Expected second largest among nine random numbers

Question

Draw 9 independent numbers uniformly from [0,1] and sort them. What is the expected value of the second largest?

Accepted Answer

The k-th smallest of n independent uniforms averages \frac{k}{n+1}, because the points cut [0,1] into n+1 equal expected gaps. The second largest of nine is the eighth smallest, so k = 8 and n = 9: E[	ext{second largest}] = \frac{8}{9+1} = \frac{8}{10} = \frac{4}{5} = 0.8. With more draws the runner-up is pushed higher, sitting at 0.8 here versus 0.6 for four draws. Takeaway. More draws crowd the top order statistics toward 1: the second largest of n averages \frac{n-1}{n+1}.

Expected second largest among nine random numbers

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