A. If X and Y are random variables with joint density function \[ f(x, y)=\left\{\begin{array}{lll}4(x+y-3) & 0

Ask by Rodriguez Moran. in Nigeria

Feb 18,2025

To find the conditional expectation of \( X \) given \( Y \), we need to follow these steps: 1. **Find the marginal density function of \( Y \)**, \( f_Y(y) \). 2. **Find the conditional density function of \( X \) given \( Y \)**, \( f_{X|Y}(x|y) \). 3. **Calculate the conditional expectation \( E[X|Y=y] \)** using the conditional density function. ### Step 1: Find the marginal density function of \( Y \) The marginal density function of \( Y \) can be found by integrating the joint density function \( f(x, y) \) over the range of \( x \): \[ f_Y(y) = \int_{-\infty}^{\infty} f(x, y) \, dx = \int_0^2 f(x, y) \, dx \] Given the joint density function: \[ f(x, y) = 4(x+y-3) \quad \text{for } 0 < x < 2 \] We need to determine the limits for \( y \). The expression \( x + y - 3 > 0 \) implies \( y > 3 - x \). Since \( 0 < x < 2 \), we have: - When \( x = 0 \), \( y > 3 \) - When \( x = 2 \), \( y > 1 \) Thus, \( y \) must be in the range \( 1 < y < 3 \). Now we can compute \( f_Y(y) \): \[ f_Y(y) = \int_0^2 4(x+y-3) \, dx \] Calculating the integral: \[ = 4 \int_0^2 (x + y - 3) \, dx \] Calculating the integral: \[ = 4 \left[ \frac{x^2}{2} + yx - 3x \right]_0^2 \] Evaluating at the limits: \[ = 4 \left[ \frac{2^2}{2} + 2y - 6 \right] = 4 \left[ 2 + 2y - 6 \right] = 4(2y - 4) = 8y - 16 \] Thus, the marginal density function of \( Y \) is: \[ f_Y(y) = 8y - 16 \quad \text{for } 1 < y < 3 \] ### Step 2: Find the conditional density function of \( X \) given \( Y \) The conditional density function \( f_{X|Y}(x|y) \) is given by: \[ f_{X|Y}(x|y) = \frac{f(x, y)}{f_Y(y)} \] Substituting the values we found: \[ f_{X|Y}(x|y) = \frac{4(x+y-3)}{8y-16} \] ### Step 3: Calculate the conditional expectation \( E[X|Y=y] \) The conditional expectation \( E[X|Y=y] \) is given by: \[ E[X|Y=y] = \int_{-\infty}^{\infty} x f_{X|Y}(x|y) \, dx = \int_0^2 x \cdot \frac{4(x+y-3)}{8y-16} \, dx \] Calculating the integral: \[ = \frac{4}{8y-16} \int_0^2 x(x+y-3) \, dx \] Calculating the integral: \[ = \frac{4}{8y-16} \int_0^2 (x^2 + xy - 3x) \, dx \] Calculating each term: 1. \( \int_0^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} \) 2. \( \int_0^2 xy \, dx = y \left[ \frac{x^2}{2} \right]_0^2 = 2y \) 3. \( \int_0^2 3x \, dx = 3 \left[ \frac{x^2}{2} \right]_0^2 = 6 \) Putting it all together: \[ = \frac{4}{8y-16} \left( \frac{8}{3} + 2y - 6 \right) = \frac{4}{8y-16} \left( 2y - \frac{10}{3} \right) \] Thus, the conditional expectation \( E[X|Y=y] \) is: \[ E[X|Y=y] = \frac{4(2y - \frac{10}{3})}{8y - 16} \] This is the final expression for the conditional expectation of \( X \) given \( Y \).