Transfer of random variables, uniqueness If $X$ is a continuous random variable with known distribution, and $Y_1= f_1(X)$, $Y_2= f_2(X)$ where $f_1$ and $f_2$ are strictly increasing functions and distribution of $Y_1$ and $Y_2$ is the same, does this imply the functions $f_1 = f_2$?
In other words, there is a unique strictly increasing function that transfers a source distribution to a destination distribution.
 A: The correct statement involves the support of $X$, namely the smallest closed set  $E\subseteq \mathbb R$ such that $X\in E$ a.s. This is the same as the support of the pushforward measure $\mu=X_{\#}\omega$. What is true is that $f_1=f_2$ on $E$, but not necessarily elsewhere (as Nate Eldredge pointed out). 
Indeed, if $a\in E$ and $f_1(a)<f_2(a)$ then (recalling that $X$ is continuous) we have either 


*

*every left neighbouhood of $a$ has positive $\mu$ measure, or

*every right neighbouhood of $a$ has positive $\mu$ measure


In the first case, pick $b<a$ such that $f_1(a)<f_2(b)$. Since $\mu((b,a))>0$, we have
$$ P(Y_1<f_1(a)) = P(X<a)>P(X<b)=P(Y_2<f_2(b)) \tag1$$
But if $Y_1$ and $Y_2$ were equally distributed, we would have
$$ P(Y_1<f_1(a)) = P(Y_2<f_1(a)) \le P(Y_2<f_2(b)) \tag2$$
The second case is similar.
A: Uniqueness can at most hold in the almost everywhere sense, consider for example $X$ uniform on $(0,1)$ and the functions $f_1$ and $f_2$ defined on $(0,1)$ by
$$
f_1(x)=\frac{x}2+\frac12\sum_{n\geqslant1}\mathbf 1_{2^nx\gt1},\qquad f_2(x)=\frac{x}2+\frac12\sum_{n\geqslant1}\mathbf 1_{2^nx\geqslant1}.
$$
The distributions of $f_1(X)$ and $f_2(X)$ coincide and the functions $f_1$ and $f_2$ are strictly increasing but $\{f_1\ne f_2\}=\{2^{-n}\mid n\geqslant1\}$.
