Push-forward of sum of two maps

Question

Let $X=R^n$ and $Y=R^m$ be two Euclidean spaces with $m<n$. Let $\varphi$ and $\phi$ are two smooth maps from $X$ to $Y$, and $\mu$ a probability measure on $X$. Is there any relationship between $(\varphi+\phi)_\#\mu$ with $\varphi_\#\mu$ and $\phi_\#\mu$ (where $\#$ denotes the push-forward of a measure)?

Answer 1

The problem is one of linear programming. To see this more clearly, let us assume for a moment that $Y$ is a finite set (rather than $\mathbb{R}^m$). Let me write $\mu f^{-1}$ instead of $f_\#\mu$. Then the problem is whether, for given probability measures $\mu_\varphi$, $\mu_\phi$, and $\mu_{\varphi+\phi}$ over $Y$ (which are "candidates" for $\mu\varphi^{-1}$, $\mu\phi^{-1}$, and $\mu(\varphi+\phi)^{-1}$, respectively), there exist nonnegative real numbers $p(u,v)$ (which are "candidates" for the values of $\mu(\{x\in X\colon\varphi(x)=u,\phi(x)=v\})$ for $(u,v)\in Y\times Y$) such that $\sum_{v\in Y}p(y,v)=\mu_\varphi(\{y\})$ and $\sum_{u\in Y}p(u,y)=\mu_\phi(\{y\})$ for all $y\in Y$, with the additional (affine) restrictions that $\sum\{p(u,v)\colon u+v=y,u\in Y,v\in Y\}=\mu_{\varphi+\phi}(y)$ for all $y\in Y$. Going back to $Y=\mathbb{R}^m$, one sees that the problem is one of infinite-dimensional linear programming.

Relevant here is the fundamental paper "The Existence of Probability Measures with Given Marginals" by Strassen (1965) in The Annals of Mathematical Statistics deals with the existence of a probability measure $\mu$ on a product space $X\times Y$ given the marginals $\mu\pi_X^{-1}$ and $\mu\pi_Y^{-1}$ (where $\pi_X$ and $\pi_Y$ are the projections from $X\times Y$ to $X$ and $Y$, respectively) plus further affine restrictions on $\mu$, as is the case here.

One simple necessary condition on the probability measures $\mu_\varphi$, $\mu_\phi$, and $\mu_{\varphi+\phi}$ is, obviously, that $$\int_Y y\mu_{\varphi+\phi}(dy)=\int_Y y\mu_{\varphi}(dy)+\int_Y y\mu_{\phi}(dy),$$ provided that (say) at least two of these three integrals exist and are finite.