# Local Lipschitzness of parameterization of Gaussians in Wasserstein space

Fix a positive integer $$n$$ and consider the $$2$$-Wasserstein space $$\mathcal{P}_2(\mathbb{R}^n)$$. Let $$X$$ be the cone of $$n\times n$$ symmetric positive semidefinite matrices with Frobenius norm and consider the map $$f:\mathbb{R}^n\times X \rightarrow \mathcal{P}_2(\mathbb{R}^n)$$ given by $$f(a,B) = N(a,B)$$ where $$N(a,B)$$ is the $$n$$-dimensional Gaussian measure with mean $$a$$ and covariance $$B$$.

Clearly the map $$f$$ is locally-Lipschitz and injective. However, is its inverse on its image in $$\mathcal{P}_2(\mathbb{R}^n)$$ also locally-Lipschitz?

$$\newcommand{\R}{\mathbb R}\newcommand{\tr}{\operatorname{tr}}$$The answer is yes.

Indeed, it is easy to see (cf. e.g. Proposition 7 or the beginning of its proof) that the Wasserstein distance between $$N(a,A)$$ and $$N(b,B)$$ is $$\begin{equation*} W_2(N(a,A),N(b,B))=\sqrt{\|a-b\|^2+W_2(N(0,A),N(0,B))^2}, \tag{10}\label{10} \end{equation*}$$ where $$\|\cdot\|$$ is the Euclidean norm. So, $$\begin{equation*} \|a-b\|\le W_2(N(a,A),N(b,B)). \tag{20}\label{20} \end{equation*}$$

Let now $$X\sim N(0,A)$$ and $$Y\sim N(0,B)$$. Then for any unit vector $$u\in\R^n=\R^{n\times1}$$ $$\begin{equation*} E\|X-Y\|^2\ge E(u^\top X-u^\top Y)^2\ge(\sqrt{u^\top A u}-\sqrt{u^\top B u})^2, \end{equation*}$$ where the last inequality holds (say, by mentioned Proposition 7) because $$u^\top X$$ and $$u^\top Y$$ are real-valued Gaussian zero-mean random variables with variances $$u^\top A u$$ and $$u^\top B u$$. So, by \eqref{10} and the definition of the $$W_2$$-distance, for any unit vector $$u\in\R^n=\R^{n\times1}$$, \begin{equation*} \begin{aligned} W_2(N(a,A),N(b,B))&\ge W_2(N(0,A),N(0,B)) \\ &\ge|\sqrt{u^\top A u}-\sqrt{u^\top B u}| \\ &=\frac{|u^\top A u-u^\top B u|}{\sqrt{u^\top A u}+\sqrt{u^\top B u}} \\ &\ge\frac{|u^\top(A-B)u|}{\sqrt{\|A\|}+\sqrt{\|B\|}} \\ &=\frac{\|A-B\|}{\sqrt{\|A\|}+\sqrt{\|B\|}} \end{aligned} \end{equation*} for some unit vector $$u\in\R^n=\R^{n\times1}$$, where $$\|M\|$$ is the spectral norm of a matrix $$M$$. So, $$\begin{equation*} \|A-B\|\le(\sqrt{\|A\|}+\sqrt{\|B\|\,})\, W_2(N(a,A),N(b,B)). \tag{30}\label{30} \end{equation*}$$

Thus, by \eqref{20} and \eqref{30}, we have the desired local Lipschitz property. $$\quad\Box$$

Remark: Inequality \eqref{20} is "exact in the limit", when $$A$$ and $$B$$ are each (close to) the zero matrix.

Inequality \eqref{30} turns into the equality when $$n=1$$ to, more generally, "exact in the limit" when $$A$$ and $$B$$ are (close to) commuting matrices of rank $$1$$ each. $$\quad\Box$$

User ABIM wrote in a comment: "@Justin_other_PhD OP claimed that $$f$$ is bi-Lipschitz but why is the upper-bound true?"

Let me answer this question as well. First of all, it is not true that $$f$$ is Lipschitz (and the OP did not claim that). Indeed, even when $$n=1$$, we have $$W_2(N(0,A),N(0,B))=|\sqrt A-\sqrt B|$$, so that there is no real $$L>0$$ such that $$W_2(N(0,A),N(0,B))\le L|A-B|$$ for all real $$A,B>0$$.

What the OP said, and what is true, is that $$f$$ is locally Lipschitz. Indeed, let $$X\sim N(0,A)$$ and $$Y:=B^{1/2}A^{-1/2}X$$. Then $$Y\sim N(0,B)$$ and hence \begin{equation*} \begin{aligned} W_2(N(0,A),N(0,B))^2&\le E\|X-Y\|^2 \\ &=\tr(A+B-B^{1/2}A^{1/2}-A^{1/2}B^{1/2}) \\ &=\tr[(A^{1/2}-B^{1/2})^2]=\|A^{1/2}-B^{1/2}\|_F^2 \\ &\le n\,\|A^{1/2}-B^{1/2}\|^2, \end{aligned} \tag{40}\label{40} \end{equation*} where $$\tr$$ denotes the trace and $$\|\cdot\|_F$$ is the Frobenius norm.

Next, for real $$a\ge0$$, $$\begin{equation*} a^{1/2}=\frac1\pi\int_0^\infty dt\,t^{-1/2}\Big(1-\frac t{t+a}\Big) \end{equation*}$$ and hence the value of the derivative of the function $$M\mapsto M^{1/2}$$ at $$A$$ at a symmetric matrix $$H$$ is \begin{equation*} \begin{aligned} (A^{1/2})'(H)&=\frac1\pi\int_0^\infty dt\,t^{1/2}(tI+A)^{-1}H(tI+A)^{-1} \\ & \le\frac1\pi\int_0^\infty dt\,t^{1/2}(t+c)^{-2}|H| =\frac1{2c^{1/2}}\,|H| \end{aligned} \end{equation*} provided that $$A\ge cI$$ for some $$c\in(0,\infty)$$ (where $$I$$ is the identity matrix), so that the operator norm of $$(A^{1/2})'$$ is $$\le\dfrac1{2c^{1/2}}$$. So, by \eqref{10} and \eqref{40}, $$\begin{equation*} W_2(N(a,A),N(b,B))\le\|a-b\|+W_2(N(0,A),N(0,B)) \\ \le \|a-b\|+\dfrac{n^{1/2}}{2c^{1/2}}\,\|A-B\| \end{equation*}$$ provided that $$A\ge cI$$ and $$B\ge cI$$ for some $$c\in(0,\infty)$$. Thus, $$f[=N]$$ is locally Lipschitz. $$\quad\Box$$

• Or just note that, more generally, taking the mean and variance of a probability distribution are continuous operations in 2-Wasserstein space... Jun 21, 2023 at 7:48
• @MartinHairer : I am not sure how to understand your comment. The question here was about Lipschitz continuity, not just continuity. Jun 21, 2023 at 11:23
• You're right, I wasn't careful in reading the question and the variance is in general only 1/2-Hölder in Wasserstein-2 space. Jun 21, 2023 at 18:23
• @MartinHairer : Thank you for your further comment. Do you have a readily available reference to the 1/2-Hölder condition in general? Jun 21, 2023 at 19:39
• @IosifPinelis I wanted to ask the same question; to me it is not obvious Jun 21, 2023 at 22:38