幾何分布

tags: probability-theory

確率関数

$$p\left(x\right)={(1-p)}^{x-1}p$$

平均

$$\begin{array}{rl}\sum _{x=1}^{\mathrm{\infty }}x{(1-p)}^{x-1}p& =\sum _{x=1}^{\mathrm{\infty }}\sum _{y=x}^{\mathrm{\infty }}{(1-p)}^{y-1}p\\ & =\sum _{x=1}^{\mathrm{\infty }}p\frac{{(1-p)}^{x-1}-0}{p}\\ & =\sum _{x=1}^{\mathrm{\infty }}{(1-p)}^{x-1}\\ & =\frac{1-0}{p}\\ & =\frac{1}{p}\end{array}$$

$x{(1-p)}^{x-1}$が

$-{(1-p)}^x$の微分であることを見つければ、次の計算も可能である。

$$\begin{array}{rl}\sum _{x=1}^{\mathrm{\infty }}x{(1-p)}^{x-1}p& =p{\{-\sum _{x=0}^{\mathrm{\infty }}{(1-p)}^x\}}^{\prime }\\ & =p{\left\{\frac{-1+p}{p}\right\}}^{\prime }\\ & =p{\{\frac{-1}{p}+1\}}^{\prime }\\ & =\frac{p}{p^2}\\ & =\frac{1}{p}\end{array}$$

分散

$x(x-1){(1-p)}^{x-2}$が

${(1-p)}^x$の2階微分であることを見つければ、次の計算ができる。

$$\begin{array}{rl}\sum _{x=1}^{\mathrm{\infty }}x(x-1){(1-p)}^{x-1}p& =\sum _{x=2}^{\mathrm{\infty }}x(x-1){(1-p)}^{x-1}p\\ & =(1-p)\sum _{x=2}^{\mathrm{\infty }}x(x-1){(1-p)}^{x-2}p\\ & =p(1-p){\left\{\sum _{x=0}^{\mathrm{\infty }}{(1-p)}^x\right\}}^{\prime \prime }\\ & =p(1-p){\left\{\frac{1-p}{p}\right\}}^{\prime \prime }\\ & =p(1-p){\{\frac{1}{p}-1\}}^{\prime \prime }\\ & =\frac{2p(1-p)}{p^3}\\ & =\frac{2(1-p)}{p^2}\end{array}$$

これより

$$V\left[X\right]=\frac{2(1-p)}{p^2}+\frac{1}{p}-\frac{1}{p^2}=\frac{1-p}{p^2}$$

モーメント母関数

$$\begin{array}{rl}\sum _{x=1}^{\mathrm{\infty }}{e}^{tx}{(1-p)}^{x-1}p& =e^t\sum _{x=1}^{\mathrm{\infty }}{\left\{e^t(1-p)\right\}}^{x-1}p\\ & =e^t\sum _{x=1}^{\mathrm{\infty }}{\left\{e^t(1-p)\right\}}^{x-1}p\\ & =p{e}^t\frac{1-0}{1-e^t(1-p)}\\ & =\frac{p{e}^t}{1-(1-p)e^t}\\ & =\frac{p}{e^{-t}-1+p}\end{array}$$