Two remarks on Normality Preserving Borel Automorphisms of R^n

Mathematics – Probability

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corrected a few typos

Scientific paper

Let $T$ be a bijective map on $\mathbb{R}^n$ such that both $T$ and $T^{-1}$ are Borel measurable. For any $\btheta \in \mathbb{R}^n$ and any real $n \times n$ positive definite matrix $\Sigma,$ let $N (\btheta, \Sigma)$ denote the $n$-variate normal (gaussian) probability measure on $\mathbb{R}^n$ with mean vector $\btheta$ and covariance matrix $\Sigma.$ Here we prove the following two results: (1) Suppose $N(\btheta_j, I)T^{-1}$ is gaussian for $0 \leq j \leq n$ where $I$ is the identity matrix and $\{\btheta_j - \btheta_0, 1 \leq j \leq n \}$ is a basis for $\mathbb{R}^n.$ Then $T$ is an affine linear transformation; (2) Let $\Sigma_j = I + \epsilon_j \mathbf{u}_j \mathbf{u}_j^{\prime},$ $1 \leq j \leq n$ where $\epsilon_j > -1$ for every $j$ and ${\mathbf{u}_j, 1 \leq j \leq n}$ is a basis of unit vectors in $\mathbb{R}^n$ with $\mathbf{u}_j^{\prime}$ denoting the transpose of the column vector $\mathbf{u}_j.$ Suppose $N(\mathbf{0}, I)T^{-1}$ and $N (\mathbf{0}, \Sigma_j)T^{-1},$ $1 \leq j \leq n$ are gaussian. Then $T(\mathbf{x}) = \sum\limits_{\mathbf{s}} 1_{E_{\mathbf{s}}} V \mathbf{s} U \mathbf{x}$ a.e. $\mathbf{x}$ where $\mathbf{s}$ 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 $\{E_{\mathbf{s}}\}$ is a collection of $2^n$ Borel subsets of $\mathbb{R}^n$ such that $\{E_{\mathbf{s}}\}$ and $\{V \mathbf{s} U (E_{\mathbf{s}})\}$ are partitions of $\mathbb{R}^n$ modulo Lebesgue-null sets and for every $j,$ $V \mathbf{s} U \Sigma_j (V \mathbf{s} U)^{-1}$ is independent of all $\mathbf{s}$ for which the Lebesgue measure of $E_{\mathbf{s}}$ is positive. The converse of this result also holds. \vskip0.1in Our results constitute a sharpening of the results of S. Nabeya and T. Kariya

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