# Expectation of Discrete Random Variable from PGF

## Theorem

Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.

Then the expectation of $X$ is the value of the first derivative of $\map {\Pi_X} s$ with respect to $s$ at $s = 1$.

That is:

- $\expect X = \map {\Pi'_X} 1$

## Proof

For ease of notation, write $\map p x$ for $\map \Pr {X = x}$.

From the definition of the probability generating function:

- $\ds \map {\Pi_X} s = \sum_{x \mathop \ge 0} \map p x s^x = \map p 0 + \map p 1 s + \map p 2 s^2 + \map p 3 s^3 + \cdots$

Differentiate this with respect to $s$:

\(\ds \map {\Pi'_X} s\) | \(=\) | \(\ds \frac \d {\d s} \sum_{k \mathop = 0}^\infty \map \Pr {X = k} s^k\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac \d {\d s} \map \Pr {X = k} s^k\) | Abel's Theorem | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty k \map \Pr {X = k} s^{k - 1}\) | Power Rule for Derivatives |

Plugging in $s = 1$ gives:

- $\ds \map {\Pi'_X} 1 = \sum_{k \mathop = 0}^\infty k \map \Pr {X = k} 1^{k - 1} = \map p 1 + 2 \map p 2 + 3 \map p 3 + \cdots$

But:

- $\ds \sum_{k \mathop = 0}^\infty k \map \Pr {X = k} 1^{k - 1} = \sum_{k \mathop = 0}^\infty k \map \Pr {X = k}$

is precisely the definition of the expectation.

$\blacksquare$

## 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 PGF and plug in $1$.

Assuming, of course, you know what the PGF is.

## Sources

- 1986: Geoffrey Grimmett and Dominic Welsh:
*Probability: An Introduction*... (previous) ... (next): $\S 4.3$: Moments: $(17)$