Mollifying a measure without changing its marginals

Is there a reasonable/canonical way to mollify a Borel probability measure without changing its marginals. Let $$\pi \in \mathcal{P}(\mathbb{R}^2)$$ with marginals $$\mu,\nu$$. I want to smooth out $$\pi$$ up to scale $$\varepsilon$$, which is equivalent to cutting it off at frequency $$|k|\approx \varepsilon^{-1}$$ in Fourier space, but I do not want to change is marginals. The resulting measure should have a smooth density. It goes without saying that the usual procedure (convolution with a smooth, positive, bump function) does not work. Of course, you can assume $$\mu,\nu$$ are absolutely continuous with smooth densities.

• Whenever this can be done $\mu$ and $\nu$ must have densities. So, you need to impose some conditions on $\mu$ and $\nu$. Commented Aug 20, 2023 at 13:44
• This goes without saying. I have edited the question to indicate this. Commented Aug 21, 2023 at 13:27
• I guess you will need a case where the marginals $\mu,\nu$ are unchanged when they are mollified. Commented Aug 21, 2023 at 13:57
• @GeraldEdgar Yes, I agree. What I had in mind is some smoothing operation that is $\mu,\nu$-dependent with the natural property that it leaves $\mu,\nu$ invariant. Maybe this is asking for too much. Commented Aug 22, 2023 at 7:52

$$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\tX}{\tilde X}\newcommand{\tY}{\tilde Y}\newcommand{\tpi}{\tilde\pi}$$Here is how this can be done in an explicit way, at least when the densities (say $$p$$ and $$q$$) of $$\mu$$ and $$\nu$$ are everywhere $$>0$$.

Let $$(X,Y)$$ be a random point in $$\R^2$$ with distribution $$\pi$$ and thus marginals $$\mu$$ and $$\nu$$.

Let $$F$$ and $$G$$ be the c.d.f.'s of $$X$$ and $$Y$$ (respectively).

Let $$\pi_\ep$$ be a distribution over $$\R^2$$ with a smooth density such that $$\pi_\ep$$ is close to $$\pi$$. Let $$(X_\ep,Y_\ep)$$ be a random point in $$\R^2$$ with distribution $$\pi_\ep$$. Let $$F_\ep$$ and $$G_\ep$$ be the c.d.f.'s of $$X_\ep$$ and $$Y_\ep$$ (respectively), so that $$F_\ep$$ and $$G_\ep$$ are close to $$F$$ and $$G$$, and hence $$F^{-1}\circ F_\ep$$ and $$G^{-1}\circ G_\ep$$ are smooth maps each close to the identity map.

Let $$\tX:=F^{-1}(F_\ep(X_\ep))$$ and $$\tY:=G^{-1}(G_\ep(Y_\ep))$$. Then the distribution (say $$\tpi$$) of the random point $$(\tX,\tY)$$ in $$\R^2$$ will have a smooth density and marginals $$\mu$$ and $$\nu$$, and $$\tpi$$ will be close to $$\pi$$.

• Thanks! Any idea how one can generalise this to higher dimensions? Commented Aug 26, 2023 at 10:11
• @almosteverywhere : I suggest you ask the question about higher dimensions in another post. For higher dimensions, I guess one could try to use Dirk's idea. However, the problem that I then envision is to even show that a feasible distribution $\pi_\epsilon$ with a finite $\Phi(\pi_\epsilon)$ exists (let alone a minimizer). Commented Aug 27, 2023 at 1:34

You can always formulate this problem as an optimization/feasibility problem and look for solutions. The conditions on the marginals gives two linear constraints: $$P_1\pi_\epsilon = \mu,\quad P_2\pi_\epsilon = \nu.$$

The closeness to a given $$\pi$$ can be formulated as another constraints (for example) $$\|\pi-\pi_\epsilon\|\leq\epsilon.$$ For smoothness you can now optimize your favorite smoothness measure $$\Phi$$ over the constraint set, i.e. you consider $$\min \Phi(\pi_\epsilon)\quad \text{s.t.}\quad P_1\pi_\epsilon = \mu,\quad P_2\pi_\epsilon = \nu,\quad \|\pi-\pi_\epsilon\|\leq\epsilon.$$