2
$\begingroup$

I was considering the following problem. Let $\{(X_i,Y_i)\}_{i=1}^n$ be i.i.d. zero-mean random vectors with covariance matrix \begin{equation} \mathrm{Cov}\{(X_1,Y_1)\}=\begin{pmatrix} 1 & \sigma\\ \sigma & 1 \end{pmatrix}. \end{equation} We assume the covariance matrix can be nearly degenerate, that is, $X$ and $Y$ can be highly Co-linear (but not exactly, since otherwise it reduces to univariate case). Now I want to know, if the multivariate Berry-Esseen Theorem still holds true. Mathematically speaking, does there exists a uniform constant $c>0$, such that \begin{equation} \sup_{(x,y)\in\mathbb{R}^2}\left|P\left(\frac{1}{\sqrt{n}}\sum X_i\leq x,\frac{1}{\sqrt{n}}\sum Y_i\leq y\right)-P\left(Z_1\leq x,Z_2\leq y\right)\right|\leq\frac{c}{\sqrt{n}}, \end{equation} where $(Z_1,Z_2)$ admits multivariate normal distribution with mean zero and covariance matrix $\mathrm{Cov}\{(X_1,Y_1)\}$, holds true for all $-1\leq\sigma\leq1$? I have found some articles about multivariate Berry-Esseen theorem, usually they assume the covariance matrix to be identity matrix in order to bound the variance of each component, but I'm not sure whether Co-linearity affect the result.

$\endgroup$
1
$\begingroup$

By Theorem 1.3 in the article of Götze you linked to, the answer is yes, because the result holds for all convex sets and by a linear transformation you can make the covariance an identity. In fact a result of Sazonov (1968) that Götze refers to already contains this.

$\endgroup$
1
  • $\begingroup$ Seems I finally got it, thank you very much! $\endgroup$ – aurora_borealis Dec 24 '19 at 2:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.