monte carlo - Simulation of GBM

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

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?