Correlation of Max and Min

Let $X_1$ and $X_2$ be independent, each uniform on $[0,1]$ , with $Y = \min(X_1, X_2)$ and $Z = \max(X_1, X_2)$ . For $y, z \in [0,1]$ , what is $\PP(Y \ge y \mid Z \le z)$ ? What is the correlation of $Y$ and $Z$ ?

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#Conditioning on the max

The max is at most $z$ exactly when both draws are, an event of probability $z^2$ . Asking in addition that the min be at least $y$ forces both draws into $[y,z]$ , an event of probability $(z-y)^2$ when $y \le z$ . The conditional probability is their ratio,

\PP(Y \ge y \mid Z \le z) = \frac{(z-y)^2}{z^2}, \qquad 0 \le y \le z, \tag*{(1)}

and it is $1$ when $y \le 0$ and $0$ when $y > z$ .

Both draws below z make the lower-left square of area z^2. Demanding both also above y shrinks it to the inner square of area (z-y)^2, so the conditional probability is the ratio of the two.

#The two end moments

With two draws the order-statistic densities are $f_Y(y) = 2(1-y)$ and $f_Z(z) = 2z$ , giving means $\E[Y] = \tfrac{1}{3}$ and $\E[Z] = \tfrac{2}{3}$ . For the variances,

\E[Y^2] = \int_0^1 y^2 \cdot 2(1-y)\,dy = \frac{1}{6}, \qquad \Var(Y) = \frac{1}{6} - \frac{1}{9} = \frac{1}{18}, \tag*{(2)}

and the reflection $Z = 1 - \min(1-X_1, 1-X_2)$ makes $\Var(Z) = \Var(Y) = \tfrac{1}{18}$ .

#The product collapses

For any two numbers the smaller times the larger is just their product, so $YZ = X_1 X_2$ at every outcome. Independence then hands over the mixed moment directly,

\E[YZ] = \E[X_1 X_2] = \E[X_1]\,\E[X_2] = \frac{1}{4}. \tag*{(3)}

The covariance and correlation follow,

\Cov(Y,Z) = \frac{1}{4} - \frac{1}{3}\cdot\frac{2}{3} = \frac{1}{36}, \qquad \Cor(Y,Z) = \frac{1/36}{\sqrt{\tfrac{1}{18}\cdot\tfrac{1}{18}}} = \frac{1/36}{1/18} = \frac{1}{2}. \tag*{(4)}

The min and the max move together because a large pair of draws lifts both ends at once, and the strength of that pull is exactly one half.