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 the sum $Z = X + Y$ is distributed as:


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

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

Now because of their independence, we have:
 * $G_{X+Y} \left({s}\right) = G_X \left({s}\right) G_Y \left({s}\right) = e^{-\lambda_1 \left({1-s}\right)} \cdot e^{-\lambda_2 \left({1-s}\right)} = e^{-\left({\lambda_1 + \lambda_2}\right) \left({1-s}\right)}$

This is the probability generating function for a $\operatorname{Poisson} \left({\lambda_1 + \lambda_2}\right)$ random variable.

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