Sum of Independent Poisson Random Variables is Poisson

Theorem
Let $X$ and $Y$ be discrete random variables with a Poisson distribution:


 * $X \sim \operatorname{Poisson} \left({\lambda_1}\right)$

and
 * $Y \sim \operatorname{Poisson} \left({\lambda_2}\right)$

Let $X$ and $Y$ be independent.

Then their sum $Z = X + Y$ is distributed as:


 * $Z \sim \operatorname{Poisson} \left({\lambda_1 + \lambda_2}\right)$

Proof
From Probability Generating Function of Poisson Distribution, we have that the probability generating functions of $X$ and $Y$ are given by:
 * $\Pi_X \left({s}\right) = e^{-\lambda_1 \left({1 - s}\right)}$
 * $\Pi_Y \left({s}\right) = e^{-\lambda_2 \left({1 - s}\right)}$

respectively.

Now because of their independence, we have:

This is the probability generating function for a discrete random variable with a Poisson distribution:
 * $\operatorname{Poisson} \left({\lambda_1 + \lambda_2}\right)$

Therefore:
 * $Z = X + Y \sim \operatorname{Poisson} \left({\lambda_1 + \lambda_2}\right)$