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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

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