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!


2 Answers 2


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?

  • 2
    $\begingroup$ That makes sense, thanks a lot (and sorry about the typos). So, do I understand correctly then that the discrete form is usually an approximation and used if there is no closed form (i.e., no exact analytical solution)? And by approaching the limit of 0 with my timestep, I can overcome the problem (somehwat) of an approximation error? $\endgroup$
    – gu7z
    Jan 28, 2013 at 13:24
  • 1
    $\begingroup$ That's absolutely right. $\endgroup$
    – SRKX
    Jan 28, 2013 at 13:29
  • $\begingroup$ One can view his equation as the Euler scheme for log(S). In that case discretization and exact soluation have no difference! The reason why one has to go in discrete steps in the way to generate W(t) from i.i.d. random variables. $\endgroup$ Jan 28, 2013 at 20:09

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

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

  • $\begingroup$ Thank's for pointing out the typos and sorry about that. I follow your point, just one question: Why does the time dependent part in the stochastic element drop out? should sigma Z_j not be dependent on the time step? $\endgroup$
    – gu7z
    Jan 28, 2013 at 13:30
  • $\begingroup$ @gu7z not it's not, by definition of the model. $\endgroup$
    – SRKX
    Jan 28, 2013 at 13:39
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    $\begingroup$ @gu7z Yes. In the formulation $W(t_j)−W(t_{j−1}) = Z_j$ the $Z_j$ is a normal distributed random variable with mean zero and variance $t_{j}-t_{j-1}$. But since I wrote that the Z_j are i.i.d. we have indeed to rescale them and write $W(t_j)−W(t_{j−1}) = \sqrt{t_{j}-t_{j-1}} Z_j$ (or otherwise assume equidistributed time stepping. I corrected that in my post! $\endgroup$ Jan 28, 2013 at 20:05

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