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