Expected value of the second largest of four

Question

Draw 4 independent numbers uniformly from [0,1] and sort them. What is the expected value of the second largest?

Accepted Answer

For n independent uniforms on [0,1], the sorted values are spread out evenly in expectation: the k-th smallest sits at \frac{k}{n+1} on average. The reason is that the n points split the interval into n+1 gaps whose expected lengths are all equal, each \frac{1}{n+1}. The second largest of four is the third smallest, so k = 3 and n = 4: E[	ext{second largest}] = \frac{3}{4+1} = \frac{3}{5} = 0.6. As a check, the largest averages \frac{4}{5} = 0.8, and the runner-up trails it by exactly one gap of \frac15. Takeaway. The k-th smallest of n uniforms averages \frac{k}{n+1}; the runner-up (second …

Expected value of the second largest of four

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