Where does the third-smallest random number land?

Question

Draw 9 independent numbers uniformly from the interval [0, 1] and sort them from smallest to largest. What is the expected value of the third-smallest number? And in general, the k-th smallest of n uniforms?

Accepted Answer

Here is the picture that makes this instant. The n random points split [0, 1] into n + 1 gaps (from 0 to the first point, between consecutive points, and from the last point to 1). By symmetry, all n + 1 gaps are exchangeable, they have the same expected length. Since the gaps sum to 1, E[	ext{each gap}] = \frac{1}{n + 1}. The k-th smallest point sits at the top of the first k gaps, so its expected position is E[X{(k)}] = \frac{k}{n + 1}. For n = 9 points and k = 3: E[X{(3)}] = \frac{3}{9 + 1} = \frac{3}{10} = 0.3. So the third-smallest of nine sits at 0.3 on average. The formula also confir…

Where does the third-smallest random number land?

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