Given $Z\perp X\mid Y$, is it true in general $Var(Z|h(X,Y))=Var(Z|h(c,Y))?$ Given random variables $X, Y, Z$:

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*If $Z\perp X\mid Y$, then I know that $Var(Z|X,Y)=Var(Z|Y)$
But is it still true in general that
$$Var(Z|h(X,Y))=Var(Z|h(c,Y))?$$
here $h$ is a real valued function and $c$ is a constant.
I feel like this should still hold because $X$ doesn't provide any additional information about $Z$. But I can't quite prove or disprove it.
 A: The expression you have proposed does not look to me to be a proper expression for the conditional variance.  Nevertheless, it is possible to effectively marginalise $X$ out of the expression for the conditional variance.  To facilitate this analysis, let $H_*=h(X,Y)$ denote the random variable formed by the function, and let $\mathscr{Y}$ denote the support of $Y$.  Then we can write the conditional density of $Z$ as:
$$\begin{align}
p(z|h_*) 
&= \frac{p(z, h_*)}{p(h_*)} \\[6pt]
&= \frac{1}{p(h_*)} \int_{\mathscr{Y}} p(z, h_*|y) p(y) \ dy \\[6pt]
&= \frac{1}{p(h_*)} \int_{\mathscr{Y}} p(z|y) p(h_*|y) p(y) \ dy. \\[6pt]
\end{align}$$
Now, consider an arbitrary measurable function $f$ operating on $Z$ and denote the support of $Z$ by $\mathscr{Z}$.  To facilitate our analysis we define the function:
$$\mu_f(y) \equiv \mathbb{E}(f(Z)|Y=y) = \int \limits_\mathscr{Z} f(z) p(z|y) \ dz.$$
Using the law of the unconscious statistician, we can then
write the expected value of this function of $Z$, conditional on the statistic of interest, as:
$$\begin{align}
\mathbb{E}(f(Z)|h(X,Y) = h_*)
&= \int \limits_\mathscr{Z} f(z) p(z|h_*) \ dz \\[6pt]
&= \int \limits_\mathscr{Z} f(z) \frac{\int_{\mathscr{Y}} p(z|y) p(h_*|y) p(y) \ dy}{\int_{\mathscr{Y}} p(h_*|y) p(y) \ dy} \ dz \\[6pt]
&= \frac{\int_{\mathscr{Y}} \Big( \int \limits_\mathscr{Z} f(z) p(z|y) \ dz \Big) p(h_*|y) p(y) \ dy}{\int_{\mathscr{Y}} p(h_*|y) p(y) \ dy} \\[6pt]
&= \frac{\int_{\mathscr{Y}} \mu_f(y) p(h_*|y) p(y) \ dy}{\int_{\mathscr{Y}} p(h_*|y) p(y) \ dy}. \\[6pt]
\end{align}$$
Since this holds for an arbitrary function $f$, the expression also holds for all the conditional moments of $Z$, including the conditional variance.  Thus, we can write the latter as:
$$\mathbb{V}(Z| h(X,Y) = h_*) = \frac{\int_{\mathscr{Y}} \mathbb{E}(Z^2|y) p(h_*|y) p(y) \ dy}{\int_{\mathscr{Y}} p(h_*|y) p(y) \ dy} - \Bigg( \frac{\int_{\mathscr{Y}} \mathbb{E}(Z|y) p(h_*|y) p(y) \ dy}{\int_{\mathscr{Y}} p(h_*|y) p(y) \ dy} \Bigg)^2.$$
A: The conjectured statement is not true. Consider $X, Y, Z$ all Bernoulli with success probability $1/2$, but with $Y = Z$. Let $h(X,Y) = XY$. Then $[Z \mid XY = 1]$ is a point-mass at 1, $[Z \mid XY = 0]$ is a Bernoulli with success probability $1/3$, and $[Z \mid h(0,Y)]$ is Bernoulli with success probability $1/2$. In any case, $\text{Var}(Z \mid XY)$ is a function of $XY$ while $\text{Var}(Z \mid h(0,Y))$ is a constant $1/4$.
