Suppose that $X_1$ and $X_2$ are independent Poisson distributed random variables with means $\lambda_1$ and $\lambda_2$. Given is that $W = X_1 + X_2$ is also Poisson distributed with mean $\lambda_1 + \lambda_2$. Use this result to show that the conditional distribution of $X_1$, given that $W=w$, is a binomial distribution with $n=w$ and $p = \lambda_1 /(\lambda_1 + \lambda_2)$.
I looked up the PDFs for the Poisson distribution and the binomial distribution. I also looked at the formula for conditional probabilities. However, I cant seem to find the clue where to start. Could anyone please help me out?
Best Answer
$$P(X_1 = x \mid W = w) = \dfrac{P(X_1 = x \text{ and } W = w)}{P(W=w)} $$ Note that if $X_1 = x$ and $W=w$ you must have $X_2 = w - x$. Now evaluate these probabilities using the Poisson distribution.