幾何分布の期待値・分散の導出

（i）期待値
期待値の定義式 $E\left(X\right)=\sum _{x=-\mathrm{\infty }}^{\mathrm{\infty }}x\cdot f\left(x\right)$ より、 $$\begin{array}{rl}E\left(X\right)& =\sum _{x=0}^{\mathrm{\infty }}x\cdot {(1-p)}^{x}p\\ & =p\sum _{x=0}^{\mathrm{\infty }}x\cdot {(1-p)}^x\end{array}$$ 「等差×等比」型の級数の公式 $\left|r\right|>1\Rightarrow \sum _{x=0}^{\mathrm{\infty }}x\cdot r^x=\frac{r}{{(1-r)}^2}$ より、 $$\begin{array}{rl}E\left(X\right)& =p\cdot \frac{1-p}{{\{1-(1-p)\}}^2}\\ & =p\cdot \frac{1-p}{p^2}\\ & =\frac{1-p}{p}\end{array}$$ $◼$

（ii）分散
2乗の期待値の定義式 $E\left(X^2\right)=\sum _{x=-\mathrm{\infty }}^{\mathrm{\infty }}{x}^2\cdot f\left(x\right)$ より、 $$\begin{array}{rl}E\left(X\right)& =\sum _{x=0}^{\mathrm{\infty }}{x}^2\cdot {(1-p)}^{x}p\\ & =p\sum _{x=0}^{\mathrm{\infty }}{x}^2\cdot {(1-p)}^x\end{array}$$ 「等差×等比」型の級数の公式 $\left|r\right|<1\Rightarrow \sum _{x=0}^{\mathrm{\infty }}{x}^2\cdot r^x=\frac{r^2+r}{{(1-r)}^3}$ より、 $$\begin{array}{rl}E\left(X^2\right)& =p\cdot \frac{{(1-p)}^2+(1-p)}{{\{1-(1-p)\}}^3}\\ & =p\cdot \frac{{(1-p)}^2+(1-p)}{p^3}\\ & =\frac{{(1-p)}^2+(1-p)}{p^2}\end{array}$$ 分散の公式 $V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2$ より、 $$\begin{array}{rl}V\left(X\right)& =\frac{{(1-p)}^2+(1-p)}{p^2}-\frac{{(1-p)}^2}{p^2}\\ & =\frac{1-p}{p^2}\end{array}$$ $◼$

（i）期待値
幾何分布の確率母関数の公式より、 $$\begin{array}{r}{G}_X\left(\theta \right)=\frac{p}{1-\theta (1-p)}\end{array}$$ 確率母関数の1階微分を求めると、商の微分公式と合成関数の微分法より、 $$\begin{array}{rl}{G}_X^{\left(1\right)}\left(\theta \right)& =-\frac{p}{{\{1-\theta (1-p)\}}^2}\cdot \frac{d}{d\theta }\{1-\theta (1-p)\}\\ & =-\frac{p}{{\{1-\theta (1-p)\}}^2}\cdot \{-(1-p)\}\\ & =\frac{p(1-p)}{{\{1-\theta (1-p)\}}^2}\end{array}$$ 1次モーメントと確率母関数の関係 $G_X^{\left(1\right)}\left(1\right)=E\left(X\right)$ より、 $$\begin{array}{rl}E\left(X\right)& =\frac{p(1-p)}{{(1-1+p)}^2}\\ & =\frac{1-p}{p}\end{array}$$ $◼$

（ii）分散
確率母関数の2階微分を求めると、 $$\begin{array}{rl}{G}_X^{\left(2\right)}\left(\theta \right)& =-2\cdot \frac{p(1-p)}{{\{1-\theta (1-p)\}}^3}\cdot \frac{d}{d\theta }\{1-\theta (1-p)\}\\ & =-2\cdot \frac{p(1-p)}{{\{1-\theta (1-p)\}}^3}\cdot \{-(1-p)\}\\ & =\frac{2p{(1-p)}^2}{{\{1-\theta (1-p)\}}^3}\end{array}$$ 2次階乗モーメントと確率母関数の関係 $G_X^{\left(2\right)}\left(1\right)=E\left\{X(X-1)\right\}$ より、 $$\begin{array}{rl}E\left\{X(X-1)\right\}& =\frac{2p{(1-p)}^2}{{(1-1+p)}^3}\\ & =\frac{2{(1-p)}^2}{p^2}\end{array}$$ 階乗モーメントを用いた分散の公式 $V\left(X\right)=E\left\{X(X-1)\right\}+E\left(X\right)-{\left\{E\left(X\right)\right\}}^2$ より、 $$\begin{array}{rl}V\left(X\right)& =\frac{2{(1-p)}^2}{p^2}+\frac{1-p}{p}-\frac{{(1-p)}^2}{p^2}\\ & =\frac{{(1-p)}^2}{p^2}+\frac{1-p}{p}\\ & =\frac{1-p}{p}(\frac{1-p}{p}+1)\\ & =\frac{1-p}{p}\left(\frac{1-p+p}{p}\right)\\ & =\frac{1-p}{p^2}\end{array}$$ $◼$

（i）期待値
幾何分布のモーメント母関数の公式より、 $$\begin{array}{r}{M}_X\left(\theta \right)=\frac{p}{1-e^{\theta }(1-p)}\end{array}$$ モーメント母関数の1階微分を求めると、商の微分公式と合成関数の微分法より、 $$\begin{array}{rl}{M}_X^{\left(1\right)}\left(\theta \right)& =-\frac{p}{{\{1-e^{\theta }(1-p)\}}^2}\cdot \frac{d}{d\theta }\{1-e^{\theta }(1-p)\}\\ & =-\frac{p}{{\{1-e^{\theta }(1-p)\}}^2}\cdot \{-e^{\theta }(1-p)\}\\ & =\frac{e^{\theta }\cdot p(1-p)}{{\{1-e^{\theta }(1-p)\}}^2}\end{array}$$ 1次モーメントとモーメント母関数の関係 $M_X^{\left(1\right)}\left(0\right)=E\left(X\right)$ より、 $$\begin{array}{rl}E\left(X\right)& =\frac{e^0\cdot p(1-p)}{{\{1-e^0(1-p)\}}^2}\\ & =\frac{1-p}{p}\end{array}$$ $◼$

（ii）分散
モーメント母関数の2階微分を求めると、商の微分公式と合成関数の微分法より、 $$\begin{array}{rl}{M}_X^{\left(2\right)}\left(\theta \right)& =e^{\theta }\cdot \frac{d}{d\theta }\left[\frac{p(1-p)}{{\{1-e^{\theta }(1-p)\}}^2}\right]+\frac{p(1-p)}{{\{1-e^{\theta }(1-p)\}}^2}\cdot \frac{d}{d\theta }{e}^{\theta }\\ & =e^{\theta }\cdot \{-2\frac{p(1-p)}{{\{1-e^{\theta }(1-p)\}}^3}\cdot \frac{d}{d\theta }\{1-e^{\theta }(1-p)\}\}+\frac{e^{\theta }\cdot p(1-p)}{{\{1-e^{\theta }(1-p)\}}^2}\\ & =\frac{2e^{2\theta }\cdot p{(1-p)}^2}{{\{1-e^{\theta }(1-p)\}}^3}+\frac{e^{\theta }\cdot p(1-p)}{{\{1-e^{\theta }(1-p)\}}^2}\end{array}$$ 2次モーメントとモーメント母関数の関係 $M_X^{\left(2\right)}\left(0\right)=E\left(X^2\right)$ より、 $$\begin{array}{rl}E\left(X^2\right)& =\frac{2e^0\cdot p{(1-p)}^2}{{\{1-e^0(1-p)\}}^3}+\frac{e^0\cdot p(1-p)}{{\{1-e^0(1-p)\}}^2}\\ & =\frac{2p{(1-p)}^2}{p^3}+\frac{p(1-p)}{p^2}\\ & =\frac{2{(1-p)}^2}{p^2}+\frac{1-p}{p}\end{array}$$ 分散の公式 $V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2$ より、 $$\begin{array}{rl}V\left(X\right)& =\frac{2{(1-p)}^2}{p^2}+\frac{1-p}{p}-\frac{{(1-p)}^2}{p^2}\\ & =\frac{{(1-p)}^2}{p^2}+\frac{1-p}{p}\\ & =\frac{1-p}{p^2}\end{array}$$ $◼$