# 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.

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What do you know? What did you try? Where are you stuck? – Did Dec 3 '11 at 21:19
This can't be true without more assumptions. Let's say $X,Y_1, Y_2$ are all uniformly distributed on $[0,1]$. Then we could take $f_1(x) = f_2(x) = x$ on $[0,1]$, but $f_1, f_2$ could differ outside $[0,1]$ without causing any problems. – Nate Eldredge Dec 3 '11 at 22:02
At least $f_1$, $f_2$ should be exactly the same at points $x$ where $Pr(x) > 0$, right? – FaMi Dec 3 '11 at 22:49
I tried setting CDF of $Y_1$, and $Y_2$ equal, which lead to $\forall x, Pr(f_1^{-1}(x)\le X \le f_2^{-1}(x))=0$. I concluded that either $f_1^{-1}(x)=f_2^{-1}(x)$ or $Pr(x)=0$ – FaMi Dec 3 '11 at 23:44

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.

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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 $\{f_1\ne f_2\}=\{2^{-n}\mid n\geqslant1\}$.

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