Expectation of Discrete Random Variable from PGF

Theorem
Let $X$ be a discrete random variable whose probability generating function is $\Pi_X \left({s}\right)$.

Then the expectation of $X$ is the value of the first derivative of $\Pi_X \left({s}\right)$ WRT $s$ at $s=1$.

That is:
 * $E \left({X}\right) = \Pi'_X \left({1}\right)$

Proof
For ease of notation, write $p \left({x}\right)$ for $\Pr \left({X = x}\right)$.

From the definition of the probability generating function:


 * $\displaystyle \Pi_X \left({s}\right) = \sum_{x \mathop \ge 0} p \left({x}\right) s^x = p \left({0}\right) + p \left({1}\right) s + p \left({2}\right) s^2 + p \left({3}\right) s^3 + \cdots$

Differentiate this WRT $s$:


 * $\displaystyle \Pi'_X \left({s}\right) = \sum_{x \mathop \ge 0} x p \left({x}\right) s^{x-1} = p \left({1}\right) + 2 p \left({2}\right) s + 3 p \left({3}\right) s^2 + \cdots$

Plugging in $s = 1$ gives:


 * $\displaystyle \Pi'_X \left({1}\right) = \sum_{x \mathop \ge 0} x p \left({x}\right) 1^{x-1} = p \left({1}\right) + 2 p \left({2}\right) + 3 p \left({3}\right) + \cdots$

But:


 * $\displaystyle \sum_{x \mathop \ge 0} x p \left({x}\right) 1^{x-1} = \sum_{x \mathop \ge 0} x p \left({x}\right)$

is precisely the definition of the expectation.

Comment
So, in order to find the expectation of a discrete random variable, then there is no need to go through the tedious process of what might be a complicated and fiddly summation.

All you need to do is differentiate the p.g.f and plug in $1$.

Assuming, of course, you know what the p.g.f is.