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Showing the Difference Between an Ito Integral & Riemann Integrals.

  1. Dec 8, 2009 #1
    Hi Everyone,
    A problem I have here. I am trying to solve a problem involving Ito Integrals and Riemann interals.

    1. The problem statement, all variables and given/known data

    [tex]\int^{T}_{0} tdW(t) = TW(T) -\int^{T}_{0} W(t)dt [/tex]

    2. Relevant equations
    I want to solve this question WITHOUT using Ito's Lemma directly.

    3. The attempt at a solution
    OK I know that in general, the Ito integral and the Riemann integral are going to be slightly different. The Ito Integral will have an extra term.

    So in my question, I know if I was to generally just integrate the Riemann integral, I would get:
    [tex]\int^{T}_{0} tdW(t) = TW(T) [/tex]
    But I know the Ito integral adds the extra:
    [tex] -\int^{T}_{0} W(t)dt [/tex]
    But how does this arise? Is it possible to show how using the mean square error ? I keep reading the mean square error when reading about this but don't really see the connection. Maybe someone could kindly help out please?

  2. jcsd
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