Question

Let the random vector \( (X, Y) \) have the joint probability density function
\[f(x, y) = \begin{cases} \frac{1}{x}, & \text{if } 0 < y < x < 1 \\0, & \text{otherwise} \end{cases}.\]
Then \( \text{Cov}(X, Y) \) is equal to:

• A. \( \frac{1}{6} \)
• B. \( \frac{1}{12} \)
• C. \( \frac{1}{18} \)
• D. \( \frac{1}{24} \)

Answer 1

The Correct Option is D

1. Marginal Distribution of \( X \): - Integrate over \( y \) to find \( f_X(x) \): \[ f_X(x) = \int_0^x f(x, y) \, dy = \int_0^x \frac{1}{x} \, dy = 1, \quad \text{for } 0 < x < 1. \] 2. Expected Values: - Compute \( \mathbb{E}[X] \): \[ \mathbb{E}[X] = \int_0^1 x f_X(x) \, dx = \int_0^1 x \cdot 1 \, dx = \frac{1}{2}. \] - Compute \( \mathbb{E}[Y] \): \[ f_Y(y) = \int_y^1 f(x, y) \, dx = \int_y^1 \frac{1}{x} \, dx = -\ln y, \quad \text{for } 0 < y < 1. \] Then: \[ \mathbb{E}[Y] = \int_0^1 y (-\ln y) \, dy = \frac{1}{4}. \] 3. Compute \( \mathbb{E}[XY] \): - Use the joint PDF: \[ \mathbb{E}[XY] = \int_0^1 \int_0^x x y f(x, y) \, dy \, dx = \int_0^1 \int_0^x x y \frac{1}{x} \, dy \, dx = \int_0^1 \int_0^x y \, dy \, dx. \] - Evaluate the inner integral: \[ \int_0^x y \, dy = \frac{x^2}{2}. \] - Evaluate the outer integral: \[ \mathbb{E}[XY] = \int_0^1 \frac{x^2}{2} \, dx = \frac{1}{6}. \] 4. Covariance: - Compute \( \text{Cov}(X, Y) \): \[ \text{Cov}(X, Y) = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y] = \frac{1}{6} - \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{6} - \frac{1}{8} = \frac{1}{24} \]