Correlation of the smaller and the larger · Problem 99

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Idea. The covariance needs $E [U V]$ , which looks hard but is easy: the minimum times the maximum is just the product of the two numbers.

Step 1: means. $P (V \leq v) = v^{2}$ , so $V$ has density $2 v$ and $E [V] = \frac{2}{3}$ . By symmetry $U$ has the distribution of $1 - V$ , so $E [U] = \frac{1}{3}$ .

Step 2: the cross moment. $U V = X Y$ , so $E [U V] = E [X] E [Y] = \frac{1}{4},$ and $Cov (U, V) = \frac{1}{4} - \frac{1}{3} \cdot \frac{2}{3} = \frac{1}{36} .$

Step 3: variances. $E [V^{2}] = \int_{0}^{1} 2 v^{3} d v = \frac{1}{2}$ , so $Var (V) = \frac{1}{2} - \frac{4}{9} = \frac{1}{18}$ . Since $U$ has the distribution of $1 - V$ , $Var (U) = \frac{1}{18}$ too.

Step 4. $corr (U, V) = \frac{1/36}{1/18} = \frac{1}{2} .$

In general. For $n$ independent uniforms, the correlation between the minimum and the maximum is $\frac{1}{n}$ : positive, because a large minimum forces a large maximum, and weaker as $n$ grows.