Let $X:\Omega\to M$ be a random variable taking values in some measurable space $M$ (not necessarily $\mathbb R$), and let $f,g:M\to M'$ be some measurable functions.

If we have the equality in distribution $f(X)=_dg(X)$, then we can't really say much about whether $f=g$; for example if $X$ is a standard Gaussian on $\mathbb R$, then we can take $f(x)=x$ and $g(x)=-x$ which are only equal at one point.

Suppose, however, that we have the equality in joint distribution $$(X,f(X))=_d(X,g(X)).$$ Can we say that $f=g$ up to a set of measure zero with respect to the distribution of $X$? Intuition suggests that this might be true because we're talking about the "joint" behaviour of $X$ with $f(X)$ or $g(X)$, but I can't really get an argument for this (or get a counterexample).

  • $\begingroup$ I think you could discern $f$ from your joint distribution by conditioning on $X$. $\endgroup$ – angryavian May 9 '17 at 1:34

Using the measurable map $(x,y)\mapsto \left(f(x),y\right)$, we can derive that $f\left(X\right)=g\left(X\right)$ almost surely (by considering the probability of the diagonal) . More formally, we have that $\left(f(X),f(X)\right)=\left(f(X),g(X)\right)$ in distribution. Let $\Delta :=\left\{(x,y)\in M'\times M'\mid x=y\right\}$. Then $$1=\mathbb P\left\{\left(f(X),f(X)\right)\in \Delta\right\}=\mathbb P\left\{\left(f(X),g(X)\right)\in \Delta\right\}.$$

Therefore, $f$ and $g$ coincide on the support of $X$. We cannot conclude anything for the values outside the support.

  • $\begingroup$ Thanks for answering. I'm not sure I follow what you mean by probability of the diagonal. Are you considering $(x,y)\mapsto (f(x),y)$ as a random variable on the product space $M\times M$ equipped with the product measure $\mu_X\otimes\mu_X$ where $\mu_X$ is the distribution of $X$? $\endgroup$ – user78270 May 9 '17 at 11:48
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    $\begingroup$ @user78270 I have edited in order to clarify. $\endgroup$ – Davide Giraudo May 9 '17 at 12:11

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