Sum of Independent Poisson Random Variables is Poisson/Proof 2

Theorem

Let $X$ and $Y$ be independent discrete random variables with:

$X \sim \Poisson {\lambda_1}$

and

$Y \sim \Poisson {\lambda_2}$

for some $\lambda_1, \lambda_2 \in \R_{> 0}$.

Then their sum $X + Y$ is distributed:

$X + Y \sim \Poisson {\lambda_1 + \lambda_2}$

Proof

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

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

and

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

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

\(\ds \map {M_{X + Y} } t\)

\(=\)

\(\ds \map {M_X} t \map {M_Y} t\)

Moment Generating Function of Linear Combination of Independent Random Variables

\(\ds \)

\(=\)

\(\ds e^{\lambda_1 \paren {e^t - 1} } e^{\lambda_2 \paren {e^t - 1} }\)

\(\ds \)

\(=\)

\(\ds e^{\paren {\lambda_1 + \lambda_2} \paren {e^t - 1} }\)

Exponential of Sum

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

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

$\blacksquare$