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Exact Solution of Geometric Brownian Motion

  1. May 3, 2012 #1
    Probably I am just confused, but why for the exact solution of the geometric brownian motion [itex]dX_t = \mu X_t dt+\sigma X_t dW_t[/itex] we have to apply Ito's lemma and manipulate the expression obtained with [itex]dlogX_t[/itex]? Couldn't we directly use the espression [itex]dX_t / X_t = dlogX_t[/itex] in the equation [itex]dX_t / X_t = \mu dt+\sigma dW_t[/itex]?
    Thank you for your help!
    Last edited: May 3, 2012
  2. jcsd
  3. May 3, 2012 #2


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    Hey Tilde90 and welcome to the forums.

    You could do this, but it's easier to have in the 1st format since you can integrate both sides which on the LHS gives you Xt - X0 and on the RHS gives you two integral expressions which you can solve using the Ito Lemma.

    You also need to remember that in one integral you have a Brownian motion measure and because of this, you need to be careful in how you calculate the integral since it is very different to how you treat normal integrals like say f(x)dx across the real line.
  4. May 4, 2012 #3
    Thank you very much chiro.

    I mistakenly believed that the "stochastic" integral were easier to calculate, as in the demonstration of the exact solution of the GBM with Ito's lemma it seems that they just integrate both terms of the SDE [itex]dlogX_t=(\mu-\frac{\sigma^2}{2})dt+\sigma dW_t[/itex].
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