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Why is order flow so persistent?

[PDF] from arxiv.orgB Toth, I Palit, F Lillo… - Arxiv preprint arXiv:1108.1632, 2011 - arxiv.org
Equity order flow is persistent in the sense that buy orders tend to be followed by buy orders
and sell orders tend to be followed by sell orders. For equity order flow this persistence is extremely
long-ranged, with positive correlations spanning thousands of orders, over time intervals ...

Looking for an Explanation for the Low Sign Span. Is Order Involved?

M Gozzi, C Geraci, C Cecchetto… - Journal of deaf …, 2011 - Oxford Univ Press
Although signed and speech-based languages have a similar internal organization of verbal
short-term memory, sign span is lower than word span. We investigated whether this is due to
the fact that signs are not suited for serial recall, as proposed by Bavelier, Newport, Hall, ...
Cited by 3 - Related articles - All 6 versions

Looking for an explanation for the low sign span. Is order involved?

C CECCHETTO, C PAPAGNO… - Journal of Deaf …, 2011 - boa.unimib.it
Gozzi, M., Geraci, C., Cecchetto, C., Perugini, M., & Papagno, C. (2011). Looking for an explanation
for the low sign span. Is order involved?. Journal of Deaf Studies and Deaf Education, 16(1),
101-107. ... There are no files associated with this item. ... Items in DSpace are ...
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Insertions of k-semi-stratifiable spaces by semi-continuous functions

K Li, S Lin… - Studia Scientiarum Mathematicarum …, 2011 - akademiai.com
... Therefore, φ(h1) ≦ φ(h2) by the definition of φ. Hence φ is order- preserving. It remains to show
that the map φ satisfies the conditions (1) and (2) in Theorem. ... (x). So φ(h) ≦ φ(h ′ ). It implies
that φ is order-preserving. For each x ∈ X, there is an n ∈ N such that x ∈ U(n, ...

[PDF] Duality Property for Positive Weak Dunford-Pettis Operators

[PDF] from hindawi.comB Aqzzouz, K Bouras… - 2011 - downloads.hindawi.com
... 1 the norm of E is order continuous and E has the Dunford-Pettis property, 2 the topological dual
E is discrete, Page 3. ... 1 the norm of E is order continuous but E does not have the AM-compactness
property nor the weak Dunford-Pettis property. In fact, consider E L2 0, 1, the ...
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