# Deriving the discretized equation of the Geometric Brownian Motion EDE

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?

• 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. May 20, 2022 at 11:27
• @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 May 20, 2022 at 17:51
• Not foolish at all. You can study the discretization by comparing it with the theoretical GBM. That's exactly how numerical methods get testet. May 20, 2022 at 17:57

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.