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?