Mean of the Poisson distribution

Statement

Let $\lambda > 0$ and let $X$ have the Poisson distribution with parameter $\lambda$. Then $$E[X] = \sum_{k=0}^{\infty} k \, e^{-\lambda} \frac{\lambda^k}{k!} = \lambda.$$

Remarks

Proof: the term with $k = 0$ is zero. For $k \ge 1$, $k \lambda^k / k! = \lambda \cdot \lambda^{k-1}/(k-1)!$. So the sum is $\lambda e^{-\lambda} \sum_{j \ge 0} \lambda^j / j! = \lambda e^{-\lambda} e^{\lambda} = \lambda$. The variance of $X$ is also $\lambda$. So the parameter $\lambda$ is the mean number of events.

Explanations

L1 human

The parameter of a Poisson distribution is its average. If a shop gets on average $4$ customers per hour, a Poisson model of the number of customers in one hour uses $\lambda = 4$.