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## Christopher Flynn

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# Blog

## Scale-invariant Hurst estimation

2019-03-28

I haven’t done much work in mathematics lately but I spent a lot of time working with fractional Brownian motion (fBm) when doing my PhD and when I was first learning to code. Fractional Brownian motion is a process which generalizes ordinary Brownian motion, a stochastic process with stationary independent Gaussian increments.

Fractional Brownian motion generalizes Brownian motion in such a way that its increments are no longer necessarily independent. The process $$(B^H_t, t \geq 0)$$ differs from Brownian motion in that its covariance function is

$$\mathbb{E}\left[B^H_t B^H_s\right] = \frac{1}{2}\left( s^{2H} + t^{2H} - |t-s|^{2H} \right)$$

for $$0 < t < s$$, where $$H \in (0, 1)$$ is known as the Hurst parameter. When $$H = \frac{1}{2}$$, the process is a Brownian motion with independent increments. If $$H \in (\frac{1}{2}, 1)$$, then the increments of the process are positively correlated. If $$H \in (0, \frac{1}{2})$$, then the increments of the process are negatively correlated.

A well-studied problem related to fBm is the estimation of the Hurst parameter from discrete samples of an fBm realization. One of the best estimators of $$H$$ uses the quadratic variation (or 2-variation) of samples $$\{B^H_{t_k}\}_{k=0}^n$$

$$V_n = \frac{1}{n}\sum_{k=1}^n \left( B_{t_k}^H - B_{t_{k-1}}^H \right)^2$$

which converges to $$(\Delta t)^{2H}$$ as $$n \to \infty$$ where $$\Delta t = t_k - t_{k-1}$$. This is a consequence of ergodic theory, specifically the Birkhoff-Khinchin theorem. The result leads to the estimator

$$\widehat{H} = \frac{\log (V_n)}{2 \log (\Delta t)}$$

This works well for standard fBm, but is not useful when the fBm process is scaled with a coefficient. For scaled fBm we can use a direction-change estimator, which I stumbled upon as a graduate student writing my dissertation (it’s included as an appendix). Here is an abridged derivation.

### Direction change estimator

Consider a discretely observed fBm process $$\{B^H_{t_k}\}_{k=0}^n$$ and let $$Z_k = B^H_{t_{k+1}} - B^H_{t_k}$$ be the fractional Gaussian noise (fGn) sequence for $$k = 0, \ldots, n-1$$. Let $$p$$ be the proportion of fGn increments which change direction from the previous increment:

$$p = \frac{1}{n-1} \sum_{k=1}^{n-1} \mathbf{1}_{{\text{sgn}(Z_k) \neq \text{sgn}(Z_{k-1})}}(k)$$

It has expected value

$$\mathbb{E}[p] = \frac{1}{n-1} \sum_{k=1}^{n-1} \mathbb{P}\left( \text{sgn}(Z_k) \neq \text{sgn}(Z_{k-1}) \right)$$

By symmetry, the probability terms are

$$\mathbb{P}\left( \text{sgn}(Z_k) \neq \text{sgn}(Z_{k-1}) \right) = \mathbb{P}\left( Z_k < 0 \mid Z_{k-1} > 0 \right)$$

The autocovariance of fGn is given by

$$\gamma_H(k) = \frac{1}{2} \left[ |k-1|^{2H} - 2k^{2H} + |k+1|^{2H} \right]$$

For $$k$$ and $$k+1$$ corresponding to the respective times $$t = k$$ and $$t = k+1$$, the unconditioned distribution of any $$Z_k = B^H_{t_{k+1}} - B^H_{t_k} \sim \mathcal{N}(0,1)$$. Thus, the multivariate distribution of $$Z_k$$ and $$Z_{k-1}$$ is

$$\left( \begin{array}{c} Z_k \\ Z_{k-1} \end{array} \right) \sim \mathcal{N}\left[\left(\begin{array}{c} 0 \\ 0 \end{array}\right) ,\left( \begin{array}{cc} \gamma_H(0) & \gamma_H(1) \\ \gamma_H(1) & \gamma_H(0) \end{array} \right)\right]$$

where $$\gamma_H(0) = 1$$ and $$\gamma_H(1) = 2^{2H-1} - 1$$.

Then it follows that the conditional distribution is

$$Z_k \mid Z_{k-1} \sim \mathcal{N}\left(\gamma_H(1) Z_{k-1}, 1-\gamma_H(1)^2\right)$$

Based on this result, the direction change probabilities can be written by the integral

$$\mathbb{P}\left( Z_k < 0 \mid Z_{k-1} > 0 \right) = \frac{1}{2\pi} \int_0^\infty \int_{-\infty}^{-\frac{\gamma_H(1) x}{\sqrt{1-\gamma_H(1)^2}}} \exp\left( -\frac{y^2+x^2}{2} \right) \,\text{d} y \,\text{d} x$$

Integration using polar coordinate transformation gives

$$\mathbb{P}\left( Z_k < 0 \mid Z_{k-1} > 0 \right) = \frac{1}{2} + \frac{1}{\pi} \arctan \left( - \frac{\gamma_H(1)}{\sqrt{1-\gamma_H(1)^2}} \right)$$

This leads to the result (after summation) of

$$\mathbb{E}[p] = \frac{1}{2} + \frac{1}{\pi} \arctan \left( - \frac{\gamma_H(1)}{\sqrt{1-\gamma_H(1)^2}} \right)$$

If we estimate $$\mathbb{E}[p]$$ with $$p$$ and solve for $$\gamma_H(1)$$ we get

$$\gamma_H(1) = \frac{-A_p}{\sqrt{1 + (-A_p)^2}}$$

where

$$A_p = \tan\left( (p-\frac{1}{2})\pi \right)$$

Solving for $$H$$ gives us the estimator

$$\widehat{H}^d = \frac{1}{2} \left[ \frac{\log \left( -\frac{A_p}{\sqrt{1 + (-A_p)^2}} + 1 \right)}{\log 2} + 1 \right]$$

which we bound by the interval $$[0, 1]$$. This estimator depends only on $$p$$ (the proportion of increments of the fBm realization where a direction change occurs from the previous increment)!

This result also provides some interesting insight on the behavior of fBm direction changes as $$H$$ approaches 0 and 1.

As $$H \to 0$$, the value of $$\gamma_H(1) \to -\frac{1}{2}$$. Therefore,

$$\lim_{H \to 0} \mathbb{E}[p] = \frac{2}{3}$$

Similarly, for $$H \to 1$$, the value of $$\gamma_H(1) \to 1$$, so

$$\lim_{H \to 1} \mathbb{E}[p] = 0$$

$$H$$