On (un)-balanced Plya urns: Analytic Combinatorics strikes again - PowerPoint PPT Presentation

On (un)-balanced Pólya urns: Analytic Combinatorics strikes again Basile Morcrette May, 30 AofA 2013, Spain Dedicated to Philippe Flajolet. 1/16

Knuth’s strings Start with m loops in a box. At each step, pick one at random, cut it and place it back in the box. Q : Average length of a string after m cuts ? Januar 10, 2011 : D.E. Knuth to P. Flajolet : “ I think you wil find [it] amusing.” Philippe thought : “ Look at the urn behind !” loop := black ball string := white ball � − 1 � 1 K = ( a 0 , b 0 ) = ( m , 0 ) 0 1 oh... unbalanced ! 2/16

Balanced Pólya urns [Flajolet–Gabarró–Pekari, 2005 ; Flajolet–Dumas–Puyhaubert, 2006] 3/16

Pólya balanced additive urns � a � b Pólya urn a , d ∈ Z , b , c ∈ N c d Initial configuration : ( a 0 , b 0 ) , a 0 black balls, b 0 white balls Balanced urn σ := a + b = c + d (deterministic total number of balls) Additive σ ≥ 0 Definition History of length n : sequence of n drawings H n , i , j x i y j z n � H ( x , y , z ) = n ! n , i , j H n , i , j is the number of histories of length n , ending in ( i , j ) , starting from the initial configuration. 3/16

Count histories - Example � 0 � 2 Take the urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 1 1 1 5 1 H ( x , y , z ) = 2 3 4 xy 1 1 2 2 4 2 3 3 4/16

Count histories - Example � 0 � 2 Take the urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 1 1 1 5 1 H ( x , y , z ) = 2 3 xy 4 1 ( xy 3 + x 2 y 2 ) z + 1 2 1 ! 2 4 2 3 3 4/16

Count histories - Example � 0 � 2 Take the urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 1 1 1 H ( x , y , z ) = 5 1 2 xy 3 4 1 ( xy 3 + x 2 y 2 ) z + 1 ! 1 2 2 4 2 3 3 4/16

Count histories - Example � 0 � 2 Take the urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 1 1 1 H ( x , y , z ) = 5 1 xy 2 3 ( xy 3 + x 2 y 2 ) z 4 1 + 1 ! 1 2 ( xy 5 + 5 x 2 y 4 + 2 x 3 y 3 ) z 2 2 4 + 2 ! 2 3 3 4/16

Count histories - Example � 0 � 2 Take the urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 1 1 1 H ( x , y , z ) = 5 1 xy 2 3 ( xy 3 + x 2 y 2 ) z 4 1 + 1 ! 1 2 ( xy 5 + 5 x 2 y 4 + 2 x 3 y 3 ) z 2 2 4 + 2 ! 2 3 + . . . 3 4/16

Combinatorics and Analytic properties � a � b of balance σ with ( a 0 , b 0 ) c d s n : total number of balls in the urn after n draws Histories of length n = s 0 s 1 . . . s n − 1 = s 0 ( s 0 + σ ) . . . ( s 0 + ( n − 1 ) σ ) Combinatorics = Probability A n : number of black balls after n draws B n : number of white balls after n draws � x i y j z n � H n , i , j H ( x , y , z ) P { A n = i , B n = j } = = s 0 s 1 . . . s n − 1 [ z n ] H ( 1 , 1 , z ) Partial Differential Equation [FGP05] ∂ z H = x a + 1 y b ∂ x H + x c y d + 1 ∂ y H 5/16

Symbolic ingredients for PDE - Balanced case Pick and replace is the symbolic pointing operator Θ x = x ∂ x ? → i x i + a y j + b + j x i + c y j + d x i y j − 6/16

Symbolic ingredients for PDE - Balanced case Pick and replace is the symbolic pointing operator Θ x = x ∂ x x i y j D → i x i + a y j + b + j x i + c y j + d − D = x a y b Θ x + x c y d Θ y 6/16

Count histories - Example � � 0 2 Take the urn with ( a 0 , b 0 ) = ( 1 , 1 ) . 1 1 1 H ( x , y , z ) = 5 1 xy 2 3 ( xy 3 + x 2 y 2 ) z 4 1 + 1 ! 1 2 ( xy 5 + 5 x 2 y 4 + 2 x 3 y 3 ) z 2 + 2 4 2 ! 2 3 + . . . 3 6/16

Symbolic ingredients for PDE - Balanced case Pick and replace is the symbolic pointing operator Θ x = x ∂ x x i y j D → i x i + a y j + b + j x i + c y j + d − D = x a y b Θ x + x c y d Θ y Iteration from the initial configuration, � D n � x a 0 y b 0 � H n , i , j x i y j = i , j D n [ x a 0 y b 0 ] z n � e D z � ◦ [ x a 0 y b 0 ] � H ( x , y , z ) = n ! = n ≥ 0 PDE proof Differentiate w.r.t. z : ∂ z H = D H 6/16

(Un)-balanced Pólya urns [M., PhD thesis, 2013] 7/16

No more histories... x 3 → 3 � − 1 � 1 starting in ( 3 , 0 ) 0 1 3 x 2 y → 3 No more : 2 1 ❆ balance xy 2 x 2 y 2 → 3 → 4 ❆ determinist total number of balls 1 2 2 2 ❆ equiprobability of y 3 xy 3 xy 3 x 2 y 3 histories 6 12 6 6 1 1 1 1 1 1 1 1 1 1 1 1 Count histories � = Prob. 3 3 3 3 3 3 3 3 4 3 3 4 7/16

Back to probability and symbolic I � P { A n = i , B n = j } x i y j p n ( x , y ) = i , j ? i j i + j x i + a y j + b + x i y j i + j x i + c y j + d − → 8/16

Back to probability and symbolic I � P { A n = i , B n = j } x i y j t i + j p n ( x , y , t ) = i , j x i y j t i + j ? i j i + j x i + a y j + b t i + j + a + b + i + j x i + c y j + d t i + j + c + d − → 8/16

Back to probability and symbolic I � P { A n = i , B n = j } x i y j t i + j p n ( x , y , t ) = i , j x i y j t i + j ? i j i + j x i + a y j + b t i + j + a + b + i + j x i + c y j + d t i + j + c + d − → Introduce two operators, integration and differentiation, � t w = x i y j t i + j x i y j w i + j dw I [ x i y j t i + j ] = i + j 0 D = x a y b t a + b Θ x + x c y d t c + d Θ y 8/16

Back to probability and symbolic I � P { A n = i , B n = j } x i y j t i + j p n ( x , y , t ) = i , j i j x i y j t i + j D ◦ I i + j x i + a y j + b t i + j + a + b + i + j x i + c y j + d t i + j + c + d − → Introduce two operators, integration and differentiation, � t w = x i y j t i + j x i y j w i + j dw I [ x i y j t i + j ] = i + j 0 D = x a y b t a + b Θ x + x c y d t c + d Θ y 8/16

Back to probability and symbolic I � P { A n = i , B n = j } x i y j t i + j p n ( x , y , t ) = i , j i j x i y j t i + j D ◦ I i + j x i + a y j + b t i + j + a + b + i + j x i + c y j + d t i + j + c + d − → Introduce two operators, integration and differentiation, � t w = x i y j t i + j x i y j w i + j dw I [ x i y j t i + j ] = i + j 0 D = x a y b t a + b Θ x + x c y d t c + d Θ y p n + 1 = D ◦ I [ p n ] = . . . = ( D ◦ I ) n + 1 [ x a 0 y b 0 t a 0 + b 0 ] –> ugly ! 8/16

Back to probability and symbolic II � P { A n = i , B n = j } x i y j t i + j p n ( x , y , t ) = i , j D = x a y b t a + b Θ x + x c y d t c + d Θ y � t p n + 1 = D ◦ I [ p n ] f ( w ) dw I [ f ( t )] = w 0 9/16

Back to probability and symbolic II � P { A n = i , B n = j } x i y j t i + j p n ( x , y , t ) = i , j D = x a y b t a + b Θ x + x c y d t c + d Θ y � t p n + 1 = D ◦ I [ p n ] f ( w ) dw I [ f ( t )] = w 0 Introduce a new Generating Function, ψ n = I [ p n ] P { A n = i , B n = j } x i y j t i + j � ψ n ( x , y , t ) = i + j i , j p n = t ∂ t ψ n and t ∂ t ψ n + 1 = D ( ψ n ) 9/16

Back to probability and symbolic II � P { A n = i , B n = j } x i y j t i + j p n ( x , y , t ) = i , j D = x a y b t a + b Θ x + x c y d t c + d Θ y � t p n + 1 = D ◦ I [ p n ] f ( w ) dw I [ f ( t )] = w 0 Introduce a new Generating Function, ψ n = I [ p n ] P { A n = i , B n = j } x i y j t i + j � ψ n ( x , y , t ) = i + j i , j p n = t ∂ t ψ n and t ∂ t ψ n + 1 = D ( ψ n ) n ψ n z n verifies Finally t ∂ t = x ∂ x + y ∂ y , thus Ψ = � ( 1 − zx a y b ) Θ x + ( 1 − zx c y d ) Θ y ◦ Ψ( x , y , z ) = x a 0 y b 0 � � for any additive urn ! 9/16

Probabilities and balance case P { A n = i , B n = j } x i y j z n � � (PGF) P ( x , y , z ) = n ≥ 0 i , j P { A n = i , B n = j } x i y j � � i + j z n ( weighted PGF) Ψ( x , y , z ) = i , j n ≥ 0 Link : P ( x , y , z ) = ∂ t Ψ( xt , yt , z ) | t = 1 H n , i , j x i y j z n � � (Histories GF) H ( x , y , z ) = n ! i , j n ≥ 0 Proposition � a b � Let be additive balanced with balance σ > 0 and ( a 0 , b 0 ) the c d starting configuration. With s 0 = a 0 + b 0 , the fonctions Ψ et H are linked by � 1 Ψ( x , y , z ) = 1 � x , y , z 1 − t � t s 0 /σ − 1 H dt σ σ 0 10/16

Solve the PDE = Solve an ODE � ( 1 − zx a y b ) Θ x + ( 1 − zx c y d ) Θ y � ◦ Ψ( x , y , z ) = x a 0 y b 0 (PDE) dx dy dw Characteristic system : x ( 1 − zx a y b ) = y ( 1 − zx c y d ) = x a 0 y b 0 First integral, coming from dy = x ( 1 − zx a y b ) dx (ODE) y ( 1 − zx c y d ) Theorem Let x = g ( y , z , u ) be the general solution of (ODE), with u integration constant, and U ( x , y , z ) (first integral) such that g ( y , z , U ( x , y , z )) = x . Then, � y g ( t , z , U ( x , y , z )) a 0 t b 0 − 1 Ψ( x , y , z ) = 1 − z g ( t , z , U ( x , y , z )) c t d dt 0 11/16

Application I : Knuth cutting loops � − 1 � 1 K = ( a 0 , b 0 ) = ( m , 0 ) 0 1 dx y ( 1 − zy ) = dw dy Characteristic system : x ( 1 − zx − 1 y ) = x m dx x − zy First order differential equation : dy = y ( 1 − zy ) y General solution : g ( y , z , u ) = 1 − zy ( u − z ln ( y )) U ( x , y , z ) = x ( 1 − zy ) First integral : + z ln ( y ) y � 1 t m − 1 ( 1 − zt ) m + 1 ( x ( 1 − z ) − z ln ( t )) m dt Ψ K ( x , 1 , z ) = 0 12/16

Application II : diagonal urns [Balaji–Mahmoud, 2006] � a � 0 BM = a , d > 0 ( a 0 , b 0 ) 0 d dy = x ( 1 − zx a ) dx First order differential equation : y ( 1 − zy d ) � − 1 / a � a / d y − d − z �� General solution : g ( y , z , u ) = + z u a y − d − z � 1 / d � First integral : U ( x , y , z ) = ( x − a − z ) 1 / a � t − d − z � − a 0 / a � y � � a / d t b 0 − 1 ( x − a − z ) Ψ BM ( x , y , z ) = + z 1 − zt d dt y − d − z 0 13/16