Joint convergence of sample cross-covariance matrices

Pub Date : 2021-03-22 DOI:10.30757/alea.v20-14
M. Bhattacharjee, A. Bose, Apratim Dey
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引用次数: 2

Abstract

Suppose $X$ and $Y$ are $p\times n$ matrices each with mean $0$, variance $1$ and where all moments of any order are uniformly bounded as $p,n \to \infty$. Moreover, the entries $(X_{ij}, Y_{ij})$ are independent across $i,j$ with a common correlation $\rho$. Let $C=n^{-1}XY^*$ be the sample cross-covariance matrix. We show that if $n, p\to \infty, p/n\to y\neq 0$, then $C$ converges in the algebraic sense and the limit moments depend only on $\rho$. Independent copies of such matrices with same $p$ but different $n$, say $\{n_l\}$, different correlations $\{\rho_l\}$, and different non-zero $y$'s, say $\{y_l\}$ also converge jointly and are asymptotically free. When $y=0$, the matrix $\sqrt{np^{-1}}(C-\rho I_p)$ converges to an elliptic variable with parameter $\rho^2$. In particular, this elliptic variable is circular when $\rho=0$ and is semi-circular when $\rho=1$. If we take independent $C_l$, then the matrices $\{\sqrt{n_lp^{-1}}(C_l-\rho_l I_p)\}$ converge jointly and are also asymptotically free. As a consequence, the limiting spectral distribution of any symmetric matrix polynomial exists and has compact support.
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样本交叉协方差矩阵的联合收敛性
假设$X$和$Y$是$p\timesn$矩阵,每个矩阵的平均值为$0$,方差为$1$,其中任何阶的所有矩都一致有界为$p,n\to\infty$。此外,条目$(X_{ij},Y_{ij})$在$i,j$上是独立的,具有公共相关性$\rho$。设$C=n^{-1}XY^*$是样本互协方差矩阵。我们证明了如果$n,p\to\infty,p/n\toy\neq0$,那么$C$在代数意义上收敛,并且极限矩仅依赖于$\rho$。具有相同$p$但不同$n$的矩阵的独立副本,例如$\{n_l\}$,不同相关性$\{\rho_l\}$,以及不同的非零$y$,例如$\{y_l\}}$也联合收敛并且渐近自由。当$y=0$时,矩阵$\sqrt{np^{-1}}(C-\rho I_p)$收敛于参数为$\rho^2$的椭圆变量。特别是,当$\rho=0$时,这个椭圆变量是圆形的,当$\ rho=1$时,它是半圆形的。如果我们取独立的$C_l$,则矩阵$\{\sqrt{n_lp^{-1}}(C_l-\rho_l I_p)\}$联合收敛,并且也是渐近自由的。因此,任何对称矩阵多项式的极限谱分布都存在,并且具有紧致支持。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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