Show that $X_t=\exp(B_t - t/2)$ converges to $0$ in probability but not uniformly integrable

Question

$X_t=\exp\left(B_t - t/2\right)$ where $B_t$ is a standard Brownian motion.

I'd like to show that $X_t$ converges in probability to $X=0$ and that $X_t$ is not uniformly integrable.

For the first affirmation, I tried using the explicit law of $X_t$ which is a log-normal if I'm not mistaken. But I have a problem:

$$P(X_t > \epsilon) = 1 - \int_0^\epsilon \frac{1}{(x\sqrt{2\pi t})} \exp\left(-\frac{(\ln(x)+t/2)^2}{2t^2}\right)dx $$ does not go to $0$ as $t$ goes to infinity for every $\epsilon>0$. Where is the problem ?

For the second affirmation, I have no idea how to show that, any help is welcome.

Thank you very much.

Answer 1

• For convergence in probability, use the fact that $B_t$ has the same distribution as $\sqrt tN$, where $N$ is standard normal. In this way, for any $\varepsilon\in(0,1)$, $$\mathbb P\left(X_t\gt \varepsilon\right)=\mathbb P\left(\exp\left(\sqrt t N-t/2\right)\gt \varepsilon\right)=\mathbb P\left(N\gt\frac{t/2+\ln\varepsilon}{\sqrt{t}} \right).$$ For $t$ large enough, $t/2+\ln \varepsilon\gt t/4$, hence $$\mathbb P\left(X_t\gt \varepsilon\right)\leqslant \mathbb P\left(N\gt\frac{\sqrt{t}}{4}\right).$$
• If a family $\left(Y_t\right)_{t \geqslant 0}$ is uniformly integrable and converges to $0$ in probability, then it converges in $\mathbb L^1$. A computation show that the expectation of $X_t$ is constant and not zero.