Sum of Independent Poisson Random Variables is Poisson/Proof 1

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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 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:

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

$\blacksquare$