I am trying to understand the solution for the following example:
Assume that passengers arrive at a bust station as a Poisson process with rate $\lambda$. a) The only bus departs after a deterministic time $T$. Let $W$ be the combined waiting time for all passengers, compute $E(W)$.
The solution says the following: If $S_1, S_2, ... $ are the arrival times in $[0,T]$, then the combined waiting time is $W = T- S_1 + T-S_2 +...$. Let $N(T)$ be the number of arrivals in $[0,t]$. We obtain the answer by conditioning on the value of $N(T)$. WE have $$E(W) = \sum_{k=0}^{\infty} E(W|N(T)=k)P(N(T)=k) = \sum_{k=0}^{\infty} k\frac{T}{2} P(N(T)=k) = \frac{T}{2}E(N(T)) $$ My question comes in the second equality, why is $E(W|N(T)=k)= k\frac{T}{2}$? I understan the expression would be $E(W|N(T)=k) = E(\sum_{l=1}^{N(T)}T-S_i|N(T)=k) = kE(T-S_i)$ but then I don't seehow the last expression becomes just $\frac{T}{2}$. Thanks for your help!
$T$ is deteministic so you have $$E[T-S_i] =T-E[S_i]$$
It works if you use the assumption that $S_i$ are uniformally distributed in $[0,T]$ in order to get $E[S_i]=T/2$.
Edit It is indeed stated in your paper, under thm 18.5 Uniformity of previous event time :
Given that $N(t) = k$, the conditional distribution of the interarrival times, $S_1,\ldots,Sk$, is distributed as order statistics of $k$ independent uniform variables. The set $$\{S_1,\ldots, S_k\} =\{U_1,\ldots, U_i\}$$, where $U_i$ are independent and uniform on $[0, t]$.