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Let $\mu$ be a probability measure on $\mathbb{R}^d$ such that $S_\mu$ is its second moment matrix: $$S_\mu=\int_{\mathbb{R}^d}xx^Td\mu(x)$$ I'm trying to prove the existence of a probability measure $\mu^\epsilon$ on $\mathbb{R}^d$ such that its second moment matrix $S_{\mu^\epsilon}=S_\mu+\epsilon I_d$ with ($\epsilon >0$) and $$W_2^2(\mu,\mu^\epsilon)\leq \epsilon,$$ where $W_2^2(.,.)$ is the 2-Wasserstein distance $$W_2^2(\mu,\nu)=\inf_{\gamma\in\Gamma(\mu,\nu)}\int_{\mathbb{R}^d}\Vert x-y\Vert^2d\gamma(x,y),$$ where $\Gamma(\mu,\nu)$ is the set of coupling of $\mu$ and $\nu$.

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  • $\begingroup$ How is your 2-Wasserstein distance defined? With respect to what metric on $\mathbb R^d$? Can you maybe just write the defining formula for $W_2^2(.,.)$? $\endgroup$ Mar 20, 2018 at 0:24
  • $\begingroup$ Is your norm $\|\cdot\|$ the Euclidean norm on $\mathbb R^d$? If so, I think your bound on the Wasserstein distance is missing the factor $d$: it should be $\epsilon d$ rather than $\epsilon$; do you have reasons to believe otherwise? $\endgroup$ Mar 20, 2018 at 2:48

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$\newcommand{\N}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathcal F} \newcommand{\X}{\mathcal X} \newcommand{\ep}{\epsilon} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \renewcommand{\c}{\circ} \newcommand{\tr}{\operatorname{tr}} \newcommand{\E}{\operatorname{\mathsf E}}$

Let us assume that the norm $\|\cdot\|$ used in the question is the Euclidean norm. It is convenient to restate the question in terms of random vectors, as follows:

Suppose that $X$ is a random vector in the space $\R^{d\times1}$ of $d\times1$ real column matrices, with the covariance matrix $S=\E XX^T$. For each real $\ep>0$, let $\X_\ep$ denote the set of all random vectors $X_\ep$ with covariance matrix $S_\ep:=S+\ep I_d$. The problem is then to show that \begin{equation} \inf_{X_\ep\in\X_\ep}\E\|X_\ep-X\|^2\le\ep. \tag{1} \end{equation}

It is easy to see that this conjecture is false for any $d\ge2$. Indeed, suppose that $X=0$, so that $S=0$ and $S_\ep=\ep I_d$. Then for any $X_\ep\in\X_\ep$ one has $\E\|X_\ep-X\|^2=\E\|X_\ep\|^2=\ep d>\ep$, so that the $\inf$ in (1) is greater than $\ep$.

However, (1) will hold if $\le\ep$ is replaced there by $\le\ep d$. Indeed, let $X_\ep:=X+Y_\ep$, where $Y_\ep$ is any zero-mean random vector with covariance matrix $\ep I_d$ and independent of $X$. Then $X_\ep\in\X_\ep$ and $\E\|X_\ep-X\|^2=\E\|Y_\ep\|^2=\ep d$, so that the $\inf$ in (1) is no greater than $\ep d$. Moreover, as it should now be clear from the previous paragraph, $\ep d$ is the best possible bound here.

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