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Ken Honda
science forum beginner
Joined: 15 May 2005
Posts: 12
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 Posted: Sun Jul 16, 2006 8:48 pm Post subject:
Brownian motion, covariance
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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
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| Posted: Sun Jul 16, 2006 9:06 pm Post subject:
Re: Brownian motion, covariance
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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.
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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 ) ?
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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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