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I was wondering if there is a way of generating a sample path of a Geometric Brownian Motion using two independent standard normal random variables instead of just one.

The exact scheme that uses one standard normal random variable

$$ \hat{S}_{t_{i+1}}= \hat{S}_{t_{i}} \text{exp}\left( (r- \frac{\sigma^2}{2})(t_{i+1}-t_i)+ \sigma \sqrt{t_{i+1}-t_i} Z \right), \ i=0, \dots, n-1$$.

I want to know if there is an exact scheme that uses multiple independent normal random variables. I am asking this specifically for a barrier-forward start type option which has a "barrier check" at say a time $t$.

The idea I had for this case is to have $Z_1$ simulate $S_{t}$ and then have an independent $Z_2$ simulate the final $S_T$ but I am not sure.

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1 Answer 1

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When you simulate a sample path of a standard Brownian motion, you are generating a sequence $(B_t)_{t \in \mathbb{\Pi}}$ where $\mathbb{\Pi} := \{t_0, ..., t_n\}$ is your time partition. You can view that sequence as $n$ draws of the same random variable, although no one could say that this isn't also 1 draw each of $n$ independent normal random variables.

This is true by definition. You can divide your sample path however you want and name/define things so that as many random variables as you wish get involved, but besides being a huge waste of time, I do not see the point.

EDIT

Say we use a Euler discretization. You split a month into a grid using 1000 time steps. For each sample path, you need $(Z_t)_{t=1,...,1000}$ where each $Z_t \sim N(0, 1/1000)$.

On your computer, you could do:

B = np.random.normal(loc=0, scale=1, size=1000 )
Z = np.sqrt(1/1000)*B

Or

B1 = np.random.normal(loc=0, scale=1, size=500 )
B2 = np.random.normal(loc=0, scale=1, size=500 )
B  = np.hstack( (B1,B2) )
 Z = np.sqrt(1/1000)*B

You can split those steps in as many vectors as you like. Each vector is a set of draws from a random normal distribution. You can treat this as many draws of 1 r.v., 500 draws each of 2 r.v., etc. It's just a question of definitions.

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  • $\begingroup$ So in the case of using just one RV per sample path, the idea is just use that same single RV per discretisation step in the sample path? Then, I can also use, say 2. I am just not sure how I can use 2 instead of 1; for example, using a different Z at each time step is clear but how can I use just 2 per sample path? $\endgroup$
    – ʎpoqou
    Commented May 2, 2020 at 19:26
  • $\begingroup$ You're probably over-thinking this. Let me give you an example in an edit. $\endgroup$
    – Stéphane
    Commented May 2, 2020 at 19:27
  • $\begingroup$ It's a big waste of time to explicit this. The only time you'd use it is if you need to patch together two simulations to save time. $\endgroup$
    – Stéphane
    Commented May 2, 2020 at 19:38
  • $\begingroup$ Except you wrote np.sqrt(1000) but shouldnt that be 1/1000 with the way you defined the variance? $\endgroup$
    – ʎpoqou
    Commented May 3, 2020 at 9:15
  • $\begingroup$ Yes, that should be 1/1000 $\endgroup$
    – Stéphane
    Commented May 4, 2020 at 2:18

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