Expected Maximum of n Uniform Draws

The problem. Draw $n$ independent random numbers, each uniform on $[0,1]$. What's the expected value of the largest one? For $n=1$ it's obviously $\frac12$. For $n=2$, is it still $\frac12$? Most people guess it stays near $\frac12$ or creeps up slowly; the actual answer is $\frac23$ for $n=2$, and in general $\frac{n}{n+1}$ — a formula that climbs toward 1 faster than intuition suggests.

Deriving it: work with the CDF, not the density

Let $M = \max(X_1, \ldots, X_n)$ where each $X_i \sim \text{Uniform}[0,1]$ independently. The trick is to first find $P(M \le m)$ rather than jumping to the density of $M$. Since $M \le m$ exactly when every $X_i \le m$, and the draws are independent:

$$P(M \le m) = P(X_1 \le m)\cdots P(X_n \le m) = m^n, \qquad 0 \le m \le 1.$$

In plain English: the max is below some threshold only if all $n$ draws individually cleared that threshold, and independence lets those probabilities multiply. Differentiating gives the density $f_M(m) = n m^{n-1}$, and then:

$$E[M] = \int_0^1 m \cdot n m^{n-1}\, dm = n\int_0^1 m^n\, dm = \frac{n}{n+1}.$$

The density of the maximum, $f_M(m) = n m^{n-1}$, is a power function that piles up more and more mass near $m=1$ as $n$ grows — drag the exponent on the explorer above to see how a higher power concentrates the curve toward the right edge, the same visual reason larger samples push their maximum closer to the true upper bound.

Worked example: n=2 and n=3 by hand

For $n=2$: $E[M] = \frac{2}{3} \approx 0.667$. Sanity check with a direct argument — among two uniform draws, by symmetry the larger one should sit, on average, two-thirds of the way up, and the smaller one one-third of the way up (they should average back to $\frac12$ together: $\frac13+\frac23=1$, average $\frac12$, correct). For $n=3$: $E[M] = \frac{3}{4} = 0.75$. Adding a third draw pushes the expected maximum up by another $\frac{1}{12}$ — each additional draw gives the max another chance to be pushed toward 1, but with diminishing returns, since $\frac{n}{n+1}$ approaches 1 asymptotically rather than reaching it.

Worked example: estimating an unknown range from data

Suppose you observe $n=9$ independent measurements, each uniformly distributed on $[0, \theta]$ for some unknown $\theta$, and the largest observed value is 18. Since $E[M] = \frac{n}{n+1}\theta = \frac{9}{10}\theta$, a natural (bias-corrected) estimate of $\theta$ is $\hat\theta = \frac{10}{9}\times 18 = 20$ — scaling the observed maximum up slightly, since on average the max of 9 draws only reaches 90% of the true ceiling. This is the classic "German tank problem" logic (estimating the size of a serial-numbered population from the largest number seen), just phrased with continuous uniforms instead of discrete labels.

The general technique: CDF-first for order statistics

Any question about the expectation of a maximum or minimum of independent draws should start from $P(M \le m) = \prod_i P(X_i \le m)$ (for a max) or the analogous complement argument for a min, rather than trying to guess the density directly. This "CDF of the extreme, then differentiate" pattern is the standard route into order-statistics problems and generalizes cleanly beyond the uniform distribution to any distribution with a known CDF.