Two remarks on Normality Preserving Borel Automorphisms of R^n

preprint2011arXivOpen access

Abstract

Let TT be a bijective map on Rn\mathbb{R}^n such that both TT and T−1T^{-1} are Borel measurable. For any $\btheta \in \mathbb{R}^n$ and any real n×nn \times n positive definite matrix Σ,Σ, let $N (\btheta, Σ)$ denote the nn-variate normal (gaussian) probability measure on Rn\mathbb{R}^n with mean vector $\btheta$ and covariance matrix Σ.Σ. Here we prove the following two results: (1) Suppose $N(\btheta_j, I)T^{-1}$ is gaussian for 0≤j≤n0 \leq j \leq n where II is the identity matrix and $\{\btheta_j - \btheta_0, 1 \leq j \leq n \}$ is a basis for Rn.\mathbb{R}^n. Then TT is an affine linear transformation; (2) Let Σj=I+εjujuj′,Σ_j = I + ε_j \mathbf{u}_j \mathbf{u}_j^{\prime}, 1≤j≤n1 \leq j \leq n where εj>−1ε_j > -1 for every jj and uj,1≤j≤n{\mathbf{u}_j, 1 \leq j \leq n} is a basis of unit vectors in Rn\mathbb{R}^n with uj′\mathbf{u}_j^{\prime} denoting the transpose of the column vector uj.\mathbf{u}_j. Suppose N(0,I)T−1N(\mathbf{0}, I)T^{-1} and N(0,Σj)T−1,N (\mathbf{0}, Σ_j)T^{-1}, 1≤j≤n1 \leq j \leq n are gaussian. Then T(x)=∑s1EsVsUxT(\mathbf{x}) = \sum\limits_{\mathbf{s}} 1_{E_{\mathbf{s}}} V \mathbf{s} U \mathbf{x} a.e. x\mathbf{x} where s\mathbf{s} runs over the set of 2n2^n diagonal matrices of order nn with diagonal entries ±1,\pm 1, U, VU,\, V are n×nn \times n orthogonal matrices and {Es}\{E_{\mathbf{s}}\} is a collection of 2n2^n Borel subsets of Rn\mathbb{R}^n such that {Es}\{E_{\mathbf{s}}\} and {VsU(Es)}\{V \mathbf{s} U (E_{\mathbf{s}})\} are partitions of Rn\mathbb{R}^n modulo Lebesgue-null sets and for every j,j, VsUΣj(VsU)−1V \mathbf{s} U Σ_j (V \mathbf{s} U)^{-1} is independent of all s\mathbf{s} for which the Lebesgue measure of EsE_{\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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