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Math Help - [SOLVED] Covariance of Brownian Motion

  1. #1
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    [SOLVED] Covariance of Brownian Motion

    Just a little confused about the following. Given that W denotes a one-dimensional standard Brownian motion, find Cov(W_t,W_s)

    Step 1: Assume that s \leq t, then
    Cov(W_t,W_s)=E[W_tW_s]=E[(W_t-W_s+W_s)W_s]=Var(W_s)=s

    Step 2: Then, in general:
    Cov(W_t,W_s)=\min \{s,t\}

    Could someone explain each step and properties used? Thanks.
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  2. #2
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    Fix 0 ≤ s ≤ t.
    Step 1:
    Cov(W(t), W(s)) = E[W(t)W(s)] − E[W(t)]E[W(s)] = E[W(t) * W(s)] as these are standard brownian motion. Hence, E(W(t)) = E(W(s)) = 0.

    E[W(t)W(s)] = E[(W(s) + W(t) − W(s))W(s)]
    = E[W^2(s)] + E[(W(t) − W(s))W(s)] = s

    Why? Since E[W(t)] = 0 for all t and since by definition of the standard Brownian motion we have E[W^2(s)] = s.

    And by independent increments property we have E[(W(t) − W(s))W(s)] = E[(W(t) − W(s))(W(s) − W(0))] = E[(W(t) − W(s)]E[(W(s) − W(0)] = 0, since
    increments have a zero mean Gaussian distribution.

    Step 2: Repeat the same for t < s. Hence, you shall see min(t,s) to be the general answer.

    Quote Originally Posted by horan View Post
    Just a little confused about the following. Given that W denotes a one-dimensional standard Brownian motion, find Cov(W_t,W_s)

    Step 1: Assume that s \leq t, then
    Cov(W_t,W_s)=E[W_tW_s]=E[(W_t-W_s+W_s)W_s]=Var(W_s)=s

    Step 2: Then, in general:
    Cov(W_t,W_s)=\min \{s,t\}

    Could someone explain each step and properties used? Thanks.
    Last edited by zigzag20; March 10th 2009 at 07:28 AM.
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  3. #3
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    I am reading the solution, and wondering in W^2(s), what does the ^ symbol stands for.
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