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?
 A: 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.
A: Let us assume we have a Poisson process with an arrival rate of $\lambda$.  After some time $t$, $N_t$ unobserved arrivals have occurred.  After some more time, say $\tau$, we observe that $N_{t+\tau}$ arrivals have occurred.   What is the distribution of $N_t$ given the observed value of $N_{t+\tau}$?
As it happens, the memoryless property of the Poisson process implies that the time of any given arrival is uniformly distributed over, in this case, $[0, t+\tau]$.  This implies that the probability $p$ that any given arrival in $[0, t+\tau]$ actually shows up in the interval $[0, t]$  is just $p = t/(t+\tau)$, the fraction of the total time that occurred before $t$, and is independent of the time of any other arrival. If we have $N_{t+\tau}$ arrivals overall, the number that arrive in $[0, t]$ is therefore distributed Binomial$(N_{t+\tau}, { t \over (t+\tau)})$
In this case, we have $t=2$, $t+\tau = 6$, and $N_{t+\tau} = 3$.  Substituting gives us the probability distribution of $Y = (N_2|N_6=3)$, which is a Binomial$(3, 1/3$) distribution.  Getting from this to $\mathbb{E}e^Y$ is a straightforward calculation.
