Expected range of n uniforms

Question

Let X1, \dots, Xn be independent U(0,1). The range is R = \maxi Xi - \mini Xi. What is E[R]? Interpret how it behaves as n grows.

Accepted Answer

Subtract the two known means. For n uniforms, E[\max] = \frac{n}{n+1} and E[\min] = \frac{1}{n+1}, so E[R] = E[\max] - E[\min] = \frac{n}{n+1} - \frac{1}{n+1} = \frac{n-1}{n+1}. Gaps picture (same answer). The n points split [0,1] into n+1 gaps of expected length \frac{1}{n+1} each. The range is the whole interval minus the two end gaps (below the min and above the max), so E[R] = 1 - \frac{2}{n+1} = \frac{n-1}{n+1}. For n = 2 this gives \frac{1}{3}; for n = 10, \frac{9}{11} \approx 0.82; and as n 	o \infty, E[R] 	o 1: with enough points the extremes crowd the endpoints and the spread fills…

Expected range of n uniforms

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