Expected Value of Max and Min

Let $X_1, X_2, \dots, X_n$ be independent random variables, each uniform on $[0,1]$ . Find the cumulative distribution function, the density, and the expected value of the maximum $Z_n = \max(X_1, \dots, X_n)$ and of the minimum $Y_n = \min(X_1, \dots, X_n)$ .

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#The maximum

The maximum is at most $z$ exactly when every draw is at most $z$ . Independence turns that into a product,

F_{Z}(z) = \PP(Z_n \le z) = \prod_{i=1}^{n} \PP(X_i \le z) = z^{\,n}, \qquad z \in [0,1]. \tag{1}

Differentiating gives the density, and integrating against it gives the mean,

f_{Z}(z) = n\,z^{\,n-1}, \qquad \E[Z_n] = \int_0^1 z \cdot n\,z^{\,n-1}\,dz = \frac{n}{n+1}. \tag{2}

#The minimum

The minimum is at least $y$ exactly when every draw is at least $y$ , so the survival function is the clean product this time,

\PP(Y_n \ge y) = \prod_{i=1}^{n} \PP(X_i \ge y) = (1-y)^{\,n}, \qquad F_{Y}(y) = 1 - (1-y)^{\,n}. \tag{3}

Differentiating and integrating as before,

f_{Y}(y) = n\,(1-y)^{\,n-1}, \qquad \E[Y_n] = \int_0^1 y \cdot n\,(1-y)^{\,n-1}\,dy = \frac{1}{n+1}. \tag{4}

With five draws the minimum crowds toward 0 and the maximum toward 1. Their densities are mirror images, and the means land at 1/6 and 5/6, symmetric about the midpoint.

#One picture ties them together

Each $1 - X_i$ is again uniform, and $Y_n = 1 - \max_i(1 - X_i)$ , so the minimum is the maximum reflected through $\tfrac{1}{2}$ . That forces $\E[Y_n] = 1 - \E[Z_n] = \tfrac{1}{n+1}$ and explains the mirror symmetry of the two densities.

The cleanest reading is the gap picture. The $n$ sorted draws split $[0,1]$ into $n+1$ intervals whose expected lengths are all equal to $\tfrac{1}{n+1}$ , so the $k$ -th smallest sits at height $\tfrac{k}{n+1}$ . The minimum is the first cut at $\tfrac{1}{n+1}$ and the maximum the last at $\tfrac{n}{n+1}$ .