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Ito-Doeblin Formula Question

  1. Aug 4, 2008 #1


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    Reading through a proof on why the higher order terms vanish and it makes this statement

    dW(t)dW(t) = dt

    where W(t) is a Brownian motion

    It is not obvious to me why this is the case, but the text seems to infer that it is because no further explanation is offered
  2. jcsd
  3. Aug 4, 2008 #2
    It may have to do with [tex] W_t=\sqrt{t} B [/tex] where [tex] B[/tex] is N(0,1).
  4. Aug 4, 2008 #3


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    [itex]\delta W_t \sim N(0,t)[/itex]. It follows that [itex]E[(\delta W_t)^2]=\delta t[/itex] and [itex]E[|\delta W_t|^3]={\rm const}\times \delta t^{3/2}[/itex]. So third and higher powers of dW are smaller order than dt on average , and therefore vanish if you sum them over a partition and let dt->0.
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