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Integral with respect to Brownian motion.

  1. Nov 15, 2012 #1
    Suppose that [itex]\sigma(t,T)[/itex] is a deterministic process, where [itex]t[/itex] varies and [itex]T[/itex] is a constant. We also have that [itex]t \in [0,T][/itex]. Also [itex]W(t)[/itex] is a Wiener process.

    My First Question

    What is [itex]\displaystyle \ \ d\int_0^t \sigma(u,T)dW(u)[/itex]? My lecture slides assert that it's equal to [itex]\sigma(t,T)dW(t)[/itex] but I'm not sure why. So my question is "Why"?

    My Second Question

    What is [itex]\displaystyle \ \ d\int_a^t \sigma(u,T)dW(u)[/itex], where [itex]a \in (0,t)[/itex].

    _________________________________

    Thanks!
     
  2. jcsd
  3. Nov 15, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    I've said it before and I will say it again: go back to the basics. What is meant by the stochastic integral? If Y(t) is a stochastic process, what do we mean by dY(t)? All this material is explained in books (admittedly, some almost unreadable), and in numerous web pages and the like. There is simply no substitute for getting the background first.

    RGV
     
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