• Two Remarks on Normality Preserving Borel Automorphisms of $\mathbb{R}^n$

• # Fulltext

https://www.ias.ac.in/article/fulltext/pmsc/123/01/0075-0084

• # Keywords

Borel automorphism; normality preserving Borel automorphism; Lebesgue partition.

• # Abstract

Let 𝑇 be a bijective map on $\mathbb{R}^n$ such that both 𝑇 and $T^{-1}$ are Borel measurable. For any $\theta\in\mathbb{R}^n$ and any real $n\times n$ positive definite matrix 𝛴 , let $N(\theta,\Sigma)$ denote the 𝑛-variate normal (Gaussian) probability measure on $\mathbb{R}^n$ with mean vector 𝜃 and covariance matrix 𝛴 . Here we prove the following two results: (1) Suppose $N(\theta_j, I)T^{-1}$ is gaussian for $0\leq j\leq n$, where 𝐼 is the identity matrix and $\{\theta_j-\theta_0,1\leq j\leq n\}$ is a basis for $\mathbb{R}^n$. Then 𝑇 is an affine linear transformation; (2) Let $\Sigma_j=I+\varepsilon_ju_j{u'}_j, 1\leq j\leq n$ where $\varepsilon_j&gt;-1$ for every 𝑗 and $\{u_j,1\leq j\leq n\}$ is a basis of unit vectors in $\mathbb{R}^n$ with ${u'}_j$ denoting the transpose of the column vector $u_j$. Suppose $N(0,I)T^{-1}$ and $N(0,\Sigma_j)T^{-1}, 1\leq j\leq n$ are gaussian. Then $T(x)=\Sigma_s 1_{E_s}(x)VsUx a.e.x$, where 𝑠 runs over the set of $2^n$ diagonal matrices of order 𝑛 with diagonal entries $\pm 1,U,V$ are $n\times n$ orthogonal matrices and $\{E_s\}$ is a collection of $2^n$ Borel subsets of $\mathbb{R}^n$ such that $\{E_s\}$ and $\{VsU(E_s)\}$ are partitions of $\mathbb{R}^n$ modulo Lebesgue-null sets and for every $j,VsU\Sigma_j(VsU)^{-1}$ is independent of all 𝑠 for which the Lebesgue measure of $E_s$ is positive. The converse of this result also holds.

Our results constitute a sharpening of the results of Nabeya and Kariya (J. Multivariate Anal. 20 (1986) 251–264) and part of Khatri (Sankhyā Ser. A 49 (1987) 395–404).

• # Author Affiliations

1. Theoretical Statistics and Mathematics Unit, Indian Statistical Institute, 7, S. J. S. Sansanwal Marg, New Delhi 110 016, India

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019

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