# Brownian motion at infinity

This is probably a standard exercise in stochastic calculus but I haven't been able to come up with a proof that relies only on a given set of results.

So my question is about proving the following statement. $B$ denotes the standard Brownian motion here. $$\limsup_{t\rightarrow\infty} B_t = \infty \qquad \text{almost surely}$$ The only tools that I have are the Borel-Cantelli lemmas.

I played with sequences of events such as $E_n=\{B_{n+1}-B_n > g(n)\}$, $E_n=\{B_n > f(n)\}$ etc. for some functions $f$ and $g$ but couldn't get the result above.

• Otherwise, by continuity of the paths of $B$, the maximum $M_\infty$ is finite with positive probability. But $M_\infty\geqslant M_t$ almost surely and, by scaling, $M_t=\sqrt{t}M_1$ in distribution hence $P(M_\infty<\infty)\leqslant P(M_1=0)$. Finally, $M_1=|B_1|$ in distribution hence $P(M_1=0)=0$, QED. (Note that several steps in this proof can be solved differently.)
– Did
May 16, 2015 at 13:45
• @Did Thanks but for this result I would have to derive some properties of $M$, which I am not supposed to do either (according to the lecturer). The main reason is that we entirely skipped the Markov property of BM and hence the running maximum of BM. He claims that application of Borel-Cantelli gives the result in the question. May 16, 2015 at 13:52
• Well, but you do know more tools than the BC lemma, don't you? Doesn't make sense to me to prove this using exclusively BC lemma only because your lecturer mentioned this result. (For example there is a nice proof using the martingale $\exp(B_t-t^2/2)$; the ingredients are optional stopping and monotone convergence theorem)
– saz
May 16, 2015 at 14:35
• @saz Do you mean $\exp{B_t-\frac{t}{2}}$? I was just working on another question which happens to be about the martingale you mentioned. But I hadn't realized that that also gives the result I want. So martingale convergence theorem gives me an integrable random variable that this martingale you proposed converges to almost surely. But how do I conclude from this that $E[\exp{B_{\infty}}]$ is either 0 or $\infty$? Just saw your edit. I will try the method you proposed. May 16, 2015 at 14:49
• @Calculon There is no such thing as $B_{\infty}$ - the Brownian motion keeps oscillating. See my answer for some more details. (Don't hesitate to ask if you don't get along with it.)
– saz
May 16, 2015 at 15:18

Hints: (This answer does not use Borel Cantelli lemma; instead it is based on basic martingale techniques.)

1. Show that for any fixed $\xi>0$, the process $$M_t^{\xi} := \exp \left( \xi B_t - \frac{1}{2} \xi^2 t \right), \qquad t \geq 0,$$ defines a martingale.
2. Fix $T>0$. For $b>0$ we define a stopping time by $\tau_b := \inf\{t>0; B_t \geq b\}$. Applying the optional stopping theorem to $(M_t^{\xi})_{t \geq 0}$ and the bounded stopping time $\tau_b \wedge T$ yields $$1 = \mathbb{E}\exp \left( \xi B_{T \wedge \tau_b} - \frac{1}{2} \xi^2 ( T \wedge \tau_b) \right).$$ Using the dominated convergence theorem, conclude that $$1 = e^{\xi b} \mathbb{E}(1_{\{\tau_b<\infty\}} e^{-\frac{1}{2} \xi^2 \tau_b}).$$
3. Letting $\xi \downarrow 0$, show that step 2 implies $$\mathbb{P}(\tau_b<\infty)=1$$ using the monotone convergence theorem.
4. Conclude from $$\left\{\limsup_{t \to \infty} B_t = \infty \right\}^c \subseteq \bigcup_{N=1}^{\infty} \{\tau_N = \infty\}$$ that $$\mathbb{P} \left( \left\{ \limsup_{t \to \infty} B_t = \infty \right\}^c \right) = 0.$$

Remark: As @Did pointed out, the claim follows also easily from the reflection principle.

• Thanks for your detailed answer. I have a few questions. When you use DCT in step 2, is this what you had in mind: $$1_{\{\tau_b < \infty\}}\exp{\left(\xi B_{T\wedge \tau_b}- \frac{1}{2}\xi^2(T\wedge \tau_b)\right)} \rightarrow 1_{\{\tau_b < \infty\}}\exp{\left(\xi B_{\tau_b}- \frac{1}{2}\xi^2\tau_b\right)}$$ as $T\rightarrow\infty$. But we only know $E\left[\exp{\left(\xi B_{T\wedge \tau_b}- \frac{1}{2}\xi^2(T\wedge \tau_b)\right)}\right] = 1$. So how do you handle $E\left[1_{\{\tau_b < \infty\}}\exp{\left(\xi B_{T\wedge \tau_b}- \frac{1}{2}\xi^2(T\wedge \tau_b)\right)}\right]$? May 16, 2015 at 18:22
• Regarding the use of MCT in step 3, we have $$1_{\{\tau_b < \infty\}}e^{-\frac{1}{2}\xi^2\tau_b} \nearrow 1_{\{\tau_b < \infty\}}$$ Since $E\left[1_{\{\tau_b < \infty\}}e^{-\frac{1}{2}\xi^2\tau_b}\right]=e^{-\xi b}$, $P\{\tau_b < \infty \} = \lim_{\xi \rightarrow 0} e^{-\xi\tau_b} = 1$ Is this how you would do it? May 16, 2015 at 18:27
• @Calculon Step 2: You want to know how to deal with $\mathbb{E}(1_{\{\tau_b = \infty\}} \dots)$, right? Note that $$\exp \left(\xi B_{T \wedge \tau_b}(\omega) - \frac{1}{2} \xi^2 (T \wedge \tau_b(\omega))^2 \right) \leq e^{\xi b} \exp \left(- \frac{1}{2} \xi^2 (T \wedge \tau_b(\omega))^2 \right) \stackrel{T \to \infty}{\to} 0$$ for any $\omega \in \{\tau_b = \infty\}$. (Here, we have used that $B_{T \wedge \tau_b} \leq b$; hence $\exp(\xi B_{T \wedge \tau_b}) \leq e^{\xi b}$ for any $\xi>0$.) Step 3: Yeah, that's correct.
– saz
May 16, 2015 at 18:46
• thanks a lot for your help. May 16, 2015 at 19:11
• @saz Hope I can revivie this question. I was wondering if we couldn't just solve this exercise by saying that $\forall k \in \mathbb{N}$ we know that $\exists n \in \mathbb{N}$ such that $W(n)>W(t)$ and thus do we know that $\{W(t),t\geq 0 \}$ is an unbounded process and thus: $\limsup_{t\rightarrow \infty} W(t) = \infty$. Does this fail because we would first have to prove that for each $k$ there exists an $n$ such that the above condition holds or is this also correct? Dec 13, 2018 at 20:12