About existence of martingale 
Question: Define $X_t=\exp (B_t-\frac{t}{2})$ to be a martingale, where $(B_t)$ is a brownian motion. Does $\lim_{t \rightarrow \infty} X_t$ exist?

I am thinking around this result (long-term behavior of trajectories): Define $\{B_t\}_{t \in [0,\infty)}$ to be a brownian motion, then $\lim_{t \rightarrow \infty} \sup \frac{B_t}{\sqrt{t}}=\infty$ and $\lim_{t \rightarrow \infty} \inf \frac{B_t}{\sqrt{t}}=-\infty$.
To fit in this result, can I rewrite $B_t-\frac{t}{2}$ as $\sqrt{t}(\frac{B_t}{\sqrt{t}}-\frac{\sqrt{t}}{2})$, then apply the result (long-term behavior of trajectories) to get infinity to get that $\lim_{t \rightarrow \infty} X_t$ exists and equal to 0?
Another approach I am thinking about is writing $X_t=\exp (B_t-\frac{t}{2})$ as $X_t=\exp (t[\frac{B_t}{t}-\frac{1}{2}])$, then apply strong law of large numbers for brownian motion. to get $X_t=\exp (-\infty)=0$, so the limit exists and is equal to 0?
I think the second approach works, but I am not sure if the first approach works. Appreciate any help.
 A: *

*The second approach works because by the strong law of large numbers $B_t/t$ converges to zero and $B_t/t-1/2$ converges to $-1/2$. Therefore, $B_t-t/2$ converges to $-\infty$. (I cannot believe that it took me a day and further hints from @daisies to see that.)


*Following the hint by @Brian Moehring the first approach should use the law of the iterated logarithm
$$\tag{1}
\limsup_{t\to\infty}\frac{B_t}{\sqrt{t}L_t}=1\,,\quad\text{ where }L_t:=
\sqrt{2\log\log t}\,.
$$
We have
\begin{align}
\limsup_{t\to\infty}e^{B_t-\frac{t}{2}}&=\limsup_{t\to\infty}\exp\Bigg(\sqrt{t}L_t\Big\{\frac{B_t}{\sqrt{t}L_t}-\frac{\sqrt{t}}{2L_t}\Big\}\Bigg)\,.
\end{align}
Using de L'Hospital it is easy to see that $t/\log\log t$ tends to $+\infty$. Therefore, $\sqrt{t}/L_t$ does the same. By (1) the limsup of the
term in the curly brackes is therefore $-\infty$ and since $\sqrt{t}L_t\to+\infty$ we have again
$$
\limsup_{t\to\infty}e^{B_t-\frac{t}{2}}=0\,.
$$
This implies $\lim_{t\to\infty}X_t=0$ because $X_t$ is non-negative.
Previous attempts gave partial answers. I leave them here because of the techniques being used which might be interesting in themselves:

*

*The limit $X_\infty$ exists. Proof. The martingale $X$ satisfies the SDE
$$
X_t=1+\int_0^tX_s\,dB_s\,.
$$
Writing $T_t=\int_0^tX_s^2\,ds$ for the quadratic variation of this stochastic integral we obtain a Brownian motion $W$ (different from $B$) such that $X$ is essentially a time changed BM:
$$\tag{1}
X_t=1+W_{T_t}\,.
$$
Since $X_t>0$ we see that $W_{T_t}>-1$ for all $t\,.$ This means that $T_t$ is bounded from above by the hitting  time
$$
\tau_{-1}=\inf\{t\ge 0: W_t=-1\}\,.
$$
This hitting time is known to be almost surely finite. Since $T_t$ is nondecreasing in $t$ it follows that
$$\tag{2}
T_\infty:=\lim_{t\to\infty}T_t=\sup_{t\ge 0}T_t\,\,\le\,\,\tau_{-1}<\infty\quad \mathbb P-a.s.
$$
Looking back at (1) this proves that $X_\infty$ exists.
$$\tag*{$\Box$}
\,
$$


*I have the feeling that the $\le$ in (2) can be shown to be an equals sign.
This would indeed mean that $X_\infty=0$ almost surely.


*The existence of the limit $X_\infty$ also follows directly from Doob's first martingale convergence theorem,  simply because $(X_t)_{0\le t<\infty}$ is a non-negative martingale. See also section 1.3 in [1] from which we can use Problem 3.16 to deduce that $(X_t)_{0\le t\le \infty}$ will be a supermartingale. In particular, for $t<\infty$, we get $E[X_\infty]\le E[X_t]\le 1\,.$ The book [1] contains many results around $X_\infty$ in that section but I haven't found anything that would allow to conclude $X_\infty=0\,.$
[1] I. Karatzas, S. Shreve, Brownian Motion and Stochastic Calculus.
