Skip to main content
MathOverflow

Return to Answer

added 19 characters in body
Source Link
Iosif Pinelis
Iosif Pinelis
  • 128k
  • 8
  • 107
  • 229

$\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 $\pi$.

$\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, 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 $\pi$.

$\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 $\pi$.

Source Link
Iosif Pinelis
Iosif Pinelis
  • 128k
  • 8
  • 107
  • 229

$\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, 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 $\pi$.