- #1
saminator910
- 95
- 2
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?
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?