# Quantitative CLT bound

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 $$Z \sim N(0,I_m)$$.

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 $$d_W\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n W_i,Z \right) = \frac{1}{\sqrt{n}} \sum_{i=1}^n d_W(W_i,Z)$$.
• 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. – passerby51 Apr 23 '19 at 16:50
• 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). – Nick Cook Sep 21 '19 at 19:05