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.

• It is not clear to me what you mean by wait times. Usually “waiting times” in a Poisson process refer to the inter-arrival times $T_{i+1}-T_i$ which are exponentially distributed with mean $1/\lambda$ but you seem to be asking about the time from arrival of customer $i$, $T_i$, until the end of the observed time $t$: $Y_i:=t-T_i$. Do I have that right? Just wanted to clarify. – Nap D. Lover Jan 23 at 18:28
• Symmetry suggests $\frac{t}2$ as the expected waiting time until $t$ – Henry Jan 23 at 18:33
• @LoveTooNap29 yes, your clarification is correct, thanks for that. – Guoyang Qin Jan 23 at 19:10
• @Henry Symmetry is a point of view. But can we derive it analytically? – Guoyang Qin Jan 23 at 19:12
• So, conditional on $n$ 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 $\mathbb{E}(\bar{Y}_n | N_t=n)=t-\frac{n+1}{2\lambda}$. The unconditional mean is then given by the LTE and is $$\mathbb{E}(\bar{Y}_{N_t})=\mathbb{E}\mathbb{E}(\bar{Y}_n | N_t=n)=\frac12 \left(t-\frac{1}{\lambda}\right).$$ I am commenting this rather than answering because I am little worried that you get a negative time whenever you observe time intervals smaller than the mean waiting time, so maybe I made a mistake... – Nap D. Lover Jan 23 at 20:45

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.

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.
• Thanks, it is very clear. This result proves that the symmetry view which says the expectation is just half the total time interval $t/2$ doesn't hold here. Do you have any comment about why this happens? – Guoyang Qin Jan 24 at 1:17
• @GuoyangQin Actually now that you mention it, I think I made a mistake. Conditional on $N_t=n$ the arrival times $T_i$ for $i=1,2,\dots,n$ are distributed like the order statistics of $n$ uniform RVs in $(0,t]$ and I'd bet for sure those means are different than what I wrote, which are the unconditionally $\Gamma(i,\lambda)$-distributed means, for instance $\mathbb{E}(T_1|N_t=1)=\mathbb{E}(U)=t/2$. You may want to un-accept this answer while I work this out and edit it. Sorry... – Nap D. Lover Jan 24 at 1:56
• It's OK. As you mentioned the order statistics of $n$, it reminds me of this I read in the book Fundamentals of Queueing Theory (5ed, Donald Gross, p43) - "Theorem 2.9 Let $N(t)$ be a Poisson process. Given that $k$ events have occurred in a time interval $[0, T]$, the times $\tau_1<\tau_2<\cdots<\tau_k$ at which the events occurred are distributed as the order statistics of $k$ independent uniform random variables on $[0,T]$." However, I didn't have a good understanding of this since I didn't understand "the order statistics of $k$". Looking forward to your revision. Thanks. – Guoyang Qin Jan 24 at 2:11