Variance of Geometric Brownian motion?

[saminator910]

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

$dX=\mu X dt + \sigma X dB$

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

$\ln X = (\mu - \frac{\sigma ^{2}}{2})t+\sigma B$

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

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

$\hat{\mu}=(\mu - \frac{\sigma ^{2}}{2})t$

But now, how to find $Var(\ln X)$

I try $Var(\ln X)=\sigma ^{2} t$

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

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

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

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

Can anyone give me an explanation of where I went wrong?

 

[Greg Bernhardt]

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