1
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ For this question to be answered, you need to explain how you sampled $r_\text{sample}$. $\endgroup$
    – SRKX
    Commented Jul 6, 2016 at 9:17

2 Answers 2

1
$\begingroup$

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)

$\endgroup$
1
$\begingroup$

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 $\sqrt{\Delta t}$.

$\endgroup$
2
  • $\begingroup$ 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
    Commented Jul 6, 2016 at 9:15
  • $\begingroup$ Right .. but as the OP asks about 1 day returns and $\Delta t = 1$ ... this is just trivial ... $\endgroup$
    – Richi Wa
    Commented Jul 6, 2016 at 12:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.