Determine $P(\lim_{t \to \infty}X_0 e^{t(1-\sigma^2/2) + \sigma W_t)}=0)$ 
Let $W_t$ be standard brownian motion and define the process $$X_t =
 X_0 e^{t(1-\sigma^2/2) + \sigma W_t}$$ where $\sigma$ has exponential
  distribution $$ P(\sigma \leq x) = 1-e^{-x}$$ for $x\geq 0$. 
  Determine $$P(\lim_{t \to \infty}X_0 e^{t(1-\sigma^2/2) + \sigma
 W_t}=0)$$


My attempt:
$$\lim_{t \to \infty}X_0 e^{t(1-\sigma^2/2) + \sigma W_t}=\lim_{t \to \infty}X_0 e^{t((1-\sigma^2/2) + \sigma W_t/t)}=X_0 e^{\lim_{t \to \infty}t((1-\sigma^2/2) + \sigma W_t/t)}$$
Hence, 
$$\lim_{t \to \infty}X_0 e^{t(1-\sigma^2/2) + \sigma W_t}=0 \iff\lim_{t \to \infty}t((1-\sigma^2/2) + \sigma W_t/t) =-\infty \iff (1-\sigma^2/2) <0$$
Then I simply calculate $P(\sigma > \sqrt{2}) = e^{-\sqrt{2}}$.
I have doubts about my final $\iff$ statement. 
 A: Just consider three cases separately:


*

*$1-\sigma^2/2<0$: Since $W_t/t \to 0$ almost surely as $t \to \infty$, we have $$\left[1- \frac{\sigma^2}{2} \right] + \sigma \frac{W_t}{t} < 0$$ for $t$ sufficiently large and so $$\lim_{t \to \infty} t \left( \left[1- \frac{\sigma^2}{2} \right] + \sigma \frac{W_t}{t} \right) = - \infty.$$

*$1-\sigma^2/2=0$: Then $$ t \left( \left[1- \frac{\sigma^2}{2} \right] + \sigma \frac{W_t}{t} \right)= \sigma W_t.$$ The limit $t \to \infty$ does (almost surely) not exist since Brownian motion keeps oscillating.

*$1-\sigma^2/2>0$: As in the first case, we get $$\left[1- \frac{\sigma^2}{2} \right] + \sigma \frac{W_t}{t} > 0$$ for $t$ sufficiently large and so $$\lim_{t \to \infty} t \left( \left[1- \frac{\sigma^2}{2} \right] + \sigma \frac{W_t}{t} \right) = \infty.$$


This shows that
$$\lim_{t \to \infty} t \left( \left[1- \frac{\sigma^2}{2} \right] + \sigma \frac{W_t}{t} \right) = - \infty \iff 1- \frac{\sigma^2}{2}< 0.$$
Note that the first equivalence in your proof holds only true if $X_0(\omega) \neq 0$ for all $\omega \in \Omega$.
