Let $\sigma_{X|Z}^2 = \min_{a} \mathbb{E}[X-aZ]^2$ denote the linear mean squared error in estimating $X$ from $Z$. Suppose $X$ and $Y$ are conditionally independent given $Z$. Then can we claim the following? $$ \sigma_{X+Y|Z}^2 = \sigma_{X|Z}^2+\sigma_{Y|Z}^2 $$ Any help is appreciated...
Edit: My strategy so far is as follows... $$ \sigma_{X+Y|Z}^2 - \sigma_{X|Z}^2 -\sigma_{Y|Z}^2 \\ = \sigma_{X}^2+\sigma_{Y}^2+2 \mathbb{E}[XY]-\frac{(\mathbb{E}[XZ+YZ])^2}{\sigma_{Z}^2}-\sigma_{X}^2+\frac{(\mathbb{E}[XZ])^2}{\sigma_{Z}^2} -\sigma_{Y}^2+\frac{(\mathbb{E}[YZ])^2}{\sigma_{Z}^2}\\ = 2 \mathbb{E}[XY]-\frac{2 \mathbb{E}[XZ] \mathbb{E}[YZ]}{\sigma_{Z}^2}$$
Having said that, I next attempted to show that $\mathbb{E}[XY]=\frac{ \mathbb{E}[XZ] \mathbb{E}[YZ]}{\sigma_{Z}^2}$. But my approach below only seems to work for jointly Gaussian RVs:
$$\mathbb{E}[XY]= \mathbb{E}[\mathbb{E}[XY|Z]]\\ = \mathbb{E}[\mathbb{E}[X|Z]\mathbb{E}[Y|Z]] \\ = \mathbb{E}\left[\frac{\mathbb{E}[XZ]}{\sigma_{Z}^2}Z \frac{\mathbb{E}[YZ]}{\sigma_{Z}^2}Z\right]=\frac{\mathbb{E}[XZ]\mathbb{E}[YZ]}{\sigma_{Z}^2}$$
(The second step above uses the conditional independence of $X$ and $Y$ given $Z$.)
My question is if this holds for general linear estimators (RVs not necessarily jointly Gaussian)? I hope the question is clearer now...
I try a counterexample to the last identity that you proposed.
Let $Y=f(Z)$ and $X$ independent of $Z$. Than I would say that $Y$ and $X$ are conditionally independent given $Z$.
Formally this means that:
$E[g(X)h(Y)|Z]=E[g(X)|Z]*E[h(Y)|Z]$
for every $g,h$ measurable. In fact:
$E[g(X)h(Y)|Z]=E[g(X)h(f(Z))|Z]=h(f(Z))*E[g(X)|Z]=E[h(f(Z))|Z]*E[g(X)|Z]=E[h(Y)|Z]*E[g(X)|Z]$
Note that we did not use the independence of $X$ over $Z$ till now. The real reason is that when $Z$ is given, $Y$ is a constant and everything, in particular $X$, is independent of a constant.
Let's try to plug in this $X$, $Y$ and $Z$ in the identity. Than we have, using the independence of $X$ from $Z$:
$E[f(Z)]=\frac{E[Z]*E[f(Z)Z]}{E[Z^2]}$
which cannot be true for every function $f$.
Hope the counterexample works.