Sum of Independent Binomial Random Variables
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Theorem
Let $X$ and $Y$ be discrete random variables with a binomial distribution:
- $X \sim \Binomial m p$
and
- $Y \sim \Binomial n p$
Let $X$ and $Y$ be independent.
Then their sum $Z = X + Y$ is distributed as:
- $Z \sim \Binomial {m + n} p$
Proof
From Probability Generating Function of Poisson Distribution, we have that the probability generating functions of $X$ and $Y$ are given by:
- $\map {\Pi_X} s = \paren {q + p s}^m$
- $\map {\Pi_Y} s = \paren {q + p s}^n$
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {q + p s}^m \paren {q + p s}^n\) | Definition of $X$ and $Y$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {q + p s}^{m + n}\) | Product of Powers |
This is the probability generating function for a discrete random variable with a binomial distribution:
- $\Binomial {m + n} p$
Therefore:
- $Z = X + Y \sim \Binomial {m + n} p$
$\blacksquare$
Sources
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 4.4$: Sums of independent random variables: Exercise $8$