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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!

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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.)

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  • $\begingroup$ 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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