# Problem with Poisson distribution

Let X and Y be two independent real random variables following two Poisson distributions that are denoted respectively by $P(\lambda_1)$ and $P(\lambda_2)$. Let $Z = X + Y$. Now:

1. Determine the generating function of $Z$. Deduce the distribution of $Z$.
2. Find the conditional distribution of $X$ given $Z = n$.

I'm concerned about the answer of part 2.

• Is this homework? If so, please tag it as such. – Fomite Nov 5 '11 at 15:59

## 1 Answer

Let $Z = X + Y$. We want to find $P(X=k|Z=n)$. For $k = 0,\ldots, n$

$P[X=k|Z=n] = P(X = k,Z = n)/P(Z = n)$

$\quad= P(X = k,X + Y = n)/P(Z = n)$

$\quad= P(X = k, Y = n − k)/P(Z = n)$

$\quad= P(X = k)\,P(Y = n − k)/P(Z = n)$

We know that $Z$ is Poisson with mean $λ_1 + λ_2$. Then

$P[X=k|Z=n] = P(X = k)\,P(Y = n − k)/P(Z = n)$

$\quad= [e^{−λ_1} λ_1^k/k!][ e^{−λ_2} λ_2^{n−k}/(n−k)!]\big/[e^{−(λ_1+λ_2)} (λ_1+λ_2)^n/n!]$

$\quad= [e^{-λ_1+λ_2}\,(n!λ_1^k\,λ_2^{n−k})]/[k!\,e^{-λ_1+λ_2}\,(n−k)!\,(λ_1+λ_2)^n]$

$\quad= (n!\,λ_1^k\,λ_2^{n−k})/k!\,(n−k)!\,(λ_1+λ_2)^n$

$\quad= [n!/k!(n-k)!] \, [λ_1^k/(λ_1+λ_2)^k] \, [λ_2^{n-k}/(λ_1+λ_2)^{n-k}]$

$\quad= [n!/k!(n-k)!] \, [λ_1/(λ_1+λ_2)]^k \, [λ_2/(λ_1+λ_2)]^{n-k}$

The binomial probability mass function looks like $$P(X=k)= \frac{n!}{k!(n-k)!} \, p^k \, (1-p)^{n-k}$$ Hence the conditional distribution of $X$ given $Z$ is a binomial distribution with parameters $n$ and $λ_1/(λ_1+λ_2)$.

• Please, use $\LaTeX$ (see help) to format your equations (use the align environment), and check their correctness. – chl Nov 9 '11 at 14:10