Sum of Poisson Random Variables

Theorem

Let $X$ and $Y$ be independent Poisson random variables with parameters $\mu$ and $\nu$ respectively.

Then $X+Y\sim \text{Po}(\mu +\nu )$.

Interpretation

If two independent sources each produce events at Poisson rates, combining them gives a Poisson with the sum of the rates. The sum of independent Poissons is Poisson.

Proof

By MGF convolution:

$$\begin{array}{rl}{\varphi }_{X+Y}(t)& ={\varphi }_X(t)\cdot {\varphi }_Y(t)\\ & =e^{\mu (e^t-1)}\cdot e^{\nu (e^t-1)}\\ & =e^{(\mu +\nu )(e^t-1)}\end{array}$$

This is the MGF of $\text{Po}(\mu +\nu )$, so $X+Y\sim \text{Po}(\mu +\nu )$.

Related

• Poisson Distribution
• Poisson-Binomial Distribution