> [!NOTE] Lemma >Let $\lambda \geq 0.$ Let $n\in \mathbb{N}.$ Then the function $p_{X}(x) = \begin{cases}\frac{e^{-\lambda}\lambda^{x}}{ x!}, & x\in \mathbb{N}, \\0, & \text{otherwise} \end{cases}$is a [[Probability Mass Function|probability mass function]]. **Proof**: We have $\sum_{k=0}^\infty p_X(k)=\sum_{k=0}^\infty\frac{\lambda^k}{k!}\cdot e^{-\lambda}=e^{-\lambda}\cdot\sum_{k=0}^\infty\frac{\lambda^k}{k!}=e^{-\lambda}\cdot e^\lambda=1$by definition of [[Real Exponential Function]]. # Applications A discrete real-valued random variable is said to have [[Poisson Distribution|Poisson distribution]] if its probability mass function is given by the above function.