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

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

Apr 23, 2023 462 likes •781 views

On (un)-balanced Plya urns: Analytic Combinatorics strikes again Basile Morcrette May, 30 AofA 2013, Spain Dedicated to Philippe Flajolet. 1/16 Knuths strings Start with m loops in a box. At each step, pick one at random, cut it and

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On (un)-balanced Pólya urns: Analytic Combinatorics strikes again

Basile Morcrette May, 30 AofA 2013, Spain Dedicated to Philippe Flajolet.

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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 K = −1 1 1

  • (a0, b0) = (m, 0)
  • h... unbalanced !

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Balanced Pólya urns

[Flajolet–Gabarró–Pekari, 2005 ; Flajolet–Dumas–Puyhaubert, 2006]

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Pólya balanced additive urns

Pólya urn

a b c d

  • a, d ∈ Z,

b, c ∈ N Initial configuration : (a0, b0), a0 black balls, b0 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(x, y, z) =

  • n,i,j

Hn,i,j xiy j zn n! Hn,i,j is the number of histories of length n, ending in (i, j), starting from the initial configuration.

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Count histories - Example

Take the urn 2 1 1

  • with (a0, b0) = (1, 1).

1 1 1 2 3 2 1 5 2 4 2 4 3 3

H(x, y, z) = xy

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Count histories - Example

Take the urn 2 1 1

  • with (a0, b0) = (1, 1).

1 1 1 2 3 2 1 5 2 4 2 4 3 3

H(x, y, z) = xy + (xy 3 + x2y 2) z 1!

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Count histories - Example

Take the urn 2 1 1

  • with (a0, b0) = (1, 1).

1 1 1 2 3 2 1 5 2 4 2 4 3 3

H(x, y, z) = xy + (xy 3 + x2y 2) z 1!

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Count histories - Example

Take the urn 2 1 1

  • with (a0, b0) = (1, 1).

1 1 1 2 3 2 1 5 2 4 2 4 3 3

H(x, y, z) = xy + (xy 3 + x2y 2) z 1! + (xy 5 + 5x2y 4 + 2x3y 3) z2 2!

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Count histories - Example

Take the urn 2 1 1

  • with (a0, b0) = (1, 1).

1 1 1 2 3 2 1 5 2 4 2 4 3 3

H(x, y, z) = xy + (xy 3 + x2y 2) z 1! + (xy 5 + 5x2y 4 + 2x3y 3) z2 2! + . . .

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Combinatorics and Analytic properties

a b c d

  • f balance σ

with (a0, b0) sn : total number of balls in the urn after n draws Histories of length n = s0s1 . . . sn−1 = s0 (s0 + σ) . . . (s0 + (n − 1)σ)

Combinatorics = Probability

An : number of black balls after n draws Bn : number of white balls after n draws P {An = i, Bn = j} = Hn,i,j s0s1 . . . sn−1 =

  • xiy jzn

H(x, y, z) [zn] H(1, 1, z)

Partial Differential Equation [FGP05]

∂zH = xa+1y b ∂xH + xcy d+1 ∂yH

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Symbolic ingredients for PDE - Balanced case

Pick and replace is the symbolic pointing operator Θx = x∂x

xiy j

?

− → i xi+a y j+b + j xi+c y j+d

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Symbolic ingredients for PDE - Balanced case

Pick and replace is the symbolic pointing operator Θx = x∂x

xiy j D − → i xi+a y j+b + j xi+c y j+d D = xay b Θx + xcy d Θy

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Count histories - Example

Take the urn

  • 2

1 1

  • with (a0, b0) = (1, 1).

1 1 1 2 3 2 1 5 2 4 2 4 3 3

H(x, y, z) = xy + (xy 3 + x2y 2) z 1! + (xy 5 + 5x2y 4 + 2x3y 3) z2 2! + . . .

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Symbolic ingredients for PDE - Balanced case

Pick and replace is the symbolic pointing operator Θx = x∂x xiy j D − → i xi+a y j+b + j xi+c y j+d D = xay b Θx + xcy d Θy Iteration from the initial configuration,

Dn xa0y b0 =

  • i,j

Hn,i,j xiy j H(x, y, z) =

  • n≥0

Dn[xa0y b0]zn n! =

  • eDz
  • [xa0y b0]

PDE proof Differentiate w.r.t. z : ∂zH = DH

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(Un)-balanced Pólya urns

[M., PhD thesis, 2013]

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No more histories...

x3 → 3 x2y → 3 xy 2 → 3 y 3 6 1 3 1 3 1 3 1 xy 3 12 1 3 1 3 1 3 2 2 x2y 2 → 4 xy 3 6 1 3 1 3 1 4 2 x2y 3 6 1 3 1 3 1 4 2 1 3 −1 1 1

  • starting in (3, 0)

No more : ❆ balance ❆ determinist total number of balls ❆ equiprobability of histories Count histories = Prob.

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Back to probability and symbolic I

pn(x, y) =

  • i,j

P {An = i, Bn = j} xiy j xiy j

?

− → i i + j xi+ay j+b + j i + j xi+cy j+d

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Back to probability and symbolic I

pn(x, y, t) =

  • i,j

P {An = i, Bn = j} xiy jti+j xiy jti+j ? − → i i + j xi+ay j+bti+j+a+b + j i + j xi+cy j+dti+j+c+d

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Back to probability and symbolic I

pn(x, y, t) =

  • i,j

P {An = i, Bn = j} xiy jti+j xiy jti+j ? − → i i + j xi+ay j+bti+j+a+b + j i + j xi+cy j+dti+j+c+d Introduce two operators, integration and differentiation, I[xiy jti+j] = t xiy jw i+j dw w = xiy j ti+j i + j D = xay bta+bΘx + xcy dtc+dΘy

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Back to probability and symbolic I

pn(x, y, t) =

  • i,j

P {An = i, Bn = j} xiy jti+j xiy jti+j D ◦ I − → i i + j xi+ay j+bti+j+a+b + j i + j xi+cy j+dti+j+c+d Introduce two operators, integration and differentiation, I[xiy jti+j] = t xiy jw i+j dw w = xiy j ti+j i + j D = xay bta+bΘx + xcy dtc+dΘy

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Back to probability and symbolic I

pn(x, y, t) =

  • i,j

P {An = i, Bn = j} xiy jti+j xiy jti+j D ◦ I − → i i + j xi+ay j+bti+j+a+b + j i + j xi+cy j+dti+j+c+d Introduce two operators, integration and differentiation, I[xiy jti+j] = t xiy jw i+j dw w = xiy j ti+j i + j D = xay bta+bΘx + xcy dtc+dΘy pn+1 = D ◦ I[pn] = . . . = (D ◦ I)n+1 [xa0y b0ta0+b0] –> ugly !

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Back to probability and symbolic II

pn(x, y, t) =

  • i,j

P {An = i, Bn = j} xiy jti+j

pn+1 = D ◦ I[pn]

D = xay bta+bΘx + xcy dtc+dΘy I[f (t)] = t f (w)dw w

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Back to probability and symbolic II

pn(x, y, t) =

  • i,j

P {An = i, Bn = j} xiy jti+j

pn+1 = D ◦ I[pn]

D = xay bta+bΘx + xcy dtc+dΘy I[f (t)] = t f (w)dw w Introduce a new Generating Function, ψn = I[pn] ψn(x, y, t) =

  • i,j

P {An = i, Bn = j} xiy j ti+j i + j

pn = t∂tψn and t∂tψn+1 = D(ψn)

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Back to probability and symbolic II

pn(x, y, t) =

  • i,j

P {An = i, Bn = j} xiy jti+j

pn+1 = D ◦ I[pn]

D = xay bta+bΘx + xcy dtc+dΘy I[f (t)] = t f (w)dw w Introduce a new Generating Function, ψn = I[pn] ψn(x, y, t) =

  • i,j

P {An = i, Bn = j} xiy j ti+j i + j

pn = t∂tψn and t∂tψn+1 = D(ψn)

Finally t∂t = x∂x + y∂y, thus Ψ =

n ψnzn verifies

  • (1 − zxay b) Θx + (1 − zxcy d) Θy
  • Ψ(x, y, z) = xa0y b0

for any additive urn !

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Probabilities and balance case

(PGF) P(x, y, z) =

  • n≥0
  • i,j

P {An = i, Bn = j} xi y j zn (weighted PGF) Ψ(x, y, z) =

  • n≥0
  • i,j

P {An = i, Bn = j} xi y j i + j zn Link :

P(x, y, z) = ∂tΨ(xt, yt, z)|t=1

(Histories GF) H(x, y, z) =

  • n≥0
  • i,j

Hn,i,j xi y j zn n!

Proposition

Let a b

c d

  • be additive balanced with balance σ > 0 and (a0, b0) the

starting configuration. With s0 = a0 + b0, the fonctions Ψ et H are linked by Ψ(x, y, z) = 1 σ 1 ts0/σ−1 H

  • x, y, z 1 − t

σ

  • dt

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Solve the PDE = Solve an ODE

(PDE)

  • (1 − zxay b) Θx + (1 − zxcy d) Θy
  • Ψ(x, y, z) = xa0y b0

Characteristic system : dx x (1 − zxay b) = dy y (1 − zxcy d) = dw xa0y b0 First integral, coming from (ODE) dx dy = x(1 − zxay b) y(1 − zxcy 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, Ψ(x, y, z) = y g(t, z, U(x, y, z))a0 tb0−1 1 − z g(t, z, U(x, y, z))c td dt

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Application I : Knuth cutting loops

K = −1 1 1

  • (a0, b0) = (m, 0)

Characteristic system : dx x(1 − zx−1y) = dy y(1 − zy) = dw xm First order differential equation : dx dy = x − zy y(1 − zy) General solution : g(y, z, u) = y 1 − zy (u − z ln(y)) First integral : U(x, y, z) = x(1 − zy) y + z ln(y)

ΨK(x, 1, z) = 1 tm−1 (1 − zt)m+1 (x(1 − z) − z ln(t))m dt

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Application II : diagonal urns

[Balaji–Mahmoud, 2006]

BM = a d

  • a, d > 0

(a0, b0) First order differential equation : dx dy = x (1 − zxa) y (1 − zy d) General solution : g(y, z, u) =

  • y −d − z

a/d ua + z −1/a First integral : U(x, y, z) =

  • y −d − z

1/d (x−a − z)1/a ΨBM(x, y, z) = y

  • (x−a − z)

t−d − z y −d − z a/d + z −a0/a tb0−1 1 − ztd dt

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Application III : Kotz–Mahmoud–Robert urn

[Kotz–Mahmoud–Robert, 2000]

KMR =

  • 1

1 1

  • (a0, b0)

ΨKMR(x, y, z) = x ta0−1 1 − zt

  • zt

1 − (1 − zt)(zU(x, y, z) − ln(1 − zt)) b0 dt where U(x, y, z) = y − zx + y(1 − zx) ln(1 − zx) yz(1 − zx) . For (a0, b0) = (0, 1), ΨKMR(1, y, z) = 1 z (1 − zt)

  • 1 − (1 − zt)
  • 1

1−z − z y(1−z) + ln(1 − z) − ln(1 − zt)

dt

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Extensions

PDE and characteristic method for urns with more than 2 colors PDE for urns with random integer coefficients

[Hosam Mahmoud’s talk on Series–Parallel graphs]

PDE for multiple drawings [Markus Kubas’ talk]

Example : random graph of n vertices

Start with n

2

  • black balls (for potential edges)

Each step : pick one ball. If it is black, choose with prob. p to turn into white (a real edge), with prob. 1 − p to discard it. −1 Bp

  • =

−1 1

  • 1Bp=1

+ −1

  • 1Bp=0

Ψ(x, y, z) verifies (x − (py + 1 − p)z) ∂xΨ + (y − zy) ∂yΨ = x(

n 2)

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Conclusion and Perspectives

general framework for additive urns exact expression of the probability generating function x asymptotics and limit distributions x understand the ordinary differential equation dx dy = x(1 − zxay b) y(1 − zxcy d)

Related topic : Counting histories in the (un)-balanced case. Classification

  • f algebraic histories generating functions (with A. Bostan and P. Dumas)

Advertissment June 26, 12 :30, PhD defense in Paris 6.

Thanks !

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