What is the expected wait times of Poisson arrivals? Suppose customers arrive at a system as a Poisson process with rate $\lambda$, given a specific time interval $[0,t]$, what is the expected wait times for those customers who arrive in this interval?
To clarify, let the number of customers arriving in $[0,t]$ be $N$, and their arrival times are $T_1, T_2, \cdots, T_N$ respectively where $0\le T_1, T_2, \cdots, T_N\le t< T_{N+1}$. For each of them, the wait time within the interval $[0,t]$ is $t-T_i$. The average wait time is,
$$\bar{W}=\frac{\sum^N_1{t-T_i}}{N}$$
Since $N$ is an r.v., so is $\bar{W}$. The expectation of $\bar{W}$ is
$$E[\bar{W}]=\sum^\infty_1 \bar{W}(N,t)\cdot p(N)$$
However $T_i$'s are r.v.'s too, it turns out to be kind of complicated for me. Could someone help me out? thanks.
 A: EDIT
Here is an updated version which matches the result from symmetry in the comment on the OP:
Let $N_t \sim Pois(\lambda t)$ and condition on the event of $n$ arrivals in time $(0,t]$, i.e. $N_t=n$. Then it is relatively well known that the arrival times, $T_i$, are distributed like the order statistics of $n$ uniformly distributed RVs on $(0,t]$. That is $T_i \sim U_{(i)}$ where $U_{(i)}$ is the $i$-th order statistics of $n$ IID $\mathcal{U}(0,t)$ RVs, $U_1,\dotsc, U_n$ (e.g. $U_{(1)}:=\min\{U_1,\dotsc, U_n\}$, $U_{(n)}:=\max\{U_1,\dotsc, U_n\}$ and $U_{(i)}$ is the $i$-th smallest number of $\{U_1,\dotsc, U_n\}$). 
Thus, we have $\mathbb{E}(T_i | N_t=n)=\mathbb{E}(U_{(i)})$, so it remains to compute these means in order to compute the expectation of $\bar{Y}_n=t-\frac{1}{n}(T_1+\dotsc +T_n).$ I will write out the full computation of $\mathbb{E}(U_{(1)})$ since by your comment these order statistics are unfamiliar but I will leave the rest for you to go through yourself. 
So, simply by the definition of minimum we have the following set equality,
$$\{U_{(1)} \geq u\}=\{U_1 \geq u, \dotsc, U_n \geq u\},$$
hence, by the fact that all the $U_i$ are IID,
$$\mathbb{P}(U_{(1)}\geq u)=\mathbb{P}(U_1 \geq u)^n=(1-u/t)^n.$$
Now we use the wonderful fact that for every non-negative random variable $X$, we may compute the expected value in an alternate fashion, as $\mathbb{E}(X)=\int \mathbb{P}(X\geq x) \mathrm{d}x,$ (omitting limits of integration for brevity), so that in this case,
$$\mathbb{E}(U_{(1)})=\int_0^t (1-u/t)^n \mathrm{d}u=\frac{t}{n+1}.$$
Using an analogous argument we can compute $\mathbb{E}(U_{(n)})=\frac{nt}{n+1}$. Then, further, I claim $\mathbb{E}(U_{(i)})=\frac{it}{n+1}$ for all $i=1,2,\dotsc,n$. Thus, we see that (after some algebra)
$$\mathbb{E}(T_1+\dotsc+T_n |N_t=n)=\mathbb{E}(U_{(1)}+\dotsc+U_{(n)})$$
$$=\frac{nt}{2},$$
from which it immediately follows that $\mathbb{E}(\bar{Y}_n | N_t=n)=t-\frac{t}{2}=\frac{t}{2}$. Now since this is just a constant independent of $N_t=n$, we finally get by the law of total expectation, 
$$\mathbb{E}(\bar{Y}_{N_t})=\mathbb{E}\mathbb{E}(\bar{Y}_{N_t} | N_t)=\frac{t}2,$$ just as user @Henry claimed by symmetry.

Original answer that is incorrect
So, conditional on $N_t=n$ customers, $Y_1+\dotsc+Y_n=nt-(T_1+\dotsc+T_n)$, hence the sample mean is $\bar{Y}_n=t-\frac{1}{n}(T_1+\dotsc+T_n)$, which has conditional expectation $\mathbb{E}(\bar{Y}_n | N_t=n)=t-\frac{n+1}{2\lambda}.$ Here, we used the fact that 
$$\mathbb{E}(T_1+\dotsc+T_n)=\frac{1}{\lambda}+\frac{2}{\lambda}+\dotsc+\frac{n}{\lambda}=\frac{1}{\lambda}(1+2+\dotsc +n)=\frac{n(n+1)}{2\lambda}.$$ That is to say, conditional on knowing $N_t$, $\mathbb{E}(\bar{Y}_{N_t} | N_t)=t-\frac{N_t+1}{2\lambda},$ (conditional expectations are, in fact, RVs!). 
Hence, upon taking the expectation with respect to $N_t$, we get, by the law of total expectation,
$$\mathbb{E}(\bar{Y}_{N_t})=\mathbb{E}\mathbb{E}(\bar{Y}_{N_t} | N_t)=t-\frac{1}{2\lambda}\mathbb{E}(N_t+1)=\frac12 \left(t-\frac{1}{\lambda}\right),$$
just using $\mathbb{E}(N_t)=\lambda t$ and linearity.
