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Suppose we have 3 stocks following GBMs.

We are given the distribution of the daily log returns which is multivariate normal.

Suppose I want to sample the stock price tomorrow ($\Delta t = 1$ day), could I just sample a return vector from this distribution and then say that the stock price tomorrow is $S_0 \cdot \exp(r_\text{sample}\Delta t)$?

I've been arguing with my friend about this and he claims I should multiply by $\sqrt{\Delta t}$? I don't understand his argument.

Is there anything wrong with what I am doing here?

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    $\begingroup$ For this question to be answered, you need to explain how you sampled $r_\text{sample}$. $\endgroup$ – SRKX Jul 6 '16 at 9:17
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The log-return of a stock over a period $\Delta t $ starting at $t=0$ is defined as: $$ r_{\Delta t} = \ln \left( \frac{S_{\Delta t}}{S_0} \right) $$ Thus you should compute $S_{\Delta t}$ as $$ S_{\Delta t} = S_0 \exp ( r_{\Delta t} ) $$ when you are given the $\Delta t $-period log-return i.e. the one which you sample as you propose above. Thus no multiplication by $\Delta t $ or its square root whatsoever.

Maybe your confusion arises from the fact that in the BS equation we traditionally use continuously compounded rates: $$ \exp ( r_{\Delta t} ) = \exp \left( \int_0^{\Delta t} r (t) dt \right) = \exp ( r \Delta t ) $$ where the last equality holds when $r (t) = r $ a constant, and in which case you should use $\Delta t \approx 1/252$ to compute daily returns if you're using annualised quantities (which is usually the case)

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You mix up several things:

if you sample from Brownian motion, then $$ B_{t+\Delta t} - B_t $$ is normally distributed with variance $\Delta t$. Thus if you sample a standard normal $Z$ (with variance 1) then you can use $$ \sqrt{\Delta t} Z $$ as sample for $B_{t+\Delta t} - B_t$ in order to get the correct variance. Recall that constant factors enter variance with the squared value.

In your question: how do you sample $r$? And if $\Delta t=1$ then it does not matter at all whether you use $\Delta t$ or $\sqrt{\Delta t}$.

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  • $\begingroup$ I totally get what you mean and obviously agree, but if you re-read your answer, you'll notice it's not very clear without prior understanding. If you have a bit of time, it would be nice to slightly rephrase it I think. $\endgroup$ – SRKX Jul 6 '16 at 9:15
  • $\begingroup$ Right .. but as the OP asks about 1 day returns and $\Delta t = 1$ ... this is just trivial ... $\endgroup$ – Richard Jul 6 '16 at 12:35

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