Expected value of conditional Poisson process I have been trying to solve this issue for quite a while. So, lets say that we have a Poisson process, $N = (N_t, t \geq 0)$ and the $\lambda = 3$. Lets say that $Y = (N_2 | N_6 = 3)$. Find $Ee^Y$. However, I am stuck on understanding the notation of $Y$ and how should I process it to continue the calculation of expected value. Should I find joint density function?

The notation '$Y = (N_2 \mid N_6 = 3)$' implies that the distribution of the random variable $Y$ is the conditional distribution of $N_2$ given $N_6=3$.
Now if $(N_t)_{t\ge 0}$ is a Poisson process with intensity parameter $\lambda(>0)$, then the following holds:

*

*$N_t\sim \text{Poisson}(\lambda t)$.


*$N_{t+s}-N_s\sim \text{Poisson}(\lambda t)$ is independent of $N_s$.
Using this information, one can find the conditional distribution of $N_s$ given $N_t$ for $0<s<t$. This turns out to be a standard distribution, and you are required to find/recall the moment generating function of this distribution.
