# Poisson Processes: Conditional Expectations

The question:
Customers arrive at a service facility according to a Poisson process {$X(t); t \ge 0$} of rate $\lambda$ customers/hour. Let X(t) be the number of customers that have arrived up to time t. Let $W_{1},W_{2},...$ be successive arrival times of the customers.
(1) Determine the conditional expectation $E[W_{3}|X(t)=5]$.
(2) Determine the conditional probability density function for $W_{2}$, given that $X(t)=5$.

(1)
So we would want to find the probability $Pr[W_{3} \le s|X(t)=5] = Pr[X(s) \ge 3|X(t)=5]$ and this is equal because if three arrivals occurred before or by time s, then the number of arrivals (X(s)) will be equal to or greater than three at time s.

So we want to find the expectation like this (I think): $E[W_{3}|X(t)=5] = \int_{0}^{t}x(\frac{d}{ds}Pr[X(s) \ge 3|X(t) = 5])dx$

$Pr[X(s) \ge 3|X(t) = 5] = 1-Pr[X(s) \le 2|X(t) = 5] \\ = 1 - \frac{Pr[X(s) \le 2, \ \ X(t)=5]}{Pr[X(t)=5]} = 1 - \frac{Pr[X(s) \le 2, \ \ 3 \le X(t)-X(s)\le5]}{Pr[X(t)=5]} = 1 - \frac{Pr[X(s) \le 2] \ Pr[ 3 \le X(t)-X(s)\le5]}{Pr[X(t)=5]}$
Now the two probabilities in the denominator must be determined:
$Pr[X(s) \le 2] = Pr[X(s)=0]+Pr[X(s)=1]+Pr[X(s)=2]$ these are all Poisson distributed with parameter $\lambda s$.
$Pr[ 3 \le X(t)-X(s)\le5] = Pr[X(t)-X(s)=3]+Pr[X(t)-X(s)=4]+Pr[X(t)-X(s)=5]$
and these are also Poisson distributed with parameter $\lambda (t-s)$. Then can I take the derivative of this and integrate? Clarifications or corrections would be very much appreciated, thanks.

(2)
For this part if part (1) is correct I believe I could get the pdf through similar methods:
$f_{W_{2}|X(t)=5}(W_{2}|X(t)=5) = \frac{d}{ds} Pr[W_{2} \le s |X(t)=5] = \frac{d}{ds} Pr[X(s) \ge 2 |X(t)=5]$.

Thanks for the help.

Edit: Fixed a typo in the question.

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there are typos in your question. For example, in (1) ask is to find $E[W_1|W(t)=5]$. It should be $E[W_1|X(t)=5]$. – jay-sun Mar 28 '13 at 18:50
You're right, thanks. I corrected the typo as well. – rhl Mar 28 '13 at 19:04
It seems that $N(t)=X(t)$. In question (1), you ask about $W_1$ and then propose a solution dealing with $W_3$. Please make up your mind. – Did Mar 28 '13 at 22:45
Sorry it was supposed to be $W_{3}$ in the expectation in the question. – rhl Mar 28 '13 at 22:49

Both questions are solved if one knows how to compute $\mathbb P(W_i\gt s\mid X(t)=j)$ for every $0\leqslant s\leqslant t$ and every integers $0\leqslant i\leqslant j$. For example, in (1), $$\mathbb E(W_3\mid X(t)=5)=\int_0^t\mathbb P(W_3\gt s\mid X(t)=5)\mathrm ds,$$ and (2) asks for $\mathbb P(W_2\gt s\mid X(t)=5)$, whose derivative is the opposite of the density of the distribution of $W_2$ conditionally on $X(t)=5$.
Now, conditionally on $[X(t)=j]$, the set $\{W_i\mid 1\leqslant i\leqslant j\}$ is distributed like $t$ times the set $\{U_i\mid 1\leqslant i\leqslant j\}$ where $(U_i)_{1\leqslant i\leqslant j}$ is an i.i.d. sample uniform on $(0,1)$. In other words, $(W_i)_{1\leqslant i\leqslant j}$ is distributed like the ordered sample $(tU_{(i)})_{1\leqslant i\leqslant j}$. Since the order statistics are well known, this yields the answer.
Why is it that the set {$W_{i}|1 \le i \le j$} is distributed t times the set of uniform RVs? And after this step I still have difficulty with seeing how ordered sample is so interchangeable with $Pr[W_{3} > s | X(t)=5]$. Thanks for the help. – rhl Mar 29 '13 at 6:35