1. Standard membertejo
    a unique loser
    LIAAA
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    30 Sep '04 17:58
    I could use some help with this exercise I just got.
    Prove that with chance zero that limit for t->infinite (Brownian motion at time t)/(sqrt (t))=0.
    I hope you understand what I mean and that you can help me.
    Thanks,
    Sander
  2. at the centre
    Joined
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    01 Oct '04 13:48
    Originally posted by tejo
    I could use some help with this exercise I just got.
    Prove that with chance zero that limit for t->infinite (Brownian motion at time t)/(sqrt (t))=0.
    I hope you understand what I mean and that you can help me.
    Thanks,
    Sander
    Are you making us do ur homework?
  3. Standard membertejo
    a unique loser
    LIAAA
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    01 Oct '04 14:00
    Originally posted by howzzat
    Are you making us do ur homework?
    I am not making anyone to do anything. I am just asking for help, because I still haven't solved this one.
  4. DonationAcolyte
    Now With Added BA
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    01 Oct '04 16:36
    Originally posted by tejo
    I could use some help with this exercise I just got.
    Prove that with chance zero that limit for t->infinite (Brownian motion at time t)/(sqrt (t))=0.
    I hope you understand what I mean and that you can help me.
    Thanks,
    Sander
    Is time discrete or continuous? Could you define Brownian motion mathematically?
  5. Standard membertejo
    a unique loser
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    01 Oct '04 17:02
    Originally posted by Acolyte
    Is time discrete or continuous? Could you define Brownian motion mathematically?
    I hope I will explain this well.

    Bt:=integral from 0 to t Ns ds.

    Ns is a completely random function of t. In other words, the continuous-time analogue of a sequence of independent identically distributed random variables.
    1.Ns is independent of Nt for t != (not equal) s
    2.The random variables Ns (s>=0) all have the same probability distribution u
    3.Expectancy of Ns = 0 (E(Nt)=0)

    So Bt has the following requirements:
    1.For any 0=t0<=t1<=...<=tn the random vaiables Bt(j+1) - Bt(j) are independent (j=0,...,n-1)
    2.Bt has stationary increments.
    3. E(Bt)=0 for all t>=0
    4. E((B1)^2)=1 also true E((Bt)^2)=t
    5. t->Bt is continuous a.s. with probability 1.

    The Brownian Motion is also called Wiener process I believe.
  6. DonationAcolyte
    Now With Added BA
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    01 Oct '04 23:061 edit
    Originally posted by tejo
    I hope I will explain this well.

    Bt:=integral from 0 to t Ns ds.

    Ns is a completely random function of t. In other words, the continuous-time analogue of a sequence of independent identically distributed random variables.
    1.Ns is indep ...[text shortened]... 1.

    The Brownian Motion is also called Wiener process I believe.
    I understand the question now. I think the following will work: Bt (for t an integer) can be turned into a discrete summation; conditions 1. and 4. imply that the summands are iid with variance 1, and then a Central Limit Theorem does the rest.
  7. Standard membertejo
    a unique loser
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    04 Oct '04 11:51
    Originally posted by Acolyte
    I understand the question now. I think the following will work: Bt (for t an integer) can be turned into a discrete summation; conditions 1. and 4. imply that the summands are iid with variance 1, and then a Central Limit Theorem does the rest.
    Thank you very much Acolyte. It worked really well.
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