> [!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 $X$ is [[Square-Integrable Discrete Real-Valued Random Variable|square-integrable]] and its [[Variance of Square-Integrable Discrete Real-Valued Random Variable|variance]] is $\lambda.$ **Proof**: By [[Expectation of Real-Valued Function of Discrete Real-Valued Random Variable]], $\begin{align} \mathbb{E}[X^{2}] &= \sum_{n\in \mathbb{N}} p_{X}(n)\cdot n^{2} \\ &= \sum_{n=0}^{\infty} n^{2} \frac{\lambda^{n}}{n!} e^{-\lambda} = \sum_{n=1}^{\infty} n \frac{\lambda^{n}}{(n-1)!} e^{-\lambda} \\ &= \sum_{n=1}^{\infty} \frac{\lambda^{n}}{(n-1)!} e^{-\lambda} + \sum_{n=1}^{\infty} (n-1) \frac{\lambda^{n}}{(n-1)!} e^{-\lambda} \\ &= \lambda e^{-\lambda} \sum_{n=1}^{\infty} \frac{\lambda^{n-1}}{(n-1)!} + \lambda^{2}e^{-\lambda} \sum_{n=2}^{\infty } \lambda^{n-2} \\ &= \lambda + \lambda^{2} \end{align}$thus $X$ is indeed square-integrable. By [[Expectation of Poisson Distribution]], $\mathbb{E}[X]=\lambda.$ Thus its variance is given by $\text{Var}(X)=\mathbb{E}[X^{2}]-(\mathbb{E}[X])^{2} = \lambda + \lambda^{2} - \lambda^{2}= \lambda.$