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:
- $X + Y \sim \Poisson {\lambda_1 + \lambda_2}$
$\blacksquare$