0
$\begingroup$

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?

  • 1
    You're confusing discrete and continuous variables. Here you want have integrals, just sums. Try to prove first that $Z\simeq \mathrm{Poisson}(2\lambda)$. – 2017-02-03

1 Answers 1

3

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 $2\lambda$: $$ P(Y=y|X+Y=z)= \frac{(\lambda^x e^{-\lambda}/x!) \times (\lambda^{z-x} e^{-\lambda}/(z-x)!)}{2^z\lambda^z e^{-2\lambda}/z!}. $$

  • 0
    PS: If you don't already 'know' that fact, you can show it via, $$\mathsf P(Z=z) ~=~ \sum_{k=0}^z \mathsf P(X=k)~\mathsf P(Y=z-k)$$ (PPS: Don't forget the binomial expansion, $(1+1)^z = \sum_{k=0}^z \binom{z}{k}$) – 2017-02-03
  • 0
    I don't think this answer is correct. See https://math.stackexchange.com/questions/2978826/potential-typo-and-independence-between-two-random-variables-that-are-related-v – 2018-10-31