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1 vote
2 answers
156 views

Wasserstein Metric Inequality

This is the exercise: This exercise shows that “spreading out” probability measures makes them closer together. Define the convolution of a measure by: for any probability density function $\phi$, let ...
Raul Bataccs's user avatar
3 votes
1 answer
123 views

Scaling property of the Wasserstein metric

I would need help with this example. Let $(S, ||\cdot||)$ denote a normed vector space over $K =\mathbb R$ or $K =\mathbb C$. Let $X$ and $Y$ be $S$-valued random vectors with $E~[~||X||~] < \infty$...
Spira's user avatar
  • 61
3 votes
1 answer
222 views

Absolutely continuous curves in Wasserstein distance and measurability.

Let $(X, d, \mu)$ be a metric measure space. Let $P^1(X)$ denote the space of probability measures on $(X,d)$, which have finite first moments, that is: \begin{equation} \nu \in P^1(X) \implies \int d(...
Kakuro's user avatar
  • 313
1 vote
0 answers
54 views

Derivative of Kantorovic Potential wrt to Measure

The Kantorovich Dual of the 1-Wasserstein distance $W_1(p,q)$ between two densities $p(x), q(x)$ is given by $$W_1(p,q) = \sup_{|f|_L\leq 1} \int f(x)(p(x)-q(x))dx$$ with $|f|_L$ denoting the ...
Jonas Metzger's user avatar
1 vote
1 answer
100 views

Extension of Kantorovich-Rubinstein inequality.

Let $(\mathcal{X}, \Sigma)$ be a Polish metric space, endowed with the Borel $\sigma$-algebra. Let $\mathscr P$ be the space of probability measures on $\mathcal X$ and $\mathscr P^1$ be defined as $$\...
ECL's user avatar
  • 2,970