Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 348 / 5

For $1\le p<\infty$, define the probability measures with finite $p$th absolute moment bywhere $x_0\in X$ is any fixed reference point. The condition is independent of the choice of $x_0$. The p-Wasserstein distance isThe infimum is over all transport plans, rather than only transport maps. The product measure gives a finite upper bound usingFor $p=1$ this is the Wasserstein distance with the metric of Euclidean space; for $p=2$ it is the second Wasserstein distance. “Bounded moment” here means a finite integral, and does not require $X$ to be bounded.

Let $D=\mathrm{diam}(X)<\infty$, and first suppose $1\le p\le q<\infty$. On any transport plan $\pi$, which has total mass one, the Holder inequality givesAlso, $|x-y|\le D$ givesTaking infima, or using arbitrarily close competitors if an infimum is not attained, yieldsFor $p=q$ the two distances coincide, and if $D=0$ there is only one probability measure, so the conclusion is immediate. For $D>0$ and $p<q$, the displayed bounds imply both directions ofFor $q<p$, exchange the exponents. Thus all finite-order p-Wasserstein distances induce the same convergence on a bounded subset of ${\mathbb{R}}^d$. Closedness of $X$ is not required for these estimates.

Write $L=W_p(\mu ,\nu )$. The assumed equality of the Monge optimal transport problem and Kantorovich optimal transport problem values givesEach interpolated pushforward measure ${\mu }_t=(P_t{)}_{\#}\mu$ has a finite $p$th absolute moment, since $|P_t(x){|}^p\le 2^{p-1}(|x{|}^p+|T^{†}(x){|}^p)$ and $(T^{†}{)}_{\#}\mu =\nu$.

For any $s,t\in [0,1]$, the common-source transport planprovides the upper boundFor the reverse bound assume $0\le s\le t\le 1$. Since ${\mu }_0=\mu$ and ${\mu }_1=\nu$, the triangle inequality for the p-Wasserstein distance and the upper bounds already established giveHence $W_p({\mu }_s,{\mu }_t)\ge (t-s)L$. Combining the bounds and using symmetry provesThis is the constant speed property of displacement interpolation. The given invertibility of $P_s$ also lets one realize the competitor as the map $P_t\circ P_s^{-1}$, assuming its inverse is measurable, but the transport plan argument proves the result without invertibility.

Let $\Delta =\{(x,y):x=y\}$. For any transport plan $\pi$ and any Borel set $A$,Both rectangles are contained in ${\Delta }^c$, soThus $\inf \mathbb{K}\ge {\sup }_A|\mu (A)-\nu (A)|$.

To attain this bound, use the dominating measure $\rho =\mu +\nu$ and Radon-Nikodym derivatives $f=d\mu /d\rho$, $g=d\nu /d\rho$. Define the common measure and residual mass bySince $\int (f-g)d\rho =0$, the positive and negative parts have equal mass. Taking $A=\{f>g\}$ givesIf $r=0$, then $\mu =\nu$ and the diagonal transport plan has zero cost. If $r>0$, put $\beta =\mu -\alpha$, $\gamma =\nu -\alpha$ and defineThe residual measures each have mass $r$, so the marginal distributions of ${\pi }^{†}$ are $\alpha +\beta =\mu$ and $\alpha +\gamma =\nu$, and its total mass is $1-r+r=1$. They are mutually singular measures, concentrated respectively on $\{f>g\}$ and $\{g>f\}$. Their product measure therefore gives no mass to $\Delta$, and $\mathbb{K}({\pi }^{†})=r$. This is a maximal coupling.

In the paper's convention for the zero-mass signed measure $\mu -\nu$, the answer isThe repository's total variation distance uses the supremum itself. The paper's formula $2{\sup }_A|\sigma (A)|$ agrees with the usual total variation norm when $\sigma$ has total mass zero, as here; it is not the usual norm for an arbitrary positive measure. All suprema above are over Borel sets.