$E(\mathbf{y}|\mathbf{x}+\mathbf{z})=g(\mathbf{x})$ almost surely, if $\mathbf{z}\perp \!\!\! \perp \{\mathbf{y},\mathbf{x}\}$ jointly?

Question

Let $\mathbf{y},\mathbf{x}$ and $\mathbf{z}$ be real-valued random vectors with possibly different dimensions. If $\mathbf{z} \perp\!\!\!\perp \{\mathbf{y},\mathbf{x}\}$ (i.e., $\mathbf{z}$ is statistically independent of $\mathbf{y}$ and $\mathbf{x}$ jointly), is it true that there exists a function $g$ such that $E(\mathbf{y}|\mathbf{x}+\mathbf{z})=g(\mathbf{x})$ almost surely?

I know that $E(\mathbf{y}|\mathbf{x}+\mathbf{z},\mathbf{z})=E(\mathbf{y}|\mathbf{x})$ almost surely. But my question is for $E(\mathbf{y}|\mathbf{x}+\mathbf{z})$ not $E(\mathbf{y}|\mathbf{x}+\mathbf{z},\mathbf{z})$.

Answer 1

No: If $x=y$ and $x,z$ are iid $N(0,1)$ (jointly normal) then $x \mid x+z$ is also normal with mean $(x+z)/2$, i.e., $$ E[x \mid x+z]= (x+z)/2. $$