Generating function of a random variable

Question

I've got the following problem:

Give the generating function of the random variable $X$ whose mass function is defined by: $$f(m) = P(X=m) = (m+1) p^2 (1-p)^m,$$ where $m$ is a positive integer (incl. $0$) and $0 < p < 1$.

How can I find this? And what's the radius of convergence?

Answer 1

Generating function $$h(s)=\sum_{m=0}^{\infty}P(X=m)s^m=\sum_{m=0}^{\infty}(m+1)p^2(1-p)^ms^m$$ $$=\frac{p^2}{(1-p)s}\sum_{m=0}^{\infty}(m+1)((1-p)s)^{m+1}$$ $$=\frac{p^2}{(1-p)s}\frac{(1-p)s}{(1-(1-p)s)^2}=\frac{p^2}{(1-(1-p)s)^2}$$

$E[X]=\sum_{m=1}^{\infty}mP(X=m)$

On differentiating starting equation, $h'(s)=\sum_{m=1}^{\infty}mP(X=m)s^{m-1}$

Thus, $h'(1)=\sum_{m=1}^{\infty}mP(X=m)=E[X]$

$Var[X]=E[X^2]-(E[X])^2$

Now, $h''(1)=\sum_{m=1}^{\infty}m(m-1)P(X=m)=\sum_{m=1}^{\infty}m^2P(X=m)-\sum_{m=1}^{\infty}mP(X=m)$ $=E[X^2]-E[X]$

It implies $E[X^2]=h''(1)+E[X]=h''(1)+h'(1)$

and hence $Var[X]=E[X^2]-(E[X])^2=h''(1)+h'(1)-(h'(1))^2$