Sum of Independent Poisson Random Variables is Poisson/Proof 2

Proof
From Moment Generating Function of Poisson Distribution, the moment generating functions $X$ and $Y$, $M_X$ and $M_Y$, are given by:


 * $\displaystyle \map {M_X} t = e^{\lambda_1 \paren {e^t - 1} }$

and


 * $\displaystyle \map {M_Y} t = e^{\lambda_2 \paren {e^t - 1} }$

As $X$ and $Y$ are independent, we may apply Moment Generating Function of Sum of Independent Random Variables, to give:

This is the moment generating function of a random variable with distribution $\Poisson {\lambda_1 + \lambda_2}$.

So, $X + Y \sim \Poisson {\lambda_1 + \lambda_2}$.