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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 $y=f(\mathbf{x}+\mathbf{z},\mathbf{z})=g(\mathbf{x})$.

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  • $\begingroup$ @LSpice, independence here is in the probability sense. $\endgroup$
    – John
    Commented Aug 28, 2022 at 1:09
  • $\begingroup$ See also mathoverflow.net/questions/429322/… concerning same circle of ideas. $\endgroup$
    – Gerry Myerson
    Commented Aug 28, 2022 at 22:56

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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 $y$.

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    $\begingroup$ @John, re, it is generally good form not to change the question after receiving an answer, but rather to ask a new question if necessary. $\endgroup$
    – LSpice
    Commented Aug 28, 2022 at 1:12
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    $\begingroup$ @John : What do you mean by "z is independent of x and y, respectively"? Which is it: (i) z is independent of (x,y) or (ii) z is independent of x and of y? $\endgroup$
    – Iosif Pinelis
    Commented Aug 28, 2022 at 1:31
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    $\begingroup$ @John : Then I think the conjecture would be true. However, at this point I will not provide details on this. Since in your original comment you said you meant z to be independent of x and of y, separately, I strongly suggest you follow the advice by LSpice and edit your post according to your original comment . $\endgroup$
    – Iosif Pinelis
    Commented Aug 28, 2022 at 13:23
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    $\begingroup$ @John : You edited it, not in accordance, but contrary to the advice of LSpice and contrary to your original comment (that you had meant z to be independent of x and of y, separately). I have now edited your post according to your original comment. Hopefully, you will keep it this way. $\endgroup$
    – Iosif Pinelis
    Commented Aug 28, 2022 at 15:13
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    $\begingroup$ @John : I saw that reply of yours. Whether another condition is "better" or not, it is important that you stick to what you said originally and not invalidate a valid answer (Carlo Beenakker's in this case). So, please restore the original condition, that z is independent of x and of y, separately. $\endgroup$
    – Iosif Pinelis
    Commented Aug 28, 2022 at 15:23

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