Let X∼Poisson(λ1)X \sim \text{Poisson}(\lambda_1)X∼Poisson(λ1) and Y∼Poisson(λ2)Y \sim \text{Poisson}(\lambda_2)Y∼Poisson(λ2) be independent. If Z=X+YZ = X+YZ=X+Y, what is the probability P(X=k∣Z=n)P(X=k | Z=n)P(X=k∣Z=n)?
(nk)(λ1λ1+λ2)k(λ2λ1+λ2)n−k\binom{n}{k} (\frac{\lambda_1}{\lambda_1+\lambda_2})^k (\frac{\lambda_2}{\lambda_1+\lambda_2})^{n-k}(kn)(λ1+λ2λ1)k(λ1+λ2λ2)n−k
λ1ke−λ1k!\frac{\lambda_1^k e^{-\lambda_1}}{k!}k!λ1ke−λ1
(nk)(λ2λ1+λ2)k(λ1λ1+λ2)n−k\binom{n}{k} (\frac{\lambda_2}{\lambda_1+\lambda_2})^k (\frac{\lambda_1}{\lambda_1+\lambda_2})^{n-k}(kn)(λ1+λ2λ2)k(λ1+λ2λ1)n−k
e−(λ1+λ2)(λ1+λ2)nn!\frac{e^{-(\lambda_1+\lambda_2)}(\lambda_1+\lambda_2)^n}{n!}n!e−(λ1+λ2)(λ1+λ2)n