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Can the bivariate Hurst exponent be higher than an average of the separate Hurst exponents?

Data Analysis, Statistics and Probability 2018-10-30 v1

Abstract

In this note, we investigate possible relationships between the bivariate Hurst exponent HxyH_{xy} and an average of the separate Hurst exponents 12(Hx+Hy)\frac{1}{2}(H_x+H_y). We show that two cases are well theoretically founded. These are the cases when Hxy=12(Hx+Hy)H_{xy}=\frac{1}{2}(H_x+H_y) and Hxy<12(Hx+Hy)H_{xy}<\frac{1}{2}(H_x+H_y). However, we show that the case of Hxy>12(Hx+Hy)H_{xy}>\frac{1}{2}(H_x+H_y) is not possible regardless of stationarity issues. Further discussion of the implications is provided as well together with a note on the finite sample effect.

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Cite

@article{arxiv.1501.02947,
  title  = {Can the bivariate Hurst exponent be higher than an average of the separate Hurst exponents?},
  author = {Ladislav Kristoufek},
  journal= {arXiv preprint arXiv:1501.02947},
  year   = {2018}
}

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9 pages