on the frequencies of patterns of rises and falls
play

On the frequencies of patterns of rises and falls Jean-Marc Luck - PowerPoint PPT Presentation

On the frequencies of patterns of rises and falls Jean-Marc Luck Institut de Physique Th eorique, Saclay (CEA & CNRS) Preprint arXiv:1309.7764 Journ ees ALEA 2014, March 1721, 2014, CIRM, Marseille 1 Outline The problem


  1. On the frequencies of patterns of rises and falls Jean-Marc Luck Institut de Physique Th´ eorique, Saclay (CEA & CNRS) Preprint arXiv:1309.7764 Journ´ ees ALEA 2014, March 17–21, 2014, CIRM, Marseille 1

  2. Outline • The problem • Probabilistic approach • Motivations • Combinatorial approach • Periodic patterns: the probabilistic route Analytical result for entropy of all families of periodic patterns • Random patterns Numerical evidence for multifractal behavior 2

  3. The problem • Consider a data series e.g. daily temperature (average, min, max) at Paris Montsouris 16 14 12 Temperature 10 8 6 4 2 0 1 4 7 10 13 16 19 22 25 28 Day (February 2014) • What is the probability of observing a given pattern ? e.g. 3 consecutive rises in the max series 3

  4. Probabilistic approach • Data modelled as i.i.d. random variables x i • Distribution of the x i can be taken to be uniform on [ 0 , 1 ] If x i > x i − 1 , there is a rise at the i -th place: ε i = + If x i < x i − 1 , there is a fall at the i -th place: ε i = − • What is the probability P n ( ε 1 , ··· , ε n ) of observing a given pattern ε 1 , ··· , ε n of n rises and falls ? 4

  5. Recursive scheme to calculate P n ( ε 1 , ··· , ε n ) • Condition on last variable Let f n ( x ) d x = Prob { ε 1 , ··· , ε n x < x n < x + d x } and Z 1 So P n ( ε 1 , ··· , ε n ) = 0 f n ( x ) d x • Linear integral recursion relation (transfer operator) Z x If ε n = + , then f n ( x ) = 0 f n − 1 ( y ) d y Z 1 If ε n = − , then f n ( x ) = x f n − 1 ( y ) d y pattern ++ − Example: x 2 + f 2 ( x ) = x 2 f 3 ( x ) = 1 − x 3 x 1 f 1 ( x ) = x , 2 , 6 − + x 3 P 3 (++ − ) = 1 x 0 8 5

  6. Motivations • Applications Null model to which real data could be compared Recent work on microarray data in genetics (Fink et al 2007) • Results Alternating patterns yield P n ∼ ( 2 / π ) n (Andr´ e 1879, 1881) How generic is exponential law P n ∼ e − α n ? α has physical interpretation of an entropy α min = ln π 2 = 0 . 451582 ··· in spin chain (Derrida & Gardner 1986) Can α be calculated for all (periodic) families of patterns ? How is α distributed for long pattern chosen at random ? 6

  7. • Technical Reminiscent of calculation of partition function � � 1 z n = r 0 , 1 r 1 , 2 ··· r n − 1 , n of open chain of n + 1 points in unit 3-dim ball ( r i − 1 , i = | x i − x i − 1 | ) (with D. Boos´ Context: Multiple scattering of waves e and J.Y. Fortin) z 1 = 6 z 2 = 51 z 3 = 62 z 0 = 1 , 5 , 35 , 35 , x 5 x 1 z 4 = 4146 z 5 = 65532 x 2 1925 , 25025 √ � � Z ( x ) = ∑ z n x n = 1 tan 3 x x 4 √ − 1 x 0 x 3 x n ≥ 0 x 3 z n ∼ ( 12 / π 2 ) n 7

  8. Combinatorial approach • Data modelled as uniform random permutation σ on n + 1 objects { 0 , 1 , ··· , n } If σ i > σ i − 1 , there is a rise at the i -th place: ε i = + If σ i < σ i − 1 , there is a fall at the i -th place: ε i = − The pattern ε 1 , ··· , ε n is the up-down signature of σ (Andr´ e 1879, 1881; MacMahon 1915, De Bruijn 1970, Viennot 1979 ...) • The probability reads P n ( ε 1 , ··· , ε n ) = A n ( ε 1 , ··· , ε n ) ( n + 1 ) ! where A n ( ε 1 , ··· , ε n ) is the number of permutations whose up-down signature is ε 1 , ··· , ε n 8

  9. Recursive scheme to calculate A n ( ε 1 , ··· , ε n ) • Condition again on last variable Let a n , j be the number of permutations whose signature is ε 1 , ··· , ε n and such that σ n = j n ∑ So A n ( ε 1 , ··· , ε n ) = a n , j j = 0 • Linear recursion relation (De Bruijn 1970, Viennot 1979, Atkinson 1985 ...) � a n , 0 = 0 , If ε n = + , then ( j = − − − − → a n , j = a n , j − 1 + a n − 1 , j − 1 1 , ··· , n ) , � a n , n = 0 , If ε n = + , then ( j = ← − − − − − − − a n , j = a n , j + 1 + a n − 1 , j 0 , ··· , n − 1 ) , This is a generalization of the boustrophedon algorithm 9

  10. Alternating permutations and boustrophedon algorithm Alternating permutations ( ε n = + − + − + −··· ) 0 0 → 1 1 ← 1 ← 0 0 → 1 → 2 → 2 5 ← 5 ← 4 ← 2 ← 0 0 → 5 → 10 → 14 → 16 → 16 61 ← 61 ← 56 ← 46 ← 32 ← 16 ← 0 Word boustrophedon (“turning ox”) introduced in this context by Millar et al (1996) Construction attributed to Seidel (1877) 10

  11. Ancient boustrophedonic inscription Gortyne Island, near Crete (5th century BC) 11

  12. Explicit correspondence between both approaches Z 1 n A n ∑ P n = ( n + 1 ) ! , P n = 0 f n ( x ) d x , A n = a n , j j = 0 • Probabilistic and combinatorial approaches complementary • Both f n ( x ) and a n , j obey linear recursion relations • Explicit correspondence given by n x j ( 1 − x ) n − j ∑ f n ( x ) = a n , j j ! ( n − j ) ! j = 0 12

  13. The most and least probable patterns Among the 2 n patterns ε 1 , ··· , ε n of length n • Two most probable patterns: the alternating ones (see below) + − + − + −··· and − + − + − + ··· α = α min = ln π P n ∼ ( 2 / π ) n , (Andr´ e 1879, 1881) 2 • Two least probable patterns: the steady ones i.e., the rising one +++ ··· and the falling one −−−··· Both routes for rising patterns f n ( x ) = x n 1 P n = Probabilistic: n ! , ( n + 1 ) ! σ = I , a n , j = δ nj , 1 A n = 1 , P n = Combinatorial: ( n + 1 ) ! α n ≈ ln n 13

  14. Periodic patterns: the probabilistic route ( ε n = + − + − + −··· ) Alternating patterns • Recursion relations Z x Z 1 f 2 k + 1 ( x ) = 0 f 2 k ( y ) d y , f 2 k ( x ) = x f 2 k − 1 ( y ) d y • Generating series F 0 ( z , x ) = ∑ F 1 ( z , x ) = ∑ f 2 k + 1 ( x ) z 2 k + 1 f 2 k ( x ) z 2 k , k ≥ 0 k ≥ 0 • Integral equations Z 1 Z x F 0 ( z , x ) = 1 + z x F 1 ( z , y ) d y , F 1 ( z , x ) = z 0 F 0 ( z , y ) d y 14

  15. • Differential equation ∂ 2 F 0 ∂ F 0 ( z , 0 ) ∂ x 2 = − z 2 F 0 , F 0 ( z , 1 ) = 1 , = 0 ∂ x • Solution F 0 ( z , x ) = cos zx F 1 ( z , x ) = sin zx cos z , cos z • Generating series for the P n Π ( z ) = ∑ P n z n = 1 z ( F 1 ( z , 1 )+ F 0 ( z , 0 ) − 1 ) n ≥ 0 • Result Π ( z ) = sin z + 1 − cos z = tan z + sec z − 1 z cos z z 15

  16. Recover thus pioneering results by Andr´ e (1879, 1881) z 2 k + 1 tan z = ∑ P 2 k z 2 k + 1 = ∑ A 2 k ( 2 k + 1 ) ! k ≥ 0 k ≥ 0 z 2 k + 2 sec z = 1 + ∑ P 2 k + 1 z 2 k + 2 = 1 + ∑ A 2 k + 1 ( 2 k + 2 ) ! k ≥ 0 k ≥ 0 • The A n are called Euler-Bernoulli numbers or Entringer numbers • Asymptotic behavior � n + 2 � 2 P n ≈ 2 π • Connection with multiple-scattering problem z n = 3 n + 1 P 2 n + 2 16

  17. p-alternating patterns: period p ≥ 2 ending with a single fall ε n = ++ − ++ − ++ −··· for p = 3 , Example: • Generating series F q ( z , x ) = ∑ f kp + q ( x ) z kp + q ( q = 0 , ··· , p − 1 ) k ≥ 0 • Integral equations Z 1 Z x F 0 ( z , x ) = 1 + z x F 1 ( z , y ) d y , F q ( z , x ) = z 0 F q − 1 ( z , y ) d y ( q � = 0 ) • Differential equation ∂ p F 0 ∂ q F 0 ( z , 0 ) ∂ x p = − z p F 0 , F 0 ( z , 1 ) = 1 , = 0 ( q � = 0 ) ∂ x q 17

  18. • Solution F q ( z , x ) = T p , q ( zx ) T p , 0 ( z ) • Result � p − 1 � 1 ∑ Π ( z ) = T p , q ( z )+ 1 − T p , 0 ( z ) zT p , 0 ( z ) q = 1 • Probabilities P n ≈ A n e − α n Smallest real positive zero z 0 of T p , 0 Entropy α = ln z 0 Other zeros at z q = z 0 ζ q for q = 1 , ··· , p − 1 Amplitudes A n periodic with period p 18

  19. Generalized hyperbolic and trigonometric functions ζ = e 2 π i / p p ≥ 2 , q = 0 , ··· , p − 1 , p − 1 z kp + q H p , q ( z ) = ∑ ( kp + q ) ! = 1 ∑ ζ − qj e ζ j z p j = 0 k ≥ 0 H ′ H ′ p , q = H p , q − 1 ( q � = 0 ) , p , 0 = H p , p − 1 p − 1 z kp + q T p , q ( z ) = ∑ ( kp + q ) ! = 1 ∑ ζ − q ( j + 1 / 2 ) e ζ j + 1 / 2 z ( − 1 ) k p k ≥ 0 j = 0 T ′ T ′ p , q = T p , q − 1 ( q � = 0 ) , p , 0 = − T p , p − 1 19

  20. Recover thus Mendes and Remmel’s (book preprint) results ... by elementary means p = 2 yields z 0 = π • 2 α = α min = ln π 2 = 0 . 451582 ··· p ≫ 1 yields z 0 ≈ ( p ! ) 1 / p • α ≈ ln p ! p ≈ ln p − 1 1.5 1 α 0.5 0 1 2 3 4 5 6 7 8 p 20

  21. General case Period p ≥ 2 p − ν rises and ν falls per period ( 1 ≤ ν ≤ p − 1 ) , end with a fall • Differential equation ∂ p F 0 ∂ x p = ( − 1 ) ν z p F 0 If ε p − q = + , then ∂ q F 0 / ∂ x q ( z , 0 ) = 0 If ε p − q = − , then ∂ q F 0 / ∂ x q ( z , 1 ) = δ q 0 • Solution F 0 ( z , x ) = ∑ or ∑ C q ( z ) H p , q ( zx ) C q ( z ) T p , q ( zx ) q q Sum over the ν indices q such that ε p − q = − Boundary conditions at x = 1 yield ν linear equations for the C q ( z ) 21

  22. • Result C q ( z ) = ... Π ( z ) = ... ∆ ( z ) , ∆ ( z ) ∆ ( z ) is ν × ν determinant Entries are generalized hyperbolic or trigonometric functions ∆ ( z ) is entire function of z p • Probabilities P n ≈ A n e − α n Smallest real positive zero z 0 of ∆ ( z ) Entropy α = ln z 0 Other zeros at z q = z 0 ζ q for q = 1 , ··· , p − 1 Amplitudes A n periodic with period p 22

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend


More recommend