Uniformly integrable martingale

Question

I have the following martingale. $M_n=\exp\left(aB_n-\frac{1}{2}a^2n\right)$ for $n\geq0$ and $a\neq0$, $B_n$ is a BM. I have to show that for $a>0$, $M_n\rightarrow0$ in probability.

Is $M_n$ uniformly integrable? I think its not but how can I prove it?

Any hints are really appreciated.

Answer 1

Hint Write $$a \cdot B_n - \frac{1}{2} a^2 n = n \cdot \left( a \cdot \frac{B_n}{n}- \frac{1}{2}a^2 \right),$$ and use that $$\frac{B_n}{n} \to 0 \quad \text{a.s.}$$ This actually even proves $M_n \to 0$ a.s.

Apply Vitali's convergence theorem in order to prove that $(M_n)_n$ is not uniformly integrable.

Answer 2

Yes, it does go to zero in probability. All you need to do is to show that its expectation raised to a fractional power goes to zero as $n\to\infty$. Since $aB_n/t \sim N(0,a^2n/t^2)$, $\exp(aB_n/t)$ is lognormal. You can do the rest (this is your homework after all).