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Ken Honda
science forum beginner

Joined: 15 May 2005
Posts: 12

Posted: Sun Jul 16, 2006 8:48 pm    Post subject: Brownian motion, covariance

Hello,
Part of the definition of standard Brownian motion X(t), as I
understand it, is that X(s), X(s+t) are normally distributed, with
cov( X(s), X(s+t) ) = s. Is there an expression for cov( X(s)^2,
X(s+t)^2 ) ?

Thank you!

KH
Stephen J. Herschkorn
science forum Guru

Joined: 24 Mar 2005
Posts: 641

Posted: Sun Jul 16, 2006 9:06 pm    Post subject: Re: Brownian motion, covariance

Ken Honda wrote:

 Quote: Part of the definition of standard Brownian motion X(t), as I understand it, is that X(s), X(s+t) are normally distributed, with cov( X(s), X(s+t) ) = s.

I have never seen that stated as part of the definition, though it is true.

 Quote: Is there an expression for cov( X(s)^2, X(s+t)^2 ) ?

Express X(s+t) as X(s) + [X(s+t) - X(s)], expand, and apply
independent increments, the bilinearity of covariance, and what you know
about the moments of a normal distribution. That will probably get you
what you want.

--
Stephen J. Herschkorn sjherschko@netscape.net
Math Tutor on the Internet and in Central New Jersey and Manhattan

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