Abstract

Let $G=(G_{1},G_{2})$ be a Gaussian vector in $R^{2}$ with $E(G_{1}G_{2})\ne 0$. Let $c_{1},c_{2}\in R^{1}$. A necessary and sufficient condition for the vector $((G_{1}+c_{1}\alpha )^{2},(G_{2}+c_{2}\alpha )^{2})$ to be infinitely divisible for all $\alpha \in R^{1}$ is that $$ \Gamma_{i,i}\ge \frac{c_{i}}{c_{j}}\Gamma_{i,j}>0\qquad\forall\,1\le i\ne j\le 2.\qquad(0.1) $$ In this paper we show that when (0.1) does not hold there exists an $0<\alpha _{0} < \infty $ such that $((G_{1}+c_{1}\alpha )^{2},(G_{2}+c_{2}\alpha )^{2})$ is infinitely divisible for all $|\alpha |\le \alpha _{0}$ but not for any $|\alpha | > \alpha _{0}$.

Highlights

  • Let η = (η1, . . . , ηn) be an Rn valued Gaussian random variable. η is said to have infinitely divisible squares if η2 := (η12, . . . , ηn2) is infinitely divisible, i.e. for any r we can find an Rn valued random vector Zr such that η2 l=aw Zr,1 + · · · + Zr,r, (1.1)where {Zr,j}, j = 1, . . . , r are independent identically distributed copies of Zr

  • Paul Levy proposed the problem of characterizing which Gaussian vectors have infinitely divisible squares

  • It is easy to see that a single Gaussian random variable has infinitely divisible squares

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Summary

Introduction

Gn) be a mean zero Gaussian random variable with strictly positive definite covariance matrix Γ = {Γi,j} = {E(GiGj)}. Gn) be a mean zero Gaussian random variable with irreducible strictly positive definite covariance matrix Γ = {Γi,j} = {E(GiGj)}. The Gaussian random variable itself minus its mean, and the local time of a related Markov process This would be a very neat result but it is not true. Note that in Theorem 1.4 we consider all (c1, c2) ∈ R2 It follows from (1.12) that when EG1G2 > 0, G2(c1, c2, α) has infinitely divisible squares for all α ∈ R1 only if c1c2 > 0. Since we can only prove Theorem 1.4 for Gaussian random variables in R2 we stick to this case, a similar analysis applies to Rn valued Gaussian random variables

Let Γ be the covariance matrix of G and
Let Then and
Set t
Proof It is clear that j p jp p!
Proof We
Now suppose that
When t
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