I am trying to obtain the discretized equation for the Geometric Brownian Motion EDE,

$$ d S_{t}=\mu S_{t} d t+\sigma S_{t} \eta_tdt \tag{1} $$

I am looking for the discretization for the case where $\mu$ and $\sigma$ are constants, $\eta_t$ is a white noise and the value of $S_t$ at the initial time, $S_0$, is known. The expected result for the discretized equation would be

$$ S_{t+d t}=S_{t}+\mu S_{t} dt+\sigma S_{t}X_t\sqrt{dt} \tag{2} $$

where $X_t$ is a standard normal variable, $X_t \sim N(0,1)$.

My attempt at a solution

I've tried integrating the whole equation ($1$) between $t$ and $t+dt$, obtaining

$$ S_{t+dt}-S_t=\mu \int_{t}^{t+d t} S_{t} d t+\sigma \int_{t}^{t+d t} S_{t} \eta(t)dt $$

The first integral of the RHS can be approximated by the area of a left-centered rectangle of height $S_t$ and width $dt$,

$$ \int_{t}^{t+d t} S_{t} d t \approx S_tdt $$

However, I am not sure how to reason the following approximation for the second integral in the RHS, in order to get the equation ($2$) from ($1$):

$$ \int_{t}^{t+d t} S_{t} \eta(t)dt \approx S_tX_t\sqrt{dt} $$

Would this approximation make any sense? Why would it be true?

  • 1
    $\begingroup$ It makes sense. The simplest thing you can do is to approximate the integrals by $S_t dt$ resp. by $S_t X_t\sqrt{dt}$. This is called Euler scheme. See the standard book by Kloeden and Platen for more advanced schemes which you may not even need for such a well-behaved equation. BTW why you solve this SDE numerically ? It has a closed form solution: GBM. $\endgroup$
    – Kurt G.
    May 20, 2022 at 11:27
  • $\begingroup$ @Kurt G It is part of a class exercise. I didn't know there was a closed-form solution. In that case, it's true that it may be a bit foolish to do it this way $\endgroup$
    – Invenietis
    May 20, 2022 at 17:51
  • 1
    $\begingroup$ Not foolish at all. You can study the discretization by comparing it with the theoretical GBM. That's exactly how numerical methods get testet. $\endgroup$
    – Kurt G.
    May 20, 2022 at 17:57

1 Answer 1


This is by no mean a full answer, in fact as Kurt G. stated in the comments this "discretization" is just the very famous Euler method. Still I wanted to make a few remarks that were too long for the comment section.

First thing one should notice is the fact that the actual SDE for the GBM is $$d S_{t}=\mu S_{t} d t+\sigma S_{t} dB_t \tag{1}$$ where $B$ is a Brownian motion.

Even though it seems reasonable to write this equation as $$d S_{t}=\mu S_{t} d t+\sigma S_{t} \dot B_tdt \tag{2}$$ where $\dot B$ is a "white noise", this is not actually correct; this is a consequence of the Wong-Zakai theorem. If we want to write the SDE as a random ODE we must use the so called "Wick product" $\diamond$ and then we could rewrite $(1)$ as $$d S_{t}=\mu S_{t} d t+\sigma S_{t} \diamond \dot B_tdt.$$

Having said that what we want to show is that

$$\int_{t}^{t+d t} S_{s} dB_s \approx S_tX_t\sqrt{dt}$$ or which is the same

$$\int_{t}^{t+d t} S_{s} dB_s \approx S_t(B_{t+dt}-B_t).$$

Now we compare the two $$S_t\int_t^{t+dt}dB_s-\int_{t}^{t+dt} S_{s} dB_s,$$ since $S$ is adapted we can write

$$\int_t^{t+dt}(S_t-S_s)dB_s,$$ taking the $L^2$ norm and using the Itô isometry we have

$$\|\int_t^{t+dt}S_t-S_sdB_s\|^2=\int_t^{t+dt}\|S_t-S_s\|^2ds$$ and this last term converges to $0$ as $dt$ vanishes.


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