# is there a large time behavoiur of (compound) Poisson processes similiar to Law of iterated logrithm for Brownian Motion

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 $\infty$?

-

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 $\mathbb P(x\leqslant X_t\leqslant y)\to0$.