Editing Geometric Brownian motion (section)

==Properties of GBM==

The above solution <math> S_t </math> (for any value of t) is a [[log-normal distribution|log-normally distributed]] [[random variable]] with [[expected value]] and [[variance]] given by<ref>{{Citation
 | title = Stochastic Differential Equations: An Introduction with Applications
 |publisher=Springer
 |author = Øksendal, Bernt K.
 | year = 2002
 | pages = 326
 | isbn = 3-540-63720-6
}}</ref>

:<math>\operatorname{E}(S_t)= S_0e^{\mu  t},</math>
:<math>\operatorname{Var}(S_t)= S_0^2e^{2\mu t} \left( e^{\sigma^2 t}-1\right).</math>

They can be derived using the fact that <math> Z_t = \exp\left(\sigma W_t - \frac{1}{2}\sigma^2 t\right) </math> is a [[martingale (probability theory)|martingale]], and that

:<math> \operatorname{E}\left[ \exp\left(2\sigma W_t - \sigma^2 t\right) \mid \mathcal{F}_s\right] = e^{\sigma^2(t - s)} \exp\left(2\sigma W_s - \sigma^2 s\right),\quad \forall 0 \leq s < t. </math>

The [[probability density function]] of <math> S_t </math> is:
: <math>f_{S_t}(s; \mu, \sigma, t) = \frac{1}{\sqrt{2 \pi}}\, \frac{1}{s \sigma \sqrt{t}}\, \exp \left( -\frac{ \left( \ln s - \ln S_0 - \left( \mu - \frac{1}{2} \sigma^2 \right) t \right)^2}{2\sigma^2 t} \right).</math>
{{Collapse top|title=Derivation of GBM probability density function}}
To derive the probability density function for GBM, we must use the [[Fokker-Planck equation]] to evaluate the time evolution of the PDF:

:<math>{\partial p\over{\partial t}} + {\partial\over{\partial S}}[\mu(t,S)p(t,S)] = {1\over{2}}{\partial^{2}\over{\partial S^{2}}}[\sigma^{2}(t,S)p(t,S)], \quad p(0,S) = \delta(S)</math>

where <math>\delta(S)</math> is the [[Dirac delta function]]. To simplify the computation, we may introduce a logarithmic transform <math>x = \log (S/S_{0})</math>, leading to the form of GBM:

:<math>dx = \left(\mu - {1\over{2}}\sigma^{2}\right)dt + \sigma dW</math>

Then the equivalent Fokker-Planck equation for the evolution of the PDF becomes:

:<math>{\partial p\over{\partial t}} + \left(\mu - {1\over{2}}\sigma^{2}\right){\partial p\over{\partial x}} = {1\over{2}}\sigma^{2}{\partial^{2}p\over{\partial x^{2}}}, \quad p(0,x) = \delta(x) </math>

Define <math>V=\mu-\sigma^{2}/2</math> and <math>D=\sigma^{2}/2</math>. By introducing the new variables <math>\xi = x-Vt</math> and <math>\tau = Dt</math>, the derivatives in the Fokker-Planck equation may be transformed as:

:<math>\begin{aligned}\partial_{t}p &= D\partial_{\tau}p - V\partial_{\xi}p \\ \partial_{x}p &= \partial_{\xi}p \\ \partial_{x}^{2}p &= \partial_{\xi}^{2}p \end{aligned}</math>

Leading to the new form of the Fokker-Planck equation:

:<math>{\partial p\over{\partial\tau}} = {\partial^{2}p\over{\partial \xi^{2}}}, \quad p(0,\xi) = \delta(\xi)</math>

However, this is the canonical form of the [[heat equation]]. which has the solution given by the [[heat kernel]]:

:<math>p(\tau,\xi) = {1\over{\sqrt{4\pi \tau}}}\exp\left(-{\xi^{2}\over{4\tau}} \right)</math>

Plugging in the original variables leads to the PDF for GBM:

:<math>p(t,S) = {1\over{S\sqrt{2\pi \sigma^{2}t}}}\exp\left\{-{\left[\log(S/S_{0})-\left( \mu - {1\over{2}}\sigma^{2}\right)t \right]^{2}\over{2\sigma^{2}t}} \right\}</math>

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When deriving further properties of GBM, use can be made of the SDE of which GBM is the solution, or the explicit solution given above can be used. For example, consider the stochastic process log(''S''<sub>''t''</sub>). This is an interesting process, because in the  Black–Scholes model it is related to the [[log return]] of the stock price. Using [[Itô's lemma]] with ''f''(''S'') = log(''S'') gives
:<math>
\begin{alignat}{2}
d\log(S) & = f'(S)\,dS + \frac{1}{2} f'' (S)S^2\sigma^2 \, dt \\[6pt]
& = \frac{1}{S} \left( \sigma S\,dW_t + \mu S\,dt\right) - \frac{1}{2}\sigma^2\,dt \\[6pt]
&= \sigma\,dW_t +(\mu-\sigma^2/2)\,dt.
\end{alignat}
</math>
It follows that <math>\operatorname{E} \log(S_t)=\log(S_0)+(\mu-\sigma^2/2)t</math>.

This result can also be derived by applying the logarithm to the explicit solution of GBM:
:<math>
\begin{alignat}{2}
\log(S_t) &=\log\left(S_0\exp\left(\left(\mu - \frac{\sigma^2}{2} \right)t + \sigma W_t\right)\right)\\[6pt]
& =\log(S_0) +\left(\mu - \frac{\sigma^2}{2} \right)t + \sigma W_t.
\end{alignat}
</math>
Taking the expectation yields the same result as above:  <math>\operatorname{E} \log(S_t)=\log(S_0)+(\mu-\sigma^2/2)t </math>.