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Homework Help: Solving the SDE [itex]dX(t) = udt + \sigma X(t)dB(t)[/itex]

  1. Nov 9, 2012 #1
    **Provided Question**

    The SDE is [itex]dX(t) = udt + \sigma X(t)dB(t)[/itex]. Find [itex]X(t)[/itex], where [itex]X(t)[/itex] is some stochastic process and [itex]B(t)[/itex] is a Wiener process. Both [itex]u[/itex] and [itex]\sigma[/itex] are constants.

    *Hint*: Multiply both sides by the "integrating factor" [itex]e^{-\sigma B(t) + \frac12 \sigma^2 t}[/itex].

    **Current Progress**

    Multiplying both sides by the appropriate integrating factor:

    \exp{\big( -\sigma B(t) + \frac12 \sigma^2t\big)}dX(t) = \exp{\big(-\sigma B(t)+\frac12\sigma^2t\big)}(udt + \sigma X(t)dB(t))

    Then set [itex]f(t,x,b):=\exp{\big( -\sigma B(t) + \frac12 \sigma^2t\big)}X(t)[/itex] and apply Ito's formula. Some of the required results before actually applying Ito's formula:

    \frac{df}{dt} = \frac12\sigma^2 X(t)e^{-\sigma B(t) + \frac12 \sigma^2t} \\
    \frac{df}{dx} = e^{-\sigma B(t) + \frac12 \sigma^2t} \\
    \frac{df}{db} = -\sigma X(t)e^{-\sigma B(t) + \frac12 \sigma^2t} \\
    \frac{d^2f}{dx^2} = 0 \\
    \frac{d^2f}{db^2} = \sigma^2 X(t) e^{-\sigma B(t) + \frac12 \sigma^2t} \\
    \frac{d^2f}{dxdb} = -\sigma e^{-\sigma B(t) + \frac12 \sigma^2t}

    Given this, we need to know the following derivatives of the quadratic variations and co-variations:

    d\langle B\rangle_t = dt \\
    d\langle X,B\rangle_t = ???

    **My Request**

    Please instruct me on what [itex]d\langle X,B\rangle_t[/itex] is equal to so that I may progress further with this problem.
  2. jcsd
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