Question

Let \( X \) and \( Y \) be continuous random variables with the joint probability density function 

where \( c \) is a positive real constant. Then \( E(X) \) equals 
 

• A. \( \frac{1}{5} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{2}{5} \)
• D. \( \frac{1}{3} \)

Hint:

When working with joint probability distributions, always normalize the distribution by integrating over the valid range and use the expected value formula for continuous random variables.

Answer 1

The Correct Option is C

Step 1: Determining the constant \( c \).
To ensure the joint probability density function is valid, we need to normalize it. The total probability must be equal to 1, so we integrate the joint pdf over the valid range of \( x \) and \( y \): \[ \int_{0}^{1} \int_{0}^{y} c x (1 - x) \, dx \, dy = 1 \] First, solve the inner integral: \[ \int_{0}^{y} x (1 - x) \, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{y} = \frac{y^2}{2} - \frac{y^3}{3} \] Now, solve the outer integral: \[ \int_{0}^{1} c \left( \frac{y^2}{2} - \frac{y^3}{3} \right) \, dy = c \left( \frac{1}{6} - \frac{1}{12} \right) = c \times \frac{1}{12} \] Set this equal to 1 to normalize the pdf: \[ c \times \frac{1}{12} = 1 \quad \Rightarrow \quad c = 12 \]
Step 2: Calculating \( E(X) \).
The expected value \( E(X) \) is given by: \[ E(X) = \int_{0}^{1} \int_{0}^{y} x f(x, y) \, dx \, dy = \int_{0}^{1} \int_{0}^{y} x \cdot 12 x (1 - x) \, dx \, dy \] Simplifying the expression inside the integral: \[ E(X) = \int_{0}^{1} \int_{0}^{y} 12 x^2 (1 - x) \, dx \, dy \] Now, solve the inner integral: \[ \int_{0}^{y} x^2 (1 - x) \, dx = \int_{0}^{y} (x^2 - x^3) \, dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0}^{y} = \frac{y^3}{3} - \frac{y^4}{4} \] Now, solve the outer integral: \[ E(X) = 12 \int_{0}^{1} \left( \frac{y^3}{3} - \frac{y^4}{4} \right) \, dy = 12 \left( \frac{1}{12} - \frac{1}{20} \right) \] Simplifying: \[ E(X) = 12 \times \frac{1}{60} = \frac{2}{5} \] Thus, the correct answer is \( \frac{2}{5} \), which corresponds to option (C).