Here's the full question
In Poisson Bus City, there is a shuttle bus that goes between Stop A and Stop B, with no stops in between. The times at which the bus arrives at Stop A are a Poisson point process with one bus arriving every five minutes on average, day and night, at which point it immediately picks up all passengers waiting. Citizens of Poisson Bus City arrive at Stop A at Poisson random times, with an average of 5 passengers arriving every minute, and board the next bus that arrives.
Suppose that you visit this city and that you arrive at Stop A at a time chosen uniformly at random from the times in a day. How many citizens of Poisson Bus City do you expect to be on the bus that you take?
Please comment on the solution below.
As I arrive in the middle of an interval, anyone who arrives at Stop A during that interval would get on the same bus as me. I've also worked out that the expected length of interarrival interval is $5$ minutes. From here, I'm thinking of using the following discussion from my textbook:
Suppose now that “events” are occurring at random points at time, and let N(t) denote the number of events that occur in the time interval [0, t].
$\dots$
[Then] N(t) has a Poisson distribution with mean $\lambda t$.
Here $\lambda = 5$ passengers/minute. So, if we let the beginning of this interval to be time $0$, then the variable I'm interested in is $N(5)$, the number of passengers arriving during the time interval $[0,5]$. According to the theory above, $N(5)$ is a Poisson random variable with mean $\lambda t = 5 \times 5 = 25$., i.e.
$$P(N(5) = k) = \frac{25^k}{e^{25}k!}$$
From here, we can calculate $E[N(5)] = \sum_{k=0}^\infty k P(N(5) = k) = \dots = 25$. Hence I should expect $25$ persons on the same bus as me.
This solution is incorrect, because this statement is wrong:
If you choose random interarrival intervals and take the average length of the intervals, then yes, you will get an expected length of $5$ minutes. However, that's not what we're doing: We're choosing a uniformly random time, not a random interarrival interval, and then looking at the interarrival time which that random time ends up in. Since longer intervals take more time, if we choose times uniformly randomly, it is more likely to choose intervals which take longer than $5$ minutes than it is to choose intervals which take less than $5$ minutes, so the expected length of the interval is actually greater than $5$ minutes.
In fact, if we choose a uniformly random time, then the time since the last bus arrived and the time since the next bus will arrive are both exponential distributions. Therefore, the expected time since the last bus arrived is $5$ minutes and the expected time since the next bus arrived is also $5$ minutes, so the expected length of the interarrival interval that we end up in is actually $10$ minutes. This is called the ``hitchhiker's paradox,'' there's a very good article about it on this page about Bitcoin mining.
Ergo, we need to find $\mathbb{E}[N(I)]$, where $I$ is a random variable representing the length of the interarrival time that we end up in. You might be tempted to do something like this: $$ \mathbb{E}[N(I)]=N(\mathbb{E}[I])=N(10) $$ and then just say the answer is $N(10)$. However, you can't just swap $\mathbb{E}$ and $N$ like this: in general, you can't just pull functions out of the expectation like this because it's not always the case that $\mathbb{E}[f(X)]=f(\mathbb{E}[X])$ for functions $f$ and random variables $X$.
Luckily for us though, in this specific case, we can show that $\mathbb{E}[N(I)]=\lambda \mathbb{E}[I]$, where $\lambda$ is the parameter of the Poisson distribution for passengers, which is as good as swapping $\mathbb{E}$ and $N$ for our purposes since all we care about is the expectation.
Let $p(t)$ be the probability distribution function of $I$. To prove $\mathbb{E}[N(I)]=\lambda \mathbb{E}[I]$, we will use the law of total expectation: $$ \begin{align*} \mathbb{E}[N(I)] &= \int_0^\infty \mathbb{E}[N(I)\mid I=t] p(t)dt \\ &= \int_0^\infty \mathbb{E}[N(t)] p(t)dt \\ &= \int_0^\infty \lambda t p(t)dt \\ &= \lambda \int_0^\infty tp(t)dt \\ &= \lambda \mathbb{E}[I] \end{align*} $$
Ergo, since $\lambda=5$, as passengers arrive every $5$ minutes, and $\mathbb{E}[I]=10$, as discussed above, the expected number of passengers we expect is $5\cdot 10=50$.