Why Geometric Brownian Motion Equations Use Different Drift Terms

Summary

The document explains why two familiar closed-form geometric Brownian motion equations appear to differ. One subtracts half the variance from the exponent’s drift term; the other does not. They correspond to stochastic differential equations with different drift parameters: the first uses the stated parameter directly, while the second equation’s parameter is shifted by half the variance when written in the standard differential form.

This distinction follows from Itô’s formula, which contributes the variance adjustment when solving for the logarithm of the process. The note also contrasts their martingale behavior: setting the drift parameter to zero in the first convention gives a martingale, whereas the second expression has an expectation that grows with time. The standard financial engineering convention is identified as the first form. The explanation is about parameterization and should be read alongside the precise stochastic differential equation being modeled.

Key ideas

• The geometric Brownian motion solution includes a negative half-variance adjustment in its exponent under the standard differential equation convention.
• The equation without that adjustment corresponds to a differential equation with a drift shifted by half the variance.
• The two forms imply different expectations when their displayed drift parameter is set to zero.
• The standard financial engineering form is the solution with the half-variance adjustment.

Tags

# What is the difference between these two equations for GBMs?


# What is the difference between these two equations for GBMs?












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.

## Answer by Drmanifold (score 6, accepted)

https://quant.stackexchange.com/a/9769

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.

## Answer by Liwei Zhang (score 0)

https://quant.stackexchange.com/a/9776

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.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.