Hence find its variance and mean.

Solution:

i) Moment Generating Function (M.G.F.)

The M.G.F. about the origin is:

$M_{0}(t)=E(e^{tx})=\sum p(x) e^{tx}= \sum_{x=0}^{\infty}\cfrac{e^{-m}m^{x}}{x!}. e^{tx}=e^{-m} \sum_{x=0}^{\infty} \cfrac{(me^{t})^{x}}{x!} = e^{-m} . e^{met}$

$\therefore M_{0}(t)=e^{m(e^{t}-1)}$

ii) Mean and Variance of Poisson Distribution

$\mu_{1}'=E(x)=\sum p_{i}x_{i}=\sum_{x=0}^{\infty} \cfrac{e^{-m}m^{x}}{x!}x=\sum_{x=1}^{\infty} \cfrac{e^{-m}m^{x}}{(x-1)!} = me^{-m} \sum_{x=1}^{\infty}\cfrac{m^{(x-1)}}{(x-1)!}$

$\mu_{1}'= me^{-m} \left[ 1+m+\cfrac{m^{2}}{2!}+\cfrac{m^{3}}{3!}+---- \right]= me^{-m} .e^{m}=m$

$\therefore$ Mean $=\mu_{1}'=m$

$\mu_{2}'=E(x^{2})= \sum p_{i}x_{i}^{2} = \sum_{x=0}^{\infty}e^{-m}.\cfrac{m^{x}}{x!}x^{2}$

$\therefore x^{2}=x+x(x-1)$

$\mu_{2}'= \sum_{x=0}^{\infty}e^{-m}.\cfrac{m^{x}}{x!}[x+x(x-1)]=\sum_{x=0}^{\infty}e^{-m}.\cfrac{m^{x}.x}{x!}+\sum_{x=0}^{\infty}e^{-m}.\cfrac{m^{x}.x(x-1)}{x!}$

$\mu_{2}'=me^{-m}\sum_{x=1}^{\infty}\cfrac{m^{x-1}}{(x-1)!}+m^{2}e^{-m}\sum_{x=2}^{\infty}\cfrac{m^{x-2}}{(x-2)!}$

$\mu_{2}'=me^{-m}.e^{m}+m^{2}e^{-m}\left[ 1+ \cfrac{m}{1!}+\cfrac{m^{2}}{2!}+----\right]$

$\mu_{2}'=me^{-m}.e^{m}+m^{2}e^{-m}.e^{m}=m+m^{2}$

$\mu_{2}=\mu_{2}'-\mu_{1}'=m+m^{2}-m^{2}=m$

$\therefore$ Variance$=\mu_{2}=m$