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.


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.

  • $\begingroup$ Thanks, based on your answer I've tried to write a complete proof. Do you think it is legitimate? $\endgroup$ – AlexanderR Mar 16 '20 at 19:13
  • $\begingroup$ By the way, if you happen to have the book and have looked at the question, I was wondering if you know what the purpose of the first few questions is? It seems like this result about the transience of three dimensional Brownian motion can actually be proved quite easily and without much preamble. In question 8 it shows that this $|B_t|^{-1}$ process is a strict local martingale, but that also doesn't need much preamble, so I am wondering if I am missing something. For example what is the point of the representation (*) in my question? Thank you so much for your help. $\endgroup$ – AlexanderR Mar 16 '20 at 19:32
  • $\begingroup$ @AlexanderR: (I got ahold of the book.) It seems to be just a collection of useful facts about multidimensional Brownian motion, not all necessarily leading up to the transience. Though of course parts 4 and 5 are certainly needed to show that $|B_t|^{-1}$ is a supermartingale. $\endgroup$ – Nate Eldredge Mar 16 '20 at 23:20
  • $\begingroup$ Ah okay that wasn't obvious to me. The fact that they said 'observe' made me think it was clear to see directly, and I thought I had a proof just by applying Jensen inequality twice with $1/x$ and $|x|$, but I realise I had one of the inequalities the wrong way round. Could you give me a clue as to how questions 4 and 5 help here? Obviously if we could upgrade the 'continuous martingale' property from question 4 to 'martingale' that would do it, but we know that's impossible. Other than that, I'm not very familiar with the relationship between supermartingale and local martingales. Thanks! $\endgroup$ – AlexanderR Mar 16 '20 at 23:44
  • $\begingroup$ @AlexanderR: Part 4 shows that $|B_{t \wedge T_\epsilon}|^{-1}$ is a continuous local martingale. Indeed, it's bounded by $1/\epsilon$ so it's a continuous nonnegative martingale. That is, for any $s<t$ and any $0<\epsilon<|x|$ we have $$E[|B_{t \wedge T_\epsilon}|^{-1} \mid \mathscr{F}_s] = |B_{s \wedge T_\epsilon}|^{-1} \tag{*}.$$ Now from part 5 you can conclude that $T_\epsilon \to \infty$ as $\epsilon \to 0$, almost surely. So let $\epsilon \to 0$ in (*) and apply Fatou's lemma. This shows that $|B_t|^{-1}$ is a supermartingale. $\endgroup$ – Nate Eldredge Mar 17 '20 at 0:41

So would this be an answer?:

Since $|B_t|^{-1}$ is a non-negative supermartingale bounded in $L^1$, we have for some non-negative $Y \in L^1$,

$$|B_t(\omega)|^{-1} \rightarrow Y(\omega)$$ for almost all $\omega \in \Omega$. But since $\limsup |B_t| = \infty$, $\liminf |B_t(\omega)|^{-1} = 0$. But also, since $|B_t(\omega)|^{-1}$ converges, $\liminf |B_t(\omega)|^{-1} = \lim |B_t(\omega)|^{-1} = Y$. So $Y=0$ a.s. So $\lim |B_t| = \infty$ a.s.

  • $\begingroup$ Yes, that works. $\endgroup$ – Nate Eldredge Mar 16 '20 at 20:14

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