Suppose we have three independent, zero mean, finite variance random variables $V,W,Z$ and where $W,Z$ are Gaussian random variables. These random variables form a new random variable $Y$ \begin{align*} Y=V+W+Z \end{align*}
I want to compute the following quantity \begin{align*} E[W(V-E[V|Y])] \end{align*}
I have the following conjecture \begin{align*} &E[W(V-E[V|Y])]\\ &=E[(W+Z)(V-E[V|Y])]\sqrt{\frac{E[W^2]}{E[W^2]+E[Z^2]}} \text{ Problematic step}\\ &=E[(Y-V)(V-E[V|Y])]\sqrt{\frac{E[W^2]}{E[W^2]+E[Z^2]}}\\ &=E[(V-E[V|Y])^2]\sqrt{\frac{E[W^2]}{E[W^2]+E[Z^2]}} \text{ by orthogonality principle} \end{align*}
In the 'Problematic Step' I am not completly sure I can add $Z$ and normalize. Is that step fine if yes how to argue it?
For example, in the case when $V$ is Gaussian this is easy to verify.
{Possible Approach:}
I was thinking of using the following approach:
In really we have to show that \begin{align*} E[W E[V|Y]]= \alpha E[(W+Z) E[V|Y]] \end{align*} where the conjecture is that $\alpha=\sqrt{\frac{E[W^2]}{E[W^2]+E[Z^2]}}$.
So, we really have to show that \begin{align*} &\int \int w E[V|Y=y] f_{W,Y} (w,y) dw dy\\ &= \alpha \int \int \int (w+z) E[V|Y=y] f_{Z,W,Y} (z,w,y) dw dy dz\\ &= \alpha \int \int s E[V|Y=y] f_{Y,S} (y,s) ds dy \end{align*} where $S=W+Z$.
but how to show the last step?? The good thing is that $S=W+Z$ is Gaussian so maybe it will make things easeir.
Thank you