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Let $X$ be a Polish space and let $(\mu_i)_{i=1}^{\infty}$ be a sequence of probability measures in the Wasserstein space $\mathcal{P}(X)$ on $X$. Let $(\beta_i)_{i=1}^{\infty}$ be a summable sequence in $(0,\infty)$. For every positive integer $k$, define the probability measures $$ \nu_k = (\sum_{1\leq i\leq k}\beta_i)^{-1} \sum_{1\leq i\leq k} \beta_i \mu_i $$ and define the probability measure $$ \nu_{\infty} = (\sum_{i=1}^{\infty}\beta_i)^{-1} \sum_{i=1}^{\infty} \beta_i \mu_i. $$

Is it true/clear that $ \lim\limits_{k\to\infty} \mathcal{W}(\nu_k,\nu_{\infty}) = 0, $ quantitatively (here $\mathcal{W}$ denotes the Wasserstein distance on $\mathcal{P}(X)$).

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It is true and clear if the metric space $X$ has a finite diameter, but false in general: Take $\beta_i=2^{-i}$ and $\mu_i$ the point mass at $3^i$.

Details: In the case $D=$diam$(X)<\infty$, write $s_k=\sum_{i \le k} \beta_i$ and $s=\sum_{i<\infty} \beta_i$. Then $$\nu_\infty=\frac{s_k}{s} \nu_k+\frac{s-s_k}{s} \gamma_k$$ for some probability measure $$\gamma_k=\frac{\sum_{i > k} \beta_i \mu_i}{s-s_k} \,.$$ Therefore $$\nu-\nu_k=\frac{s-s_k}{s} (\gamma_k-\nu_k)$$ whence $$\mathcal{W}(\nu_k,\nu_{\infty}) \le \frac{s-s_k}{s} D$$ because the diameter of the Wasserstein metric on $\mathcal{P}(X)$ is $D$.

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  • $\begingroup$ You mean it is obvious buy controlling $\mathcal{W}(\nu_k,\nu_{\infty})\leq \|\nu_k-\nu_{\infty}\|_{TV}\leq \operatorname{diam}(X)\sum_{s\leq k} |\beta_s - \frac{\beta_s}{\sum_{j\leq k}(\beta_j)}|\|\mu_s\| + \sum_{s>k} \|\beta_k\|\to 0$? The issue is that we don't have this control if $X$ is of unbounded diameter? $\endgroup$
    – BLBA
    Oct 26 at 7:12
  • $\begingroup$ @BLBA I get a different formula- I added some detail to my answer. $\endgroup$ Oct 28 at 17:46
  • $\begingroup$ thanks for the added details; they helped a lot. $\endgroup$
    – BLBA
    Nov 3 at 12:03
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Yet another answer, but just a comment first: which Wasserstein distance are you talking about? There is a whole family $\mathcal W_p$ of p-th Wasserstein distances, built upon the family of costs $c_p(x,y)=d(x,y)^p$ for any $p\in[1,\infty)$. (I presume you are talking here about the $\mathcal W_1$ distance.)

In any case, it is well-known that convergence in the Wasserstein distance $\mathcal W_p$ is equivalent to weak convergence of measures together with convergence of the $p$-moments $$ \int_X d(x,x_0)^pd\nu _k(x)\to \int_X d(x,x_0)^p d\nu_\infty(x) $$ (for some - and thus any - arbitrary $x_0\in X$). For a precise reference, see Theorem 6.9 in Villani's "big book" [1]. So, just expanding a little bit and slightly sharpening Yuval Pere's answer: in your setting it is pretty clear that $\nu_k\to \nu$ weakly, so the only thing left to check is that the $p$-th moments converge as well. In bounded spaces this is automatically guaranteed by the weak convergence, since $d(x_0,\cdot)^p$ is a bounded continuous function. Of course Yuval's counter-example gives a sequence such that the moments do NOT converge (actually for any $p\geq 1$), but there could also be slightly intermediate situations where the supports of $\nu_k$ can also "escape at infinity", while still keeping convergence of the moments and thus guaranteeing convergence in Wasserstein distance. So in other words, the real issue here is not really boundedness of the domain, but rather $p$-th moments are the key to a full "if and only if" characterization.

[1] Villani, C. (2009). Optimal transport: old and new (Vol. 338, p. 23). Berlin: Springer.

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  • $\begingroup$ This is a very helpful, simple, but insightful answer. Thanks leo. $\endgroup$
    – BLBA
    Nov 3 at 12:02

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