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:

• $G_X\left(s\right) = e^{-\lambda_1\left(1-s\right)}$
• $G_Y\left(s\right) = e^{-\lambda_2\left(1-s\right)}$

respectively.

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:

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

$\blacksquare$