Lesson 2

Formally, for \(\lambda \ge 0\), we say that a random variable \(\alpha\) follows a Poisson distribution (and write \(\alpha \sim \operatorname{Poisson}(\lambda)\)) if for any \(k \in \mathbb{Z}_{\ge 0}\) it holds that \[\Pr[\alpha=k] = \frac{\lambda^ke^{-\lambda}}{k!}\ ,\] where for \(\lambda=k=0\) we use the convention \(0^{0}=1\). The fact that the sum of these probabilities equals one follows from the expansion of the function \(e^{\lambda}\) into its Taylor series \(\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}\). This same series helps compute the expected value of this random variable: \[\begin{align*}\operatorname{E}[\alpha]&=\sum_{k=0}^{\infty}k\Pr[\alpha=k]= \\&=\sum_{k=0}^{\infty}k \cdot \frac{\lambda^ke^{-\lambda}}{k!}=\\&=e^{-\lambda}\lambda \cdot \sum_{k=1}^{\infty}\frac{\lambda^{k-1}}{(k-1)!}=&\text{(\(l=k-1\))}\\&=e^{-\lambda}\lambda \cdot \sum_{l=0}^{\infty}\frac{\lambda^l}{l!}=&\text{(Taylor series)}\\&=e^{-\lambda}\lambda e^{\lambda}=\lambda \ .\end{align*}\] This aligns with the real-world phenomenon we are modeling: we expected the mean value of our random variable to be \(\lambda\).