Sum of Independent Poisson Random Variables is Poisson/Proof 1

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

respectively.

Now because of their independence, we have:

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

Therefore:
 * $X + Y \sim \Poisson {\lambda_1 + \lambda_2}$