Under the P-measure, a geometric Brownian motion can be specified using the following SDE:

$$dS_t=\mu S_tdt+\sigma S_tdW_t^P$$

and its Euler discretization is

$$S_{t+\Delta t}=S_t + \mu S_t \Delta t + \sigma S_t \sigma \sqrt{\Delta t}Zt$$

Under the Q-measure, should the drift $\mu$ be substituted by $r$ or by $r-\frac{\sigma^2}{2}$?


2 Answers 2


$r-\frac{\sigma^2}{2}$ for the drift only applies to the log-returns. The Euler discretisation simply discretises the SDE directly. You'd use the risk-free rate for you drift under the risk-neutral measure for your question.

For your reference:

enter image description here

Please read the wikipedia for more details.

  • $\begingroup$ Is it clear that the discretised discounted process also has the martingale property? $\endgroup$
    – g g
    Jul 21, 2015 at 7:24
  • $\begingroup$ I have to admit I don't know. We usually talk about continuous process. Maybe you should open a new post for your question? $\endgroup$
    – SmallChess
    Jul 21, 2015 at 8:37
  • 2
    $\begingroup$ if you discretise the log process and use $r-0.5 \sigma^2$ it has the martingale property. $\endgroup$
    – Mark Joshi
    Jul 21, 2015 at 8:55


let $\theta_t$ be an adapted procee such that the solution of SDE $$dL_t=-L_t\, \theta_t \,dW_t , \, L_0=1$$ is a Martingale.We set $Q{{|}_{\mathcal{F}_t}}=L_t\,P{{|}_{\mathcal{F}_t}}$,then $$W_{t}^{Q}=W_{t}^{P}+\int_{0}^{t} \theta_s\,ds$$ is a standard wiener process under Q measure.


Now we assume $\{S_t\}_{t\geq0}$ be a Geometric Brownian Motion. let $\theta=\frac{\mu-r}{\sigma}$ and $dL_t=-L_t\, \theta \,dW_t$. Then

$$W_{t}^{Q}=W_{t}^{P}+\left(\frac{\mu-r}{\sigma}\right)t$$ is a standard wiener process under Q measure. As a result $$dW_{t}^{Q}=dW_{t}^{P}+\frac{\mu-r}{\sigma}dt$$ and $$ dS_t= \mu S_t dt+\sigma S_t dW_t^P=\mu S_t dt+\sigma \,S_t(dW_t^Q-\frac{\mu-r}{\sigma}dt)=r S_t dt+\sigma S_tdW_t^Q$$

Euler scheme

First let $x_t=\ln S_t$, by application of Ito's lemma, we have $$d{{x}_{t}}=\left( r-\frac{1}{2}{{\sigma }^{2}} \right)dt+\sigma d{{W}_{t}}$$ and $$x_{t+\Delta t}=x_t+\left( r-\frac{1}{2}{{\sigma }^{2}} \right)\Delta t+\sigma (W_{t+\Delta t}-W_t)$$ but we know ${{W}_{t+\Delta t}}-{{W}_{t}}\sim N(0,\Delta t)$ then $$x_{t+\Delta t}=x_t+\left( r-\frac{1}{2}{{\sigma }^{2}} \right)\Delta t+\sigma \sqrt{\Delta t }Z$$ where $Z$ is a standard normal stochastic Variable.


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