Discussion Forum

Hello, 

I have a question regarding problem 3d from the 6th exercise sheet. 

I have been able to show that $\mathbb{E}[(X - \mathbb{E}[X \vert Z])(Y - \mathbb{E}[Y \vert Z])] = 0 \implies \mathbb{E}_z[Z^2] \mathbb{E}_{xy}[XY] = \mathbb{E}_{xz}[XZ] \mathbb{E}_{yz}[YZ]$, but not as shown in the solution (my reasoning was like this: $\mathbb{E}[(X - \mathbb{E}[X \vert Z])(Y - \mathbb{E}[Y \vert Z])] = 0 \implies \mathbb{E}[XY \vert Z] = \mathbb{E}[X \vert Z] \mathbb{E}[Y \vert Z] \implies \int \int xy p(x, y \vert x) dx dy = \int x p(x \vert z) dx \int y p(y \vert z) dy \implies z^2 p(z) \int \int xy p(x, y, z) dx dy =  \int x, z p(x, z) dx \int y z p(y, z) dy$ which we can integrate to get the conclusion). However, from the way the solution was written, I got the feeling that this should have been more direct, so my question is, is there an easier way of showing that Eq. (35) from the solution pdf holds? 

Thank you!