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Autocorrelation of a wiener process

  1. Dec 29, 2012 #1
    Let [itex]\phi(t)[/itex] be a Brownian Walk (Wiener Process), where [itex]\phi\in[0,2\pi)[/itex]. As such we work with the variable [itex]z(t)=e^{i\phi(t)}[/itex]. I would like to calculate

    [itex]E(z(t)z(t+\tau))[/itex]

    This is equal to [itex]E(e^{i\phi(t)+i\phi(t+\tau)})[/itex] and I know that
    [itex]E(e^{i\phi(t)})=e^{-\frac{1}{2}\sigma^{2}(t)}[/itex], where the mean is 0 and [itex]\sigma^{2}(t)=2Dt[/itex].
    However, I have been stuck a week on how to proceed, any thoughts?

    Thank you :)

    Aim For Clarity
     
  2. jcsd
  3. Dec 29, 2012 #2

    chiro

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    Science Advisor

    Hey aimforclarity.

    Have you made an attempt to partition this whole thing in terms of independent processes so that you can calculate E[XY] = E[X]E[Y]?
     
  4. Dec 29, 2012 #3
    ok, so can I say [itex]<e^{\phi(t)+\phi(t+\tau)}>=<e^{\phi(t)+\phi(t)+\phi(\tau)}>[/itex] since [itex]W(t+\Delta t)=W(t)+W(\Delta t)[/itex]. Next we can say [itex]<e^{\phi(t)+\phi(t)+\phi(\tau)}>=<e^{2\phi(t)}><e^{\phi(\tau)}>[/itex] because [itex]<\phi(t)\phi(\tau)>=0[/itex]
     
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