# Proof that $p$-th total variation of a brownian motion is $0$ while $p>2$

The p-th total variation is defined as $$|f|_{p,TV}=\sup_{\Pi_n}\lim_{||\Pi_n||\to n}\sum^{n-1}_{i=0}|f(x_{i+1}-f(x_{i})|^p$$

And I know how to calculate the first total variation of the standard Brownian motion. But when dealing with high order TV, there are some problem.

At first we assume that p is even.

First I define $$\xi_i=|B_{\frac{i+1}{n}}-B_{\frac{i}{n}}|^p$$ then we can get that $$E[\xi_i]=\left(\frac1n\right)^{\frac p2}(p-1)!!$$ and $$E[\xi_i^2]=\left(\frac1n\right)^{p}(2p-1)!!$$

Next, we define $V_n=\sum^{n-1}_{i=0}\xi_i$

Then we have$$E[V_n]=\sum^{n-1}_{i=0}\left(\frac 1n\right)^{\frac p2}(p-1)!!$$

But there's something wrong in the following step, when calculating $E[V_n^2]$ \begin{align} E[V_n^2] &= E\left[\sum^{n-1}_{i=0}\xi_i\sum^{n-1}_{j=0}\xi_j\right]\\ &=E\left[\sum^{n-1}_{i=0}\sum^{n-1}_{j=0}\xi_i\xi_j\right]\\ &=\sum^{n-1}_{i=0}\sum^{n-1}_{j=0}E\left[\xi_i\xi_j\right]\\ &=\sum^{n-1}_{i=0}E[\xi_i^2]+\sum_{i\neq j}E[\xi_i]E[\xi_j]\\ &=\left(\frac1n\right)^{p-1}(2p-1)!!+n(n-1)\left(\frac1n\right)^{p}\left[(p-1)!!\right]^2 \end{align}

And then the question is I have no idea that how to deal with this awesome equation. Do I need to brute it out or if there is any method more efficient to prove it?

• Yes, there is some more efficient way to prove this. Do you know the total variation for $p=2$? (And what about $p$? Is $p \in (0,2)$ or $p>2$ or...? Please add this information to your question.) – saz Mar 5 '15 at 8:05
• @saz sorry I forgot that. p>2 in this case – zhshr Mar 5 '15 at 14:59
• I see. And what about $p=2$, do you know the total variation for this particular case? – saz Mar 5 '15 at 15:17
• @saz I only know the case for standard brownian motion but I have no idea how to apply that for this problem – zhshr Mar 5 '15 at 15:33

## 2 Answers

It is widely known that for $p=2$ the quadratic variation

$$S_{\Pi} := \sum_{t_i \in \Pi} |B_{t_{i+1}}-B_{t_i}|^2$$

converges to $T$ in $L^2$ as $|\Pi| \to 0$. Here $\Pi = \{0=t_0<\ldots< t_n \leq T\}$ denotes a partition of the interval $[0,T]$ with mesh size

$$|\Pi| = \max_i |t_{i+1}-t_i|.$$

For a proof of this statement see any book on Brownian motion (e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 9). Since $L^2$-convergence implies almost sure convergence of a subsequence, we may assume that

$$S^{\Pi_n} \to T \qquad \text{almost surely} \tag{1}$$

for a (given) sequence of partitions $(\Pi_n)_n$ satisfying $|\Pi_n| \to 0$. Now, as $p>2$,

\begin{align*} \sum_{t_i \in \Pi_n} |B_{t_{i+1}}-B_{t_i}|^p &= \sum_{t_i \in \Pi_n} |B_{t_{i+1}}-B_{t_i}|^{p-2} |B_{t_{i+1}}-B_{t_i}|^2 \\ &\leq \sup_{s,t \leq T, |s-t| \leq |\Pi_n|} |B_t-B_s|^{p-2} S_{\Pi_n}. \tag{2}\end{align*}

Since $(B_t)_{t \geq 0}$ is a Brownian motion, we know that $t \mapsto B_t(\omega)$ is continuous almost surely; hence, $[0,T] \ni t \mapsto B_t(\omega)$ is uniformly continuous. This implies in particular

$$\sup_{s,t \leq T, |s-t| \leq |\Pi_n|} |B_t-B_s| \stackrel{n \to \infty}{\to} 0 \qquad \text{almost surely}$$

as the mesh size $|\Pi_n|$ tends to $0$. Combining $(1)$ and $(2)$ yields

$$\lim_{n \to \infty} \sum_{t_i \in \Pi_n} |B_{t_{i+1}}-B_{t_i}|^p = 0.$$

This finishes the proof.

Remark: For a more general statement see this question.

• Ah got it. I knew how to prove quadratic variation is 0 with finite total variation. But what I was missing is \begin{align*} \sum_{t_i \in \Pi_n} |B_{t_{i+1}}-B_{t_i}|^p &= \sum_{t_i \in \Pi_n} |B_{t_{i+1}}-B_{t_i}|^{p-2} |B_{t_{i+1}}-B_{t_i}|^2 \\ &\leq \sup_{s,t \leq T, |s-t| \leq |\Pi_n|} |B_t-B_s|^{p-2} S_{\Pi_n}. \tag{1}\end{align*} It holds not only for p=2 – zhshr Mar 5 '15 at 16:27
• @user2841003 Yes. If you find the answer helpful, you can upvote/accept it by clicking on the tick/arrow next to it. – saz Mar 5 '15 at 16:54
• @saz doesn't this only show that the $p$-variation is 0 along a subsequence of $\Pi_n$? In other words, doesn't your proof only show convergence in probability to 0? – user217285 Jul 5 '18 at 4:09
• @Nitin Yeah, that's true... to get rid of this problem one could use that the Brownian motion has (with probability $1$) $\beta$-Hölder continuous sample paths for any $\beta<1/2$... choosing $\beta \in (1/p,1/2)$ should give the desired a.s. convergence of the total variation. – saz Jul 5 '18 at 17:45

Note that $$2$$-variation is not the same as quadratic variation. For the quadratic variation you take the limit as the partition gets finer, whereas for $$p$$-variation you take the supremum over all partitions. In particular, the Brownian motion has finite quadratic variation on any finite interval but infinite $$2$$-variation (link).

Now coming to your question and assuming the interval $$[0,1]$$, we have that $$\|B\|_{p-\text{var}} \geq |B_1|,$$ because the supremum is larger or equal to taking the partition that only contains $$0$$ and $$1$$.

So, the $$p$$-variation of Brownian motion for any $$p\geq 1$$ is clearly not converging to zero in any sense.