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Define the exponential Brownian martingale as $N_t = \exp\left\{a W_t - \frac12 a^2 t \right\}$ which is a martingale with respect to the natural filtration of $W$ which stands for a standard Brownian motion.

I now want to prove that $N_t \rightarrow 0$ a.s. for $t \rightarrow \infty$. As $N_t$ is a martingale it is by definition in $L^1$ hence Doob's martingale convergence theorem tells us that there exists a $N_\infty \in L^1$ such that $N_t \rightarrow N_\infty$ a.s., but how could I prove that $N_\infty \equiv 0$? Any help is greatly appreciated!

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  • $\begingroup$ Much simpler is to show that $aW_t-\frac12a^2t\to-\infty$ almost surely, when $t\to\infty$. $\endgroup$
    – Did
    Apr 19, 2015 at 0:02
  • $\begingroup$ Could you give a hint how to show this? Seems like we should have that $W_{t} \leq \frac12 a t$ almost surely, but don't know how I could prove this. $\endgroup$
    – user155670
    Apr 19, 2015 at 0:07

1 Answer 1

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As @Did already pointed out, it is much easier to show that

$$a W_t - \frac{1}{2} a^2 t \to - \infty \qquad \text{almost surely as $t \to \infty$.}$$


Recall that the process $$B_t := \begin{cases} t W_{\frac{1}{t}}, & t>0, \\ 0, & t=0 \end{cases}$$ defines a Brownian motion; in particular

$$\lim_{t \to 0} t W_{\frac{1}{t}} = \lim_{t \to 0} B_t = 0.$$

This implies

$$\lim_{t \to \infty} \frac{W_t}{t}= \lim_{s \to 0} s W_{\frac{1}{s}} = 0.$$

Hence,

$$a W_t - \frac{1}{2} a^2 t = t \underbrace{\left( a \frac{W_t}{t} -\frac{a^2}{2} \right)}_{\to - \frac{a^2}{2}} \to - \infty$$

almost surely as $t \to \infty$.

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  • $\begingroup$ I feel like something is off: Essentially, in the last line it is implied that $\alpha t\frac{W_t}{t}\to 0$, because there's a $t$ outside the parenthesis. But wouldn't that imply that $W_t\to 0$?. $\endgroup$ Dec 4, 2019 at 21:57
  • $\begingroup$ @antonzm No, this is not implied by the last line. The last line is saying that $$a \frac{W_t}{t} - \frac{a^2}{2}$$ is strictly smaller than zero for large $t$, and therefore $$t \left( a \frac{W_t}{t} - \frac{a^2}{2} \right)$$ converges to $-\infty$ as $t \to \infty$. $\endgroup$
    – saz
    Dec 5, 2019 at 7:24
  • $\begingroup$ Thanks!, now it is entirely clear. @saz $\endgroup$ Dec 5, 2019 at 12:52
  • $\begingroup$ I was thinking on another way of doing it with the Law of Iterated Logarithms. Could you please take a look to see if it's correct please? :D $\endgroup$ Dec 5, 2019 at 12:55
  • $\begingroup$ @antonzm As it is currently written, it is not correct. Note that the liminf should be $-\infty$ not $+\infty$; in particular, you need to estimate $\liminf (\alpha W(t)-\frac{1}{2} \alpha ^2 t)$ from above, not from below. $\endgroup$
    – saz
    Dec 5, 2019 at 13:46

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