I have a very fundamental question regarding simulation of DRIFTED geometric brownian motion.

We have the standard Blackos Scholes model:

$dS(t)=r S(t)dt+\sigma S(t) dW^{\mathbb{P}}(t)$, where $W^{\mathbb{P}}(t)$ is the standard Wiener process under probability measure $\mathbb{P}$.

If we want to simulate this, using constant $\Delta t$, we use the recursive formula:

$S_{t+1}=S_te^{(r-\frac{\sigma^2}{2})\Delta t+\sigma \sqrt{\Delta t} Z_t }$, where $Z_t \sim N(0,1)$.

Now assume that we want to change the drift such that:

$W^{\mathbb{Q}}(t) = W^{\mathbb{P}}(t) - \int_0^t \theta_s ds$ is a brownian motion under $\mathbb{Q}$ such that:

$dS(t)=(r + \sigma \theta )S(t)dt+\sigma S(t) dW^{\mathbb{Q}}(t)$

Now this is where I become unsure. If I want to simulate the drifted process, is it just fine to use the similar method as:

$S_{t+1}=S_te^{(r+\sigma \theta-\frac{\sigma^2}{2})\Delta t+\sigma \sqrt{\Delta t} Z_t }$, where $Z_t \sim N(0,1)$.

OR is it that I have to use $Z_t \sim N(\theta, 1)$?

Im not that strong in the change of measure part, so thats why I'm a bit unsure. Would appreciate for help.


  • $\begingroup$ Is there a typo in your equations for $S_{t+1} = ...$? Don't you mean $S_t$ instead of $S_0$? If not, then you won't simulate GBM paths even if you use the right drift. $\endgroup$
    – Quantuple
    May 30, 2016 at 9:29
  • $\begingroup$ Then the answer provided by Mark Joshi is spot on ;) $\endgroup$
    – Quantuple
    May 30, 2016 at 9:43
  • $\begingroup$ And the likelihood function is also only dependent on $Z_t \sim N(0,1)$ ? $\endgroup$
    – Elekko
    May 30, 2016 at 9:59
  • $\begingroup$ I don't know what you mean by that but yes, $Z_t \sim N(\theta,1)$ makes absolutely no sense $\endgroup$
    – Quantuple
    May 30, 2016 at 10:09
  • $\begingroup$ Ok, because I was reading this aueb.gr/pympe/hermis/hermis-volume-7/ACTUARIAL-RISK-MODELS/…, on equation (15), they define the $Z$ with non standard-normal densities. And then when they define the likelihood ratio (Radon-Nikodym derivative) I assumed that they used the "drifted mean" in the random $Z_t$. $\endgroup$
    – Elekko
    May 30, 2016 at 10:13

2 Answers 2


in the new measure, the stock has drift $r + \sigma \theta$ so yes you just proceed with that drift as you say. If $\theta$ is time dependent, it gets more complicated.

  • $\begingroup$ You mean $Z_t \sim N(0,1)$? How is the time dependance case then? $\endgroup$
    – Elekko
    May 29, 2016 at 23:37
  • $\begingroup$ you have to use the integral of theta against time instead of multiplying it, assuming it is deterministic. $\endgroup$
    – Mark Joshi
    May 30, 2016 at 4:37

Note that, under measure $Q$, the dynamics is of the form \begin{align*} dS_t = S_t \big[(r+ \sigma \theta_t) dt + \sigma dW_t^Q \big]. \end{align*} Then, for $\Delta>0$ sufficiently small, \begin{align*} S_{t+\Delta} &= S_te^{\left(r-\frac{1}{2}\sigma^2\right)\Delta + \sigma \int_t^{t+\Delta} \theta_s ds + \sigma \left(W_{t+\Delta}^Q-W_t^Q\right)}\\ &\approx S_te^{\left(r-\frac{1}{2}\sigma^2 + \sigma \theta_t\right)\Delta + \sigma \sqrt{\Delta} Z}, \end{align*} assuming that $\theta_t$ is continuous, where $Z\sim N(0, 1)$.


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