Does this property hold for a Geometric Brownian Motion?

Question

Assume that $B_t$ is a Brownian motion. Let $\mu, \sigma$ be real numbers greater than zero. Do we then have

$$P\left(\liminf\limits_{t\rightarrow \infty}e^{\mu t+\sigma B_t}>1\right)=1?$$

I am not sure how to start to figure this out.

Answer 1

Yes, in fact $\lim_{t \rightarrow \infty} e^{\mu t + \sigma B_t} = \infty$ almost surely. We can show $\lim_{t \rightarrow \infty} \frac{B_t}{t} = 0$ a.s. so $$\mu t + \sigma B_t = t\left(\mu + \sigma \frac{B_t}{t}\right)$$ converges to $\infty$ almost surely (since $\mu > 0$). This implies $\lim_{t \rightarrow \infty} e^{\mu t + \sigma B_t} = \infty$ a.s. and in particular $$P\left( \liminf_{t \rightarrow \infty} e^{\mu t + \sigma B_t} > 1 \right) = 1.$$