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Expected Maximum of n Uniforms

Asked atJane StreetAkuna Capital

Let $U_1, \dots, U_n$ be independent $\text{Uniform}(0,1)$ random variables, and let $M = \max_i U_i$ .

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

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Via the CDF. For $0 \le x \le 1$ ,

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

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

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

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

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