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Variance of Geometric Brownian motion?

  1. Jul 5, 2014 #1
    I am trying to derive the Probability distribution of Geometric Brownian motion, and I don't know how to find the variance.

    start with geometric brownian motion

    [itex]dX=\mu X dt + \sigma X dB[/itex]

    I use ito's lemma working towards the solution, and I get this.

    [itex]\ln X = (\mu - \frac{\sigma ^{2}}{2})t+\sigma B[/itex]

    Now, it seems to me that from here I can treat this as a standard drift diffusion which follows

    [itex]N'(x,t)=\displaystyle\frac{1}{\sigma\sqrt{2 \pi t}}\exp(-\frac{(x-\mu t)^{2}}{2\sigma ^{2} t})[/itex]

    [itex]\hat{\mu}=(\mu - \frac{\sigma ^{2}}{2})t[/itex]

    But now, how to find [itex]Var(\ln X)[/itex]

    I try [itex]Var(\ln X)=\sigma ^{2} t[/itex]

    In theory, since the random varable can be written [itex]X=X_{0}e^{Y}[/itex], where [itex]Y=(\mu - \frac{\sigma ^{2}}{2})t+\sigma B[/itex]. We can describe the natural log of [itex]\frac{X}{X_{0}}[/itex] the same way.

    [itex]N'(\ln \frac{X}{X_{0}},t)=\displaystyle\frac{1}{\sigma\sqrt{2 \pi t}}\exp(-\frac{(\ln X- \ln X_{0}-(\mu - \frac{\sigma ^{2} }{2})t)^{2}}{2\sigma ^{2}t})[/itex]

    Apparently it yields a log-normal distribution for [itex]X[/itex]. According to wikipedia, this is the end result... Notice the extra X in the denominator.

    [itex]f_{X_t}(X; \mu, \sigma, t) =\displaystyle \frac{1}{X \sigma \sqrt{2 \pi t}}\, \, \exp \left( -\frac{ \left( \ln X - \ln X_0 - \left( \mu - \frac{1}{2} \sigma^2 \right) t \right)^2}{2\sigma^2 t} \right)[/itex]

    Can anyone give me an explanation of where I went wrong?
     
  2. jcsd
  3. Jul 15, 2014 #2
    I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
     
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