POINT
【関連記事】
- 確率分布の期待値・分散
ベルヌーイ分布
母比率（信頼区間・検定） - Notes_JP\begin{aligned}
\begin{cases}
\, P(X = 1) = p \\
\, P(X = 0) = 1 - p
\end{cases}
\end{aligned}
\begin{cases}
\, P(X = 1) = p \\
\, P(X = 0) = 1 - p
\end{cases}
\end{aligned}
期待値
\begin{aligned}
&\mu = E[X] \\
& = \sum_{x \in \{0,1\}} x P(X = x) \\
&= 0\cdot(1-p) + 1\cdot p \\
&=p
\end{aligned}
&\mu = E[X] \\
& = \sum_{x \in \{0,1\}} x P(X = x) \\
&= 0\cdot(1-p) + 1\cdot p \\
&=p
\end{aligned}
分散
\begin{aligned}
V[X]
&= E[(X - \mu)^{2}] \\
&= E[X^{2}] -2\mu E[X] + E[X]^{2} \\
&= E[X^{2}] -\mu^{2}
\end{aligned}
を使えば，V[X]
&= E[(X - \mu)^{2}] \\
&= E[X^{2}] -2\mu E[X] + E[X]^{2} \\
&= E[X^{2}] -\mu^{2}
\end{aligned}
\begin{aligned}
&E[X^{2}] \\
& = \sum_{x \in \{0,1\}} x^{2} P(X = x) \\
&= 0\cdot(1-p) + 1\cdot p \\
&=p
\end{aligned}
より&E[X^{2}] \\
& = \sum_{x \in \{0,1\}} x^{2} P(X = x) \\
&= 0\cdot(1-p) + 1\cdot p \\
&=p
\end{aligned}
\begin{aligned}
V[X]
&= p - p^{2} \\
&= p(1-p)
\end{aligned}
となる．V[X]
&= p - p^{2} \\
&= p(1-p)
\end{aligned}
二項分布
$X\sim B(n, p)$\begin{aligned}
& P(X = x) =
\binom{n}{x}
p^{x} (1 - p)^{n - x} \\
&(x = 0, 1, ..., n)
\end{aligned}
& P(X = x) =
\binom{n}{x}
p^{x} (1 - p)^{n - x} \\
&(x = 0, 1, ..., n)
\end{aligned}
$X_{i}\sim Ber(p)$とし，$\{X_{i}\}_{i}$が互いに独立だとすると
\begin{aligned}
X = \sum_{i=1}^{n} X_{i} \sim B(n, p)
\end{aligned}
X = \sum_{i=1}^{n} X_{i} \sim B(n, p)
\end{aligned}
期待値
$\{X_{i}\}_{i}$が互いに独立で$X_{i}\sim Ber(p)$とする．$ E[X_{i}] = p$だから，$X = \sum_{i=1}^{n} X_{i} \sim B(n, p)$の期待値は\begin{aligned}
E[X]
&= np
\end{aligned}
E[X]
&= np
\end{aligned}
\begin{aligned}
E[X]
&= \sum_{x=0}^{n} x P(X = x) \\
&= \sum_{x=0}^{n} x
\binom{n}{x}
p^{x} (1 - p)^{n - x} \\
&= \sum_{x=0}^{n} \binom{n}{x}
p \frac{\mathrm{d} p^{x} }{\mathrm{d} p} (1 - p)^{n - x} \\
&= p \frac{\mathrm{d} }{\mathrm{d} p}
\biggl[\underbrace{\sum_{x=0}^{n} \binom{n}{x} p^{x} q^{n - x}}_{=(p + q)^{n}}\biggr] \Biggl|_{p + q = 1} \\
&= np
\end{aligned}
E[X]
&= \sum_{x=0}^{n} x P(X = x) \\
&= \sum_{x=0}^{n} x
\binom{n}{x}
p^{x} (1 - p)^{n - x} \\
&= \sum_{x=0}^{n} \binom{n}{x}
p \frac{\mathrm{d} p^{x} }{\mathrm{d} p} (1 - p)^{n - x} \\
&= p \frac{\mathrm{d} }{\mathrm{d} p}
\biggl[\underbrace{\sum_{x=0}^{n} \binom{n}{x} p^{x} q^{n - x}}_{=(p + q)^{n}}\biggr] \Biggl|_{p + q = 1} \\
&= np
\end{aligned}
分散
$\{X_{i}\}_{i}$が互いに独立で$X_{i}\sim Ber(p)$とする．$ V[X_{i}] = p(1-p)$だから，$X = \sum_{i=1}^{n} X_{i} \sim B(n, p)$の分散は\begin{aligned}
V[X] &= \sum_{i=1}^{n} V[X_{i}] \\
&= np(1-p)
\end{aligned}
V[X] &= \sum_{i=1}^{n} V[X_{i}] \\
&= np(1-p)
\end{aligned}
幾何分布
確率$p$で成功する独立な試行を，成功するまで繰り返すとき，はじめて成功するときの回数を$X$とする．\begin{aligned}
& P(X = x)
= (1- p)^{x - 1} p \\
&(x=1,2,...)
\end{aligned}
& P(X = x)
= (1- p)^{x - 1} p \\
&(x=1,2,...)
\end{aligned}
注：初めて成功するまでに失敗した回数を$X$とすることもある
【関連記事】事象が起こるまでの試行回数 - Notes_JP
