a)Fred visits Blotchville again. He finds that the city has installed an electronic display at the bus stop, showing the time when the previous bus arrived. The times between arrivals of buses are still independent Exponentials with mean 10 minutes. Fred waits for the next bus, and then records the time between that bus and the previous bus. On average, what length of time between buses does he see?
(b) Fred then visits Blunderville, where the times between buses are also 10 minutes on average, and independent. Yet to his dismay, he finds that on average he has to wait more than 1 hour for the next bus when he arrives at the bus stop! How is it possible that the average Fred-to-bus time is greater than the average bus-to-bus time even though Fred arrives at some time between two bus arrivals? Explain this intuitively, and construct a specific discrete distribution for the times between buses showing that this is possible
Can you help me to solve the problems? I did this so far, but I think that I am wrong:
a) I think the answer to the question is in the question itself: $10$ minutes $$T \sim \mathrm{Expo}(1/10) \implies E[T]=10$$
b) Let $X$ be the time Fred arrival. Assume, that Fred come in a uniformly random time between buses. $$E[T \mid T \ge t] = t + E[T],$$ where $X \sim \mathrm{Unif}(0,t)$ and $E[T \mid T \ge t]$ is the average time between buses given that the bus hadn't arrive yet in time $t$. If Fred has to wait more than $1$ hour, then $t+E[T] \ge 60$ $=>$ $t \ge 50$.
(a) You are correct but should explain your reasoning more carefully.
(b) It is unclear what you are trying to show here. You do not need conditional expectations. Are you still assuming the distribution is exponential in this part?