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 $=\mu_{1}'=m$ …