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How can we show that this martingale $$ e^{aW_{t} - \frac{1}{2}a^2t}$$ converges to $0$ as $ t \rightarrow \infty$ using law of iterated logarithm, for $a \neq 0$.

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You could try the following steps: (1) translate the desired result in terms of $aW_t-\frac12a^2t$, (2) deduce from the law of iterated logarithm an upper bound on $aW_t$ for $t$ large enough, (3) conclude. – Did Feb 18 '12 at 0:54
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Assume that $a\neq0$. Note that

$$\exp\left(aW_t-\frac{1}{2}a^2t\right) =\exp\left(t\left(\frac{aW_t}{t}-\frac{1}{2}a^2\right)\right).$$

From the law of the iterated logarithm, it follows as a special case that $W_t/t$ converges almost surely to zero as $t\to\infty$. Therefore, as $a\neq0$, $\frac{aW_t}{t}-\frac{1}{2}a^2$ converges almost surely to $-\frac{1}{2}a^2$, which is negative. As a consequence, the argument to the exponential in the above converges to minus infinity, and so the exponential itself converges to zero.

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The law of the iterated logarithm is overkill here: the fact that $W_t/t \to 0$ a.s. is just the strong law of large numbers. – Nate Eldredge Apr 13 '12 at 1:16

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