> [!NOTE] Theorem > Let $\lambda\geq 0.$ Let $X$ be a [[Discrete random variables|discrete real-valued random]] that has a [[Poisson Distribution|poisson distribution]] with parameter $\lambda.$ Then the [[Expectation of Discrete Real-Valued Random Variable|expectation]] of $X$ is $\lambda.$ **Proof**: We have $\begin{align} \mathbb{E}[X] &= \sum_{n\in \mathbb{N}} p_{X}(n)\cdot n \\ &= \sum_{n\in \mathbb{N}} n \frac{\lambda^{n}}{n!} e^{-\lambda} \\ & = \lambda e^{-\lambda}\sum_{n\in \mathbb{N^{+}}} \frac{\lambda^{n-1}}{(n-1)!} \\ & = \lambda e^{-\lambda}\cdot e^{\lambda} \\ &= \lambda. \end{align}$since by [[Real Exponential Function]], $e^{\lambda} = \sum_{n\in \mathbb{N}} \frac{\lambda^{n}}{n!}.$