derrida retaux model from discrete to continuous time
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DerridaRetaux model: from discrete to continuous time Michel Pain - PowerPoint PPT Presentation

DerridaRetaux model: from discrete to continuous time Michel Pain (LPSM DMA) joint work with Yueyun Hu and Bastien Mallein Les Probabilits de demain 14 June 2019 Discrete-time DerridaRetaux model Defjnition: Start with a nonnegative


  1. Derrida–Retaux model: from discrete to continuous time Michel Pain (LPSM – DMA) joint work with Yueyun Hu and Bastien Mallein Les Probabilités de demain 14 June 2019

  2. Discrete-time Derrida–Retaux model

  3. Defjnition: Start with a nonnegative random variable X 0 and, for any where X n is an independent copy of X n . Defjnition glass theory. Re-introduced by Derrida–Retaux (2014) for studying the depinning transition. n 0, X n 1 X n X n 1 1/12 ⊲ Introduced by Collet–Eckmann–Glaser–Martin (1984), motivated by spin

  4. Defjnition: Start with a nonnegative random variable X 0 and, for any where X n is an independent copy of X n . Defjnition glass theory. transition. n 0, X n 1 X n X n 1 1/12 ⊲ Introduced by Collet–Eckmann–Glaser–Martin (1984), motivated by spin ⊲ Re-introduced by Derrida–Retaux (2014) for studying the depinning

  5. Defjnition glass theory. transition. 1/12 ⊲ Introduced by Collet–Eckmann–Glaser–Martin (1984), motivated by spin ⊲ Re-introduced by Derrida–Retaux (2014) for studying the depinning ⊲ Defjnition: Start with a nonnegative random variable X 0 and, for any n ≥ 0, � � X n + � X n + 1 = X n − 1 + where � X n is an independent copy of X n .

  6. Defjnition on a tree 0 X n i.i.d. copies of X 0 0 0 3 0 0 1 0 0 4 1 3 0 2 0 0 n 2/12 Construction of X n on a binary tree:

  7. Defjnition on a tree 0 b a X n i.i.d. copies of X 0 0 2 0 0 3 3 1 0 0 3 2/12 3 n 0 0 2 0 1 0 4 0 0 0 1 0 Construction of X n on a binary tree: ( a + b − 1 ) +

  8. Defjnition on a tree 0 3 3 0 0 2 0 5 0 0 1 i.i.d. copies of X 0 X n a b 1 0 2/12 1 n 0 0 2 0 3 4 0 0 0 0 1 0 0 3 Construction of X n on a binary tree: ( a + b − 1 ) +

  9. Defjnition on a tree 5 3 0 0 2 0 0 0 1 1 4 0 i.i.d. copies of X 0 X n a b 3 0 2/12 0 n 0 0 2 0 3 1 4 0 0 0 1 0 0 3 0 Construction of X n on a binary tree: ( a + b − 1 ) +

  10. Defjnition on a tree 1 0 0 2 0 0 5 0 4 3 0 3 i.i.d. copies of X 0 X n a b X n 3 1 2/12 4 n 0 0 2 0 3 1 0 0 0 0 1 0 0 3 0 0 Construction of X n on a binary tree: ( a + b − 1 ) +

  11. 2 X 0 or Phase transition a.s. 0 n probability X n and 0 F , then 2 X 0 X 0 2 X 0 (Subcritical) If F n X n 2 n and 0 F , then 2 X 0 X 0 2 X 0 (Supercritical) If a.s. Theorem (Collet–Eckmann–Glaser–Martin 1984): Assume that X 0 2 n 3/12 E [ X n ] Free energy: F ∞ := lim ∈ [ 0 , ∞ ] . n →∞

  12. 2 X 0 or Phase transition a.s. 0 n probability X n and 0 F , then 2 X 0 X 0 2 X 0 (Subcritical) If F n 2 n X n and 0 F , then 2 X 0 X 0 2 X 0 (Supercritical) If 2 n 3/12 E [ X n ] Free energy: F ∞ := lim ∈ [ 0 , ∞ ] . n →∞ Theorem (Collet–Eckmann–Glaser–Martin 1984): Assume that X 0 ∈ N a.s.

  13. Phase transition and 0 n probability X n and 0 F , then 2 X 0 X 0 2 X 0 (Subcritical) If a.s. 2 n X n 3/12 2 n E [ X n ] Free energy: F ∞ := lim ∈ [ 0 , ∞ ] . n →∞ Theorem (Collet–Eckmann–Glaser–Martin 1984): Assume that X 0 ∈ N a.s. � X 0 2 X 0 � � 2 X 0 � � 2 X 0 � ⊲ (Supercritical) If E > E or E = ∞ , then F ∞ > 0 − − n →∞ F ∞ . − →

  14. Phase transition and probability X n and a.s. 2 n X n 3/12 2 n E [ X n ] Free energy: F ∞ := lim ∈ [ 0 , ∞ ] . n →∞ Theorem (Collet–Eckmann–Glaser–Martin 1984): Assume that X 0 ∈ N a.s. � X 0 2 X 0 � � 2 X 0 � � 2 X 0 � ⊲ (Supercritical) If E > E or E = ∞ , then F ∞ > 0 − − n →∞ F ∞ . − → � X 0 2 X 0 � � 2 X 0 � ≤ E < ∞ , then ⊲ (Subcritical) If E F ∞ = 0 − − − − − − → 0 . n →∞

  15. (d) 1 p c 1 2 p c 1 2 Free energy near criticality by 0 , F p exp K o 1 p . Theorem (Chen–Dagard–Derrida–Hu–Lifshits–Shi 2019+): If is supported 0 and 5 x 3 2 x dx , then as p p c F p exp 1 p o 1 Conjecture (Derrida–Retaux 2014): If p c 1 4/12 inf p Consider X 0 p 0 p for each p 0 1 . Let F p denote the free energy and p c 0 1 p c F p 0 . p F p p c 1 0 1 2 ⊲ Let µ be a probability measure on ( 0 , ∞ ) , in the supercritical phase.

  16. p c 1 2 p c 1 2 Free energy near criticality and exp K o 1 p . Theorem (Chen–Dagard–Derrida–Hu–Lifshits–Shi 2019+): If is supported by x 3 2 x 0 0 , F dx , then as p p c F p exp 1 p o 1 p Conjecture (Derrida–Retaux 2014): If p c 4/12 0 . (d) Let F p denote the free energy and p c inf p 0 1 F 5 p p F p p c 1 0 1 2 p c 1 ⊲ Let µ be a probability measure on ( 0 , ∞ ) , in the supercritical phase. = ( 1 − p ) δ 0 + p µ for each ⊲ Consider X 0 p ∈ [ 0 , 1 ] .

  17. p c 1 2 p c 1 2 Free energy near criticality x 3 2 x . Theorem (Chen–Dagard–Derrida–Hu–Lifshits–Shi 2019+): If is supported by and 0 , then as p dx o 1 p c F p exp 1 p o 1 p K 4/12 1 (d) p exp 1 0 p c 5 Conjecture (Derrida–Retaux 2014): If p c 0 , F p F ∞ ( p ) ⊲ Let µ be a probability measure on ( 0 , ∞ ) , in the supercritical phase. µ = δ 2 p c = 1 = ( 1 − p ) δ 0 + p µ for each ⊲ Consider X 0 p ∈ [ 0 , 1 ] . ⊲ Let F ∞ ( p ) denote the free energy and p c := inf { p ∈ [ 0 , 1 ] : F ∞ ( p ) > 0 } .

  18. p c 1 2 Free energy near criticality , then as p is supported by and 0 x 3 2 x dx p c . F p exp 1 p o 1 Theorem (Chen–Dagard–Derrida–Hu–Lifshits–Shi 2019+): If 4/12 p c 0 p 1 1 5 (d) F ∞ ( p ) ⊲ Let µ be a probability measure on ( 0 , ∞ ) , in the supercritical phase. µ = δ 2 p c = 1 = ( 1 − p ) δ 0 + p µ for each ⊲ Consider X 0 p ∈ [ 0 , 1 ] . ⊲ Let F ∞ ( p ) denote the free energy and p c := inf { p ∈ [ 0 , 1 ] : F ∞ ( p ) > 0 } . � � − K + o ( 1 ) Conjecture (Derrida–Retaux 2014): If p c > 0 , F ∞ ( p ) = exp ( p − p c ) 1 / 2

  19. Free energy near criticality p c 1 0 . 5 1 0 1 4/12 (d) p F ∞ ( p ) ⊲ Let µ be a probability measure on ( 0 , ∞ ) , in the supercritical phase. µ = δ 2 p c = 1 = ( 1 − p ) δ 0 + p µ for each ⊲ Consider X 0 p ∈ [ 0 , 1 ] . ⊲ Let F ∞ ( p ) denote the free energy and p c := inf { p ∈ [ 0 , 1 ] : F ∞ ( p ) > 0 } . � � − K + o ( 1 ) Conjecture (Derrida–Retaux 2014): If p c > 0 , F ∞ ( p ) = exp ( p − p c ) 1 / 2 Theorem (Chen–Dagard–Derrida–Hu–Lifshits–Shi 2019+): If µ is supported � ∞ by N ∗ and x 3 2 x µ ( dx ) < ∞ , then as p ↓ p c � � F ∞ ( p ) = exp − . ( p − p c ) 1 / 2 + o ( 1 )

  20. Behavior at criticality 0 0, what does the subtree bringing mass to the root look like? Given X n n 2 4 0 X n , then 0 2 X 0 X 3 Conjecture (Chen–Derrida–Hu–Lifshits–Shi 2017): If n c X n , then 0 2 X 0 X 3 Theorem (Chen–Derrida–Hu–Lifshits–Shi 2017): If 0 in probability. Recall that X n 5/12 � X 0 2 X 0 � � 2 X 0 � ⊲ Critical case for X 0 ∈ N : E = E < ∞ .

  21. Behavior at criticality c 0, what does the subtree bringing mass to the root look like? Given X n n 2 4 0 X n , then 0 2 X 0 X 3 Conjecture (Chen–Derrida–Hu–Lifshits–Shi 2017): If n 0 X n , then 0 2 X 0 X 3 Theorem (Chen–Derrida–Hu–Lifshits–Shi 2017): If 5/12 � X 0 2 X 0 � � 2 X 0 � ⊲ Critical case for X 0 ∈ N : E = E < ∞ . ⊲ Recall that X n → 0 in probability.

  22. Behavior at criticality Conjecture (Chen–Derrida–Hu–Lifshits–Shi 2017): If 0, what does the subtree bringing mass to the root look like? Given X n n 2 4 0 X n , then 0 2 X 0 X 3 5/12 X 3 � X 0 2 X 0 � � 2 X 0 � ⊲ Critical case for X 0 ∈ N : E = E < ∞ . ⊲ Recall that X n → 0 in probability. � 0 2 X 0 � < ∞ , then ⊲ Theorem (Chen–Derrida–Hu–Lifshits–Shi 2017): If E P ( X n > 0 ) ≤ c n .

  23. Behavior at criticality X 3 0, what does the subtree bringing mass to the root look like? Given X n X 3 5/12 � X 0 2 X 0 � � 2 X 0 � ⊲ Critical case for X 0 ∈ N : E = E < ∞ . ⊲ Recall that X n → 0 in probability. � 0 2 X 0 � < ∞ , then ⊲ Theorem (Chen–Derrida–Hu–Lifshits–Shi 2017): If E P ( X n > 0 ) ≤ c n . � 0 2 X 0 � ⊲ Conjecture (Chen–Derrida–Hu–Lifshits–Shi 2017): If E < ∞ , then P ( X n > 0 ) ∼ 4 n 2 .

  24. Behavior at criticality X 3 X 3 5/12 � X 0 2 X 0 � � 2 X 0 � ⊲ Critical case for X 0 ∈ N : E = E < ∞ . ⊲ Recall that X n → 0 in probability. � 0 2 X 0 � < ∞ , then ⊲ Theorem (Chen–Derrida–Hu–Lifshits–Shi 2017): If E P ( X n > 0 ) ≤ c n . � 0 2 X 0 � ⊲ Conjecture (Chen–Derrida–Hu–Lifshits–Shi 2017): If E < ∞ , then P ( X n > 0 ) ∼ 4 n 2 . ⊲ Given X n > 0, what does the subtree bringing mass to the root look like?

  25. Continuous-time Derrida–Retaux model

  26. X t is the remaining paint at the root. Defjnition painting the branches with a quantity 1 t X t remaining paint in common. When two painters meet, they put their of paint per unit of branch length. Then, painters climb down the tree, Initial condition : a nonnegative random variable X 0 . cording to the law of X 0 . with i.i.d. amount of paint chosen ac- Initially : painters start on the leaves tributed lifetimes). nary tree with i.i.d. exponentially dis- Consider a Yule tree of height t (bi- 6/12 For t > 0, X t is defjned using a painting procedure :

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