# Transforming two smooth densities to the same density

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 $$f$$.

• Maybe you just want to transform between two finite samples, in an optimal transport way? Sep 6, 2022 at 19:19

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, $$Q_1 = Q_2$$.