├── README.md ├── .DS_Store ├── ex1 ├── main.pdf ├── .gitignore └── main.tex ├── ex2 ├── main.pdf ├── .gitignore └── main.tex ├── ex3 ├── main.pdf ├── .gitignore └── main.tex ├── ex4 ├── main.pdf ├── .gitignore └── main.tex ├── ex5 ├── main.pdf ├── .gitignore └── main.tex └── ex6 ├── main.pdf ├── .gitignore └── main.tex /README.md: -------------------------------------------------------------------------------- 1 | # 概率论与数理统计习题课课件 -------------------------------------------------------------------------------- /.DS_Store: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/luoyt14/ProbabilityExercise/HEAD/.DS_Store -------------------------------------------------------------------------------- /ex1/main.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/luoyt14/ProbabilityExercise/HEAD/ex1/main.pdf -------------------------------------------------------------------------------- /ex2/main.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/luoyt14/ProbabilityExercise/HEAD/ex2/main.pdf -------------------------------------------------------------------------------- /ex3/main.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/luoyt14/ProbabilityExercise/HEAD/ex3/main.pdf -------------------------------------------------------------------------------- /ex4/main.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/luoyt14/ProbabilityExercise/HEAD/ex4/main.pdf -------------------------------------------------------------------------------- /ex5/main.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/luoyt14/ProbabilityExercise/HEAD/ex5/main.pdf -------------------------------------------------------------------------------- /ex6/main.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/luoyt14/ProbabilityExercise/HEAD/ex6/main.pdf -------------------------------------------------------------------------------- /ex1/.gitignore: -------------------------------------------------------------------------------- 1 | ## Core latex/pdflatex auxiliary files: 2 | *.aux 3 | *.lof 4 | *.log 5 | *.lot 6 | *.fls 7 | *.out 8 | *.toc 9 | *.fmt 10 | *.fot 11 | *.cb 12 | *.cb2 13 | .*.lb 14 | 15 | ## Intermediate documents: 16 | *.dvi 17 | *.xdv 18 | *-converted-to.* 19 | # these rules might exclude image files for figures etc. 20 | # *.ps 21 | # *.eps 22 | # *.pdf 23 | 24 | ## Generated if empty string is given at "Please type another file name for output:" 25 | .pdf 26 | 27 | ## Bibliography auxiliary files (bibtex/biblatex/biber): 28 | *.bbl 29 | *.bcf 30 | *.blg 31 | *-blx.aux 32 | *-blx.bib 33 | *.run.xml 34 | 35 | ## Build tool auxiliary files: 36 | *.fdb_latexmk 37 | *.synctex 38 | *.synctex(busy) 39 | *.synctex.gz 40 | *.synctex.gz(busy) 41 | *.pdfsync 42 | 43 | ## Build tool directories for auxiliary files 44 | # latexrun 45 | latex.out/ 46 | 47 | ## Auxiliary and intermediate files from other packages: 48 | # algorithms 49 | *.alg 50 | *.loa 51 | 52 | # achemso 53 | acs-*.bib 54 | 55 | # amsthm 56 | *.thm 57 | 58 | # beamer 59 | *.nav 60 | *.pre 61 | *.snm 62 | *.vrb 63 | 64 | # changes 65 | *.soc 66 | 67 | # comment 68 | *.cut 69 | 70 | # cprotect 71 | *.cpt 72 | 73 | # elsarticle (documentclass of Elsevier journals) 74 | *.spl 75 | 76 | # endnotes 77 | *.ent 78 | 79 | # fixme 80 | *.lox 81 | 82 | # feynmf/feynmp 83 | *.mf 84 | *.mp 85 | *.t[1-9] 86 | *.t[1-9][0-9] 87 | *.tfm 88 | 89 | #(r)(e)ledmac/(r)(e)ledpar 90 | *.end 91 | *.?end 92 | *.[1-9] 93 | *.[1-9][0-9] 94 | *.[1-9][0-9][0-9] 95 | *.[1-9]R 96 | *.[1-9][0-9]R 97 | *.[1-9][0-9][0-9]R 98 | *.eledsec[1-9] 99 | *.eledsec[1-9]R 100 | *.eledsec[1-9][0-9] 101 | *.eledsec[1-9][0-9]R 102 | *.eledsec[1-9][0-9][0-9] 103 | *.eledsec[1-9][0-9][0-9]R 104 | 105 | # glossaries 106 | *.acn 107 | *.acr 108 | *.glg 109 | *.glo 110 | *.gls 111 | *.glsdefs 112 | 113 | # gnuplottex 114 | *-gnuplottex-* 115 | 116 | # gregoriotex 117 | *.gaux 118 | *.gtex 119 | 120 | # htlatex 121 | *.4ct 122 | *.4tc 123 | *.idv 124 | *.lg 125 | *.trc 126 | *.xref 127 | 128 | # hyperref 129 | *.brf 130 | 131 | # knitr 132 | *-concordance.tex 133 | # TODO Comment the next line if you want to keep your tikz graphics files 134 | *.tikz 135 | *-tikzDictionary 136 | 137 | # listings 138 | *.lol 139 | 140 | # makeidx 141 | *.idx 142 | *.ilg 143 | *.ind 144 | *.ist 145 | 146 | # minitoc 147 | *.maf 148 | *.mlf 149 | *.mlt 150 | *.mtc[0-9]* 151 | *.slf[0-9]* 152 | *.slt[0-9]* 153 | *.stc[0-9]* 154 | 155 | # minted 156 | _minted* 157 | *.pyg 158 | 159 | # morewrites 160 | *.mw 161 | 162 | # nomencl 163 | *.nlg 164 | *.nlo 165 | *.nls 166 | 167 | # pax 168 | *.pax 169 | 170 | # pdfpcnotes 171 | *.pdfpc 172 | 173 | # sagetex 174 | *.sagetex.sage 175 | *.sagetex.py 176 | *.sagetex.scmd 177 | 178 | # scrwfile 179 | *.wrt 180 | 181 | # sympy 182 | *.sout 183 | *.sympy 184 | sympy-plots-for-*.tex/ 185 | 186 | # pdfcomment 187 | *.upa 188 | *.upb 189 | 190 | # pythontex 191 | *.pytxcode 192 | pythontex-files-*/ 193 | 194 | # tcolorbox 195 | *.listing 196 | 197 | # thmtools 198 | *.loe 199 | 200 | # TikZ & PGF 201 | *.dpth 202 | *.md5 203 | *.auxlock 204 | 205 | # todonotes 206 | *.tdo 207 | 208 | # vhistory 209 | *.hst 210 | *.ver 211 | 212 | # easy-todo 213 | *.lod 214 | 215 | # xcolor 216 | *.xcp 217 | 218 | # xmpincl 219 | *.xmpi 220 | 221 | # xindy 222 | *.xdy 223 | 224 | # xypic precompiled matrices 225 | *.xyc 226 | 227 | # endfloat 228 | *.ttt 229 | *.fff 230 | 231 | # Latexian 232 | TSWLatexianTemp* 233 | 234 | ## Editors: 235 | # WinEdt 236 | *.bak 237 | *.sav 238 | 239 | # Texpad 240 | .texpadtmp 241 | 242 | # LyX 243 | *.lyx~ 244 | 245 | # Kile 246 | *.backup 247 | 248 | # KBibTeX 249 | *~[0-9]* 250 | 251 | # auto folder when using emacs and auctex 252 | ./auto/* 253 | *.el 254 | 255 | # expex forward references with \gathertags 256 | *-tags.tex 257 | 258 | # standalone packages 259 | *.sta -------------------------------------------------------------------------------- /ex2/.gitignore: -------------------------------------------------------------------------------- 1 | ## Core latex/pdflatex auxiliary files: 2 | *.aux 3 | *.lof 4 | *.log 5 | *.lot 6 | *.fls 7 | *.out 8 | *.toc 9 | *.fmt 10 | *.fot 11 | *.cb 12 | *.cb2 13 | .*.lb 14 | 15 | ## Intermediate documents: 16 | *.dvi 17 | *.xdv 18 | *-converted-to.* 19 | # these rules might exclude image files for figures etc. 20 | # *.ps 21 | # *.eps 22 | # *.pdf 23 | 24 | ## Generated if empty string is given at "Please type another file name for output:" 25 | .pdf 26 | 27 | ## Bibliography auxiliary files (bibtex/biblatex/biber): 28 | *.bbl 29 | *.bcf 30 | *.blg 31 | *-blx.aux 32 | *-blx.bib 33 | *.run.xml 34 | 35 | ## Build tool auxiliary files: 36 | *.fdb_latexmk 37 | *.synctex 38 | *.synctex(busy) 39 | *.synctex.gz 40 | *.synctex.gz(busy) 41 | *.pdfsync 42 | 43 | ## Build tool directories for auxiliary files 44 | # latexrun 45 | latex.out/ 46 | 47 | ## Auxiliary and intermediate files from other packages: 48 | # algorithms 49 | *.alg 50 | *.loa 51 | 52 | # achemso 53 | acs-*.bib 54 | 55 | # amsthm 56 | *.thm 57 | 58 | # beamer 59 | *.nav 60 | *.pre 61 | *.snm 62 | *.vrb 63 | 64 | # changes 65 | *.soc 66 | 67 | # comment 68 | *.cut 69 | 70 | # cprotect 71 | *.cpt 72 | 73 | # elsarticle (documentclass of Elsevier journals) 74 | *.spl 75 | 76 | # endnotes 77 | *.ent 78 | 79 | # fixme 80 | *.lox 81 | 82 | # feynmf/feynmp 83 | *.mf 84 | *.mp 85 | *.t[1-9] 86 | *.t[1-9][0-9] 87 | *.tfm 88 | 89 | #(r)(e)ledmac/(r)(e)ledpar 90 | *.end 91 | *.?end 92 | *.[1-9] 93 | *.[1-9][0-9] 94 | *.[1-9][0-9][0-9] 95 | *.[1-9]R 96 | *.[1-9][0-9]R 97 | *.[1-9][0-9][0-9]R 98 | *.eledsec[1-9] 99 | *.eledsec[1-9]R 100 | *.eledsec[1-9][0-9] 101 | *.eledsec[1-9][0-9]R 102 | *.eledsec[1-9][0-9][0-9] 103 | *.eledsec[1-9][0-9][0-9]R 104 | 105 | # glossaries 106 | *.acn 107 | *.acr 108 | *.glg 109 | *.glo 110 | *.gls 111 | *.glsdefs 112 | 113 | # gnuplottex 114 | *-gnuplottex-* 115 | 116 | # gregoriotex 117 | *.gaux 118 | *.gtex 119 | 120 | # htlatex 121 | *.4ct 122 | *.4tc 123 | *.idv 124 | *.lg 125 | *.trc 126 | *.xref 127 | 128 | # hyperref 129 | *.brf 130 | 131 | # knitr 132 | *-concordance.tex 133 | # TODO Comment the next line if you want to keep your tikz graphics files 134 | *.tikz 135 | *-tikzDictionary 136 | 137 | # listings 138 | *.lol 139 | 140 | # makeidx 141 | *.idx 142 | *.ilg 143 | *.ind 144 | *.ist 145 | 146 | # minitoc 147 | *.maf 148 | *.mlf 149 | *.mlt 150 | *.mtc[0-9]* 151 | *.slf[0-9]* 152 | *.slt[0-9]* 153 | *.stc[0-9]* 154 | 155 | # minted 156 | _minted* 157 | *.pyg 158 | 159 | # morewrites 160 | *.mw 161 | 162 | # nomencl 163 | *.nlg 164 | *.nlo 165 | *.nls 166 | 167 | # pax 168 | *.pax 169 | 170 | # pdfpcnotes 171 | *.pdfpc 172 | 173 | # sagetex 174 | *.sagetex.sage 175 | *.sagetex.py 176 | *.sagetex.scmd 177 | 178 | # scrwfile 179 | *.wrt 180 | 181 | # sympy 182 | *.sout 183 | *.sympy 184 | sympy-plots-for-*.tex/ 185 | 186 | # pdfcomment 187 | *.upa 188 | *.upb 189 | 190 | # pythontex 191 | *.pytxcode 192 | pythontex-files-*/ 193 | 194 | # tcolorbox 195 | *.listing 196 | 197 | # thmtools 198 | *.loe 199 | 200 | # TikZ & PGF 201 | *.dpth 202 | *.md5 203 | *.auxlock 204 | 205 | # todonotes 206 | *.tdo 207 | 208 | # vhistory 209 | *.hst 210 | *.ver 211 | 212 | # easy-todo 213 | *.lod 214 | 215 | # xcolor 216 | *.xcp 217 | 218 | # xmpincl 219 | *.xmpi 220 | 221 | # xindy 222 | *.xdy 223 | 224 | # xypic precompiled matrices 225 | *.xyc 226 | 227 | # endfloat 228 | *.ttt 229 | *.fff 230 | 231 | # Latexian 232 | TSWLatexianTemp* 233 | 234 | ## Editors: 235 | # WinEdt 236 | *.bak 237 | *.sav 238 | 239 | # Texpad 240 | .texpadtmp 241 | 242 | # LyX 243 | *.lyx~ 244 | 245 | # Kile 246 | *.backup 247 | 248 | # KBibTeX 249 | *~[0-9]* 250 | 251 | # auto folder when using emacs and auctex 252 | ./auto/* 253 | *.el 254 | 255 | # expex forward references with \gathertags 256 | *-tags.tex 257 | 258 | # standalone packages 259 | *.sta -------------------------------------------------------------------------------- /ex3/.gitignore: -------------------------------------------------------------------------------- 1 | ## Core latex/pdflatex auxiliary files: 2 | *.aux 3 | *.lof 4 | *.log 5 | *.lot 6 | *.fls 7 | *.out 8 | *.toc 9 | *.fmt 10 | *.fot 11 | *.cb 12 | *.cb2 13 | .*.lb 14 | 15 | ## Intermediate documents: 16 | *.dvi 17 | *.xdv 18 | *-converted-to.* 19 | # these rules might exclude image files for figures etc. 20 | # *.ps 21 | # *.eps 22 | # *.pdf 23 | 24 | ## Generated if empty string is given at "Please type another file name for output:" 25 | .pdf 26 | 27 | ## Bibliography auxiliary files (bibtex/biblatex/biber): 28 | *.bbl 29 | *.bcf 30 | *.blg 31 | *-blx.aux 32 | *-blx.bib 33 | *.run.xml 34 | 35 | ## Build tool auxiliary files: 36 | *.fdb_latexmk 37 | *.synctex 38 | *.synctex(busy) 39 | *.synctex.gz 40 | *.synctex.gz(busy) 41 | *.pdfsync 42 | 43 | ## Build tool directories for auxiliary files 44 | # latexrun 45 | latex.out/ 46 | 47 | ## Auxiliary and intermediate files from other packages: 48 | # algorithms 49 | *.alg 50 | *.loa 51 | 52 | # achemso 53 | acs-*.bib 54 | 55 | # amsthm 56 | *.thm 57 | 58 | # beamer 59 | *.nav 60 | *.pre 61 | *.snm 62 | *.vrb 63 | 64 | # changes 65 | *.soc 66 | 67 | # comment 68 | *.cut 69 | 70 | # cprotect 71 | *.cpt 72 | 73 | # elsarticle (documentclass of Elsevier journals) 74 | *.spl 75 | 76 | # endnotes 77 | *.ent 78 | 79 | # fixme 80 | *.lox 81 | 82 | # feynmf/feynmp 83 | *.mf 84 | *.mp 85 | *.t[1-9] 86 | *.t[1-9][0-9] 87 | *.tfm 88 | 89 | #(r)(e)ledmac/(r)(e)ledpar 90 | *.end 91 | *.?end 92 | *.[1-9] 93 | *.[1-9][0-9] 94 | *.[1-9][0-9][0-9] 95 | *.[1-9]R 96 | *.[1-9][0-9]R 97 | *.[1-9][0-9][0-9]R 98 | *.eledsec[1-9] 99 | *.eledsec[1-9]R 100 | *.eledsec[1-9][0-9] 101 | *.eledsec[1-9][0-9]R 102 | *.eledsec[1-9][0-9][0-9] 103 | *.eledsec[1-9][0-9][0-9]R 104 | 105 | # glossaries 106 | *.acn 107 | *.acr 108 | *.glg 109 | *.glo 110 | *.gls 111 | *.glsdefs 112 | 113 | # gnuplottex 114 | *-gnuplottex-* 115 | 116 | # gregoriotex 117 | *.gaux 118 | *.gtex 119 | 120 | # htlatex 121 | *.4ct 122 | *.4tc 123 | *.idv 124 | *.lg 125 | *.trc 126 | *.xref 127 | 128 | # hyperref 129 | *.brf 130 | 131 | # knitr 132 | *-concordance.tex 133 | # TODO Comment the next line if you want to keep your tikz graphics files 134 | *.tikz 135 | *-tikzDictionary 136 | 137 | # listings 138 | *.lol 139 | 140 | # makeidx 141 | *.idx 142 | *.ilg 143 | *.ind 144 | *.ist 145 | 146 | # minitoc 147 | *.maf 148 | *.mlf 149 | *.mlt 150 | *.mtc[0-9]* 151 | *.slf[0-9]* 152 | *.slt[0-9]* 153 | *.stc[0-9]* 154 | 155 | # minted 156 | _minted* 157 | *.pyg 158 | 159 | # morewrites 160 | *.mw 161 | 162 | # nomencl 163 | *.nlg 164 | *.nlo 165 | *.nls 166 | 167 | # pax 168 | *.pax 169 | 170 | # pdfpcnotes 171 | *.pdfpc 172 | 173 | # sagetex 174 | *.sagetex.sage 175 | *.sagetex.py 176 | *.sagetex.scmd 177 | 178 | # scrwfile 179 | *.wrt 180 | 181 | # sympy 182 | *.sout 183 | *.sympy 184 | sympy-plots-for-*.tex/ 185 | 186 | # pdfcomment 187 | *.upa 188 | *.upb 189 | 190 | # pythontex 191 | *.pytxcode 192 | pythontex-files-*/ 193 | 194 | # tcolorbox 195 | *.listing 196 | 197 | # thmtools 198 | *.loe 199 | 200 | # TikZ & PGF 201 | *.dpth 202 | *.md5 203 | *.auxlock 204 | 205 | # todonotes 206 | *.tdo 207 | 208 | # vhistory 209 | *.hst 210 | *.ver 211 | 212 | # easy-todo 213 | *.lod 214 | 215 | # xcolor 216 | *.xcp 217 | 218 | # xmpincl 219 | *.xmpi 220 | 221 | # xindy 222 | *.xdy 223 | 224 | # xypic precompiled matrices 225 | *.xyc 226 | 227 | # endfloat 228 | *.ttt 229 | *.fff 230 | 231 | # Latexian 232 | TSWLatexianTemp* 233 | 234 | ## Editors: 235 | # WinEdt 236 | *.bak 237 | *.sav 238 | 239 | # Texpad 240 | .texpadtmp 241 | 242 | # LyX 243 | *.lyx~ 244 | 245 | # Kile 246 | *.backup 247 | 248 | # KBibTeX 249 | *~[0-9]* 250 | 251 | # auto folder when using emacs and auctex 252 | ./auto/* 253 | *.el 254 | 255 | # expex forward references with \gathertags 256 | *-tags.tex 257 | 258 | # standalone packages 259 | *.sta -------------------------------------------------------------------------------- /ex4/.gitignore: -------------------------------------------------------------------------------- 1 | ## Core latex/pdflatex auxiliary files: 2 | *.aux 3 | *.lof 4 | *.log 5 | *.lot 6 | *.fls 7 | *.out 8 | *.toc 9 | *.fmt 10 | *.fot 11 | *.cb 12 | *.cb2 13 | .*.lb 14 | 15 | ## Intermediate documents: 16 | *.dvi 17 | *.xdv 18 | *-converted-to.* 19 | # these rules might exclude image files for figures etc. 20 | # *.ps 21 | # *.eps 22 | # *.pdf 23 | 24 | ## Generated if empty string is given at "Please type another file name for output:" 25 | .pdf 26 | 27 | ## Bibliography auxiliary files (bibtex/biblatex/biber): 28 | *.bbl 29 | *.bcf 30 | *.blg 31 | *-blx.aux 32 | *-blx.bib 33 | *.run.xml 34 | 35 | ## Build tool auxiliary files: 36 | *.fdb_latexmk 37 | *.synctex 38 | *.synctex(busy) 39 | *.synctex.gz 40 | *.synctex.gz(busy) 41 | *.pdfsync 42 | 43 | ## Build tool directories for auxiliary files 44 | # latexrun 45 | latex.out/ 46 | 47 | ## Auxiliary and intermediate files from other packages: 48 | # algorithms 49 | *.alg 50 | *.loa 51 | 52 | # achemso 53 | acs-*.bib 54 | 55 | # amsthm 56 | *.thm 57 | 58 | # beamer 59 | *.nav 60 | *.pre 61 | *.snm 62 | *.vrb 63 | 64 | # changes 65 | *.soc 66 | 67 | # comment 68 | *.cut 69 | 70 | # cprotect 71 | *.cpt 72 | 73 | # elsarticle (documentclass of Elsevier journals) 74 | *.spl 75 | 76 | # endnotes 77 | *.ent 78 | 79 | # fixme 80 | *.lox 81 | 82 | # feynmf/feynmp 83 | *.mf 84 | *.mp 85 | *.t[1-9] 86 | *.t[1-9][0-9] 87 | *.tfm 88 | 89 | #(r)(e)ledmac/(r)(e)ledpar 90 | *.end 91 | *.?end 92 | *.[1-9] 93 | *.[1-9][0-9] 94 | *.[1-9][0-9][0-9] 95 | *.[1-9]R 96 | *.[1-9][0-9]R 97 | *.[1-9][0-9][0-9]R 98 | *.eledsec[1-9] 99 | *.eledsec[1-9]R 100 | *.eledsec[1-9][0-9] 101 | *.eledsec[1-9][0-9]R 102 | *.eledsec[1-9][0-9][0-9] 103 | *.eledsec[1-9][0-9][0-9]R 104 | 105 | # glossaries 106 | *.acn 107 | *.acr 108 | *.glg 109 | *.glo 110 | *.gls 111 | *.glsdefs 112 | 113 | # gnuplottex 114 | *-gnuplottex-* 115 | 116 | # gregoriotex 117 | *.gaux 118 | *.gtex 119 | 120 | # htlatex 121 | *.4ct 122 | *.4tc 123 | *.idv 124 | *.lg 125 | *.trc 126 | *.xref 127 | 128 | # hyperref 129 | *.brf 130 | 131 | # knitr 132 | *-concordance.tex 133 | # TODO Comment the next line if you want to keep your tikz graphics files 134 | *.tikz 135 | *-tikzDictionary 136 | 137 | # listings 138 | *.lol 139 | 140 | # makeidx 141 | *.idx 142 | *.ilg 143 | *.ind 144 | *.ist 145 | 146 | # minitoc 147 | *.maf 148 | *.mlf 149 | *.mlt 150 | *.mtc[0-9]* 151 | *.slf[0-9]* 152 | *.slt[0-9]* 153 | *.stc[0-9]* 154 | 155 | # minted 156 | _minted* 157 | *.pyg 158 | 159 | # morewrites 160 | *.mw 161 | 162 | # nomencl 163 | *.nlg 164 | *.nlo 165 | *.nls 166 | 167 | # pax 168 | *.pax 169 | 170 | # pdfpcnotes 171 | *.pdfpc 172 | 173 | # sagetex 174 | *.sagetex.sage 175 | *.sagetex.py 176 | *.sagetex.scmd 177 | 178 | # scrwfile 179 | *.wrt 180 | 181 | # sympy 182 | *.sout 183 | *.sympy 184 | sympy-plots-for-*.tex/ 185 | 186 | # pdfcomment 187 | *.upa 188 | *.upb 189 | 190 | # pythontex 191 | *.pytxcode 192 | pythontex-files-*/ 193 | 194 | # tcolorbox 195 | *.listing 196 | 197 | # thmtools 198 | *.loe 199 | 200 | # TikZ & PGF 201 | *.dpth 202 | *.md5 203 | *.auxlock 204 | 205 | # todonotes 206 | *.tdo 207 | 208 | # vhistory 209 | *.hst 210 | *.ver 211 | 212 | # easy-todo 213 | *.lod 214 | 215 | # xcolor 216 | *.xcp 217 | 218 | # xmpincl 219 | *.xmpi 220 | 221 | # xindy 222 | *.xdy 223 | 224 | # xypic precompiled matrices 225 | *.xyc 226 | 227 | # endfloat 228 | *.ttt 229 | *.fff 230 | 231 | # Latexian 232 | TSWLatexianTemp* 233 | 234 | ## Editors: 235 | # WinEdt 236 | *.bak 237 | *.sav 238 | 239 | # Texpad 240 | .texpadtmp 241 | 242 | # LyX 243 | *.lyx~ 244 | 245 | # Kile 246 | *.backup 247 | 248 | # KBibTeX 249 | *~[0-9]* 250 | 251 | # auto folder when using emacs and auctex 252 | ./auto/* 253 | *.el 254 | 255 | # expex forward references with \gathertags 256 | *-tags.tex 257 | 258 | # standalone packages 259 | *.sta -------------------------------------------------------------------------------- /ex5/.gitignore: -------------------------------------------------------------------------------- 1 | ## Core latex/pdflatex auxiliary files: 2 | *.aux 3 | *.lof 4 | *.log 5 | *.lot 6 | *.fls 7 | *.out 8 | *.toc 9 | *.fmt 10 | *.fot 11 | *.cb 12 | *.cb2 13 | .*.lb 14 | 15 | ## Intermediate documents: 16 | *.dvi 17 | *.xdv 18 | *-converted-to.* 19 | # these rules might exclude image files for figures etc. 20 | # *.ps 21 | # *.eps 22 | # *.pdf 23 | 24 | ## Generated if empty string is given at "Please type another file name for output:" 25 | .pdf 26 | 27 | ## Bibliography auxiliary files (bibtex/biblatex/biber): 28 | *.bbl 29 | *.bcf 30 | *.blg 31 | *-blx.aux 32 | *-blx.bib 33 | *.run.xml 34 | 35 | ## Build tool auxiliary files: 36 | *.fdb_latexmk 37 | *.synctex 38 | *.synctex(busy) 39 | *.synctex.gz 40 | *.synctex.gz(busy) 41 | *.pdfsync 42 | 43 | ## Build tool directories for auxiliary files 44 | # latexrun 45 | latex.out/ 46 | 47 | ## Auxiliary and intermediate files from other packages: 48 | # algorithms 49 | *.alg 50 | *.loa 51 | 52 | # achemso 53 | acs-*.bib 54 | 55 | # amsthm 56 | *.thm 57 | 58 | # beamer 59 | *.nav 60 | *.pre 61 | *.snm 62 | *.vrb 63 | 64 | # changes 65 | *.soc 66 | 67 | # comment 68 | *.cut 69 | 70 | # cprotect 71 | *.cpt 72 | 73 | # elsarticle (documentclass of Elsevier journals) 74 | *.spl 75 | 76 | # endnotes 77 | *.ent 78 | 79 | # fixme 80 | *.lox 81 | 82 | # feynmf/feynmp 83 | *.mf 84 | *.mp 85 | *.t[1-9] 86 | *.t[1-9][0-9] 87 | *.tfm 88 | 89 | #(r)(e)ledmac/(r)(e)ledpar 90 | *.end 91 | *.?end 92 | *.[1-9] 93 | *.[1-9][0-9] 94 | *.[1-9][0-9][0-9] 95 | *.[1-9]R 96 | *.[1-9][0-9]R 97 | *.[1-9][0-9][0-9]R 98 | *.eledsec[1-9] 99 | *.eledsec[1-9]R 100 | *.eledsec[1-9][0-9] 101 | *.eledsec[1-9][0-9]R 102 | *.eledsec[1-9][0-9][0-9] 103 | *.eledsec[1-9][0-9][0-9]R 104 | 105 | # glossaries 106 | *.acn 107 | *.acr 108 | *.glg 109 | *.glo 110 | *.gls 111 | *.glsdefs 112 | 113 | # gnuplottex 114 | *-gnuplottex-* 115 | 116 | # gregoriotex 117 | *.gaux 118 | *.gtex 119 | 120 | # htlatex 121 | *.4ct 122 | *.4tc 123 | *.idv 124 | *.lg 125 | *.trc 126 | *.xref 127 | 128 | # hyperref 129 | *.brf 130 | 131 | # knitr 132 | *-concordance.tex 133 | # TODO Comment the next line if you want to keep your tikz graphics files 134 | *.tikz 135 | *-tikzDictionary 136 | 137 | # listings 138 | *.lol 139 | 140 | # makeidx 141 | *.idx 142 | *.ilg 143 | *.ind 144 | *.ist 145 | 146 | # minitoc 147 | *.maf 148 | *.mlf 149 | *.mlt 150 | *.mtc[0-9]* 151 | *.slf[0-9]* 152 | *.slt[0-9]* 153 | *.stc[0-9]* 154 | 155 | # minted 156 | _minted* 157 | *.pyg 158 | 159 | # morewrites 160 | *.mw 161 | 162 | # nomencl 163 | *.nlg 164 | *.nlo 165 | *.nls 166 | 167 | # pax 168 | *.pax 169 | 170 | # pdfpcnotes 171 | *.pdfpc 172 | 173 | # sagetex 174 | *.sagetex.sage 175 | *.sagetex.py 176 | *.sagetex.scmd 177 | 178 | # scrwfile 179 | *.wrt 180 | 181 | # sympy 182 | *.sout 183 | *.sympy 184 | sympy-plots-for-*.tex/ 185 | 186 | # pdfcomment 187 | *.upa 188 | *.upb 189 | 190 | # pythontex 191 | *.pytxcode 192 | pythontex-files-*/ 193 | 194 | # tcolorbox 195 | *.listing 196 | 197 | # thmtools 198 | *.loe 199 | 200 | # TikZ & PGF 201 | *.dpth 202 | *.md5 203 | *.auxlock 204 | 205 | # todonotes 206 | *.tdo 207 | 208 | # vhistory 209 | *.hst 210 | *.ver 211 | 212 | # easy-todo 213 | *.lod 214 | 215 | # xcolor 216 | *.xcp 217 | 218 | # xmpincl 219 | *.xmpi 220 | 221 | # xindy 222 | *.xdy 223 | 224 | # xypic precompiled matrices 225 | *.xyc 226 | 227 | # endfloat 228 | *.ttt 229 | *.fff 230 | 231 | # Latexian 232 | TSWLatexianTemp* 233 | 234 | ## Editors: 235 | # WinEdt 236 | *.bak 237 | *.sav 238 | 239 | # Texpad 240 | .texpadtmp 241 | 242 | # LyX 243 | *.lyx~ 244 | 245 | # Kile 246 | *.backup 247 | 248 | # KBibTeX 249 | *~[0-9]* 250 | 251 | # auto folder when using emacs and auctex 252 | ./auto/* 253 | *.el 254 | 255 | # expex forward references with \gathertags 256 | *-tags.tex 257 | 258 | # standalone packages 259 | *.sta -------------------------------------------------------------------------------- /ex6/.gitignore: -------------------------------------------------------------------------------- 1 | ## Core latex/pdflatex auxiliary files: 2 | *.aux 3 | *.lof 4 | *.log 5 | *.lot 6 | *.fls 7 | *.out 8 | *.toc 9 | *.fmt 10 | *.fot 11 | *.cb 12 | *.cb2 13 | .*.lb 14 | 15 | ## Intermediate documents: 16 | *.dvi 17 | *.xdv 18 | *-converted-to.* 19 | # these rules might exclude image files for figures etc. 20 | # *.ps 21 | # *.eps 22 | # *.pdf 23 | 24 | ## Generated if empty string is given at "Please type another file name for output:" 25 | .pdf 26 | 27 | ## Bibliography auxiliary files (bibtex/biblatex/biber): 28 | *.bbl 29 | *.bcf 30 | *.blg 31 | *-blx.aux 32 | *-blx.bib 33 | *.run.xml 34 | 35 | ## Build tool auxiliary files: 36 | *.fdb_latexmk 37 | *.synctex 38 | *.synctex(busy) 39 | *.synctex.gz 40 | *.synctex.gz(busy) 41 | *.pdfsync 42 | 43 | ## Build tool directories for auxiliary files 44 | # latexrun 45 | latex.out/ 46 | 47 | ## Auxiliary and intermediate files from other packages: 48 | # algorithms 49 | *.alg 50 | *.loa 51 | 52 | # achemso 53 | acs-*.bib 54 | 55 | # amsthm 56 | *.thm 57 | 58 | # beamer 59 | *.nav 60 | *.pre 61 | *.snm 62 | *.vrb 63 | 64 | # changes 65 | *.soc 66 | 67 | # comment 68 | *.cut 69 | 70 | # cprotect 71 | *.cpt 72 | 73 | # elsarticle (documentclass of Elsevier journals) 74 | *.spl 75 | 76 | # endnotes 77 | *.ent 78 | 79 | # fixme 80 | *.lox 81 | 82 | # feynmf/feynmp 83 | *.mf 84 | *.mp 85 | *.t[1-9] 86 | *.t[1-9][0-9] 87 | *.tfm 88 | 89 | #(r)(e)ledmac/(r)(e)ledpar 90 | *.end 91 | *.?end 92 | *.[1-9] 93 | *.[1-9][0-9] 94 | *.[1-9][0-9][0-9] 95 | *.[1-9]R 96 | *.[1-9][0-9]R 97 | *.[1-9][0-9][0-9]R 98 | *.eledsec[1-9] 99 | *.eledsec[1-9]R 100 | *.eledsec[1-9][0-9] 101 | *.eledsec[1-9][0-9]R 102 | *.eledsec[1-9][0-9][0-9] 103 | *.eledsec[1-9][0-9][0-9]R 104 | 105 | # glossaries 106 | *.acn 107 | *.acr 108 | *.glg 109 | *.glo 110 | *.gls 111 | *.glsdefs 112 | 113 | # gnuplottex 114 | *-gnuplottex-* 115 | 116 | # gregoriotex 117 | *.gaux 118 | *.gtex 119 | 120 | # htlatex 121 | *.4ct 122 | *.4tc 123 | *.idv 124 | *.lg 125 | *.trc 126 | *.xref 127 | 128 | # hyperref 129 | *.brf 130 | 131 | # knitr 132 | *-concordance.tex 133 | # TODO Comment the next line if you want to keep your tikz graphics files 134 | *.tikz 135 | *-tikzDictionary 136 | 137 | # listings 138 | *.lol 139 | 140 | # makeidx 141 | *.idx 142 | *.ilg 143 | *.ind 144 | *.ist 145 | 146 | # minitoc 147 | *.maf 148 | *.mlf 149 | *.mlt 150 | *.mtc[0-9]* 151 | *.slf[0-9]* 152 | *.slt[0-9]* 153 | *.stc[0-9]* 154 | 155 | # minted 156 | _minted* 157 | *.pyg 158 | 159 | # morewrites 160 | *.mw 161 | 162 | # nomencl 163 | *.nlg 164 | *.nlo 165 | *.nls 166 | 167 | # pax 168 | *.pax 169 | 170 | # pdfpcnotes 171 | *.pdfpc 172 | 173 | # sagetex 174 | *.sagetex.sage 175 | *.sagetex.py 176 | *.sagetex.scmd 177 | 178 | # scrwfile 179 | *.wrt 180 | 181 | # sympy 182 | *.sout 183 | *.sympy 184 | sympy-plots-for-*.tex/ 185 | 186 | # pdfcomment 187 | *.upa 188 | *.upb 189 | 190 | # pythontex 191 | *.pytxcode 192 | pythontex-files-*/ 193 | 194 | # tcolorbox 195 | *.listing 196 | 197 | # thmtools 198 | *.loe 199 | 200 | # TikZ & PGF 201 | *.dpth 202 | *.md5 203 | *.auxlock 204 | 205 | # todonotes 206 | *.tdo 207 | 208 | # vhistory 209 | *.hst 210 | *.ver 211 | 212 | # easy-todo 213 | *.lod 214 | 215 | # xcolor 216 | *.xcp 217 | 218 | # xmpincl 219 | *.xmpi 220 | 221 | # xindy 222 | *.xdy 223 | 224 | # xypic precompiled matrices 225 | *.xyc 226 | 227 | # endfloat 228 | *.ttt 229 | *.fff 230 | 231 | # Latexian 232 | TSWLatexianTemp* 233 | 234 | ## Editors: 235 | # WinEdt 236 | *.bak 237 | *.sav 238 | 239 | # Texpad 240 | .texpadtmp 241 | 242 | # LyX 243 | *.lyx~ 244 | 245 | # Kile 246 | *.backup 247 | 248 | # KBibTeX 249 | *~[0-9]* 250 | 251 | # auto folder when using emacs and auctex 252 | ./auto/* 253 | *.el 254 | 255 | # expex forward references with \gathertags 256 | *-tags.tex 257 | 258 | # standalone packages 259 | *.sta 260 | 261 | -------------------------------------------------------------------------------- /ex6/main.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt]{article} 2 | \usepackage{xeCJK} 3 | \usepackage{caption} 4 | \setCJKmainfont{KaiTi} 5 | %\setmainfont{Times New Roman} 6 | \setCJKfamilyfont{hei}{SimHei} %黑体 hei 7 | \newcommand{\hei}{\CJKfamily{hei}} % 黑体 8 | \usepackage{amsmath} 9 | \usepackage{amsthm} 10 | \usepackage{amssymb} 11 | \usepackage{tikz} 12 | \usepackage{enumerate} 13 | \usepackage{fontspec} 14 | \usepackage{diagbox} 15 | \usepackage{amsfonts} 16 | 17 | \newcommand{\numpy}{{\tt numpy}} % tt font for numpy 18 | \newcommand*{\dif}{\mathop{}\!\mathrm{d}} 19 | 20 | \topmargin -.5in 21 | \textheight 9in 22 | \oddsidemargin -.25in 23 | \evensidemargin -.25in 24 | \textwidth 7in 25 | 26 | \begin{document} 27 | %\newfontfamily{\Hei}{SimHei} 28 | % ========== Edit your name here 29 | \author{罗雁天} 30 | \title{期末考题整理} 31 | \maketitle 32 | 33 | \medskip 34 | 35 | % ========== Begin answering questions here 36 | \begin{enumerate} 37 | \item {\hei (15分)对于掷两颗骰子的随机试验。 38 | \begin{enumerate}[(a)] 39 | \item 写出样本空间$\Omega$; 40 | \item 记事件$A$为点数之和是奇数，事件$D$为至少出现一个1点，求$P(A\cup D)$; 41 | \item 记事件$B$为某刻骰子出现奇数点，事件$C$为另一颗骰子出现奇数点。问$A,B,C$之间相互独立吗？说明理由。 42 | \end{enumerate} 43 | } 44 | \begin{proof}[解] 45 | \begin{enumerate}[(a)] 46 | \item \begin{equation*}\Omega=N^2,N=\{1,2,3,4,5,6\}\end{equation*} 47 | \item \begin{equation*} 48 | \begin{aligned} 49 | &P(A)=\frac{|A|}{|\Omega|}=\frac{18}{36}=\frac{1}{2},P(D)=\frac{|D|}{|\Omega|}=\frac{11}{36} \\ 50 | &P(AD)=\frac{|AD|}{|\Omega|}=\frac{6}{36}=\frac{1}{6} \\ 51 | &\mbox{所以},P(A\cup D)=P(A)+P(D)-P(AD)=\frac{23}{36} 52 | \end{aligned} 53 | \end{equation*} 54 | \item 不独立。\begin{equation*} 55 | P(ABC)=P(\emptyset)=0\neq P(A)P(B)P(C) 56 | \end{equation*} 57 | \end{enumerate} 58 | \end{proof} 59 | 60 | \item {\hei (10分)连续地做某项试验，每次试验只有成功和失败两种结果。已知当第$k$次试验成功时，第$k+1$次试验成功的概率为$1/2$；当第$k$次试验失败时，第$k+1$次试验成功的概率为$3/4$。如果第一次试验成功的概率为$1/2$, 61 | \begin{enumerate}[(a)] 62 | \item 求出第九次试验成功的概率； 63 | \item 若记$X$为首次获得成功所需的试验次数，求$P(X=9)$； 64 | \end{enumerate}} 65 | \begin{proof}[解] 66 | \begin{enumerate}[(a)] 67 | \item 由全概率公式： 68 | \begin{equation*} 69 | \begin{aligned} 70 | P(A_k)&=P(A_k|A_{k-1})P(A_{k-1})+P(A_k|\overline{A_{k-1}})P(\overline{A_{k-1}}) \\ 71 | &=\frac{1}{2}P(A_{k-1})+\frac{3}{4}P(\overline{A_{k-1}}) \\ 72 | &=\frac{1}{2}P(A_{k-1})+\frac{3}{4}[1-P(A_{k-1})] \\ 73 | &=\frac{3}{4}-\frac{1}{4}P(A_{k-1}) 74 | \end{aligned} 75 | \end{equation*} 76 | 求解上述递推式得： 77 | \begin{equation*} 78 | P(A_k)=\frac{3}{5}-\frac{1}{10}\left(-\frac{1}{4}\right)^{k-1} 79 | \end{equation*} 80 | 所以\begin{equation*} 81 | P(A_9)=\frac{3}{5}-\frac{1}{5\times 2^{17}} 82 | \end{equation*} 83 | \item 由乘法公式： 84 | \begin{equation*} 85 | \begin{aligned} 86 | P(X=9)&=P(\overline{A_{1}}\overline{A_{2}}\cdots \overline{A_{8}}A_9) \\ 87 | &=P(\overline{A_{1}})P(\overline{A_{2}}|\overline{A_{1}})\cdots P(A_9|\overline{A_{1}}\overline{A_{2}}\cdots \overline{A_{8}}) \\ 88 | &=\frac{3}{2^{17}} 89 | \end{aligned} 90 | \end{equation*} 91 | \end{enumerate} 92 | \end{proof} 93 | \item {\hei (15分)von Neumann曾断言，用任何一枚硬币都可做出公平的抉择，具体办法是：把这枚硬币独立投掷两次(称为一轮投掷)，如果得到“正反”或“反正”，试验就结束，否则就要重复此过程。记掷一次硬币得到正面的概率为$p\in (0,1)$并设试验结束时的掷币轮数为$X$。 94 | \begin{enumerate}[(a)] 95 | \item 求$X$的概率分布列； 96 | \item 求$X$的数学期望； 97 | \item 证明试验以“正反”结束和“反正”结束的可能性相同。 98 | \end{enumerate}} 99 | \begin{proof}[解] 100 | \begin{enumerate}[(a)] 101 | \item $X$的概率分布列为(本质上就是参数为$2pq$的几何分布)： 102 | \begin{equation*} 103 | P(X=n)=(p^2+q^2)^{n-1}pq+(p^2+q^2)^{n-1}qp=2pq(1-2pq)^{n-1} 104 | \end{equation*} 105 | 其中$q=1-p$ 106 | \item 几何分布的期望：$E(X)=\frac{1}{2pq}$ 107 | \item \begin{equation*} 108 | \begin{aligned} 109 | &P(\mbox{以“正反”结束})=\sum_{n\geq 1}(p^2+q^2)^{n-1}pq=\frac{pq}{1-p^2-q^2}=\frac{1}{2} \\ 110 | &P(\mbox{以“反正”结束})=\sum_{n\geq 1}(p^2+q^2)^{n-1}qp=\frac{qp}{1-p^2-q^2}=\frac{1}{2} 111 | \end{aligned} 112 | \end{equation*} 113 | 所以，试验以“正反”结束和“反正”结束的可能性相同 114 | \end{enumerate} 115 | \end{proof} 116 | 117 | \item {\hei 假设一批灯泡的使用寿命年数服从$\lambda=1$的指数分布。从这批灯泡中任取两个分别标记为1号和2号，先点亮1号灯泡并开始计时，等其熄灭后立即点亮2号灯泡，两灯泡的总时间记为$Y$年。将1号灯泡的使用寿命记做$X$年。 118 | \begin{enumerate}[(a)] 119 | \item 求给定$X=x>0$条件下$Y$的条件密度函数； 120 | \item 求$X$和$Y$的联合密度函数； 121 | \item 判断$X$和$Y$的独立性并说明理由； 122 | \item 已知$X=2$的条件下，求$Y$值在均方误差最小意义下的最优预测。 123 | \end{enumerate}} 124 | \begin{proof}[解] 125 | \begin{enumerate}[(a)] 126 | \item $Y-X$是2号灯的寿命，且$X$与$Y-X$相互独立。当$y\ge x >0$时， 127 | \begin{equation*} 128 | F(y|x)=P(Y\le y|X=x)=P(Y-X\le y0}$ 131 | \item $p(x,y)=p(x)p(y|x)=e^{-x}e^{x-y}I_{y\ge x>0}=e^{-y}I_{y\ge x>0}$ 132 | \item 不独立。因为： 133 | \begin{equation*} 134 | \begin{aligned} 135 | &p(x)=e^{-x},x\ge 0\\ 136 | &p(y)=\int_{0}^yp(x,y)dx=\int_{0}^{y}e^{-y}dx=ye^{-y},y\ge 0 \\ 137 | &p(x,y)=e^{-y}I_{y\ge x>0}\ne e^{-x}ye^{-y}=p(x)p(y) 138 | \end{aligned} 139 | \end{equation*} 140 | 所以，不独立。 141 | \item $\hat{y}=\arg\min E[(Y-y)^2|X=2]$. 142 | \begin{equation*} 143 | \begin{aligned} 144 | E[(Y-y)^2|X=2]&=\int_{2}^{+\infty}(u-y)^2e^{2-u}du \\ 145 | &=y^2-2(2+1)y+2^2+2(2+1) \\ 146 | &=y^2-6y+10 147 | \end{aligned} 148 | \end{equation*} 149 | 所以$\hat{y}=3$ 150 | \end{enumerate} 151 | \end{proof} 152 | 153 | 154 | \item {\hei 已知随机变量$X$和$Y$服从二元正态分布$(X,Y)\sim N(0,1,4,9,1/2)$，令$U=\frac{X}{2}+\frac{Y}{3},V=\frac{X}{2}-\frac{Y}{3}$ 155 | \begin{enumerate}[(a)] 156 | \item 求随机变量$U$和$V$的协方差； 157 | \item 求随机变量$U$和$V$的联合分布； 158 | \item 判断$U$和$V$的独立性并说明理由； 159 | \end{enumerate}} 160 | \begin{proof}[解] 161 | \begin{enumerate}[(a)] 162 | \item $Cov(U,V)=Cov(\frac{X}{2}+\frac{Y}{3},\frac{X}{2}-\frac{Y}{3})=Var(X)/4-Var(Y)/9=0$ 163 | \item 易得：\begin{equation*} 164 | \begin{aligned} 165 | &E(U)=E(\frac{X}{2}+\frac{Y}{3})=\frac{1}{3},E(V)=E(\frac{X}{2}-\frac{Y}{3})=-\frac{1}{3} \\ 166 | &Var(U)=E(U^2)-E(U)^2=3 \\ 167 | &Var(V)=E(V^2)-E(V)^2=1 \\ 168 | \end{aligned} 169 | \end{equation*} 170 | 根据正态分布在线性变换下的不变性，有$(U,V)\sim N(1/3,-1/3,3,1,0)$ 171 | 172 | {\hei $\bigstar$ 本小问也可以用矩阵法解决，更简单。} 173 | \item 独立。正态分布下，独立和不相关等价。 174 | \end{enumerate} 175 | \end{proof} 176 | 177 | \item {\hei 一批产品中含有废品。从中作放回抽样$n$次，每次随机抽取一件检查，结果发现抽到废品$m$次。 178 | \begin{enumerate}[(a)] 179 | \item 求这批产品废品率的矩估计； 180 | \item\label{b} 求废品率的最大似然估计； 181 | \item 判断(b)中估计量的相合性和无偏性并说明理由； 182 | \item 构造一个废品率的置信区间所需要的近似服从标准正态分布的枢轴量。 183 | \end{enumerate}} 184 | \begin{proof}[解] 185 | 设废品率为$p$，则总体$X\sim b(1,p)$。 186 | \begin{enumerate}[(a)] 187 | \item 因为$p=EX$，所以矩估计$\hat{p}_M=\bar{x}=m/n$ 188 | \item 似然函数为$L(p)=p^m(1-p)^{n-m}$，对数似然函数为$\ln L(p)=m\ln p+(n-m)\ln (1-p)$ 189 | 190 | 似然方程为： 191 | \begin{equation} 192 | \frac{m}{p}-\frac{n-m}{1-p}=0 193 | \end{equation} 194 | 所以，$\hat{p}=m/n$ 195 | \item 由大数定理可知满足相合性。令$x_i$为第$i$次抽到次品的示性函数，则$\hat{p}=\sum_{i=1}^{n}\frac{x_i}{n}$，所以$E(\hat{p})=E(\sum_{i=1}^{n}\frac{x_i}{n})=p$，所以也是无偏的。 196 | \item 由中心极限定理，枢轴量$\frac{m-np}{\sqrt{np(1-p)}}\sim N(0,1)$ 197 | \end{enumerate} 198 | 199 | \end{proof} 200 | \end{enumerate} 201 | 202 | 203 | \end{document} 204 | -------------------------------------------------------------------------------- /ex5/main.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt]{article} 2 | \usepackage{xeCJK} 3 | \usepackage{caption} 4 | \setCJKmainfont{KaiTi} 5 | %\setmainfont{Times New Roman} 6 | \setCJKfamilyfont{hei}{SimHei} %黑体 hei 7 | \newcommand{\hei}{\CJKfamily{hei}} % 黑体 8 | \usepackage{amsmath} 9 | \usepackage{amsthm} 10 | \usepackage{tikz} 11 | \usepackage{enumerate} 12 | \usepackage{fontspec} 13 | \usepackage{diagbox} 14 | \usepackage{amsfonts} 15 | 16 | \newcommand{\numpy}{{\tt numpy}} % tt font for numpy 17 | \newcommand*{\dif}{\mathop{}\!\mathrm{d}} 18 | 19 | \topmargin -.5in 20 | \textheight 9in 21 | \oddsidemargin -.25in 22 | \evensidemargin -.25in 23 | \textwidth 7in 24 | 25 | \begin{document} 26 | %\newfontfamily{\Hei}{SimHei} 27 | % ========== Edit your name here 28 | \author{罗雁天} 29 | \title{习题整理} 30 | \maketitle 31 | 32 | \medskip 33 | 34 | % ========== Begin answering questions here 35 | \begin{enumerate} 36 | \item {\hei $X,Y$相互独立，均服从参数为$\lambda$的指数分布，求$E[(X-Y)^2|Xy \right\} 177 | \end{equation} 178 | 请计算 179 | \begin{equation}\nonumber 180 | \underset{y\to \infty }{\mathop{\lim }}\,P\left\{ N\left( y \right)\ge E\left[ N\left( y \right) \right] \right\} 181 | \end{equation}} 182 | \begin{proof}[解] 183 | 由题意知 184 | \begin{equation} 185 | \begin{aligned} 186 | P\left\{ N\left( y \right)=k \right\}&=P\left\{ {{X}_{1}}\le y,{{X}_{2}}\le y,\cdots ,{{X}_{k-1}}\le y,{{X}_{k}}>y \right\} \\ 187 | & ={{\left[ {{F}_{X}}\left( y \right) \right]}^{k-1}}\left[ 1-{{F}_{X}}\left( y \right) \right] 188 | \end{aligned} 189 | \end{equation} 190 | 即$ N\left( y \right)\sim Ge\left( p \right) $，其中$ p = 1-{{F}_{X}}\left( y \right) $，所以$ E\left[ N\left( y \right) \right]=\displaystyle{\frac{1}{p}} $，则 191 | \begin{equation} 192 | \begin{aligned} 193 | P\left\{ N\left( y \right)\ge E\left[ N\left( y \right) \right] \right\}&=\sum\limits_{k=\left\lfloor 1/p \right\rfloor +1}^{\infty }{{{\left( 1-p \right)}^{k-1}}p} \\ 194 | & =p\sum\limits_{k=\left\lfloor 1/p \right\rfloor +1}^{\infty }{{{\left( 1-p \right)}^{k-1}}} \\ 195 | & = p\cdot \frac{{{\left( 1-p \right)}^{\left\lfloor \frac{1}{p} \right\rfloor}}}{p} \\ 196 | & ={{\left( 1-p \right)}^{\left\lfloor \frac{1}{p} \right\rfloor}} 197 | \end{aligned} 198 | \end{equation} 199 | 其中$ \left\lfloor \cdotp \right\rfloor $表示向下取整。当$ y\longrightarrow\infty $时，$ p\longrightarrow 0 $，所以 200 | \begin{equation} 201 | \underset{y\to \infty }{\mathop{\lim }}\,P\left\{ N\left( y \right)\ge E\left[ N\left( y \right) \right] \right\}=\underset{p\to 0}{\mathop{\lim }}\,{{\left( 1-p \right)}^{\left\lfloor \frac{1}{p} \right\rfloor}} 202 | \end{equation} 203 | 而由于 204 | \begin{equation} 205 | {{\left( 1-p \right)}^{\frac{1}{p}}}\le {{\left( 1-p \right)}^{\left\lfloor \frac{1}{p} \right\rfloor }}<{{\left( 1-p \right)}^{\frac{1}{p}-1}} 206 | \end{equation} 207 | 并且 208 | \begin{equation} 209 | \underset{p\to 0}{\mathop{\lim }}\,{{\left( 1-p \right)}^{\frac{1}{p}}}=\underset{p\to 0}{\mathop{\lim }}\,{{\left( 1-p \right)}^{\frac{1}{p}-1}}=\frac{1}{\text{e}} 210 | \end{equation} 211 | 因此 212 | \begin{equation} 213 | \underset{y\to \infty }{\mathop{\lim }}\,P\left\{ N\left( y \right)\ge E\left[ N\left( y \right) \right] \right\} = \frac{1}{\text{e}} 214 | \end{equation} 215 | \end{proof} 216 | 217 | \end{enumerate} 218 | 219 | 220 | \end{document} 221 | -------------------------------------------------------------------------------- /ex1/main.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt]{article} 2 | \usepackage{xeCJK} 3 | \usepackage{caption} 4 | \setCJKmainfont{KaiTi} 5 | %\setmainfont{Times New Roman} 6 | \setCJKfamilyfont{hei}{SimHei} %黑体 hei 7 | \newcommand{\hei}{\CJKfamily{hei}} % 黑体 8 | \usepackage{amsmath} 9 | \usepackage{amsthm} 10 | \usepackage{tikz} 11 | \usepackage{enumerate} 12 | \usepackage{fontspec} 13 | 14 | \newcommand{\numpy}{{\tt numpy}} % tt font for numpy 15 | 16 | \topmargin -.5in 17 | \textheight 9in 18 | \oddsidemargin -.25in 19 | \evensidemargin -.25in 20 | \textwidth 7in 21 | 22 | \begin{document} 23 | %\newfontfamily{\Hei}{SimHei} 24 | % ========== Edit your name here 25 | \author{罗雁天} 26 | \title{习题课1} 27 | \maketitle 28 | 29 | \medskip 30 | 31 | % ========== Begin answering questions here 32 | \begin{enumerate} 33 | 34 | \item {\hei 在单位圆内随机挑选一条弦，请问弦长大于圆内接等边三角形边长的概率是多大。} 35 | \begin{proof}[解] 36 | 本题可以从三个方面考虑： 37 | \begin{itemize} 38 | \item 如果固定住弦的一个端点A，考察另外一个端点B，那么样本空间就是圆周，等概指的 是B在圆周上均匀分布。那么服从要求的B必然落在以A为端点的圆内接等边三角形中A的对 边所对应的劣弧中。这一段劣弧的长度恰为圆周长度的1/3，因此所求概率为1/3。 39 | \item 如果考察弦的中点O，以单位圆盘作为样本空间，等概指的是O在圆盘上均匀分布。那 么服从要求的O必然落在半径为1/2的单位圆的同心圆内。由于两个圆面积比为1/4，因此所 求概率为1/4。 40 | \item 同样考察弦的中点O，不过以与该弦垂直的半径作为样本空间，等概指的是O在该半径 上均匀分布。那么服从要求的O 必然落在靠近圆心的一半上。因此所求概率为1/2。 41 | \end{itemize} 42 | 三种角度三个答案，看似矛盾实际却很合理。样本空间不同导致概率模型本身存在差 异，出现不同的结果也就不奇怪了。 43 | \end{proof} 44 | 45 | %\item {\hei 假定某赌徒携带k元赌资进入赌场，赌博规则很简单，每赢一局则赢1元，否则输1元。假定每局赌博，赌徒赢的概率都是p，且各局间相互独立。试问，赌徒将所带赌资全部输光，被迫离开赌场的概率有多大?} 46 | %\begin{proof}[解] 47 | % 设事件$A_k$表示赌徒拥有$k$元初始赌本并最终输光，事件$W$表示赌徒赢得一局。那么有： 48 | % \begin{equation} 49 | % P(A_k)=P(A_k|W)P(W)+P(A_k|W^c)P(W^c) 50 | % \end{equation} 51 | % 注意到$P(A_k|W)=P(A_{k+1}),P(A_k|W^c)=P(A_{k-1})$，我们有： 52 | % \begin{equation} 53 | % P(A_k)=pP(A_{k+1})+(1-p)P(A_{k-1}) 54 | % \end{equation} 55 | % 其中$p=P(W)$表示赌徒赢一局的概率。通过此递推式我们可以得到： 56 | % \begin{equation} 57 | % P(A_k)=a+b\left(\frac{1-p}{p}\right)^k 58 | % \end{equation} 59 | % 其中$a,b$为确定性的参数，由初值决定。 60 | % 61 | % 注意到$P(A_0)=1$，因此我们可以得到$a+b=1$；由于$0\le P(A_k)\le 1$。因此我们有如下结论： 62 | % \begin{enumerate}[i)] 63 | % \item 如果$p<0.5$(大多数赌场都满足这一条件)，那么$b=0$，因此$P(A_k)\equiv 1$。这一点不难理解，如 果赌徒赢面小，那么输光应该是肯定的。 64 | % \item 如果$p>0.5$(这种情况几乎不会出现)，那么 65 | % \begin{equation} 66 | % P(A_k)=\left(\frac{1-p}{p}\right)^k+a\left(1-\left(\frac{1-p}{p}\right)^k\right) 67 | % \end{equation} 68 | % \item 如果$p=0.5$(赌场绝对公平)，那么$P(A_k)\equiv1$。这一点 69 | % 很让人惊讶。即使在绝对公平的赌场内，赌徒输光也几乎是肯定的。 70 | % \end{enumerate} 71 | %\end{proof} 72 | 73 | \item {\hei 考虑医疗诊断问题，假设对于某种疾病，诊断的正确率为$p$。也就是说，如果就诊者确实患有该病，则医生能够以概率$p$做出准确诊断；如果就诊者实际没有患该病，医生做出正确判断的概率也是$p$。假设疾病自身的发病率是$q$。现已知某就诊者被医生诊断为患病，则其实际患该病的概率是多少？从中能得到什么结论呢？} 74 | \begin{proof}[解] 75 | 设事件$A$表示就诊者实际患病，$D$表示就诊者被诊断为患病，那么由Bayesian公式： 76 | \begin{equation} 77 | P(A|D)=\frac{P(D|A)P(A)}{P(D)}=\frac{P(D|A)P(A)}{P(D|A)P(A)+P(D|A^c)P(A^c)} 78 | \end{equation} 79 | 带入$P(A)=q,P(A^c)=1-q,P(D|A)=p,P(D|A^c)=1-p$得： 80 | \begin{equation} 81 | P(A|D)=\frac{pq}{pq+(1-p)(1-q)} 82 | \end{equation} 83 | 同理也可以得到： 84 | \begin{equation} 85 | P(A|D^c)=\frac{(1-p)q}{(1-p)q+p(1-q)} 86 | \end{equation} 87 | 下面我们通过一些有趣的计算来进一步认识这一问题。如果疾病的发病率很低($q=0.01$)，医生的医术值得信任，诊断的正确率很高($p=0.99$)，那么在诊断患病的条件下，我们有： 88 | \begin{equation} 89 | P(A|D)=\frac{0.99\times 0.01}{0.99\times 0.01+0.01\times 0.99}=\frac{1}{2} 90 | \end{equation} 91 | 也就是说，真实患病的概率只有$50\%$。不难发现，条件概率的计算中，疾病本身较低的发病率使得医生判断的准确性大打折扣。尝试将发病率$q$升高至0.5(例如普通感冒)，那么在医生医术保持不变的情况下，有： 92 | \begin{equation} 93 | P(A|D)=\frac{0.99\times 0.5}{0.99\times 0.5+0.01\times 0.5}=0.99 94 | \end{equation} 95 | 可见，对于普通常见病，医生的水平能够得到充分的体现。但如果进一步降低发病率$(q=0.001)$，即所谓“疑难杂症”，如果维持医生水平不变，那么： 96 | \begin{equation} 97 | P(A|D)=\frac{0.99\times 0.001}{0.99\times 0.001+0.01\times 0.999}=0.09 98 | \end{equation} 99 | 诊断的实际准确率连$10\%$都不到，误报(实际没病，诊断有病)的概率达到了$90\%$。看起来，即使是非常称职的医生，当面临疑难杂症的时候也难免误报，患者应给予充分的理解，并在得知诊断结果之后保持冷静，争取用复诊的方法进一步确定是否患病。另一方面： 100 | \begin{equation} 101 | P(A|D^c)=\frac{0.01\times 0.001}{0.01\times 0.001+0.99\times 0.999}\approx10^{-5}\ll 1 102 | \end{equation} 103 | 这说明，尽管误报的概率很高，但是漏报(实际患病，诊断无病)的概率却很低。所以，如果医生真的非常称职且水平很高，那么漏报性质的误诊率是能够充分降低的，即使面对的是十分罕见的疑难杂症。 104 | \end{proof} 105 | 106 | 107 | \item {\hei （匹配问题）$n$个人随机地选取帽子，试问至少有一人戴上了自己帽子的概率是多少。} 108 | \begin{proof}[解] 109 | 设$B_k$表示第$k$个人戴对的所有可能结果构成的集合。则至少有一人戴对的所有可能结果可以描述为: 110 | \begin{equation} 111 | B_1\cup B_2\cup\cdots\cup B_n 112 | \end{equation} 113 | 我们的任务是计算这个集合的概率。由于： 114 | \begin{equation} 115 | P(B_k)=\frac{(n-1)!}{n!}=\frac{1}{n},P(B_k\cap B_j)=\frac{(n-2)!}{n!}=\frac{1}{n(n-1)},\cdots 116 | \end{equation} 117 | 所以根据容斥原理，我们有： 118 | \begin{equation} 119 | \begin{aligned} 120 | P(B_1\cup B_2\cup\cdots\cup B_n)=&\sum_{k=1}^{n}P(B_k)-\sum_{j0 $，所以$ \left| {h}'\left( y \right) \right|={h}'\left( y \right) $。因此，当$ 0\le y\le 1 $时， 130 | \begin{equation} 131 | \begin{aligned} 132 | {{p}_{Y}}\left( y \right)&={{p}_{X}}\left[ h\left( y \right) \right]\cdot\left|{h}'\left( y \right)\right| \\ 133 | & =\frac{1 }{\lambda \left( 1-y \right)}\lambda{{e}^{-\lambda \left[ -\frac{\ln \left( 1-y \right)}{\lambda } \right]}} \\ 134 | & =1 135 | \end{aligned} 136 | \end{equation} 137 | 所以，随机变量$ Y $的概率密度函数为 138 | \begin{equation} 139 | \begin{aligned} 140 | {{p}_{Y}}\left( y \right)=\left\{ \begin{matrix} 141 | 1,&0\le y\le 1 \\ 142 | 0,&\text{others} \\ 143 | \end{matrix} \right. 144 | \end{aligned} 145 | \end{equation} 146 | 即$ Y\sim U\left( 0,1 \right) $。随机变量$ {Y_2} $与之类似，也服从均匀分布，证明略。 147 | \end{proof} 148 | 149 | 150 | 151 | \item {\hei 设连续随机变量$ X $服从参数为$ \lambda $的指数分布，另一离散型随机变量$ Y $的可能取值为全体正整数，其分布列为$ {{P}_{Y}}\left( Y=k \right)={{P}_{X}}\left[ \left( k-1 \right)\Delta \le X\le k\Delta \right] $,其中$ k=1,2,\cdots $，常数$ \Delta >0 $。那么，随机变量$ Y $服从哪种常用的概率分布。} 152 | \begin{proof}[解] 153 | 指数函数的分布函数为 154 | \begin{equation} 155 | \begin{aligned} 156 | F\left( x \right)=\left\{ \begin{matrix} 157 | 1-{{e}^{-\lambda x}},&x\ge 0 \\ 158 | 0,&x<0 \\ 159 | \end{matrix} \right. 160 | \end{aligned} 161 | \end{equation} 162 | 从而，$ Y $的概率分布为 163 | \begin{equation} 164 | \begin{aligned} 165 | {{P}_{Y}}\left( Y=k \right)& ={{P}_{X}}\left[ \left( k-1 \right)\Delta \le X\le k\Delta \right] \\ 166 | & =F\left( k\Delta \right)-F\left[ \left( k-1 \right)\Delta \right] \\ 167 | & ={{\left( {{e}^{-\lambda \Delta }} \right)}^{k-1}}\left( 1-{{e}^{-\lambda \Delta }} \right) 168 | \end{aligned} 169 | \end{equation} 170 | 令$ p=1-{{e}^{-\lambda \Delta }} $，则易知$ 00) $个细菌为事件$ A_k $，产生的细菌全是甲类细菌没有乙类细菌为事件$ B $，则 181 | \begin{equation} 182 | \begin{aligned} 183 | P\left( B \right)& =\sum\limits_{k=1}^{+\infty }{P\left( {{A}_{k}} \right)P\left( B\left| {{A}_{k}} \right. \right)} \\ 184 | & =\sum\limits_{k=1}^{+\infty }{\frac{{{\lambda }^{k}}}{k!}{{e}^{-\lambda }}C_{k}^{k}{{\left( \frac{1}{2} \right)}^{k}}} \\ 185 | & ={{e}^{-\lambda }}\left( {{e}^{\frac{\lambda }{2}}}-1 \right) \\ 186 | \end{aligned} 187 | \end{equation} 188 | (2) 所求概率为 189 | \begin{equation} 190 | \begin{aligned} 191 | P\left( {{A}_{2}}\left| B \right. \right)& =\frac{P\left( {{A}_{2}} \right)P\left( B\left| {{A}_{2}} \right. \right)}{P\left( B \right)} \\ 192 | & =\frac{\frac{{{\lambda }^{2}}}{2!}{{e}^{-\lambda }}C_{2}^{2}{{\left( \frac{1}{2} \right)}^{2}}}{{{e}^{-\lambda }}\left( {{e}^{\frac{\lambda }{2}}}-1 \right)} \\ 193 | & =\frac{{{\lambda }^{2}}}{8\left( {{e}^{\frac{\lambda }{2}}}-1 \right)} \\ 194 | \end{aligned} 195 | \end{equation} 196 | \end{proof} 197 | 198 | 199 | \item {\hei 设随机变量$ X $取值于$ [0,1] $，若$ P\left( x \leq X \leq y\right) $只与长度$ y-x $有关（对一切$ 0\leq x \leq y\leq1 $），求$ X $的概率分布。} 200 | \begin{proof}[解] 201 | 记$ P\left( x\leq X \leq y\right)=f\left( y - x\right) $，则对$ x = 0 $，$ \forall y\in \left[ 0,1 \right] $，有 202 | \begin{equation} 203 | P\left( 0\leq X \leq y\right)=f\left( y\right) 204 | \end{equation} 205 | 对$ \forall {y_1},{y_2} \in \left[ 0,1 \right] $且$ {y_1}<{y_2} $,有 206 | \begin{equation} 207 | P\left( 0\leq X \leq {y_1}+{y_2} \right)=P\left( 0\leq X < {y_1} \right)+P\left( {y_1} \leq X \leq {y_1}+{y_2} \right) 208 | \end{equation} 209 | 即有 210 | \begin{equation} 211 | f \left( {y_1}+{y_2} \right) = f \left( {y_1} \right) + f \left( {y_2} \right) 212 | \end{equation} 213 | 因此 214 | \begin{equation} 215 | f \left( {y} \right) = C y 216 | \end{equation} 217 | 由$ f \left( 1 \right) = f \left( 1-0 \right) = P\left( 0 \leq X \leq 1 \right) = 1 $ 推得$ C = 1 $，所以$ f \left( {x} \right) = x $，即 218 | \begin{equation} 219 | P\left( 0 \leq X \leq x \right)=f \left( x \right) , x \in \left[ 0,1 \right] 220 | \end{equation} 221 | 因而$ X $服从均匀分布，即$ X \sim U\left( 0,1 \right) $。 222 | \end{proof} 223 | 224 | 225 | 226 | \item {\hei 设$ X $ 为伯努利试验中第一个游程(连续的成功或失败)的长，求$ X $ 的概率分布及其期望。} 227 | \begin{proof}[解] 228 | 设每次试验中成功的概率为$ p $，则失败的概率为$ 1-p $，则随机变量$ X $ 的概率分布为 229 | \begin{equation} 230 | P \left( X = k \right) = p^k \left( 1 - p \right) + p {\left( 1 - p \right)}^k , k = 1,2,\cdots 231 | \end{equation} 232 | 期望为 233 | \begin{equation} 234 | \begin{aligned} 235 | E \left( X \right) & = \sum\limits_{k=1}^{+\infty }{\left[ k p^k \left( 1 - p \right) + kp {\left( 1 - p \right)}^k \right] }\\ 236 | & = \frac{p}{1-p} + \frac{1-p}{p} 237 | \end{aligned} 238 | \end{equation} 239 | \end{proof} 240 | 241 | 242 | %\item {\hei 食品厂把印有水浒108将之一的画卡作为赠券装入某种儿童食品袋中，每袋一卡，打开包装袋后获得每张画卡的概率是相同的。求：(1)为收集其中的$ r $张画卡所需购买的食品袋数$ X_r $ 的数学期望；(2)为集齐水浒108将，平均要购买多少袋？} 243 | %\begin{proof}[解] 244 | % (1)将收集齐第$ j-1 $张赠券之后到收集齐第$ j $张赠券时所购买的食品袋数记为$ Y_j $，易知 245 | % \begin{equation} 246 | % {X_r} = {Y_1} + {Y_2} + \cdots + {Y_r} 247 | % \end{equation} 248 | % 且$ Y_j $服从几何分布 249 | % \begin{equation} 250 | % P\left( {Y_i} = k \right) ={{\left( 1-\frac{N-i+1}{N} \right)}^{k-1}}\frac{N-i+1}{N},k=1,2,\cdots 251 | % \end{equation} 252 | % 其中$ N $代表画卡的总张数，此题中$ N=108 $，那么 253 | % \begin{equation} 254 | % E\left( {{Y}_{i}} \right)=\frac{N}{N-j+1},j=1,2,\cdots ,r 255 | % \end{equation} 256 | % 因此 257 | % \begin{equation} 258 | % \begin{aligned} 259 | % E\left( {{X}_{r}} \right)& =E\left( {{Y}_{1}} \right)+E\left( {{Y}_{2}} \right)+\cdots +E\left( {{Y}_{r}} \right) \\ 260 | % & =N\left( \frac{1}{N}+\cdots +\frac{1}{N-r+1} \right) \\ 261 | % \end{aligned} 262 | % \end{equation} 263 | % (2) 由上一问知 264 | % \begin{equation} 265 | % \begin{aligned} 266 | % E\left( {{X}_{108}} \right)& =108\times \left( 1+\frac{1}{2}+\cdots +\frac{1}{108} \right) \\ 267 | % & \approx 108\times \left( \ln 108+0.5772 \right) \\ 268 | % & \approx 568.01 269 | % \end{aligned} 270 | % \end{equation} 271 | % 其中，$ 0.5772 $是欧拉常数的近似值。所以，要集齐所有画卡平均要购买569袋该食品。 272 | %\end{proof} 273 | 274 | 275 | %\item {\hei 蛤蛤蛤} 276 | %\begin{proof}[解] 277 | % 278 | %\end{proof} 279 | 280 | 281 | \end{enumerate} 282 | 283 | 284 | 285 | 286 | \end{document} 287 | -------------------------------------------------------------------------------- /ex4/main.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt]{article} 2 | \usepackage{xeCJK} 3 | \usepackage{caption} 4 | \setCJKmainfont{KaiTi} 5 | %\setmainfont{Times New Roman} 6 | \setCJKfamilyfont{hei}{SimHei} %黑体 hei 7 | \newcommand{\hei}{\CJKfamily{hei}} % 黑体 8 | \usepackage{amsmath} 9 | \usepackage{amsthm} 10 | \usepackage{tikz} 11 | \usepackage{enumerate} 12 | \usepackage{fontspec} 13 | %\usepackage{setspace} 14 | 15 | \newcommand{\numpy}{{\tt numpy}} % tt font for numpy 16 | %\renewcommand{\baselinestretch}{1.0} 17 | 18 | \topmargin -.5in 19 | \textheight 9in 20 | \oddsidemargin -.25in 21 | \evensidemargin -.25in 22 | \textwidth 7in 23 | 24 | \begin{document} 25 | %\newfontfamily{\Hei}{SimHei} 26 | % ========== Edit your name here 27 | \author{闻健} 28 | \title{习题课4} 29 | \maketitle 30 | 31 | \medskip 32 | 33 | % ========== Begin answering questions here 34 | \begin{enumerate} 35 | 36 | %\item {\hei 设随机变量$X\sim N(0,\sigma^2),Y=[X]$，即$Y$是$X$向下取整所得（例如：$[1.2]=1,[-2.3]=-3$），计算$Y$的期望。} 37 | %\begin{proof}[解] 38 | % 设$X$的分布函数为$f_X(x)$，$Y$是整数集上的离散随机变量，我们有： 39 | % \begin{equation} 40 | % P(Y=n)=\int_{n}^{n+1}f_X(x)dx 41 | % \end{equation} 42 | % 考虑到$f_X(x)$是偶函数，所以: 43 | % \begin{equation} 44 | % \begin{aligned} 45 | % E(Y)&=\sum_{n=-\infty}^{+\infty}n\times P(y=n) \\ 46 | % &=\sum_{n=-\infty}^{+\infty}n\times\int_{n}^{n+1}f_X(x)dx \\ 47 | % &=\int_{1}^{2}f_X(x)dx+2\int_{2}^{3}f_X(x)dx+\cdots\\ 48 | % &-\int_{0}^{1}f_X(x)dx-2\int_{1}^{2}f_X(x)dx-3\int_{2}^{3}f_X(x)dx-\cdots \\ 49 | % &=-\int_{0}^{+\infty}f_X(x)dx \\ 50 | % &=-\frac{1}{2} 51 | % \end{aligned} 52 | % \end{equation} 53 | %\end{proof} 54 | % 55 | %\item {\hei 在单位圆上随机取两个点构成一条弦，试计算从原点到该弦的距离所服从的概率密度函数。} 56 | %\begin{proof}[解] 57 | % 设两点对圆心的张角为$\theta$，则$\theta\sim U(0,\pi)$，我们有： 58 | % \begin{equation} 59 | % f_\theta(\theta)=\frac{1}{\pi} 60 | % \end{equation} 61 | % 设从原点到弦的距离为$r$，则$r=\cos\left(\displaystyle{\frac{\theta}{2}}\right)$。由于在$\theta\in[0,\pi],r\in[0,1]$时，$r$是$\theta$的单调函数，于是： 62 | % \begin{equation} 63 | % f_r(r)=f_\theta(\theta)\left|\frac{d\theta}{dr}\right|=\frac{2}{\pi\sqrt{1-r^2}},r\in[0,1] 64 | % \end{equation} 65 | % 66 | % {\hei 注意}：这里的随机取点认为是两者所对的圆心角(取范围$[0,\pi]$)服从均匀分布，若选用其他方式，例如弦中点到圆心的距离服从均匀分布，则会得到不同的结果，这与Bertrand悖论的道理是相同的，即不同的样本空间会导致不同的结果。 67 | %\end{proof} 68 | % 69 | % 70 | % 71 | % 72 | %\item {\hei 某城市共有$N$辆车，车牌号从$1$到$N$($N$充分大)，若随机地(可重复)记下$n$辆车的车牌号，其最大号码为$\xi$，求$E(\xi)$。} 73 | %\begin{proof}[解] 74 | % \begin{equation} 75 | % P\left\{ \xi =m \right\}=\frac{{{m}^{n}}-{{\left( m-1 \right)}^{n}}}{{{N}^{n}}},m=1,2,\cdots ,N 76 | % \end{equation} 77 | % \begin{equation} 78 | % E\left( \xi \right)=\sum\limits_{m=1}^{N}{m\cdot \frac{{{m}^{n}}-{{\left( m-1 \right)}^{n}}}{{{N}^{n}}}}=N-\sum\limits_{m=1}^{N-1}{\frac{{{m}^{n}}}{{{N}^{n}}}} 79 | % \end{equation} 80 | % 81 | % 当$N$充分大的时候， 82 | % \begin{equation} 83 | % \sum\limits_{m=1}^{N-1}{\frac{{{m}^{n}}}{{{N}^{n}}}}\text{=}N\sum\limits_{m=0}^{N-1}{\frac{1}{N}\frac{{{m}^{n}}}{{{N}^{n}}}\approx N\int_{0}^{1}{{{x}^{n}}dx}}=N\frac{1}{n+1} 84 | % \end{equation} 85 | % 所以， 86 | % \begin{equation} 87 | % E\left( \xi \right)=N-\sum\limits_{m=1}^{N-1}{\frac{{{m}^{n}}}{{{N}^{n}}}}\approx \frac{n}{n+1}N 88 | % \end{equation} 89 | %\end{proof} 90 | % 91 | % 92 | %\item{\hei 某海港每天早上对停泊船只供给净水，初始价为每吨$a$元，不够用续供则要加$50\%$的附加费；若用不完造成浪费则每吨加收资源费$a/4$元。设某轮船的净水用量是服从密度函数为$p\left( x \right)$的随机变量，为节约用水总开支，求该轮船的最佳首次供水量$y$。} 93 | %\begin{proof}[解] 94 | % 设实际用水量为$x$,总开支为$g\left( x \right)$，则有 95 | % \begin{equation} 96 | % \begin{aligned} 97 | % g\left( x \right)=\left\{ \begin{matrix} 98 | % ay+\displaystyle{\frac{a}{4}}\left( y-x \right),&x0 $，所以$ \left| {h}'\left( y \right) \right|={h}'\left( y \right) $。因此，当$ 0\le y\le 1 $时， 130 | % \begin{equation} 131 | % \begin{aligned} 132 | % {{p}_{Y}}\left( y \right)&={{p}_{X}}\left[ h\left( y \right) \right]\cdot\left|{h}'\left( y \right)\right| \\ 133 | % & =\frac{1 }{\lambda \left( 1-y \right)}\lambda{{e}^{-\lambda \left[ -\frac{\ln \left( 1-y \right)}{\lambda } \right]}} \\ 134 | % & =1 135 | % \end{aligned} 136 | % \end{equation} 137 | % 所以，随机变量$ Y $的概率密度函数为 138 | % \begin{equation} 139 | % \begin{aligned} 140 | % {{p}_{Y}}\left( y \right)=\left\{ \begin{matrix} 141 | % 1,&0\le y\le 1 \\ 142 | % 0,&\text{others} \\ 143 | % \end{matrix} \right. 144 | % \end{aligned} 145 | % \end{equation} 146 | % 即$ Y\sim U\left( 0,1 \right) $。随机变量$ {Y_2} $与之类似，也服从均匀分布，证明略。 147 | %\end{proof} 148 | % 149 | % 150 | % 151 | %\item {\hei 设连续随机变量$ X $服从参数为$ \lambda $的指数分布，另一离散型随机变量$ Y $的可能取值为全体正整数，其分布列为$ {{P}_{Y}}\left( Y=k \right)={{P}_{X}}\left[ \left( k-1 \right)\Delta \le X\le k\Delta \right] $,其中$ k=1,2,\cdots $，常数$ \Delta >0 $。那么，随机变量$ Y $服从哪种常用的概率分布。} 152 | %\begin{proof}[解] 153 | % 指数函数的分布函数为 154 | % \begin{equation} 155 | % \begin{aligned} 156 | % F\left( x \right)=\left\{ \begin{matrix} 157 | % 1-{{e}^{-\lambda x}},&x\ge 0 \\ 158 | % 0,&x<0 \\ 159 | % \end{matrix} \right. 160 | % \end{aligned} 161 | % \end{equation} 162 | % 从而，$ Y $的概率分布为 163 | % \begin{equation} 164 | % \begin{aligned} 165 | % {{P}_{Y}}\left( Y=k \right)& ={{P}_{X}}\left[ \left( k-1 \right)\Delta \le X\le k\Delta \right] \\ 166 | % & =F\left( k\Delta \right)-F\left[ \left( k-1 \right)\Delta \right] \\ 167 | % & ={{\left( {{e}^{-\lambda \Delta }} \right)}^{k-1}}\left( 1-{{e}^{-\lambda \Delta }} \right) 168 | % \end{aligned} 169 | % \end{equation} 170 | % 令$ p=1-{{e}^{-\lambda \Delta }} $，则易知$ 00) $个细菌为事件$ A_k $，产生的细菌全是甲类细菌没有乙类细菌为事件$ B $，则 181 | % \begin{equation} 182 | % \begin{aligned} 183 | % P\left( B \right)& =\sum\limits_{k=1}^{+\infty }{P\left( {{A}_{k}} \right)P\left( B\left| {{A}_{k}} \right. \right)} \\ 184 | % & =\sum\limits_{k=1}^{+\infty }{\frac{{{\lambda }^{k}}}{k!}{{e}^{-\lambda }}C_{k}^{k}{{\left( \frac{1}{2} \right)}^{k}}} \\ 185 | % & ={{e}^{-\lambda }}\left( {{e}^{\frac{\lambda }{2}}}-1 \right) \\ 186 | % \end{aligned} 187 | % \end{equation} 188 | % (2) 因为产生甲、乙两类细菌的机会是相等的，故所求概率等于 189 | % \begin{equation} 190 | % \begin{aligned} 191 | % P\left( {{A}_{2}}\left| B \right. \right)& =\frac{P\left( {{A}_{2}} \right)P\left( B\left| {{A}_{2}} \right. \right)}{P\left( B \right)} \\ 192 | % & =\frac{\frac{{{\lambda }^{2}}}{2!}{{e}^{-\lambda }}C_{2}^{2}{{\left( \frac{1}{2} \right)}^{2}}}{{{e}^{-\lambda }}\left( {{e}^{\frac{\lambda }{2}}}-1 \right)} \\ 193 | % & =\frac{{{\lambda }^{2}}}{8\left( {{e}^{\frac{\lambda }{2}}}-1 \right)} \\ 194 | % \end{aligned} 195 | % \end{equation} 196 | %\end{proof} 197 | % 198 | % 199 | %\item {\hei 设随机变量$ X $取值于$ [0,1] $，若$ P\left( x \leq X \leq y\right) $只与长度$ y-x $有关（对一切$ 0\leq x \leq y\leq1 $），求$ X $的概率分布。} 200 | %\begin{proof}[解] 201 | % 记$ P\left( x\leq X \leq y\right)=f\left( y - x\right) $，则对$ x = 0 $，$ \forall y\in \left[ 0,1 \right] $，有 202 | % \begin{equation} 203 | % P\left( 0\leq X \leq y\right)=f\left( y\right) 204 | % \end{equation} 205 | % 对$ \forall {y_1},{y_2} \in \left[ 0,1 \right] $且$ {y_1}<{y_2} $,有 206 | % \begin{equation} 207 | % P\left( 0\leq X \leq {y_1}+{y_2} \right)=P\left( 0\leq X < {y_1} \right)+P\left( {y_1} \leq X \leq {y_1}+{y_2} \right) 208 | % \end{equation} 209 | % 即有 210 | % \begin{equation} 211 | % f \left( {y_1}+{y_2} \right) = f \left( {y_1} \right) + f \left( {y_2} \right) 212 | % \end{equation} 213 | % 因此 214 | % \begin{equation} 215 | % f \left( {y} \right) = C y 216 | % \end{equation} 217 | % 由$ f \left( 1 \right) = f \left( 1-0 \right) = P\left( 0 \leq X \leq 1 \right) = 1 $ 推得$ C = 1 $，所以$ f \left( {x} \right) = x $，即 218 | % \begin{equation} 219 | % P\left( 0 \leq X \leq x \right)=f \left( x \right) , x \in \left[ 0,1 \right] 220 | % \end{equation} 221 | % 因而$ X $服从均匀分布，即$ X \sim U\left( 0,1 \right) $。 222 | %\end{proof} 223 | % 224 | % 225 | % 226 | %\item {\hei 设$ X $ 为伯努利试验中第一个游程(连续的成功或失败)的长，求$ X $ 的概率分布及其期望。} 227 | %\begin{proof}[解] 228 | % 设每次试验中成功的概率为$ p $，则失败的概率为$ 1-p $，则随机变量$ X $ 的概率分布为 229 | % \begin{equation} 230 | % P \left( X = k \right) = p^k \left( 1 - p \right) + p {\left( 1 - p \right)}^k , k = 1,2,\cdots 231 | % \end{equation} 232 | % 期望为 233 | % \begin{equation} 234 | % \begin{aligned} 235 | % E \left( X \right) & = \sum\limits_{k=1}^{+\infty }{\left[ k p^k \left( 1 - p \right) + kp {\left( 1 - p \right)}^k \right] }\\ 236 | % & = \frac{p}{1-p} + \frac{1-p}{p} 237 | % \end{aligned} 238 | % \end{equation} 239 | %\end{proof} 240 | 241 | 242 | %\item {\hei 食品厂把印有水浒108将之一的画卡作为赠券装入某种儿童食品袋中，每袋一卡，打开包装袋后获得每张画卡的概率是相同的。求：(1)为收集其中的$ r $张画卡所需购买的食品袋数$ X_r $ 的数学期望；(2)为集齐水浒108将，平均要购买多少袋？} 243 | %\begin{proof}[解] 244 | % (1)将收集齐第$ j-1 $张赠券之后到收集齐第$ j $张赠券时所购买的食品袋数记为$ Y_j $，易知 245 | % \begin{equation} 246 | % {X_r} = {Y_1} + {Y_2} + \cdots + {Y_r} 247 | % \end{equation} 248 | % 且$ Y_j $服从几何分布 249 | % \begin{equation} 250 | % P\left( {Y_i} = k \right) ={{\left( 1-\frac{N-i+1}{N} \right)}^{k-1}}\frac{N-i+1}{N},k=1,2,\cdots 251 | % \end{equation} 252 | % 其中$ N $代表画卡的总张数，此题中$ N=108 $，那么 253 | % \begin{equation} 254 | % E\left( {{Y}_{i}} \right)=\frac{N}{N-j+1},j=1,2,\cdots ,r 255 | % \end{equation} 256 | % 因此 257 | % \begin{equation} 258 | % \begin{aligned} 259 | % E\left( {{X}_{r}} \right)& =E\left( {{Y}_{1}} \right)+E\left( {{Y}_{2}} \right)+\cdots +E\left( {{Y}_{r}} \right) \\ 260 | % & =N\left( \frac{1}{N}+\cdots +\frac{1}{N-r+1} \right) \\ 261 | % \end{aligned} 262 | % \end{equation} 263 | % (2) 由上一问知 264 | % \begin{equation} 265 | % \begin{aligned} 266 | % E\left( {{X}_{108}} \right)& =108\times \left( 1+\frac{1}{2}+\cdots +\frac{1}{108} \right) \\ 267 | % & \approx 108\times \left( \ln 108+0.5772 \right) \\ 268 | % & \approx 568.01 269 | % \end{aligned} 270 | % \end{equation} 271 | % 其中，$ 0.5772 $是欧拉常数的近似值。所以，要集齐所有画卡平均要购买569袋该食品。 272 | %\end{proof} 273 | 274 | 275 | %\item {\hei 蛤蛤蛤} 276 | %\begin{proof}[解] 277 | % 278 | %\end{proof} 279 | 280 | 281 | \item {\hei 随机变量$ X $的特征函数为$ \varphi (t) $，通过构造随机变量来证明：$ {{\left| \varphi \left( t \right) \right|}^{2}} $也是特征函数。} 282 | \begin{proof}[解] 283 | 假设随机变量$ X_{1} $和$ X_{2} $，分别与$ X $和$ -X $同分布，并且相互独立，则随机变量$ X_{1}+X_{2} $的特征函数为$ {{\varphi }_{{{X}_{1}}+{{X}_{2}}}}\left( t \right)=\varphi \left( t \right)\cdot \varphi \left( -t \right)=\varphi \left( t \right)\cdot \overline{\varphi \left( t \right)}={{\left| \varphi \left( t \right) \right|}^{2}} $。 284 | \end{proof} 285 | 286 | 287 | \item {\hei 设$ \varphi(t) $表示独立同分布随机变量序列$ \left\{ {{X}_{k}} \right\} $的特征函数，$ {\xi} $为与$ \left\{ {{X}_{k}} \right\} $独立的随机变量，它的概率分布为$ P\left\{ \xi =n \right\}={{p}_{n}} $，其中$ n $为正整数。求随机变量$ Y=\sum\limits_{k=1}^{\xi }{{{X}_{k}}} $的特征函数。} 288 | \begin{proof}[解] 289 | \begin{equation} 290 | \begin{aligned} 291 | {{\varphi }_{Y}}\left( t \right)&=E\left( {{\text{e}}^{itY}} \right) \\ 292 | & =E\left[ E\left( {{\text{e}}^{itY}}\left| \xi \right. \right) \right] \\ 293 | & =\sum\limits_{n=1}^{\infty }{E\left[ {{\text{e}}^{it\left( {{X}_{1}}+{{X}_{2}}+\cdots +{{X}_{n}} \right)}}\left| \xi =n \right. \right]P\left\{ \xi =n \right\}} \\ 294 | & =\sum\limits_{n=1}^{\infty }{E\left[ {{\text{e}}^{it\left( {{X}_{1}}+{{X}_{2}}+\cdots +{{X}_{n}} \right)}} \right]P\left\{ \xi =n \right\}} \\ 295 | & =\sum\limits_{n=1}^{\infty }{{{\varphi }^{n}}\left( t \right){{p}_{n}}} 296 | \end{aligned} 297 | \end{equation} 298 | \end{proof} 299 | 300 | 301 | \item {\hei 设$ \left\{ {{X}_{n}} \right\} $为独立的随机变量序列，且具有相同的特征函数$ \varphi \left( t \right)=1+iat+o\left( t \right) $，其中$ a $为常数。证明：$ \displaystyle{\frac{1}{n}}\sum\limits_{k=1}^{n}{{{X}_{k}}}\stackrel{P}{\longrightarrow} a $。} 302 | \begin{proof}[解] 303 | 相同的特征函数意味着序列$ \left\{ {{X}_{n}} \right\} $服从相同的分布，且 304 | \begin{equation} 305 | E\left( {{X}_{n}} \right)=\frac{{\varphi }'\left( 0 \right)}{i}=a<\infty, 306 | \end{equation} 307 | 故依据辛钦大数定律，对任意的$ \varepsilon>0 $， 308 | \begin{equation} 309 | \underset{n\to \infty }{\mathop{\lim }}\,P\left\{ \left| \frac{1}{n}\sum\limits_{k=1}^{n}{{{X}_{k}}}-\frac{1}{n}\sum\limits_{k=1}^{n}{E\left( {{X}_{k}} \right)} \right|<\varepsilon \right\}=\underset{n\to \infty }{\mathop{\lim }}\,P\left\{ \left| \frac{1}{n}\sum\limits_{k=1}^{n}{{{X}_{k}}}-a \right|<\varepsilon \right\}=1, 310 | \end{equation} 311 | 即$ \displaystyle{\frac{1}{n}}\sum\limits_{k=1}^{n}{{{X}_{k}}} $依概率收敛于$ a $。 312 | \end{proof} 313 | 314 | 315 | %\item {\hei 利用特征函数证明教材98页的定理2.4.1。} 316 | %\begin{proof}[解] 317 | % 事件A发生的次数为$ X \sim b\left( n,{{p}_{n}} \right) $，其特征函数为 318 | % \begin{equation} 319 | % {{\varphi }_{n}}\left( t \right)={{\left[ 1+{{p}_{n}}\left( {{\text{e}}^{it}}-1 \right) \right]}^{n}}={{\left[ 1+\frac{n{{p}_{n}}\left( {{\text{e}}^{it}}-1 \right)}{n} \right]}^{n}}, 320 | % \end{equation} 321 | % 当$ n \longrightarrow \infty $时，则 322 | % \begin{equation} 323 | % \underset{n\to \infty }{\mathop{\lim }}\,{{\varphi }_{n}}\left( t \right)={{\left[ 1+\frac{\lambda \left( {{\text{e}}^{it}}-1 \right)}{n} \right]}^{n}}={{\text{e}}^{\lambda \left( {{\text{e}}^{it}}-1 \right)}}, 324 | % \end{equation} 325 | % 由唯一性定理可知泊松定理成立。 326 | %\end{proof} 327 | 328 | 329 | \item {\hei 独立随机变量序列$ \left\{ {{X}_{k}} \right\} $，其中$ k $为正整数，服从参数为$ {{k}^{r}} $的泊松分布。证明：当$ r < 1 $时，$ \left\{ {{X}_{k}} \right\} $服从大数定律。} 330 | \begin{proof}[解] 331 | 泊松分布的方差为$ \operatorname{var}\left( {{X}_{k}} \right)={{k}^{r}} $，当$ r < 1 $时， 332 | \begin{equation} 333 | \frac{1}{{{n}^{2}}}\operatorname{Var}\left( \sum\limits_{k=1}^{n}{{{X}_{k}}} \right)=\frac{1}{{{n}^{2}}}\sum\limits_{k=1}^{n}{{{k}^{r}}}\le \frac{n}{{{n}^{2}}}{{n}^{r}}={{n}^{r-1}} \longrightarrow 0, 334 | \end{equation} 335 | 即马尔可夫条件成立，故$ \left\{ {{X}_{k}} \right\} $服从大数定律。 336 | \end{proof} 337 | 338 | 339 | \item {\hei 航空公司为了减小损失，会对一些航班进行超售管理。假设某航班的执飞飞机共有310个座位，根据历史数据，该航班的乘客不能按时登机的概率为0.03。那么如果航空公司实际售出314张票，求所有按时登机的乘客都有座位的概率。} 340 | \begin{proof}[解] 341 | 设按时登机的乘客数为$ Y_n $，则 342 | \begin{equation} 343 | Y_n \sim b(314,0.97), E(Y_n ) = 304.58, \operatorname{Var}\left( {{Y}_{n}} \right) = 9.1374 . 344 | \end{equation} 345 | 所求概率为 346 | \begin{equation} 347 | P\left( {{Y}_{n}}\le 310 \right)\approx \Phi \left( \frac{310+0.5-304.58}{\sqrt{9.1374}} \right)\approx 0.9749 . 348 | \end{equation} 349 | \end{proof} 350 | 351 | 352 | \item {\hei 设$ n $重伯努利试验中，事件$ A $在每次试验中出现的概率为$ p\left( 00.99 355 | \end{equation} 356 | 的$ n $的值。} 357 | \begin{proof}[解] 358 | (1)切比雪夫不等式：\\ 359 | \begin{equation} 360 | P\left( \left| \frac{X}{n}-p \right|<\frac{\sqrt{\operatorname{Var}\left( X \right)}}{2} \right) \geq 1-\frac{\frac{\operatorname{Var}\left( X \right)}{{{n}^{2}}}}{{{\left( \frac{\sqrt{\operatorname{Var}\left( X \right)}}{2} \right)}^{2}}}=1-\frac{4}{{{n}^{2}}}>0.99, 361 | \end{equation} 362 | 所以$ n > 20 $。\\ 363 | (2)中心极限定理：\\ 364 | \begin{equation} 365 | P\left( \left| \frac{X}{n}-p \right|<\frac{\sqrt{\operatorname{Var}\left( X \right)}}{2} \right)=P\left( \frac{\left| \frac{X}{n}-p \right|}{\frac{\sqrt{\operatorname{Var}\left( X \right)}}{n}}<\frac{n}{2} \right)=2\Phi \left( \frac{n}{2} \right)-1>0.99 366 | \end{equation} 367 | 所以，$ \Phi \left( \displaystyle{\frac{n}{2}} \right)>0.995 $，则$ n>5.152 $。 368 | \end{proof} 369 | 370 | 371 | \item {\hei 试分别利用切比雪夫不等式和中心极限定理，求以不小于0.99的概率保证用频率估计事件发生概率的误差不大于0.01的独立伯努利试验次数。} 372 | \begin{proof}[解] 373 | 记事件发生次数为$ X $，所需试验次数为$ n $。因为事件发生概率未知，所以事件发生次数的方差应取最大值，即假设$ \operatorname{Var}\left( X \right)=npq=0.25n $。在此假设下，\\ 374 | (1)切比雪夫不等式：\\ 375 | \begin{equation} 376 | P\left( \left| \frac{X}{n}-p \right|<0.01 \right)\ge 1-\frac{\frac{0.25}{n}}{{{\left( 0.01 \right)}^{2}}}>0.99 377 | \end{equation} 378 | 所以，$ n > 250000 $。\\ 379 | (2)中心极限定理：\\ 380 | \begin{equation} 381 | P\left( \left| \frac{X}{n}-p \right|<0.01 \right)=P\left( \frac{\left| \frac{X}{n}-p \right|}{\sqrt{\frac{0.25}{n}}}<\frac{0.01}{\sqrt{\frac{0.25}{n}}} \right)=2\Phi \left( \frac{0.01}{\sqrt{\frac{0.25}{n}}} \right)-1>0.99 382 | \end{equation} 383 | 所以，$ \Phi \left( \displaystyle{\frac{0.01}{\sqrt{\frac{0.25}{n}}}} \right)>0.995 $，则$ n>16587.24 $。 384 | \end{proof} 385 | 386 | 387 | \item {\hei 设$ \left\{ {{X}_{n}} \right\} $为独立的随机变量序列，且$ P\left( {{X}_{n}}=\pm {{n}^{r}} \right)=\displaystyle{\frac{1}{2}} $，其中$ n $为正整数。利用中心极限定理证明：若$ r\geq \displaystyle{\frac{1}{2}} $，$ \left\{ {{X}_{n}} \right\} $不服从大数定律。} 388 | \begin{proof}[解] 389 | $ {{X}_{n}} $的期望和方差分别为 390 | \begin{equation} 391 | E\left( {{X}_{n}} \right)=0,\operatorname{Var}\left( {{X}_{n}} \right)={{n}^{2r}}. 392 | \end{equation} 393 | 对$ \delta > 0 $， 394 | \begin{equation} 395 | \begin{aligned} 396 | \frac{1}{B_{n}^{2+\delta }}\sum\limits_{k=1}^{n}{E\left( {{\left| {{X}_{k}}-{{\mu }_{k}} \right|}^{2+\delta }} \right)} &=\frac{\sum\limits_{k=1}^{n}{{{k}^{r\left( 2+\delta \right)}}}}{{{\left( \sum\limits_{k=1}^{n}{{{k}^{2r}}} \right)}^{\frac{2+\delta }{2}}}} \\ 397 | & =\frac{{{\left( 2r+1 \right)}^{\frac{2+\delta }{2}}}}{r\left( 2+\delta \right)+1}\cdot \frac{{{n}^{r\left( 2+\delta \right)+1}}}{{{n}^{\left( 2r+1 \right)\frac{2+\delta }{2}}}} \\ 398 | & =\frac{{{\left( 2r+1 \right)}^{\frac{2+\delta }{2}}}}{r\left( 2+\delta \right)+1}\cdot \frac{1}{{{n}^{\frac{\delta }{2}}}} \longrightarrow 0, 399 | \end{aligned} 400 | \end{equation} 401 | 即$ {{X}_{n}} $满足李雅普诺夫条件，所以可以使用中心极限定理。如果$ r\geq \displaystyle{\frac{1}{2}} $，则对任意$ \varepsilon>0 $，有 402 | \begin{equation} 403 | \frac{{{\varepsilon }^{2}}{{n}^{2}}}{B_{n}^{2}}=\frac{{{\varepsilon }^{2}}{{n}^{2}}}{\sum\limits_{k=1}^{n}{{{k}^{2r}}}}\le \frac{{{\varepsilon }^{2}}{{n}^{2}}}{\sum\limits_{k=1}^{n}{k}}=\frac{2{{\varepsilon }^{2}}{{n}^{2}}}{n\left( n+1 \right)} \longrightarrow 2{{\varepsilon }^{2}} 404 | \end{equation} 405 | 当$ n $充分大时，$ \displaystyle{\frac{\varepsilon n}{{{B}_{n}}}}<\sqrt{3}\varepsilon $，且有 406 | \begin{equation} 407 | \begin{aligned} 408 | P\left\{ \frac{1}{n}\left| \sum\limits_{k=1}^{n}{{{X}_{k}}} \right|<\varepsilon \right\} &=P\left\{ \frac{{{B}_{n}}}{n}\cdot \frac{1}{{{B}_{n}}}\left| \sum\limits_{k=1}^{n}{{{X}_{k}}} \right|<\varepsilon \right\} \\ 409 | & =P\left\{ \frac{1}{{{B}_{n}}}\left| \sum\limits_{k=1}^{n}{{{X}_{k}}} \right|<\frac{\varepsilon n}{{{B}_{n}}} \right\} \\ 410 | & \le P\left\{ \frac{1}{{{B}_{n}}}\left| \sum\limits_{k=1}^{n}{{{X}_{k}}} \right|<\sqrt{3}\varepsilon \right\} . 411 | \end{aligned} 412 | \end{equation} 413 | 因此， 414 | \begin{equation} 415 | \begin{aligned} 416 | \underset{n\to \infty }{\mathop{\lim }}\,P\left\{ \frac{1}{n}\left| \sum\limits_{k=1}^{n}{{{X}_{k}}} \right|<\varepsilon \right\} 417 | & \le \underset{n\to \infty }{\mathop{\lim }}\,P\left\{ \frac{1}{{{B}_{n}}}\left| \sum\limits_{k=1}^{n}{{{X}_{k}}} \right|<\sqrt{3}\varepsilon \right\} \\ 418 | & =\frac{1}{2\pi }\int_{-\sqrt{3}\varepsilon }^{\sqrt{3}\varepsilon }{{{\text{e}}^{-\frac{1}{2}{t}^{2}}}}\text{d}t \\ 419 | & <1 . 420 | \end{aligned} 421 | \end{equation} 422 | 故不服从大数定律。 423 | \end{proof} 424 | 425 | 426 | \item {\hei 用特征函数法直接证明棣莫弗-拉普拉斯中心极限定理。} 427 | \begin{proof}[解] 428 | 记$ {{S}_{n}}\sim b\left( n,p \right) $，则$ {{S}_{n}} $的特征函数为$ {{\varphi }_{n}}\left( t \right)={{\left( q+p{{\text{e}}^{it}} \right)}^{n}} $，由此$ Y_{n}^{*}=\displaystyle{\frac{{{S}_{n}}-np}{\sqrt{npq}}} $的特征函数为 429 | \begin{equation} 430 | \begin{aligned} 431 | {{f}_{n}}\left( t \right)&={{\left( q+p{{\text{e}}^{\frac{it}{\sqrt{npq}}}} \right)}^{n}}{{\text{e}}^{-\frac{npit}{\sqrt{npq}}}} \\ 432 | & ={{\left( q{{\text{e}}^{-\frac{pit}{\sqrt{npq}}}}+p{{\text{e}}^{\frac{qit}{\sqrt{npq}}}} \right)}^{n}} \\ 433 | & ={{\left[ q\left( 1-\frac{pit}{\sqrt{npq}}-\frac{p{{t}^{2}}}{2nq} \right)+p\left( 1+\frac{qit}{\sqrt{npq}}-\frac{q{{t}^{2}}}{2np} \right)+o\left( \frac{{{t}^{2}}}{n} \right) \right]}^{n}} \\ 434 | & ={{\left[ 1-\frac{{{t}^{2}}}{2n}+o\left( \frac{{{t}^{2}}}{n} \right) \right]}^{n}}\longrightarrow {{\text{e}}^{-\frac{1}{2}{{t}^{2}}}} 435 | \end{aligned} 436 | \end{equation} 437 | 由定理 4.2.6 即证得棣莫弗-拉普拉斯中心极限定理。 438 | \end{proof} 439 | 440 | 441 | %\item {\hei 蛤蛤蛤} 442 | %\begin{proof}[解] 443 | % 444 | %\end{proof} 445 | 446 | 447 | %\item {\hei 蛤蛤蛤} 448 | %\begin{proof}[解] 449 | % 450 | %\end{proof} 451 | 452 | 453 | %\item {\hei 蛤蛤蛤} 454 | %\begin{proof}[解] 455 | % 456 | %\end{proof} 457 | 458 | 459 | %\item {\hei 蛤蛤蛤} 460 | %\begin{proof}[解] 461 | % 462 | %\end{proof} 463 | 464 | \end{enumerate} 465 | 466 | 467 | 468 | 469 | \end{document} 470 | -------------------------------------------------------------------------------- /ex3/main.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt]{article} 2 | \usepackage{xeCJK} 3 | \usepackage{caption} 4 | \setCJKmainfont{KaiTi} 5 | %\setmainfont{Times New Roman} 6 | \setCJKfamilyfont{hei}{SimHei} %黑体 hei 7 | \newcommand{\hei}{\CJKfamily{hei}} % 黑体 8 | \usepackage{amsmath} 9 | \usepackage{amsthm} 10 | \usepackage{tikz} 11 | \usepackage{enumerate} 12 | \usepackage{fontspec} 13 | \usepackage{diagbox} 14 | \usepackage{amsfonts} 15 | 16 | \newcommand{\numpy}{{\tt numpy}} % tt font for numpy 17 | \newcommand*{\dif}{\mathop{}\!\mathrm{d}} 18 | 19 | \topmargin -.5in 20 | \textheight 9in 21 | \oddsidemargin -.25in 22 | \evensidemargin -.25in 23 | \textwidth 7in 24 | 25 | \begin{document} 26 | %\newfontfamily{\Hei}{SimHei} 27 | % ========== Edit your name here 28 | \author{罗雁天} 29 | \title{习题课3} 30 | \maketitle 31 | 32 | \medskip 33 | 34 | % ========== Begin answering questions here 35 | \begin{enumerate} 36 | 37 | %\item {\hei 再次考虑“匹配”问题。$n$个人随机地选取帽子，试问恰好戴上了自己帽子的人数的期望和方差是多少。} 38 | % 39 | %\begin{proof}[解] 40 | % 带上自己帽子的人数$X$满足: 41 | % \begin{equation} 42 | % X=H_1+H_2+\cdots+H_n,\quad H_k=\left\{ 43 | % \begin{array}{cc} 44 | % 1 & \mbox{第$k$个人戴对帽子}\\ 45 | % 0 & \mbox{第$k$个人戴错帽子} 46 | % \end{array} 47 | % \right. 48 | % \end{equation} 49 | % 则由\begin{equation} 50 | % P(H_k=1)=1-P(H_k=0)=\frac{1}{n} 51 | % \end{equation} 52 | % 得到\begin{equation} 53 | % E(H_k)=1\times P(H_k=1)+0\times P(H_k=0)=\frac{1}{n} 54 | % \end{equation} 55 | % 从而\begin{equation} 56 | % \begin{aligned} 57 | % E(X)&=E(H_1+H_2+\cdots+H_n)\\ 58 | % &=E(H_1)+E(H_2)+\cdots+E(H_n) \\ 59 | % &=n\times \frac{1}{n} \\ 60 | % &=1 61 | % \end{aligned} 62 | % \end{equation} 63 | % 即戴上自己帽子的人数的期望值为1。也就是说，平均只有一个人戴上自己的帽子。相比于上次习题课得到的人数分布的结果，这里给出的期望能够更加直观地表明恰好匹配(戴上自己的帽子)是较为困难的事情。 64 | % 65 | % 接下来求方差，首先求$E(X^2)$如下： 66 | % \begin{equation} 67 | % \label{eq1} 68 | % \begin{aligned} 69 | % E(X^2)&=E\left((H_1+H_2+\cdots+H_n)^2\right) \\ 70 | % &=E\left(\sum_{k=1}^{n}H_k^2+\sum_{i=1}^{n}\sum_{j\ne i}H_iH_j\right) \\ 71 | % &=\sum_{k=1}^nE(H_k^2)+\sum_{i=1}^{n}\sum_{j\ne i}E(H_iH_j) 72 | % \end{aligned} 73 | % \end{equation} 74 | % 根据我们的定义，我们可以得到： 75 | % \begin{equation} 76 | % E(H_k^2)=1\times P(H_k=1)+0\times P(H_k=0)=\frac{1}{n} 77 | % \end{equation} 78 | % $H_iH_j$表示第$i,j$两人均带对帽子，因此： 79 | % \begin{equation} 80 | % P(H_iH_j=1)=1-P(H_iH_j=0)=\frac{1}{n(n-1)} 81 | % \end{equation} 82 | % 因此： 83 | % \begin{equation} 84 | % E(H_iH_j)=1\times P(H_iH_j=1)+0\times P(H_iH_j=0)=\frac{1}{n(n-1)} 85 | % \end{equation} 86 | % 带入式(\ref{eq1})中得： 87 | % \begin{equation} 88 | % E(X^2)=n\times\frac{1}{n}+n(n-1)\times\frac{1}{n(n-1)}=2 89 | % \end{equation} 90 | % 因此： 91 | % \begin{equation} 92 | % Var(X)=E(X^2)-\left[E(X)\right]^2=1 93 | % \end{equation} 94 | %\end{proof} 95 | 96 | \item {\hei n对夫妇共2n人待在一个房间内，现从中随机挑选m个人走出房间，问房间内还剩下的未被拆散的夫妇数目的均值。} 97 | \begin{proof}[解] 98 | 房间内剩下的未被拆散的夫妇数目$X$满足 99 | \begin{equation} 100 | X=C_1+C_2+\cdots+C_n,\quad C_k=\left\{ 101 | \begin{array}{cc} 102 | 1 & \mbox{第$k$对夫妇在房间}\\ 103 | 0 & \mbox{第$k$对夫妇不在房间} 104 | \end{array} 105 | \right. 106 | \end{equation} 107 | 则由\begin{equation} 108 | P(C_k=1)=1-P(C_k=0)=\frac{\binom{2n-2}{m}}{\binom{2n}{m}}=\left(1-\frac{m}{2n}\right)\left(1-\frac{m}{2n-1}\right) 109 | \end{equation} 110 | 得到\begin{equation} 111 | E(C_k)=1\times P(C_k=1)+0\times P(C_k=0)=\left(1-\frac{m}{2n}\right)\left(1-\frac{m}{2n-1}\right) 112 | \end{equation} 113 | 从而\begin{equation} 114 | \begin{aligned} 115 | E(X)&=E(C_1+C_2+\cdots+C_n)\\ 116 | &=E(C_1)+E(C_2)+\cdots+E(C_n) \\ 117 | &=n\left(1-\frac{m}{2n}\right)\left(1-\frac{m}{2n-1}\right) 118 | \end{aligned} 119 | \end{equation} 120 | \end{proof} 121 | 122 | \item {\hei 某商场发行$n$种购物券，每一次在该商场购物，即可获得一张种类随机选取的购物券。该商场规定，如果能够收集齐所有的购物券，那么就可以得到该商场的大奖。问获得大奖所需要的购物次数的期望是多少。} 123 | 124 | \begin{proof}[解] 125 | 收集齐所有的购物券所需要的购物次数$X$可以写为 126 | \begin{equation} 127 | X=C_1+C_2+\cdots+C_n 128 | \end{equation} 129 | 其中$C_k$表示在收集到$k-1$种购物券的前提下，收集到第$k$种购物券所需要的购物次数。由于手中已经有$k-1$种购物券，因此收集到新购物券的可能只有$n-k+1$种，所以$C_k$是几何分布随机变量，参数为$p=(n-k+1)/n$，因此我们有： 130 | \begin{equation} 131 | E(C_k)=\frac{1}{p}=\frac{n}{n-k+1} 132 | \end{equation} 133 | 所以： 134 | \begin{equation} 135 | \begin{aligned} 136 | E(X)&=E(C_1+C_2+\cdots+C_n)\\ 137 | &=E(C_1)+E(C_2)+\cdots+E(C_n) \\ 138 | &=1+\frac{n}{n-1}+\frac{n}{n-2}+\cdots+\frac{n}{1} \\ 139 | &=n\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right) 140 | \end{aligned} 141 | \end{equation} 142 | 由微积分的知识可以知道$E(X)\approx n\ln n$，可见，当$n$较大时，如果想收集到所有的购物券，则需要付出数倍于购物券数目的购物次数(以30张购物券为例，$\ln(30)\approx3.4$) 。这正是商家的精明之处。 143 | \end{proof} 144 | 145 | \item {\hei 司机在一年发生事故的次数满足$\lambda$的泊松分布，而$\lambda$服从参数为$\mu$的指数分布，问某一司机上一年不发生事故，今年也不发生事故的概率。} 146 | \begin{proof}[解] 147 | 假设该司机在一年发生事故的次数为$X$，则有： 148 | \begin{equation} 149 | \begin{aligned} 150 | P(X=k)&=\int_{0}^{+\infty}P(X=k|\lambda=x)f_{\lambda}(x)dx \\ 151 | &=\int_{0}^{+\infty}\frac{x^ke^{-x}}{k!}\mu e^{-\mu x}dx \\ 152 | &=\frac{\mu}{k!}\int_{0}^{+\infty}x^ke^{-(\mu+1)x}dx \\ 153 | &=\frac{\mu}{k!}\frac{\Gamma(k+1)}{(\mu+1)^{k+1}} \\ 154 | &=\frac{\mu}{(\mu+1)^{k+1}} 155 | \end{aligned} 156 | \end{equation} 157 | 因此，司机连续两年不发生事故的概率为(两年发生事故相互独立)： 158 | \begin{equation} 159 | P=\left(\frac{\mu}{\mu+1}\right)^2 160 | \end{equation} 161 | \end{proof} 162 | 163 | \item {\hei 设$R$罐中有$n$个红球，$H$罐中有$n$个黑球。每次操作都从两个罐中各随机取出一球，交换放置($R$罐中取出的放入$H$罐，$H$罐中取出的放入$R$罐)，问$k$次操作后，$R$罐中的红球数目的均值。} 164 | 165 | \begin{proof}[解] 166 | $k$次操作后$R$罐的红球数目满足： 167 | \begin{equation} 168 | X=R_1+R_2+\cdots+R_n,\quad R_t=\left\{ 169 | \begin{array}{cc} 170 | 1 & \mbox{第$t$个红球仍在$R$罐}\\ 171 | 0 & \mbox{第$t$个红球不在$R$罐} 172 | \end{array} 173 | \right. 174 | \end{equation} 175 | 对于操作开始前$R$罐中原有的每一个红球而言，其经过$k$次操作仍回到$R$罐意味着被选中的次数为偶数(从来未被选中意味着选中次数为0，仍为偶数)。所以： 176 | \begin{equation} 177 | P(R_t=1)=1-P(R_t=0)=\sum_{m=2i}\binom{k}{m}\left(\frac{1}{n}\right)^m\left(1-\frac{1}{n}\right)^{k-m} 178 | \end{equation} 179 | 根据二项式定理，我们可以计算出偶数次项的和可以计算如下： 180 | \begin{equation} 181 | \sum_{m=2i}\binom{k}{m}a^mb^{k-m}=\frac{1}{2}\left(\left(b+a\right)^k+\left(b-a\right)^k\right) 182 | \end{equation} 183 | 所以： 184 | \begin{equation}\begin{aligned} 185 | P(R_t=1)&=\frac{1}{2}\left(\left(1-\frac{1}{n}+\frac{1}{n}\right)^k+\left(1-\frac{1}{n}-\frac{1}{n}\right)^k\right) \\ 186 | &=\frac{1}{2}\left(1+\left(1-\frac{2}{n}\right)^k\right) 187 | \end{aligned} 188 | \end{equation} 189 | 从而 190 | \begin{equation} 191 | \begin{aligned} 192 | E(X)&=E(R_1+R_2+\cdots+R_n) \\ 193 | &=E(R_1)+E(R_2)+\cdots+E(R_n) \\ 194 | &=\frac{n}{2}\left(1+\left(1-\frac{2}{n}\right)^k\right) 195 | \end{aligned} 196 | \end{equation} 197 | \end{proof} 198 | 199 | %\item {\hei 设随机变量$X\sim N(0,\sigma^2),Y=[X]$，即$Y$是$X$的整数部分（例如：$[1.2]=1,[-2.3]=-3$），计算$Y$的期望} 200 | %\begin{proof}[解] 201 | % 设$X$的分布函数为$f_X(x)$，$Y$是整数集上的离散随机变量，我们有： 202 | % \begin{equation} 203 | % P(Y=n)=\int_{n}^{n+1}f_X(x)dx 204 | % \end{equation} 205 | % 考虑到$f_X(x)$是偶函数，所以: 206 | % \begin{equation} 207 | % \begin{aligned} 208 | % E(Y)&=\sum_{n=-\infty}^{+\infty}n\times P(y=n) \\ 209 | % &=\sum_{n=-\infty}^{+\infty}n\times\int_{n}^{n+1}f_X(x)dx \\ 210 | % &=\int_{1}^{2}f_X(x)dx+2\int_{2}^{3}f_X(x)dx+\cdots\\ 211 | % &-\int_{0}^{1}f_X(x)dx-2\int_{1}^{2}f_X(x)dx-3\int_{2}^{3}f_X(x)dx-\cdots \\ 212 | % &=-\int_{0}^{+\infty}f_X(x)dx \\ 213 | % &=-\frac{1}{2} 214 | % \end{aligned} 215 | % \end{equation} 216 | %\end{proof} 217 | \item {\hei $X,Y$独立都服从$\mathcal{N}(0,1)$，求$E[(X-3Y)^2|2X+Y=3]$} 218 | \begin{proof}[解] 219 | 根据定义有： 220 | \begin{equation} 221 | E[(X-3Y)^2|2X+Y=3]=\frac{\iint_{2x+y=3}(x-3y)^2p(x,y)\dif x\dif y}{\iint_{2x+y=3}p(x,y)\dif x\dif y} 222 | \end{equation} 223 | 首先计算分母如下： 224 | \begin{equation} 225 | \begin{aligned} 226 | \iint_{2x+y=3}p(x,y)\dif x\dif y&=\iint_{2x+y=3}p(x)p(y)\dif x\dif y \\ 227 | &=\iint_{2x+y=3}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right)\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{y^2}{2}\right)\dif x\dif y \\ 228 | &=\iint_{2x+y=3}\frac{1}{2\pi}\exp\left(-\frac{x^2+y^2}{2}\right)\dif x\dif y \\ 229 | &\mbox{令}x=1+t,y=1-2t \\ 230 | &=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\exp\left[-\frac{(1+t)^2+(1-2t)^2}{2}\right]\dif t \\ 231 | &=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\exp\left[-\frac{5t^2-2t+2}{2}\right]\dif t \\ 232 | &\mbox{令}x=t-\frac{1}{5}\\ 233 | &=\exp\left(\frac{9}{10}\right)\cdot \frac{1}{2\pi}\int_{-\infty}^{+\infty}\exp\left[-\frac{5x^2}{2}\right]\dif t \\ 234 | &=\exp\left(\frac{9}{10}\right)\cdot \frac{1}{\sqrt{10\pi}} 235 | \end{aligned} 236 | \end{equation} 237 | 然后计算分子如下： 238 | \begin{equation} 239 | \begin{aligned} 240 | \iint_{2x+y=3}(x-3y)^2p(x,y)\dif x\dif y&=\iint_{2x+y=3}(x-3y)^2p(x,y)\dif x\dif y\\ 241 | &=\frac{1}{2\pi}\iint_{2x+y=3}(x-3y)^2\exp\left(-\frac{x^2+y^2}{2}\right)\dif x\dif y \\ 242 | &\mbox{令}x=1+t,y=1-2t \\ 243 | &=\frac{1}{2\pi}\int_{-\infty}^{+\infty}(7t-2)^2\exp\left[-\frac{5t^2-2t+2}{2}\right]\dif t \\ 244 | &\mbox{令}x=t-\frac{1}{5}\\ 245 | &=\exp\left(\frac{9}{10}\right)\cdot \frac{1}{2\pi}\int_{-\infty}^{+\infty}(7x-\frac{3}{5})^2\exp\left[-\frac{5x^2}{2}\right]\dif t \\ 246 | &=\exp\left(\frac{9}{10}\right)\cdot \frac{1}{\sqrt{10\pi}}\times\frac{254}{25} 247 | \end{aligned} 248 | \end{equation} 249 | 所以： 250 | \begin{equation} 251 | E[(X-3Y)^2|2X+Y=3]=\frac{254}{25} 252 | \end{equation} 253 | \end{proof} 254 | 255 | \item {\hei 设有$k$种不同的优惠券，每次收集到第$i$种优惠券的概率为$p_i$，$\sum_{i=1}^kp_i=1$，且每次收集之间是相互独立的。如果收集了$n$张优惠券，那么优惠券的种类的期望是多少？} 256 | \begin{proof}[解] 257 | 优惠券的种类数$X$满足: 258 | \begin{equation} 259 | X=H_1+H_2+\cdots+H_k,\quad H_i=\left\{ 260 | \begin{array}{cc} 261 | 1 & \mbox{收集到第$i$种优惠券}\\ 262 | 0 & \mbox{没有收集到第$i$种优惠券} 263 | \end{array} 264 | \right. 265 | \end{equation} 266 | 那么我们有： 267 | \begin{equation} 268 | P(H_i=1)=1-P(H_i=0)=1-(1-p_i)^n 269 | \end{equation} 270 | 所以： 271 | \begin{equation} 272 | E(H_i)=1\times P(H_i=1)+0\times P(H_i=0)=1-(1-p_i)^n 273 | \end{equation} 274 | 从而： 275 | \begin{equation} 276 | \begin{aligned} 277 | E(X)&=E(H_1+H_2+\cdots+H_k)\\ 278 | &=E(H_1)+E(H_2)+\cdots+E(H_k) \\ 279 | &=\sum_{i=1}^{k}\left(1-(1-p_i)^n\right) 280 | \end{aligned} 281 | \end{equation} 282 | \end{proof} 283 | 284 | %\item {\hei 在单位圆上随机取两个点构成一条弦，试计算从原点到该弦的距离所服从的概率密度函数} 285 | %\begin{proof}[解] 286 | % 设两点对圆心的张角为$\theta$，则$\theta\sim U(0,\pi)$，我们有： 287 | % \begin{equation} 288 | % f_\theta(\theta)=\frac{1}{\pi} 289 | % \end{equation} 290 | % 设从原点到弦的距离为$r$，则$r=\cos\left(\frac{\theta}{2}\right)$。由于在$\theta\in[0,\pi],r\in[0,1]$时，$r$是$\theta$的单调函数，于是： 291 | % \begin{equation} 292 | % f_r(r)=f_\theta(\theta)\left|\frac{d\theta}{dr}\right|=\frac{2}{\pi\sqrt{1-r^2}},r\in[0,1] 293 | % \end{equation} 294 | % 295 | % {\hei 注意}：这里的随机取点认为是两者所对的圆心角(取范围$[0,\pi]$)服从均匀分布，若选用其他方式，例如弦中点到圆心的距离服从均匀分布，则会得到不同的结果，这与Bertrand悖论的道理是相同的，即不同的样本空间会导致不同的结果。 296 | %\end{proof} 297 | 298 | \item {\hei 假设$\lambda$服从以$\alpha,\beta$为参数的Gamma分布，即$\lambda\sim \Gamma(\alpha,\beta)$。在给定$\lambda$的条件下，$x$服从以$\lambda$为参数的Poisson的分布，即$x|\lambda\sim Poisson(\lambda)$。试问，在给定$x$的条件下，$\lambda$的分布是什么？} 299 | \begin{proof}[解] 300 | 由于 $\lambda \sim \Gamma(\alpha,\beta)$, 我们有: 301 | \begin{equation} 302 | p(\lambda)=\dfrac{\beta^{\alpha}\lambda^{\alpha-1}e^{-\beta\lambda}}{\Gamma(\alpha)} 303 | \end{equation} 304 | 又由于 $x|\lambda \sim Poisson(\lambda)$, 因而: 305 | \begin{equation} 306 | p(x|\lambda)=\frac{\lambda^x}{x!}e^{-\lambda} 307 | \end{equation} 308 | 所以，根据贝叶斯公式: 309 | \begin{equation} 310 | \begin{aligned} 311 | p(\lambda|x)&=\dfrac{p(x|\lambda)p(\lambda)}{p(x)} \\ 312 | &=\dfrac{p(x|\lambda)p(\lambda)}{\int_{0}^{+\infty}p(x|\lambda)p(\lambda)\mathrm{d}\lambda} \\ 313 | &=\dfrac{\frac{\lambda^x}{x!}e^{-\lambda}\cdot\dfrac{\beta^{\alpha}\lambda^{\alpha-1}e^{-\beta\lambda}}{\Gamma(\alpha)}}{\int_{0}^{+\infty}\frac{\lambda^x}{x!}e^{-\lambda}\cdot\dfrac{\beta^{\alpha}\lambda^{\alpha-1}e^{-\beta\lambda}}{\Gamma(\alpha)}\mathrm{d}\lambda} \\ 314 | &=\dfrac{\lambda^{(\alpha+x-1)}e^{-(\beta+1)\lambda}}{\int_{0}^{+\infty}\lambda^{(\alpha+x-1)}e^{-(\beta+1)\lambda}\mathrm{d}\lambda} \\ 315 | &=\dfrac{\lambda^{(\alpha+x-1)}e^{-(\beta+1)\lambda}}{\int_{0}^{+\infty}\left(\frac{u}{\beta+1}\right)^{(\alpha+x-1)}e^{-u}\mathrm{d}\left(\frac{u}{\beta+1}\right)} \qquad (u=(\beta+1)\lambda) \\ 316 | &=\dfrac{\lambda^{(\alpha+x-1)}e^{-(\beta+1)\lambda}}{\frac{1}{(\beta+1)^{\alpha+x}}\int_{0}^{+\infty}u^{(\alpha+x-1)}e^{-u}\mathrm{d}u} \\ 317 | &=\dfrac{(\beta+1)^{\alpha+x}\lambda^{(\alpha+x-1)}e^{-(\beta+1)\lambda}}{\Gamma(\alpha+x)} 318 | \end{aligned} 319 | \end{equation} 320 | 所以, $\lambda|x \sim \Gamma(\lambda|\alpha+x,\beta+1)$ 321 | \end{proof} 322 | 323 | \item {\hei 公交站起点站等可能发出$a,b$两班汽车，其中$a$停$m$站，$b$停$n$站，车上人数服从参数为$\lambda$的泊松分布，每名乘客在各站下车的概率相同，如果该站没有乘客下车，则公交车不停站。求一辆从起点站开出的公交车停站的期望。} 324 | \begin{proof}[解] 325 | 一辆从起点站开出的公交车停站的次数$X$的期望满足： 326 | \begin{equation} 327 | E(X)=\frac{1}{2}E(X_a)+\frac{1}{2}E(X_b) 328 | \end{equation} 329 | 其中，$X_a$表示如果开出的是$a$车停站的次数，$X_b$表示如果开出的是$b$车停站的次数。对$a$车和$b$车的讨论类似，我们这里这讨论$a$车的情况。我们有： 330 | \begin{equation} 331 | X_a=X_{a1}+X_{a2}+\cdots+X_{am},\quad X_{at}=\left\{ 332 | \begin{array}{cc} 333 | 1 & \mbox{第$t$站有人下车}\\ 334 | 0 & \mbox{第$t$站没有人下车} 335 | \end{array} 336 | \right. 337 | \end{equation} 338 | 由于人数$K$服从参数为$\lambda$的泊松分布，则有： 339 | \begin{equation} 340 | P(X_{at}=1|K=k)=1-P(X_{at}=0)=1-\left(1-\frac{1}{m}\right)^k 341 | \end{equation} 342 | 所以： 343 | \begin{equation} 344 | \begin{aligned} 345 | E(X_{at}|K=k)&=1\times P(X_{at}=1|K=k)+0\times P(X_{at}=0|K=k) \\ 346 | &=1-\left(1-\frac{1}{m}\right)^k 347 | \end{aligned} 348 | \end{equation} 349 | 因此： 350 | \begin{equation} 351 | \begin{aligned} 352 | E(X_{at})&=E(E(X_{at}|K=k)) \\ 353 | &=\sum_{k=0}^{+\infty}E(X_{at}|K=k)P(K=k) \\ 354 | &=\sum_{k=0}^{+\infty}\left(1-\left(1-\frac{1}{m}\right)^k\right)\frac{\lambda^ke^{-\lambda}}{k!} \\ 355 | &=1-e^{-\frac{\lambda}{m}} 356 | \end{aligned} 357 | \end{equation} 358 | 所以我们可以得到： 359 | \begin{equation} 360 | \begin{aligned} 361 | E(X_a)&=E(X_{a1}+X_{a2}+\cdots+X_{am}) \\ 362 | &=E(X_{a1})+E(X_{a2})+\cdots+E(X_{am}) \\ 363 | &=m\left(1-e^{-\frac{\lambda}{m}}\right) 364 | \end{aligned} 365 | \end{equation} 366 | 同理我们可以得到： 367 | \begin{equation} 368 | E(X_b)=n\left(1-e^{-\frac{\lambda}{n}}\right) 369 | \end{equation} 370 | 所以： 371 | \begin{equation} 372 | E(X)=\frac{1}{2}\left(m\left(1-e^{-\frac{\lambda}{m}}\right)+n\left(1-e^{-\frac{\lambda}{n}}\right)\right) 373 | \end{equation} 374 | \end{proof} 375 | 376 | \item {\hei 设随机变量$X,Y$均服从均值为$0$，方差为$1$的正态分布，且相互独立，试求：$E\left[X^2+Y^2|\cos\left(\frac{X}{Y}\right)\right]$} 377 | \begin{proof}[解] 378 | 根据课后习题中的结论，我们有： 379 | \begin{equation} 380 | X^2+Y^2\mbox{与}\frac{X}{Y}\mbox{相互独立} 381 | \end{equation} 382 | 因此， 383 | \begin{equation} 384 | X^2+Y^2\mbox{与}\cos\left(\frac{X}{Y}\right)\mbox{相互独立} 385 | \end{equation} 386 | 所以： 387 | \begin{equation} 388 | \begin{aligned} 389 | E\left[X^2+Y^2|\cos\left(\frac{X}{Y}\right)\right]&=E\left[X^2+Y^2\right] \\ 390 | &=E\left[X^2\right]+E\left[Y^2\right] \\ 391 | &=2 392 | \end{aligned} 393 | \end{equation} 394 | \end{proof} 395 | 396 | \item {\hei 篮球赛时长$n$分钟，某球员每一分钟有一次投篮机会，投中概率为$p$，并且教练规定，若一次投篮不中，下一分钟不得投篮，需将机会交给队友，请计算一场球赛中该队员投中次数的均值} 397 | \begin{proof}[解] 398 | 该队员投中次数$X$满足 399 | \begin{equation} 400 | X=C_1+C_2+\cdots+C_n,\quad C_k=\left\{ 401 | \begin{array}{cc} 402 | 1 & \mbox{第$k$分钟投中}\\ 403 | 0 & \mbox{第$k$分钟未投或者未投中} 404 | \end{array} 405 | \right. 406 | \end{equation} 407 | 引入随机变量$Y_k(k=1,2,\cdots,n)$: 408 | \begin{equation} 409 | Y_k=\left\{ 410 | \begin{array}{cc} 411 | 1 & \mbox{第$k$分钟投了}\\ 412 | 0 & \mbox{第$k$分钟未投} 413 | \end{array} 414 | \right. 415 | \end{equation} 416 | 我们有： 417 | \begin{equation} 418 | \begin{aligned} 419 | P(Y_k=1)&=P(Y_k=1|Y_{k-1}=1)P(Y_{k-1}=1)+P(Y_k=1|Y_{k-1}=0)P(Y_{k-1}=0) \\ 420 | &=p\times P(Y_{k-1}=1)+1\times P(Y_{k-1}=0) \\ 421 | &=p\times P(Y_{k-1}=1)+1- P(Y_{k-1}=1) 422 | \end{aligned} 423 | \end{equation} 424 | 由于$P(Y_1=1)=1$，求解上述递推式我们可以得到： 425 | \begin{equation} 426 | P(Y_k=1)=\frac{1}{2-p}\left[1-(p-1)^k\right] 427 | \end{equation} 428 | 因此： 429 | \begin{equation} 430 | \begin{aligned} 431 | P(C_k=1)&=P(C_k=1|Y_k=1)P(Y_k=1)+P(C_k=1|Y_k=0)P(Y_k=0) \\ 432 | &=p\times P(Y_k=1) + 0\times P(Y_k=0) \\ 433 | &=\frac{p}{2-p}\left[1-(p-1)^k\right] 434 | \end{aligned} 435 | \end{equation} 436 | 所以： 437 | \begin{equation} 438 | \begin{aligned} 439 | E(C_k)&=1\times P(C_k=1) + 0\times P(C_k=0) \\ 440 | &=P(C_k=1) \\ 441 | &=\frac{p}{2-p}\left[1-(p-1)^k\right] 442 | \end{aligned} 443 | \end{equation} 444 | 从而\begin{equation} 445 | \begin{aligned} 446 | E(X)&=E(C_1+C_2+\cdots+C_n)\\ 447 | &=E(C_1)+E(C_2)+\cdots+E(C_n) \\ 448 | &=\sum_{k=1}^{n}\frac{p}{2-p}\left[1-(p-1)^k\right] \\ 449 | &=\frac{np}{2-p}-\frac{p(p-1)}{(2-p)^2}+\frac{p(p-1)^{n+1}}{(2-p)^2} 450 | \end{aligned} 451 | \end{equation} 452 | \end{proof} 453 | 454 | \item {\hei 设$U_1,U_2,\cdots$为一相互独立的$(0,1)$上均匀分布的随机变量序列，求$E[N]$。其中： 455 | \begin{equation*} 456 | N=\min\left\{n:\sum_{i=1}^nU_i>1\right\} 457 | \end{equation*}} 458 | \begin{proof}[解] 459 | 我们将得到一个更一般的结果。对于$x\in [0,1]$，令： 460 | \begin{equation} 461 | N(x)=\min\left\{n:\sum_{i=1}^nU_i>x\right\} 462 | \end{equation} 463 | 再令： 464 | \begin{equation} 465 | m(x)=E[N(x)] 466 | \end{equation} 467 | 即$N(x)$是部分和$\sum_{i=1}^{n}U_i$超过$x$的最小指标$n$，$m(x)$是$N(x)$的期望值。将$U_1$作为条件，我们有： 468 | \begin{equation} 469 | m(x)=E\left[E[N(x)|U_1=y]\right]=\int_{0}^{1}E[N(x)|U_1=y]dy 470 | \end{equation} 471 | 对于条件期望$E[N(x)|U_1=y]$，我们有： 472 | \begin{equation} 473 | E[N(x)|U_1=y]=\left\{ 474 | \begin{array}{cc} 475 | 1, & y>x \\ 476 | 1+m(x-y), & y\le x 477 | \end{array} 478 | \right. 479 | \end{equation} 480 | 代入上式可得： 481 | \begin{equation} 482 | \begin{aligned} 483 | m(x)&=1+\int_{0}^xm(x-y)dy \\ 484 | &=1+\int_{0}^xm(u)du\qquad(\mbox{做变量替换}u=x-y) 485 | \end{aligned} 486 | \end{equation} 487 | 对上式两边求微分可以得到： 488 | \begin{equation} 489 | m'(x)=m(x) 490 | \end{equation} 491 | 求解上述微分方程可以得到： 492 | \begin{equation} 493 | m(x)=ke^x 494 | \end{equation} 495 | 又由于$m(0)=1$可以得到$k=1$，因此： 496 | \begin{equation} 497 | m(x)=e^x 498 | \end{equation} 499 | 所以，原问题$E[N]=m(1)=e$ 500 | \end{proof} 501 | 502 | \item {\hei (平面上的随机徘徊)设在平面坐标系的原点放一质点，质点在平面上作如下的随机徘徊。 503 | \begin{enumerate}[1)] 504 | \item 每一步质点移动一个单位的距离，且前进方向与$x$轴的夹角$\theta$在$(0,2\pi)$上均匀分布； 505 | \item 每一步质点移动一个单位的距离，前进方向只有上、下、左、右四种情况且概率相等。 506 | \end{enumerate} 507 | 假设每秒运动一次，以第$n$秒时质点所在位置与原点的距离为半径画圆，请计算两种情况下圆面积的均值。} 508 | \begin{proof}[解] 509 | 用$(X_i,Y_i)$表示第$i$秒坐标的变化量， 510 | \begin{enumerate}[1)] 511 | \item 对于此问，我们有： 512 | \begin{equation} 513 | X_i=\cos \theta_i\qquad Y_i=\sin \theta_i 514 | \end{equation} 515 | 其中$\theta_i,i=1,2,\cdots,n$为相互独立且在$(0,2\pi)$上均匀分布的随机变量。经过$n$秒之后，质点的位置为$\left(\sum_{i=1}^nX_i,\sum_{i=1}^nY_i\right)$，则质点所在位置与原点距离的平方$R^2$可以计算如下： 516 | \begin{equation} 517 | \begin{aligned} 518 | R^2&=\left(\sum_{i=1}^nX_i\right)^2+\left(\sum_{i=1}^nY_i\right)^2 \\ 519 | &=\sum_{i=1}^n\left(X_i^2+Y_i^2\right)+\sum_{i=1}^n\sum_{j\ne i}\left(X_iX_j+Y_iY_j\right) \\ 520 | &=n+\sum_{i=1}^n\sum_{j\ne i}\left(\cos \theta_i\cos \theta_j + \sin \theta_i\sin\theta_j\right) 521 | \end{aligned} 522 | \end{equation} 523 | 由于$\theta_i与\theta_j$相互独立并且： 524 | \begin{equation} 525 | \begin{aligned} 526 | E[\cos\theta_i]&=\frac{1}{2\pi}\int_{0}^{2\pi}\cos\theta_i\dif \theta_i=0 \\ 527 | E[\sin\theta_i]&=\frac{1}{2\pi}\int_{0}^{2\pi}\sin\theta_i\dif \theta_i=0 528 | \end{aligned} 529 | \end{equation} 530 | 因此，对于圆面积$S=\pi R^2$，我们有： 531 | \begin{equation} 532 | E[S]=\pi E[R^2]=n\pi 533 | \end{equation} 534 | \item 对于此问，质点每一步的移动都是上、下、左、右四个方向之一，因此我们可以得到$X_i$和$Y_i$的联合分布为如表\ref{tab1}所示。 535 | \begin{table}[!h] 536 | \centering 537 | \caption{\label{tab1}$X,Y$的联合分布} 538 | \begin{tabular}{|c|c|c|c|} 539 | \hline 540 | \diagbox{X}{P}{Y} & 0 & 1 & -1 \\ 541 | \hline 542 | 0 & 0 & $\frac{1}{4}$ & $\frac{1}{4}$ \\ 543 | \hline 544 | 1 & $\frac{1}{4}$ & 0 & 0 \\ 545 | \hline 546 | -1 & $\frac{1}{4}$ & 0 & 0 \\ 547 | \hline 548 | \end{tabular} 549 | \end{table} 550 | 551 | 经过$n$秒之后，质点的位置为$\left(\sum_{i=1}^nX_i,\sum_{i=1}^nY_i\right)$，则质点所在位置与原点距离的平方$R^2$可以计算如下： 552 | \begin{equation} 553 | \begin{aligned} 554 | R^2&=\left(\sum_{i=1}^nX_i\right)^2+\left(\sum_{i=1}^nY_i\right)^2 \\ 555 | &=\sum_{i=1}^n\left(X_i^2+Y_i^2\right)+\sum_{i=1}^n\sum_{j\ne i}\left(X_iX_j+Y_iY_j\right) \\ 556 | &=n+\sum_{i=1}^n\sum_{j\ne i}\left(X_iX_j+Y_iY_j\right) 557 | \end{aligned} 558 | \end{equation} 559 | 下面我们计算$E[X_iX_j]$: 560 | \begin{equation} 561 | \begin{aligned} 562 | E[X_iX_j]&=0\times P(X_iX_j=0)+(-1)\times P(X_iX_j=-1)+1\times P(X_iX_j=1) \\ 563 | &=0\times\frac{3}{4}+(-1)\times\frac{1}{8}+1\times\frac{1}{8} \\ 564 | &=0 565 | \end{aligned} 566 | \end{equation} 567 | 同理我们有$E[Y_iY_j]=0$，因此因此，对于圆面积$S=\pi R^2$，我们有： 568 | \begin{equation} 569 | E[S]=\pi E[R^2]=n\pi 570 | \end{equation} 571 | \end{enumerate} 572 | 综上所述，两种情况下圆面积的均值均为$n\pi$ 573 | \end{proof} 574 | 575 | \item {\hei 已知$(X_1,X_2)\sim \mathcal{N}(0,0,\sigma_1^2,\sigma_2^2,\rho)$，设$Y_1=\cos X_1,Y_2=\cos X_2$，试求$Y_1,Y_2$的协方差} 576 | \begin{proof}[解] 577 | 首先我们证明一个结论，方便之后的计算。对于任意的正态分布$X\sim\mathcal{N}(0,\sigma^2)$ 578 | \begin{equation} 579 | \begin{aligned} 580 | g_X(t)=E(e^{jtX})&=\int_{-\infty}^{+\infty}e^{jtx}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}\dif x \\ 581 | &=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{+\infty}e^{-\frac{x^2-2\sigma^2jtx}{2\sigma^2}}\dif x \\ 582 | &\mbox{令}y=x-\sigma^2jt \\ 583 | &=e^{-\frac{1}{2}\sigma^2t^2}\cdot\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{+\infty}e^{-\frac{y^2}{2\sigma^2}}\dif y \\ 584 | &=e^{-\frac{1}{2}\sigma^2t^2} 585 | \end{aligned} 586 | \end{equation} 587 | 利用此结论，计算$E[Y_1]$如下： 588 | \begin{equation} 589 | \label{eq1} 590 | \begin{aligned} 591 | E[Y_1]=E[\cos X_1]&=E\left[\frac{e^{jX_1}+e^{-jX_1}}{2}\right] \\ 592 | &=\frac{1}{2}E\left[e^{jX_1}\right]+\frac{1}{2}E\left[e^{-jX_1}\right] \\ 593 | &=\frac{1}{2}g_{X_1}(1)+\frac{1}{2}g_{X_1}(-1) \\ 594 | &=\frac{1}{2}e^{-\frac{1}{2}\sigma_1^2}+\frac{1}{2}e^{-\frac{1}{2}\sigma_1^2} \\ 595 | &=e^{-\frac{1}{2}\sigma_1^2} 596 | \end{aligned} 597 | \end{equation} 598 | 同理我们可以得到$E[Y_2]$: 599 | \begin{equation} 600 | E[Y_2]=e^{-\frac{1}{2}\sigma_2^2} 601 | \end{equation} 602 | 下面计算$E[Y_1Y_2]$: 603 | \begin{equation} 604 | \label{eq2} 605 | \begin{aligned} 606 | E[Y_1Y_2]&=E[\cos X_1\cos X_2] \\ 607 | &=E\left[\frac{1}{2}\left(\cos(X_1+X_2)+\cos(X_1-X_2)\right)\right] \\ 608 | &=\frac{1}{2}E\left[\cos(X_1+X_2)\right]+\frac{1}{2}E\left[\cos(X_1-X_2)\right] 609 | \end{aligned} 610 | \end{equation} 611 | 可以证明： 612 | \begin{equation} 613 | \begin{aligned} 614 | X_1+X_2 &\sim \mathcal{N}(0,\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2) \\ 615 | X_1-X_2 &\sim \mathcal{N}(0,\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2) 616 | \end{aligned} 617 | \end{equation} 618 | 所以根据式(\ref{eq1})的计算方式可以得到： 619 | \begin{equation} 620 | \begin{aligned} 621 | E\left[\cos(X_1+X_2)\right]&=e^{-\frac{1}{2}(\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2)} \\ 622 | E\left[\cos(X_1-X_2)\right]&=e^{-\frac{1}{2}(\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2)} \\ 623 | \end{aligned} 624 | \end{equation} 625 | 代入式(\ref{eq2})得： 626 | \begin{equation} 627 | E[Y_1Y_2]=\frac{1}{2}e^{-\frac{1}{2}(\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2)}+\frac{1}{2}e^{-\frac{1}{2}(\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2)} 628 | \end{equation} 629 | 所以$Y_1,Y_2$的协方差为： 630 | \begin{equation} 631 | \begin{aligned} 632 | Cov(Y_1,Y_2)&=E[Y_1Y_2]-E[Y_1]E[Y_2] \\ 633 | &=\frac{1}{2}e^{-\frac{1}{2}(\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2)}+\frac{1}{2}e^{-\frac{1}{2}(\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2)}-e^{-\frac{1}{2}\sigma_1^2}\cdot e^{-\frac{1}{2}\sigma_2^2} \\ 634 | &=\frac{1}{2}e^{-\frac{1}{2}(\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2)}+\frac{1}{2}e^{-\frac{1}{2}(\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2)}-e^{-\frac{1}{2}(\sigma_1^2+\sigma_2^2)} 635 | \end{aligned} 636 | \end{equation} 637 | \end{proof} 638 | 639 | 640 | \end{enumerate} 641 | 642 | 643 | \end{document} 644 | --------------------------------------------------------------------------------