Expectation of Poisson Distribution

Theorem

Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.

Then the expectation of $X$ is given by:

$E \left({X}\right) = \lambda$

Proof 1

From the definition of expectation:

$\displaystyle E \left({X}\right) = \sum_{x \mathop \in \operatorname{Im} \left({X}\right)} x \Pr \left({X = x}\right)$

By definition of Poisson distribution:

$\displaystyle E \left({X}\right) = \sum_{k \mathop \ge 0} k \frac 1 {k!} \lambda^k e^{-\lambda}$

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle E \left({X}\right)\)	\(=\)	\(\displaystyle \)	\(\displaystyle \lambda e^{-\lambda} \sum_{k \mathop \ge 1} \frac 1 {\left({k-1}\right)!} \lambda^{k-1}\)	\(\displaystyle \)	\(\displaystyle \)	the $k = 0$ term vanishes
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \lambda e^{-\lambda} \sum_{j \mathop \ge 0} \frac {\lambda^j} {j!}\)	\(\displaystyle \)	\(\displaystyle \)	putting $j = k - 1$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \lambda e^{-\lambda} e^{\lambda}\)	\(\displaystyle \)	\(\displaystyle \)	Taylor Series Expansion for Exponential Function
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \lambda\)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Proof 2

From the Probability Generating Function of Poisson Distribution, we have:

$\Pi_X \left({s}\right) = e^{-\lambda \left({1-s}\right)}$

From Expectation of Discrete Random Variable from PGF, we have:

$E \left({X}\right) = \Pi'_X \left({1}\right)$

We have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \Pi'_X \left({s}\right)\)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac {\mathrm d} {\mathrm ds} e^{-\lambda \left({1-s}\right)}\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \lambda e^{- \lambda \left({1-s}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	Derivatives of PGF of Poisson Distribution

Plugging in $s = 1$:

$\Pi'_X \left({1}\right) = \lambda e^{- \lambda \left({1-1}\right)} = \lambda e^0$

Hence the result from Exponential of Zero and One: $e^0 = 1$.

$\blacksquare$

Sources

• Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction (1986)... (previous)... (next): $\S 2.4$: Expectation: Exercise $10$