Let $X$ and $Y$ be two positive independent random variables, and let $Z =_d X/Y$ where $=_d$ stands for "has the same distribution of". Let $\psi(\cdot)$ a function such that $\psi(z)=\frac{\psi(x)}{y}$, where $x$, $y$ and $z$ are three positive real number such that $z = x/y$. I want to evaluate the following expectation: \begin{equation*} E\{(Z\psi(Z))^2\}. \end{equation*} This is my try: \begin{equation*} \begin{split} E\{(Z\psi(Z))^2\} & = \int_z (z\psi(z))^2 p_Z(z)dz \\ &= \int_x \frac{x^2}{y^2}\psi\left( \frac{x}{y}\right) ^2 \frac{1}{y} p_Z(x)dx\\ &=\int_x x^2\psi\left( x\right)^2 \frac{1}{y} p_Z(x)dx \end{split} \end{equation*} Now, using the ratio distribution formula enter link description here, $p_Z(z)$ can be expressed as: \begin{equation*} p_Z(z) = \int_y y p_X(yz)p_Y(y)dy, \quad y>0. \end{equation*} Then, we have: \begin{equation*} \begin{split} E\{(Z\psi(Z))^2\} & = \int_x (x\psi( x))^2 \frac{1}{y} p_Z(x)dx\\ &= \int_x (x\psi( x))^2 \frac{1}{y} \left[ \int_y y p_X(x)p_Y(y)dy \right] dx\\ &= ? \end{split} \end{equation*}
Van anyone help me with it? Thanks!
Maybe I found an error in my previous derivation. This is my new try.
Using the ratio distribution formula, $p_Z(z)$ can be expressed as: \begin{equation*} p_Z(z) = \int_y y p_X(zy)p_Y(y)dy, \quad y>0. \end{equation*} \begin{equation*} \begin{split} E\{(Z\psi(Z))^2\} & = \int_z (z\psi(z))^2 p_Z(z)dz \\ &= \int_z (z\psi(z))^2 \left[ \int_y y p_X(zy)p_Y(y)dy \right] dz\\ &= \int_z \left[ \int_y (z\psi(z))^2y p_X(zy)p_Y(y)dy \right] dz\\ &= \int_y \left[ \int_z (z\psi(z))^2y p_X(zy)p_Y(y)dz \right] dy \quad \mathrm{ (from \; Fubini's \; Theorem)}\\ &=\int_y \left[ \int_x \left( \frac{x}{y}\psi\left( \frac{x}{y}\right)\right)^2 y p_X(x)p_Y(y)\frac{dx}{y}\right] dy \quad \mathrm{from}\; z=x/y\\ &=\int_y \left[ \int_x \frac{1}{y^4} (x \psi(x))^2 p_X(x)dx\right] p_Y(y) dy \\ &= E\{Y^{-4}\}E\{(X \psi(X))^2\} \end{split} \end{equation*}
Is this correct?