# Does X = Y in distribution and X being Y-measurable imply Y is X-measurable?

Suppose $X,Y$ are random variables taking values in some Borel space, $X \overset {d}{=} Y$, and $X$ is $Y$-measurable.

It follows from the fact that $X$ is $Y$-measurable that there exists a measurable $f$ such that $X = f(Y)$ a.s.

Is it the case that there exists a measurable $g$ such that $Y = g(X)$ a.s.? It seems plausible, and is true for finite-valued, discrete random variables.

EDIT: I've thought about it a bit, and one consequence is $X \overset {d}{=} f(X)$, which may or may not be helpful.

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No.

Let $Y$ be uniformly distributed on $[-1,1]$, let $f(x) = 2|x|-1$ and let $X = f(Y)$. Then $X$ is also uniformly distributed on $[-1,1]$, but $Y$ is not $X$-measurable, since, intuitively, by looking at $X$ you cannot tell the sign of $Y$.

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Thanks! Looks like I was going down the wrong track. The bigger question I was trying to ask was this one: math.stackexchange.com/questions/686220/… – Roy D. Feb 22 '14 at 17:48

There's an "accepted" answer and it's good, so normally I might not bite, but I have a slight preference for this one:

Suppose $\alpha$ is uniformly distributed in the interval $[0,2\pi)$, so that $X=(\cos\alpha,\sin\alpha)$ is uniformly distributed on the circle.

Then let $Y=(\cos(2\alpha),\sin(2\alpha))$. Then $Y$ is also uniformly distributed on the circle.

Certainly if you know $X$ you can find $Y$, but not vice-versa. The mapping $X\mapsto Y$ is two-to-one.

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