probability - Related to conditional expectation of the product of two independent random variables

I have two random variables $X$ and $Y$. The random variable have non-negative support meaning that their PDF is defined only for non negative values. I know the PDF's of $X$ and $Y$. I want to find the expectation of $XY^j$ (where $j$ is a real number) given that $XY^j

Answer

To be clear, you are looking for $\mathsf E(XY^j\mid XY^j

I get:

$$\begin{align}\mathsf E(XY^j\mid XY^j\dfrac{\mathsf E (XY^j~\mathbf 1_{XY^j\\[1ex]~=~&\dfrac{\mathsf E (X~\mathsf E(Y^j~\mathbf 1_{Y<(w/X)^{1/j}}\mid X)) }{ \mathsf P(Y<(w/X)^{1/j})} &\text{only if}~&j > 0
\\[1ex] ~=~& \bbox[gainsboro]{\dfrac{\int\limits_0^\infty~x~\mathsf E( Y^j~\mathbf 1_{Y<(w/x)^{1/j}}\mid X=x)~f_X(x)~\mathrm d x }{\int\limits_0^\infty~\mathsf P(Y<(w/x)^{1/j}\mid X=x)~f_X(x)~\mathrm d x }}
\\[1ex] ~=~& \dfrac{\int\limits_0^\infty~x~f_X(x)\int\limits_0^{(w/x)^{1/j}} y^j~f_{Y\mid X}(y\mid x)~\mathrm d~y~\mathrm d x }{\int\limits_0^\infty~f_X(x)\int\limits_0^{(w/x)^{1/j}} f_{Y\mid X}(y\mid x)~\mathrm d~y~\mathrm d x }

\end{align}$$