I am working on the following problem

Given the solution to the Geometric Brownian Motion $$S_t=S(0)\exp\Big[(\mu-\frac{1}{2}\sigma^2)t+\sigma B_t\Big]$$ Where $\{B_t:t\geq 0\}$ is a Brownian motion.

a) Show that for $\mu>\frac{1}{2}\sigma^2$, we have that $S(t)\rightarrow\infty$ as $t\rightarrow\infty$.

b) Show that for $\mu<\frac{1}{2}\sigma^2$, we have that $S(t)\rightarrow 0$ as $t\rightarrow\infty$.

I know multiple things that will probably help me in finding the solution, but I'm not able to connect the dots. For question a), we know that the term in the exponent can be split into an exponent with a strictly positive term, and an exponent with the Brownian Motion.

I can't figure out why the exponent with the Brownian motion would go to infinity in question a), but not in question b). In other words, I cannot see why we would have that $$S(0)\exp\Big[ct+\sigma B_t\Big]\rightarrow\infty ,\quad c>0$$But also $$S(0)\exp\Big[dt+\sigma B_t\Big]\rightarrow\infty ,\quad d<0$$ What is the convergence behavior of Brownian Motion in the exponent?

Moreover, we know that the definition of convergence in probability of some sequence $(X_n)$ is that $$\forall\epsilon>0:\mathbb{P}(|X_n-X|>\epsilon)\rightarrow 0$$

Any tips are highly appreciated!

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    $\begingroup$ Your solution $S_t$ is incorrect. It should read $S_t=S_0\exp\left[(\mu-0.5\sigma^2)t +\sigma B_t\right]$. Now does the limiting behavior make more sense? $\endgroup$ – Nap D. Lover Mar 4 at 18:04
  • $\begingroup$ @LoveTooNap29 Ah, thanks for the correction. It makes slightly more sense that way but the thing that is confusing me is what exactly happens with the term $$exp(\sigma B_t)$$ as $t\rightarrow\infty$? I imagine this term does not converge at all but then again I am not sure. $\endgroup$ – Charlie Shuffler Mar 4 at 18:42
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    $\begingroup$ It isn't necessary that that term alone converge. We just care about the combination of the terms $\exp\left(ct + \sigma B_t\right)$ (where $c < 0$). So perhaps you could start by analysing what happens with $ct + \sigma B_t$ as $t\to \infty$, if $c <0$. $\endgroup$ – Minus One-Twelfth Mar 4 at 19:17
  • $\begingroup$ @MinusOne-Twelfth I see. I think I am able to solve the problem now. Could you explain why or why not we only have convergence in probability and not also convergence almost surely? $\endgroup$ – Charlie Shuffler Mar 5 at 8:00
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    $\begingroup$ Possibly this page can be of help: math.stackexchange.com/questions/1241149/… $\endgroup$ – Minus One-Twelfth Mar 5 at 8:34

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