Solution
Waiting for Erlang Buses
Show the problem again
A bus line's inter-arrival gaps are independent, and each gap is the sum of two independent exponential random variables with mean 1 minute apiece (an Erlang-2 gap, mean 2 minutes). You arrive at a time chosen uniformly over a very long stretch of the schedule, with no knowledge of the history. What is your expected wait for the next bus, in minutes? Give a reduced fraction.
Worked solution
A uniformly random arrival time lands in a gap with probability proportional to the gap's length, so the gap you land in is size-biased, with mean $E[G^2]/E[G]$. Given its length, your arrival is uniform within it, so the expected wait is half of that: $$E[W] = \frac{E[G^2]}{2E[G]}.$$ For Erlang-2 with unit-rate stages, $E[G] = 2$ and $\mathrm{Var}(G) = 1+1 = 2$, so $E[G^2] = 2 + 4 = 6$. Therefore $E[W] = 6/(2\cdot2) =$ **3/2** minutes. Compare: a perfectly regular schedule (gap always 2) would give 1 minute, and exponential gaps with mean 2 would give 2. Variability in the gaps increases the expected wait even when the mean gap is fixed.
Source: Inspection paradox / residual life in renewal theory, as in Feller vol. 2 and Ross, Stochastic Processes. Statement written for AxiomIQ.