For $1\leq p<\infty$, define the <probability measures> with finite $p$th <absolute moment> by
$$
\mathcal P_p(X)=\left\{\mu\in\mathcal P(X):\int_X|x-x_0|^p\,d\mu(x)<\infty\right\},
$$
where $x_0\in X$ is any fixed reference point. The condition is independent of the choice of $x_0$. The <p-Wasserstein distance> is
$$
\boxed{d_{W_p}(\mu,\nu)=W_p(\mu,\nu)
=\left(\inf_{\pi\in\Pi(\mu,\nu)}\int_{X\times X}|x-y|^p\,d\pi(x,y)\right)^{1/p}.}
$$
The infimum is over all <transport plans>, rather than only <transport maps>. The <product measure> gives a finite upper bound using
$$
|x-y|^p\leq2^{p-1}\bigl(|x-x_0|^p+|y-x_0|^p\bigr).
$$
For $p=1$ 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.

Solution

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

{title2=$W_p(\mu,\nu)$}

For $1\leq p<\infty$ and <probability measures> with finite $p$th <absolute moments>,
$$
W_p(\mu,\nu)=\left(\inf_{\pi\in\Pi(\mu,\nu)}\int\rho(x,y)^p\,d\pi(x,y)\right)^{1/p},
$$
where $\rho$ is the ground metric and $\pi$ ranges over <transport plans>. For $p=2$ this is the <second Wasserstein distance>.

p-Wasserstein distance

{title2=$W_2(\mu,\nu)$}

For probability measures with finite second moments on a metric space,
$$
W_2(\mu,\nu)^2=\inf\{\mathbb E[\rho(X,Y)^2]:(X,Y)\text{ couples }\mu,\nu\}.
$$


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 Incoming links: Second Wasserstein distance