# What is the distribution of an arrival time in a Poisson processes conditioned on the value of another arrival time?

The problem: Customer arrival times are modeled by a Poisson process with rate $\lambda = 10$. $X(t)$ refers to the number of customers who have arrived by time $t$, where $0 < t \le 1$. Assume that $\{S_i : i = 1,...,n\}$ represents the arrival times, so customer $i$ arrives at time $S_i$.

Determine the distribution of $S_1$ given that the fifth customer arrived exactly one hour after the bank was open.

My attempt: I'm not really sure how to solve this because I only know how arrival times are distributed when conditioned on $N(t) = n$. I know that $S_5 = 1$ implies that $X(1) = 5$. I first tried doing:

$$Pr(S_1 = s | S_5 = 1) = \frac{Pr(S_1 = s \bigcap S_5 = 1)}{Pr(S_5 = 5)}$$

Then:

$$\frac{Pr(S_1 = s \bigcap S_5 = 1)}{Pr(S_5 = 5)} = \frac{Pr(S_1 = s \bigcap X(1) = 5)}{Pr(S_5 = 1)}$$

I know that the lower probability has a gamma distribution with parameters $\alpha = 5$ and $\beta = 10$. But I'm not sure what to do with the top one or whether this is even the correct approach at all.

The key problem I have is: how can I solve for $Pr(S_1 = s \bigcap S_5 = 1)$? I know that if they were interarrival times, I could use the product since they're independent variables. But $S_1$ and $S_5$ are dependent variables.

You are dealing with probability denisities rather than mass, otherwise okay.

Note that (in a Poisson Process) the wait time (inter-arival time) before the arrival of the first customer will be independent of the subseqent wait time until the arrival of the fifth customer.   Further, the later has identical distribution as the wait time before the fourth arrival.

That is, $S_1, S_5-S_1$ are independent, and $S_5-S_1$ and $S_4$ have identical distributions, so:

\begin{align}f(S_1=t\mid S_5=1) &= \dfrac{f(S_1=t)~f(S_5-S_1=1-t)}{f(S_5=1)} \\[1ex] &= \dfrac{f(S_1=t)~f(S_4=1-t)}{f(S_5=1)}\end{align}

Now, $f(S_k=\tau) = \dfrac{\lambda^k \tau^{k-1} e^{-\lambda \tau}}{(k-1)!}\mathbf 1_{\tau\in[0;\infty), k\in[1;\infty){\cap}\Bbb N}$, which is the probability density function for an Erlang distribution, as you noted.   So you will obtain a rather neat result.   Do you recognise what it is?

• Thank you for your answer! Can you clarify, though, on what you mean by "dealing with probability densities rather than mass"? Also, why is $Pr(S_1 = t \bigcap S_5 = 1) = Pr(S_1 = t) Pr(S_5 - S_1 = 1 - t)$? I can tell that the interval of time between $S_1$ and $S_5$ is of length 1 - t, but I don't understand how that works in terms of probability concepts. Commented Mar 7, 2018 at 1:00
• The inter-arrival times are continuous random variables, so they have no probability mass at any discrete point (in time). Please avoid dividing by zero. $$\Pr(S_n=t)=0$$ Poisson processes are memoryless so the interarrival time until future arrivals does not depend on the time of any past arrivals. $$f(S_5-S_1=1-t\mid S_1=t)=f(S_5-S_1=1-t)$$ Commented Mar 7, 2018 at 1:19
• Oh, I see. I knew that a Poisson process had independent increments, but until now I hadn't been able to grasp the meaning mathematically. As for what the result is, I only got $\frac{(1-s)^3}{4}$, which I don't immediately recognize. Does it have significance? Commented Mar 7, 2018 at 3:44
• That should be $4(1-s)^3~\mathbf 1_{0\lt t\lt 1}$ , the distribution of the first (or least) order statistic for a set of four independent samples from a uniform distribution over $(0;1)$. It is the probability that one from the four samples occurs at time $t$ and the other three occur between $t$ and $1$. Commented Mar 7, 2018 at 4:47
• Woops, it looks like you're right--I made a tiny arithmetic error in the cancellation of the factorials. Hm, that's interesting. I haven't worked enough with order statistics to recognize the formula, but that will all change this spring break. All I currently know by heart are the formulas $S_{X_{(1)}} (x) = [S_X (x)]^n$ and $F_{X_{(n)}} (x) = [F_X (x)]^n$. Commented Mar 7, 2018 at 5:00