Consider an independent collection of random variables $W_i, i=1,\dots,n.$ and let $Z \sim N(0,1)$. Roughly speaking, we know that $W_i$ are close in distribution to $Z$, say each is itself a sum of $d_i$ independent (bounded) variables hence satisfies a Berry-Esseen type CLT of the form $$ d_W(W_i,Z) \lesssim \frac{1}{\sqrt{d_i}} $$ where $d_W$ is the $L_1$ Wasserstein distance. One would then hope that $\frac1{\sqrt n} \sum_{i=1}^n W_i$ is itself close to a standard normal in distribution. Are there quantitative bounds on how close this approximation is? Namely, can we get a bound on $$ \delta_n := d_W\Big(\frac1{\sqrt n} \sum_{i=1}^n W_i, Z\Big) $$ in terms of $d_W( W_i,Z)$? What sort of growth is allowed/required for $\max_i d_i$ and $\min_i d_i$ relative to $n$ to ensure that $\delta_n = o(1)$ as $n \to \infty$.

EDIT: We can assume that $W_i$ are zero-mean. A more general question is this: Is there a distance $d$ that metrizes the weak convergence (of probability measures) for which we have something close to $$ d \Big(\frac1{\sqrt n} \sum_{i=1}^n W_i, Z\Big) \le C \frac{\max_i d(W_i,Z)}{\sqrt{n}}, $$ for independent zero mean $W_i$. To see that this is plausible, consider the case where $W_i$ have equal variance, say $\mathbb E W_i^2 = 1$. Then a bound of the form $\lesssim \frac1{\sqrt{n}}\max_i \mathbb E |W_i|^3$ holds for Kolomogrov distance, maybe even for $d_W$. The question is whether equal variance $\mathbb E W_i^2 =1$ assumption can be relaxed and 3rd moment bounds can be replaced with the distance itself (the latter might be doable if $W_i$ are sufficiently concentrated, say they are sub-Gaussians). It might not be possible if Berry and Esseen bounds for non-identically distributed variables are tight.

A bonus point is if everything can be done in higher dimensions, say $W_i \in \mathbb R^m$ and $Z \sim N(0,I_m)$.


1 Answer 1


Of course an easy bound is $\delta_n \leq \frac{1}{\sqrt{n}} \sum_{i=1}^n d_W(W_i,Z)$. I don't think you can get better than this without some additional assumptions. For instance, I think if $W_i \sim N(\mu_i,1)$ then $d_W(W_i,Z) = \mu_i$. Then in this case we would have $d_W\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n W_i,Z \right) = \frac{1}{\sqrt{n}} \sum_{i=1}^n d_W(W_i,Z)$.

  • $\begingroup$ Thanks. In your example, the distance would be $|\mu_i|$ so the LHS could still be potentially much smaller. Let us say W_i are zero mean as well. I have also updated the question. $\endgroup$
    – passerby51
    Apr 23, 2019 at 16:50
  • $\begingroup$ For similar reasons I think you'll also need to assume the $W_i$ all have variance 1. For instance if $W_i\sim N(0, \sigma^2)$ for all $i$ then $d_W( \frac1{\sqrt{n}} \sum_{i=1}^n W_i, Z) = d_W(W_1, Z)$ for all $n$ ($\sqrt{n}$ better than in Jon's example with varying means but still $\sqrt{n}$ off of what you want). $\endgroup$
    – Nick Cook
    Sep 21, 2019 at 19:05

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