Recall the Geometric Brownian Motion $X={\rm e}^{\mu W+\left(\sigma-\frac{\mu^2}{2}\right)t}$. If $\sigma<\frac{\mu^2}{2}$, then $X$ tends to 0 almost surely. But if we consider the following case, $$X=\exp\left\{\int_0^t\mu(t'){\rm d} W+\int_0^t\sigma(t')-\frac{\mu^2(t')}{2}{\rm d}t'\right\},$$ and also assume that $\sigma(t)<\frac{\mu^2(t)}{2}$ for all the $t>0$ ($\mu$ and $\sigma$ are assumed to be good enough), do we also have the almost decay property? I mean, $X$ tends to $0$, almost surely? It looks like right, but how would the proof look like? I'm not really sure how to approach it at the moment. Any help is appreciated. Many thanks!


The process $$Y_t = \int_0^t \mu(s) dW_s$$ is a time-changed Wiener process: if $$\phi(t) = \int_0^t (\mu(s))^2 ds$$ and $\phi^{-1}$ denotes the inverse function, then $$U_t = Y_{\phi^{-1}(t)}$$ is a Wiener process. (Indeed, it is a Gaussian process with the correct covariance structure.)

Thus, assuming that $\lim_{t\to \infty} \phi(t) = \infty$, your question is equivalent to the following one: does $$U_t + \int_0^{\phi^{-1}(t)} \sigma(s) ds - \frac{t}{2}$$ drift to $-\infty$ with probability one? The answer is given in terms of the law of the iterated logarithm: it does if $$ \lim_{t \to \infty} \frac{1}{\sqrt{2 t \log \log t}} \biggl(\int_0^{\phi^{-1}(t)} \sigma(s) ds - \frac{t}{2}\biggr) < -1 , $$ and it does not if the limit is greater than $-1$. (I suppose more refined answers can be given, but I doubt there is a simple "if and only if" condition.)

  • $\begingroup$ Thanks very much. I try to understand your idea in this way: to make $X$ tends to $0$, $\int_0^t\mu(t'){\rm d} W+\int_0^t\sigma(t')-\frac{\mu^2(t')}{2}{\rm d}t'$ should tend to $-\infty$. And then, we consider $f(t)\times\left[\frac{\int_0^t\mu(t'){\rm d} W}{f(t)}+\frac{\int_0^t\sigma(t')-\frac{\mu^2(t')}{2}{\rm d}t'}{f(t)}\right]$. If we can find a $f(t)$ such that $f(t)$ tends to $+\infty$ and $\frac{\int_0^t\mu(t'){\rm d} W}{f(t)}$ tends to some constant $a$, then we solve $a+\frac{\int_0^t\sigma(t')-\frac{\mu^2(t')}{2}{\rm d}t'}{f(t)}<0$ to find the condition on $\mu(t)$ and $\sigma(t)$. $\endgroup$ – beyond_th Apr 15 at 20:49
  • $\begingroup$ So now it is clear that $\sigma(t)<\frac{\mu^2(t)}{2}$ is not enough. You are right. Let me guess, the transform $U_t = Y_{\phi^{-1}(t)}$ is used to deal with $\frac{\int_0^t\mu(t'){\rm d} W}{f(t)}$, right? $\endgroup$ – beyond_th Apr 15 at 20:55
  • $\begingroup$ @beyond_th: I am afraid I did not understand your comments. Time-change allows one to use standard tools available for the Wiener process, like the law of the iterated logarithm. $\endgroup$ – Mateusz Kwaśnicki Apr 16 at 18:19
  • $\begingroup$ Yes. I agree with you. Without the time-change, it is difficult to estimate $\frac{\int_0^t\mu(t'){\rm d}W}{f(t)}$, as in my previous comments. Anyway, thanks very much! Your answer indeed helped me and it is very inspiring! Best wishes! $\endgroup$ – beyond_th Apr 16 at 19:18

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