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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

1 Answer 1

up vote 1 down vote accepted

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.

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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
    
This is a basic result on Poisson processes (which can be proved directly). Poisson processes are often generated using this, by first drawing the (Poisson) size of the population, then drawing the (uniform) locations of the individuals. Which book are you following? –  Did Mar 29 '13 at 6:37
    
An Introduction to Stochastic Modeling 4th edition, by Pinsky and Karlin. Personally, I'm not finding it to be too great unfortunately. –  rhl Mar 29 '13 at 6:45
    
And they don't mention the "Poisson total population + i.i.d. uniform individuals" representation? As I said, many practitioners know and use this, for simulation purposes. –  Did Mar 29 '13 at 6:46
    
They do talk about the wait times being uniformly distributed over an interval, but in particular I don't understand why the set of waiting times is distributed like t times the uniform set. –  rhl Mar 29 '13 at 6:51

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