Expected time of the second alarm among three

Question

Three independent alarms are armed. Each fires at an exponential time with rate \lambda = 2 per hour (mean 0.5 hours). What is the expected time at which the second alarm fires?

Accepted Answer

Watch the alarms fire in order. The wait until the first is the minimum of three exponentials, which is exponential with rate 3\lambda and mean \frac{1}{3\lambda}. By memorylessness, after the first fires the two survivors restart, so the extra wait to the second is the minimum of two exponentials, mean \frac{1}{2\lambda}. The second alarm's time is the sum of those two stages: E[	ext{second}] = \frac{1}{3\lambda} + \frac{1}{2\lambda} = \frac{1}{6} + \frac{1}{4} = \frac{5}{12} \approx 0.417 	ext{ hours}. Takeaway. For n independent exponentials of rate \lambda, the k-th event time averages …

Expected time of the second alarm among three

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