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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.

Calculon
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    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 '15 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. – Calculon May 16 '15 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 '15 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. – Calculon May 16 '15 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 '15 at 15:18
  • @saz thank you very much. – Calculon May 16 '15 at 15:20
  • @saz sorry to bother you again. Could you perhaps give me an outline of the proof of the idea you mentioned? something like a rough sketch if it won't be too much trouble. – Calculon May 16 '15 at 17:24
  • @Calculon Actually, that's what I did in my answer below ... is there anything wrong about it? – saz May 16 '15 at 17:32
  • @saz Oh no, for some reason your answer didn't appear in my browser. I am just seeing it now. – Calculon May 16 '15 at 17:45
  • @Did In your proof, you seem to use the reflection principle. Doesn't that mean you implicitly use that any first passage time is finite almost surely? So implicitly what we want to prove? – mathsquestion88 Jan 17 '17 at 15:21
  • @mathsquestion88 I am not quite sure how the reflection principle is supposed to require that any first passage time is almost surely finite, nor how the strategy outlined in my comment uses the reflection principle. If you are mentioning this because of the identity in distribution $M_1=|B_1|$, note that it can easily be bypassed since the goal here is to justify that $P(T<1)=1$ where $T=\inf{t\mid B_t>0}$ is the entrance time in $(0,\infty)$. To show this, several approaches are available, for example, one can show that $P(T<1)=P(T<\infty)$, by homogeneity, and discretize $(B_t)$. – Did Jan 17 '17 at 15:56
  • @Did Alright. Thanks. – mathsquestion88 Jan 17 '17 at 16:06

1 Answers1

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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.

saz
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  • 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 $E\left[1_{{\tau_b < \infty}}\exp{\left(\xi B_{T\wedge \tau_b}- \frac{1}{2}\xi^2(T\wedge \tau_b)\right)}\right]$? – Calculon May 16 '15 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?

     – Calculon May 16 '15 at 18:27
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    @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 '15 at 18:46
  • thanks a lot for your help. – Calculon May 16 '15 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? – Charlie Shuffler Dec 13 '18 at 20:12
  • @S.Crim Note that the existence of $n \in \mathbb{N}$ such that $W(n)>W(t)$ does not imply $\limsup_{t \to \infty} W(t)=\infty$... consider for instance $W(t) := (1+1/t)^t$. – saz Dec 13 '18 at 20:26
  • @saz Hmm.. Then I guess I have some things mixed up. I was under the impression that sequences/processes without an upward bound automatically had that the limsup of the sequence was equal to infinity. – Charlie Shuffler Dec 13 '18 at 20:40
  • @S.Crim Well, perhaps there are typos in your previous comment ... but the way it' s currntly written, it does not guarantee $\limsup_t W_t = \infty$. – saz Dec 13 '18 at 20:42
  • @saz Alright, I'll make sure to look into it again. Thanks for the quick responses. I appreciate the help as always! – Charlie Shuffler Dec 13 '18 at 20:44