ポアソン分布(Poisson distribution)の積率母凾数(moment-generating function)と期待値(expected value)・分散(variance)

ポアソン分布

$$\begin{array}{rcl} \displaystyle Po(\lambda)&=&\displaystyle \frac{\lambda^{x}}{x!}\mathrm{e}^{-\lambda}\\ \end{array}$$

積率母凾数

$$\begin{array}{rcl} \displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\ &=&\displaystyle \sum_{x=0}^{\infty}(\mathrm{e}^{tx})\frac{\lambda^{x}}{x!}\mathrm{e}^{-\lambda}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}(\mathrm{e}^{tx})\frac{\lambda^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}\frac{\mathrm{e}^{tx}\lambda^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}\frac{(\mathrm{e}^{t}\lambda)^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\mathrm{e}^{\mathrm{e}^{t}\lambda} \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\sum_{x=0}^{\infty}\frac{a^{x}}{x!}=\mathrm{e}^a}\\ &=&\displaystyle \mathrm{e}^{\mathrm{e}^{t}\lambda-\lambda}=\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)} \end{array}$$

期待値・分散

$$\begin{array}{rcl} \displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_x(t)|_{t=0}\\ &=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\ &=&\displaystyle E[X^m]\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^{(1)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t }(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)}) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} s }(\mathrm{e}^{s})\frac{ \mathrm{d}s}{ \mathrm{d}t} \right\}|_{t=0} \,\dotso\,s=\lambda(\mathrm{e}^{t}-1),\frac{ \mathrm{d}s}{ \mathrm{d}t}=\lambda\mathrm{e}^{t}\\ &=&\displaystyle \left\{ (\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)})(\lambda\mathrm{e}^{t}) \right\}|_{t=0} \,\dotso\,\frac{ \mathrm{d} }{ \mathrm{d} x }\mathrm{e}^{x}=\mathrm{e}^{x}\\ &=&\displaystyle \left\{ \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+t}) \right\}|_{t=0}\\ &=&\displaystyle \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{0}-1)+0})\\ &=&\displaystyle \lambda\mathrm{e}^0 \,\dotso\,a^0=1\\ &=&\lambda \,\dotso\,a^0=1\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^{(2)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d}^2 }{ \mathrm{d} t^2 }(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)}) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t } \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+t}) \right\}|_{t=0} \,\dotso\,E[X]の展開から．\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} s }(\lambda\mathrm{e}^{s})\frac{ \mathrm{d}s}{ \mathrm{d}t} \right\}|_{t=0} \,\dotso\,s=\lambda(\mathrm{e}^{t}-1)+t,\frac{ \mathrm{d}s}{ \mathrm{d}t}=\lambda\mathrm{e}^{t}+1\\ &=&\displaystyle \left\{ (\lambda\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+1})(\lambda\mathrm{e}^{t}+1) \right\}|_{t=0} \,\dotso\,\frac{ \mathrm{d} }{ \mathrm{d} x }C\mathrm{e}^{x}=C\mathrm{e}^{x}\\ &=&\displaystyle (\lambda\mathrm{e}^{\lambda(\mathrm{e}^{0}-1)+1})(\lambda\mathrm{e}^{0}+1)\\ &=&\displaystyle \lambda\mathrm{e}^0(\lambda+1) \,\dotso\,a^0=1\\ &=&\lambda(\lambda+1) \,\dotso\,a^0=1\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle V[X]&=&\displaystyle E[X^2]-E[X]^2\\ &=&\lambda(\lambda+1)-\lambda^2\\ &=&\lambda^2+\lambda-\lambda^2\\ &=&\lambda\\ \end{array}$$