{"id":1874,"date":"2020-07-12T11:00:00","date_gmt":"2020-07-12T02:00:00","guid":{"rendered":"https:\/\/develop-chronos.com\/statistics-top\/statistics\/?p=1874"},"modified":"2020-08-11T11:12:28","modified_gmt":"2020-08-11T02:12:28","slug":"dgeod-mv","status":"publish","type":"post","link":"https:\/\/develop-chronos.com\/statistics-top\/statistics\/dgeod-mv","title":{"rendered":"\u5e7e\u4f55\u5206\u5e03\u306e\u671f\u5f85\u5024\u30fb\u5206\u6563\u306e\u6c42\u3081\u65b9\u3010\u8a3c\u660e\u4ed8\u304d\u3067\u89e3\u8aac\u3011"},"content":{"rendered":"\n

\u5b66\u7fd2\u30ec\u30d9\u30eb\uff1a\u5927\u5b66\u751f\u3000\u96e3\u6613\u5ea6\uff1a\u2605\u2605\u2606\u2606\u2606<\/span><\/p>\n\n\n\n\n\n\n\n

\u3053\u306e\u8a18\u4e8b\u3067\u306f\u5e7e\u4f55\u5206\u5e03\u306e\u671f\u5f85\u5024\u30fb\u5206\u6563\u3092\u8a3c\u660e\u4ed8\u304d\u3067\u89e3\u8aac\u3057\u3066\u3044\u304d\u307e\u3059\u3002\u671f\u5f85\u5024\u30fb\u5206\u6563\u306e\u6c42\u3081\u65b9\u304c\u5206\u304b\u3089\u306a\u3044\u65b9\u306f\u662f\u975e\u304a\u8aad\u307f\u304f\u3060\u3055\u3044\u3002\u305d\u306e\u4ed6\u306e\u5e7e\u4f55\u5206\u5e03\u306e\u57fa\u672c\u60c5\u5831\u306f\uff1c\u5e7e\u4f55\u5206\u5e03\uff1e<\/a>\u306e\u8a18\u4e8b\u3092\u304a\u8aad\u307f\u304f\u3060\u3055\u3044\u3002<\/p>\n\n\n\n

\u3000<\/p>\n\n\n\n

\u5e7e\u4f55\u5206\u5e03\u306e\u671f\u5f85\u5024\u30fb\u5206\u6563<\/h2>\n\n\n\n
\u671f\u5f85\u5024\u3068\u5206\u6563<\/strong><\/div>\n
\u5e7e\u4f55\u5206\u5e03\\(X\\sim Geo(p)\\)\u306b\u5f93\u3046\u78ba\u7387\u5909\u6570\u306e\u671f\u5f85\u5024\u30fb\u5206\u6563\u306f\u6b21\u306e\u3088\u3046\u306b\u306a\u308a\u307e\u3059\u3002\n
\\begin{align}\n\\mathrm{E}[X]=\\frac{1}{p},\\ \\ \\ \\mathrm{Var}[X]=\\frac{1-p}{p^{2}}\n\\end{align}<\/div>\n<\/div>\n\n\n\n

\u671f\u5f85\u5024\u30fb\u5206\u6563\u3092\u6c42\u3081\u308b\u969b\u306b\u306f\uff1c\u671f\u5f85\u5024\u306e\u5b9a\u7fa9\uff1e<\/a>\u304a\u3088\u3073\uff1c\u5206\u6563\u306e\u5b9a\u7fa9\uff1e<\/a>\u3092\u4f7f\u7528\u3059\u308b\u306e\u3067\u3001\u899a\u3048\u3066\u3044\u306a\u3044\u65b9\u306f\u8a3c\u660e\u3092\u8aad\u3080\u524d\u306b\u4e00\u5ea6\u3001\u76ee\u3092\u901a\u3057\u3066\u304a\u3044\u3066\u304f\u3060\u3055\u3044\u3002<\/p>\n\n\n\n

\u3000<\/p>\n\n\n\n

\u3000<\/p>\n\n\n\n

\u8a3c\u660e<\/h4>\n\n\n\n

\u78ba\u7387\u5909\u6570\u304c\\(X\\sim Geo(p)\\)\u306b\u5f93\u3063\u3066\u3044\u308b\u3068\u3057\u307e\u3059\u3002\u3053\u306e\u3068\u304d\u3001\\(X\\)\u306e\u78ba\u7387\u95a2\u6570\u306f

\\begin{align}
f(x)= p(1-p)^{x-1}
\\end{align}<\/div>
\u3068\u306a\u308a\u307e\u3059\u3002\u3053\u306e\u3053\u3068\u306f<\/a>\uff1c\u5e7e\u4f55\u5206\u5e03\u306e\u57fa\u672c\u60c5\u5831\uff1e<\/a>\u3092\u304a\u8aad\u307f\u304f\u3060\u3055\u3044\u3002
\u3000\u307e\u305a\u3001\u671f\u5f85\u5024\u3092\u6c42\u3081\u3066\u3044\u304d\u307e\u3059\u3002\u307e\u305a\u3001\u671f\u5f85\u5024\u306e\u5b9a\u7fa9\u304b\u3089
\\begin{align}
\\mathrm{E}[X] &= \\sum_{x=0}^{\\infty}xf(x) \\\\
&= \\sum_{x=0}^{\\infty}x\\cdot\\ p(1-p)^{x-1} \\\\
&= p\\sum_{x=0}^{\\infty}x\\cdot\\ (1-p)^{x-1} \\\\
\\end{align}<\/div>
\u3068\u306a\u308a\u307e\u3059\u3002\u3053\u3053\u3067\\(1\/(1-x)\\)\u306b\u3064\u3044\u3066\u30de\u30af\u30ed\u30fc\u30ea\u30f3\u5c55\u958b\u3092\u884c\u3044\u307e\u3059\u3002\u30de\u30af\u30ed\u30fc\u30ea\u30f3\u5c55\u958b\u306f\u95a2\u6570\\(f(x)\\)\u306b\u5bfe\u3057\u3066
\\begin{align}
f(x) = \\sum_{k=0}^{\\infty} \\frac{f^{(k)}(0)}{k!}x^{k}
\\end{align}<\/div>
\u3068\u3044\u3046\u5f62\u306b\u5909\u5f62\u3059\u308b\u3053\u3068\u3067\u3059\u3002\u3053\u3053\u3067\\(f^{(i)}(x)\\)\u306f\\(f(x)\\)\u3092\\(i\\)\u56de\u5fae\u5206\u3057\u305f\u3082\u306e\u3092\u8868\u3057\u307e\u3059\u3002\\(f(x)=1\/(1-x)\\)\u306b\u3064\u3044\u3066\u30de\u30af\u30ed\u30fc\u30ea\u30f3\u5c55\u958b\u3092\u8003\u3048\u308b\u3068\u304d\u3001
\\begin{align}
f^{(1)}(0) = 1,\\ \\ f^{(2)}(0) = 2!,\\ \\ f^{(3)} = 3!,\\ \\ \\cdots\\ \\ ,f^{(n)}(0) = n!
\\end{align}<\/div>
\u3068\u306a\u308b\u3053\u3068\u3092\u7528\u3044\u308b\u3068\u3001
\\begin{align}
\\frac{1}{1-x} &= \\sum_{k=0}^{\\infty} \\frac{k!}{k!}x^{k} = \\sum_{k=0}^{\\infty} x^{k}
\\end{align}<\/div>
\u304c\u6210\u7acb\u3057\u307e\u3059\u3002\u3053\u306e\u5f0f\u306e\u4e21\u8fba\u3092\\(x\\)\u3067\u5fae\u5206\u3059\u308b\u3068
\\begin{align}
\\frac{1}{(1-x)^{2}} &= \\sum_{k=0}^{\\infty} kx^{k-1}
\\end{align}<\/div>
\u3068\u306a\u308a\u307e\u3059\u3002\u3053\u306e\u5f0f\u306b\u304a\u3044\u3066\\(x=1-p,\\ x=k\\)\u3067\u7f6e\u304d\u63db\u3048\u308b\u3068\u3001
\\begin{align}
\\frac{1}{p^{2}} &= \\sum_{x=0}^{\\infty} x(1-p)^{x-1}
\\end{align}<\/div>
\u3068\u306a\u308b\u3053\u3068\u304b\u3089\u3001\u6c42\u3081\u305f\u3044\u671f\u5f85\u5024\u306f
\\begin{align}
\\mathrm{E}[X] = p\\cdot \\frac{1}{p^{2}} = \\frac{1}{p}
\\end{align}<\/div>
\u3068\u306a\u308a\u307e\u3059\u3002\u5c11\u3005\u30c6\u30af\u30cb\u30ab\u30eb\u306a\u6c42\u3081\u65b9\u3067\u3059\u304c\u3001\u300c\u3053\u3093\u306a\u6c42\u3081\u65b9\u3082\u3042\u308b\u3093\u3060\u300d\u3068\u3044\u3046\u7a0b\u5ea6\u306b\u8a8d\u8b58\u3057\u3066\u304a\u304f\u3050\u3089\u3044\u3067\u3044\u3044\u3068\u601d\u3044\u307e\u3059\u3002
\u3000\u3042\u3068\u306f\u5206\u6563\u3092\u6c42\u3081\u307e\u3057\u3087\u3046\u3002\uff1c\u5206\u6563\u306e\u5b9a\u7fa9\uff1e<\/a>\u306e\u8a18\u4e8b\u304b\u3089\u5206\u6563\u306f
\\begin{align}
\\mathrm{Var}[X] &= \\mathrm{E}[X^{2}]-\\mathrm{E}[X]^{2}
\\end{align}<\/div>
\u3068\u8868\u3059\u3053\u3068\u304c\u3067\u304d\u308b\u306e\u3067\u3001\\(\\mathrm{E}[X^{2}]\\)\u3092\u6c42\u3081\u308c\u3070\u3088\u3044\u3053\u3068\u304c\u308f\u304b\u308a\u307e\u3059\u3002\u3088\u3063\u3066
\\begin{align}
\\mathrm{E}[X^{2}] &= \\sum_{x=0}^{n}x^{2}f(x) \\\\
&= \\sum_{x=0}^{\\infty}x^{2}\\cdot\\ p(1-p)^{x-1} \\\\
&= p\\sum_{x=0}^{\\infty}x^{2}\\cdot\\ (1-p)^{x-1} \\\\
\\end{align}<\/div>
\u3068\u306a\u308a\u307e\u3059\u3002\u3053\u3053\u3067\\(1\/(1-x)\\)\u306b\u3064\u3044\u3066\u30de\u30af\u30ed\u30fc\u30ea\u30f3\u5c55\u958b\u3092\u884c\u3063\u305f\u7d50\u679c\u3001
\\begin{align}
\\frac{1}{1-x} &= \\sum_{k=0}^{\\infty} \\frac{k!}{k!}x^{k} = \\sum_{k=0}^{\\infty} x^{k}
\\end{align}<\/div>
\u3068\u306a\u308a\u3001\u4e21\u8fba\u3092\u5fae\u5206\u3059\u308b\u3068
\\begin{align}
\\frac{1}{(1-x)^{2}} &= \\sum_{k=0}^{\\infty} kx^{k-1}
\\end{align}<\/div>
\u3068\u306a\u308a\u307e\u3057\u305f\u3002\u3053\u306e\u4e21\u8fba\u306b\\(x\\)\u3092\u304b\u3051\u308b\u3068
\\begin{align}
\\frac{x}{(1-x)^{2}} &= \\sum_{k=0}^{\\infty} kx^{k}
\\end{align}<\/div>
\u304c\u6210\u308a\u7acb\u3061\u307e\u3059\u3002\u3053\u306e\u5f0f\u306e\u4e21\u8fba\u3092\u3082\u3046\u4e00\u5ea6\u5fae\u5206\u3059\u308b\u3068
\\begin{align}
\\frac{x+1}{(1-x)^{3}} &= \\sum_{k=0}^{\\infty} k^{2}x^{k-1}
\\end{align}<\/div>
\u3068\u306a\u308a\u3001\\(x=1-p,\\ x=k\\)\u3067\u7f6e\u304d\u63db\u3048\u308b\u3068\u3001
\\begin{align}
\\frac{2-p}{p^{3}} &= \\sum_{x=0}^{\\infty} x^{2}(1-p)^{x-1}
\\end{align}<\/div>
\u304c\u6210\u308a\u7acb\u3061\u307e\u3059\u3002\u3086\u3048\u306b\u3001\u6c42\u3081\u305f\u3044\u5206\u6563\u306f
\\begin{align}
\\mathrm{Var}[X] = p\\cdot\\frac{2-p}{p^{3}}-\\frac{1}{p^{2}} = \\frac{1-p}{p^{2}}
\\end{align}<\/div>
\u3068\u306a\u308a\u307e\u3059\u3002<\/p>\n\n\n\n

\u25a1<\/p>\n","protected":false},"excerpt":{"rendered":"

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