I'm currently learn the Statistics 110: Probability by Joe Blitzstein at Harvard. There is a SP and HW 6. My question is based on 1-d of HW6.
Fred arrive at a uniform instant between the previous bus arrival and the next bus arrival. The average length of that interval between buses is 10 minutes. Since Fred arrive at any time in that interval, the average waiting time can be though as 5 minutes.
I understand this wrong intuition based on length-biasing. But what exact average waiting time and distribution of waiting time?
I try to use bus arriving time as prior. But stuck with starting point... Is there a good strategy to get this answer?
When in doubt, refer to a specific example. Here is one which may clarify some confusion you're having.
Q: You arrive at the bus stop at 1:15pm and there is no bus in sight. If the last bus to pick up passengers departed at 1:00pm, what is the probability that you wait no more than $t$ minutes for the next bus to arrive?
A: Let $T$ be the random variable representing the amount of time you wait the next bus to arrive, and let $X$ the random variable denoting the amount of time between the departure of the 1:00pm bus and the arrival of the very next bus. Then $$P(T\leq t)=P(X\leq t+15|X>15)=\frac{P(15<X\leq t+15)}{P(X>15)}$$ Since $X\sim \text{Expo}\Big(\frac{1}{10}\Big)$ we have $$P(T \leq t)=\frac{\int_{15}^{t+15}\frac{1}{10}e^{-x/10}dx}{\int_{15}^{\infty}\frac{1}{10}e^{-x/10}dx}=1-e^{-t/10}$$ This shows that $T\sim \text{Expo}\Big(\frac{1}{10}\Big)$ as well. Would it matter if the previous bus departed at 12:00pm instead of 1:00pm? What if the bus you're waiting for is the first bus to arrive at the station? The answer is resounding no. The amount of time you wait for the next bus to arrive will always be exponentially distributed with mean $10$ minutes regardless of how much time elapsed since the arrival (or non$-$arrival) of the previous bus. This is what meant when we say that the exponential distribution is memoryless.
Hope this helps.