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Suppose that I have $n$ dependent random variables $X_1,\ldots,X_n$ with $\mathbb{E}[X_i]=0, \mathbb{E}[X_i^2]=1$, where we have the following bounds on the Kolmogorov distance from a normal distribution: $$|P(X_i\leq x)-\Phi(x)|\leq \rho(x)\text{ for all }i=1,\ldots,n.$$ Here $\rho(x)$ is a generic function of $x$ that goes to zero as $|x|\to \infty$ and $\Phi$ is the CDF of the standard normal distribution.

Now what can we say about the distribution of the functions of $X_i$? For example, suppose that we have a Lipschitz function $f(X_1,\ldots,X_n) = \sqrt{\sum_{i=1}^n X_i^2}$, let $Z_i$ be standard Gaussian distributions such that $\mathbb{E}[Z_iZ_j]=\mathbb{E}[X_iX_j]$. What can we say about \begin{align}|P(f(X_1,\ldots,X_n)\leq x)- P(f(Z_1,\ldots,Z_n)\leq x)|\quad ?\hspace{20mm}(1)\end{align}

The continuous mapping theorem tells us that as $\rho\to 0$, the above quantity also converges to $0$. However I am interested in trying to find explicit convergence rates for $(1)$. Stein's method most likely plays a big role here, but most papers I've found are concerned with limiting distributions, eg. https://arxiv.org/pdf/1507.08688.pdf.

A similar question can be found in: Berry-Esseen bound in 2 dimensions for linear combinations, but I am hoping for a better answer/references if possible.

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  • $\begingroup$ Is the maximum value of $\rho$ small, or do you want to take advantage of the fact that it decreases? Is the first inequality supposed to be multidimensional, with $x$ a vector in $\mathbb R^n$? If not, won't you run into trouble even with $\rho=0$? $\endgroup$
    – Will Sawin
    Commented Apr 16, 2021 at 19:29
  • $\begingroup$ I intended for the first inequality to be univariate and true for all $i$. An example of $\rho(x)$ could be like $\varepsilon/(1+x^2)$. $\endgroup$
    – 61plus
    Commented Apr 16, 2021 at 20:12
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    $\begingroup$ That's not good enough. For example we can take $X_1$ and $X_2$ to be Gaussian but $X_1=X_2 $ half the time and $X_1=-X_2$ half the time. Then $\mathbb E[X_1X_2]=0$ but $X_1,X_2$ are very different from the standard Gaussians $Z_1,Z_2$ with the same mean and variance. $\endgroup$
    – Will Sawin
    Commented Apr 16, 2021 at 20:47
  • $\begingroup$ I see what is going on now. The continuous mapping theorem will require some form of convergence of joint distribution as well. Thank you for the counterexample. $\endgroup$
    – 61plus
    Commented Apr 16, 2021 at 21:17

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