Suppose $E(X|Y, Z=1) = E(X|Y, Z=0)$, is $E(X|Z=1) = E(X|Z=0)?$

Question

Suppose $X, Y, Z$ are random variables. Assume $E(X|Y, Z=1) = E(X|Y, Z=0)$ holds. Is it true that $E(X|Z=1) = E(X|Z=0)?$

Here's my attempt: \begin{align*} E(X|Z=1) &= E(E(X|Y, Z=1)|Z=1)\\ &= E(E(X|Y, Z=0)|Z=1)\\ &\neq E(E(X|Y, Z=0)|Z=0) \end{align*}

Therefore, $E(X|Z=1) \neq E(X|Z=0)$. Is the above correct?

Answer 1

Consider $X = Y$ be a standard Gaussian, and let $Z$ be a noisy observation of whether $Y > 0$. (i.e. say $Z = 1_{Y > 0} + Z'$ where $Z' \sim Bernoulli(1/3)$, and addition is to be thought of XORing).

Then, the conditional expectations you gave are equal (they are both $Y$), but certainly, the claim made in the question is false.

Answer 2

I think the equality may hold. Please correct me if you find errors.

I would claim that :

$$ E[E[X|Y,Z=a]]=E[X|Z=a] \ [1]$$

for a fixed value $a$.

Intuitively, this is the tower property where instad of the original measure we have the measure conditioned on the event $Z=a$. We can check it in a more explicit way. First write the conditional expectation as a function of the density:

$E[X|Y,Z=a]=\int dx x p(x|Y,Z=a)$

and than take expectation values:

$E[E[X|Y,Z=a]]=\int dy p(y|z=a)\int dx x p(x|y,z=a)$

Now using that:

$\int p(x|y,z=a)p(y|z=a)dy=p(x|z=a)$

we have:

$E[E[X|Y,Z=a]]=\int dx x p(x|z=a)=E[X|Z=a]$

, therefore showing [1], if no mistakes are there. Once we have [1], the fact that:

$$ E[X|Z=1]=E[X|Z=0]$$

follows by taking expectation values of the original condition.