91Ó°ÊÓ

The conditional probability density function of X given Y = y is given by: \[f_{X|Y}(x | y) = \frac{f_{XY}(x, y)}{f_Y(y)}\] In order to find \(f_Y(y)\), we need to marginalize the joint density function \(f_{XY}(x, y)\) with respect to x: \[f_Y(y) = \int_{-y}^{y} f_{XY}(x, y) \, dx\] Now, substitute the given joint density function and integrate: \[ f_Y(y) = \int_{-y}^{y}\frac{\left(y^{2}-x^{2}\right)}{8} e^{-y} \, dx \] To solve the integral, first we factor out the constants: \[ f_Y(y) = \frac{1}{8} e^{-y} \int_{-y}^{y}\left(y^{2}-x^{2}\right) \, dx \] Now, evaluate the integral: \[ f_Y(y) = \frac{1}{8} e^{-y} \left[ y^2x - \frac{x^3}{3} \right]_{-y}^{y} \] \[ f_Y(y) = \frac{1}{8} e^{-y} \left( \frac{8y^3}{3} \right) \] Now, simplify the expression: \[ f_Y(y) = e^{-y} \left(\frac{y^3}{3}\right) \] Now that we have both \(f_{XY}(x, y)\) and \(f_Y(y)\), we can find the conditional probability density function \(f_{X|Y}(x | y)\): \[ f_{X|Y}(x | y) = \frac{f_{XY}(x, y)}{f_Y(y)} = \frac{\frac{\left(y^{2}-x^{2}\right)}{8} e^{-y}}{e^{-y} \left(\frac{y^3}{3}\right)} = \frac{3\left(y^{2}-x^{2}\right)}{8y^3} \]

Now we'll compute the expected value of X given Y = y using the conditional probability density function: \[ E[X | Y = y] = \int_{-y}^y x \cdot f_{X|Y}(x | y) \, dx \] First, substitute the conditional probability density function into the integral: \[ E[X | Y = y] = \int_{-y}^y x \cdot \frac{3\left(y^{2}-x^{2}\right)}{8y^3} \, dx \] Now, simplify the expression inside the integral: \[ E[X | Y = y] = \frac{3}{8y^3}\int_{-y}^y x\left(y^{2}-x^{2}\right) \, dx \] To compute the integral, note that the integrand is an odd function, and we are integrating over a symmetric interval. Thus, the integral evaluates to 0: \[ E[X | Y = y] = \frac{3}{8y^3} \cdot 0 = 0 \] Thus, the expected value of X given Y = y is 0.