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I have just finished proving that the quadratic variation of any Brownian motion on $[0,t]$ is $t$. That is if $\mathcal{P}$ is a partition of $[0,t]$ then

$$ \lim_{\Delta t\to 0}\sum_{t_k \leq t} |B_{t_{k+1}}- B_{t_k}|^2 = t \hspace{8mm} \text{in }\hspace{4mm} L^2$$

In Stochastic Differential Equations by Oksendal, it states that if quadratic variation of a stochastic process is a.s. positive, then the total variation of the process is almost surely $\infty$. Where does this fact come from? Is the proof very intensive? It doesn't seem to appear on the wikipedia pages for https://en.wikipedia.org/wiki/P-variation, https://en.wikipedia.org/wiki/Total_variation or https://en.wikipedia.org/wiki/Quadratic_variation#Finite_variation_processes.

For my specific problem I have been able to show that $$ \mathbb{E}\left[\sum_{t_k\leq t} |B_{t_{k+1}}- B_{t_k}|\right] = \sum_{t_k\leq t} \sqrt{\Delta t_k}$$ from properties of Brownian motion by noting that $B_{t_{k+1}}-B_{t_k}\sim \mathcal{N}(0, \Delta t_{k})$ and a a property of the expected value of the absolute value of normally distributed random variable (https://en.wikipedia.org/wiki/Normal_distribution#Moments). The right hand side can be shown to diverge, but this doesn't tell us about any of the sample paths $\sum_{t_k\leq t} |B_{t_{k+1}}(\omega)- B_{t_k}(\omega)|$.

Andrew Shedlock
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1 Answers1

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This is a general property of continuous functions.

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 $$V^2_f[a,b]:=\lim_{\|P\|\rightarrow0}\sum^{n_P}_{k=1}|f(x_k)-f(x_{k-1})|^2=0$$ whereto limit is taken over all partitions $P=\{a=x_0<\ldots<x_n=b\}$ of $[a,b]$ such that $\|P\|=\max_k(x_k-x_{k-1})\rightarrow0$.

Here is a short proof:

By uniform continuity, given $\varepsilon>0$, there is $\delta>0$ such that $$|f(x)-f(y)|<\varepsilon\quad\text{whenever}\quad|x-y|<\delta$$

For any partition $P=\{a=t_0<\ldots <t_{n_P}=b\}$ of $[a,b]$ such that $\max_{1\leq k\leq n_P}(x_k-x_{k-1})<\delta$,

$$ \begin{align} V^2_f[a,b]&:=\sum^{n_P}_{k=1}|f(x_k)-f(x_{k-1})|^2\leq\max_{1\leq j\leq n}|f(x_j)-f(x_{j-1})|\sum^{n_P}_{k=1}|f(x_k)-f(x_{k-1})|\\ &\leq\Big(\max_{1\leq j\leq n_P}|f(x_j)-f(x_{j-1})|\Big)\,V_f[a,b]<\varepsilon\,V_f[a,b] \end{align} $$ Consequently $$ V^2_f[a,b]=\lim_{\|P\|\rightarrow0}\sum^{n_P}_{k=1}|f(x_k)-f(x_{k-1})|^2=0$$

From the Lemma above, if $f$ is continuous function of finite positive quadratic variation $V^2_f[a,b]$, then $V_f[a,b]=\infty$

Mittens
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  • One note on your proof, since $V_f^2[a,b]$ is the supremum we don't know of the existence of a partition. But we could definitely find a partition such that $V_f^2[a,b] \leq \epsilon + \sum_{k=1}^{n_P} |f(x_k) - f(x_{k-1})|^2$. It doesn't change the result, just double checking – Andrew Shedlock Jul 26 '20 at 02:47
  • @AndrewShedlock: The supremum is taken over all partitions $P$ of $[a,b]$. Of course, if $\beta< V^2_f[a,b]$, then there is a partition $P$ of $[a,b]$ such that $\beta<\sum^{n_P}{k=1}|f(x_j)-f(x{j-1})|^2$. – Mittens Jul 26 '20 at 03:02
  • To then finish my original question, since we know that almost all paths of a brownian are continuous and that a.e. path has positive quadratic variation. Hence a.e. path must have infinite total variation. – Andrew Shedlock Jul 26 '20 at 03:05
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    @AndrewShedlock: yes. typically paths of Brownian motion are continuous and so that have infinite variation on any bounded interval $[a,b]$. There are other Levy processes and more generally , local martingales with that property. That is one reason stochastic integration was developed. Lebesgue-Stielthes integration only handles finite variation drivers. – Mittens Jul 26 '20 at 03:09