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Let $X$ and $Y$ be two random variables with first order moments, i.e. $E[|X|]$, $E[|Y|]<+\infty$. Assume further that

$$E\left[|X-Y|\right]<\varepsilon.$$

Set $Law(X)=\mu$ and $Law(Y)=\nu$, it is clear that $\mu$ and $\nu$ are close in the Prokhorov metric, see

https://en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric

for definition. Denote by $\rho(\cdot,\cdot)$ the Prokhorov metric. My question is how to estimate $\rho(\mu,\nu)$. For example, could we show that $\rho(\mu,\nu)<\varepsilon$ or $\rho(\mu,\nu)<\sqrt{\varepsilon}$? Thanks for the reply!

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$\sqrt{\varepsilon}$ works. Assume that for some set $A$ we have $\mu(A)=a$ and $\nu(A^{\sqrt{\varepsilon}})<a-\sqrt{\varepsilon}$. Then with probability more then $\sqrt{\varepsilon}$ we have $X\in A$, $Y\notin A^{\sqrt{\varepsilon}}$. In this case $|X-Y|\geqslant \sqrt{\varepsilon}$ for sure. Hence expectation of $|X-Y|$ is more than $\varepsilon$.

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  • $\begingroup$ Thanks a lot for your reply. Indeed, that will work even we replace the Prokhorov metric by the Wasserstein metric. $\endgroup$
    – CodeGolf
    Jan 12, 2016 at 14:12
  • $\begingroup$ Kantorovich optimal transportation distance (is it the same thing that you call Wasserstein metric?) may be defined as supremum of integrals $E(f(X)-f(Y))$ for 1-Lipschitz functions $f$, clearly this does not exceed $E(|X-Y|)<\varepsilon$. $\endgroup$ Jan 12, 2016 at 14:22
  • $\begingroup$ Exactly. It follows from the dual formulation. $\endgroup$
    – CodeGolf
    Jan 12, 2016 at 19:56

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