Let X and Y be independent random variable each Poisson distributed with common parameter $\lambda$. Find the conditional probability $$P(Y=y|X+Y=z) \text{ for } y=0,1,…,z.$$
my method is:
let
$Z=X+Y$ and $X1=X$ ,after finding the joint pdf of $Z$ and $X$, $f_{X Z}(x,z)=\frac{\lambda^{2x+2y}\times e^{-2y}}{x!(x+y)!}$, and i will know the marginal PDF of $Z$,
but I am stuck here, I dont know how to do this,$f_{Z}(z)=\int^z_0\frac{\lambda^{2x+2y}\times e^{-2y}}{x!(x+y)!}dx$, or is there another method to solve this question?
Best Answer
From the definition of conditional probability and independence of $X$ and $Y$: $$\begin{align} P(Y=y\mid X+Y=z) &= \frac{P(Y=y \text{ and }X+Y=z)}{P(X+Y=z)} \\[1ex] &= \frac{P(Y=y \text{ and }Y=z-x)}{P(X+Y=z)} \\[1ex] &= \frac{P(Y=y)\times P(Y=z-x)}{P(X+Y=z)} && \text{because, independence} \end{align}$$ Now use the fact that $X+Y$ will be a Poisson r.v. with mean $2\lambda$: $$ P(Y=y|X+Y=z)= \frac{(\lambda^x e^{-\lambda}/x!) \times (\lambda^{z-x} e^{-\lambda}/(z-x)!)}{2^z\lambda^z e^{-2\lambda}/z!}. $$