I have a problem that I'm stuck on that has to do with recalling the time of arrival of a Poisson process. Here's the question:
Buses arrive at a stop in a way that the next bus's arrival is exponentially distributed. You recall a trip you made last year on that bus. You don't remember exactly when the bus arrived, but you know that it must have arrived somewhere between 9 AM and 10 AM. Given this memory, what is the distribution of the random variable describing the exact time at which the bus arrived?
My original guess was that the exact time is uniformly distributed between 9 AM and 10 AM, but I feel like that is too simple of an answer and that I'm missing something here.
Then I started thinking, what is being implied in the statement "you know it must have arrived somewhere between 9 AM and 10 AM". How would you "know" that? Is it because you know how long it takes to get from the bus stop to your destination? In that can we parameterize another exponential distribution with information relating to the time you reached your destination? Or am I way off here.
I think there are two cases here. First, consider the case where you were sitting at the bus stop and watched the busses go by. And you remember that only one bus arrived between 9 AM and 10 AM. Conditional on this fact, you get a uniform distribution. Let $T$ be the time until the bus arrives. And let $u$ be the time for which you waited at the stop (here one hour). Because of the memory-less property of the exponential distribution, we can consider the process starting at 9 AM. Take the probability that the arrival happened between $t$ and $t+\delta t$ for some $t<u$ and $\delta t \to 0$. And let $X$ be the random variable denoting the number of events that happened in the $u$ interval. The conditional probability we need:
$$P(T \in (t, t+\delta t) | X=1) = \frac{P(T \in (t, t+\delta t) \;\&\; X=1)}{P(X=1)}$$
Let $T_1$ be the time from that arrival to the next one. We know only one buss arrived, so we must have: $T_1>u-t$. So, we can replace the numerator of the expression above:
$$P(T \in (t, t+\delta t) |X =1) = \frac{P(T \in (t, t+\delta t) \;\&\; T_1>u-t)}{P(X=1)}$$
Since the exponential distribution is memory-less, we get:
\begin{align}P(T \in (t, t+\delta t) |X =1) = \frac{P(T \in (t, t+\delta t) P(T_1>u-t)}{P(X=1)}\delta t \\ = \frac{\lambda e^{-\lambda t} e^{-\lambda(u-t)}}{e^{-\lambda u}\lambda u}\delta t\\ =\frac{\delta t}{u}\end{align}
Where in the denominator for the second step, we used the fact that the number of arrivals in an interval is Poisson distributed.
Hence, we get the PDF of the uniform distribution.
Next, consider the case where you boarded the bus between 9 AM and 10 AM. In this case, you don't know if there was a bus arrival sometime after you boarded. Here, we just get a truncated exponential distribution. The probability we're after now is
\begin{align}P(T \in (t, t+\delta t) | T<u) =\frac{P(T \in (t, t+\delta t) \; \& \; T<u)}{P(T<u)}\\ \frac{P(T \in (t, t+\delta t) )}{P(T<u)}\\ =\frac{\lambda e^{-\lambda t}}{1-e^{-\lambda u}} \end{align}