finding a probability generating function and mean and variance

Question

given a function $Pr(X=k)=$ ($5\over11$)($6\over11$)$^{k-1}$ for $k=1,2,...$

I've been tasked with finding out the probability generating function $G_{X}(s)$ for $|s|\lt$ $11\over6$ And the mean and variance of X.

So to begin with; the Generating function $G_{X}(s)= s^{k}Pr(X=k)=s^{k}$$5\over11$($6\over11$)$^{k-1}$

So $G_{X}(s)$ = $s^{k}$($5\over6$)($6\over11$)$^{k}$ = ($5\over6$)[$s$($6\over11$) + $s^{2}$($6\over11$)$^{2}$+.....] = ($5\over6$)$\sum_{k=1}^{n}($$6s\over 11$)$^{k}$ = $11s \over (11-6s)$ Since it is an infinite geometric sequence

Provided my method initially above is correct, would i use E{x} = G'(1) and Var{x}=$G''(1)$+$G'(1)$-$(G''(1))^{2}$

Answer 1

You appear to have made a small miscalculation early on. It helps to not put values in until the very end.

Let $p:=\frac{5}{11},\,q:=1-p$ so, if $|s|<1/q$,$$G_X(s)=\sum_{k\ge1}pq^{k-1}s^k=ps\sum_{j\ge0}(qs)^j=\frac{ps}{1-qs}=\frac{p/q}{1-qs}-\frac{p}{q}.$$Separating out that additive constant will make differentiating easier:$$G^\prime(s)=\frac{p}{(1-qs)^2},\,G^{\prime\prime}(s)=\frac{2pq}{(1-qs)^3}.$$So$$\mu=G^\prime(1)=\frac1p=\frac{11}{5},\,\sigma^2=G^{\prime\prime}(1)+\mu-\mu^2=\frac{2q}{p^2}+\frac1p-\frac{1}{p^2}=\frac{2q+p-1}{p^2}=\frac{q}{p^2}=\frac{66}{25}.$$