Pólya Urns An analytic combinatorics approach Basile Morcrette Algorithms project, INRIA Rocquencourt. LIP6, UPMC CALIN Seminar 07/02/2012 1/35
Outline 1. Urn model 2. An exact approach boolean formulas 3. Singularity analysis family of k -trees 4. Saddle-point method preferential growth models 5. Towards other urn models unbalanced, with random entries 2/35
1. Urns models ◮ an urn containing balls of two colours ◮ rules for urn evolution � 1 � 0 0 1 3/35
Balanced Pólya urns � � α β α, δ ∈ Z , β, γ ∈ N γ δ Balanced urn : α + β = γ + δ (deterministic total number of balls) A given initial configuration ( a 0 , b 0 ) : a 0 balls • (counted by x ) b 0 balls ◦ (counted by y ) Definition History of length n : a sequence of n evolutions ( n rules, n drawings) H n , a , b x a y b z n � H ( x , y , z ) = n ! n , a , b H n , a , b : number of histories of length n , beginning in the configuration ( a 0 , b 0 ) , and ending in ( a , b ) 4/35
Combinatorics of histories - Example � � 1 0 We consider this urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 0 1 H ( x , y , z ) = xy 5/35
Combinatorics of histories - Example � � 1 0 We consider this urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 0 1 H ( x , y , z ) = xy 5/35
Combinatorics of histories - Example � � 1 0 We consider this urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 0 1 H ( x , y , z ) = xy ( xy 2 + x 2 y ) z + 5/35
Combinatorics of histories - Example � � 1 0 We consider this urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 0 1 H ( x , y , z ) = xy ( xy 2 + x 2 y ) z + 5/35
Combinatorics of histories - Example � � 1 0 We consider this urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 0 1 H ( x , y , z ) = xy ( xy 2 + x 2 y ) z + 5/35
Combinatorics of histories - Example � � 1 0 We consider this urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 0 1 H ( x , y , z ) = xy ( xy 2 + x 2 y ) z + 5/35
Combinatorics of histories - Example � 1 � 0 We consider this urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 0 1 H ( x , y , z ) = xy ( xy 2 + x 2 y ) z + ( 2 xy 3 + 2 x 2 y 2 + 2 x 3 y ) z 2 + 2 5/35
Combinatorics of histories - Example � 1 � 0 We consider this urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 0 1 H ( x , y , z ) = xy ( xy 2 + x 2 y ) z + ( 2 xy 3 + 2 x 2 y 2 + 2 x 3 y ) z 2 + 2 + . . . 5/35
Various behaviours Problem : Understand the urn composition after n steps, and asymptot- ically when n tends to ∞ . 6/35
Various behaviours Problem : Understand the urn composition after n steps, and asymptot- ically when n tends to ∞ . 0 1 0 � 1 � � 2 � 0 1 0 − 1 2 0 1 1 2 0 0 1 Pólya urn Preferential growth urn Triangular 3 × 3 urn 6/35
Probabilistic results � α � Ratio ρ = α − γ β Urn γ δ α + β ◮ Small urns : ρ � 1 2 Gaussian limit law [Smythe96] [Janson04] ◮ Large urns : ρ > 1 2 Non gaussian laws [Mahmoud] [Janson04] [Chauvin–Pouyanne–Sahnoun11] Tools : • embedding in continuous time [Jan04] [ChPoSa11] • martingales, central limit theorem 7/35
Balanced urns and analysis ◮ First steps : [Flajolet–Gabarro–Pekari05], Analytic urns ◮ [Flajolet–Dumas–Puyhaubert06], on urns with negative coefficients, and triangular cases ◮ [Kuba–Panholzer–Hwang07], unbalanced urns Analytic approach : theorem [FlDuPu06] H = X a 0 Y b 0 � ˙ � � � α β ( a 0 , b 0 ) X α + 1 Y β Urn and = ⇒ X = γ δ α + β = γ + δ with X γ Y δ + 1 ˙ Y = 8/35
Isomorphism proof Differenciate = Pick x . . . x ) + . . . + ( xx . . . ✁ ∂ x [ xx . . . x ] = ( ✁ xx . . . x ) + ( x ✁ x ) x ∂ x [ xx . . . x ] = ( xx . . . x ) + ( xx . . . x ) + . . . + ( xx . . . x ) Let D = x α + 1 y β ∂ x + x γ y δ + 1 ∂ y Then D [ x a y b ] = ax a + α y b + β + bx a + γ y b + δ 9/35
Counting histories - Example � 1 � 0 Take the urn and ( a 0 , b 0 ) = ( 1 , 1 ) . 0 1 H ( x , y , z ) = xy ( xy 2 + x 2 y ) z + ( 2 xy 3 + 2 x 2 y 2 + 2 x 3 y ) z 2 + 2 + . . . 9/35
Isomorphism proof Differenciate = Pick x . . . x ) + . . . + ( xx . . . ✁ ∂ x [ xx . . . x ] = ( ✁ xx . . . x ) + ( x ✁ x ) x ∂ x [ xx . . . x ] = ( xx . . . x ) + ( xx . . . x ) + . . . + ( xx . . . x ) Let D = x α + 1 y β ∂ x + x γ y δ + 1 ∂ y Then D [ x a y b ] = ax a + α y b + β + bx a + γ y b + δ � D n [ x a 0 y b 0 ] = H n , a , b x a y b a , b D n [ x a 0 y b 0 ] z n � H ( x , y , z ) = n ! n ≥ 0 9/35
Isomorphism proof Let ( X ( t ) , Y ( t )) be solu- tion of Differenciate = Pick � ˙ X α + 1 Y β X = X ( t = 0 ) = x x ∂ x [ xx . . . x ] = ( xx . . . x )+( xx . . . x )+ . . . +( xx . . . x ) X γ Y δ + 1 ˙ Y = Y ( t = 0 ) = y D = x α + 1 y β ∂ x + x γ y δ + 1 ∂ y D [ x a y b ] = ax a + α y b + β + bx a + γ y b + δ ∂ t ( X a Y b ) aX a − 1 ˙ XY b + bX a Y b − 1 ˙ � D n [ x a 0 y b 0 ] = H n , a , b x a y b = Y aX a + α Y b + β + bX a + γ Y b + δ a , b = D n [ x a 0 y b 0 ] z n � H ( x , y , z ) = n ! ∂ n t ( X a Y b ) = D n � x a y b � n ≥ 0 x → X y → Y X ( t ) a 0 Y ( t ) b 0 � z n � ∂ n n ! = X ( t + z ) a 0 Y ( t + z ) b 0 � H ( X ( t ) , Y ( t ) , z ) = t n ≥ 0 Then t = 0, and it’s over !! 9/35
2. Urns and random trees ◮ Motivation : quantify the fraction of tautologies among all logic formulas having only one logic operator : implication. [Mailler11] ◮ Probabilistic model : uniform growth in leaves (BST model) ◮ choose randomly a leave ◮ replace it by a binary node and two leaves 10/35
2. Urns and random trees ◮ Motivation : quantify the fraction of tautologies among all logic formulas having only one logic operator : implication. [Mailler11] ◮ Probabilistic model : uniform growth in leaves (BST model) ◮ choose randomly a leave ◮ replace it by a binary node and two leaves 10/35
2. Urns and random trees ◮ Motivation : quantify the fraction of tautologies among all logic formulas having only one logic operator : implication. [Mailler11] ◮ Probabilistic model : uniform growth in leaves (BST model) ◮ choose randomly a leave ◮ replace it by a binary node and two leaves 10/35
2. Urns and random trees ◮ Motivation : quantify the fraction of tautologies among all logic formulas having only one logic operator : implication. [Mailler11] ◮ Probabilistic model : uniform growth in leaves (BST model) ◮ choose randomly a leave ◮ replace it by a binary node and two leaves 10/35
2. Urns and random trees ◮ Motivation : quantify the fraction of tautologies among all logic formulas having only one logic operator : implication. [Mailler11] ◮ Probabilistic model : uniform growth in leaves (BST model) ◮ choose randomly a leave ◮ replace it by a binary node and two leaves 10/35
A 3 × 3 urn model 3 colors, Corresponding with rules : urn : ∇ → • ∇ 0 1 0 • → × × 0 − 1 2 × → × × 0 0 1 Generating function of histories � � � � 1 H ( y , z ) = exp ln + ( y − 1 ) z 1 − z z counts the length of history, y counts the number of • balls. 11/35
Poisson Law in subs-trees Let U k , n be the number of left sub-trees of of size k directly hanging on the right branch of a random tree of size n . Theorem ◮ U 1 , n converges in law, U 1 , n − n →∞ U 1 , → � 2 n � ◮ U 1 ∼ P oisson ( 1 ) , with rate of convergence O . n ! 12/35
Poisson Law in subs-trees Let U k , n be the number of left sub-trees of of size k directly hanging on the right branch of a random tree of size n . Theorem ◮ U 1 , n converges in law, U 1 , n − n →∞ U 1 , → � 2 n � ◮ U 1 ∼ P oisson ( 1 ) , with rate of convergence O . n ! Generalisation With a ( k + 2 ) × ( k + 2 ) urn Theorem ◮ U k , n converges in law, U k , n − n →∞ U k , → � 1 ( 2 k ) n � � ◮ U k ∼ P oisson � , with rate of convergence O . n ! k 12/35
3. An urn for k -trees Motivation : model of graphs [Panholzer–Seitz 2010] Definition A k -tree T is ◮ either a k -clique ◮ or there exists a vertex f with a k -clique as neighbor and T \ f is a k -tree Ordered : distinguishable children. Increasing : vertices labelled in apparition order 13/35
3. An urn for k -trees Motivation : model of graphs [Panholzer–Seitz 2010] Definition A k -tree T is ◮ either a k -clique ◮ or there exists a vertex f with a k -clique as neighbor and T \ f is a k -tree Ordered : distinguishable children. Increasing : vertices labelled in apparition order increasing ordered 1-tree (or PORT) 13/35
3. An urn for k -trees Motivation : model of graphs [Panholzer–Seitz 2010] Definition A k -tree T is ◮ either a k -clique ◮ or there exists a vertex f with a k -clique as neighbor and T \ f is a k -tree Ordered : distinguishable children. Increasing : vertices labelled in apparition order increasing ordered 2-tree 13/35
Urn Model 14/35
Urn Model For 1-trees � 0 � 2 1 1 14/35
Urn Model For 1-trees � 0 � 2 1 1 For k -trees � k − 1 � 2 k 1 14/35
Urn Model For 1-trees � 0 � 2 1 1 For k -trees � k − 1 � 2 k 1 d dz H ( x , z ) H ( x , z ) k ( H ( x , z ) − x + 1 ) 2 = 1 14/35
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