The two equations commonly found online for GBM are:

$\begin{matrix} S_{ t }=S_{ 0 }\exp\left( \left( \mu -\frac { \sigma ^{ 2 } }{ 2 } \right) t+\sigma W_{ t } \right) \\ S_{ t }=S_{ 0 }\exp\left(\mu t+\sigma W_{ t } \right) \end{matrix}$

I found the first one on Wikipedia, and the second one in a Columbia university PDF about simulation of GBMs, Page 4.


2 Answers 2


The first the solution to: $$dS_t = S_t\left[\mu dt +\sigma dW_t\right]$$ The second is the solution to: $$ dS_t = S_t\left[\left(\mu+\frac{\sigma^2}{2}\right)dt + \sigma dW_t\right]$$

The difference is that the first one is a martingale when $\mu$ is equal to zero while the second one is not: $$ \mathbb{E}[S_0 \exp(\sigma W_t)]= S_0\exp\left(\frac{\sigma^2}{2}t\right)$$

The one usually used in a financial engineering context is the first one.

  • $\begingroup$ Can you explain which one do you mean by the first and which one do you mean by the second? In my question the equation on the top is the first equation. The equation on the bottom is the second equation. $\endgroup$ Dec 22, 2013 at 13:35

The standard form of a geometric Brownian motion is $dS_t = S_t(\mu dt+\sigma dB_t)$, where B is a BM and $\mu$ and $\sigma$ are two real numbers. When you write this process in closed form: it is $S_t = S_0 exp(\mu t + \sigma B_t - \sigma ^2 t/2)$. The process $(S_t e^{-\mu t}, t\geq 0)$ is a martingale.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.