Variance of Discrete Random Variable from P.G.F.

Theorem

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

Then the variance of $X$ can be easily obtained from the value of the second derivative of $\Pi_X \left({s}\right)$ WRT $s$ at $x=1$:

$\operatorname{var} \left({X}\right) = \Pi''_X \left({1}\right) + \mu - \mu^2$

where $\mu = E \left({x}\right)$ is the expectation of $X$.

Proof

From the definition of the probability generating function:

$\displaystyle \Pi_X \left({s}\right) = \sum_{x \ge 2} p \left({x}\right) s^x$

Differentiating this twice WRT $s$ gives us:

$\displaystyle \Pi''_X \left({s}\right) = \sum_{x \ge 2} x \left({x-1}\right) p \left({x}\right) s^{x-2}$

from Differentiation of Power Series‎.

But it also holds when you include $x = 0$ and $x = 1$ in the sum, as in both cases the term evaluates to zero and therefore vanishes.

So:

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

Plugging in $s = 1$ gives:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \Pi''_X \left({1}\right)\)	\(=\)	\(\displaystyle \sum_{x \ge 0} x \left({x-1}\right) p \left({x}\right) 1^{x-2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{x \ge 0} x^2 p \left({x}\right) - \sum_{x \ge 0} x p \left({x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle E \left({X^2}\right) - E \left({X}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

The result follows from the definition of variance:

$\operatorname{var} \left({X}\right) = E \left({X^2}\right) - \left({E \left({X}\right)}\right)^2$

after a little algebra.

$\blacksquare$

Comment

So, in order to find the variance 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 twice, and plug in $1$.

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

Sources

• Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction (1986): $\S 4.3 \ (20)$