Calculating a Geometric Brownian Motion Price Decline Probability

Summary

The document shows how to find the probability that an asset following geometric Brownian motion finishes below a specified percentage of its starting price at a fixed time. Rather than attempting to calculate a particular Brownian path, it uses the closed-form solution of the stochastic differential equation. The solution expresses the log price relative to its initial value as a normal random variable, with drift adjusted by half the variance and volatility scaled by the square root of time.

For the stated decline threshold, the event is rewritten as an inequality for a standard normal variable. The cutoff is calculated from the logarithm of the target price ratio, the adjusted drift, volatility, and elapsed time; evaluating the standard normal cumulative distribution function then yields the probability. The response gives the method but not a numerical result. It assumes constant drift and volatility and the geometric Brownian motion model, so it does not address jumps, changing parameters, or the probability of crossing the threshold at any time before the horizon.

Key ideas

• The geometric Brownian motion solution gives the asset price as an exponential function of drift and Brownian motion.
• At a fixed time, the Brownian term can be represented in distribution by a standard normal variable scaled by the square root of time.
• A percentage decline event can be converted into a threshold for that standard normal variable using the logarithm of the target price ratio.
• The standard normal cumulative distribution function gives the probability of finishing below the target at the specified time.
• The method assumes constant drift and volatility and concerns the terminal price, not an earlier path crossing.

Tags

# Probability of geometric brownian motion taking a certain value


# 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.

## Answer by loyd.f (score 4, accepted)

https://quant.stackexchange.com/a/37981

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.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.