So we have an asset whose price follows a GMB:

$dS_t = \mu S_t dt + \sigma S_t d W_t$

and want to know the probability that it drops 5% or more at time $t = 2$, given that $\mu = 0.04$ and $\sigma = 0.2$. I think (thanks Wikipedia) that it should be solved like this:

  • first pass is finding the value of $S_2$ (question: how do I compute $W_t$?)
  • somehow taking advantage that $S_t$ is log-normally distributed (I'm not sure how to use standard normal CDF tables)

Disclaimer: I know this must be super simple, but have not found the solution and don't know anyone that can help.

  • 1
    $\begingroup$ It's not clear what you mean by 'drops 5% or more at time t=2'. GBM is continuous. Drops 5% or more in what time period? $\endgroup$ – jwg Jan 29 '18 at 12:25
  • $\begingroup$ I have to translate the statement from Italian, however I think we are not interested in any value before $t = 2$, nor in any value after; just exactly at $t = 2$ $\endgroup$ – Raffaele Jan 29 '18 at 12:35
  • $\begingroup$ OK, so let's say the process starts at $S_0$ and we require $S_2 \le 0.95 S_0$. What is the probability distribution of $S_2$? $\endgroup$ – noob2 Jan 29 '18 at 12:56
  • $\begingroup$ If I'm supposed to know, I'm afraid I don't. There's nothing more in the original problem statement $\endgroup$ – Raffaele Jan 29 '18 at 13:06
  • $\begingroup$ Can't understand the downvotes, plus it's just stupid to not tell what's wrong with my question because I can't guess how to improve neither this very question nor my attitude. $\endgroup$ – Raffaele Jan 29 '18 at 13:43

Given that the solution of this SDE is,

$$S_t = S_0e^{\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t},$$

which is equal in law to:

$$S_t = S_0e^{\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma \sqrt{t}Z},$$

where $Z\sim \mathcal{N}(0,1)$. You have:

$$\mathbb{P}\left(\frac{S_2}{S_0}-1\leq-0.05\right) = \mathbb{P}\left(Z \leq \frac{\log(0.95)- 2\left(\mu-\frac{\sigma^2}{2}\right)}{\sqrt{2} \sigma}\right),$$

quantity that you can calculate given the table of the normal law.

  • $\begingroup$ Up and thanks for stopping by! Would you mind explaining why $\sigma W_t$ becomes $\sigma \sqrt{t} Z$ and especially the last transformation? $\endgroup$ – Raffaele Jan 29 '18 at 17:09
  • $\begingroup$ Because $W_t$ is Brownian Motion. You need to refresh your memory on BM, GBM, etc. $\endgroup$ – noob2 Jan 29 '18 at 17:35
  • $\begingroup$ It sounded like @Raffaele wanted the first time hitting model (I.e, what he meant by first pass). Can you confirm that this is not the case? en.m.wikipedia.org/wiki/First-hitting-time_model $\endgroup$ – David Addison Jan 29 '18 at 18:26
  • $\begingroup$ Confirm that is not the case. This answer is the solution to my problem, even if I'm still struggling to understand how to get there $\endgroup$ – Raffaele Jan 29 '18 at 18:30
  • 3
    $\begingroup$ You will need basic notions in stochastic calculus to understand how to compute the solution of such SDEs, it is explained here by the way: en.wikipedia.org/wiki/… Secondly, by definition of the BM, $W_t \sim \mathcal{N}(0,t)$, and you know that, if $Z\sim \mathcal{N}(0,1)$, then $\mu+\sigma Z \sim \mathcal{N}(\mu,\sigma^2)$, hence $\sqrt{t}Z\overset{\mathcal{L}}{=} W_t$. For the last equality, it's just that, for $Y \sim \mathcal{N}(\mu,\sigma^2)$, you have $\mathbb{P}(Y\leq a)=\mathbb{P}\left(Z\leq \frac{a-\mu}{\sigma}\right)$. $\endgroup$ – loyd.f Jan 29 '18 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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