conditional expectation for Poisson

Question

Suppose that $X_t$ is a Poisson process with parameter $\lambda= 3$. Find $E(X_2 + X_4 | X_3)$?

I do not know how to approach this problem. Thanks.

Answer 1

To begin with, you can determine the conditional distribution of $X_2$ given $X_3$ as follows: if $n \geq m$, then \begin{align} & P[X_2 = m | X_3 = n] \\ = & \frac{P(X_2 = m, X_3 = n)}{P(X_3 = n)} \\ = & \frac{P(X_2 = m, X_3 - X_2 = n - m)}{P(X_3 = n)} \\ = & \frac{P(X_2 = m)P(X_3 - X_2 = n - m)}{P(X_3 = n)} \quad \text{(by incremental independence)}\\ = & \frac{e^{-6}\frac{6^m}{m!} \times e^{-3}\frac{3^{n - m}}{(n - m)!}}{e^{-9}\frac{9^n}{n!}} \quad \text{(by } X_t - X_s \sim \text{ Poisson}(\lambda(t - s)))\\ = & \binom{n}{m}\left(\frac{2}{3}\right)^m\left(\frac{1}{3}\right)^{n - m} , \quad m = 0, 1, \ldots, n. \end{align} If $n < m$, the conditional probability is $0$. Therefore, $$X_2 | X_3 \sim \text{Binomial}(X_3, 2/3),$$ which leads to $E[X_2 | X_3] = \frac{2}{3}X_3$. $E[X_4 | X_3]$ can be handled similarly.