Suppose $X,Y, Z$ are random variables, and without loss of generality let's assume $X=f(Y, Z)$. Is the following statement true in general?
$$ E[X|Y]=0 \iff E[X|Y,Z]=0$$
I seem to have a proof (might be flawed). But that conclusion just doesn't seem natural to me.
- $\Leftarrow$: by the law of total expectation, $$E[X|Y]=E_Z [E[X|Y,Z]=E_Z[0]=0$$
- $\Rightarrow$: by the law of total probability, $$E[X|Y,Z]=\int_x xp(x|y,z)dx=\int_x \frac{xp(x|y)}{p(z)}dx=0$$
The $E[X|Y,Z]=0 \Rightarrow E[X|Y]=0$ direction is true, as you proved.
The other direction isn't. An easy way to construct counterexamples is to use degenerate random variables (r.v.s that are constant with probability $1$).
So let $X=Y+Z$ and $Y=1$ and $Z:\left(\genfrac{}{}{0pt}{}{-3/2}{1/2}\ \genfrac{}{}{0pt}{}{-1/2}{1/2}\right)$. Thus $Y$, being constant, is indepentent of both $X$ and $Z$, and we can say $X=1+Z$ (almost surely).
Then $E[X\mid Y]=E[1+Z]=0$. However, $E[X\mid Y,Z]=E[1+Z \mid Z]=1+Z \neq 0$.
Paul's example in the comments is nicer as it doesn't rely on degenerate cases, however I wanted to note an easy way to test the truthfulness of your statements.