# Sum of Independent Poisson Random Variables is Poisson

## 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 1

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:

 $\ds \map {\Pi_{X + Y} } s$ $=$ $\ds \map {\Pi_X} s \, \map {\Pi_Y} s$ PGF of Sum of Independent Discrete Random Variables $\ds$ $=$ $\ds e^{-\lambda_1 \paren {1 - s} } e^{-\lambda_2 \paren {1 - s} }$ Definition of $X$ and $Y$ $\ds$ $=$ $\ds e^{-\paren {\lambda_1 + \lambda_2} \paren {1 - s} }$ Exponential of Sum

This is the probability generating function for a discrete random variable with a Poisson distribution:

$\Poisson {\lambda_1 + \lambda_2}$

Therefore:

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

$\blacksquare$

## Proof 2

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

$\blacksquare$