An analytic-combinatoric proof of Polya’s recurrencetheorem

Brian Burns

November 20, 2016

Abstract

We provide a direct proof of Polya’s famous recurrence theorem based on asymptotictechniques and the Cauchy integral formula. We also provide an introduction to thebasic notions from the theory of analytic combinatorics to show how one might “guess”a proof of Polya’s famous theorem.

1 Introduction.

“A drunk man returns to the bar, but a drunk bird may never return home”. The originof this confusing phrase is in the behavior of a simple random walk (SRW) on the latticeZd. The SRW on Zd is the path a particle traces on Zd if it is placed at the origin at time0 and if, at each second, it randomly chooses to move one unit length in the 2d availabledirections. More formally, a SRW is a sequence of random variables Yn∞n=1, with

Yn =n∑i=1

Xi

where the Xi are independently, identically distributed in −1, 1d.A SRW on Zd is said to be recurrent if it eventually returns to the origin with probability

1, and transient if not. An incredible theorem due to Polya [1] says that once d hits 3, spacebecomes too “big” for the SRW on Zd to be recurrent:

Theorem 1.1 (Polya’s recurrence theorem). The SRW on Zd is recurrent if d = 1, 2and is transient for d ≥ 3.

Remarkably, this theorem can be proven with essentially no combinatorics; complex analysisand asymptotic estimates of integrals stand in its stead. The purpose of this note is toexplain how, and to explain a correspondence between “counting” problems and “calculus”problems which greatly generalizes and illuminates the techniques used in the proof.

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2 Analytic combinatorics.

A remarkable theory of analytic combinatorics due to computer scientists Philippe Flajoletand Robert Sedgewick provides a systematic framework for the often ad-hoc use of generatingfunctions in combinatorial problems [2]. To a countable class of combinatorial objects C andassociated size function |·| 1, one assigns an ordinary generating function

C(z) =∑c∈C

z|c| =∞∑n=0

(#c ∈ C|n = |c|)zn.

This method of baking a combinatorial problem into a function is well-known, with its originsin the work of Euler. The insight of Sedgewick and Flajolet, however, is twofold:

• Combinatorial constructions “play nice” with this mapping: combinatorial sentencesare mapped to algebraic or differential equations, which may then be solved.

• The behavior of the function C(z) near its singularities can determine asymptoticproperties of the combinatorial objects in C as their size grows large.2

We acquiant the reader with the first principle via a few examples.

2.1 Examples of combinatorial sentences.

Given combinatorial classes A and B with size functions |·|A and |·|B, we may create thecombinatorial class A OR B whose set of objects is A ∪ B and whose size function is givenby

|c| =

|c|A c ∈ A|c|B c ∈ B.

Similarly we may create the combinatorial class A AND B whose set of objects is thecartesian product A× B and whose size function is given by

∣∣(a, b)∣∣ = |a|A + |b|B. If A andB have associated generating functions A(z) and B(z), it is easy to see that A OR B andA AND B have associated generating functions A(z) +B(z) and A(z)B(z) respectively.

We can iterate these basic constructions to make more complicated combinatorial classes.For example, consider SEQ A, the set of all (possibly empty) sequences of elements from A,whose size function is

∣∣(a1, . . . an)∣∣ = |a1|A+ · · ·+ |an|A. Note that a sequence of length n is

just an element of the n-fold cartestian product A× · · · × A, and therefore letting ε denotethe empty sequence (with size 0), we have

SEQ A = ε OR A OR (A AND A) OR (A AND A AND A) OR . . .

and therefore SEQ A has associated generating function

1 + A(z) + A(z)2 + · · · = 1

1− A(z).

1think of C, for example, as the set of all binary trees with |T | = number of nodes in the tree T .2This idea actually goes back to Hardy and Ramanujan in [3].

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To show this is not an exercise in mindless generalization, take the following two examples.A binary tree is either empty, or is a node with two trees. In addition, a simple combinatorialargument shows that the nth Fibonacci number is given by the number of ways to make a2× n tile out of 2× 1 and 1× 2 tiles, see [2]. The size of a tree is the number of nodes, andthe size of a tile is its horizontal length. Therefore we have two combinatorial classes T andF satisfying

T = ε OR (N AND T AND T ), F = SEQ G

where N is one-element combinatorial class containing a single node of size one, and G isthe two-element combinatorial class containing a vertical tile of size 1 and a horizontal tileof size 2. Therefore we get equations in generating functions

T (z) = 1 + zT (z)2, F (z) =1

1− (z + z2).

Solving the quadratic3 in T and applying partial fraction decomposition to F give the for-mulae

T (z) =1−√

1− 4z

2z, F (z) =

1√5

(1

1− (1+√5

2)z− 1

1− (1−√5

2)

)Applying Newton’s binomial formula to T and expanding F in geometric series give the well-known formulae for the number of binary trees and Fibonacci numbers of size n respectively:

Tn =1

n+ 1

(2n

n

), Fn =

1√5

(1 +√

5

2

)n

−

(1−√

5

2

)n .

3 The generating function Ω(~z, t).

3.1 Caveats.

In the proof that follows, we will deal with combinatorial classes whose elements have manysizes rather than just one, and we allow some of the sizes to be negative. This is to keeptrack of the motion of the random walk, whose position in Zd has d integer coordinates, notall of which of course must be positive.

Therefore instead of dealing with a single-variable power series in z, on a d-dimensionallattice our generating function will be a many-variable Laurent series in zα := zα1

1 . . . zαdd .

Moreover since we are dealing with a random variable, the coefficients in our generatingfunctions will be real-valued probabilities rather than integer-valued counts. Despite thesemodifications, analogues of the statements in section 2 carry over essentially verbatim.

3.2 SRW ordinary generating function.

With this in mind, we try to bake the behavior of a SRW into a generating function. To dothis, first note a SRW in Zd is a sequence of simple moves : stepping once forward in time

3We throw out one solution as a result of monotonicity considerations.

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and randomly stepping backwards or forwards in one of the d directions available. Thissimple move therefore has probabilistic specification

M = T AND (S1 OR S2 OR . . .Sd), where

• T is the one-element combinatorial class consisting of a single step forward in timewith generating function T (t, z1, . . . , zd) = t, and

• Si is the two-element combinatorial class containing a backwards move and a forwardsmove in the ith direction, each with probability 1

2dand therefore generating function

Si(t, z1, . . . zd) = zi + z−1i .

By the discussion in section one, our simple move thus is associated to the generating function

M(t, z1, . . . , zd) =t

2d

d∑i=1

zi + z−1i .

Since an SRW is a sequence of such simple moves, we have probabilistic specification SRW =SEQ M, so therefore the simple random walk on Zd has associated generating function

Ω(z1, . . . , zd, t) =1

1−(

t2d

∑di=1 zi + z−1i

) .The coefficient of zαtn in the Laurent series expansion of Ω about the origin4 is the probabilitythat the simple random walk is at position (α1, . . . , αd) at time t = n. Note we may alsotake coefficients in time and space alone rather than together. For example with d = 3,

[z11z−32 z53 ]Ω(~z, t)

is a function of t alone, whose power series expansion∑

n antn has the following property:

an is the chance that at time n, the random walk is at position (1,−3, 5).

4 Pure walks.

With this in mind, consider a special type of finite random walk: one which starts at theorigin and returns at time n, but never returns beforehand. Call such a walk a “pure” walk,and let pn be the probability that a random walk of size n is pure. Recurrence is equivalentto the assertion

∑∞n=0 pn = 1, so we take this line of attack.

Associate to the combinatorial class of pure walks the probability generating functionP (t) =

∑∞n=0 pnt

n with the convention that p0 = 0. Note that a random walk with position(0, . . . , 0) is a sequence of pure walks. So as before, we have

1

1− P (z)= [z

~0]Ω(~z, t) =⇒ P (z) = 1− 1

[z~0]Ω(~z, t).

4Which from hereon out we will write as [zαtn]Ω(~z, t).

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From Abel’s theorem on the convergence of power series or the monotone convergence the-orem applied to the counting measure on N, we may take t ↑ 1 to get

∞∑n=0

pn = limt↑1

P (t) = 1− 1

[z~0]Ω(~z, 1).

Since recurrence is equivalent to∑

n pn = 1, recurrence occurs if and only if

∞ = [z~0]Ω(~z, 1) = [z

~0]1

1− 12d

∑di=1(zi + z−1i )

=∞∑n=0

[z~0]

d∑i=0

zi + z−1i2d

n

.

With this in mind let

wn = [z~0]

d∑i=0

zi + z−1i2d

n

.

We seek to put an estimate on these numbers to see if the sum converges or diverges.

5 Probabilities as Cauchy integrals.

If γ is a positively-oriented circle around the origin, the fundamental calculation from com-plex analysis

1

2πi

∮γ

zn dz =

1 z = −1

0 otherwise

applied to the multinomial expansion of (∑d

i=1 zi + z−1i )n gives

wn =1

(2πi)d

∮γ

. . .

∮γ

1

z1 . . . zd

d∑i=0

zi + z−1i2d

n

dz1 . . . dzd

where we do d iterated contour integrals around the origin. Parametrizing the contour byγ(t) = eit and using cos(t) = 1

2(eit + e−it), we get

wn =1

(2π)d

∫[0,2π]d

1

d

d∑i=1

cos(ti)

n

dt1 . . . dtd.

6 Laplace’s Method.

It’s helpful to now stop and examine the situation we find ourselves in. The function f(~t) :=1d

∑di=1 cos(ti) is bounded by 1, so as n→∞ the only meaningful contributions to the above

integral will come from portions of space where f is very near 1 - other contributions willdecay exponentially in n. There are two of these portions of space, one localized about thepoint (0, . . . , 0) and one localized about the point (π, . . . , π); we’ll consider the first of themand multiply by two at the end to get the right answer.

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Since the above integral is essentially determined by the local behavior about ~t = 0, weblithely expand in a power series and notice the coincidence

1

d

d∑i=1

cos(ti) ≈ 1− 1

2d(t21 + · · ·+ t2d) ≈ e−

12(t21+...t

2d).

Therefore to approximate the integral formula for wn, we should be able to replace f(~t) bya Gaussian and call it a day. This is precisely the idea behind Laplace’s method, which isused to estimate extremely localized or extremely oscillatory integrals which crop up in bothreal and complex analysis. For a more detailed discussion, see, for example [4]: the proofis simply Fourier inversion. With this in mind we invoke Laplace’s method to obtain, asn→∞ the asymptotic estimate

wn ∼2

(2π)2

∫Rd

e−n2(t21+···+t2d)dt1 . . . dtd =

dd/2

2d−1πd/2n−d/2 = Cdn

−d/2.

Then by the comparison test for series convergence and our earlier discussion, we get

SRW Recurrent ⇐⇒∞∑n=1

n−d/2 diverges

which holds if and only if the dimension d is 1 or 2.

References

[1] Polya, G. “Uber eine Aufgabe der Wahrscheinlichkeitsrechnung betreffend die Irrfahrt imStrassennetz.” Math. Ann., 84:149-160, 1921.

[2] Philippe F., Sedgewick, R. Analytic Combinatorics. Cambridge University Press, 2009.

[3] Hardy, G. H. and Ramanujan S., “Asymptotic formulae in combinatory analysis”, Proc.London Math. Soc., 17(2) (1918) 75?115

[4] Bruijn, N. G. De. Asymptotic Methods in Analysis. Amsterdam. North-Holland Publish-ing Co., 1958.

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