Sum of independent Poisson random variables

Statement

Let $X_1, \dots, X_n$ be independent random variables with $X_i \sim \operatorname{Poisson}(\lambda_i)$ and $\lambda_i > 0$. Then $$X_1 + \dots + X_n \sim \operatorname{Poisson}(\lambda_1 + \dots + \lambda_n).$$

Remarks

Proof with generating functions: $G_{X_i}(z) = e^{\lambda_i (z - 1)}$, so the sum has $G(z) = e^{(\lambda_1 + \dots + \lambda_n)(z - 1)}$. This is the generating function of $\operatorname{Poisson}(\lambda_1 + \dots + \lambda_n)$. A direct proof for $n = 2$ uses the binomial theorem: $\sum_{k=0}^{m} e^{-\lambda_1} \frac{\lambda_1^k}{k!} e^{-\lambda_2} \frac{\lambda_2^{m-k}}{(m-k)!} = e^{-(\lambda_1 + \lambda_2)} \frac{(\lambda_1 + \lambda_2)^m}{m!}$. Conversely, given $X_1 + X_2 = m$, the conditional law of $X_1$ is $\operatorname{Bin}(m, \lambda_1 / (\lambda_1 + \lambda_2))$.

Explanations

L1 human

If calls come to two help desks as independent Poisson counts, the total number of calls is again a Poisson count. Its mean is the sum of the two means.