Given that $X$ is a Poisson process, find the expectations.

Question

Okay, so $X$ is a Poisson process with intensity $\lambda>0$. We have arrival times ${W_1,W_2,\dots}$ and are given that $X(T)=5$.

We have to find:

1. E$[W_1|X(T)=5]$
2. E$[W_4|X(T)=5]$
3. E$[W_1-W_4|X(T)=5]$

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So, I understand that up to time $t$ there are $5$ arrivals. I think that I want to calculate the density functions of $W_1|X(T)=5$ and $W_4|X(T)=5$ by finding the CDF, but I'm not sure where to start with that. My book doesn't actually show how to do a conditional expectation with Poisson processes, it just shows the theory. I'd really appreciate help with this example.

Answer 1

#OrderStatistics

It is a property of Poisson events that, on condition that it happens in a given period, the time of that appearance will be uniformly distributed over that given period.

It is also a property of events in a Poisson process that they are mutually independent (and otherwise indistinguishable).

Therefore on condition that a certain number happen within a period, the times of appearance will be distributed as the order statistics of so many independent uniform distributed events.

1. $\mathsf E(W_1\mid X(T)=5)$ is the expectation of the first arrival time given that it, and the next four (but not any more) arrive within period $(0;5]$.

$$\begin{align}\mathsf E(W_1\mid X(T)=5) ~=~&5\int_0^5 w~f_U(w)~\Big(1-F_U(w)\Big)^4\operatorname d w & : U\sim\mathcal{U}[0;5] \\[1ex] ~=~& \frac 1{5^4}\int_0^5 w~(5-w)^4\operatorname d w \\[1ex] ~=~& \frac{5}{6} \\[1ex] ~=~& 0.8\dot{\overline 3} \end{align}$$

2. $\mathsf E(W_4\mid X(T)=5)$ is the expectation of the fourth arrival time given that it, the next, and the previous three (but not any more) arrive within period $(0;5]$.

$$\begin{align}\mathsf E(W_4\mid X(T)=5) ~=~& \binom{5}{3,1,1} \int_0^5 w~f_U(w)~F_U(w)^3~\Big(1-F_U(w)\Big)\operatorname d w & : U\sim\mathcal{U}[0;5] \end{align}$$

3. $\mathsf E(W_4-W_1\mid X(T)=5)$ is the expected difference between the fourth and first arrival times.   This is immediately obvious (via Linearity of Expectation).