# 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 $XY^j<w$ ($w$ is some positive value). Can I find the expectation as follows $$\int_0^{\infty}E\left[xY^j|X=x,Y<\left(\frac{w}{x}\right)^{1/j}\right]f_X(x)dx$$ Is my strategy right or there are mistakes? I will be very thankful for your response. Thanks in advance.

• looks good to me. – BGM Sep 23 '16 at 3:08
• @BGM I posted a question about which I have some confusion (math.stackexchange.com/questions/2094428/…). I have lot of confusion about this but I have not received any comments on my strategy. Please comment whether my strategy is right or wrong in the above link question. I will be very thankful to you. – Frank Moses Jan 14 '17 at 1:36

To be clear, you are looking for $\mathsf E(XY^j\mid XY^j<w)$ ?
\begin{align}\mathsf E(XY^j\mid XY^j<w) ~=~& \dfrac{\mathsf E (XY^j~\mathbf 1_{XY^j<w}) }{ \mathsf P(XY^j<w)} \\[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}