Question on two independent random variables under Poisson distribution

Question

$X$, $Y$ are independent random variables, $X$ ~ Poiss$(λ)$, $Y$ ~ Poiss$(μ)$.

How to find:

a) $P( X > 0 | X+Y )$

b) $E( X | X+Y) $ ?

Answer 1

If $X$ and $Y$ are independent Poisson Random Variables with respective means $\lambda_1$ and $\lambda_2$, the conditional PMF of $X$ given $X+Y=n$ can be obtained as follows:

Let $Z=X+Y$. We want to find $p_{X|Z=n}(k)$.

For $k=0,1,\ldots ,n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p_{X|Z=n}(k)=\frac{P(X=k,\ Z=n)}{P(Z=n)}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{P(X=k,\ X+Y=n)}{P(Z=n)}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{P(X=k,\ Y=n-k)}{P(Z=n)}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{P(X=k)P(Y=n-k)}{P(Z=n)}$

We know that $Z$ is Poisson with mean $\lambda_1 +\lambda_2$.

$$ p_{X|Z=n}(k)= \frac{\frac{e^{-\lambda_1}\lambda_1^k}{k!}\cdot\frac{e^{-\lambda_2}\lambda_2^{n-k}}{(n-k)!}}{\frac{e^{-(\lambda_1+\lambda_2)}(\lambda_1+\lambda_2)^n}{n!}}$$

$$= {n\choose k}\bigg(\frac{\lambda_1}{\lambda_1+\lambda_2}\bigg)^{k}\bigg(\frac{\lambda_2}{\lambda_1+\lambda_2}\bigg)^{n-k}$$ Hence, the conditional distribution of $X$ given $X+Y=n$ is a binomial distribution with parameters $n$ and $\frac{\lambda_1}{\lambda_1+\lambda_2}$

Therefore, the conditional expected value of $X$ given $X+Y=n$ will be$$E(X|X+Y=n)=\frac{\lambda_1n}{\lambda_1+\lambda_2}$$

Answer 2

HINT Since $X,Y$ are both integer and non-negative, I would try to find the reverse, i.e. $\mathbb{P}[X=0|X+Y=n]$.

More generally, for the second one, consider $\mathbb{P}[X=k|X+Y=n]$ for $0 \le k \le n$.