Where does the k-th smallest uniform land?

Question

Let X1, \dots, X5 be independent U(0,1) and sort them as X{(1)} \le \dots \le X{(5)}. What are E[X{(3)}] (the median) and E[X{(4)}] (the second largest)? Generalize to the k-th smallest of n uniforms.

Accepted Answer

The gaps picture. Dropping n points on [0,1] cuts it into n + 1 segments. By symmetry all segments are exchangeable, so each has expected length \frac{1}{n+1}. The k-th smallest point sits at the top of the first k gaps, so E[X{(k)}] = \frac{k}{n+1}. Apply with n = 5. The median is the 3rd point: E[X{(3)}] = \frac{3}{6} = \frac{1}{2}. The second largest is the 4th point: E[X{(4)}] = \frac{4}{6} = \frac{2}{3}. Confirm with the density. The k-th order statistic of n uniforms is 	ext{Beta}(k,\, n-k+1), whose mean is exactly \frac{k}{n+1}. For X{(4)} with n = 5 that is 	ext{Beta}(4, 2), mean \f…

Where does the k-th smallest uniform land?

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