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Suppose $(B_t)_{t\geq 0}$ is a Brownian motion and that $S_t = \exp (B_t-\frac{t}{2})$. By the martingale convergence theorem, $S_t\to S_\infty$, some random constant, a.s..

It seems that we should have $\mathbb{P}[S_\infty=0]=1$, but I'm not sure how to prove it. What's the best approach here?

Thank you.

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Maybe writing $S_t = \exp(t[B_t / t - 1/2])$ and noting that $B_t / t \to 0$ almost surely. –  Shai Covo May 5 '11 at 15:35

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up vote 3 down vote accepted

By the strong law of large numbers, $B_t/t \to 0$ a.s. Thus $$B_t - \frac{t}{2} = t\left(\frac{B_t}{t} - \frac{1}{2}\right) \to -\infty.$$ So $S_t \to 0$ a.s.

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And the same is true for any Levy process $X$ with ${\rm E}(X_1)=0$. –  Shai Covo May 5 '11 at 15:43

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