Problem
Assume $X \sim \text{Poisson}(\lambda)$ and $Y \sim \text{Poisson}(\mu)$, and that $X, Y$ are independent. Let $n$ be a positive integer. Show that $(X \vert X+Y = n) \sim \text{Binomial}(n, p)$ with $p = \frac{\lambda}{\lambda + \mu}$.
Solution
First, let's write out all of the facts in full form:
- $X \sim \text{Poisson}(\lambda)$ means $f_X(x) = e^{-\lambda} \frac{\lambda^x}{x!}$
- $Y \sim \text{Poisson}(\mu)$ means $f_Y(y) = e^{-\mu} \frac{\mu^y}{y!}$
- $X, Y$ are independent means:
- $P(X=x, Y=y) = P(X=x)P(Y=y)$ and $f_{X,Y}(x,y) = f_x(x) f_y(y)$
- so $\text{Poisson}(\lambda) \text{Poisson}(\mu) = \text{Poisson}(\lambda + \mu)$
- $\text{Binomial}(n,p) = \binom{n}{x} p^x(1-p)^{n-x}$
Let's apply the definition of conditional probability to $P(X=x \vert X+Y=n)$:
$P(X=x \vert X+Y=n) \\ = \frac{P(X=x, X+Y=n)}{P(X+Y=n)} $
And let's apply what is perhaps the key observation in solving this problem: $P(X+Y=n) = P(Y=n-x)$. We can do this because we are given that $X+Y=n$. (Note: I still feel a bit uneasy about this).
$\dfrac{P(X=x, X+Y=n)}{P(X+Y=n)} \\ = \dfrac{P(X=x, Y=n-x)}{P(X+Y=n)} \\ = \dfrac{P(X=x)P(Y=n-x)}{P(X+Y=n)} \\ = \dfrac{f_X(x) f_Y(n-x)}{f_{X+Y}(n)} \\ = \dfrac { e^{-\lambda} \frac{\lambda ^ x}{x!} e^{-\mu} \frac{\mu^{(n-x)}}{(n-x)!} } { e^{-(\lambda + \mu)} \frac{(\lambda + \mu) ^ n}{n!} } $
From here it's a matter of algebraic manipulation to make this look like $\text{Binomial}(n, p)$ with $p = \frac{\lambda}{\lambda + \mu}$:
$\dfrac { e^{-\lambda} \frac{\lambda ^ x}{x!} e^{-\mu} \frac{\mu^{(n-x)}}{(n-x)!} } { e^{-(\lambda + \mu)} \frac{(\lambda + \mu) ^ n}{n!} } \\ = \dfrac { e^{-(\lambda + \mu)} \frac{\lambda ^ x}{x!} \frac{\mu^{(n-x)}}{(n-x)!} } { e^{-(\lambda + \mu)} \frac{(\lambda + \mu) ^ n}{n!} } \\ = \dfrac { \frac{\lambda ^ x}{x!} \frac{\mu^{(n-x)}}{(n-x)!} } { \frac{(\lambda + \mu) ^ n}{n!} } \\ = \frac{n!}{(n-x)! x!} \dfrac { \lambda^x \mu^{n-x} } { (\lambda + \mu)^n } \\ = \binom{n}{x} \dfrac { \lambda^x \mu^{n-x} } { (\lambda + \mu)^n } \\ = \binom{n}{x} \dfrac { \lambda^x \mu^{n-x} } { (\lambda + \mu)^x (\lambda + \mu)^{n-x} } \\ = \binom{n}{x} (\dfrac{\lambda}{\lambda + \mu})^x (\dfrac{\mu}{\lambda + \mu})^{n-x} \\ = \binom{n}{x} (\dfrac{\lambda}{\lambda + \mu})^x (\dfrac{\lambda + \mu}{\lambda + \mu} - \dfrac{\lambda}{\lambda + \mu})^{n-x} \\ = \binom{n}{x} (\dfrac{\lambda}{\lambda + \mu})^x (1 - \dfrac{\lambda}{\lambda + \mu})^{n-x} $
now setting $p = \frac{\lambda}{\lambda + \mu}$ we finally get $ \binom{n}{x} p^x(1-p)^{n-x} = \text{Binomial}(n,p)$
Sources
- 36-700 hw3 problem 2