Question about orthogonality principle

Question

Orthogonality principle states the following:

let $V$ and $W$ be any random variables then for any function $f(W)$ \begin{align} E[(V-E[V|W])f(W)]=0 \end{align}

My question is wether the following is also true let let $g(V,W)$ be some function of $V,W$ then for any $f(W)$ we have that \begin{align} E \left[ \left(g(V,W)-E[g(V,W)|W]\right) f(W) \right]=0 \end{align}

I strongly believe this is true and a simple substitution into the fist equation should prove it. But I would like confirmation from you guys.

Answer 1

The identity

$$\mathbb{E}[(g(V,W)-\mathbb{E}(g(V,W) \mid W)) f(W)]=0$$

is a direct consequence of the tower property. Indeed: Conditioning on $\sigma(W)$ yields

$$\begin{align*} \mathbb{E}[(g(V,W)-\mathbb{E}(g(V,W) \mid W)) f(W)) &= \mathbb{E} \bigg[ \mathbb{E} \big[ (g(V,W)-\mathbb{E}(g(V,W) \mid W)) f(W)) \mid W \big] \bigg] \\ &\stackrel{\ast}{=} \mathbb{E} \bigg[ f(W) \underbrace{[\mathbb{E}(g(V,W) \mid W)-\mathbb{E}(g(V,W) \mid W)]}_{0} \bigg] \\ &= 0. \end{align*}$$

In $(\ast)$ we used that $f(W)$ is measurable with respect to $\sigma(W)$, and that we can therefore pull it outside the conditional expectation.