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Math Help - Martingales with brownian motion

  1. #1
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    Dundee
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    Martingales with brownian motion

    Hi,

    I've to prove that a couple of equations are martingales. yet i don't really know where to start (well i know that if i can show them as driftless or as an itos integral that would answer the questions.)

    I was hoping someone could give me suggestions on where to start - rather than simply answer the questions (i do want to learn it)

    note i've used | to mean integral from 0 to t
    and X(t) is the Brownian Motion

    the equations are:

    Y(t) = [1 - X(t)|sin(kX(t)).dk ] * e^(t/2)

    Y(t) = sqrt(t)X(t) - | X(s)/2sqrt(s).ds

    Y(t) = wX1(t) + sqrt(1-w^2)*X2(t)

    I think I have to integrate equations 1 and 2 (in respect to k and s) but i'm getting into a muddle. Any suggestions?
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  2. #2
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    Re: Martingales with brownian motion

    with respect to the first equation i'm thinking the following:
    focusing on X int Sin (kX) .dk
    as the integral is in respect to k, simple integration gives
    X *[1/X* -cos(kX)] = -cos(kX) as over 0_1
    -cos(1X) + cos(0X) = 1 - cos(X)
    Plugging in to formula
    Y(t) = [1 - 1 - cos(X) ] * e^(t/2)
    gives
    Y(t) = e^(t/2)cos(X)
    as there is no drift term (.... .dt) this must mean its a martingale.
    but i don't think i can perform simple integration on a function containing Brownian motion.
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  3. #3
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    Re: Martingales with brownian motion

    The second equation i'm thinking of using a change of measure.
    so z= sqrt(s)
    dz = 1/2sqrt(s)

    simplifies formula to Y(t) = sqrt(t)X(t) - int X(z^2).dz
    integrating Y(t) = sqrt(t)X(t) - (z^2)X(z^2) int X(z^2).dz
    or
    Y(t) = sqrt(t)X(t) sX(s)[0_t] - integral Sqrt(s).dX
    so
    Y(t) = (sqrt(t) - t)X(t) - integral Sqrt(s).dX
    as integral is an ito must mean martingale?
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