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