# Double checking solution (Poisson process)

I am just double checking to see if my solutions are correct for these two questions. Assume that for both customers arrive at rate lambda. I am unsure about the second one so confirmation would be great. Thanks!

1) Amy is the first customer to arrive after 12 NOON and Frank is the second. What is the expected amount of time between the times that they arrive?

SOLUTION: E[S_1] where S_1 is the sojourn time between arrivals and is exponentially distributed. Therefore E[S_1] = 1/lambda

2) You are told that Amy and Frank arrived between 12 AM noon and 2:00 PM and that nobody else arrived between 12:00 and 2:00. Given this information what is the expected amount of time between the times that they arrived?

SOLUTION: This would be E[S_2 | X(2) = 2] = E[W_3 | X(2) = 2] - E[W_2 | X(2) = 2] = 1/lambda Note: W_i is the time of the ith event happening

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How do you jump to $1/\lambda$ in #2? That step is wrong. The answer does not depend on lambda, – Douglas Zare Mar 25 '11 at 5:17
@DouglasZare E[W_3|X(2) = 2] is 2 + 1/lambda and E[W_2|X(2) = 2] is 2. Subtracting both terms gives 1/lambda – icobes Mar 25 '11 at 5:18
Care to accept an answer? – Did Nov 13 '11 at 10:18

You want to find ${\rm E}[W_2 - W_1 |X(t) = 2]$. Given $X(t)=2$, $W_1$ and $W_2$ are distributed as $U_{1:2}:=\min \{ U_1 ,U_2 \}$ and $U_{2:2}:=\max \{ U_1 ,U_2 \}$, respectively, where $U_1$ and $U_2$ are independent uniform$[0,t]$ random variables. So, you need to find ${\rm E}[U_{2:2}-U_{1:2}]$, which is equal to ${\rm E}[U_{2:2}] - {\rm E}[U_{1:2}]$. This is an elementary question in order statistics, and the answer should be $${\rm E}[W_2 - W_1 |X(t) = 2] = {\rm E}[U_{2:2}] - {\rm E}[U_{1:2}] = \frac{{2t}}{3} - \frac{t}{3}.$$ To confirm this, you can use the general formula $${\rm E}[U_{i:n} ] = \frac{{n!}}{{(i - 1)!(n - i)!}}\int_0^t {x[F(x)]^{i - 1} [1 - F(x)]^{n - i} f(x)\,{\rm d}x} ,$$ where $U_{1:n} \leq U_{2:n} \leq \cdots \leq U_{n:n}$ are $n$ order statistics from the uniform$[0,t]$ distribution, and $F$ and $f$ are the cdf and pdf of that distribution, which are given by $F(x)=x/t$ and $f(x)=1/t$, for $x \in [0,t]$.