First four moments of the Poisson distribution

First four moments of Poisson distribution

The r th moment about origin is given by $${{{\mu }'}_{r}}=E({{x}^{r}})=\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}$$ When r=1 we get $${{{\mu }'}_{1}}=\sum\limits_{x=0}^{\infty }{x\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}$$ $$={{e}^{-\lambda }}\lambda \sum\limits_{x=0}^{\infty }{x\frac{{{\lambda }^{x-1}}}{x(x-1)!}}$$ $$=\lambda {{e}^{-\lambda }}\left[ 1+\lambda +\frac{{{\lambda }^{2}}}{2!}+\frac{{{\lambda }^{3}}}{3!}+\ldots \right]$$ $$=\lambda {{e}^{-\lambda }}.{{e}^{\lambda }}$$ $${{{\mu }'}_{1}}=E(X)=\lambda$$ Hence mean of the Poisson distribution is lambda When r=2 we get $${{{\mu }'}_{2}}=\sum\limits_{x=0}^{\infty }{{{x}^{2}}\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}$$ $$=\sum\limits_{x=0}^{\infty }{\left\{ x(x-1)+x \right\}\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}$$ $$=\sum\limits_{x=0}^{\infty }{x(x-1)\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}+\sum\limits_{x=0}^{\infty }{x\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}$$ $$=\sum\limits_{x=0}^{\infty }{x(x-1)\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x(x-1)(x-2)!}}+\sum\limits_{x=0}^{\infty }{x\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x(x-1)!}}$$ $$=\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{(x-2)!}}+\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{(x-1)!}}$$ $$={{e}^{-\lambda }}{{\lambda }^{2}}\sum\limits_{x=0}^{\infty }{\frac{{{\lambda }^{x-2}}}{(x-2)!}}+E(X)$$ $$={{\lambda }^{2}}{{e}^{-\lambda }}\left[ 1+\lambda +\frac{{{\lambda }^{2}}}{2!}+\frac{{{\lambda }^{3}}}{3!}+\ldots \right]+\lambda$$ $$={{\lambda }^{2}}{{e}^{-\lambda }}.{{e}^{\lambda }}+\lambda$$ $${{{\mu }'}_{2}}=E({{X}^{2}})={{\lambda }^{2}}+\lambda$$ Hence mean of the Poisson distribution is $${{\mu }_{2}}={{{\mu }'}_{2}}-{{{\mu }'}_{1}}^{2}={{\lambda }^{2}}+\lambda -{{\lambda }^{2}}=\lambda$$ Thus the mean and the variance of Poisson distribution are same i.e $\lambda =\lambda $. When r=3 we get $${{{\mu }'}_{3}}=\sum\limits_{x=0}^{\infty }{{{x}^{3}}\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}$$ $$=\sum\limits_{x=0}^{\infty }{\left\{ x(x-1)(x-2)+3x(x-1)+x \right\}\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}$$ $$=\sum\limits_{x=0}^{\infty }{x(x-1)(x-2)\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}+3\sum\limits_{x=0}^{\infty }{x(x-1)\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}+\sum\limits_{x=0}^{\infty }{x\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}$$ $$\begin{align} & =\sum\limits_{x=0}^{\infty }{x(x-1)(x-2)\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x(x-1)(x-2)(x-3)!}}+\sum\limits_{x=0}^{\infty }{x(x-1)\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x(x-1)(x-2)!}} \\ & +\sum\limits_{x=0}^{\infty }{x\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x(x-1)!}} \\ \end{align}$$ $$=\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{(x-3)!}}+\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{(x-2)!}}+\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{(x-1)!}}$$ $$={{\lambda }^{3}}\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x-3}}}{(x-3)!}}+3{{\lambda }^{2}}\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x-2}}}{(x-2)!}}+\lambda \sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x-1}}}{(x-1)!}}$$ $$\begin{align} & ={{\lambda }^{2}}{{e}^{-\lambda }}\left[ 1+\lambda +\frac{{{\lambda }^{2}}}{2!}+\frac{{{\lambda }^{3}}}{3!}+\ldots \right]++3{{\lambda }^{2}}{{e}^{-\lambda }}\left[ 1+\lambda +\frac{{{\lambda }^{2}}}{2!}+\frac{{{\lambda }^{3}}}{3!}+\ldots \right] \\ & +\lambda \ {{e}^{-\lambda }}\left[ 1+\lambda +\frac{{{\lambda }^{2}}}{2!}+\frac{{{\lambda }^{3}}}{3!}+\ldots \right] \\ \end{align}$$ $$={{\lambda }^{3}}{{e}^{-\lambda }}.{{e}^{\lambda }}+3{{\lambda }^{2}}{{e}^{-\lambda }}.{{e}^{\lambda }}+\lambda \ {{e}^{-\lambda }}.{{e}^{\lambda }}$$ $$={{\lambda }^{3}}+3{{\lambda }^{2}}+\lambda \ $$ $${{{\mu }'}_{2}}=E({{X}^{2}})={{\lambda }^{3}}+3{{\lambda }^{2}}+\lambda \ $$ Hence the third central moment of the Poisson distribution is $${{\mu }_{3}}={{{\mu }'}_{3}}-3{{{\mu }'}_{2}}{{{\mu }'}_{1}}+2{{{\mu }'}_{1}}^{3}={{\lambda }^{2}}+\lambda -{{\lambda }^{2}}=\lambda$$ When r=4 we get $${{{\mu }'}_{4}}=\sum\limits_{x=0}^{\infty }{{{x}^{4}}\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}$$ $$=\sum\limits_{x=0}^{\infty }{\left\{ x(x-1)(x-2)(x-3)+6x(x-1)(x-2)+7x(x-1)+x \right\}\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}$$ $$\begin{align} & =\sum\limits_{x=0}^{\infty }{x(x-1)(x-2)(x-3)\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}+6\sum\limits_{x=0}^{\infty }{x(x-1)(x-2)\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}} \\ & +7\sum\limits_{x=0}^{\infty }{x(x-1)\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}}+\sum\limits_{x=0}^{\infty }{x\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x!}} \\ \end{align}$$ $$\begin{align} & =\sum\limits_{x=0}^{\infty }{x(x-1)(x-2)(x-3)\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x(x-1)(x-2)(x-3)(x-4)!}} \\ & +6\sum\limits_{x=0}^{\infty }{x(x-1)(x-2)\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x(x-1)(x-2)(x-3)!}} \\ & +7\sum\limits_{x=0}^{\infty }{x(x-1)\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x(x-1)(x-2)!}}+\sum\limits_{x=0}^{\infty }{x\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{x(x-1)!}} \\ \end{align}$$ $$=\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{(x-4)!}}+6\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{(x-3)!}}+7\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{(x-2)!}+}\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x}}}{(x-1)!}}$$ $$={{\lambda }^{4}}\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x-4}}}{(x-4)!}}+6{{\lambda }^{3}}\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x-3}}}{(x-3)!}}+7{{\lambda }^{2}}\sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x-2}}}{(x-2)!}}+\lambda \sum\limits_{x=0}^{\infty }{\frac{{{e}^{-\lambda }}{{\lambda }^{x-1}}}{(x-1)!}}$$ $$\begin{align} & ={{\lambda }^{4}}{{e}^{-\lambda }}\left[ 1+\lambda +\frac{{{\lambda }^{2}}}{2!}+\frac{{{\lambda }^{3}}}{3!}+\ldots \right]+6{{\lambda }^{3}}{{e}^{-\lambda }}\left[ 1+\lambda +\frac{{{\lambda }^{2}}}{2!}+\frac{{{\lambda }^{3}}}{3!}+\ldots \right] \\ & +7{{\lambda }^{2}}\ {{e}^{-\lambda }}\left[ 1+\lambda +\frac{{{\lambda }^{2}}}{2!}+\frac{{{\lambda }^{3}}}{3!}+\ldots \right]+\lambda \ {{e}^{-\lambda }}\left[ 1+\lambda +\frac{{{\lambda }^{2}}}{2!}+\frac{{{\lambda }^{3}}}{3!}+\ldots \right] \\ \end{align}$$ $$={{\lambda }^{4}}{{e}^{-\lambda }}.{{e}^{\lambda }}+6{{\lambda }^{3}}{{e}^{-\lambda }}.{{e}^{\lambda }}+7{{\lambda }^{2}}\ {{e}^{-\lambda }}.{{e}^{\lambda }}+\lambda \ {{e}^{-\lambda }}.{{e}^{\lambda }}$$ $$={{\lambda }^{4}}+6{{\lambda }^{3}}+7{{\lambda }^{2}}+\lambda \ $$ $${{{\mu }'}_{4}}={{\lambda }^{4}}+6{{\lambda }^{3}}+7{{\lambda }^{2}}+\lambda \ $$ Hence the third central moment of the Poisson distribution is $${{\mu }_{3}}={{{\mu }'}_{4}}-4{{{\mu }'}_{3}}{{{\mu }'}_{1}}+6{{{\mu }'}_{2}}{{{\mu }'}_{1}}^{2}-3{{{\mu }'}_{1}}^{4}=3{{\lambda }^{2}}+\lambda$$ The indices of skewness and kurtosis of Poisson distribution are given respectively, by $${{\alpha }_{3}}=\sqrt{{{\beta }_{1}}}={{\left( \frac{\mu _{3}^{2}}{\mu _{2}^{3}} \right)}^{\frac{1}{2}}}={{\left( \frac{{{\lambda }^{2}}}{{{\lambda }^{3}}} \right)}^{\frac{1}{2}}}=\frac{1}{\sqrt{\lambda }}$$ and $${{\alpha }_{4}}={{\beta }_{4}}=\left( \frac{{{\mu }_{4}}}{\mu _{2}^{2}} \right)=\left( \frac{3{{\lambda }^{2}}+\lambda }{{{\lambda }^{2}}} \right)=3+\frac{1}{\lambda }$$