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Let $\alpha > 0$, $M_t=\exp(\alpha B_t - \alpha^2 t/2)$ for all $t \ge 0$. Show that $M_t \to 0$ a.s.

I have started a class in stochastic Calculus and we are using the text by Steele. I have been staring at this text book for days and I just can't seem to understand this. I guess I don't have sufficient background in probability theory. Apparently we are supposed to use the inversion property of Brownian motion for the main tool here but I still cannot see it. I know that what we are trying to show is this. $$P \left ( \lim_{t \to \infty} M_t = 0 \right) = 1$$

We have a main tool called Doob's Maximal inequality which is supposed to be used for probabilistic purposes but I'm not sure if it fits in. I start by trying to take a limit.

$$\lim_{t \to \infty }\exp(\alpha B_t - \alpha^2 t/2) = \lim_{t \to \infty } \frac{ \exp(\alpha B_t) }{ \exp(\alpha^2 t/2) }$$

I know the point is that we need to show somehow that the denominator increased "faster" than the numerator, so this limit goes to zero. However I really am at a loss for how to proceed. I don't see how to fit probability into it. I feel that I have had zero examples about how this works I just don't understand the technique.

Thanks for any help!

jeffery_the_wind
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  • Use Martingale ConvergenceTheorem. – Kavi Rama Murthy Feb 08 '21 at 11:49
  • You do know that you are dealing with a martingale, right? We have a result for martingales that allows these kind of results, it is called the Martingale convergence theorem. Can you attempt to apply it? – Sarvesh Ravichandran Iyer Feb 08 '21 at 11:49
  • @TeresaLisbon Yes, sorry I didn't write that. – jeffery_the_wind Feb 08 '21 at 11:51
  • See here: https://math.stackexchange.com/questions/1241149/convergence-of-exponential-brownian-martingale-to-zero-almost-surely –  Feb 08 '21 at 11:51
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    Does this answer your question? Convergence of exponential Brownian martingale to zero almost surely –  Feb 08 '21 at 11:53
  • @KaviRamaMurthy This is my main issue with this class. The prof says, use this thm or that thm. OK so I have the Martingale Convergence Theorem that says $M_t \to M_\infty$ a.s, but that point is that I don't know that $M_\infty = 0$, that is what I'm trying to show. So this doesn't help me. – jeffery_the_wind Feb 08 '21 at 11:54
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    @jeffery_the_wind Is the attached answer helpful to you? See , to "understand" what $M_{\infty}$ could be, you need to look at the "long-term" behaviour of the martingale itself. The attached answer uses time-inversion to actually convert the problem of looking at "long-term" behaviour, to looking at the process "near zero" where things are easier to analyse. Then the rest is pretty clear (to me : I can understand if you are confused, please ask if you have any issues). I can see you are having issues integrating class material into problems : you can speak more about this to me. – Sarvesh Ravichandran Iyer Feb 08 '21 at 12:14
  • Thank you @TeresaLisbon I am going through that answer. I can see that using the inversion is a neat trick. Of course it isn't so intuitive since $M_t$ is a martingale $E[M_t]=E[M_0]=1$. But I need to stare at it for a few minutes I will get back to you. – jeffery_the_wind Feb 08 '21 at 13:08
  • @jeffery_the_wind Sure, please take your time. – Sarvesh Ravichandran Iyer Feb 08 '21 at 13:10
  • @TeresaLisbon ok yes as long as I accept the properties of the inverted Brownian motion then yes the rest of the math I agree with. However in this case it seems that all we have proven is that $P( \lim_{t \to \infty} \alpha B_t - \alpha^2 t/2 = -\infty ) = 1$. It seems we still need an additional property to bring the exponential function inside the limit. This may be a common property of probability theory but I am not familiar. (Thanks so much for your comments) – jeffery_the_wind Feb 08 '21 at 13:15
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    @jeffery_the_wind It's called an "$\omega$-by-$\omega$" argument : basically, the set of $\omega$ in the sample space such that $\alpha B_t(\omega) - \alpha \frac{t^2}2 \to -\infty$ has prob. $1$. For each such $\omega$, we can take the exponential and use a property of the exponential to see that $e^{\alpha B_t(\omega) - \alpha \frac{t^2}2} \to e^{-\infty} = 0$ for each such $\omega$ in that set, and therefore with prob. $1$ as well. This basically says : look at each element of the sample space and imagine that it occurred. Then work from there. – Sarvesh Ravichandran Iyer Feb 08 '21 at 13:20
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    The "$\omega$-by-$\omega$" argument works very well in providing intuition for problems regarding stochastic processes, as you can see from this example. – Sarvesh Ravichandran Iyer Feb 08 '21 at 13:21
  • Let us continue this discussion in chat. – jeffery_the_wind Feb 08 '21 at 14:50

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