Question

Let \( X \) and \( Y \) be continuous random variables with the joint probability density function
\[ f(x, y) = \begin{cases} c x (1 - x), & 0 < x < y < 1 \\ 0, & \text{otherwise} \end{cases} \] where \( c \) is a positive real constant. Then \( E(X) \) equals

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

Hint:

When dealing with joint probability distributions, always ensure to normalize the distribution and use the correct limits for integration to compute the expected values.

Answer 1

The Correct Option is A

Step 1: Find the value of constant \( c \).
The joint probability density function \( f(x, y) \) is given by: \[ f(x, y) = \begin{cases} c x (1 - x), & 0 < x < y < 1 \\ 0, & \text{otherwise} \end{cases} \] To find the constant \( c \), we use the fact that the total probability must be 1. Therefore, we integrate the joint pdf over the entire range of \( x \) and \( y \): \[ \int_0^1 \int_0^y c x (1 - x) \, dx \, dy = 1 \] First, integrate with respect to \( x \): \[ \int_0^y x (1 - x) \, dx = \int_0^y (x - x^2) \, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^y = \frac{y^2}{2} - \frac{y^3}{3} \] Now, integrate with respect to \( y \): \[ c \int_0^1 \left( \frac{y^2}{2} - \frac{y^3}{3} \right) \, dy = 1 \] Perform the integration: \[ c \left[ \frac{y^3}{6} - \frac{y^4}{12} \right]_0^1 = 1 \] \[ c \left( \frac{1}{6} - \frac{1}{12} \right) = 1 \] \[ c \times \frac{1}{12} = 1 \quad \Rightarrow \quad c = 12 \]

Step 2: Find \( E(X) \).
The expected value of \( X \) is given by: \[ E(X) = \int_0^1 \int_0^y x f(x, y) \, dx \, dy \] Substitute the expression for \( f(x, y) \): \[ E(X) = \int_0^1 \int_0^y x \cdot 12 x (1 - x) \, dx \, dy = 12 \int_0^1 \int_0^y x^2 (1 - x) \, dx \, dy \] Integrate with respect to \( x \): \[ \int_0^y x^2 (1 - x) \, dx = \int_0^y (x^2 - x^3) \, dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^y = \frac{y^3}{3} - \frac{y^4}{4} \] Now, integrate with respect to \( y \): \[ E(X) = 12 \int_0^1 \left( \frac{y^3}{3} - \frac{y^4}{4} \right) \, dy \] Perform the integration: \[ E(X) = 12 \left[ \frac{y^4}{12} - \frac{y^5}{20} \right]_0^1 = 12 \left( \frac{1}{12} - \frac{1}{20} \right) \] \[ E(X) = 12 \times \frac{8}{60} = \frac{96}{60} = \frac{8}{5} = \frac{2}{5} \]

Final Answer: \( \frac{2}{5} \)