Suppose in a biased random walk, $r(i,n)$ is the probability that a particle appears at position $i$ at time $n$. The corresponding probability generating function is $$ R(z,s)=\sum_{n=0}^\infty \sum_{i=-\infty}^\infty r(i,n)z^i s^n. $$ Suppose by some means I obtain the probability generating function, and I want to know $$ \mu(n)=\sum_{i=-\infty}^\infty r(i,n)i $$ and $$ \sigma^2(n)=\sum_{i=-\infty}^\infty r(i,n)\left[i-\mu(n)\right]^2 $$
My question is: how to get from $R(z,s)$ to $\mu(n)$ and $\sigma^2(n)$ (presumably by differentiation and other operations)?
Formally :
$$(\frac{\partial}{\partial z}R)(z,s)=\sum_{n=0}^{\infty}\sum_{i=-\infty}^{\infty}r(i,n)iz^{i-1}s^n $$
$$(\frac{\partial^k}{\partial s^k}(\frac{\partial}{\partial z}R))(z,s)=\sum_{n=k}^{\infty}\sum_{i=-\infty}^{\infty}r(i,n)iz^{i-1}\frac{n!}{(n-k)!}s^{n-k} $$
So that :
$$(\frac{\partial^k}{\partial s^k}(\frac{\partial}{\partial z}R))(1,0)=\sum_{n=k}^{\infty}\sum_{i=-\infty}^{\infty}r(i,n)i\frac{n!}{(n-k)!}1^{i-1}0^{n-k}=k!\mu(k) $$
Whence :
$$\mu(k)=\frac{1}{k!}(\frac{\partial^k}{\partial s^k}(\frac{\partial}{\partial z}R))(1,0) $$
For the standard deviation it should be mostly the same, remark (by the same formal argument) that :
$$\sum_{i=-\infty}^{\infty}r(i,k)i^2=\frac{1}{k!}(\frac{\partial^k}{\partial s^k}(\frac{\partial}{\partial z}z(\frac{\partial}{\partial z}R)))(1,0) $$
$$\sum_{i=-\infty}^{\infty}r(i,k)=\frac{1}{k!}(\frac{\partial^k}{\partial s^k}R)(1,0)$$
Since :
$$\sigma^2(n)=\sum_{i=-\infty}^{\infty}r(i,n)[i-\mu(n)]^2=\sum_{i=-\infty}^{\infty}r(i,n)i^2-\sum_{i=-\infty}^{\infty}r(i,n)\mu(n)^2$$
You get :
$$\sigma^2(k)=\frac{1}{k!}(\frac{\partial^k}{\partial s^k}(\frac{\partial}{\partial z}z(\frac{\partial}{\partial z}R)))(1,0)- \frac{1}{k!}(\frac{\partial^k}{\partial s^k}R)(1,0)[\frac{1}{k!}(\frac{\partial^k}{\partial s^k}(\frac{\partial}{\partial z}R))(1,0)]^2$$