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Can someone please point me to a reference/answer me this question?

From the law of iterated logarithm, we see that Brownian motion with drift converge to $\infty$ or $-\infty$.

For a Poisson processes $N_t$ with rate $\lambda$, is there a similiar thing? As an exercise, I am consider what happens if the Brownian motion exponential martingale (plus an additional drift r) is replaced by a Poisson one. More explicitly, that is

$\exp(aN_t-(e^a-1)\lambda t + rt)$

how would this behave as $t$ tend to $\infty$?

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up vote 2 down vote accepted

Since $N_t/t\to\lambda$ almost surely, $X_t=\exp(aN_t-(e^a-1)\lambda t + rt)=\exp(\mu t+o(t))$ almost surely, with $\mu=(1+a-\mathrm e^a)\lambda+r$. If $\mu\ne0$, this yields that $X_t\to0$ or that $X_t\to+\infty$ almost surely, according to the sign of $\mu$.

If $\mu=0$, the central limit theorem indicates that $N_t=\lambda t+\sqrt{\lambda t}Z_t$ where $Z_t$ converges in distribution to a standard normal random variable $Z$, hence $X_t=\exp(a\sqrt{\lambda t}Z_t)$ and $X_t$ diverges in distribution (except in the degenerate case $a=r=0$) in the sense that, for every positive $x\leqslant y$, $\mathbb P(X\leqslant x)\to\frac12$ and $\mathbb P(X_t\geqslant y)\to\frac12$, hence $\mathbb P(x\leqslant X_t\leqslant y)\to0$.

Note: The LIL is a much finer result than all those above.

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