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A question about brownian motion

  1. Jun 28, 2010 #1
    for a brownian motion W(t)
    W(t_i+1)-W(t_i) is normal distribution with mean 0 and variance t_i+1-t_i

    so this means var(W(t_i+1)-W(t_i))=var(W(t_i+1))-var(W(t_i))=t_i+1-t_i

    I don't think the above equation satisfies because W(t_i+1) and W(t_i) are not independent. Any comment? thanks
    Last edited: Jun 28, 2010
  2. jcsd
  3. Jun 28, 2010 #2
    I think you miss out the covariance term in [tex] Var(X-Y) = Var(X) + Var(Y) - 2Cov(X,Y)[/tex] and note that in Brownian motion [tex] Cov(W_{s},W_{t}) = \min(s,t) [/tex]
  4. Jun 29, 2010 #3
    Yeah, you can't just subtract variances like that.
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