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I tried to solve this problem in the following way.

Suppose $\{B_t | t\in [0,1]\}$ is our Brownian motion. Define,

$$f_n(w) = \sum_{k=1}^{2^n} \bigg|B_{\frac{k}{2^n}}(w)-B_{\frac{k-1}{2^n}}(w)\bigg|.$$

Firstly, I showed that $f_n \leq f_{n+1}$ using triangle inequality.

We have to show that $f_n \xrightarrow{a.s} \infty$.

For that I want to prove first the following,

$\mathbb{P}(f_n\geq\alpha)\rightarrow 1\ \forall \alpha > 0....(*)$. Then from the definition of almost sure convergence we can conclude our desired result.

But I couldn't able to show the $(*)$. How to show that?

Cantor_Set
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3 Answers3

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Let $Z$ be a standard normal, so that $E|Z|=\sqrt{2/\pi}$. $\;$Then $$E(f_n)= 2^{n/2}\cdot E|Z| = 2^{n/2} \cdot \sqrt{2/\pi} $$ and $$\text{Var}(f_n) \le 1 \,,$$ so by Chebyshev's inequality, $$P[f_n \le E(f_n)/2] \le \frac{4}{2^{n} \cdot 2/\pi}\,.$$

Yuval Peres
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  • What form of Chebyshev inequality you used? Isn't it inner inequality sign is reverse? – Cantor_Set Apr 14 '22 at 12:07
  • @Cantor_Set: Notice that If $m=E[Y]>0$, then ${Y<m/2}={Y-m<-m/2}\subset{|Y-m|>m/2}$. YuvalPeres then applies the usual Chebyshev inequality. – Mittens Apr 14 '22 at 12:20
  • @OliverDíaz yeah yeah got it!...nice solution – Cantor_Set Apr 14 '22 at 12:21
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    @Cantor_Yes: Yuval's was a nice solution. Notice that by Borel-Cantelly, it follows that $P[{f_n\leq E[f_n]/2,, \text{i.o.}}]=0$ and so, for a.a. $\omega\in \Omega$, there is $n_\omega\in\mathbb{N}$ such that $f_n(\omega)>E[f_n]/2$ for all $n\geq n_\omega$. – Mittens Apr 14 '22 at 12:25
  • Isn't it $Var(f_n)= 2^n Var(|Z|)$? @OliverDíaz – Cantor_Set Apr 14 '22 at 12:31
  • @Cantor_Set: no, it is not. Notice $X_j:=B_{j2^{-n}}-B_{(j-1)2^{-n}}$, $1\leq j\leq n$ are iid Normal with mean $0$ and variance $2^{-n}$ thus $var(f_n)=1$ indeed. – Mittens Apr 14 '22 at 13:35
  • @OliverDíaz yeah yeah got it... actually he wrote that $Z$ is standard normal. That's why ... it's my bad! – Cantor_Set Apr 14 '22 at 13:56
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    @Cantor set, I hope it is clear now. Let me know if not. Thanks to Oliver Diaz for clarifying. Do you know how to accept an answer? – Yuval Peres Apr 14 '22 at 17:31
  • It looks to me that $var(f_n)<1$ instead of $=1$, am I wrong? – puppyHuazai Jul 26 '22 at 14:06
  • Right, corrected. – Yuval Peres Jul 26 '22 at 21:43
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Brownnian motion is continuous and has positive finite quadratic variation, in fact $$\lim_n\sum|B_{t^n_{j-1}}-B_{t^n_j}|^2=T$$ where the supremum is taken partitions $P_n\subset P_{n-1}$ of $[0,T]$ with $\operatorname{mesh}(P_n)\xrightarrow{n\rightarrow\infty}0$. This implies that Brownian motion has infinite variation over $[0,T]$. This follows from a slight adaptation to the following Lemma

Lemma: Suppose $f\not\equiv0$ is continuous on $[a,b]$ and of finite variation, that is $V_f[a,b]=\sup_{P}\sum^{n_p}_{k=1}|f(x_k)-f(x_{k-1})|<\infty$, where supremum is taken over all partitions $P$ of $[a,b]$. Then $$V^2_f[a,b]:=\lim_{\|P\|\rightarrow0}\sum^{n_P}_{k=1}|f(x_k)-f(x_{k-1})|^2=0$$ where as before, the supremum is taken over partitions $P$ of $[a,b]$ such that $\|P\|\rightarrow0$.

A detailed proof of the facts mentioned in my posting can be found in Protter, P. E., Stochastic Integration and Differential Equations, 2nd edition, Springer-Verlag Berlin Heidelberg 2004, pp. 18-19.

Mittens
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Hints: Let $(X_i)$ be i.i.d. standard normal variates. Then $f_n$ has the same distribution as $2^{-n/2} \sum\limits_{k=1}^{2^{n}} |X_i|$. By CLT we have $f_n-2^{n/2}E|X_1| \to \sqrt {Var (|X_1|)}X_1$ in distribution. Thus, $f_n \to \infty$ in probability. Since $(f_n)$ is increasing it follows that $f_n \to \infty$ almost surely.

Kavi Rama Murthy
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