Question

For two random variables X and Y having the joint probability density function \( f(x,y) = \frac{1}{3}(x+y); 0 \le x \le 1, 0 \le y \le 2 \), then cov(X, Y) is

• A. \( -\frac{1}{9} \)
• B. \( -\frac{1}{81} \)
• C. \( \frac{2}{3} \)
• D. \( -\frac{5}{9} \)

Hint:

When calculating multiple integrals for covariance, be systematic. Calculate E(X), E(Y), and E(XY) separately and carefully. Double-check your integration limits and the final arithmetic, as small errors are common.

Answer 1

The Correct Option is B

Step 1: Concept Overview:
Covariance measures the joint variability of random variables X and Y, calculated as Cov(X, Y) = E(XY) - E(X)E(Y). We will determine E(X), E(Y), and E(XY) from the joint PDF.

Step 2: Calculation Steps:
1. Compute \( E(X) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} x f(x,y) \,dy\,dx \) 2. Compute \( E(Y) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} y f(x,y) \,dy\,dx \) 3. Compute \( E(XY) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} xy f(x,y) \,dy\,dx \) 4. Calculate \( \text{Cov}(X,Y) = E(XY) - E(X)E(Y) \)

Step 3: Detailed Calculations:
1. Calculating E(X): \[ E(X) = \int_{0}^{1} \int_{0}^{2} x \frac{1}{3}(x+y) \,dy\,dx = \frac{1}{3} \int_{0}^{1} \left[ x^2y + \frac{xy^2}{2} \right]_{y=0}^{y=2} \,dx \]\ \[ = \frac{1}{3} \int_{0}^{1} (2x^2 + 2x) \,dx = \frac{1}{3} \left[ \frac{2x^3}{3} + x^2 \right]_{0}^{1} = \frac{1}{3} \left( \frac{2}{3} + 1 \right) = \frac{1}{3} \left( \frac{5}{3} \right) = \frac{5}{9} \]\ 2. Calculating E(Y): \[ E(Y) = \int_{0}^{1} \int_{0}^{2} y \frac{1}{3}(x+y) \,dy\,dx = \frac{1}{3} \int_{0}^{1} \left[ \frac{xy^2}{2} + \frac{y^3}{3} \right]_{y=0}^{y=2} \,dx \]\ \[ = \frac{1}{3} \int_{0}^{1} \left( 2x + \frac{8}{3} \right) \,dx = \frac{1}{3} \left[ x^2 + \frac{8x}{3} \right]_{0}^{1} = \frac{1}{3} \left( 1 + \frac{8}{3} \right) = \frac{1}{3} \left( \frac{11}{3} \right) = \frac{11}{9} \]\ 3. Calculating E(XY): \[ E(XY) = \int_{0}^{1} \int_{0}^{2} xy \frac{1}{3}(x+y) \,dy\,dx = \frac{1}{3} \int_{0}^{1} \int_{0}^{2} (x^2y+xy^2) \,dy\,dx \]\ \[ = \frac{1}{3} \int_{0}^{1} \left[ \frac{x^2y^2}{2} + \frac{xy^3}{3} \right]_{y=0}^{y=2} \,dx = \frac{1}{3} \int_{0}^{1} \left( 2x^2 + \frac{8x}{3} \right) \,dx = \frac{1}{3} \left[ \frac{2x^3}{3} + \frac{4x^2}{3} \right]_{0}^{1} = \frac{1}{3} \left( \frac{2}{3} + \frac{4}{3} \right) = \frac{1}{3}(2) = \frac{2}{3} \]\ 4. Calculating Cov(X,Y): \[ \text{Cov}(X,Y) = E(XY) - E(X)E(Y) = \frac{2}{3} - \left(\frac{5}{9}\right)\left(\frac{11}{9}\right) = \frac{2}{3} - \frac{55}{81} = \frac{54-55}{81} = -\frac{1}{81} \]\
Step 4: Result:
The covariance cov(X, Y) is \( -\frac{1}{81} \).