負の二項分布の期待値・分散の導出

（i）期待値
負の二項分布の確率母関数の公式より、 $$\begin{array}{r}{G}_X\left(\theta \right)=\frac{p^n}{{\{1-\theta (1-p)\}}^n}\end{array}$$ 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、 $$\begin{array}{rl}{G}_X^{\left(1\right)}\left(\theta \right)& =-n\cdot \frac{p^n}{{\{1-\theta (1-p)\}}^{n+1}}\cdot \frac{d}{d\theta }\{1-\theta (1-p)\}\\ & =-\frac{n{p}^n}{{\{1-\theta (1-p)\}}^{n+1}}\cdot \{-(1-p)\}\\ & =\frac{n{p}^n(1-p)}{{\{1-\theta (1-p)\}}^{n+1}}\end{array}$$ 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、 $$\begin{array}{rl}E\left(X\right)& =\frac{n{p}^n(1-p)}{{\{1-(1-p)\}}^{n+1}}\\ & =\frac{n{p}^n(1-p)}{p^{n+1}}\\ & =\frac{n(1-p)}{p}\end{array}$$ $◼$

（ii）分散
確率母関数の2階微分を求めると、 $$\begin{array}{rl}{G}_X^{\left(2\right)}\left(\theta \right)& =-\frac{n(n+1)p^n(1-p)}{{\{1-\theta (1-p)\}}^{n+2}}\cdot \frac{d}{d\theta }\{1-\theta (1-p)\}\\ & =-\frac{n(n+1)p^n(1-p)}{{\{1-\theta (1-p)\}}^{n+2}}\cdot \{-(1-p)\}\\ & =\frac{n(n+1)p^n{(1-p)}^2}{{\{1-\theta (1-p)\}}^{n+2}}\end{array}$$ 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、 $$\begin{array}{rl}E\left\{X(X-1)\right\}& =\frac{n(n+1)p^n{(1-p)}^2}{{\{1-(1-p)\}}^{n+2}}\\ & =\frac{n(n+1)p^n{(1-p)}^2}{p^{n+2}}\\ & =\frac{n(n+1){(1-p)}^2}{p^2}\end{array}$$ 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、 $$\begin{array}{rl}V\left(X\right)& =\frac{n(n+1){(1-p)}^2}{p^2}+\frac{n(1-p)}{p}-\frac{n^2{(1-p)}^2}{p^2}\\ & =\frac{n(1-p)}{p^2}\{(n+1)(1-p)+p-n(1-p)\}\\ & =\frac{n(1-p)}{p^2}(n-np+1-p+p-n+np)\\ & =\frac{n(1-p)}{p^2}\cdot 1\\ & =\frac{n(1-p)}{p^2}\end{array}$$ $◼$

（i）期待値
負の二項分布のモーメント母関数の公式より、 $$\begin{array}{r}{M}_X\left(\theta \right)=\frac{p^n}{{\{1-e^{\theta }(1-p)\}}^n}\end{array}$$ 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、 $$\begin{array}{rl}{M}_X^{\left(1\right)}\left(\theta \right)& =-n\cdot \frac{p^n}{{\{1-e^{\theta }(1-p)\}}^{n+1}}\cdot \frac{d}{d\theta }\{1-e^{\theta }(1-p)\}\\ & =-\frac{n{p}^n}{{\{1-e^{\theta }(1-p)\}}^{n+1}}\cdot \{-e^{\theta }(1-p)\}\\ & =\frac{e^{\theta }\cdot n{p}^n(1-p)}{{\{1-e^{\theta }(1-p)\}}^{n+1}}\end{array}$$ 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、 $$\begin{array}{rl}E\left(X\right)& =\frac{e^0\cdot n{p}^n(1-p)}{{\{1-e^0(1-p)\}}^{n+1}}\\ & =\frac{n{p}^n(1-p)}{p^{n+1}}\\ & =\frac{n(1-p)}{p}\end{array}$$ $◼$

（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、 $$\begin{array}{rl}{M}_X^{\left(2\right)}\left(\theta \right)& =e^{\theta }n{p}^n(1-p)\cdot \frac{d}{d\theta }\left[\frac{1}{{\{1-e^{\theta }(1-p)\}}^{n+1}}\right]+\frac{n{p}^n(1-p)}{{\{1-e^{\theta }(1-p)\}}^{n+1}}\cdot \frac{d}{d\theta }{e}^{\theta }\\ & =[-\frac{e^{\theta }\cdot n(n+1)p^n(1-p)}{{\{1-e^{\theta }(1-p)\}}^{n+2}}\cdot \frac{d}{d\theta }\{1-e^{\theta }(1-p)\}]+\frac{e^{\theta }\cdot n{p}^n(1-p)}{{\{1-e^{\theta }(1-p)\}}^{n+1}}\\ & =\frac{e^{\theta }\cdot n(n+1)p^n{(1-p)}^2}{{\{1-e^{\theta }(1-p)\}}^{n+2}}+\frac{e^{\theta }\cdot n{p}^n(1-p)}{{\{1-e^{\theta }(1-p)\}}^{n+1}}\end{array}$$ 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、 $$\begin{array}{rl}E\left(X^2\right)& =\frac{e^0\cdot n(n+1)p^n{(1-p)}^2}{{\{1-e^0(1-p)\}}^{n+2}}+\frac{e^0\cdot n{p}^n(1-p)}{{\{1-e^0(1-p)\}}^{n+1}}\\ & =\frac{n(n+1){(1-p)}^2}{p^2}+\frac{n(1-p)}{p}\end{array}$$ 分散の公式 $V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2$ より、 $$\begin{array}{rl}V\left(X\right)& =\frac{n(n+1){(1-p)}^2}{p^2}+\frac{n(1-p)}{p}-\frac{{(1-p)}^2}{p^2}\\ & =\frac{n(1-p)}{p^2}\end{array}$$ $◼$

負の二項分布の確率変数 $Y$ を互いに独立に幾何分布に従う確率変数 $X_i(i=1,2,\cdots ,n)$ の和と考えると、 $$\begin{array}{r}Y=X_1+X_2+\cdots +X_n\end{array}$$ また、幾何分布の期待値と分散の公式より、 $$\begin{array}{c}E\left(X\right)=\frac{1-p}{p}\\ V\left(X\right)=\frac{1-p}{p^2}\end{array}$$ 期待値の性質 $E\left(\sum _{i=1}^n{X}_i\right)=\sum _{i=1}^{n}E\left(X_i\right)$ より、 $$\begin{array}{r}E\left(Y\right)=E\left(\sum _{i=1}^n{X}_i\right)=\sum _{i=1}^n\frac{1-p}{p}=\frac{n(1-p)}{p}\end{array}$$ 分散の性質 $V\left(\sum _{i=1}^n{X}_i\right)=\sum _{i=1}^{n}V\left(X_i\right)$ より、 $$\begin{array}{r}V\left(Y\right)=V\left(\sum _{i=1}^n{X}_i\right)=\sum _{i=1}^n\frac{1-p}{p^2}=\frac{n(1-p)}{p^2}\end{array}$$ $◼$