# Probability of geometric brownian motion taking a certain value

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.

• 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? – jwg Jan 29 '18 at 12:25
• 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 $t = 2$ – Raffaele Jan 29 '18 at 12:35
• 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 $S_2$? – noob2 Jan 29 '18 at 12:56
• If I'm supposed to know, I'm afraid I don't. There's nothing more in the original problem statement – Raffaele Jan 29 '18 at 13:06
• 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. – 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.

• Up and thanks for stopping by! Would you mind explaining why $\sigma W_t$ becomes $\sigma \sqrt{t} Z$ and especially the last transformation? – Raffaele Jan 29 '18 at 17:09
• Because $W_t$ is Brownian Motion. You need to refresh your memory on BM, GBM, etc. – noob2 Jan 29 '18 at 17:35
• 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 – David Addison Jan 29 '18 at 18:26
• 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 – Raffaele Jan 29 '18 at 18:30
• 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 $\mathbb{P}(Y\leq a)=\mathbb{P}\left(Z\leq \frac{a-\mu}{\sigma}\right)$. – loyd.f Jan 29 '18 at 18:46