Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\infty$?

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.