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For random variables $X_1,\dots,X_n$ with common mean $\mathbb{E}[X_i]=\mu$ and common bounds $a\leq X_i\leq b$, we have the very useful Hoeffding's inequality: $$\mathbb{P}\left(\left|\mu -\frac1n\sum_iX_i \right|\geq t\right)\leq 2\exp\left(-\frac{2nt^2}{(b-a)^2}\right).$$

I am interested in the following case: let $\nu_1,\dots,\nu_n$ be random measures (that is $\nu_i:\Omega \to \mathscr{P}(\mathbb{R})$), such that $\mathbb{E}[\nu_i]=\nu$ (where $\mathbb{E}(\nu_i):= \sum_{\omega\in \Omega} \mathbb{P}(\omega)\nu_i(\omega)$). Moreover, we have a similar boundedness assumption, $\text{supp}(\nu_i)\subseteq[a,b]$. Is there a similar concentration inequality such as Hoeffding's that can be applied in this case?

In particular, I am interested in whether one can say $$\mathbb{P}\left(\mathcal{W}\left(\nu,\frac1n\sum_i\nu_i \right)\geq t\right)\leq 2\exp\left(-\frac{2nt^2}{(b-a)^2}\right),$$ where $\mathcal{W}$ is the Wasserstein distance.

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  • $\begingroup$ Stein's method has been flexible for Wasserstein eg. for Xi see (1) in "Stein’s method for normal approximation in Wasserstein distances..." arxiv.org/pdf/1905.13615.pdf They show bounds for all p and so perhaps you can push that to exponential bounds via the Linfty norm approximation by Lp and that the Xi are bounded. These are wrt to Gaussian but by dividing by 1/srqr(n), it seems reasonable to get the concentration bound. $\endgroup$ Commented Oct 31, 2022 at 23:10
  • $\begingroup$ I believe that my answer here might answer your question: mathoverflow.net/a/407994 $\endgroup$ Commented Nov 10, 2022 at 15:50

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