# lim sup and lim infs of Brownian Motion: $B_t/\sqrt{t}$ as $t \to \infty$ or as $t \to 0$.

Below is my question. Q7.9 is what I'm stuck on. I've done Q7.8; I included it in the picture because I'll use it in Q7.9, and it gives a definition that I'll use.

Update: This question is now solved, and I've added the details below.

What I've done so far is this:

By using time-inversion, $(tB_{1/t})_{t\ge0}$, and $(-B_t)_{t\ge0}$, sign-inversion of Brownian motion, we have that the four random variables in question are all equal to each other. Further, we see that the first is $\mathcal{F}_{0^+}$ measurable, since

$$\limsup_{t \to 0} \frac{B_t}{\sqrt{t}} = \lim_{s \to 0} \sup_{t \le s} \frac{B_s}{\sqrt{s}}, \ \text{and} \ \sup_{t \le s} \frac{B_s}{\sqrt{s}}$$

is $\mathcal{F}_s$ measurable for all $s > 0$, and thus the limit is $\mathcal{F}_{0^+}$ measurable. Hence by Blumenthal's $0$-$1$ law, it is almost surely constant. Hence we now have that the four random variables in question are equal to each other and almost surely constant. Since $\mathbb{P}(B_{t'} > 0) = 1/2$ for all $t' > 0$, by the Markov property, we have that the almost sure constant must be at least $0$.

I'm stuck on showing that this constant is in fact $+\infty$. I wanted to use the scaling property, $(cB_{t/c^2})_{t\ge0}$, of Brownian motion, but the issue is that this gives

$$\frac{cB_{t/c^2}}{\sqrt{t}} = \frac{B_{t/c^2}}{\sqrt{t/c^2}} = \frac{B_s}{\sqrt{s}}$$

where $s = t/c^2$. When considering just $B_t$ or $B_t/t$, we get a factor $c$ or $1/c$ out the front, and so, since this must hold for all $c$, we know that it must be $0$ or $\infty$. (We can then show which it is.) However, we don't get this nice property when using $B_t/\sqrt{t}$.

Solution: Define $A^x_t$ and $A^x$ as follows: $$A^x_t = \left\{ \sup_{s \le t} \frac{B_s}{\sqrt{s}} \le x \right\}, ~ A^x = \left\{ \lim_{t \downarrow 0}\sup_{s \le t} \frac{B_s}{\sqrt{s}} \le x \right\} = \left\{ \limsup_{t \downarrow 0} \frac{B_s}{\sqrt{s}} \le x \right\}.$$ Observe that $B_t/\sqrt{t} \sim N(0,1)$. Thus, since $\sup_{s \le t} {B_s}/{\sqrt{s}} \ge {B_t}/{\sqrt{t}}$, $$P \left( \sup_{s \le t} \frac{B_s}{\sqrt{s}} \le x \right) \le P \left( \frac{B_t}{\sqrt{t}} \le x \right) = \Phi(x),$$ where $\Phi$ is the cdf for the standard normal. We want to show that $P(A^x) = 0$ for every $x \in \mathbb{R}$ ($\therefore x \neq \infty$); by Blumenthal's $0$-$1$ law, it is enough to show that $P(A^x) < 1$. Now, $A^x_{1/n} \downarrow A^x$ as $n \to \infty$, so by monotone convergence, $$P(A^x) = \lim_{n \to \infty}P(A^x_{1/n}) \le \Phi(x) < 1, \ x \in \Bbb R.$$ Thus $P(A^x) < 1$, ie $P(A^x) = 0$, for all $x \in \mathbb{R}$. Thus the almost sure constant must be $+\infty$.

Thank you to Jay.H for helping me with this!

## 2 Answers

Try to use the fact that $B_t/\sqrt{t}$ has the same distribution as $B_1$ and the fact that $\sup_{s\le t }B_s/\sqrt{s}$ is monotonic w.r.t. $t$

[Warning: more details below]

For any constant $C>0$, let $$A = (\lim_{t\to 0} \sup_{0<s\le t}\frac{B_s}{\sqrt{s}}<C )$$ We want to show that $P(A)=0$. Since $A\in {\mathcal F}_{0+}$, we only have to show that $P(A)<1$.

Let $$A_n = (\sup_{0<s\le \frac{1}{n}}\frac{B_s}{\sqrt{s}}<C)$$ Then $$P(A_n) \le P(\frac{B_{1/n}}{\sqrt{1/n}}<C) = P(B_1<C)$$ Since $A_n \uparrow A$, we have $$P(A) = \lim_{n\to \infty} P(A_n) \le P(B_1<C) <1$$ Done.

• Do you mean $\limsup_{t \to 0} \frac{B_t}{\sqrt{t}} = \lim_{ t \to 0} \sup_{s \le t} \frac{B_s}{\sqrt{s}}$? Commented Nov 28, 2015 at 16:38
• Ok, yeah, I hadn't thought about the fact that $B_t/\sqrt{t} \sim N(0,1)$. I'd like to interpret $$\limsup_{t \to 0} \frac{B_t}{\sqrt{t}} = \lim_{s \to 0} \sup_{t \le s} \frac{B_s}{\sqrt{s}}$$ as sampling for each $t \le s$, and then the result would follow; however, continuity doesn't allow that. But indeed $\sup_{s\le t }B_s/\sqrt{s}$ is monotonic with $t$, but it's decreasing as $t$ decreases. Thank you for your answer; I'll have a think about it. Commented Nov 28, 2015 at 16:39
• Ok, so I've got somewhere. It's not really based on what you said (!), but what you said made me think of it. I've put it in as "Update 1". Can you check it? It seems a bit dodgy to me! Commented Nov 28, 2015 at 17:11
• I am not convinced of the first equation, which involves some kind of mapping of the whole path in some interval. Can you elaborate why $$\mathbb{P}\left(\sup_{s \le t}\frac{B_s}{\sqrt{s}} = x\right) = \mathbb{P}\left(\sup_{s \le t}\frac{B_{s/c^2}}{\sqrt{s/c^2}} = x\right)$$ Commented Nov 28, 2015 at 17:54
• It uses the Brownian scaling property, $(\tilde{B}_t)_{t\ge0} = (cB_{t/c^2})_{t\ge0}$ is a Brownian motion if and only if $(B_t)_{t\ge0}$ is. Also, $$\frac{\tilde{B}}{\sqrt{t}} = \frac{cB_{t/c^2}}{\sqrt{t}} = \frac{B_{t/c^2}}{\sqrt{t/c^2}} = \frac{B_s}{\sqrt{s}}.$$ So in any expression exclusively about the distribution of the Brownian motion -- not the actual path itself -- we can replace $B$ by $\tilde{B}$. Commented Nov 28, 2015 at 17:59

I am just a student so I hope my answer is correct.

We want to prove that $$\lim_{t\rightarrow \infty } |\frac{W_t}{\sqrt{t}}|$$ is not finite.
Since we know by the law of iterated logarithm that $$\lim_{t\rightarrow \infty } |\frac{W_t}{\sqrt{2tlog(log(t)))}}|=1$$
Now: $$\lim_{t\rightarrow \infty } |\frac{W_t}{\sqrt{t}}|=\lim_{t\rightarrow \infty } |\frac{W_t}{\sqrt{2tlog(log(t)))}}||\sqrt{2log(log(t)))}|=1\cdot \infty =\infty$$
The two last equalities given by the simple limit arithmetic.

• Sure, this is correct, but you are merely stating that $W_t$ is approximately $\sqrt{2 t \log\log t}$, so is larger than $t$. You're assuming a much stronger statement than the one I want(ed) to prove. Eg, if I had some sequence $(x_n)_{n\ge1}$ with certain properties and ask you: "Show that $x_n \to 0$ as $n \to \infty$." I wouldn't accept as a solution, "We know that $x_n n \to 0$, so clearly $x_n \to 0$ too." Rather, you need to prove these statements assuming/using only weaker ones. Hopefully this is clear :) Commented Jul 14, 2023 at 16:06