# 幾何分布
###### tags: `probability-theory`
## 確率関数

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

## 平均

$$
\begin{align}
\sum_{x=1}^{\infty} x \left(1-p\right)^{x-1} p
&= \sum_{x=1}^{\infty} \sum_{y=x}^{\infty} \left(1-p\right)^{y-1} p \notag \\
&= \sum_{x=1}^{\infty} p \frac{\left(1-p\right)^{x-1}-0}{p} \notag \\
&= \sum_{x=1}^{\infty} \left(1-p\right)^{x-1} \notag \\
&= \frac{1-0}{p} \notag \\
&= \frac{1}{p}
\end{align}
$$

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

$$
\begin{align}
\sum_{x=1}^{\infty} x \left(1-p\right)^{x-1} p
&= p\left\{-\sum_{x=0}^{\infty} \left(1-p\right)^x \right\}^{\prime} \notag \\
&= p\left\{\frac{-1+p}{p}\right\}^{\prime} \notag \\
&= p\left\{\frac{-1}{p}+1\right\}^{\prime} \notag \\
&= \frac{p}{p^2} \notag \\
&= \frac{1}{p}
\end{align}
$$

## 分散

$x\left(x-1\right)\left(1-p\right)^{x-2}$が$\left(1-p\right)^x$の2階微分であることを見つければ、次の計算ができる。

$$
\begin{align}
\sum_{x=1}^{\infty} x \left(x-1\right)\left(1-p\right)^{x-1} p
&= \sum_{x=2}^{\infty} x \left(x-1\right)\left(1-p\right)^{x-1} p \notag \\
&= \left(1-p\right)\sum_{x=2}^{\infty} x \left(x-1\right)\left(1-p\right)^{x-2} p \notag \\
&= p\left(1-p\right)\left\{\sum_{x=0}^{\infty} \left(1-p\right)^x \right\}^{\prime\prime} \notag \\
&= p\left(1-p\right)\left\{\frac{1-p}{p}\right\}^{\prime\prime} \notag \\
&= p\left(1-p\right)\left\{\frac{1}{p}-1\right\}^{\prime\prime} \notag \\
&= \frac{2p\left(1-p\right)}{p^3} \notag \\
&= \frac{2\left(1-p\right)}{p^2}
\end{align}
$$

これより

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

## モーメント母関数

$$
\begin{align}
\sum_{x=1}^{\infty} e^{tx}\left(1-p\right)^{x-1} p
&= e^t \sum_{x=1}^{\infty} \left\{e^t\left(1-p\right)\right\}^{x-1} p \notag \\
&= e^t \sum_{x=1}^{\infty} \left\{e^t\left(1-p\right)\right\}^{x-1} p \notag \\
&= p e^t \frac{1-0}{1-e^t\left(1-p\right)} \notag \\
&= \frac{p e^t}{1-\left(1-p\right)e^t} \notag \\
&= \frac{p}{e^{-t}-1+p} \notag \\
\end{align}
$$