# Can a random function of a normal variable be independent of it

I have two normal variables $X,Y$.

Is it possible to have $X = f(Y,Z)$ for some non-trivial function $f$, an independent normal variable $Z$ and that $X$ will be independent of $Y$?

Thanks.

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Here you mean $Z$ is independent of both $X$ and $Y$, right? –  Patrick Li Jan 3 '13 at 0:59

You're not being that precise about what's independent of what. However, let's assume you mean the following: $Y$ and $Z$ are independent random variables and are normally distributed. Is there a non-trivial function $f$ such that $X=f(Y,Z)$ is both normally distributed and independent of $Y$? The answer is yes. Suppose $Z$ has mean $E[Z]=0$. Then any function $f(y,z)=g(y)z$, where $g(y)=\pm 1$ everywhere, meets your condition: regardless of the value of $Y$, $X$ has the same (normal) distribution.

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I suppose $g(y) = \pm 1$ does not meet the "non-trivial"-part of the question. –  TMM Jan 3 '13 at 1:28
@TMM: It doesn't have to be a constant function... it could be $\text{sgn}(y)$, or a periodic function that alternates between $+1$ and $-1$, or whatever. –  mjqxxxx Jan 3 '13 at 4:32