91Ó°ÊÓ

Since we know f(x, y), we can find the conditional pdf f(x | Y = y) using the following formula: \[f(x | Y=y) = \frac{f(x,y)}{f_Y(y)}\] where \(f_Y(y)\) is the marginal pdf of Y, which is found by integrating the joint density f(x, y) with respect to x from 0 to infinity. So let's find the marginal pdf of Y first: \[f_Y(y)=\int_{0}^{\infty}f(x,y)dx = \int_{0}^{\infty}\frac{e^{-x/y}e^{-y}}{y}dx\] Now we can find the conditional pdf f(x | Y = y).

To find the marginal pdf of Y, we need to evaluate the integral of the joint pdf with respect to x: \[f_Y(y)=\int_{0}^{\infty}\frac{e^{-x/y}e^{-y}}{y}dx\] We can use substitution to solve this integral: Let \(u = -\frac{x}{y}\) Then, \(du = -\frac{dx}{y}\) and \(dx = -y \cdot du\) Now, the integral can be rewritten as: \[f_Y(y) = -e^{-y}\int_{0}^{\infty} e^{uy} \cdot y \cdot du\] Now, the integral is easier to evaluate: \[f_Y(y) = -e^{-y}[-y \cdot e^{uy} |_0^\infty]= e^{-y} y\] Now we have the marginal pdf of Y.

Now that we have both the joint pdf and the marginal pdf of Y, we can find the conditional pdf: \[f(x|Y=y) = \frac{f(x,y)}{f_Y(y)}\] Plug in the values \(f(x,y)=\frac{e^{-x/y}e^{-y}}{y}\) and \(f_Y(y) = e^{-y}y\), we get: \[f(x|Y=y) = \frac{\frac{e^{-x/y}e^{-y}}{y}}{e^{-y}y} = \frac{e^{-x/y}}{y}\] Now we have the conditional pdf of X given Y = y.

Finally, we can compute the conditional expectation of \(X^2\) given \(Y = y\): \[E[X^2 | Y=y] = \int_{0}^{\infty} x^2 f(x | Y=y) dx\] Substitute the conditional pdf \(f(x|Y=y) = \frac{e^{-x/y}}{y}\) into the integral, we get: \[E[X^2 | Y=y] = \int_{0}^{\infty} x^2 \frac{e^{-x/y}}{y} dx\] Now, use integration by parts twice to solve this integral: Let \(u = x^2\), \(dv = \frac{e^{-x/y}}{y}dx\) Integrating dv, we get \(v=-ye^{-x/y}\). Differentiating u, we get \(du=2x \cdot dx\). Integration by parts formula: \(\int{u \cdot dv} = u \cdot v - \int{v \cdot du}\) By applying integration by parts twice for \(E[X^2 | Y=y]\), we get: \[E[X^2 | Y=y]= \left. x^2(-ye^{-x/y})\right|_0^\infty + y^2 \int_{0}^{\infty} 2x e^{-x/y} dx\] Now we need to compute the integral in the last term. Apply integration by parts one more time: \(\int{u \cdot dv} = u \cdot v - \int{v \cdot du}\) Let \(u = x\), \(dv = 2e^{-x/y}dx\) \(v = -2ye^{-x/y}\), \(du = dx\) \(\int_{0}^{\infty} 2x e^{-x/y} dx = -2y^2 \int_{0}^{\infty} e^{-x/y} dx = -2y^2 (-y\left.\left(e^{-x/y}\right)\right|_{0}^{\infty})\) After evaluating the integrals, we have: \[E[X^2 | Y=y] = 0 + y^2 \cdot 2y = 2y^3\] Thus, the conditional expectation of \(X^2\) given \(Y=y\) is \(2y^3\).