Let $X\sim \operatorname{Poi}(\lambda)$ and $Y\sim \operatorname{Poi}(\mu)$, find $P(X>0|X+Y)$.

Question

Let $X\sim \operatorname{Poi}(\lambda)$ and $Y\sim \operatorname{Poi}(\mu)$, find $P(X>0|X+Y)$.

My attempt

We have $P(X>0|X+Y)=\frac{P(\{X>0\}\cap \{X+Y=k\})}{P(X+Y=k)}$. What is $\{X>0\}\cap \{X+Y=k\}$?

Answer 1

Note that the probability in your title does not make sense, you should be conditioning on $X+Y=k$ which I guess is what you are getting at with your attempt. If $X,Y$ are independent then $X+Y \sim \text{Poisson}(\lambda + \mu)$ and

\begin{align*} P(X>0|X+Y=k) &= 1-P(X=0|X+Y=k) \\ &= 1 - \frac{P(X=0, X+Y=k)}{P(X+Y=k)}\\ &= 1 - \frac{P(X=0, Y=k)}{P(X+Y=k)}\\ &= 1 - \frac{P(X=0)P(Y=k)}{P(X+Y=k)}\\ &= 1 - \frac{e^{-\lambda} \frac{\mu^k e^{-\mu}}{k!} }{\frac{(\lambda+\mu)^k e^{-\mu-\lambda}}{k!}}\\ &= 1- \left ( \frac{\mu}{\lambda+\mu}\right)^k \end{align*}