Simulation of GBM

Question

I have a question regarding the simulation of a GBM. I have found similar questions here but nothing which takes reference to my specific problem:

Given a GBM of the form

$dS(t) = \mu S(t) dt + \sigma S(t) dW(t)$

it is clear that this SDE has a closed form solution in

$S(t) = S(0) exp ([\mu - \frac{1}{2}\sigma^2]t + \sigma W(t))$

for a given $S(0)$.

Now, I have found sources claiming that in order to simulate the whole trajectory of the GBM, one needs to convert it to its discrete form (e.g., a similar question here or Iacus: "Simulation and Inference for Stochastic Differential Equations", 62f.). Yet, in Glasserman: "Monte Carlo Methods in Fin. Eng.", p. 94, I find that

$S(t_{i+1}) = S(t_i) exp ([\mu - \frac{1}{2}\sigma^2](t_{i+1}-t_i) + \sigma\sqrt{t_{i+1}-t_i} Z_{i+1})$

where $i=0,1,\cdots, n-1$ and $Z_1,Z_2,\cdots,Z_n$ are independent standard normals is an exact method (i.e., has no apprximation error from discretization).

I really don't understand what the difference between the two is, or put differently, if the exact method lets me simulate the whole trajectory, why would I bother converting it to the discrete form?

Maybe I'm just not seeing the point here but I'm really confused and grateful for any help!

Answer 1

For completeness, let's restate that the discrete case goes like this:

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

with $Z_t \sim \mathcal{N}(0,1)$

What you are doing in your case is to use the exact solution of the SDE to model the movement between two points of $S$.

Essentially, you are doing the same thing with the 2 approaches.

Actually, if you choose a $\Delta t$ small enough, you shall have almost no difference.

Your question can be reversed: if you can simply simulate the path using the discrete version, why would you care about solving the SDE to get the closed-form formula?

Answer 2

Note: There is a typo in your third equations. Instead of $S(u)$ it should be $S(t_{i})$ and in place of $S(t)$ there should be $S(t_{i+1})$.

In fact, given $S(t_{i})$ we have that

$$S(t_{i+1}) = S(t_{i}) \exp\left( (\mu - \frac{1}{2} \sigma^2) (t_{i+1} - t_{i}) + \sigma (W(t_{i+1}) - W(t_{i})) \right)$$

is the exact solution of the SDE. Hence, the discretization is exact (which is a special case here).

Note that $W(t_{i+1})$ is not independent of $W(t_{i})$ but $W(t_{i+1})-W(t_{i})$ is independet from $W(t_{i})-W(t_{i-1})$. So in order to simulate the discrete points $S(t_{j})$ for different $j$ you use the representation above with i.i.d. random variable $Z_{j}$ with $W(t_{j})-W(t_{j-1}) = \sqrt{t_{j}-t_{j-1}} Z_{j}$ and not the representation

$$S(t_{i+1}) = S(0) \exp\left( (\mu - \frac{1}{2} \sigma^2) t_{i+1} + \sigma W(t_{i+1}) \right)$$.