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I have seen in class that for some reasons I forgot, the Brownian Motion has a Multivariate normal distribution, but I am unable to prove it easily. Could someone tell me why it's true? From what I understand, I have to take a finite linear combination of values of Brownian motion at different times, and check that it's normally distributed. Could someone help me on this? thanks

edit : the definition I start from is the one from wikipedia : http://en.wikipedia.org/wiki/Brownian_motion#Mathematics points 1 to 4 what I'm trying to prove is that Y = a1*B1 + … + ak*Bk is normally distributed, where Bi are values of the Brownian motion at time Ti

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  • $\begingroup$ Brownian motion by the usual definition has normally distributed increments. What was the definition given in the class? $\endgroup$
    – user0820
    Oct 8, 2013 at 22:37
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    $\begingroup$ @Proba : i've been given the usual defition that you mention, the issue I have is that a given set of normally distributed function doesn't automatically make it a multivariate normal distribution $\endgroup$
    – lezebulon
    Oct 8, 2013 at 22:40
  • $\begingroup$ I think it would be great if you could expand you question a little bit with the definitions you were given in the class. How did you define Brownian motion? You should also be more precise with what you mean by "normal distribution of Brownian motion"? Is it higher dimensional Brownian motion or finite number of increments of one-dimensional Brownian motion? $\endgroup$
    – user0820
    Oct 8, 2013 at 22:46
  • $\begingroup$ @Proba : I edited the question $\endgroup$
    – lezebulon
    Oct 8, 2013 at 22:51

1 Answer 1

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You can choose $\lambda_q,\dots,\lambda_k$ so that $$Y = a_1B_1 + \dots + a_kB_k = \lambda_1B_1 + \lambda_2(B_2-B_1) + \dots + \lambda_k(B_k-B_{k-1}).$$ But from the definition of Brownian motion, you know that $B_1, B_2-B_1, \dots, B_k-B_{k-1}$ are normally distributed and independent, so a linear combination of them is again normally distributed.

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  • $\begingroup$ Sorry, but how do you choose $\lambda_i$? $\endgroup$
    – 0xbadf00d
    Dec 16, 2022 at 19:32
  • $\begingroup$ @0xbadf00d $\lambda_k = a_k$ and the others are determined inductively I guess $\endgroup$
    – Sebastian Bechtel
    Aug 1 at 12:38

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