Let $y\in \mathbb{R}$ and $\mathbf{x},\mathbf{z}\in\mathbb{R}^p$ be *random variable* and *random vectors*. Assume $y=f(\mathbf{x}+\mathbf{z},\mathbf{z})$ for some function $f$.
Is the following statement correct?
If $\mathbf{z}\perp \!\!\! \perp y$ and $\mathbf{z}\perp \!\!\! \perp\mathbf{x}$, where $\perp \!\!\! \perp$ means *statistical independence*, then there exists a function $g$ such that $y=f(\mathbf{x}+\mathbf{z},\mathbf{z})=g(\mathbf{x})$.



Independent input feature z can be removed: if y=f(x+z,z), then y=g(x)?
A simple counter-example: take $p=1$, $x=+1$ or $x=-1$ with probability 1/2, $z=+1$ or $z=-1$ with probability 1/2, independently of $x$.


Define $y=xz$ [or, if you wish, $f(x+z,z)=(x+z)z-z^2$]; one has $y=+1$ or $-1$ with probability 1/2, independently of either $x$ or $z$; hence, there is no function $g(x)$ such that $y=g(x)$, although $z$ is independent of $x$ and $y$.


