Let \( X_1, X_2 \sim \mathrm{Uniform}(0,1) \) be independent. Define:

\[ Y = \min(X_1, X_2), \quad Z = \max(X_1, X_2) \]

Part 1: Conditional Probability \( P(Y \ge y \mid Z \le z) \)

We are interested in computing:

\[ P(Y \ge y \mid Z \le z), \quad \text{for } 0 \le y \le z \le 1 \]

Step 1: Joint Probability

Let’s compute the joint event:

\[ P(Y \ge y, Z \le z) = P(y \le X_1, X_2 \le z, X_1 \ne X_2) \]

Since \( X_1 \) and \( X_2 \) are iid, we can write:

\[ P(Y \ge y, Z \le z) = \text{Area of the square } [y, z]^2 = (z - y)^2 \]

Step 2: Marginal Probability

\[ P(Z \le z) = P(X_1 \le z, X_2 \le z) = z^2 \]

Final Result

\[ P(Y \ge y \mid Z \le z) = \frac{P(Y \ge y, Z \le z)}{P(Z \le z)} = \frac{(z - y)^2}{z^2} \]

Part 2: Correlation \( \mathrm{Corr}(Y, Z) \)

We use known results from order statistics of Uniform(0,1).

For two iid Uniform(0,1) variables:

  • \( \mathbb{E}[Y] = \frac{1}{3} \)
  • \( \mathbb{E}[Z] = \frac{2}{3} \)
  • \( \mathrm{Var}(Y) = \mathrm{Var}(Z) = \frac{1}{18} \)
  • \( \mathbb{E}[YZ] = \frac{1}{4} \)

So:

\[ \mathrm{Cov}(Y, Z) = \mathbb{E}[YZ] - \mathbb{E}[Y] \cdot \mathbb{E}[Z] = \frac{1}{4} - \frac{1}{3} \cdot \frac{2}{3} = \frac{1}{4} - \frac{2}{9} = \frac{1}{36} \]

Now compute the correlation:

\[ \mathrm{Corr}(Y, Z) = \frac{\mathrm{Cov}(Y, Z)}{\sqrt{\mathrm{Var}(Y)\mathrm{Var}(Z)}} = \frac{1/36}{\sqrt{(1/18)(1/18)}} = \frac{1/36}{1/18} = \boxed{\frac{1}{2}} \]

Final Answers

  1. Conditional probability:

\[ \boxed{P(Y \ge y \mid Z \le z) = \frac{(z - y)^2}{z^2}} \]

  1. Correlation:

\[ \boxed{\mathrm{Corr}(Y, Z) = \frac{1}{2}} \]

Reference