Long-Run Geometric Brownian Motion Behavior from Brownian Scaling
Summary
The document asks how to establish the long-run behavior of a geometric Brownian motion. The responses write the process as an exponential whose log contains a deterministic term proportional to time and a Brownian term. Using the Brownian strong law, which says Brownian motion divided by time tends to zero almost surely, the sign of the deterministic log-growth rate determines whether the process tends to zero or infinity when that rate is nonzero.
A second response examines the probability that the process exceeds a fixed positive threshold and obtains limiting probabilities according to the same drift threshold. At the boundary, that probability tends to one half for each fixed threshold, which does not by itself establish the claimed pathwise behavior. The original question’s stated cases are inconsistent at equality: the boundary case is separately described as having no limit, rather than tending to infinity. The key almost-sure argument applies directly away from the boundary; Brownian fluctuation behavior is needed to analyze equality rigorously.
Key ideas
- The logarithm of geometric Brownian motion separates into deterministic growth and a Brownian fluctuation term.
- Brownian motion divided by time tends to zero almost surely, so the deterministic log-growth rate controls behavior away from the boundary.
- A positive net log-growth rate leads to divergence, while a negative one leads to decay toward zero.
- At the boundary, fixed-threshold exceedance probabilities alone do not prove pathwise convergence or nonconvergence.
- The document’s stated equality case requires separate analysis beyond the basic strong-law argument.
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# Asymptotic behavior property of geometric Brownian Motion proof
# Asymptotic behavior property of geometric Brownian Motion proof
Online I found the asymptotic behavior property of geometric Brownian Motion $X_t$as:
If $\mu$ (drift parameter) is $\ge$ $\sigma^2/2$ where $\sigma$ is the volatility parameter, then $X_t \rightarrow \infty$ as $t \rightarrow \infty$
If $\mu < \sigma^2/2$, then $X_t \rightarrow 0$ as $t \rightarrow \infty$
If $\mu = \sigma^2/2$, then $X_t$ has no limit as $t \rightarrow \infty$
While this makes sense, how would the proof look like for this property? I'm not really sure how to approach it at the moment. Any help is appreciated.
## Answer by Gordon (score 6)
https://quant.stackexchange.com/a/24830
For any $\alpha > 0$, \begin{align*} \lim_{t\rightarrow\infty}P\left(e^{\big(u-\frac{\sigma^2}{2}\big) t +\sigma W_t} > \alpha \right) &= \lim_{t\rightarrow\infty}P\left(\big(u-\frac{\sigma^2}{2}\big) t +\sigma W_t > \ln \alpha \right)\\ &=\lim_{t\rightarrow\infty}P\left(\frac{W_t}{\sqrt{t}} > \frac{\ln\alpha- \big(u-\frac{\sigma^2}{2}\big) t }{\sigma \sqrt{t}} \right)\\ &=\lim_{t\rightarrow\infty}\Phi\left(\frac{\big(u-\frac{\sigma^2}{2}\big) t -\ln\alpha}{\sigma \sqrt{t}} \right)\\ &= \begin{cases} 1, &\mbox{ if } u>\frac{\sigma^2}{2},\\ \frac{1}{2}, &\mbox{ if } u=\frac{\sigma^2}{2},\\ 0, &\mbox{ if } u < \frac{\sigma^2}{2}. \end{cases} \end{align*} The conclusion now follows immediately.
> Edit based on comments below.
Let $X_t = e^{\big(u-\frac{\sigma^2}{2}\big) t +\sigma W_t}$. Note that \begin{align*} \left(\omega:\, \lim_{t\rightarrow \infty} X_t = \infty \right) &= \cap_{m=1}^{\infty}\cup_{n=1}^{\infty} \cap_{t\ge n}(\omega:\,X_t > m). \end{align*} The almost surely convergence, for the case where $u>\frac{\sigma^2}{2}$, follows immediately.
## Answer by Tobias (score 4)
https://quant.stackexchange.com/a/24829
Write $X_t = \exp(\mu B_t + (\mu - \frac {\sigma^2}{2})t )$. To prove the statement you can use the law of large numbers for Brownian motion which states that $\lim_{t \to \infty} \frac {B_t}{t} = 0$. Then rewrite $X_t$ as $$X_t = \exp(t (\mu \frac {B_t}{t} + (\mu - \frac{\sigma^2}{2})).$$ Using these two properties, you can analyze the convergence behaviour of $X_t$.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.