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Let $n\in\mathbb{N}$ and $\sigma>0$ be fixed. I have a certain class $\mathcal{C}$ of random variables I am interested in analyzing. This contains

  1. $\vec X\sim \mathcal{N}(0, \sigma^2I_n)$
  2. Sums of (independent) random variables in $\mathcal{C}$, and
  3. Convolutions of (independent) random variables in $\mathcal{C}$. By this, I mean that if one views the random variables $\vec X$ as a polynomials $\sum_{i = 0}^n \vec X_i y^i$, the operation is polynomial multiplication in $\mathbb{R}[y]/(y^n-1)$.

Practically, authors in my applied area have observed that the resulting random variables look Gaussian. This can be qualitatively justified by noting that the first two points preserve Gaussianity, and the third (by the CLT) should yield something Gaussian-like.

My issue is with trying to quantify this. By standard quantitative versions of the CLT (Berry Esseen theorem), one can get that a random variable in $\mathcal{C}$ is at most $O(1/\sqrt{n})$ (and perhaps improve it to $O(1/n)$) from a Gaussian in the Kolmogorov Smirnov metric. These rates hold in much more generality though, where there is not such a strong assumption that the underlying random variables are related to Gaussians.

Can one obtain better rates in this setting? I would also be interested in rates in other metrics, in particular in the total variation, but if random variables in $\mathcal{C}$ are still quite far from Gaussian this is of second-order concern.

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  • $\begingroup$ Where you speak of "convolutions of random variables,", I wonder whether the things whose convolutions you speak of are actually the density functions rather than the random variables themselves? $\endgroup$ Commented Jan 26 at 22:56
  • $\begingroup$ @MichaelHardy The random vectors $\vec X, \vec X'$ are viewed as polynomials $\vec X \sim \sum_{i} \vec X_i y^i$, and then multiplied in $\mathbb{R}[y]/(y^n-1)$. As an example, the first coordinate of $\vec X\ast \vec X'$ is $\sum_{i = 0}^n \vec X_i \vec X'_{n-i}$ (up to a potential off-by-one error in the indices). $\endgroup$ Commented Jan 26 at 22:59
  • $\begingroup$ I think you ought to explain that in the question. $\endgroup$ Commented Jan 27 at 1:54

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