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I know that for standard Brownian motion, the total variation sampled at intervals of length $\Delta$ converges to $V(\Delta) = C \Delta^{-1/2}$ for some constant $C$. I wish to use this fact to study whether or not my data behaves as Brownian motion by calculating $V(\Delta)$ for many $\Delta$ in order to estimate $\beta$ in $V(\Delta) = C \Delta ^{-\beta}$, and verify that it is close to $\frac{1}{2}$.

One thing that I'm unsure about however, is how this total variation would behave if it is not standard Brownian motion. For example, if it is fractal Brownian motion with exponent $\alpha$, would I expect that my estimated $\hat{\beta}$ be close to $\alpha$ (or $1- \alpha$)? Or does this not apply to fractal brownian motion in general?

What is some good literature that discusses the behavior total variation of fractals (and specifically fractal Brownian motion)?

I have tried googling extensively but it always just leads me to normal Brownian motion.

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    $\begingroup$ I've more often heard it called "fractional Brownian motion", not "fractal", in case that helps your search queries. $\endgroup$ Jul 11, 2017 at 22:02
  • $\begingroup$ What do you mean by the "total variation" here? $\endgroup$
    – R W
    Jul 12, 2017 at 0:54
  • $\begingroup$ For a process $Z(t)$ we sample it at times $\Delta, 2\Delta,...n \Delta,...$ to partition the entire interval it's defined on. Then take the total variation as being equal to $V(\Delta) = \sum_i |Z(i \Delta) - Z((i-1)\Delta|$ $\endgroup$
    – Patty
    Jul 12, 2017 at 2:37
  • $\begingroup$ But then, for the standard Brownian, $V(\Delta)$ behaves as $\Delta^{-1/2}$, not as $\Delta^{1/2}$ ... $\endgroup$ Jul 12, 2017 at 13:11
  • $\begingroup$ Yes sorry that is a dumb typo on my part. Of course it is to a negative exponent since it increases as $\Delta$ decreases. $\endgroup$
    – Patty
    Jul 12, 2017 at 19:19

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I presume that you are considering sample paths on the time interval $[0,1]$, so that the number of samples is $n=1/\Delta$. The fractional Brownian motion has stationary increments, whence $$ \mathbf E V(\Delta) = n \mathbf E |Z(\Delta)| ;. $$ Since $Z(\Delta)$ is Gaussian with variance $\Delta^{2\alpha}$, $$ \mathbf E |Z(\Delta)| = C \Delta^\alpha $$ for a constant $C$, and therefore $$ \mathbf E V(\Delta) = C \Delta^{\alpha-1} \;. $$

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