Five independent exponential lifetimes each have rate 1. What is the expected value of the third shortest of the five?

Five independent exponential lifetimes each have rate 1. What is the expected value of the third shortest of the five?

Approach: Use memorylessness so that after each failure the surviving lifetimes restart, making the gaps between consecutive order statistics independent exponentials whose rate counts the survivors.

47/60. The memoryless property of the exponential distribution means the wait for the first of five failures is Exp(5), the extra wait for the second is Exp(4), and the extra wait for the third is Exp(3), with those gaps independent. Summing the means over the order statistics gives E[X_(3)] = 1/5 + 1/4 + 1/3 = (12 + 15 + 20)/60 = 47/60 = 0.7833. The same argument gives E[X_(5)] = 1/5 + 1/4 + 1/3 + 1/2 + 1 = 137/60. That total is the harmonic number H_5, and the gaps widen as the pool of survivors shrinks, so the final failure alone contributes 1 of that 2.283.

Follow-up: What is the variance of the third order statistic?

Key concepts: order statistics, memoryless property, exponential distribution, harmonic number.