{"paper":{"title":"Can the bivariate Hurst exponent be higher than an average of the separate Hurst exponents?","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"physics.data-an","authors_text":"Ladislav Kristoufek","submitted_at":"2015-01-13T10:36:46Z","abstract_excerpt":"In this note, we investigate possible relationships between the bivariate Hurst exponent $H_{xy}$ and an average of the separate Hurst exponents $\\frac{1}{2}(H_x+H_y)$. We show that two cases are well theoretically founded. These are the cases when $H_{xy}=\\frac{1}{2}(H_x+H_y)$ and $H_{xy}<\\frac{1}{2}(H_x+H_y)$. However, we show that the case of $H_{xy}>\\frac{1}{2}(H_x+H_y)$ 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."},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1501.02947","kind":"arxiv","version":1},"verdict":{"id":null,"model_set":{},"created_at":null,"strongest_claim":"","one_line_summary":"","pipeline_version":null,"weakest_assumption":"","pith_extraction_headline":""},"references":{"count":0,"sample":[],"resolved_work":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57","internal_anchors":0},"formal_canon":{"evidence_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"author_claims":{"count":0,"strong_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"builder_version":"pith-number-builder-2026-05-17-v1"}