to \infty}P(\cup_{n\geq m} (A_n^c\cap A_{n+1}))+\lim_{n\to \infty}A_n \leq \lim_{m\to \infty}\sum_{m}^{\infty}P(A_n^c\cap A_{n+1})=0$$ 66 | 67 | So $P(A_n i.o.)=0$. 68 | \end{proof} 69 | 70 | \end{document} -------------------------------------------------------------------------------- /Week6/What's_fair_about_a_fair_game/whats_fair_about_a_fair_game.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/What's_fair_about_a_fair_game/whats_fair_about_a_fair_game.pdf -------------------------------------------------------------------------------- /Week6/Zero-One Law Of First Order Logic 金弘义/zero_one_law_of_fo.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week6/Zero-One Law Of First Order Logic 金弘义/zero_one_law_of_fo.pdf 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8 | \usepackage{float} 9 | \usepackage{mdframed} 10 | 11 | \usepackage{indentfirst} 12 | \setlength{\parindent}{2em} 13 | 14 | \providecommand{\abs}[1]{\lvert#1\rvert} 15 | \providecommand{\norm}[1]{\lVert#1\rVert} 16 | \providecommand{\ud}[1]{\underline{#1}} 17 | 18 | \newmdtheoremenv{thm}{Theorem} 19 | \newmdtheoremenv{lemma}[thm]{Lemma} 20 | \newmdtheoremenv{fact}[thm]{Fact} 21 | \newmdtheoremenv{cor}[thm]{Corollary} 22 | \newtheorem{eg}{Example} 23 | \newtheorem{ex}{Exercise} 24 | \newmdtheoremenv{defi}{Definition} 25 | \newenvironment{sol} 26 | {\par\vspace{3mm}\noindent{\it Solution}.} 27 | {\qed \\ \medskip} 28 | 29 | \newcommand{\ov}{\overline} 30 | \newcommand{\ca}{{\cal A}} 31 | \newcommand{\cb}{{\cal B}} 32 | \newcommand{\cc}{{\cal C}} 33 | \newcommand{\cd}{{\cal D}} 34 | \newcommand{\ce}{{\cal E}} 35 | \newcommand{\cf}{{\cal F}} 36 | \newcommand{\ch}{{\cal H}} 37 | \newcommand{\cl}{{\cal L}} 38 | \newcommand{\cm}{{\cal M}} 39 | \newcommand{\cp}{{\cal P}} 40 | \newcommand{\cs}{{\cal S}} 41 | \newcommand{\cz}{{\cal Z}} 42 | \newcommand{\eps}{\varepsilon} 43 | \newcommand{\ra}{\rightarrow} 44 | \newcommand{\la}{\leftarrow} 45 | \newcommand{\Ra}{\Rightarrow} 46 | \newcommand{\dist}{\mbox{\rm dist}} 47 | \newcommand{\bn}{{\mathbb N}} 48 | \newcommand{\bz}{{\mathbb Z}} 49 | 50 | \newcommand{\expe}{{\mathsf E}} 51 | \newcommand{\pr}{{\mathsf{Pr}}} 52 | 53 | 54 | \setlength{\parindent}{0pt} 55 | %\setlength{\parskip}{2ex} 56 | \newenvironment{proofof}[1]{\bigskip\noindent{\itshape #1. }}{\hfill$\Box$\medskip} 57 | \usepackage{amsthm,amsmath,amssymb} 58 | 59 | \theoremstyle{definition} 60 | \newtheorem{problem}{Problem} 61 | \newtheorem*{problem*}{Problem} 62 | 63 | \pagenumbering{gobble} 64 | 65 | \begin{document} 66 | 67 | \title{A Trivial Idea of Exercise 4.1} 68 | \date{Apr. 28, 2020} 69 | 70 | \maketitle 71 | \paragraph{}By imitating the method we used in the proof of Lemma 4.2, we can easily get the proof of this exercise. 72 | \paragraph{}By fixing $I_2\in \mathcal{I}_2$ and $I_3\in \mathcal{I}_3$, the two measures on $\sigma(\mathcal{I}_1)$ agree on $\mathcal{I}_1$, and they have the same total mass: 73 | \begin{align*} 74 | &\mathbb{P}(I_2\cap I_3) \\&= \mathbb{P}(\Omega\cap I_2\cap I_3)\\ &=\mathbb{P}(\Omega)\mathbb{P}(I_2)\mathbb{P}(I_3)\\&=\mathbb{P}(I_2)\mathbb{P}(I_3) 75 | \end{align*} 76 | \paragraph{}Hence, they agree on $\sigma(\mathcal{I}_1)$ 77 | \paragraph{}By fixing $H_1 \in \sigma(\mathcal{I}_1)$ and $I_3 \in \mathcal{I}_3$, the two measures on $\sigma(\mathcal{I}_2)$ agree on $\mathcal{I}_2$, and they have the same total mass: 78 | \begin{align*} 79 | &\mathbb{P}(H_1\cap I_3) \\&=\mathbb{P}(H_1)\mathbb{P}(I_3) 80 | \end{align*} 81 | \paragraph{}Similarly, by fixing $H_1 \in \sigma(\mathcal{I}_1)$ and $H_2 \in \sigma(\mathcal{I}_2)$, the two measures agree on $\sigma(\mathcal{I}_3)$. 82 | \paragraph{}Then we conclude that $\sigma(\mathcal{I}_1)$ $\sigma(\mathcal{I}_1)$ $\sigma(\mathcal{I}_3)$ are independent. 83 | \\ 84 | \paragraph{}We need $\Omega \in \mathcal{I}_k$ because we need the equation for the total mass in each case. Otherwise, consider one $\pi$-system being a set of measure 0. 85 | \end{document} 86 | 87 | -------------------------------------------------------------------------------- /Week7/Exercise4.1_郭睿涵/tmp: -------------------------------------------------------------------------------- 1 | 2 | -------------------------------------------------------------------------------- /Week7/Independence of multiple pi-systems and their conditions - 陆嘉馨/Independence of multiple pi-systems and their conditions - 陆嘉馨.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week7/Independence of multiple pi-systems and their conditions - 陆嘉馨/Independence of multiple pi-systems and their conditions - 陆嘉馨.pdf -------------------------------------------------------------------------------- /Week7/Independence of pi-system(E4.1)_傅凌玥，唐泽/Independence of pi-system(E4.1).pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week7/Independence of pi-system(E4.1)_傅凌玥，唐泽/Independence of pi-system(E4.1).pdf -------------------------------------------------------------------------------- /Week7/Integral-of-positive-simple-functions-赖睿航/integral-of-positive-simple-functions.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week7/Integral-of-positive-simple-functions-赖睿航/integral-of-positive-simple-functions.pdf -------------------------------------------------------------------------------- /Week7/Note for Lebesgue integration-陈彤/Note for Lebesgue integration.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week7/Note for Lebesgue integration-陈彤/Note for Lebesgue integration.pdf -------------------------------------------------------------------------------- /Week7/Notes of Integration 刘成锴/Integration.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week7/Notes of Integration 刘成锴/Integration.pdf -------------------------------------------------------------------------------- /Week7/P(A_k,i.o.) of throw a coin(E4.4)_傅凌玥/P(Ak,i.o.) of throw a coin(E4.4).pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week7/P(A_k,i.o.) of throw a coin(E4.4)_傅凌玥/P(Ak,i.o.) of throw a coin(E4.4).pdf -------------------------------------------------------------------------------- /Week7/P(A_k,i.o.) of throw a coin(E4.4)_傅凌玥/P(Ak,i.o.) of throw a coin(E4.4).tex: -------------------------------------------------------------------------------- 1 | \input{/Users/fulingyue/Desktop/def} 2 | 3 | \title{$P(A_k,i.o.)$ of Throw a Coin(E4.4)} 4 | \author{Fu Lingyue} 5 | \date{\today} 6 | 7 | \begin{document} 8 | \maketitle 9 | 10 | Throw a coin is a basic problem of probability, but mathematicians are not satisfied with the case where p is equal to $1/2$. In the textbook, the author puts forward a problem as follows. 11 | \begin{theorem} 12 | Suppose that a coin with probability $p$ of heads is tossed repeatedly. Let $A_k$ be the event that a sequence of $k$ (or more) consecutive heads occurs amongst tosses numbered $2^k,2^{k+1},…,2^{k+1}-1$. Prove that 13 | \begin{equation} 14 | P(A_k,i.o) = 15 | \begin{cases} 16 | 1, & \text{if } p \geq 1/2,\\ 17 | 0, & \text{otherwise.} 18 | 19 | \end{cases} 20 | \end{equation} 21 | 22 | \end{theorem} 23 | 24 | 25 | \begin{proof} 26 | 1) When $p < \frac{1}{2}$, we use BC1 to prove. Let $B_n$ be the event that there are k heads starting from nth toss. Thus 27 | $$A_k = \bigcup_{n=2^k}^{2^{k+1}-k}B_n.$$ 28 | 29 | In this way, we obtain(by inclusion-exclusion principle) 30 | $$ 31 | P(A_k) \leq \Sigma_{n=2^k}^{2^{k+1}-k}P(B_n) \leq 2^kp^k. 32 | $$ 33 | For $p<1/2$, 34 | $$\Sigma_kP(A_k) \leq \frac{2p}{1-2p} \leq \infty.$$ 35 | 36 | According BC1, we get $P(A_k,i.o.) = 0$ when $p<1/2.$ 37 | 38 | 2) When $p\geq 1/2$, use BC2 to prove. According to hint, we can firstly let $E^k_i$ be the event that there are $k$ consecutive heads beginning at toss numbered $2^k+(i-1)k$. Then $i$ is between $1$ and $2^k/k$. That is, the beginning of $k$ consecutive heads are $2^k, 2^k + k, \dots ,2^{k} + 2^{k}-k(i.e.,2^{k+1}-k)$. These events $E_i^k$ are independent, and it is clear that 39 | $$\{E_i^k,i.o.\} \Rightarrow \{A^k,i.o.\}.$$ 40 | 41 | For we have 42 | \begin{equation} 43 | \begin{aligned} 44 | \Sigma_k\Sigma_{i=1}^{2^k/k}P(E_i^k) &\geq \Sigma_k(2^k/k - 1)p^k \\ 45 | &= \Sigma_k\frac{1}{k}\frac{1}{2}^{k-1} - \frac{p}{1-p}\\ 46 | &\geq \Sigma_k\frac{1}{k} - \frac{p}{1-p} = +\infty 47 | \end{aligned} 48 | \end{equation} 49 | 50 | According BC2, we get $P(E_i^k,i.o.) =1$. Thus $P(A_k,i.o.) = 1$ when $p\geq 1/2.$ 51 | \end{proof} 52 | \end{document} 53 | -------------------------------------------------------------------------------- /Week7/P(A_k,i.o.) of throw a coin(E4.4)_傅凌玥/P(Ak,i.o.) of throw a coin(E4.4)_傅凌玥.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week7/P(A_k,i.o.) of throw a coin(E4.4)_傅凌玥/P(Ak,i.o.) of throw a coin(E4.4)_傅凌玥.pdf -------------------------------------------------------------------------------- /Week7/Proof of MON 马浩博/discussion.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week7/Proof of MON 马浩博/discussion.pdf -------------------------------------------------------------------------------- /Week8/A proof of a μ-integrable lemma and h(fμ)=(hf)μ_吴润哲/A proof of a mu-integrable lemma and h(f mu)=(hf)mu using the standard machine.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week8/A proof of a μ-integrable lemma and h(fμ)=(hf)μ_吴润哲/A proof of a mu-integrable lemma and h(f mu)=(hf)mu using the standard machine.pdf -------------------------------------------------------------------------------- /Week8/DOM's dominating function is necessary 李孜睿/DOM's dominating function is necessary 李孜睿.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week8/DOM's dominating function is necessary 李孜睿/DOM's dominating function is necessary 李孜睿.pdf -------------------------------------------------------------------------------- /Week8/DOM's dominating function is necessary 李孜睿/DOM's dominating function is necessary 李孜睿.tex: -------------------------------------------------------------------------------- 1 | % This is a template for lecture notes. 2 | \documentclass{article} 3 | \usepackage[UTF8]{ctex} 4 | \usepackage{amssymb} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{geometry} 8 | \usepackage{booktabs} 9 | \usepackage{bm} 10 | \usepackage{tcolorbox} 11 | \CTEXoptions[today=old] 12 | %Some commonly used notations 13 | %\geometry{a4paper,bottom = 3cm,left = 3cm, right = 3cm} 14 | 15 | %for reference 16 | \usepackage{hyperref} 17 | \usepackage[capitalise]{cleveref} 18 | \crefname{enumi}{}{} 19 | 20 | \newtheorem{theorem}{Theorem} 21 | \newtheorem{lemma}[theorem]{Lemma} 22 | \newtheorem{proposition}[theorem]{Proposition} 23 | \newtheorem{corollary}[theorem]{Corollary} 24 | \newtheorem{fact}[theorem]{Fact} 25 | \newtheorem{definition}[theorem]{Definition} 26 | \newtheorem{remark}[theorem]{Remark} 27 | \newtheorem{question}[theorem]{Question} 28 | \newtheorem{answer}[theorem]{Answer} 29 | \newtheorem{exercise}[theorem]{Exercise} 30 | \newtheorem{example}[theorem]{Example} 31 | %\newenvironment{proof}{\noindent \textbf{Proof:}}{$\Box$} 32 | \newtheorem{observation}[theorem]{Observation} 33 | 34 | %to use newcommand for convenience 35 | \newcommand\field{\mathbb{F}} 36 | \newcommand\Real{\mathbb{R}} 37 | \newcommand\Q{\mathbb{Q}} 38 | \newcommand\Z{\mathbb{Z}} 39 | \newcommand\complex{\mathbb{C}} 40 | \newenvironment{myproof}{\ignorespaces\paragraph{Proof:}}{\hfill $\square$\par\noindent} 41 | %this is how we define operators. 42 | \DeclareMathOperator{\rank}{rank} % rank 43 | 44 | \title{DOM's dominating function is necessary} 45 | \author{李孜睿 518030910424} 46 | \date{\today} 47 | 48 | 49 | \begin{document} 50 | \maketitle 51 | \begin{tcolorbox} 52 | \begin{theorem} 53 | \text{DOM(Lebesgue's Dominated Convergence Theorem)} 54 | \end{theorem} 55 | 56 | $f_n \rightarrow f(\text{a.e.})$ and $|f_n(s)|\leq g(s),\forall s\in S,\forall n\in\mathbb{N}$, $\mu(g)<\infty$ $\Rightarrow$ $\mu(|f_n-f|)\rightarrow 0$ which also implies $\mu(f_n)\rightarrow\mu(f)$. 57 | \end{tcolorbox} 58 | 59 | 60 | 61 | I will prove the necessity of the dominating function $g\in\mathcal{L}^1(S,\Sigma,\mu)$ by giving a counterexample. 62 | 63 | 64 | 65 | Assume $S=(0, 1]$, $\Sigma=\mathcal{B}(S)$, and $\mu=\text{Leb}(S)$. Define $f=0$ and 66 | 67 | $$ 68 | f_n(s)=\left\{ 69 | \begin{array}{lr} 70 | n, s\in(0,\frac1n]\\ 71 | 0, s\in(\frac1n,1] 72 | \end{array} 73 | \right. 74 | $$ 75 | 76 | Clearly, $f_n\rightarrow f(\text{a.e.})$. But $\mu(|f_n-f|)=n\times\mu((0,\frac1n])=1\Rightarrow\mu(|f_n-f|)\rightarrow 1$, contradicting to DOM's conclusion $\mu(|f_n-f|)\rightarrow 0$. 77 | \end{document} -------------------------------------------------------------------------------- /Week8/Integrals over subsets_董海辰/Integrals over subsets_董海辰.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week8/Integrals over subsets_董海辰/Integrals over subsets_董海辰.pdf -------------------------------------------------------------------------------- /Week8/Integrals over subsets_董海辰/Integrals over subsets_董海辰.tex: -------------------------------------------------------------------------------- 1 | \input{/Users/oscar/Documents/LaTeX_Templates/HW.tex} 2 | 3 | \title{Integrals over subsets} 4 | \date{\today} 5 | \author{董海辰 518030910417} 6 | 7 | \begin{document} 8 | \maketitle 9 | 10 | \begin{thm}{}{} 11 | In measure space $(S, \Sigma, \mu)$, $A \in \Sigma$. Show that: 12 | \begin{enumerate} 13 | \item $(A, \Sigma\cap 2^A)$ is a $\sigma $-algebra. 14 | \item $\forall f \in (m\Sigma)^+$, 15 | $$\mu_{\Sigma\cap 2^A}(f) = \mu(f\cdot 1_A).$$ 16 | \end{enumerate} 17 | \end{thm} 18 | 19 | \begin{proof}[Proof] 20 | ~ 21 | \begin{enumerate} 22 | \item We have: 23 | \begin{itemize} 24 | \item $A \in \Sigma \implies A \in \Sigma\cap 2^A$. 25 | \item $\forall F \in \Sigma \cap 2^A $, 26 | \begin{math} 27 | A \setminus F &= (S \setminus(S \setminus A )) \setminus F & \\ 28 | &= S \setminus (F \cup (S \setminus A)) \in \Sigma \\ 29 | &= (S \setminus F ) \cap A \in 2^A 30 | .\end{math} 31 | Thus $A \setminus F \in \Sigma \cap 2^A$. 32 | \item $\forall (F_i)_{i\in \mathbb{N} }$ where $F_i \in \Sigma \cap 2^A \subseteq \Sigma$. Then $\bigcup_{i} F_i \in \Sigma$, and $\bigcup_{i} F_i \in 2^A$, thus, $\bigcup_{i} F_i \in \Sigma \cap 2^A$. 33 | \end{itemize} 34 | 35 | Therefore, $(A, \Sigma\cup 2^A)$ is a $\sigma $-algebra. 36 | \item We will show this with the standard machine. 37 | \begin{itemize} 38 | \item Let $f = 1_B, B \in \Sigma \cap 2^A$. We have $\mu(f \cdot 1_A) = \mu(1_B \cdot 1_A) = \mu(A\cap B)$. Meanwhile, 39 | $$\mu|_{\Sigma \cap 2^A}(1_B) = \mu|_{\Sigma \cap 2^A}((B\setminus A) \cup (B\cap A)) = \mu|_{\Sigma \cap 2^A}(B\cap A) + 0 = \mu(A \cap B).$$ 40 | 41 | Thus $\mu(f\cdot 1_A) = \mu|_{\Sigma \cap 2^A}(f)$ for all $f = 1_B$. 42 | 43 | \item Then we can let $f \in SF^+$. Let $f = \sum ^n b_i 1_{B_i}$. Then 44 | \begin{math} 45 | \mu(f\cdot 1_A) = \mu((\sum ^n b_i 1_{B_i})1_A) &= \mu(\sum ^n b_i 1_{B_i \cap A}) \\ 46 | &= \sum ^n b_i\mu(1_{B_i \cap A}) \\ 47 | &= \sum ^n b_i \mu|_{\Sigma \cap 2^A}(1_{B_i}) = \mu|_{\Sigma \cap 2^A}(\sum ^n b_i 1_{B_i}) 48 | .\end{math} 49 | \item Finally we let $f \in (m\Sigma)^+$, $\mu(f) = \sup \{\mu(h): h \in SF^+, h\le f\} $. 50 | 51 | \begin{math} 52 | \mu(f \cdot 1_A) &= \sup \{\mu(h): h \in SF^+, h \le f\cdot 1_A\} \\ 53 | &= \sup \{\mu(g\cdot 1_A): g \in SF^+, g \cdot 1_A \le f \cdot 1_A\} \\ 54 | &= \sup \{\mu|_{\Sigma \cap 2^A}(g): g \in SF^+, g \cdot 1_A \le f \cdot 1_A\} \\ 55 | &= \sup \{\mu|_{\Sigma \cap 2^A}(g): g \in SF^+, g\le f\} = \mu|_{\Sigma \cap 2^A}(f) 56 | .\end{math} 57 | \end{itemize} 58 | 59 | Therefore, for all $f \in (m\Sigma)^+$, the result of integral over a subset is equal to the result of the integral in restricted measure space. 60 | \end{enumerate} 61 | \end{proof} 62 | 63 | 64 | \end{document} 65 | -------------------------------------------------------------------------------- /Week8/More to say on Scheffe's Lemma-毛昕渝/More to say on Scheffe's Lemma.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week8/More to say on Scheffe's Lemma-毛昕渝/More to say on Scheffe's Lemma.pdf -------------------------------------------------------------------------------- /Week8/More to say on Scheffe's Lemma-毛昕渝/ref.bib: -------------------------------------------------------------------------------- 1 | @book{tao2011introduction, 2 | title={An introduction to measure theory}, 3 | author={Tao, Terence}, 4 | volume={126}, 5 | year={2011}, 6 | publisher={American Mathematical Society Providence, RI} 7 | } -------------------------------------------------------------------------------- /Week8/Notes on the proof of Radon-Nikodym theorem_金弘义/proof of Radon-Nikodym theorem.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week8/Notes on the proof of Radon-Nikodym theorem_金弘义/proof of Radon-Nikodym theorem.pdf -------------------------------------------------------------------------------- /Week8/Relationship_between_convergences_张志成/Relationship_between_convergences.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week8/Relationship_between_convergences_张志成/Relationship_between_convergences.pdf -------------------------------------------------------------------------------- /Week8/The Necessity of Continuity Constraints on Convergence in Distribution_寻之扬/The Necessity of Continuity Constraints on Convergence in Distribution.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week8/The Necessity of Continuity Constraints on Convergence in Distribution_寻之扬/The Necessity of Continuity Constraints on Convergence in Distribution.pdf -------------------------------------------------------------------------------- /Week8/pointwise convergence and convergence in probability_董海辰/pointwise convergence and convergence in probability.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week8/pointwise convergence and convergence in probability_董海辰/pointwise convergence and convergence in probability.pdf -------------------------------------------------------------------------------- /Week9/Lebesgue Covering Dimension-寻之扬/Lebesgue Covering Dimension of One-dimensional Euclidean Space Equals to One.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/Week9/Lebesgue Covering Dimension-寻之扬/Lebesgue Covering Dimension of One-dimensional Euclidean Space Equals to One.pdf -------------------------------------------------------------------------------- /Week9/Lebesgue Covering Dimension-寻之扬/Lebesgue Covering Dimension of One-dimensional Euclidean Space Equals to One.tex: -------------------------------------------------------------------------------- 1 | \input{D:/template} 2 | \title{Lebesgue Covering Dimension of One-dimensional Euclidean Space Equals to One} 3 | \author{Zhiyang Xun} 4 | \date{May 6, 2020} 5 | \begin{document} 6 | \maketitle 7 | \begin{tcolorbox} 8 | \begin{problem} 9 | Suppose real line is covered by a family of open intervals $(I_k)$. Is there always a family of open intervals $(J_k)$ satisfying $I_k\supseteq J_k$ such that every point on the real line is covered once or twice? 10 | \end{problem} 11 | \end{tcolorbox} 12 | 13 | The answer is yes. We can prove a stronger result: We can always find a family of open intervals $(J_k)$ that covers every point once or twice. In addition, for each $J_k$, either $J_k = I_k$ or $J_k = \emptyset$. 14 | 15 | For simplicity, we will regard a family of open intervals as a set consisting of open intervals. For example, $(J_k) \subseteq (I_k) \cup \{\emptyset\}$. 16 | 17 | Suppose $F_j$ is a family of open intervals that covers the real line. Let $F_0 := (I_k)$. When $j > 0$, $F_j \subseteq F_{j-1}$ and every point in $[-j, +j]$ is covered by $F_j$ once or twice. 18 | 19 | If for each $j \in \N$, such $F_j$ exists, then it is easy to check that \[ 20 | \bigcap_{j = 0}^{\infty} F_j \cup \{\emptyset\} 21 | \] 22 | is the $(J_k)$ we are looking for. 23 | 24 | To verify the existence of $F_j$, we can give an inductive proof. 25 | 26 | {Basis Step:} $F_0 = (I_k)$, so $F_0$ exits. 27 | 28 | {Inductive Hypothesis:} $F_{j-1}$ exists. 29 | 30 | {Inductive Step:} Define 31 | \begin{align*} 32 | S := &\{\alpha \in F_{j-1} | \alpha \cap [-j, +j] \neq \emptyset\} \\ 33 | T := &\bigcup_{I \in (F_{j-1} \setminus S)} I. 34 | \end{align*} 35 | Choose an interval $(l_1, r_1) \in S$ such that \[ \forall p \in (R \setminus T),\ l_1 \leq p. 36 | \] 37 | Again, choose an interval $(l_2, r_2) \in S$ such that \[ 38 | \forall p \in (R \setminus T),\ r_2 \geq p. 39 | \] 40 | Since $S$ covers $[-j, +j]$, it has a finite subcover $S_f$ on $[-j, +j]$. Let \[ 41 | S' := S_f \cup \{[l_1, r_1] , [l_2, r_2]\}. 42 | \] 43 | Obviously $S'$ is also a finite subcover of $S$ on $[-j, +j]$. 44 | 45 | Now we are going to remove some ``bad intervals'' from $S'$. Each time, we choose a bad interval and eliminate it. Since $S'$ has finite intervals, the elimination process will end after finite steps. We call an interval $I_r$ as a ``bad interval'' if and only if we can find two other intervals $I_s, I_t \in S'$ such that \[ 46 | I_r \subseteq I_s \cup I_t. 47 | \] 48 | 49 | Denote the $S'$ after the whole elimination process by $S^*$. Every point in $[-j, +j]$ is covered once or twice by $S^*$. That's because if one point is covered for more than twice, we can continue our elimination process. 50 | 51 | $S^* \cup (F_{j-1} \setminus S)$ is the $F_j$ we want. Since every point in $[-j, +j]$ is only covered by intervals from $S^*$, every point is covered for no more than twice. From the construction of $S^*$, we know $S^* \cup (F_{j-1} \setminus S)$ covers the whole real line. Hence, we proved the existence $F_j$. 52 | 53 | This finishes the proof. 54 | 55 | \end{document} -------------------------------------------------------------------------------- /example/example.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/example/example.pdf -------------------------------------------------------------------------------- /example/example.tex: -------------------------------------------------------------------------------- 1 | % This is a template for lecture notes. 2 | \documentclass{article} 3 | \usepackage[UTF8]{ctex} 4 | \usepackage{amssymb} 5 | \usepackage{amsmath} 6 | \usepackage{amsthm} 7 | \usepackage{geometry} 8 | \usepackage{booktabs} 9 | \usepackage{bm} 10 | \usepackage{tcolorbox} 11 | \CTEXoptions[today=old] 12 | %Some commonly used notations 13 | %\geometry{a4paper,bottom = 3cm,left = 3cm, right = 3cm} 14 | 15 | %for reference 16 | \usepackage{hyperref} 17 | \usepackage[capitalise]{cleveref} 18 | \crefname{enumi}{}{} 19 | 20 | \newtheorem{theorem}{Theorem} 21 | \newtheorem{lemma}[theorem]{Lemma} 22 | \newtheorem{proposition}[theorem]{Proposition} 23 | \newtheorem{corollary}[theorem]{Corollary} 24 | \newtheorem{fact}[theorem]{Fact} 25 | \newtheorem{definition}[theorem]{Definition} 26 | \newtheorem{remark}[theorem]{Remark} 27 | \newtheorem{question}[theorem]{Question} 28 | \newtheorem{answer}[theorem]{Answer} 29 | \newtheorem{exercise}[theorem]{Exercise} 30 | \newtheorem{example}[theorem]{Example} 31 | %\newenvironment{proof}{\noindent \textbf{Proof:}}{$\Box$} 32 | \newtheorem{observation}[theorem]{Observation} 33 | 34 | %to use newcommand for convenience 35 | \newcommand\field{\mathbb{F}} 36 | \newcommand\Real{\mathbb{R}} 37 | \newcommand\Q{\mathbb{Q}} 38 | \newcommand\Z{\mathbb{Z}} 39 | \newcommand\complex{\mathbb{C}} 40 | 41 | %this is how we define operators. 42 | \DeclareMathOperator{\rank}{rank} % rank 43 | 44 | \title{Notes on Whatever You Want 1} 45 | \author{Whoever You Are} 46 | \date{\today} 47 | 48 | \begin{document} 49 | \maketitle 50 | \section{Basic Logic for Ancient Greeks} 51 | \begin{theorem}\label{thm:mortal} 52 | All men are mortal. 53 | \end{theorem} 54 | 55 | \begin{fact}\label{fact:socrates} 56 | Socrates is a man. 57 | \end{fact} 58 | %use \cref instead of \ref 59 | Combine \cref{thm:mortal} and \cref{fact:socrates} 60 | and we get 61 | \begin{corollary} 62 | Socrates is mortal. 63 | \end{corollary} 64 | \end{document} -------------------------------------------------------------------------------- /sheep/吴润哲/dp.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | #define N 55 3 | using namespace std; 4 | 5 | typedef long double ld; 6 | 7 | double p, q; 8 | ld f1[N<<1][N<<1], f2[N<<1][N<<1], g[N<<1][N<<1]; 9 | 10 | void Gauss(int n) 11 | { 12 | for(int i = 1; i <= n; i++) 13 | { 14 | for(int j = i+1; j <= n; j++) 15 | { 16 | assert(g[i][i] > 0); 17 | ld x = g[j][i] / g[i][i]; 18 | for(int k = i; k <= n+1; k++) 19 | g[j][k] -= x * g[i][k]; 20 | } 21 | } 22 | for(int i = 1; i <= n; i++) 23 | for(int j = n+1; j >= i; j--) 24 | { 25 | assert(g[i][i] > 0); 26 | g[i][j] /= g[i][i]; 27 | } 28 | for(int i = n; i >= 1; i--) 29 | { 30 | ld x = g[i][n+1]; 31 | for(int j = i-1; j; j--) 32 | { 33 | g[j][n+1] -= g[j][i]*x; 34 | g[j][i] = 0; 35 | } 36 | } 37 | } 38 | 39 | int main() 40 | { 41 | scanf("%lf%lf",&p,&q); 42 | 43 | f1[1][0] = f2[1][0] = 1; 44 | 45 | for(int d = 2; d < N; d++) 46 | { 47 | f1[d][0] = f2[d][0] = d; 48 | 49 | for(int a = 1; a < d; a++) 50 | { 51 | int b = d - a; 52 | 53 | f1[a][b] = max(f1[a][b], (long double)a); 54 | for(int c = 1; c < b; c++) 55 | { 56 | ld tmp = 1.0 * a * p / (a * p + c * q) * f2[a+1][c-1] + 1.0 * c * q / (a * p + c * q) * f2[a-1][c+1]; 57 | if(tmp > f1[a][b]) 58 | { 59 | f1[a][b] = tmp; 60 | } 61 | } 62 | } 63 | 64 | for(int a = 1; a < d; a++) 65 | { 66 | int b = d - a; 67 | f2[a][b] = f1[a][b]; 68 | } 69 | 70 | for(int l = 0; l < d; l++) 71 | for(int r = l+2; r <= d; r++) 72 | { 73 | int n = r - l + 1; 74 | for(int i = 0; i <= n+1; i++) 75 | for(int j = 0; j <= n+1; j++) 76 | g[i][j] = 0; 77 | g[1][1] = 1; 78 | g[n][n] = 1; g[n][n+1] = 1; 79 | 80 | for(int a = l+1; a < r; a++) 81 | { 82 | int b = d - a; 83 | g[a-l+1][a-l+1] = 1; 84 | g[a-l+1][a+1-l+1] = -1.0 * a * p / (a * p + b * q); 85 | g[a-l+1][a-1-l+1] = -1.0 * b * q / (a * p + b * q); 86 | } 87 | Gauss(n); 88 | for(int a = l+1; a < r; a++) 89 | { 90 | int b = d - a; 91 | ld pro = g[a-l+1][n+1]; 92 | f2[a][b] = max(f2[a][b], pro * f1[r][d-r] + (1-pro) * f1[l][d-l]); 93 | } 94 | 95 | } 96 | 97 | } 98 | 99 | printf("%lf\n", (double)f2[21][15]); 100 | 101 | // for(int i = 1; i <= 30; i++, puts("")) 102 | // for(int j = 1; j <= 30; j++) 103 | // { 104 | // if(i*p-j*q <= 0) 105 | // printf("(%lf)\t", (double)(f2[i][j] - f2[i][j-1])); 106 | // else 107 | // printf("%lf\t", (double)(f2[i][j] - f2[i][j-1])); 108 | // } 109 | 110 | // puts(""); 111 | 112 | // for(int i = 1; i <= 30; i++, puts("")) 113 | // for(int j = 1; j <= 30; j++) 114 | // printf("%lf\t", i*p-j*q); 115 | 116 | } -------------------------------------------------------------------------------- /sheep/吴润哲/sheep.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/sheep/吴润哲/sheep.pdf -------------------------------------------------------------------------------- /sheep/吴润哲/sheet.xlsx: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/sheep/吴润哲/sheet.xlsx -------------------------------------------------------------------------------- /sheep/吴润哲/simulator.cpp: -------------------------------------------------------------------------------- 1 | #include 2 | using namespace std; 3 | double p, q; int a, b; 4 | 5 | default_random_engine dre(time(NULL)); 6 | uniform_real_distribution dis2(0.0, 1.0); 7 | 8 | int simul() 9 | { 10 | int cur_a = a, cur_b = b; 11 | while(cur_a && cur_b) 12 | { 13 | while(cur_a * p <= cur_b * q) cur_b --; 14 | 15 | if(!cur_a || !cur_b) break; 16 | 17 | // uniform_int_distribution dis1(0, cur_a + cur_b - 1); 18 | // int choose = dis1(dre) < cur_a ? 1 : 0; 19 | // double pro = dis2(dre); 20 | 21 | // if(choose == 1) // black 22 | // { 23 | // if(pro < p) 24 | // cur_a++, cur_b--; 25 | // } 26 | // else 27 | // { 28 | // if(pro < q) 29 | // cur_a--, cur_b++; 30 | // } 31 | 32 | if(dis2(dre) < cur_a * p / (cur_a * p + cur_b * q)) 33 | cur_a++, cur_b--; 34 | else 35 | cur_a--, cur_b++; 36 | 37 | } 38 | return cur_a ? cur_a : 0; 39 | } 40 | int main() 41 | { 42 | // scanf("%lf%lf%d%d",&p,&q,&a,&b); 43 | 44 | // p = 1, q = 1; a = 21, b = 15; 45 | p = 0.5, q = 0.8; a = 21, b = 15; 46 | 47 | printf("(%lf)\n", a*p-b*q); 48 | 49 | int T = 20000000; long long ans = 0; 50 | for(int i = 1; i <= T; i++) 51 | ans += simul(); 52 | printf("%lf\n", 1.0 * ans / T); 53 | 54 | } -------------------------------------------------------------------------------- /sheep/唐泽/魔法羊的一条死路和两个猜想.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/sheep/唐泽/魔法羊的一条死路和两个猜想.pdf -------------------------------------------------------------------------------- /sheep/唐泽/魔法羊的一条死路和两个猜想.tex: -------------------------------------------------------------------------------- 1 | % !Mode:: "TeX:UTF-8" 2 | 3 | \documentclass{article} 4 | \usepackage[UTF8]{ctex} 5 | \usepackage{amssymb} 6 | \usepackage{amsmath} 7 | \usepackage{amsthm} 8 | 9 | \usepackage{geometry} 10 | \geometry{a4paper,left = 3cm, right = 3cm, bottom = 2cm} 11 | 12 | \CTEXoptions[today=old] 13 | %Some commonly used notations 14 | \newcommand\Real{\mathbb{R}} 15 | \newcommand\rational{\mathbb{Q}} 16 | \newcommand\Rational{\mathbb{Q}^*} 17 | \newcommand\integer{\mathbb{Z}} 18 | \newcommand\Interger{\mathbb{Z}^*} 19 | 20 | \title{魔法羊的一条死路和两个猜想} 21 | \author{唐泽} 22 | \date{\today} 23 | 24 | \begin{document} 25 | 26 | \maketitle 27 | 28 | 定义1：$[i,j]$为$i$只黑羊，$j$只白羊的状态 29 | 30 | 转移概率：$[i,j]$ 31 | $\begin{cases} 32 | [i+1,j-1] \ (Possibility=\frac{ip}{i+j})\\ 33 | [i,j] \ (Possibility=\frac{i(1-p)+j(1-q)}{i+j})\\ 34 | [i-1,j+1] \ (Possibility=\frac{jq}{i+j}) 35 | \end{cases}$ 36 | 37 | 定义2：$f(i,j)$为$i$只黑羊，$j$只白羊，剩下黑羊数量的最大期望 38 | 39 | 定义3：$F(i,j)$为$i$只黑羊，$j$只白羊，本次不可移除白羊，剩下黑羊数量的最大期望 40 | 41 | 于是有： 42 | $$f(i,j)=max_{0\le k\le j}\{F(i,k)\}$$ 43 | $$F(i,j)=\frac{ip}{i+j}f(i+1,j-1)+\frac{i(1-p)+j(1-q)}{i+j}f(i,j)+\frac{jq}{i+j}f(i-1,j+1)$$ 44 | 45 | 边界条件： 46 | $$f(i,0)=i$$ 47 | $$f(0,j)=0$$ 48 | 49 | 但这几个式子里有个max，无法通过Gauss消元求解，所以这条路目前到此为止了。 50 | 51 | 于是我想尝试一些感性的想法。看到有变化的两个转移概率看上去像正反馈，变化对方的优势会随着个数增多而变大。于是就有了下面两个猜想： 52 | 53 | 结论1：$\forall i\exists_1 k ((\forall j\ge k,f(i,j)=f(i,k)) \wedge (\forall j\frac{jq}{i+j}$,这个条件约束了白羊数量$j$有个上限，最大值必然在这个上限之前到来。$j$足够大后函数值不变是由于取F函数的前缀max. 56 | 57 | 定义4：将结论1中$i$所对应的唯一的$k$定义为$g(i)$ 58 | 59 | 猜想1（白羊单调性）：$\forall i\forall j_1\frac{jq}{i+j}$.这本可以用C++模拟的，但我和毛昕渝交流后发现他已经对于这个模型模拟过了，我就不再做一遍了。 66 | \end{document} 67 | 68 | -------------------------------------------------------------------------------- /sheep/杨宗翰/sheep-yzh.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/sheep/杨宗翰/sheep-yzh.pdf -------------------------------------------------------------------------------- /sheep/杨宗翰/sheep.tex: -------------------------------------------------------------------------------- 1 | \documentclass[11pt,a4paper,oneside]{article} 2 | 3 | \usepackage[utf8]{inputenc} 4 | \usepackage[english]{babel} 5 | \usepackage{olymp,clrscode3e} 6 | \usepackage{amsmath,graphicx,epigraph,bnf,enumerate,comment,listings,fontspec,color,indentfirst} 7 | \usepackage{subcaption} 8 | \usepackage{listings} 9 | \usepackage{ctex} 10 | \usepackage{CJK} 11 | 12 | \newcommand{\chr}[1]{\mbox{`\texttt{#1}'}} 13 | \newcommand{\txt}[1]{\mbox{``\texttt{#1}''}} 14 | 15 | \renewcommand{\contestname}{ 16 | Magical Sheep, MS107 Probability Homework, Yang Zonghan, \today 17 | } 18 | 19 | \lstset{ 20 | breaklines=true, 21 | numbers=left, 22 | basicstyle = \fontspec{Consolas}, 23 | commentstyle=\color{gray}, 24 | frame=shadowbox 25 | } 26 | \setCJKmainfont[BoldFont={黑体}, ItalicFont={楷体}, BoldItalicFont={黑体}]{华文中宋} 27 | \begin{document} 28 | \title{\textbf{Magical Sheep}} 29 | \author{Yang Zonghan(518030910435)} 30 | \date{\today} 31 | \maketitle 32 | \setlength{\parindent}{2em} 33 | \section{Problem} 34 | 草原上有2000只黑羊和2000只白羊，它们不会死亡和新生。每天都有一只羊被随机选中，该随机选择与以前的选择无关。 这只被选中的羊会找到一只与之异色的羊（如果还有的话）并对其鸣叫。一只黑羊对准一只白羊鸣叫会使白羊以概率p发生变色（变为黑羊）；一只白羊对准一只黑羊鸣叫会使黑羊以概率q发生变色（变为白羊）。管理者每天可以在神奇的羊变色现象发生前进场移走任意数量的白羊。管理者的目标是使得最后所剩黑羊数量的期望值尽可能大。譬如，头天就移走所有白羊，可以保证最后保住2000只黑羊。试针对一些不同的p,q取值来给管理者建议管理策略以及对最好可能的黑羊遗留数量做出估计。 35 | 36 | \section{Intuitive thoughts} 37 | 假设某次操作完后，有 $n$ 只黑羊和 $m$ 只白羊时，那么此时黑羊变为白羊的概率是 $\frac {np} {n+m}$，白羊变为黑羊的概率是 $\frac {mq} {n+m}$，其余的 $\frac {n(1 - p) + m(1 - q)} {n+m}$ 什么都不会发生。这三个事件是互斥的，因此在 $\frac {np} {n+m} > \frac {mq} {n+m}$ 时，期望黑羊是变多的。因此，我以为每次将白羊移至这样的不等式成立即在期望下使黑羊变多。 38 | 39 | \section{Further thoughts} 40 | 但是上述讨论并不完备，因为羊的数量是有限的。在无限的情况下，期望能够趋近取得，但是实际中并不是这样的——也许是风险非常大的：上述过程并没有保证黑羊的最小值。举例来说，我们抛一枚均匀的硬币，正面获得一块钱，反面失去一块钱，而玩家钱数有限，这样等价于 1d random walk，在无限步是有 $1$ 的概率走到过 $0$ 点的；而只要走到过 $0$，那么玩家就一无所有了。从这个角度来看，上述策略并不是最优策略——至少需要一些风险规避的手段。 41 | 42 | 注意到 $2000+2000$ 是个相对小的值，我们可以通过动态规划来解出最优值。设 $f_{i,j}$ 表示某天在移动羊之前有 $i$ 只黑羊，$j$ 只白羊，期望下能够在最后保留的黑羊的数量；$g_{i,j}$ 表示某天在移动羊之后有 $i$ 只黑羊，$j$ 只白羊，期望下能够在最后保留的黑羊的数量。那么有如下两个式子： 43 | \begin{align} 44 | f_{i,j} &= \max \{g_{i,k} | k \leq j\} \\ 45 | g_{i,j} &= \Big ( \frac i {i + j} \big(p f_{i - 1, j + 1} + (1 - p) f_{i, j}\big) + \frac j {i + j} \big(q f_{i + 1, j - 1} + (1 - q) f_{i, j}\big) \Big ) 46 | \end{align} 47 | 初始条件为 \begin{align} 48 | f_{i,0} = g_{i,0} = i \\ 49 | f_{0,i} = g_{0,i} = 0 50 | \end{align} 51 | 52 | 很不幸的是，上述动态规划转移是有环的，因此递推求解是困难的。但是我们可以使用迭代最短路的方式来进行迭代求解。注意到这是一个按 $i+ j$ 分层的分层图，第 $n$ 层只有 $n$ 个需要迭代的变量，因此直观感受下在 $O(n ^ 2)$ 次更新内应该能迭代出解，总更新次数应该在 $O(n ^ 3)$ 左右，在现代计算机上大约需要几分钟的时间。我不确定这个复杂度对不对（毕竟迭代复杂度是玄学），但是可以试一试。 53 | 54 | 代码实现稍微有点复杂，写完后如果复杂度对能跑出结果的，我再向您报告结果好了。 55 | 56 | \end{document} -------------------------------------------------------------------------------- /sheep/杨宗翰/sheep2-yzh.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/sheep/杨宗翰/sheep2-yzh.pdf -------------------------------------------------------------------------------- /sheep/杨宗翰/sheep2.tex: -------------------------------------------------------------------------------- 1 | \documentclass[11pt,a4paper,oneside]{article} 2 | 3 | \usepackage[utf8]{inputenc} 4 | \usepackage[english]{babel} 5 | \usepackage{olymp,clrscode3e} 6 | \usepackage{amsmath,graphicx,epigraph,bnf,enumerate,comment,listings,fontspec,color,indentfirst} 7 | \usepackage{subcaption} 8 | \usepackage{listings} 9 | \usepackage{ctex} 10 | \usepackage{CJK} 11 | 12 | \newcommand{\chr}[1]{\mbox{`\texttt{#1}'}} 13 | \newcommand{\txt}[1]{\mbox{``\texttt{#1}''}} 14 | 15 | \renewcommand{\contestname}{ 16 | Magical Sheep (2), MS107 Probability Homework, Yang Zonghan, \today 17 | } 18 | 19 | \lstset{ 20 | breaklines=true, %代码过长则换行 21 | numbers=left, %行号在左侧显示 22 | basicstyle = \fontspec{Consolas}, 23 | commentstyle=\color{gray}, %注释颜色 24 | frame=shadowbox%用方框框住代码块 25 | } 26 | \setCJKmainfont[BoldFont={黑体}, ItalicFont={华文楷体}, BoldItalicFont={黑体}]{华文中宋} 27 | \begin{document} 28 | \title{\textbf{Magical Sheep(2)}} 29 | \author{Yang Zonghan(518030910435)} 30 | \date{\today} 31 | \setlength{\parindent}{2em} 32 | 33 | 吴润哲哥哥说得非常对，我来补充几句。 34 | 35 | \section{$O(n)$ solving Markov Chain} 36 | 首先，$O(n ^ 3)$ 的高斯消元是可以优化的。本质上，消元解决的是 Markov 链问题，可以抽象为 $$ 37 | \begin{cases} 38 | x_1 = L \\ 39 | x_n = R \\ 40 | x_i = a_i x_{i- 1} + b_i x_{i + 1}, &i \in (1, n) 41 | \end{cases} 42 | $$ 而消元方程组的系数是形如下的，只有 $O(n)$ 个： 43 | $$\begin{bmatrix} 44 | 1 & 0 & 0 & 0 &\cdots & 0 & L\\ 45 | a_2 & -1 & b_2 & 0 & \cdots & 0 & 0 \\ 46 | 0 & a_3 & -1 & b_3 &\cdots 0& & 0 \\ 47 | \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 48 | 0 & 0 & \cdots & a_{n - 1} & -1 & b_{n - 1} & 0\\ 49 | 0 & 0 & \cdots & 0 & 0 & 1 & R 50 | \end{bmatrix}$$ 因此，这个矩阵消元复杂度可以减少为 $O(n ^ 2)$ 51 | 52 | 但是其实我们可以做到 $O(n)$。对于方程 $$ 53 | x_i = a_i x_{i- 1} + b_i x_{i + 1} 54 | $$ 可以变形为 $$ 55 | x_{i + 1} = \frac {x_i - a_i x_{i - 1}} {b_{i}} 56 | $$ 57 | 58 | 因此对于这样的 Markov 链我们可以用 $O(n)$ 的时间递推，用 $x_1, x_2$ 将所有项表达出来。算出 $x_n$ 的表达我们就能带入计算得到 $x_2$，进而得到所有值了。 59 | 60 | \section{About monotonicity} 61 | 62 | 如果我们承认了决策意义上的单调性，也即，承认了 $i + j = \text{const}$ 时，在 $i$ 由小到大时 $f_{i, j}$ 的决策存在一个分割点，前半段决策为取 $\max$，后半段决策为取 Markov 链更新，那么我们可以一层更新做到 $O(n)$，即用上述更新方法倒着维护值，直到取 $\max$ 更优为止。总复杂度 $O(n ^ 2)$。（当然其实不承认这种单调性也是能 $O(n ^ 2)$ 做的，随时维护随时取 $\max$ 即可，正确性应该不依赖于这个单调性） 63 | 64 | 但是这个单调性，直觉上很对，并且实践证明很对，但囿于我的水平不行，没法给出一个完整的解释。 65 | 66 | 尽管这个 $f_{i, j}$ 的本身的单调性非常优秀, 67 | \begin{gather*} 68 | f_{i, j} \leq f_{i + 1, j} \\ 69 | f_{i, j} \leq f_{i, j + 1} 70 | \end{gather*} 71 | \begin{center} 72 | \includegraphics[width=100mm]{sheepsheet.jpg} 73 | \end{center} 74 | 75 | 并且我们可以在 $\frac i {i + j}$ 充分小时，可以知道某个前缀必然是取 $\max$：对于上图的表格中的任意一列，即 $i = \text{const}$ 时，当 $p, q > 0$ 时，取 $i + j \rightarrow \infty$，有 76 | \begin{gather*} 77 | \frac {ip} {ip + jq} \rightarrow 0 \\ 78 | \frac {jq} {ip + jq} \rightarrow 1 79 | \end{gather*} 80 | 如果 $f_{i, j}$ 的决策为取 $\max$ 而 $f_{i + 1, j - 1}$ 取 Markov 链更新，那么 81 | $$ f_{i + 1, j - 1} \rightarrow f_{i, j} = f_{i, j - 1} \leq f_{i + 1, j - 2}$$ 82 | 83 | 这是一个矛盾，因此每一列充分高的位置都是取 $\max$ 的。同理可证充分高的每一列充分右的位置是取 Markov 链更新的。这两种情况对应白羊太多和黑羊太多的情况。 84 | 85 | 但我，我还是不会证明决策点对于每一行每一列唯一。 86 | 87 | 而且，根据吴润哲哥哥的结果，这样的分割点\emph{大概}就是 $ip - jq \leq 0$——和直觉分析相符的结果。但其中存在几个畸变点。这几个畸变点我想了很久，也没有办法给出解释。 88 | 89 | 也许只好寄希望于吴老师或其他助教、同学来解答这些疑惑了。或者我想，学习了本学期的内容或许我就能说出其中的道理了吧。 90 | \end{document} -------------------------------------------------------------------------------- /sheep/毛昕渝/The Mabinogion sheep problem-mxy.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/sheep/毛昕渝/The Mabinogion sheep problem-mxy.pdf -------------------------------------------------------------------------------- /sheep/毛昕渝/The Mabinogion sheep problem-mxy.tex: -------------------------------------------------------------------------------- 1 | 2 | \documentclass[11pt]{article} 3 | % This is a template for homework of Abstract Algebra. 4 | \usepackage[UTF8]{ctex} 5 | \usepackage{amssymb} 6 | \usepackage{amsmath} 7 | \usepackage{amsthm} 8 | \usepackage{geometry} 9 | \usepackage{mathrsfs} 10 | \usepackage{listings} 11 | \usepackage{graphics} 12 | \usepackage{booktabs} 13 | \usepackage{bm} 14 | \usepackage{tcolorbox} 15 | \usepackage{texlogos} 16 | \geometry{a4paper,left = 3cm, right = 3cm} 17 | 18 | \newcommand \cpp{C\raisebox{0.5ex}{\tiny\textbf{++}}} 19 | 20 | \title{问题讨论\\魔法羊(The Mabinogion sheep problem)} 21 | \author{毛昕渝} 22 | 23 | \date{\today} 24 | \begin{document} 25 | \maketitle 26 | \begin{tcolorbox} 27 | \textbf{问题描述} 28 | 29 | 草原上有$N$只黑羊和$N$只白羊，它们不会死亡和新生。每天都有一只羊被随机选中，该随机选择与以前的选择无关。 这只被选中的羊会找到一只与之异色的羊（如果还有的话）并对其鸣叫。一只黑羊对准一只白羊鸣叫会使白羊以概率$p$发生变色（变为黑羊）；一只白羊对准一只黑羊鸣叫会使黑羊以概率$q$发生变色（变为白羊）。管理者每天可以在神奇的羊变色现象发生前进场移走任意数量的白羊。管理者的目标是使得最后所剩黑羊数量的期望值尽可能大。试针对一些不同的$p,q$ 取值来给管理者建议管理策略以及对最好可能的黑羊遗留数量做出估计。 30 | \end{tcolorbox} 31 | 32 | 首先$p = 0$或$q = 0$的情形, 最优策略是显然的: 33 | \begin{itemize} 34 | \item $p = 0$. 移走所有白羊, 剩下$N$只黑羊. 35 | \item $q = 0$. 什么也不做,最后得到$2N$只黑羊. 36 | \end{itemize} 37 | 38 | 以下讨论$p \neq 0, q \neq 0$的情形. 我们从一个简单问题开始: 如果某一天有$w$ 只白羊和$b$只黑羊, 魔法变色发生后黑羊的期望数量是多少? 记黑羊的期望数量为$E$, 容易得出, 39 | \begin{equation} 40 | E = b + \frac{b}{w+b} p - \frac{w}{w + b}q. 41 | \end{equation} 42 | 根据规则, 我们可以任意减少白羊的数量$w$; 从(1)可以看出, 如果我们想要黑羊的数量增加, 就需要 43 | \begin{equation} 44 | E > b \iff w < \frac{p}{q}b. 45 | \end{equation} 46 | 于是我们可以得到如下的策略: 47 | \vspace{0.5cm} 48 | \begin{tcolorbox} 49 | \textbf{策略A} 50 | 51 | 为了让最后的黑羊尽可能多, 我们在(2)的条件下让$w$尽可能大. 换言之, 在变色发生前将白羊的数量变为$min(w,\left \lceil \frac{p}{q}b\right \rceil - 1)$. 52 | \end{tcolorbox} 53 | 这个策略可以看作是\textbf{局部的贪心}, 并不一定就是问题的最优解. 幸运的是, 对于$p = q = 1$ 的情况, 策略A 就是最优解. 《概率与鞅》一书中给出了证明, 并且指出在这一策略下, 最终黑羊的期望数量满足 54 | \begin{equation} 55 | V(N,N,1,1) \to 2N + \frac{\pi}{4} - \sqrt{\pi N}, \text{ as } N \to \infty, 56 | \end{equation} 57 | 其中$V(N,M,p,q)$表示采用策略A, 从$N$只黑羊$M$ 只白羊的局面开始, 最终得到黑羊的数量; $p$是白羊变色的概率, $q$是黑羊变色的概率. 例如, $N = 10000$时, 最终大约得到19824只黑羊. 58 | 59 | 对于一般的情形, 假设$p < q < 1$, 我们只需要考虑一个新的问题: 白羊变色的概率为$p' := \frac{p}{q}$, 黑羊变色的概率为$q' := 1$, 但\textbf{每天变色事件发生的概率仅为}$q$. 在这个新的问题中(1)式仍然成立, 而且\textbf{变色以一定概率发生不会对策略造成影响}, 所以这两个问题应该有相同的答案(最终黑羊的数量). 换言之, 60 | \begin{equation} 61 | V(N,M,p,q) = V(N,M,\frac{p}{q},1). 62 | \end{equation} 63 | 因此, 我们只研究$q = 1$的情况就足够了. 64 | 类似地, 如果$q < p < 1$, 研究$p = 1$的情形也足够了. 65 | 66 | 对$p < q$的过程，我写了一个\cpluspluslogo 程序模拟这个过程, 结果符合(3) 和(4) 的预测. 此外, 对$p$的不同取值进行模拟并用MATLAB分析发现, $V(N,N,p,1)$ 与$p$ 呈\textbf{线性关系}. 显然, $V(N,N,0,1) = N$, 再结合(3) 的结果, 可以得出 67 | \begin{equation} 68 | V(N,N,p,1) \to p(N + \frac{\pi}{4} - \sqrt{\pi N}) + N, \text{ as } N \to \infty. 69 | \end{equation} 70 | 但是我不会证明这个线性关系. 71 | \vspace{0.5cm} 72 | \begin{tcolorbox} 73 | \textbf{结论} 74 | 75 | 记$OP(N,N,p,q)$表示最优策略下最终黑羊的数量,(4)和(5)给出了一个下界, 即 76 | \begin{equation} 77 | OP(N,N,p,q) \geq \frac{p}{q}(N + \frac{\pi}{4} - \sqrt{\pi N}) + N, \text{ as } N \to \infty. 78 | \end{equation} 79 | 特别地, 当$p = q$时, (6)中的等号成立. 80 | 这表明, 当$p,q$比较接近时, 策略A是一个优秀的策略. (5)给出了使用A策略的结果估计, 但没有严格证明. 81 | \end{tcolorbox} 82 | \end{document} 83 | -------------------------------------------------------------------------------- /sheep/董海辰/sheep-董海辰.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/sheep/董海辰/sheep-董海辰.pdf -------------------------------------------------------------------------------- /sheep/钱程阳/数据.xlsx: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SJTU-ACM-Class-2018/lecture-notes-of-probability/7bc588fe311e008e9cb8e3dc8a0df1b2784e9bc0/sheep/钱程阳/数据.xlsx --------------------------------------------------------------------------------