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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 

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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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