I'm attempting Exercise 5.33 of Le Gall's Brownian motion, Martingales and Stochastic Calculus.


Let $B\_t$ be a 3-dimensional Brownian motion starting from $x$.


Part 6 asks me to show that
$$|B\_t| = |x| + \beta\_t +\int\_0^t\dfrac{ds}{|B\_s|} \quad (\*)$$ 


where $$\beta\_t = \sum\_{i=1}^3 \int\_0^t \dfrac{B^i\_s}{|B^i\_s|} dB^i\_s$$.


I have shown this by Ito's formula, and I have shown that $\beta\_t$ is a 1-dimensional Brownian motion (by calculating the quadratic variation and using Levy's characterisation).


Now part 7 says 


Show that $|B\_t| \rightarrow \infty$ as $t \rightarrow \infty$ a.s. (Hint observe that $|B\_t|^{-1}$ is a non-negative supermartingale.)


So I'm not sure show to show the transience. I feel like it could follow from (\*), since if $|B\_t|$ does not tend to infinity then the integral on the right hand side must tend to infinity, but then $|B\_t|$ must tend to infinity to balance this (not sure how to make this rigorous).


My other idea is to say since $|B\_t|^{-1}$ is an $L^1$ bounded supermartingale it converges a.s., and to show $|B\_\infty|^{-1} = 0$, but again, I'm not sure exactly how to do this.


Thanks in advance.



Transience of 3-dimensional Brownian motion
You're on the right track near the end. As a non-negative supermartingale, $|B\_t|^{-1}$ converges almost surely; call the limit $X$. On the event $\{X \ne 0\}$, we have $|B\_t|$ converging to the finite limit $1/X$. But intuitively it is absurd for a Brownian motion to do that (it is trying to "wiggle", not "settle down") and so you should be able to show that the probability of $|B\_t|$ converging to a finite limit is 0. 


As one way to show this, you presumably know that for a one-dimensional Brownian motion $b\_t$, we have $\limsup\_{t \to \infty} b\_t = +\infty$ and $\liminf\_{t \to \infty} b\_t =-\infty$ almost surely. Since $|B\_t| = \sqrt{(b\_t^1)^2 + (b\_t^2)^2 + (b\_t^3)^2} \ge |b\_t^1|$, we see that $\limsup\_{t \to \infty} |B\_t| = +\infty$ almost surely. 


