I am looking for an example of the following: Find a bijective, differentiable function $f$ and continuous probability density functions $q_1\ne q_2$ such that $f_*q_1=p=f_*q_2$, where $f_*$ is the pushforward density and $p$ is continuous as well. What if continuity is strengthened to differentiability?

Edit: Intuitively this seems impossible, just by continuity considerations; e.g. pick a neighborhood where $q_1$ and $q_2$ differ, and invoke bijectivity of $f$.

  • $\begingroup$ Maybe you just want to transform between two finite samples, in an optimal transport way? $\endgroup$
    – Henry.L
    Sep 6 at 19:19

1 Answer 1


This is impossible if $f$ is injective, without further assumptions such as bijective, differentiable, etc. Let $Q_1,Q_2$ be probability measures on a measurable space $(\Omega, \mathcal{F})$, and assume $f_* Q_1 = f_* Q_2$ for some injective (bimeasurable) $f : (\Omega,\mathcal{F}) \to (\Xi,\mathcal{G})$. For any $A\in \mathcal{F}$, definitions give

$$ Q_1(A) = f_* Q_1 (f(A)) = f_* Q_2 (f(A)) = Q_2(A). $$

Thus, $Q_1 = Q_2$.


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