Non-monotone convergence in the quadratic Wasserstein distance

TL;DR

This paper provides a simple counter-example demonstrating that the quadratic Wasserstein distance does not always decrease monotonically when considering n-fold normalized convolutions of two measures, challenging an existing assumption.

Contribution

It presents the first explicit counter-example to the presumed monotonicity of quadratic Wasserstein distance under convolution.

Findings

Quadratic Wasserstein distance can increase under convolution.

Counter-example disproves the monotonicity assumption.

Challenges previous beliefs in mass transport theory.

Abstract

We give an easy counter-example to Problem 7.20 from C. Villani's book on mass transport: in general, the quadratic Wasserstein distance between $n$-fold normalized convolutions of two given measures fails to decrease monotonically.

Equations

$${\mathcal{T}}(\mu,\nu):=\inf_{(X,Y)}{\mathbb{E}}\bigl{[}\|X-Y\|^{2}\bigr{]}$$

$$\mathcal{T}(aX+bY,a{X}^{\prime }+b{Y}^{\prime })\le a^2\mathcal{T}(X_{,}{X}^{\prime })+b^2\mathcal{T}(Y,Y^{\prime }).$$

$$S_m:=\frac{1}{\sqrt{m}}\sum _{i=1}^m{X}_i,m\in \mathbb{N}.$$

$${\mathcal{T}}\big{(}\mu^{(2m)},\nu^{(2m)}\big{)}\leq{\mathcal{T}}\big{(}\mu^{(m)},\nu^{(m)}\big{)},\qquad m\in{\mathbb{N}},$$

$${\mathcal{T}}\big{(}\mu^{(m+1)},\nu^{(m+1)}\big{)}\leq{\mathcal{T}}\big{(}\mu^{(m)},\nu^{(m)}\big{)}$$

$${\mathcal{T}}\big{(}\mu^{(3)},\nu^{(3)}\big{)}\leq\frac{1}{3}\bigl{[}2{\mathcal{T}}\big{(}\mu^{(2)},\nu^{(2)}\big{)}+{\mathcal{T}}(\mu,\nu)\bigr{]}\leq{\mathcal{T}}(\mu,\nu).$$

$$\underset{n\to \infty }{\lim }\sqrt{n}\mathcal{T}({\mu }_n\ast {\mu }_n,{\nu }_n\ast {\nu }_n)=\frac{2}{\sqrt{2\pi }},$$

$${\mathcal{T}}(\mu_{2}\ast\mu_{2},\nu_{2}\ast\nu_{2})=0.625<\frac{2}{3}\leq{\biggr{(}\frac{\sqrt{2}}{\sqrt{3}}\biggl{)}}^{2}{\mathcal{T}}(\mu_{2}\ast\mu_{2}\ast\mu_{2},\nu_{2}\ast\nu_{2}\ast\nu_{2}).$$

$$p_{n,k}=\left(\genfrac{}{}{0ex}{}{4n-2}{k+2n-1}\right)\frac{1}{2^{4n-2}}.$$

$$\begin{split}\tau_{n}(2k+2)&=\frac{1}{4}p_{n,k}+\frac{1}{4}p_{n,k+2}+\frac{1}{2}p_{n,k+1}\\ &=\frac{1}{4}(p_{n,k}-p_{n,k+1})+\frac{1}{4}(p_{n,k+2}-p_{n,k+1})+\sigma_{n}(2k+2)\\ &=\frac{1}{4}p_{n,k}\Bigl{(}1-\frac{p_{n,k+1}}{p_{n,k}}\Bigr{)}+\frac{1}{4}p_{n,k+1}\Bigl{(}\frac{p_{n,k+2}}{p_{n,k+1}}-1\Bigr{)}+\sigma_{n}(2k+2).\end{split}$$

$$0\le 1-\frac{p_{n,k+1}}{p_{n,k}}=1-\frac{\left(\genfrac{}{}{0ex}{}{4n-2}{k+2n}\right)}{\left(\genfrac{}{}{0ex}{}{4n-2}{k+2n-1}\right)}=1-\frac{2n-k-1}{k+2n}=\frac{2k+1}{2n+k}\le \frac{2k+1}{2n},$$

$$\mathcal{T}({\sigma }_n,{\tau }_n)\le 2\sum _{k=0}^{2n-1}\frac{2^2}{4}{p}_{n,k}\frac{2k+1}{2n+k}\le \sum _{k=0}^{2n-1}{p}_{n,k}\frac{2k+1}{n}.$$

$$\underset{n\to \infty }{\lim }\frac{1}{2\sqrt{n}}\sum _{k=0}^{2n-1}(2k)p_{n,k}={\int }_0^{\infty }\frac{x}{\sqrt{2\pi }}{e}^{-x^2/2}dx.$$

$$\underset{n\to \infty }{\mathrm{limsup}}\sqrt{n}\mathcal{T}({\sigma }_n,{\tau }_n)\le \frac{2}{\sqrt{2\pi }}\approx \mathrm{0.79788.}$$

$$\tau_{n}(2k+2)=\frac{1}{4}p_{n,k+1}\Bigl{(}\frac{p_{n,k}}{p_{n,k+1}}-1\Bigr{)}+\frac{1}{4}p_{n,k+2}\Bigl{(}1-\frac{p_{n,k+1}}{p_{n,k+2}}\Bigr{)}+\sigma_{n}(2k+2)$$

$$\tau_{n}(2k+2)=\frac{1}{4}p_{n,k+1}\Bigl{(}\frac{p_{n,k}}{p_{n,k+1}}-1\Bigr{)}+\sigma_{n}(2k+2)$$

$$\frac{p_{n,k}}{p_{n,k+1}}-1=\frac{\left(\genfrac{}{}{0ex}{}{4n-2}{k+2n-1}\right)}{\left(\genfrac{}{}{0ex}{}{4n-2}{k+2n}\right)}-1=\frac{k+2n}{2n-k-1}-1=\frac{2k+1}{2n-k-1}\ge \frac{2k+1}{2n}$$

$$\mathcal{T}({\sigma }_n,{\tau }_n)\ge \sum _{k=0}^{2n-2}{p}_{n,k+1}\frac{2k+1}{n},$$

$$\underset{n\to \infty }{\mathrm{liminf}}\sqrt{n}\mathcal{T}({\sigma }_n,{\tau }_n)\ge \frac{2}{\sqrt{2\pi }}.$$

$$\underset{n\to \infty }{\lim }\mathcal{T}({\mu }_n^{(p)},{\nu }_n^{(p)})=0.$$

Full text

Financial support from the Austrian Science Fund (FWF) under grant P 15889, from the Vienna Science and Technology Fund (WWTF) under grant MA13, from the European Union under grant HPRN-CT-2002-00281 is gratefully acknowledged. Furthermore this work was financially supported by the Christian Doppler Research Association (CDG) via PRisMa Lab. The authors gratefully acknowledge a fruitful collaboration and continued support by Bank Austria Creditanstalt (BA-CA) and the Austrian Federal Financing Agency (ÖBFA) through CDG

We use the terminology and notation from [5]. For Borel measures $\mu$, $\nu$ on ${\mathbb{R}}^{d}$ we define the quadratic Wasserstein distance

[TABLE]

where $\|\mathchoice{\mskip 1.5mu}{\mskip 1.5mu}{}{}\cdot\mathchoice{\mskip 1.5mu}{\mskip 1.5mu}{}{}\|$ is the Euclidean distance on ${\mathbb{R}}^{d}$ and the pairs $(X,Y)$ run through all random vectors defined on some common probabilistic space $(\Omega,\mathcal{F},{\mathbb{P}})$, such that $X$ has distribution $\mu$ and $Y$ has distribution $\nu$. By a slight abuse of notation we define ${\mathcal{T}}(U,V):={\mathcal{T}}(\mu,\nu)$ for two random vectors $U$, $V\mathchoice{\mskip-1.5mu}{\mskip-1.5mu}{}{}$, such that $U$ has distribution $\mu$ and $V$ has distribution $\nu$. The following theorem (see [5, Proposition 7.17]) is due to Tanaka [4].

[TABLE]

For a sequence of i.i.d. random vectors $(X_{i})_{i\in{\mathbb{N}}}$ we define the normalized partial sums

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If $\mu$ denotes the law of $X_{1}$, we write $\mu^{(m)}$ for the law of $S_{m}$. Clearly $\mu^{(m)}$ equals, up to the scaling factor $\sqrt{m}$, the $m$-fold convolution $\mu\ast\mu\ast\dots\ast\mu$ of $\mu$.

We shall always deal with measures $\mu$, $\nu$ with vanishing barycenter. Given two measures $\mu$ and $\nu$ on ${\mathbb{R}}^{d}$ with finite second moments, we let $(X_{i})_{i\in{\mathbb{N}}}$ and $(X^{\prime}_{i})_{i\in{\mathbb{N}}}$ be i.i.d. sequences with law $\mu$ and $\nu$, respectively, and denote by $S_{m}$ and $S_{m}^{\prime}$ the corresponding normalized partial sums. From Theorem 1 we obtain

[TABLE]

from which one may quickly deduce a proof of the Central Limit Theorem (compare [5, Ch. 7.4] and the references given there).

However, we can not deduce from Theorem 1 that the inequality

[TABLE]

holds true for all $m\in{\mathbb{N}}$. Specializing to the case $m=2$, an estimate, which we can obtain from Tanaka’s Theorem, is

[TABLE]

This contains some valid information, but does not imply (1). It was posed as Problem 7.20 of [5], whether inequality (1) holds true for all probability measures $\mu$, $\nu$ on ${\mathbb{R}}^{d}$ and all $m\in{\mathbb{N}}$.

The subsequent easy example shows that the answer is no, even for $d=1$ and symmetric measures. We can choose $\mu=\mu_{n}$ and $\nu=\nu_{n}$ for sufficiently large $n\geq 2$, as the proposition (see also Remark 1) shows.

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while ${\mathcal{T}}(\mu_{n}\ast\mu_{n}\ast\mu_{n},\nu_{n}\ast\nu_{n}\ast\nu_{n})\geq 1$ for all $n\in{\mathbb{N}}$.

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For the proof of (2), fix $n\in{\mathbb{N}}$, define $\sigma_{n}=\mu_{n}\ast\mu_{n}$ and $\tau_{n}=\nu_{n}\ast\nu_{n}$, and note that $\sigma_{n}$ and $\tau_{n}$ are supported by the even numbers. For $k=-(2n-1),\dots,(2n-1)$ we denote by $p_{n,k}$ the probability of the point $2k$ under $\sigma_{n}$, i.e.

[TABLE]

We define $p_{n,k}=0$ for $|k|\geq 2n$. We have $\tau_{n}=\sigma_{n}\ast\rho$, where $\rho$ is the distribution giving probability $\frac{1}{4}$, $\frac{1}{2}$, $\frac{1}{4}$ to $-2$, [math], $2$, respectively. We deduce that for $0\leq k\leq 2n-2$,

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Notice that $p_{n,k}\geq p_{n,k+1}$ for $0\leq k\leq 2n-1$. The term in the first parentheses is therefore non-negative. It can easily be calculated and estimated via

[TABLE]

for $0\leq k\leq 2n-1$.

Following [5] we know that the quadratic Wasserstein distance ${\mathcal{T}}$ can be given by a cyclically monotone transport plan $\pi=\pi_{n}$. We define the transport plan $\pi$ via an intuitive transport map $T$. It is sufficient to define $T$ for $0\leq k\leq 2n-1$, since it acts symmetrically on the negative side. $T$ moves mass $\frac{1}{4}p_{n,k}\frac{2k+1}{2n+k}$ from the point $2k$ to $2k+2$ for $k\geq 1$. At $k=0$ the transport $T$ moves $\frac{1}{8n}p_{n,0}$ to every side, which is possible, since there is enough mass concentrated at [math].

By equation (3) we see that the transport $T$ moves $\sigma_{n}$ to $\tau_{n}$, since, for $1\leq k\leq 2n-2$, the first terms corresponds to the mass, which arrives from the left and is added to $\sigma_{n}$, and the second term to the mass, which is transported away: summing up one obtains $\tau_{n}$. For $k=2n-1$, mass only arrives from the left. At $k=0$ mass is only transported away. By the symmetry of the problem around [math] and by the quadratic nature of the cost function (the distance of the transport is $2$, hence cost $2^{2}$), we finally have

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By the Central Limit Theorem and uniform integrability of the function $x\mapsto x_{+}:=\max(0,x)$ with respect to the binomial approximations, we obtain

[TABLE]

Hence

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In order to obtain equality we start from the local monotonicity of the respective transport maps on non-positive and non-negative numbers. It easily follows that the given transport plan is cyclically monotone and hence optimal (see [5, Ch. 2]). The subsequent equality allows also to consider estimates from below. Rewriting (3) yields

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for $0\leq k\leq 2n-3$, and

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for $k=2n-2$. Furthermore,

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for $0\leq k\leq 2n-2$. This yields by a reasoning similar to the above that

[TABLE]

hence

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∎

[TABLE]

When entropy is considered instead of the quadratic Wasserstein distance the corresponding question on monotonicity was answered affirmatively in the recent paper [1].

One may also formulate a variant of Problem 7.20 as given in (1) by replacing the measure $\nu$ through a log-concave probability distribution. This would again generalize problem 1.