Finding the expectation and variance from a probability generating function

Question

For the probability generating function $G_X(z) = \frac{p}{1-qz}$ where $q=1-p$, use $G_X(z)$ to deduce that $$E[X] = \frac{1-p}{p}\quad \text{ and } \quad Var(X) = \frac{1-p}{p^2}$$

Do I need to differentiate with respect to $z$ for the expectation?

Answer 1

Yes, you need to differentiate with respect to $z$ and evaluate at $z=1$. Specifically $$G'(z)=\frac{d}{dz}\left(\frac{p}{1-qz}\right)=\frac{pq}{(1-qz)^2}$$ and so at $z=1$ $$E[X]=G'(1)=\frac{pq}{(1-q)^2}=\frac{p(1-p)}{p^2}=\frac{1-p}{p}$$ Now the variance is given by $$Var(X)=G''(1)+G'(1)-[G'(1)]^2$$ Can you take it from here? For more details, see here (note also the notation $1^-$ instead of $1$).