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Let $\xi$ be a random vector taking values in $\mathbb{R}^d$. Is there an upper bound on the $p$-Wasserstein distance $\mathcal{W}_p(\xi,a\,\xi)$ for some constant $a \neq 0$?

I have seen that if $p=1$ or $p=2$ and $\xi$ is Gaussian, there even exists a quite manageable upper bound ($p=1$) or explicit formula ($p=2$) for such a distance, but I wonder if it is possible to get an upper bound for a general distribution (and even for general $p \geqslant 1)$.

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There is a bound, the coupling $(\xi, a \xi)$ leads to the upper bound \begin{align*} W_p(\xi, a \xi)^p &\leq \mathbb{E}[|\xi - a\xi|^p] = |(1-a)|^p \mathbb{E}[|\xi|^p]\\ \Rightarrow ~~~~ W_p(\xi, a\xi) &\leq |1-a| (\mathbb{E}[|\xi|^p])^{1/p}. \end{align*}

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    $\begingroup$ Actually by the Brenier-McCann theorem this gives the exact value for $p>1$: since the map $x\mapsto T(x)=ax$ is the gradient of a convex function it is automatically optimal in the Monge formulation of the transportation problem between $\mu=\mathcal L(\xi)$ and $\nu=\mathcal L(a\xi)=T\#\mu$ $\endgroup$
    – leo monsaingeon
    Commented Jun 15, 2021 at 16:58
  • $\begingroup$ Many thanks, Steve and Leo, for your useful feedback! I guess then that such an upper bound easily generalizes to the case in which $a$ is instead a matrix (for instance, a square matrix in $\mathbb{R}^d$) by simply using the consistent matrix norm, that is, $W_p(\xi, a\xi) \leq \|I-a\| \mathbb{E}[\|\xi\|^p]^{1/p}$, right? Thanks again (and sorry if my questions are too obvious, I am a novice in this topic). $\endgroup$
    – Juan Miguel Morales González
    Commented Jun 16, 2021 at 11:59

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