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Expected Maximum of n Uniforms

考察公司Jane StreetAkuna Capital

题目

Let U1,…,UnU_1, \dots, U_n be independent Uniform(0,1)\text{Uniform}(0,1) random variables, and let M=max⁡iUiM = \max_i U_i.

Find E[M]E[M], and give the intuition for the answer in one sentence.

解析

先自己试一试，卡住了再打开提示。

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Via the CDF. For 0≤x≤10 \le x \le 1,

P(M≤x)=P(all Ui≤x)=xn,P(M \le x) = P(\text{all } U_i \le x) = x^{n},

so the density is fM(x)=nxn−1f_M(x) = n x^{n-1} and

E[M]=∫01x⋅nxn−1 dx=nn+1.E[M] = \int_0^1 x \cdot n x^{n-1}\,dx = \frac{n}{n+1}.

Via the tail formula (faster to say out loud):

E[M]=∫01P(M>x) dx=∫01(1−xn)dx=1−1n+1=nn+1.E[M] = \int_0^1 P(M > x)\,dx = \int_0^1 \left(1 - x^n\right) dx = 1 - \frac{1}{n+1} = \frac{n}{n+1}.

The intuition. The nn points cut (0,1)(0,1) into n+1n+1 gaps whose expected lengths are equal by symmetry, so the largest point sits exactly one gap — 1/(n+1)1/(n+1) — below 1.

Follow-ups worth knowing. E[min⁡]=1/(n+1)E[\min] = 1/(n+1); the kk-th order statistic is Beta(k,n+1−k)\text{Beta}(k, n+1-k) with mean k/(n+1)k/(n+1); and Var(M)=n/[(n+1)2(n+2)]\text{Var}(M) = n / \left[(n+1)^2 (n+2)\right], which shrinks like n−2n^{-2}.