Advanced Topics in Markov Chains

Advanced Topics in Markov chains

J.M. Swart May 16, 2012

Abstract This is a short advanced course in Markov chains, i.e., Markov processes with discrete space and time. The ﬁrst chapter recalls, without proof, some of the basic topics such as the (strong) Markov property, transience, recurrence, periodicity, and invariant laws, as well as some necessary background material on martingales. The main aim of the lecture is to show how topics such as harmonic functions, coupling, Perron-Frobenius theory, Doob transformations and intertwining are all related and can be used to study the properties of concrete chains, both qualitatively and quantitatively. In particular, the theory is applied to the study of ﬁrst exit problems and branching processes. 2

Notation

N natural numbers {0, 1,...} N+ positive natural numbers {1, 2,...} NN ∪ {∞} Z integers ZZ ∪ {−∞, ∞} Q rational numbers R real numbers R extended real numbers [−∞, ∞] C complex numbers B(E) Borel-σ-algebra on a topological space E 1A indicator function of the set A A ⊂ BA is a subset of B, which may be equal to B Ac complement of A A\B set diﬀerence A closure of A int(A) interior of A (Ω, F, P) underlying probability space ω typical element of Ω E expectation with respect to P σ(...) σ-ﬁeld generated by sets or random variables

kfk∞ supremumnorm kfk∞ := supx |f(x)| µ ν µ is absolutely continuous w.r.t. ν fk gk lim fk/gk = 0 fk ∼ gk lim fk/gk = 1 o(n) any function such that o(n)/n → 0

O(n) any function such that supn o(n)/n ≤ ∞ δx delta measure in x µ ⊗ ν product measure of µ and ν ⇒ weak convergence of probability laws Contents

0 Preliminaries 5 0.1 Stochastic processes ...... 5 0.2 Filtrations and stopping times ...... 6 0.3 Martingales ...... 7 0.4 Martingale convergence ...... 9 0.5 Markov chains ...... 10 0.6 Kernels, operators and linear algebra ...... 14 0.7 Strong Markov property ...... 15 0.8 Classiﬁcation of states ...... 17 0.9 Invariant laws ...... 19

1 Harmonic functions 23 1.1 (Sub-) harmonicity ...... 23 1.2 Random walk on a tree ...... 29 1.3 Coupling ...... 34 1.4 Convergence in total variation norm ...... 37

2 Eigenvalues 41 2.1 The Perron-Frobenius theorem ...... 41 2.2 Subadditivity ...... 43 2.3 Spectral radius ...... 44 2.4 Existence of the leading eigenvector ...... 45 2.5 Uniqueness of the leading eigenvector ...... 46 2.6 The Jordan normal form ...... 48 2.7 The spectrum of a nonnegative matrix ...... 51 2.8 Convergence to equilibrium ...... 53 2.9 Conditioning to stay inside a set ...... 55

3 Intertwining 61 3.1 Intertwining of Markov chains ...... 61 3.2 Markov functionals ...... 65 3.3 Intertwining and coupling ...... 69 3.4 First exit problems revisited ...... 72

4 Branching processes 75 4.1 The branching property ...... 75

3 4 CONTENTS

4.2 Generating functions ...... 78 4.3 The survival probability ...... 81 4.4 First moment formula ...... 84 4.5 Second moment formula ...... 86 4.6 Supercritical processes ...... 89 4.7 Trimmed processes ...... 90

A Supplementary material 97 A.1 The spectral radius ...... 97 Chapter 0

Preliminaries

0.1 Stochastic processes

Let I be a (possibly infinite) interval in Z. By definition, a stochastic process with discrete time is a collection of random variables X = (Xk)k∈I , defined on some underlying probability space (Ω, F, P) and taking values in some measurable space (E, E). We call the random function

I 3 k 7→ Xk(ω) ∈ E the sample path of the process X. The sample path of a discrete-time stochastic process is in fact itself a random variable X = (Xk)k∈I , taking values in the product space (EI , E I ), where

I E := {x = (xk)k∈I : xk ∈ E ∀k ∈ I} is the space of all functions x : I → E and E I denotes the product-σ-field. It is well-known that a probability law on (EI , E I ) is uniquely characterized by its finite-dimensional marginals, i.e., even if I is infinite, the law of the sample path X is uniquely determined by the finite dimensional distributions P (Xk,...,Xk+n) ∈ · ({k, . . . , k + n} ⊂ I). of the process. Conversely, if (E, E) is a Polish space equipped with its Borel-σ- field, then by the Daniell-Kolmogorov extension theorem, any consistent collection of probability measures on the finite-dimensional product spaces (EJ , E J ), with J ⊂ I a finite interval, uniquely defines a probability measure on (EI , E I ). Polish

5 6 CHAPTER 0. PRELIMINARIES spaces include many of the most commonly used spaces, such as countable spaces equipped with the discrete topology, Rd, separable Banach spaces, and much more. Moreover, open or closed subsets of Polish spaces are Polish, as are countable carthesian products of Polish spaces, equipped with the product topology.

0.2 Filtrations and stopping times

As before, let I be an interval in Z. A discrete filtration is a collection of σ-fields (Fk)k∈I such that Fk ⊂ Fk+1 for all k, k + 1 ∈ I. If X = (Xk)k∈I is a stochastic process, then X Fk := σ {Xj : j ∈ I, j ≤ k} (k ∈ I) is a filtration, called the filtration generated by X. For any filtration (Fk)k∈I , we set [ F∞ := σ Fk . k∈I X In particular, F∞ = σ((Xk)k∈I ).

A stochastic process X = (Xk)k∈I is adapted to a filtration (Fk)k∈I if Xk is Fk- X measurable for each k ∈ I. Then (Fk )k∈I is the smallest filtration that X is X adapted to, and X is adapted to a filtration (Fk)k∈I if and only if Fk ⊂ Fk for all k ∈ I.

Let (Fk)k∈I be a ﬁltration. An Fk- stopping time is a function τ :Ω → I ∪ {∞} such that the {0, 1}-valued process k 7→ 1{τ≤k} is Fk-adapted. Obviously, this is equivalent to the statement that

{τ ≤ k} ∈ Fk (k ∈ I).

If (Xk)k∈I is an E-valued stochastic process and A ⊂ E is measurable, then the ﬁrst entrance time of X into A

τA := inf{k ∈ I : Xk ∈ A} X with inf ∅ := ∞ is an Fk -stopping time. More generally, the same is true for the ﬁrst entrance time of X into A after σ

τσ,A := inf{k ∈ I : k > σ, Xk ∈ A}, where σ is an Fk-stopping time. Deterministic times are stopping times (w.r.t. any ﬁltration). Moreover, if σ, τ are Fk-stopping times, then also σ ∨ τ, σ ∧ τ 0.3. MARTINGALES 7

are Fk-stopping times. If f : I ∪ {∞} → I ∪ {∞} is measurable and f(k) ≥ k for all k ∈ I, and τ is an Fk-stopping time, then also f(τ) is an Fk-stopping time.

If X = (Xk)k∈I is an Fk-adapted stochastic process and τ is an Fk-stopping time, then the stopped process

ω 7→ Xk∧τ(ω)(ω)(k ∈ I) is also an Fk-adapted stochastic process. If τ < ∞ a.s., then moreover ω 7→ Xτ(ω)(ω) is a random variable. If τ is an Fk-stopping time defined on some filtered probability space (Ω, F, (Fk)k∈I , P) (with Fk ⊂ F for all k ∈ I), then the σ-field of events observable before τ is defined as Fτ := A ∈ F∞ : A ∩ {τ ≤ k} ∈ Fk ∀k ∈ I .

Exercise 0.1 If (Fk)k∈I is a ﬁltration and σ, τ are Fk-stopping times, then show that Fσ∧τ = Fσ ∧ Fτ .

Exercise 0.2 Let (Fk)k∈I be a ﬁltration, let X = (Xk)k∈I be an Fk-adapted X stochastic process and let τ be an Fk -stopping time. Let Yk := Xk∧τ denote the stopped process Show that the ﬁltration generated by Y is given by

Y X Fk = Fk∧τ k ∈ I ∪ {∞} . In particular, since this formula holds also for k = ∞, one has

X Fτ = σ (Xk∧τ )k∈I ,

X i.e., Fτ is the σ-algebra generated by the stopped process.

0.3 Martingales

By deﬁnition, a real stochastic process M = (Mk)k∈I , where I ⊂ Z is an interval, is an Fk-submartingale with respect to some ﬁltration (Fk)k∈I if M is Fk-adapted, E[|Mk|] < ∞ for all k ∈ I, and

E[Mk+1|Fk] ≥ Mk ({k, k + 1} ⊂ I). (0.1) We say that M is a supermartingale if the reverse inequality holds, i.e., if −M is a submartingale, and a martingale if equality holds in (0.1), i.e., M is both a 8 CHAPTER 0. PRELIMINARIES submartingale and a supermartingale. By induction, it is easy to show that (0.1) holds more generally when k, k + 1 are replaced by more general times k, m ∈ I with k ≤ m.

0 0 If M is an Fk-submartingale and (Fk)k≥0 is a smaller ﬁltration (i.e., Fk ⊂ Fk for all k ∈ I) that M is also adapted to, then

0 0 0 E[Mk+1|Fk] = E E[Mk+1|Fk]|Fk] ≥ E[Mk|Fk] = Mk ({k, k + 1} ⊂ I), which shows that M is also an Fk-submartingale. In particular, a stochastic process M is a submartingale with respect to some ﬁltration if and only if it is a M submartingale with respect to its own ﬁltration (Fk )k∈I . In this case, we simply say that M is a submartingale (resp. supermartingale, martingale).

Let (Fk)k∈I be a filtration and let (Fk−1)k∈I be the filtration shifted one step to left, where we set Fk−1 := {∅, Ω} if k − 1 6∈ I. Let X = (Xk)k∈I be a real Fk- adapted stochastic process such that E[|Xk|] < ∞ for all k ∈ I. By definition, a compensator of X w.r.t. the filtration (Fk)k∈I is an Fk−1-adapted real process K = (Kk)k∈I such that E[|Kk|] < ∞ for all k ∈ I and (Xk − Kk)k∈I is an Fk- martingale. It is not hard to show that K is a compensator if and only if K is Fk−1-adapted, E[|Kk|] < ∞ for all k ∈ I and Kk+1 − Kk = E Xk+1 Fk] − Xk ({k, k + 1} ⊂ I). T It follows that any two compensators must be equal up to an additive k∈I Fk−1- measurable random constant. In particular, if I = N, then because of the way we have defined F−1, such a constant must be deterministic. In this case, it is customary to put K0 := 0, i.e., we call

n X Kn := E Xk Fk−1] − Xk−1 (n ≥ 0) k=1 the (unique) compensator of X with respect to the ﬁltration (Fk)k∈N. We note that X is a submartingale if and only if its compensator is a.s. nondecreasing. The proof of the following basic fact can be found in, e.g., [Lach12, Thm 2.4].

Proposition 0.3 (Optional stopping) Let I ⊂ Z be an interval, (Fk)k∈I a ﬁltration, let τ be an Fk-stopping time and let (Mk)k∈I be an Fk-submartingale. Then the stopped process (Mk∧τ )k∈I is an Fk-submartingale.

The following proposition is a special case of [Lach12, Prop. 2.1]. 0.4. MARTINGALE CONVERGENCE 9

Proposition 0.4 (Conditioning on events up to a stopping time) Let I ⊂ Z be an interval, (Fk)k∈I a ﬁltration, let τ be an Fk-stopping time and let (Mk)k∈I be an Fk-submartingale. Then E Mk Fk∧τ ≥ Mk∧τ (k ∈ I).

0.4 Martingale convergence

If F, Fk (k ≥ 0) are σ-fields, then we say that Fk ↑ F if Fk ⊂ Fk+1 (k ≥ 0) and S F = σ( k≥0 Fk). Note that this is the same as saying that (Fk)k≥0 is a filtration and F = F∞, as we have defined it above. Similarly, if F, Fk (k ≥ 0) are σ-fields, T then we say that Fk ↓ F if Fk ⊃ Fk+1 (k ≥ 0) and F = k≥0 Fk.

Exercise 0.5 Let (Fk)k∈N be a ﬁltration and let τ be an Fk-stopping time. Show that Fk∧τ ↑ Fτ as k ↑ ∞.

The following proposition says that conditional expectations are continuous w.r.t. convergence of σ-fields. A proof can be found in, e.g., [Lach12, Prop. 4.12], [Chu74, Thm 9.4.8] or [Bil86, Thms 3.5.5 and 3.5.7]. Proposition 0.6 (Continuity in σ-field) Let X be a real random variable defined on a probability space (Ω, F, P) and let F∞, Fk ⊂ F (k ≥ 0) be σ-fields. Assume that E[|X|] < ∞ and Fk ↑ F∞ or Fk ↓ F∞. Then 1 E[X | Fk] −→ E[X | F∞] a.s. and in L -norm. k→∞

Note that if Fk ↑ F and E[|X|] < ∞, then Mk := E[X | Fk] deﬁnes a martingale. Proposition 0.6 says that such a martingale always converges. Conversely, we would like to know for which martingales (Mk)k≥0 there exists a ﬁnal element X such that Mk = E[X | Fk] . This leads to the problem of martingale convergence. Since each submartingale is the sum of a martingale and a nondecreasing compensator and since nondecreasing functions always converge, we may more or less equivalently ask the same question for submartingales. For a proof of the following fact we refer to, e.g., [Lach12, Thm 4.1].

Proposition 0.7 (Submartingale convergence) Let (Mk)k∈N be a submartingale such that supn≥0 E[Mk ∨ 0] < ∞. Then there exists a random variable M∞ with E[|M∞|] < ∞ such that

Mk −→ M∞ a.s. k→∞ 10 CHAPTER 0. PRELIMINARIES

In particular, this implies that nonnegative supermartingales converge almost surely. The same is not true for nonnegative submartingales: a counterexample is one-dimensional random walk reﬂected at the origin.

In general, even if M is a martingale, it need not be true that E[M∞] ≥ E[M0] (a counterexample is random walk stopped at the origin). We recall that a collection of random variables (Xk)k∈I is uniformly integrable if lim sup E |Xk|1{|Xk|≥n} = 0. n→∞ k∈I

1 Suﬃcient for this is that supk∈I E[ψ(|Xk|)] < ∞, where ψ : [0, ∞) → [0, ∞) is nonnegative, increasing, convex, and satisﬁes limr→∞ ψ(r)/r = ∞. Possible choices are for example ψ(r) = r2 or ψ(r) = (1 + r) log(1 + r) − r. It is well-known that uniform integrability and a.s. convergence of a sequence of real random variables imply convergence in L1-norm. For submartingales, the following result is known [Lach12, Thm 4.8].

Proposition 0.8 (Final element) In addition to the assumptions of Proposi- tion 0.7, assume that (Mk)k∈N is uniformly integrable. Then E |Mk − M∞| −→ 0 a.s. k→∞ and E[M∞|Fk] ≥ Mk for all k ≥ 0. If M is a martingale, then E[M∞|Fk] = Mk for all k ≥ 0.

Note that in particular, if M is a martingale, then Proposition 0.8 says that Mk = E[M∞|Fk], which shows that all information about the martingale M is hidden in its ﬁnal element M∞.

Combining Propositions 0.8 and 0.3, we see that if τ is an Fk-stopping time such that τ < ∞ a.s., (Mk)k∈N is an Fk-submartingale, and (Mk∧τ )k∈N is uniformly integrable, then E[Mτ ] = limk→∞ E[Mk∧τ ] ≥ M0.

There also exist convergence results for ‘backward’ martingales (Mk)k∈{−∞,...,0}.

0.5 Markov chains

Proposition 0.9 (Markov property) Let (E, E) be a measurable space, let I ⊂ Z be an interval and let (Xk)k∈I be an E-valued stochastic process. For each n ∈ I, 1By the De la Valle-Poussin theorem, this condition is in fact also necessary. 0.5. MARKOV CHAINS 11

− + X set I := {k ∈ I : k ≤ n} and I := {k ∈ I : k ≥ n}, and let F := σ((X ) − ) n n n k k∈In be the ﬁltration generated by X. Then the following conditions are equivalent.

(i) (X ) − ∈ A, (X ) + ∈ B X P k k∈In k k∈In n = (X ) − ∈ A X (X ) + ∈ B X a.s. P k k∈In n P k k∈In n − + for all A ∈ E In , B ∈ E In , n ∈ I.

X I+ (ii) (X ) + ∈ B F = (X ) + ∈ B X a.s. for all B ∈ E n , n ∈ I. P k k∈In n P k k∈In n X (iii) P Xn+1 ∈ C Fn = P Xn+1 ∈ C Xn a.s. for all C ∈ E, {n, n + 1} ⊂ I.

Remarks Property (i) says that the past and future are conditionally independent given the present. Property (ii) says that the future depends on the past only through the present, i.e., after we condition on the present, conditioning on the whole past does not give any extra information. Property (iii) says that it suﬃces to check (ii) for single time steps.

X Proof of Proposition 0.9 Set G := σ((X ) + ). If (i) holds, then, for any n k k∈In X X A ∈ Fn and B ∈ Gn , we have E 1AP[B |Xn] = E E 1AP[B | Xn] Xn = E E[1A | Xn]P[B | Xn] = E P[A | Xn]P[B | Xn] = E P[A ∩ B | Xn] = P[A ∩ B], where we have pulled the σ(Xn)-measurable random variable P[B | Xn] out of the X conditional expectation. Since this holds for arbitrary A ∈ Fn and since P[B |Xn] X is Fn -measurable, it follows from the deﬁnition of conditional expectations that

Since this holds for any C ∈ σ(Xn), it follows from the deﬁnition of conditional expectations that

P[A ∩ B | Xn] = P[A | Xn]P[B | Xn] a.s. 12 CHAPTER 0. PRELIMINARIES

To see that (iii) is suﬃcient for (ii), one ﬁrst proves by induction that

X P Xn+1 ∈ C1,...,Xn+m ∈ Cm Fn = P Xn+1 ∈ C1,...,Xn+m ∈ Cm Xn , and then uses that these sort events uniquely determine conditional probabilities X of events in Gn .

If a process X = (Xk)k∈I satisﬁes the equivalent conditions of Proposition 0.9, then we say that X has the Markov property. For processes with countable state spaces, there is an easier formulation.

Proposition 0.10 (Markov chains) Let I ⊂ Z be an interval and let X = (Xk)k∈I be a stochastic process taking values in a countable space S. Then X has the Markov property if and only if for each {k, k + 1} ⊂ I there exists a probability kernel Pk,k+1(x, y) on S such that

P[Xk = xk,...,Xk+n = xk+n] (0.2) = P[Xk = xk]Pk,k+1(xk, xk+1) ··· Pk+n−1,k+n(xk+n−1, xk+n) for all {k, . . . , k + n} ⊂ I, xk, . . . , xk+n ∈ S.

Proof See, e.g., [LP11, Thm 2.1].

If I = N, then Proposition 0.10 shows that the finite dimensional distributions, and hence the whole law of a Markov chain X are defined by its initial law P[X0 ∈ · ] and its transition probabilities Pk,k+1(x, y). If the initial law and transition probabilities are given, then it is easy to see that the probability laws defined by (0.2) are consistent, hence by the Daniell-Kolmogorov extension theorem, there exists a Markov chain X, unique in distribution, with this initial law and transition probabilies. We note that conversely, a Markov chain X determines its transition probabilities Pk,k+1(x, y) only for a.e. x ∈ S w.r.t. the law of Xk. If it is possible to choose the transition kernels Pk,k+1’s in such a way that they do not depend on k, then we say that the Markov chain is homogeneous. We are usually not interested in the problem to find Pk,k+1 given X, but typically we start with a given probability kernel P on S and are interested in all Markov chains that have P as their transition probability in each time step, and arbitrary initial law. Often, the word Markov chain is used in this more general sense. Thus, we often speak of ‘the Markov chain with state space S that jumps from x to y with probability. . . ’ without specifying 0.5. MARKOV CHAINS 13 the initial law. From now on, we use the convention that all Markov chains are homogeneous, unless explicitly stated otherwise. Moreover, if we don’t mention the initial law, then we mean the process started in an arbitrary initial law. We note from Proposition 0.9 (i) that the Markov property is symmetric under 0 time reversal, i.e., if (Xk)k∈I has the Markov property and I := {−k : k ∈ I}, then 0 0 0 0 the process X = (Xk)k∈I0 defined by Xk := X−k (k ∈ I ) also has the Markov property. It is usually not true, however, that X0 is homogeneous if X is. An exception are stationary processes, which leads to the useful concept of reversible laws.

Exercise 0.11 (Stopped Markov chain) Let X = (Xk)k≥0 be a Markov chain with countable state space S and transition kernel P , let A ⊂ S and let τA := inf{k ≥ 0 : Xk ∈ A} be the ﬁrst entrance time of A. Let Y be the stopped process

Yk := Xk∧τA (k ≥ 0). Show that Y is a Markov chain and determine its transition kernel. If we replace τA by the second entrance time of A, is Y then still Markov?

By deﬁnition, a random mapping representation of a probability kernel P on a countable state space S is a probability space (E, E, µ) together with a measurable function f : S ×E → S such that if Z1 is a random variable taking values in (E, E) with law µ, then

P (x, y) = P[f(x, Z1) = y](x, y ∈ S).

If (Zk)k≥1 are i.i.d. random variables with values in (E, E) and common law µ and X0 is a random variable taking values in S, independent of the (Zk)k≥1, then setting, inductively, Xk := f(Xk−1,Zk)(k ≥ 1) deﬁnes a Markov chain with transition kernel P and initial state X0. Random mapping representations are essential for simulating Markov chains on a computer. In addition, they have plenty of theoretical applications, for example for coupling Markov chains with diﬀerent initial states. (See Section 1.3 for an introduction to coupling.) One can show that each probability kernel has a random mapping representation, but such a representation is far from unique. Often, the key to a good proof is choosing the right one. We note that it is in general not true that functions of Markov chains are themselves Markov chains. An exception is the case when X P f(Xk+1) ∈ · |Fk ] depends only on f(Xk). In this case, we say that f(X) is an autonomous Markov chain. 14 CHAPTER 0. PRELIMINARIES

Lemma 0.12 (Autonomous Markov functional) Let X = (Xk)k∈I be a Mar- kov chain with countable state space S and transition kernel P . Let S0 be a countable set and let f : S → S0 be a function. Assume that there exists a probability kernel P 0 on S0 such that X P 0(x0, y0) = P (x, y0)(x ∈ S, f(x) = x0). y: f(y)=y0

0 Then f(X) := (f(Xk))k∈I is a Markov chain with state space S and transition kernel P 0.

0.6 Kernels, operators and linear algebra

Let X = (Xk)k∈I be a stochastic process taking values in a countable space S, and let P be a probability kernel on S. Then it is not hard to see that X is a Markov process with transition kernel P (and arbitrary initial law) if and only if

X P[Xk+1 = y | Fk ] = P (Xk, y) a.s. (y ∈ S, {k, k + 1} ⊂ I),

X where (Fk )k∈I is the ﬁltration generated by X. More generally, one has

X n P[Xk+n = y | Fk ] = P (Xk, y) a.s. (y ∈ S, n ≥ 0, {k, k + n} ⊂ I), where P n denotes the n-th power of the transition kernel P . Here, if K,L are probability kernels on S, then we define their product as X KL(x, z) := K(x, y)L(y, z)(x, z ∈ S), y∈S which is again a probability kernel on S. Then Kn is defined as the product of n 0 times K with itself, where K (x, y) := 1{x=y}. We may associate each probability kernel on S with a linear operator, acting on bounded real functions f : S → R, defined as X Kf(x) := K(x, y)f(y)(x ∈ S). y∈S Then KL is just the composition of the operators K and L, and for each bounded f : S → R, one has

X n E[f(Xk+n) | Fk ] = P f(Xk) a.s. (n ≥ 0, {k, k + n} ⊂ I), (0.3) 0.7. STRONG MARKOV PROPERTY 15 and this formula holds more generally provided the expectations are well-deﬁned (e.g., if E[|f(Xk+n)|] < ∞ or f ≥ 0). If µ is a probability measure on S and K is a probability kernel on S, then we may deﬁne a new probability measure µK on S by X µK(y) := µ(x)K(x, y)(y ∈ S). x∈S

In this notation, if X is a Markov process with transition kernel P and initial law n P[X0 ∈ · ] = µ, then P[Xn ∈ · ] = µP is its law at time n. We may view transition kernels as (possibly inﬁnite) matrices that act on row vectors µ or column vectors f by left and right multiplication, respectively.

0.7 Strong Markov property

We assume that the reader is familiar with some basic facts about Markov chains, as taught in the lecture [LP11]. In particular, this concerns the strong Markov property, which we formulate now.

Let X = (Xk)k≥0 be a Markov chain with countable state space S and transition kernel P . As usual, it goes without saying that X is homogeneous (i.e., we use the same P in each time step) and when we don’t mention the initial law, we mean the process started in an arbitrary initial law. Often, it is notationally convenient to assume that our process X is always the same, while the dependence on the initial law is expressed in the choice of the probability measure on our underlying probability space. More precisely, we assume that we have a measurable space (Ω, F) and a collection x X = (Xk)k≥0 of measurable maps Xk :Ω → S, as well as a collection (P )x∈S of probability measures on (Ω, F), such that under the measure Px, the process X is x a Markov chain with initial law P [X0 = x] = 1 and transition kernel P . In this P x set-up, if µ is any probability measure on S, then under the law P := x∈S µ(x)P , the process X is distributed as a Markov chain with initial law µ and transition kernel P .

x X If X, P, P are as just described and (Fk )k≥0 is te ﬁltration generated by X, then it follows from Proposition 0.9 (ii) and homogeneity that

X Xn P (Xn+k)k≥0 ∈ · Fn = P (Xk)k≥0 ∈ · a.s. (0.4) 16 CHAPTER 0. PRELIMINARIES

Here, for ﬁxed n ≥ 0, we consider (Xn+k)k≥0 as a random variable taking values in SN (i.e., this is the process Y deﬁned by Yk := Xn+k (k ≥ 0)). Since SN is a nice (in particular Polish) space, we can choose a regular version of the conditional probability on the left-hand side of (0.4), i.e., this is a random probability measure on SN. Since Xn is random, the same is true for the right-hand side. In words, formula (0.4) says that given the process up to time n, the process after time n is distributed as the process started in Xn. The strong Markov property extends this to stopping times.

Proposition 0.13 (Strong Markov property) Let X, P, Px be as deﬁned above. X Then, for any Fk -stopping time τ such that τ < ∞ a.s., one has

X Xτ P (Xτ+k)k≥0 ∈ · Fτ = P (Xk)k≥0 ∈ · a.s. (0.5)

Proof This follows from [LP11, Thm 2.3].

Remark 1 Even if P[τ = ∞] > 0, formula (0.5) still holds a.s. on the event {τ < ∞}. Remark 2 Homogeneity is essential for the strong Markov property, at least in the (useful) formulation of (0.5). Since this is closely related to formula (0.4), we also mention the following useful principle here. Proposition 0.14 (What can happen must eventually happen) Let X = (Xk)k≥0 be a Markov chain with countable state space S. Let B ⊂ SN be measurable x and set ρ(x) := P [(Xk)k≥0 ∈ B]. Then the event (Xn+k)k≥0 ∈ B for inﬁnitely many n ≥ 0 ∪ ρ(Xn) −→ 0 n→∞ has probability one.

Proof Let A denote the event that (Xn+k)k≥0 ∈ B for some n ≥ 0. Then by Proposition 0.6,

Xn X ρ(Xn) = P (Xk)k≥0 ∈ B = P (Xn+k)k≥0 ∈ B Fn X X ≤ A Fn −→ A F∞ = 1A a.s. P n→∞ P c In particular, this shows that ρ(Xn) → 0 a.s. on the event A . Applying the same argument to Am := {(Xn+k)k≥0 ∈ B for some n ≥ m}, we see that the event Am ∪ {ρ(Xn) → 0} has probability one for each m. Letting m ↑ ∞ and observing that Am ↓ {(Xn+k)k≥0 ∈ B for inﬁnitely many n ≥ 0}, the claim follows. 0.8. CLASSIFICATION OF STATES 17

0.8 Classiﬁcation of states

Let X be a Markov chain with countable state space S and transition kernel P . n For each x, y ∈ S, we write x y if P (x, y) > 0 for some n ≥ 0. Note that x y z implies x z. Two states x, y are called equivalent if x y and y x. It is easy to see that this deﬁnes an equivalence relation on S. A Markov chain / transition kernel is called irreducible if all states are equivalent. A state x is called recurrent if

x P [Xk = x for some k ≥ 1] = 1, otherwise it is called transient. If two states are equivalent and one of them is recurrent (resp. transient), then so is the other. Fix x ∈ S, let τ0 := 0 and let

τk := inf{n > τk−1 : Xn = x} (k ≥ 1) be the times of the k-th visit to x after time zero. Consider the process started in X0 = x. If x is recurrent, then τ1 < ∞ a.s. It follows from the strong Markov property that τ2 − τ1 is equally distributed with and independent of τ1. By induction, (τk − τk−1)k≥1 are i.i.d. In particular, τk < ∞ for all k ≥ 1, i.e., the process returns to x inﬁnitely often. On the other hand, if x is transient, then by the same sort of argument we see P that the number Nx = k≥1 1{Xk=x} of returns to x is geometrically distributed

x n n P [Nx = n] = θ (1 − θ) where θ := P [Xk = x for some k ≥ 1].

x In particular, E [Nx] < ∞ if and only if x is transient. Lemma 0.15 (Recurrent classes are closed) Let X be a Markov chain with countable state space S and transition kernel P . Assume that S0 ⊂ S is an equivalence class of recurrent states. Then P (x, y) = 0 for all x ∈ S0, y ∈ S\S0.

Proof Imagine that P (x, y) > 0 for some x ∈ S0, y ∈ S\S0. Then, since S0 is an equivalence class, y 6 x, i.e., the process cannot return from y to x. Since P (x, y) > 0, this shows that the process started in x has a positive probability never to return to x, a contradiction.

A state x is called positively recurrent if

x E [inf{n ≥ 1 : Xn = x}] < ∞. 18 CHAPTER 0. PRELIMINARIES

Recurrent states that are not positively recurrent are called null recurrent. If two states are equivalent and one of them is positively recurrent (resp. null recurrent), then so is the other. From this, it is easy to see that a ﬁnite equivalence class of states can never be null recurrent. The following lemma is an easy application of the principle ‘what can happen must happen’ (Proposition 0.14).

Lemma 0.16 (Finite state space) Let X = (Xk)k≥0 be a Markov chain with ﬁnite state space S and transition kernel P . Let Spos denote the set of all positively recurrent states. Then P[Xk ∈ Spos for some k ≥ 0] = 1.

By deﬁnition, the period of a state x is the greatest common divisor of {n ≥ 1 : P (x, x) > 0}. Equivalent states have the same period. States with period one are called aperiodic. If x is an aperiodic state, then there exists an m ≥ 1 such that P n(x, x) > 0 for all n ≥ m. Irreducible Markov chains are called aperiodic if one, and hence all states have period one. If X = (Xk)k≥0 is an irreducible Markov 0 chain with period n, then Xk := Xkn (k ≥ 0) deﬁnes an aperiodic Markov chain 0 0 X = (Xk)k≥0. The following example is of special importance. Lemma 0.17 (Recurrence of one-dimensional random walk) The Markov 1 chain X with state space Z and transition kernel P (k, k − 1) = P (k, k + 1) = 2 is null recurrent.

Proof Note that this Markov chain is irreducible and has period two, as it takes value alternatively in the even and odd integers. Using Stirling’s formula, it is not hard to show that (see [LP11, Example 2.9]) 1 P 2k(0, 0) ∼ √ as k → ∞. πk 0 In particular, this shows that the expected number of returns to the origin E [N0] = P∞ 2k k=1 P (0, 0) is inﬁnite, hence X is recurrent. To prove that X is not positively recurrent, we jump a bit ahead. By Theo- rem 0.18 (a) and (b) below, an irreducible Markov chain is positively recurrent if and only if it has an invariant law. It is not hard to check that any invariant measure for X must be inﬁnite, hence X has no invariant law, so it cannot be positively recurrent.

We will later see that, more generally, random walks on Zd are recurrent in dimensions d = 1, 2 and transient in dimensions d ≥ 3. 0.9. INVARIANT LAWS 19

0.9 Invariant laws

By deﬁnition, an invariant law for a Markov process with transition kernel P and countable state space S is a probability measure µ on S that is invariant under left-multiplication with P , i.e., µP = µ, or, written out per coordinate, X µ(y)P (y, x) = µ(x)(x ∈ S). y∈S

More generally, a (possibly inﬁnite) measure µ on S satisfying this equation is called an invariant measure. A probability measure µ on S is an invariant law if and only if the process (Xk)k≥0 started in the initial law P[X0 ∈ · ] = µ is (strictly) stationary. If µ is an invariant law, then there also exists a stationary process X = (Xk)k∈Z, unique in distribution, such that X is a Markov process with transition kernel P and P[Xk ∈ · ] = µ for all k ∈ Z (including negative times). A detailed proof of the following theorem can be found in [LP11, Thms 2.10 and 2.26].

Theorem 0.18 (Invariant laws) Let X be a Markov chain with countable state space S and transition kernel P . Then

(a) If µ is an invariant law and x is not positively recurrent, then µ(x) = 0.

(b) If S0 ⊂ S is an equivalence class of positively recurrent states, then there exists a unique invariant law µ of X such that µ(x) > 0 for all x ∈ S0 and µ(x) = 0 for all x ∈ S\S0.

x −1 µ(x) = E inf{k ≥ 1 : Xk = x} . (0.6)

Sketch of proof For any x ∈ S, deﬁne µ(x) as in (0.6), with 1/∞ := 0. Since consecutive return times are i.i.d., it is not hard to prove that

n 1 X x lim P [Xk = x] = µ(x), (0.7) n→∞ n k=1 i.e., the process started in x spends a µ(x)-fraction of its time in x. As a result, it is not hard to show that if x is transient or null-recurrent, then the process started 20 CHAPTER 0. PRELIMINARIES

in any initial law satisﬁes P[Xn = x] → 0 for n → ∞, hence no invariant law can give positive probability to such a state. On the other hand, if S0 ⊂ S is an equivalence class of positively recurrent states, then one can check that (0.7) holds more generally for the process started in any initial law on S0. It follows that for any such process, the C´esaro-limit

n 1 X µ = lim P[Xk ∈ · ] n→∞ n k=1 exists and does not depend on the initial law. In particular, if we start in an invariant law, then this limit must be µ, which proves uniqueness. It is not hard to check that any such C´esaro-limit must be an invariant law, from which we obtain existence.

Remark Using Lemma 0.15, it is not hard to prove that a general invariant law of the process is a convex combination of invariant laws that are concentrated on one equivalence class of positively recurrent states.

Theorem 0.19 (Convergence to invariant law) Let X be an irreducible positively recurrent Markov chain with invariant law µ. Then the process started in any initial law satisﬁes

n−1 1 X P[Xk = x] −→ µ(x)(x ∈ S). n n→∞ k=0 If X is moreover aperiodic, then one also has that

P[Xk = x] −→ µ(x)(x ∈ S). k→∞ On the other hand, if X is a Markov chain such that all states of X are transient or null recurrent, then the process started in any initial law satisﬁes

P[Xk = x] −→ 0 (x ∈ S). k→∞

Proof See [LP11, Thm 2.26].

If µ is an invariant law and X = (Xk)k∈Z is a stationary process such that P[Xk ∈ · ] = µ for all k ∈ Z, then by the symmetry of the Markov property w.r.t. time 0 0 0 reversal, the process X = (Xk)k∈Z deﬁned by Xk := X−k (k ∈ Z) is also a 0.9. INVARIANT LAWS 21

Markov process. By stationarity, X0 is moreover homogeneous, i.e., there exists 0 0 0 a transition kernel P such that the transition probabilities Pk,k+1 of X satisfy 0 0 P (x, y) = Pk,k+1(x, y) for a.e. x w.r.t. µ. In general, it will not be true that P 0 = P . We say that µ is a reversible law if µ is invariant and in addition, the stationary processes X and X0 are equal in law. One can check that this is equivalent to the detailed balance condition

µ(x)P (x, y) = P (x, y)µ(y)(x, y ∈ S), which says that the process X started in P[X0 ∈ · ] = µ satisfies P[X0 = x, X1 = y] = P[X0 = y, X1 = x]. More generally, a (possibly infinite) meaure µ on S satisfying detaied balance is called an reversible measure. If µ is reversible measure and we define a (semi-) inner product of real functions f : S → R by X hf, giµ := f(x)g(x)µ(x), x∈S then P is self-adjoint w.r.t. this inner product:

hf, P giµ = hP f, giµ. 22 CHAPTER 0. PRELIMINARIES Chapter 1

Harmonic functions

1.1 (Sub-) harmonicity

Let X be a Markov chain with countable state space S and transition kernel P . As we have seen, an invariant law of X is a vector that is invariant under left- multiplication with P . Harmonic functions 1 are functions that are invariant under right-multiplication with P . More precisely, we will say that a function h : S → R is subharmonic for X if X P (x, y)|h(y)| < ∞ (x ∈ S), y and X h(x) ≤ P (x, y)h(y)(x ∈ S). y We say that h is superharmonic if −h is subharmonic, and harmonic if it is both subharmonic and superharmonic. Lemma 1.1 (Harmonic functions and martingales) Assume that h is subharmonic for the Markov chain X = (Xk)k≥0 and that E[|h(Xk)|] < ∞ (k ≥ 0). Then Mk := h(Xk)(k ≥ 0) defines a submartingale M = (M(Xk))k≥0 w.r.t. to the X filtration (Fk )k≥0 generated by X. 1Historically, the term harmonic function was first used, and is still commonly used, for a smooth function f : U → R, defined on some open domain U ⊂ Rd, that solves the Laplace Pd ∂2 equation 2 f(x) = 0. This is basically the same as our definition, but with our Markov i=1 ∂xi chain X replaced by Brownian motion B = (Bt)t≥0. Indeed, a smooth function f solves the Laplace equation if and only if (f(Bt))t≥0 is a local martingale.

23 24 CHAPTER 1. HARMONIC FUNCTIONS

Proof This follows by writing (using (0.3)),

X X E h(Xk+1) Fk = P (Xk, y)h(y) ≥ h(Xk)(k ≥ 0). y

We will say that a state x is an absorbing state or trap for a Markov chain X if P (x, x) = 1.

Lemma 1.2 (Trapping probability) Let X be a Markov chain with countable state space S and transition kernel P , and let z ∈ S be a trap. Then the trapping probability x h(x) := P Xk = z for some k ≥ 0 is a harmonic function for X.

Proof Since 0 ≤ h ≤ 1, integrability is not an issue. Now

x h(x) = P Xk = z for some k ≥ 0 X x x = P Xk = z for some k ≥ 0 X1 = y P [X1 = y] y X y X = P (x, y)P Xk = z for some k ≥ 0 = P (x, y)h(y). y y

Lemma 1.3 (Trapping estimates) Let X be a Markov chain with countable state space S and transition kernel P , and let T := {z ∈ S : z is a trap}. Assume that the chain gets trapped a.s., i.e., P[∃n ≥ 0 s.t. Xn ∈ T ] = 1 (regardless of the initial law). Let z ∈ T and let h : S → [0, 1] be a subharmonic function such that h(z) = 1 and h ≡ 0 on T \{z}. Then

x h(x) ≤ P Xk = z for some k ≥ 0 If h is superharmonic, then the same holds with the inequality sign reversed.

Proof Since h is subharmonic, Mk := h(Xk) is a submartingale. Since h is bounded, M is uniformly integrable. Therefore, by Propositions 0.7 and 0.8, Mk → M∞ a.s. and in L1-norm, where M∞ is some random variable such that x E [M∞] ≥ M0 = h(x). Since the chain gets trapped a.s., we have M∞ = h(Xτ ), 1.1. (SUB-) HARMONICITY 25

where τ := inf{k ≥ 0 : Xk ∈ T } is the trapping time. Since h(z) = 1 and h ≡ 0 x x on T \{z}, we have M∞ = 1{Xτ =z} and therefore P [Xτ = z] = E [M∞] ≥ h(x). If h is superharmonic, the same holds with the inequality sign reversed.

Remark 1 If S0 ⊂ S is a ‘closed’ set in the sense that P(x, y) = 0 for all x ∈ S0, y ∈ S\S0, then define φ : S → (S\S0) ∪ {∗} by φ(x) := ∗ if x ∈ S0 and φ(x) := x 0 if x ∈ S\S . Now (φ(Xk))k≥0 is a Markov chain that gets trapped in ∗ if and only if the original chain enters the closed set S0. In this way, Lemma 1.3 can easily be generalized to Markov chains that eventually get ‘trapped’ in one of finitely many equivalence classes of recurrent states. In particular, this applies when S is finite. Remark 2 Lemma 1.3 tells us in particular that, provided that the chain gets trapped a.s., the function h from Lemma 1.2 is the unique harmonic function satisfying h(z) = 1 and h ≡ 0 on T \{z}. For a more general statement of this type, see Exercise 1.7 below. Remark 3 Even in situations where it is not feasable to calculate trapping probabilities exactly, Lemma 1.3 can sometimes be used to derive lower and upper bounds for these trapping probabilities. The following transformation is usually called an h-transform or Doob’s h-transform. Following [LPW09], we will simply call it a Doob transform.2

Lemma 1.4 (Doob transform) Let X be a Markov chain with countable state space S and transition kernel P , and let h : S → [0, ∞) be a nonnegative harmonic function. Then setting S0 := {x ∈ S : h(x) > 0} and P (x, y)h(y) P h(x, y) := (x, y ∈ S0) h(x) deﬁnes a transition kernel P h on S0.

Proof Obviously P h(x, y) ≥ 0 for all x, y ∈ S0. Since X X P h(x, y) = h(x)−1 P (x, y)h(y) = h(x)−1P h(x) = 1 (x ∈ S0), y∈S0 y∈S0

P h is a transition kernel. 2The term h-transform is somewhat inconvenient for several reasons. First of all, having mathematical symbols in names of chapters or articles causes all kinds of problems for referencing. Secondly, if one performs an h-transform with a function g, then should one speak of a g-transform or an h-transform? The situation becomes even more confusing when there are several functions around, one of which may be called h. 26 CHAPTER 1. HARMONIC FUNCTIONS

Proposition 1.5 (Conditioning on the future) Let X = (Xk)k≥0 be a Markov chain with countable state space S and transition kernel P , and let z ∈ S be a 0 0 trap. Set S := {y ∈ S : y z} and assume that P[X0 ∈ S ] > 0. Then, under the conditional law P[(Xk)k≥0 ∈ · Xm = z for some m ≥ 0 , the process X is a Markov process in S0 with Doob-transformed transition kernel P h, where x h(x) := P Xm = z for some m ≥ 0 satisﬁes h(x) > 0 if and only if x ∈ S0.

Proof Using the Markov property (in its strong form (0.4)), we observe that P Xn+1 = y (Xk)0≤k≤n = (xk)0≤k≤n,Xm = z for some m ≥ 0 = P Xn+1 = y (Xk)0≤k≤n = (xk)0≤k≤n,Xm = z for some m ≥ n + 1 P Xn+1 = y, Xm = z for some m ≥ n + 1 (Xk)0≤k≤n = (xk)0≤k≤n = P Xm = z for some m ≥ n + 1 (Xk)0≤k≤n = (xk)0≤k≤n y P (xn, y)P [Xm = z for some m ≥ 0] h = x = P (xn, y) P n [Xm = z for some m ≥ 1]

0 for each (xk)0≤k≤n and y such that P[(Xk)0≤k≤n = (xk)0≤k≤n] > 0 and xn, y ∈ S .

Remark At ﬁrst sight, it is surprising that conditioning on the future may preserve the Markov property. What is essential here is that being trapped in x is a tail event, i.e., an event measurable w.r.t. the tail-σ-algebra \ T := σ(Xk,Xk+1,...). k≥0

Similarly, if we condition a Markov chain (Xk)0≤k≤n that is defined on finite time interval on its final state Xn, then under the conditional law, (Xk)0≤k≤n is still Markov, although no longer homogeneous.

Exercise 1.6 (Suﬃcient conditions for integrability) Let h : S → R be any function. Assume that E[|h(X0)|] < ∞ and there exists a constant K < ∞ such P that y P (x, y)|h(y)| ≤ K|h(x)|. Show that E[|h(Xk)|] < ∞ (k ≥ 0).

Exercise 1.7 (Boundary conditions) Let X be a Markov chain with countable state space S and transition kernel P , and let T := {z ∈ S : z is a trap}. Assume 1.1. (SUB-) HARMONICITY 27

that the chain gets trapped a.s., i.e., P[∃n ≥ 0 s.t. Xn ∈ T ] = 1 (regardless of the initial law). Show that for each real function φ : T → R there exists a unique bounded harmonic function h : S → R such that h = φ on T . Hint: take h(x) := E[φ(Xτ )], where τ := inf{k ≥ 0 : Xk ∈ T } is the trapping time.

Exercise 1.8 (Conditions for getting trapped) If we do not know a priori that a Markov chain eventually gets trapped, then the following fact is often useful. Let X be a Markov chain with countable state space S and transition kernel P , and let h : S → [0, 1] be a sub- or superharmonic function. Assume that for all ε > 0 there exists a δ > 0 such that

x P |h(X1) − h(x)| ≥ δ ≥ δ whenever ε ≤ h(x) ≤ 1 − ε.

Show that limk→∞ h(Xk) ∈ {0, 1} a.s. Hint: use martingale convergence to prove that limk→∞ h(Xk) exists and use the principle ‘what can happen must happen’ (Proposition 0.14) to show that the limit cannot take values in (0, 1).

Exercise 1.9 (Trapping estimate) Let X,S,,P and h be as in Excercise 1.8. Assume that h is a submartingale and there is a point z ∈ S such that h(z) = 1 and supx∈S\{z} h(x) < 1. Show that x h(x) ≤ P [Xk = z for some k ≥ 0].

Exercise 1.10 (Compensator) Let X = (Xk)k≥0 be a Markov chain with countable state space S and transition kernel P , and let f : S → R be a function such P that y P (x, y)|f(y)| < ∞ for all x ∈ S. Assume that, for some given initial law, the process X satisﬁes E[|f(Xk)|] < ∞ for all k ≥ 0. Show that the compensator of (f(Xk))k≥0 is given by

n−1 X Kn = P f(Xk) − f(Xk) (n ≥ 0). k=0

Exercise 1.11 (Expected time till absorption: part 1) Let X be a Markov chain with countable state space S and transition kernel P , and let T := {z ∈ S : x z is a trap}. Let τ := inf{k ≥ 0 : Xk ∈ T } and assume that E [τ] < ∞ for all x ∈ S. Show that the function

x f(x) := E [τ] satisﬁes P f(x) − f(x) = −1 (x ∈ S\T ) and f ≡ 0 on T . 28 CHAPTER 1. HARMONIC FUNCTIONS

Exercise 1.12 (Expected time till absorption: part 2) Let X be a Markov chain with countable state space S and transition kernel P , let T := {z ∈ S : z is a trap}, and set τ := inf{k ≥ 0 : Xk ∈ T }. Assume that f : S → [0, ∞) satisﬁes P f(x) − f(x) ≤ −1 (x ∈ S\T ) and f ≡ 0 on T . Show that

x E [τ] ≤ f(x)(x ∈ S).

Hint: show that the compensator K of (f(Xk))k≥0 satisﬁes Kn ≤ −(n ∧ τ).

Exercise 1.13 (Absorption of random walk) Consider a random walk W = (Wk)k≥0 on Z that jumps from x to x + 1 with probability p and to x − 1 with the remaining probability q := 1 − p, where 0 < p < 1. Fix n ≥ 1 and set τ := inf k ≥ 0 : Wk ∈ {0, n} . Calculate, for each 0 ≤ x ≤ n, the probability P[Wτ = n].

Exercise 1.14 (First ocurrence of a pattern: part 1)

Let (Xk)k≥0 be i.i.d. Bernoulli random variables 1 with P[Xk = 0] = P[Xk = 1] = 2 (k ≥ 0). Set 000 τ110 := inf k ≥ 0 : (Xk,Xk+1,Xk+2) = (1, 1, 0) , 100 001 and deﬁne τ010 similarly. Calculate P[τ010 < τ110]. ~ Hint: Set Xk := (Xk,Xk+1,Xk+2)(k ≥ 0). Then ~ (Xk)k≥0 is a Markov chain with transition prob- 010 abilities as in the picture on the right. Now the problem amounts to calculating the trapping probabilities for the chain stopped at τ010 ∧ τ110. 101

110 011

111

Exercise 1.15 (First ocurrence of a pattern: part 2) In the set-up of the previous exercise, calculate E[τ110] and E[τ111]. Hints: you need to solve a system 1.2. RANDOM WALK ON A TREE 29 of linear equations. To ﬁnd the solution, it helps to use Theorem 0.18 (c) and the fact that the uniform distribution is an invariant law. In the case of τ111, it also x helps to observe that E [τ111] depends only on the number of ones at the end of x.

1.2 Random walk on a tree

In this section, we study random walk on an infinite tree in which every vertex has three neighbors. Such random walks have many interesting properties. At present they are of interest to us because they have many different bounded harmonic functions. As we will see in the next section, the situation for random walks on Zd is quite different.

Let T2 be an infinite tree, (i.e., a connected graph without cycles) in which each vertex has degree 3 (i.e., there are three edges incident to each vertex). We will be interested in the Markov chain whose state space are the vertices of T2 and that jumps in each step with equal probabilities to one of the three neighboring sites. We first need a convenient way to label vertices in such a tree. Consider a finitely generated group with generators a, b, c satisfying a = a−1, b = b−1 and c = c−1. More formally, we can construct such a group as follows. Let G be the set of all finite sequences x = x(1) ··· x(n) where n ≥ 0 (we allow for the empty sequence ∅), x(i) ∈ {a, b, c} for all 1 ≤ i ≤ n, and x(i) 6= x(i + 1) for all 1 ≤ i ≤ i + 1 ≤ n. We define a product on V by concatenation, where we apply the rule that any two a’s, b’s or c’s next to each other cancel each other, inductively, till we obtain an element of G. So, for example,

(abacb)(cab) = abacbcab, (abacb)(bab) = abacbbab = abacab, and (abacb)(bcb) = abacbbcb = abaccb = abab.

With these rules, G is a group with unit element ∅, the empty sequence, and inverse (x(1) ··· x(n))−1 = x(n) ··· x(1). Note that G is not abelian, i.e., the group product is not commutative. We can make G into a graph by drawing an edge between two elements x, y ∈ G if x = ya, x = yb, or x = yc. It is not hard to see that the resulting graph is an inﬁnite tree in which each vertex has degree 3; see Figure 1.1.3 We let

3This is a special case of a much more general construction. Let G be a ﬁnitely generated group and let ∆ ⊂ G be a ﬁnite, symmetric (in the sense that a ∈ ∆ implies a−1 ∈ ∆) set of elements that generates G. Draw a vertex between two elements a, b ∈ G if a = cb for some 30 CHAPTER 1. HARMONIC FUNCTIONS

|x| = |x(1) ··· x(n)| := |n| denote the length of an element x ∈ G. It is not hard to see that this is the same as the graph distance of x to the ‘origin’ ∅, i.e., the length of the shortest path connecting x to ∅.

aca aba

ac ab

acb a abc

bca cbc

∅ bc cb bcb cba b c

ba ca bac bab cab cac

Figure 1.1: The regular tree T2

Let X = (Xk)k≥0 be the Markov chain with state space G and transition probabilities 1 P (x, xa) = P (x, xb) = P (x, xc) = (x ∈ G), 3 i.e., X jumps in each step to a uniformly chosen neighboring vertex in the graph. We call X the nearest neighbor random walk on G.

We observe that if X is such a random walk on G, then |X| = (|Xk|)k≥0 is a c ∈ ∆ (or equivalently, by the symmetry of ∆, if b = c0a for some c0 ∈ ∆). The resulting graph is called the left Cayley graph associated with G and ∆. This is a very general method of making graphs with some sort of translation-invariant structure. 1.2. RANDOM WALK ON A TREE 31

Markov chain with state space N and transition probabilities given by 1 2 Q(n, n − 1) := and Q(n, n + 1) := (n ≥ 1), 3 3 and Q(0, 1) := 1. For each x = x(1) ··· x(n) ∈ G, let us write x(i) := ∅ if i > n. The following lemma shows that the random walk X is transient and walks away to inﬁnity in a well-deﬁned ‘direction’.

Lemma 1.16 (Transience) Let X be the random walk on G described above, + started in any initial law. Then there exists a random variable X∞ ∈ {a, b, c}N such that lim Xn(i) = X∞(i) a.s. (i ∈ +). n→∞ N

Proof We may compare |X| to a random walk Z = (Zk)k≥0 on Z that jumps from n to n − 1 or n + 1 with probabilities 1/3 and 2/3, respectively. Such a random walk has independent increments, i.e., (Zk − Zk−1)k≥1 are i.i.d. random variables that take the values −1 and +1 with probabilities 1/3 and 2/3. Therefore, by the strong law of large numbers, (Zn − Z0)/n → 1/3 a.s. and therefore Zn → ∞ a.s. In particular Z visits each state only ﬁnitely often, which shows that all states are transient. It follows that the process Z started in Z0 = 0 has a positive probability of not returning to 0. Since Zn → ∞ a.s. and since |X| has the same dynamics as Z as long as it is in N+, this shows that the process started in X0 = a satisﬁes

a 1 P |Xk| ≥ 1 ∀k ≥ 1 = P Zk ≥ 1 ∀k ≥ 1 > 0.

This shows that a is a transient state for X. By irreducibility, all states are transient and |Xk| → ∞ a.s., which is easily seen to imply the lemma.

We are now ready to prove the existence of a many bounded harmonic functions for the Markov chain X. Let

∂G := x ∈ {a, b, c}N+ : x(i) 6= x(i + 1) ∀i ≥ 1 .

Elements in ∂G correspond to different ways of walking to infinity. Note that ∂G is an uncountable set. In fact, if we identify elements of ∂G with points in [0, 1] written in base 3, then ∂G corresponds to a sort of Cantor set. We equip ∂G with the product-σ-field, which we denote by B(∂G). (Indeed, one can check that this is the Borel-σ-field associated with the product topology.) 32 CHAPTER 1. HARMONIC FUNCTIONS

Proposition 1.17 (Bounded harmonic functions) Let φ : ∂G → R be bounded and measurable, let X be the random walk on the tree G described above, and let X∞ be as in Lemma 1.16. Then

x h(x) := E φ(X∞) (x ∈ G) deﬁnes a bounded harmonic function for X. Moreover, the process started in an arbitrary initial law satisﬁes

h(Xn) −→ φ(X∞) a.s.. n→∞

Proof It follows from the Markov property (in the form (0.4)) that

x X y X h(x) = E [φ(X∞)] = P (x, y)E [φ(X∞)] = P (x, y)h(y)(x ∈ G), y y which shows that h is harmonic. Since khk∞ ≤ kφk∞, the function h is bounded. Moreover, by (0.4) and Proposition 0.6,

Xn X X h(Xn) = [φ(X∞)] = [φ(X∞) | Fn ] −→ [φ(X∞) | F∞] = φ(X∞) a.s. E E n→∞ E

For example, in Figure 1.2, we have drawn a few values of the harmonic function

x h(x) := P [X∞(1) = a](x ∈ G). Although Proposition 1.17 proves that each bounded measurable function φ on ∂G yields a bounded harmonic function for the process X, we have not actually shown that diﬀerent φ’s yield diﬀerent h’s.

Lemma 1.18 (Many bounded harmonics) Let µ be the probability measure ∅ on ∂G deﬁned by µ := P [X∞ ∈ · ]. Let φ, ψ : ∂G → R be bounded and measurable and let

x x h(x) := E φ(X∞) and g(x) := E ψ(X∞) (x ∈ G). Then h = g if and only if φ = ψ a.s. w.r.t. µ.

x Proof Let us deﬁne more generally µx = P [X∞ ∈ · ]. Since

X n z n µx(A) = P (x, z)P [X∞ ∈ · ] ≤ P (x, y)µy(A) z 1.2. RANDOM WALK ON A TREE 33

23 23 24 24

5 5 6 6 23 23 24 2 24 3

1 1 24 24 1 1 3 1 12 12 1 1 24 1 1 24 6 6

1 1 12 12 1 1 1 1 24 24 24 24

Figure 1.2: A bounded harmonic function

(x, y ∈ G, n ≥ 0,A ∈ B(∂G)) and P is irreducible, we see that µx µy for all x, y ∈ G, hence the measures (µx)x∈G are all equivalent. Thus, if φ = ψ a.s. w.r.t. µ, then they are a.s. equal w.r.t. to µx for each x ∈ G, and therefore Z Z h(x) = φdµx = ψdµx = g(x)(x ∈ G).

On the other hand, if the set {φ 6= ψ} has positive probability under µ, then by Proposition 1.17 ∅ lim h(Xn) 6= lim g(Xn) > 0, P n→∞ n→∞ which shows that there must exist x ∈ G with h(x) 6= g(x).

Exercise 1.19 (Escape probability) Let Z = (Zk)k≥0 be the Markov chain with state space Z that jumps in each step from n to n − 1 with probability 1/3 1 and to n + 1 with probability 2/3. Calculate P [Zk ≥ 1 ∀k ≥ 0]. Hint: ﬁnd a suitable harmonic function for the process stopped at zero. 34 CHAPTER 1. HARMONIC FUNCTIONS

Exercise 1.20 (Independent increments) Let (Yk)k≥1 be i.i.d. and uniformly distributed on {a, b, c}. Deﬁne (Xn)n≥0 by the random group product (in the group G) Xn := Y1 ··· Yn (n ≥ 1), with X0 := ∅. Show that X is the Markov chain with transition kernel P as deﬁned above.

1.3 Coupling

d Pd For any x = (x(1), . . . , x(d)) ∈ Z , let |x|1 := i=1 |x(i)| denote the ‘L1-norm’ of d x. Set ∆ := {x ∈ Z : |x|1 = 1}. Let (Yk)k≥1 be i.i.d. and uniformly distributed on ∆ and let n X Xn := Yk (n ≥ 1), k=1 with X0 := 0. (Here we also use the symbol 0 to denote the origin 0 = (0,..., 0) ∈ Zd.) Then, just as in Excercise 1.20, X is a Markov chain, that jumps in each time step from its present position x to a uniformly chosen position in x + ∆ = {x + y : y ∈ ∆}. We call X the symmetric nearest neighbor random walk on the integer lattice Zd. Sometimes X is also called simple random walk. Let P denote its transition kernel. We will be interested in bounded harmonic functions for P . We will show that in contrast to the random walk on the tree, the random walk on the integer lattice has very few bounded harmonic functions. Indeed, all such functions are constant. We will prove this using coupling, which is a technique of much more general interest, with many applications. Usually, when we talk about a random variable X (which may be the path of a process X = (Xk)k≥0), we are not so much interested in the concrete probability space (Ω, F, P) that X is defined on. Rather, all that we usually care about is the law P[X ∈ · ] of X. Likewise, when we have in mind two random variables X and Y (for example, one binomially and the other normally distributed, or X and Y may be two Markov chains with possibly different initial states or transition kernels), then we usually do not a priori know what their joint distribution is, even if we know there individual distributions. A coupling of two random variables X and Y , in the most general sense of the word, is a way to construct X and Y together on one underlying probability space (Ω, F, P). More precisely, if X and Y are random variables defined on different underlying probability spaces, then a coupling of X and Y is a pair of random variables (X0,Y 0) defined on one 1.3. COUPLING 35 underlying probability space (Ω, F, P), such that X0 is equally distributed with X and Y 0 is equally distributed with Y . Equivalently, since the laws of X and Y are all we really care about, we may say that a coupling of two probability laws µ, ν defined on measurable spaces (E, E) and (F, F), respectively, is a probability measure ρ on the product space (E × F, E ⊗ F) such that the first marginal of ρ is µ and its second marginal is ν. Obviously, a trivial way to couple any two random variables is to make them independent, but this is usually not what we are after. A typical coupling is designed to compare two random variables, for example by showing that they are close, or one is larger than the other. The next excercise gives a simple example.

Exercise 1.21 (Monotone coupling) Let X be uniformly distributed on [0, λ] and let Y be exponentially distributed with mean λ > 0. Show that X and Y can be coupled such that X ≤ Y a.s. (Hint: note that this says that you have to construct a probability measure on [0, λ] × [0, ∞) that is concentrated on {(x, y): x ≤ y} and has the ‘right’ marginals.) Use your coupling to prove that E[Xα] ≤ E[Y α] for all α > 0.

Now let ∆ ⊂ Zd be as defined at the beginning of this section and let P be the transition kernel on Zd defined by 1 P (x, y) := 1 (x, y ∈ d). 2d {y − x ∈ ∆} Z We are interested in bounded harmonic functions for P , i.e., bounded functions h : Zd → R such that P h = h. It is somewhat inconvenient that P is periodic.4 In light of this, we define a ‘lazy’ modification of our transition kernel by

1 1 Plazy(x, y) := 2 P (x, y) + 2 1{x=y}.

1 1 Obviously, Plazyf = 2 P f + 2 f, so a function h is harmonic for P if and only if it is harmonic for Plazy.

Proposition 1.22 (Coupling of lazy walks) Let Xx and Xy be two lazy random d walks, i.e., Markov chains on Z with transition kernel Plazy, and initial states x y d x y X0 = x and X0 = y, x, y ∈ Z . Then X and X can be coupled such that

x y ∃n ≥ 0 s.t. Xk = Xk ∀k ≥ n a.s.

4 d Indeed, the Markov chain with transition kernel P takes values alternatively in Zeven := {x ∈ d Pd d d Pd Z : i=1 x(i) is even} and Zodd := {x ∈ Z : i=1 x(i) is odd}. 36 CHAPTER 1. HARMONIC FUNCTIONS

Proof We start by choosing a suitable random mapping representation. Let (Uk)k≥1,(Ik)k≥1, and (Wk)k≥1 be collections of i.i.d. random variables, each collection independent of the others, such that for each k ≥ 1, Uk is uniformly distributed on {0, 1}, Ik is uniformly distributed on {1, . . . , d}, and Wk is uniformly distributed d on {−1, +1}. Let ei ∈ Z be deﬁned as ei(j) := 1{i=j}. Then we may construct x x X inductively by setting X0 := x and

x x Xk = Xk−1 + UkWkeIk (k ≥ 1).

Note that this says that Uk decides if we jump at all, Ik decides which coordinate jumps, and Wk decides whether up or down. y y To construct also X on the same probability space, we deﬁne inductively X0 := y and

( y y x X + (1 − Uk)WkeI if X (Ik) 6= X (Ik), Xy = k−1 k k−1 k−1 (k ≥ 1). k y y x Xk−1 + UkWkeIk if Xk−1(Ik) = Xk−1(Ik),

x y Note that this says that X and X always select the same coordinate Ik ∈ {1, . . . , d} that is allowed to move. As long as Xx and Xy differ in this coordinate, they jump at different times, but after the first time they agree in this cordinate, they always increase or decrease this coordinate by the same amount at the same time. In particular, these rules ensure that

x y x y Xk (i) = Xk (i) for all k ≥ τi := inf{n ≥ 0 : Xn (i) = Xn(i)}.

x y Since (Xk ,Xk )k≥0 is deﬁned in terms of i.i.d. random variables (Uk,Ik,Wk)k≥1 by a random mapping representation, the joint process (Xx,Xy) is clearly a Markov chain. We have already seen that Xx, on its own, is also a Markov chain, with the right transition kernel Plazy. It is straightforward to check that y x y y P[Xk = z | (Xk ,Xk )] = Plazy(Xk , z) a.s. In particular, this transition probability y y depends only on Xk , hence by Lemma 0.12, X is an autonomous Markov chain with transition kernel Plazy.

In view of this, our claim will follow provided we show that τi < ∞ a.s. for each i = 1, . . . , d. Fix i and deﬁne inductively σ0 := 0 and

σk := inf{k > σk−1 : Ik = i}.

Consider the diﬀerence process

D := Xx − Xy (k ≥ 0). k σk σk 1.4. CONVERGENCE IN TOTAL VARIATION NORM 37

Then D = (Dk)k≥0 is a Markov process on Z that in each step jumps from z to z + 1 or z − 1 with equal probabilities, except when it is in zero, which is a trap. In other words, this says that D is a simple random walk stopped at the ﬁrst time it hits zero. By Lemma 0.17, there a.s. exists some (random) k ≥ 0 such that Dk = 0 and hence τi = σk < ∞ a.s.

As a corollary of Proposition 1.22, we obtain that all bounded harmonic functions for nearest-neighbor random walk on the d-dimensional integer lattice are constant.

Corollary 1.23 (Bounded harmonic functions are constant) Let P (x, y) = −1 (2d) 1{|x−y|1=1} be the transition kernel of nearest-neighbor random walk on the d-dimensional integer lattice Zd. If h : Zd → R is bounded and satisﬁes P h = h, then h is constant.

Proof Couple Xy and Xy as in Proposition 1.22. Since h is harmonic and bounded, x y (h(Xk ))k≥0 and (h(Xk ))k≥0 are martingales. It follows that

x y h(x) − h(y) = E[h(Xk )] − E[h(Xk )] x y x y = E[h(Xk ) − h(Xk )] ≤ 2khk∞P[Xk 6= Xk ] −→ 0 k→∞ for each x, y ∈ Zd, proving that h is constant.

Remark Actually, a much stronger statement than Corollary 1.23 is true: for nearest-neighbor random walk on Zd, all nonnegative harmonic functions are constant. This is called the strong Liouville property, see [Woe00, Corollary 25.5]. In general, the problem of ﬁnding all positive harmonic functions for a Markov chain leads to the (rather diﬃcult) problem of determining the Martin boundary of a Markov chain.

1.4 Convergence in total variation norm

In this section we turn our attention away from harmonic functions and instead show another application of coupling. We will use coupling to give a proof of the statement in Theorem 0.19 (stated without proof in the Introduction) that any aperiodic, irreducible, positively recurrent Markov chain is ergodic, in the sense that regardless of the initial state, its law at time n converges to the invariant law as n → ∞. 38 CHAPTER 1. HARMONIC FUNCTIONS

Recall that the total variation distance between two probability measures µ, ν on a countable set S is deﬁned as

kµ − νkTV := max µ(A) − ν(A) A⊂S

The following lemma gives another formula for k · kTV.

Lemma 1.24 (Total variation distance) For any probability measures µ, ν on a countable set S, one has

X X X kµ−νkTV = µ(x)−ν(x) = ν(x)−µ(x) = µ(x)−ν(x) . x: µ(x)≥ν(x) x: µ(x)<ν(x) x∈S

Proof Set S+ := {x ∈ S : µ(x) > ν(x)}, S0 := {x ∈ S : µ(x) = ν(x)}, and S− := {x ∈ S : µ(x) < ν(x)}. Deﬁne a ﬁnite measure ρ by ρ(x) := |µ(x) − ν(x)|. Then

µ(A) − ν(A) = ρ(A ∩ S+) − ρ(A ∩ S−). It follows that

−ρ(S−) ≤ µ(A) − ν(A) ≤ ρ(S+) where either inequality may be an equality for a suitable choice of A (A = S− or A = S+, respectively). Here

X ρ(S+) − ρ(S−) = µ(x) − ν(x) = 1 − 1 = 0, x∈S so X X µ(x) − ν(x) = ρ(S+) = ρ(S−) = ν(x) − µ(x) . x: µ(x)≥ν(x) x: µ(x)<ν(x)

Theorem 1.25 (Convergence to invariant law) Let X be an irreducible, aperiodic, positively recurrent Markov chain with transition kernel P , state space S, and invariant law µ. Then the process started in any initial law satisﬁes

[Xn ∈ ·] − µ −→ 0. P TV n→∞ 1.4. CONVERGENCE IN TOTAL VARIATION NORM 39

Proof We take the existence of an invariant law µ as proven. Uniqueness will follow from our proof. Let X and X be two independent Markov chains with transition kernel P , where X is started in an arbitrary initial law and P[X0 ∈ · ] = µ. It is easy to see that the joint process (X, X) = (Xk, Xk)k≥0 is a Markov process with state space S × S. Let is denote its transition kernel by P2, i.e., by independence, P2 (x, x), (y, y) = P (x, y)P (x, y)(x, x, y, y ∈ S).

We claim that P2 is irreducible. Fix x, x, y, y ∈ S. Since P is irreducible and n aperiodic, it is not hard to see that there exists an m1 ≥ 1 such that P (x, y) > 0 n for all n ≥ m1. Likewise, there exists an m2 ≥ 1 such that P (x, y) > 0 for all n ≥ m2. Choosing n ≥ m1 ∨ m2, we see that

n n n P2 (x, x), (y, y) = P (x, y)P (x, y) > 0, proving that P2 is irreducible. By Theorem 0.18 (a) and (b), an irreducible Markov chain is positively recurrent if and only if it has an invariant law. Obviously, the product measure µ ⊗ µ is an invariant law for P2, so P2 is positively recurrent. In particular, this proves that the stopping time τ := inf{k ≥ 0 : Xk = Xk} 0 0 is a.s. finite and has, in fact, finite expectation. Let X = (Xk)k≥0 be the process defined by ( X if k < τ, 0 k Xk := Xk if k ≥ τ. It is not hard to see that X0 is a Markov chain with transition kernel P and initial 0 0 law P[X0 ∈ · ] = P[X0 ∈ · ], hence X is equal in law with X. Now

0 0 P[Xn ∈ · ] − µ TV = sup P[Xk ∈ A] − µ(A) = sup P[Xk ∈ A] − P[Xk ∈ A] A⊂S A⊂S 0 = sup P Xk ∈ A, Xk 6∈ A or Xk 6∈ A, Xk ∈ A A⊂S 0 ≤ P Xk 6= Xk = P[k < τ] −→ 0. k→∞

Exercise 1.26 (Periodic kernels) Show that the probability kernel P2 in the proof of Theorem 1.25 is not irreducible if P is periodic. 40 CHAPTER 1. HARMONIC FUNCTIONS Chapter 2

Eigenvalues

2.1 The Perron-Frobenius theorem

In this section, we recall the classical Perron-Frobenius theorem about the leading eigenvalue and eigenvector of a nonnegative matrix. This theorem is usually viewed as a part of linear algebra and proved in that context. We will see that the theorem has close connections with Markov chains and some of its statements can even be proved by probabilistic methods. Let S be a ﬁnite set (S = {1, . . . , n} in the traditional formulation of the Perron- Frobenius theorem) and let A : S × S → C be a function. We view such functions as matrices, equipped with the usual matrix product, or, equivalently, we identify S S P A with the linear operator A : C → C given by Af(x) := y∈S A(x, y)f(y), where, CS denotes the linear space consisting of all functions f : S → C. We say that A is real if A(x, y) ∈ R, and nonnegative if A(x, y) ≥ 0, for all x, y ∈ S. Note that probability kernels are nonnegative matrices. A nonnegative matrix A is called irreducible if for each x, y ∈ S there exists an n ≥ 1 such that An(x, y) > 0. For probability kernels, this coincides with our earlier deﬁnition of irreducibility. We let spec(A) denote the spectrum of A, i.e., the collection of (possibly complex) eigenvalues of A, and we let ρ(A) denote its spectral radius

ρ(A) := sup{|λ| : λ ∈ spec(A)}.

If k · k is any norm on CS, then we deﬁne the associated operator norm kAk of A as S kAk := sup{kAfk : f ∈ C , kfk = 1}.

41 42 CHAPTER 2. EIGENVALUES

It is well-known that for any such operator norm

ρ(A) = lim kAnk1/n. (2.1) n→∞

The following version of the Perron-Frobenius theorem can be found in [Gan00, Section 8.3] (see also, e.g., [Sen73, Chapter 1]). In this section, we will give our own proof, with a distinctly probabilistic ﬂavor, of this theorem and the remark below it.

Theorem 2.1 (Perron-Frobenius) Let S be a ﬁnite set and let A : CS → CS be a linear operator whose matrix is nonnegative and irreducible. Then

(i) There exist an f : S → [0, ∞), unique up to multiplication by a positive constant, and a unique α ∈ R such that Af = αf and f is not identically zero.

(ii) f(x) > 0 for all x ∈ S.

(iii) α = ρ(A) > 0.

(iv) The algebraic multiplicity of α is one. In particular, if A is written in its Jordan normal form, then α corresponds to a block of size 1 × 1.

Remark We deﬁne periodicity for nonnegative matrices in the same way as for probability kernels. If A is moreover aperiodic, then there exists some n ≥ 1 such that An(x, y) > 0 for all x, y ∈ S (i.e., the same n works for all x, y). Now Perron’s theorem [Gan00, Section 8.2] implies that all other eigenvalues λ of A satisfy |λ| < α. If A is not aperiodic, then it is easy to see that this statement fails in general. (This is stated incorrectly in [DZ98, Thm 3.1.1 (b)].)

We call the constant α and function f from Theorem 2.1 the Perron-Frobenius eigenvalue and eigenfunction of A, respectively. We note that if A†(x, y) := A(y, x) denotes the transpose of A, then A† is also nonnegative and irreducible. It is well- known that the spectra of a matrix and its transpose agree: spec(A) = spec(A†), and therefore also ρ(A) = ρ(A†), which implies that the Perron-Frobenius eigenvalues of A and A† are the same. The same is usually not true for the corresponding Perron-Frobenius eigenvectors. We call eigenvectors of A and A† also right and left eigenvectors, respectively, and write φA := A†φ, consistent with earlier notation for probability kernels. 2.2. SUBADDITIVITY 43

2.2 Subadditivity

By deﬁnition, a function f : N+ → R is subadditive if

f(n + m) ≤ f(n) + f(m)(n, m ≥ 1).

The following simple lemma has many applications. Proofs can be found in many places, e.g. [Lig99, Thm B.22].

Lemma 2.2 (Fekete’s lemma) If f : N+ → R is subadditive, then the limit

1 1 lim f(n) = inf f(n) n→∞ n n≥1 n exists in [−∞, ∞).

Proof Note that we can always extend f to a subadditive function f : N → R by setting f(0) = 0. Fix m ≥ 1 and for each n ≥ 0 write n = km(n)m + rm(n) where km(n) ≥ 0 and 0 ≤ rm(n) < m, i.e., km(n) is n/m rounded oﬀ to below and rm(n) is the remainder. Setting sm := sup1≤r

f(n) fk (n)m + r (n) k (n)f(m) + s f(m) = m m ≤ m m −→ , n km(n)m + rm(n) km(n)m n→∞ m which proves that f(n) f(m) lim sup ≤ (m ≥ 1). n→∞ n m Taking the inﬁmum over m we conclude that

f(n) f(m) lim sup ≤ inf . n→∞ n m≥1 m

This shows in particular that the limit superior is less or equal than the limit infe- rior, hence the limit exists. Moreover, the limit (which equals the limit superior) is given by the inﬁmum. Since f takes values in R, the inﬁmum is clearly less than +∞. 44 CHAPTER 2. EIGENVALUES

2.3 Spectral radius

Let S be a ﬁnite set and let L(CS, CS) denote the space of all linear operators A : CS → CS. Let k · k be any norm on CS and let kAk denote the associated operator norm of an operator A ∈ L(CS, CS). Then obviously

S kAfk = kA(f/kfk)k kfk ≤ kAk kfk (f ∈ C ), where the inequality between the left- and right-hand sides holds also for f = 0 even though in this case the intermediate steps are not deﬁned. It follows that kABfk ≤ kAk kBk kfk, which in turn shows that

S S kABk ≤ kAk kBk A, B ∈ L(C , C ) . (2.2)

Exercise 2.3 (Operator norm induced by supremumnorm) If k · k∞ denotes the supremumnorm on CS, then show that the associated operator norm on L(CS, CS) is given by X kAk∞ = sup |A(x, y)|. x∈S y∈S

Lemma 2.4 (Spectral radius) For each operator A ∈ L(CS, CS), the limit

ρ(A) := lim kAnk1/n n→∞ exists in [0, ∞). For each ε > 0 there exists a Cε < ∞ such that

n n n ρ(A) ≤ kA k ≤ Cε ρ(A) + ε (n ≥ 0). (2.3)

Proof It follows from (2.2) that kAn+mk ≤ kAnk kAmk. In other words, this says that the function n 7→ log kAnk is subadditive. By Lemma 2.2, it follows that the limit 1 1 lim log kAnk = inf log kAnk n→∞ n n≥1 n exists in [−∞, ∞). Applying the exponential function to both sides of this equation, we see that the limit

ρ(A) := lim kAnk1/n = inf kAnk1/n n→∞ n≥1 2.4. EXISTENCE OF THE LEADING EIGENVECTOR 45 exists in [0, ∞). In particular, this shows that ρ(A) ≤ kAnk1/n (n ≥ 1) and for each ε > 0 there exists an N ≥ 0 such that kAnk1/n ≤ ρ(A) + ε (n ≥ N). Raising both sides of our inequalities to the n-th power and choosing n −n Cε := 1 ∨ sup kA k(ρ(A) + ε) 0≤n

Exercise 2.5 (Choice of the norm) Show that the limit ρ(A) from Lemma 2.4 does not depend on the choice of the norm on CS. Hint: you can use the fact that on a ﬁnite-dimensional space, all norms are equivalent. In particular, if k · k and k · k0 are two diﬀerent norms on the space L(CS, CS), then there exist constants 0 < c < C < ∞ such that ckAk ≤ kAk0 ≤ CkAk for all A ∈ L(CS, CS).

2.4 Existence of the leading eigenvector

Let A be a nonnegative real matrix with coordinates indexed by a ﬁnite set S. The Perron-Frobenius theorem tells us that if A is irreducible, then A has a unique nonnegative eigenvector with eigenvalue ρ(A). In this section, we will prove the existence of such an eigenvector.

Lemma 2.6 (Existence of eigenvector) Let S be a ﬁnite set, let A : CS → CS be a linear operator whose matrix is nonnegative, and let ρ(A) be its spectral radius as deﬁned in Lemma 2.4. Then there exists a function h : S → [0, ∞) that is not identically zero such that Ah = ρ(A)h.

Proof We will treat the cases ρ(A) = 0 and ρ(A) > 0 separately. We start with the latter. In this case, for each z ∈ [0, 1/ρ(A)), let us deﬁne a function fz : S → [0, ∞) by ∞ X n n fz := z A 1, n=0 where 1 denotes the function on S that is identically one. Note that for each z < 1/ρ(A), this sequence is absolutely summable in any norm on CS by (2.3). It follows that ∞ X n n+1 −1 −1 Afz = z A 1 = z fz − 1 (0 < z < ρ(A) ). (2.4) n=0 46 CHAPTER 2. EIGENVALUES

By Excercise 2.3 and the nonnegativity of A (and hence of An for any n ≥ 0),

n X n n kA 1k∞ = sup A (x, y) = kA k∞. x∈S y∈S P Let kfk1 := x∈S |f(x)| denote the `1-norm of a function f : S → C. Then, by nonnegativity and (2.3) ∞ ∞ ∞ ∞ X n n X n n X n n X n n kfzk1 = z kA 1k1 ≥ z kA 1k∞ = z kA k∞ ≥ z ρ(A) . (2.5) n=0 n=0 n=0 n=0 S Let hz := fz/kfzk1. Since hz takes values in the compact set [0, 1] , we can choose −1 zn ↑ ρ(A) such that hzn → h where h is a nonnegative function h with khk1 = 1. Now (2.4) tells us that −1 −1 Ahzn = zn hzn − kfzn k1 . −1 By (2.5), kfzk1 → ∞ as z ↑ ρ(A) , so letting n → ∞ we obtain a nonnegative function h such that Ah = ρ(A)h and khk1 = 1. We are left with the case ρ(A) = 0. We will show that this implies An = 0 for some n ≥ 1. Indeed, by the ﬁniteness of the state space, if A|S| 6= 0, then by nonnegativity there must exist x0, . . . , xn ∈ S with x0 = xn and A(xk−1, xk) > 0 for nm m Qn each k = 1, . . . , n. But then A (x0, x0) ≥ η where η := k=1 A(xk−1, xk) > 0, which is easily seen to imply that ρ(A) ≥ η1/n > 0. Thus, we see that ρ(A) = 0 implies An = 0 for some n. Let m := inf{n ≥ 1 : An1 = 0}. Then h := Am−11 6= 0 while Ah = 0 = ρ(A)h.

2.5 Uniqueness of the leading eigenvector

In the previous section, we have shown that each nonnegative matrix A has a nonnegative eigenvector with eigenvalue ρ(A). In this section, we will show that if A is irreducible, then such an eigenvector is unique. We will use probabilistic methods. Our proof will be based on the following observation. Lemma 2.7 (Generalized Doob transform) Let S be a ﬁnite set and let A : CS → CS be a linear operator whose matrix is nonnegative. Assume that h : S → [0, ∞) is not identically zero, α > 0, and Ah = αh. Then setting S0 := {x ∈ S : h(x) > 0} and A(x, y)h(y) Ah(x, y) := (x, y ∈ S0) αh(x) deﬁnes a transition kernel Ah on S0. 2.5. UNIQUENESS OF THE LEADING EIGENVECTOR 47

Proof Obviously Ah(x, y) ≥ 0 for all x, y ∈ S0. Since X X Ah(x, y) = (αh(x))−1 A(x, y)h(y) = (αh(x))−1Ah(x) = 1 (x ∈ S0), y∈S0 y∈S0

Ah is a transition kernel.

We need one more preparatory lemma. Lemma 2.8 (Eigenfunctions of positively recurrent chains) Let P be a probability kernel on a ﬁnite or countably inﬁnite space S. Assume that the Markov chain X with transition kernel P is irreducible and positively recurrent. Let f : S → (0, ∞) be a bounded function and let α > 0 be a constant. Assume that P f = αf. Then α = 1 and f is constant.

Proof Since P is irreducible and positively recurrent, by Theorem 0.18 (b), it has a unique invariant law µ, which satisﬁes µ(x) > 0 for all x ∈ S. Now µf = µP f = µαf which by the fact that f > 0 proves that α = 1 and hence f is harmonic. Let X denote the Markov chain with transition kernel P , let Px denote the law x x of the chain started in X0 = x, and let E denote expectation w.r.t. P . Let δx denotes the delta measure in x, i.e., the probability measure on S such that δx({y}) := 1{x=y}. Then

n−1 n−1 n−1 1 X k 1 X k 1 X x f(x) = P f(x) = δxP f = E [f(Xk)] −→ µf (x ∈ S), n n n n→∞ k=0 k=0 k=0 where in the last step we have used Theorem 0.19. In particular, since the limit does not depend on x, the function f is constant.

We can now prove parts (i)–(iii) of the Perron-Frobenius theorem.

Proposition 2.9 (Perron-Frobenius) Let S be a ﬁnite set and let A : CS → CS be a linear operator whose matrix is nonnegative and irreducible. Then

(i) There exist an f : S → [0, ∞), unique up to multiplication by a positive constant, and a unique α ∈ R such that Af = αf and f is not identically zero.

(ii) f(x) > 0 for all x ∈ S.

(iii) α = ρ(A) > 0. 48 CHAPTER 2. EIGENVALUES

Proof Assume that f : S → [0, ∞) is not identically zero and satisﬁes Af = αf for some α ∈ R. Since f is nonnegative and not identically zero, we can choose x ∈ S such that f(x) > 0. Since A is nonnegative, Af(x) ≥ 0 and therefore α ≥ 0. In fact, since A is irreducible, we can choose some n ≥ 1 such that An(x, x) > 0 and hence Anf(x) > 0, so we must have α > 0. Moreover, by irreducibility, for each y ∈ S there is an m such that Am(y, x) > 0 and hence Amf(y) > 0, which shows that f(y) > 0 for all y ∈ S.

Now let h : S → [0, ∞) and β ∈ R be another function and real constant such that h is not identically zero and Ah = βh. We will show that this implies f = ch for some c > 0 and α = β. In particular, choosing h as in Lemma 2.6, this then implies the statements of the proposition. By what we have already proved, β > 0 and h(y) > 0 for all y ∈ S. Let Ah be the probability kernel on S0 = S (since h > 0 everywhere) deﬁned in Lemma 2.7. Note that Ah is irreducible since A is and since h > 0 everywhere. Set g := f/h. We observe that X A(x, y)h(y) f(y) Af(x) α Ahg(x) = = = g(x). βh(x) h(y) βh(x) β y∈S Since Ah is positively recurrent by the ﬁniteness of the state space, Lemma 2.8 implies that α/β = 1 and g(x) = c (x ∈ S) for some c > 0.

2.6 The Jordan normal form

In this section, we recall some facts from linear algebra. Let V be a ﬁnite- dimensional linear vector space over the complex numbers and let L(V,V ) denote the space of linear operators A : V → V . If {e1, . . . , ed} is a basis for V , then each vector φ ∈ V can in a unique way be written as

d X φ = φ(i)ei, i=1 where the φ(i) are complex numbers, the coordinates of φ with respect to the basis {e1, . . . , ed}. There exist complex numbers A(i, j) such that for any φ ∈ V , the coordinates of Aφ are given by

d X Aφ(i) = A(i, j)φ(j)(i = 1, . . . , d). j=1 2.6. THE JORDAN NORMAL FORM 49

We call (A(i, j))1≤i,j≤d the matrix of A w.r.t. basis {e1, . . . , ed}. We warn the reader that whether a matrix is nonnegative (i.e., wether its entries A(x, y) are all nonnegative) very much depends on the choice of the basis.1 The spectrum of A is the set of eigenvalues of A, i.e., spec(A) := λ ∈ C : ∃0 6= φ ∈ V s.t. Aφ = λφ .

The function C 3 z 7→ pA(z) deﬁned by

pA(z) := det(A − z) is the characteristic polynomial of A. We may write

n Y ma(λi) pA(z) = (λi − z) , i=1 where spec(A) = {λ1, . . . , λn}, with λ1, . . . , λn all diﬀerent, and ma(λ) is the algebraic multiplicity of the eigenvalue λ. We need to distinguish the algebraic multiplicity of an eigenvalue λ from its geometric multiplicity mg(λ), which is the dimension of the corresponding eigenspace {φ ∈ V : Aφ = λφ}. By deﬁni- tion, φ ∈ V is a generalized eigenvector of A with eigenvalue λ if (A − λ)kφ = 0 for some k ≥ 1. Then the algebraic multiplicity ma(λ) of λ is the dimension of the generalized eigenspace {φ ∈ V :(A − λ)kφ = 0 for some k ≥ 1}. Note that mg(λ) ≤ ma(λ).

It is possible to ﬁnd a basis {e1, . . . , ed} such that the matrix of A w.r.t. this basis has the Jordan normal form , i.e., the matrix of A has the block diagonal form   λk 1  B  1  ......   ..    A =  .  , with Bk =  .  .  .. 1  Bd λk We can read oﬀ both the algebraic and geometric multiplicity of an eigenvalue λ from the Jordan normal form of A. Indeed, ma(λ) is simply the number of times

1The statement that the matrix of a linear operator A is nonnegative should not be confused with the statement that A is nonnegative definite. The latter concept, which is defined only for linear spaces that are equipped with an inner product hφ, ψi, means that hφ, Aφi is real and nonnegative for all φ. One can show that A is nonnegative definite if and only if there exists an orthonormal basis of eigenvectors of A and all eigenvalues are nonnegative. Operators whose matrix is nonnegative w.r.t. some basis, by contrast, need not be diagonalizable and their eigenvalues can be negative or even complex. 50 CHAPTER 2. EIGENVALUES

λ occurs on the diagonal, while mg(λ) is the number of blocks having λ on the diagonal. Note that

 λ 1   1   λ  . .  .. ..   0   0         .   .  =  .  .  .. 1   .   .  λ 0 0 Each block in the Jordan normal form of A corresponds to an invariant subspace of generalized eigenvectors that contains exactly one eigenvector of A. The subspaces corresponding to all blocks with a given eigenvalue span the generalized eigenspace corresponding to this eigenvalue. To give a concrete example: if A has the Jordan normal form   λ1 1  λ1 1     λ1     λ1 1     λ1     λ2     λ3 1  λ3 then the eigenspace and generalized eigenspace corresponding to the eigenvalue λ1 consist of all vectors of the forms  φ(1)   φ(1)   0   φ(2)       0   φ(3)       φ(4)   φ(4)    and    0   φ(5)       0   0       0   0  0 0 repectively.

n 1/n Recall that in Lemma 2.4, we deﬁned ρ(A) as ρ(A) := limn→∞ kA k . The next lemma relates ρ(A) to the spectrum of A. This fact is well-known, but for completeness we include a proof in Appendix A.1.

Lemma 2.10 (Gelfand’s formula) Fix any norm on V and let kAk denote the associated operator norm of an operator A ∈ L(V,V ). Let ρ(A) be deﬁned as in 2.7. THE SPECTRUM OF A NONNEGATIVE MATRIX 51

Lemma 2.4. Then

ρ(A) = sup{|λ| : λ ∈ spec(A)} A ∈ L(V,V ).

2.7 The spectrum of a nonnegative matrix

We note that in the proof of Proposition 2.9, which contains the main conclusions of the Perron Frobenius theorem, we have used very little linear algebra. In fact, we have not even very often used the fact that our state space S is finite. We used the finiteness of S several times in the proof of Lemma 2.6 (existence of the leading eigenvalue), for example when we used that k1k1 < ∞ or when we used that any sequence of functions that is bounded in `1-norm has a convergent subsequence. Later, in the proof of Proposition 2.9, we needed the finiteness of S when we used that any irreducible transition kernel is positively recurrent. In both of these places, our arguments can sometimes be replaced by more ad hoc calculations to show that certain infinite dimensional nonnegative matrices also have a Perron- Frobenius eigenvector that is unique up to a multiplicative constant, in a suitable space of functions. In the present section, we will use the finiteness of S in a more essential way to prove the missing part (iv) of Theorem 2.1, as well as the remark below it. In particular, we will use that each finite-dimensional square matrix can be brought in its Jordan normal form.

Lemma 2.11 (Spectrum and Doob transform) Let S be a ﬁnite set and let A : CS → CS be a linear operator whose matrix is nonnegative. Assume that h : S → (0, ∞) and α > 0 satisfy Ah = αh, and let Ah be the probability kernel deﬁned in Lemma 2.7. Then Ah is irreducible, resp. apriodic if and only if A has these properties. Moreover, a function f : S → C is an eigenvector, resp. generalized eigenvector of A corresponding to some eigenvalue λ ∈ C if and only if f/h is an eigenvector, resp. generalized eigenvector of Ah corresponding to the eigenvalue λ/α.

Proof Let us use the symbol h to denote the linear operator on CS that maps a function f into hf. Similarly, let h−1 denote pointwise multiplication with the function 1/h(x). Then Lemma 2.7 says that Ah = α−1h−1Ah. Recall that f is an eigenvector, resp. generalized eigenvector of A corresponding to the eigenvalue λ if and only if (A − λ)f = 0, resp. (A − λ)kf = 0 for some k ≥ 1. Now 52 CHAPTER 2. EIGENVALUES

Ah − λ/α = α−1h−1Ah − α−1h−1λh = α−1h−1(A − λ)h and therefore

(Ah − λ/α)k = α−1h−1(A − λ)h ··· α−1h−1(A − λ)h = α−kh−1(A − λ)kh. | {z } k times

It follows that (Ah−λ/α)(f/h) = α−1h−1(A−λ)f is zero if and only if (A−λ)f = 0 and more generally (Ah − λ/α)k(f/h) = α−kh−1(A − λ)kf is zero if and only if (A − λ)kf = 0.

Exercise 2.12 (Non-diagonalizable kernel) Give an example of a probability kernel P that is not diagonalizable. Hint: It is easy to give an example of a nonnegative matrix that is not diagonalizable. Now apply Lemma 2.11.

Remark Regarding the excercise above: I do not know how to give an example such that P is irreducible.

Proposition 2.13 (Spectrum of a nonnegative matrix) Let S be a ﬁnite set, let A : CS → CS be a linear operator whose matrix is nonnegative and irreducible, and let α = ρ(A) be its Perron-Frobenius eigenvalue. Then the algebraic multiplicity of α is one. If A is moreover aperiodic, then all other eigenvalues λ of A satisfy |λ| < α.

Proof By Lemma 2.11 it suffices to prove the proposition for the case that A is a probability kernel. Writing P instead of A to remind us of this fact, we first observe that since the constant function 1 is nonnegative and satisfies P 1 = 1, by Proposition 2.9, this is the Perron-Frobenius eigenfunction of P and its eigenvalue is 1. In particular, α = ρ(P ) = 1. Let µ be the unique invariant law of P , which satisfies µ(x) > 0 for all x ∈ S.2 Set V := {f ∈ CS : µf = 0}. Since f ∈ V implies µP f = µf = 0, we see that P maps the space V into itself, i.e., V is an invariant subpace of P . Note that for any f ∈ CS, we have f − (µf)1 ∈ V . We observe that for any f ∈ CS, f − (µf)1 ∈ V . In fact, since f − c1 6∈ V for all c 6= µf, each function in CS can in a unique way be written as a linear combination of a function in V and the constant function 1, i.e., the space V has codimension one.

Let P denote the restriction of the linear operator P to the space V . We claim V that 1 is not an eigenvalue of P V . To prove this, imagine that f ∈ V satisﬁes 2Note that, actually, µ is the left Perron-Frobenius eigenvector of P . 2.8. CONVERGENCE TO EQUILIBRIUM 53

P f = f. Then, by the reasoning in Lemma 2.8,

n−1 1 X f(x) = P kf(x) −→ µf = 0 (x ∈ S). n n→∞ k=0 If P is aperiodic, a stronger conclusion can be drawn. In this case, if 0 6= f ∈ V satisﬁes P f = λf for some λ ∈ C, then by Theorem 0.19, n n n x λ f(x) = P f(x) = δxP f = [f(Xn)] −→ µf = 0 (x ∈ S), E n→∞ which is possible only if |λ| < 1.

Now imagine that f ∈ CS (not necessarily f ∈ V ) is a generalized eigenvector of P corresponding to some eigenvalue λ, i.e., (P − λ)kf = 0. Then we may write f in a unique way as f = g + c1 where g ∈ V and c ∈ C. Now 0 = (P − λ)kf = (P − λ)kg + c(1 − λ)k1, where (P − λ)kg ∈ V by the fact that V is an invariant subspace. Since each function in CS can in a unique way be written as a linear combination of a function in V and the constant function 1, this implies that both (P − λ)kg = 0 and c(1 − λ)k1 = 0. We now distinguish two cases.

If λ = 1, then since we have just proved that 1 6∈ spec(P V ), the operator P V has no generalized eigenvectors corresponding to the eigenvalue λ, and hence g = 0. This shows that the constant function 1 is (up to constant multiples) the only generalized eigenvector of P corresponding to the eigenvalue 1, i.e., the algebraic multiplicity of the eigenvalue 1 is 1. If λ 6= 1, then we must have c = 0 and hence f ∈ V . In this case, if P is aperiodic, then by what we have just proved we must have |λ| < 1.

Exercise 2.14 (Periodic kernel) Let P be the probability kernel on {0, . . . , n− 1} deﬁned by

P (x, y) = 1{y = x mod(n)} (0 ≤ x, y < n). Show that P is irreducible but not aperiodic. Determine the spectrum of P .

2.8 Convergence to equilibrium

Now that the Perron Frobenius theorem is proved, we can start to reap some of its useful consequences. The next proposition says that for irreducible, apriodic Markov chains with ﬁnite state space, convergence to the invariant law always happens exponentially fast. 54 CHAPTER 2. EIGENVALUES

Proposition 2.15 (Exponential convergence to equilibrium) Let S be a ﬁnite set and let P be an irreducible, aperiodic probability kernel on S. Then the Perron-Frobenius eigenvalue of P is ρ(P ) = 1 and its associated eigenvector is the constant function 1. Set

γ := sup |λ| : λ ∈ spec(P ), λ 6= 1 , and let µ denote the unique invariant law of P . Then γ < 1 and for any norm on S C and ε > 0, there exists a constant Cε < ∞ such that

n n S kP f − (µf)1k ≤ Cε(γ + ε) kf − (µf)1k (n ≥ 0, f ∈ C ).

Proof Obviously P 1 = 1 and 1 is a positive function, so by Proposition 2.9, this is the Perron-Frobenius eigenfunction of P and its eigenvalue is 1. As in the proof of Proposition 2.13, we set V := {f ∈ CS : µf = 0}. We observe that V is an S invariant subspace of P and f − (µf)1 ∈ V for all f ∈ C . Let P V denote the restriction of the linear operator P to V .

By Lemma 2.4, for any norm on CS and its associated operator norm, and for any S ε > 0, there exists a Cε < ∞ such that for any f ∈ C , kP nf − (µf)1k = kP n(f − (µf)1)k n n ≤ kP V k kf − (µf)k ≤ Cε ρ(P V ) + ε kf − (µf)k. As we have seen near the end of the proof of Proposition 2.13, if f is an eigenvector of P corresponding to an eigenvalue λ 6= 1, then we must have f ∈ V . By

Proposition 2.13, the algebraic multiplicity of 1 is 1, hence 1 6∈ spec(P V ). It follows that

spec(P )\{1} = spec(P V ), and therefore γ = ρ(P V ). By Proposition 2.13, all eigenvalues λ 6= 1 of P satisfy |λ| < 1, hence γ < 1.

Remark If γ is as in Proposition 2.15, then the positive quantity 1−γ is called the absolute spectral gap of the operator P . If X = (Xk)k≥0 denotes the Markov chain x with transition kernel P , then Proposition 2.15 says that the expectation E [f(Xn)] of any function f at time n converges to its equilibrium value µf exponentially fast: x −(η − ε)n E [f(Xn)] − µf ≤ Cεe , where η = log(1 − γ).

Here Cε can even be chosen uniformly in x (if we take k · k to be the supremumnorm). 2.9. CONDITIONING TO STAY INSIDE A SET 55

2.9 Conditioning to stay inside a set

Let X be a Markov chain with countable state space S and transition kernel P , and let S0 ⊂ S be a ﬁnite set of states such that

• P (x, y) > 0 for some x ∈ S, y 6∈ S, i.e., it is possible leave S0.

• The restriction of P to S0 is irreducible, i.e., it is possible to go from any state x in S0 to any other state y ∈ S0 without leaving S0.

We will be interested in the process X started in any initial state in S0 and conditioned not to leave S0 till time n, in the limit that n → ∞. Obviously, our assumptions imply that starting from anywhere in S0, there is a positive probability to leave S0 at some time in the future. Since S0 is ﬁnite, this probability is uniformly bounded from below, so by the principle ‘what can happen must happen’ (Proposition 0.14), we have that

0 P Xk ∈ S ∀k ≥ 0 = 0.

0 0 0 Let P := P S0 denote the restriction of P to S , i.e., the matrix (P (x, y))x,y∈S 0 and its associated linear operator acing on CS . Since P 0 is nonnegative and irreducible, it has a unique Perron-Frobenius eigenvalue α and associated left and right eigenfunctions η and h, unique up to multiplicative constants, i.e., η and h are strictly positive functions on S0 such that

ηP 0 = αη and P 0h = αh.

We choose our normalization of η and h such that

π(x) := η(x)h(x)(x ∈ S0) is a probability measure on S0. It turns out that π is the invariant law of the Doob transformed probability kernel (P 0)h deﬁned as in Lemma 2.7.

Lemma 2.16 (Invariant law of Doob transformed kernel) Let S0 be a ﬁnite 0 0 set and let A : CS → CS be linear operator whose matrix is irreducible and nonnegative. Let α, η and h be its Perron-Frobenius eigenvalue and left and right eigenvectors, respectively, normalized such that π(x) := η(x)h(x)(x ∈ S0) is a probability law on S0. Then π is the invariant law of the probability kernel Ah deﬁned in Lemma 2.7. 56 CHAPTER 2. EIGENVALUES

Proof Note that since Ah is irreducible, its invariant law is unique. Now X A(y, x)h(x) πAh(x) = η(y)h(y) αh(y) y∈S0 X = α−1h(x) η(y)A(y, x) = α−1(ηA)(x)h(x) = π(x), y∈S0 which proves that π is an invariant law.

For simplicity, we assume below that P 0 is not only irreducible but also aperiodic. Theorem 2.17 (Process conditioned not to leave a set) Let X be a Markov chain with countable state space S and transition kernel P , and let S0 ⊂ S be 0 ﬁnite. Assume that P := P S0 is irreducible and aperiodic. Let α and η, h denote its Perron-Frobenius eigenvalue and left and right eigenvectors, respectively, P P normalized such that x∈S0 η(x) = 1 and x∈S0 η(x)h(x) = 1. Set 0 τ := inf k ≥ 0 : Xk 6∈ S . Then, for any x ∈ S0, lim α−n x[n < τ] = h(x)(x ∈ S0). (2.6) n→∞ P Moreover, for each m ≥ 1 and x ∈ S0,

x x h (Xk)0≤k≤m ∈ · n < τ =⇒ (Xk )0≤k≤m ∈ · , P n→∞ P where Xh denotes the Markov chain with state space S0 and Doob transformed transition kernel P 0(x, y)h(y) P h(x, y) := (x, y ∈ S0). αh(x)

Proof We observe that, setting x0 := x, n x X X Y X 0n 0n P [n < τ] = ··· P (xk−1, xk) = P (x0, xn) = P 1(x). 0 0 0 x1∈S xn∈S k=1 xn∈S As in the proof of Lemma 2.11, we let h and h−1 also denote the operators that correspond to multiplication with these functions. Then P h = α−1h−1P 0h and therefore −n x −n 0n −n h −1n α P [n < τ] = α P 1(x) = α αhP h 1(x) = h(x)(P h)nh−1(x) −→ h(x)πh−1 = h(x), n→∞ 2.9. CONDITIONING TO STAY INSIDE A SET 57 where X X X πh−1 = π(x)h−1(x) = η(x)h(x)h−1(x) = η(x) = 1 x0∈S0 x0∈S0 x∈S0 is the average of the function h−1 with respect to the invariant law π of P h. To prove the remaining statement of the theorem, we observe that for any 0 ≤ 0 0m m < n, xm+1 ∈ S , and (x0, . . . , xm) ∈ S such that P (X0,...,Xm) = (x0, . . . , xm) > 0, we have

x0 P Xm+1 = xm+1 (X0,...,Xm) = (x0, . . . , xm), n < τ P P Qn 0 ··· 0 P (x , x ) xm+2∈S xn∈S k=1 k−1 k = P P Qm Qn 0 0 0 0 ··· 0 0 ( P (xk−1, xk))( P (x , x )) xm+1∈S xn∈S k=1 k=m+1 k−1 k P (x , x )P 0n−m−11(x ) = m m+1 m+1 . 0n−m P 1(xm)

This shows that conditional on the event n < τ, the process (X0,...,Xn) is a time-inhomogeneous Markov chain with transition kernel in the (m + 1)-th step given by

P (x, y)P 0n−m+11(y) P (x, y)α−(n−m−1)P 0n−m+11(y) P (n) (x, y) = = . m,m+1 P 0n−m1(x) αα−(n−m)P 0n−m1(x)

Since we have already shown that α−nP 0n1 → h as n → ∞, the result follows.

Exercise 2.18 (Behavior at typical times) Let 0 ≤ mn ≤ n be such that 0 mn → ∞ and n − mn → ∞ as n → ∞. Show that for any x ∈ S , P Xmn ∈ · n < τ = π, where π = ηh is the invariant law of the Doob transformed probability kernel P h.

Excercise 2.18 shows that conditional on the unlikely event that n < τ where n is large, most of the time up to n, the process X is approximately distributed according to the invariant law π of the Doob transformed probability kernel P h. We will see below that at the time n, the situation is diﬀerent. 58 CHAPTER 2. EIGENVALUES

Let X be a Markov chain with countable state space S and transition kernel P , and let S0 ⊂ S be ﬁnite. By deﬁnition, we say that a probability measure ρ on S0 is a quasi-stationary law3 of P on S0 if

0 P[X0 ∈ · ] = ρ implies P[X1 ∈ · | X1 ∈ S ] = ρ. It seems that quasi-stationary laws were ﬁrst introduced by Darroch and Seneta in [DS67].

Proposition 2.19 (Quasi stationary law) Let X be a Markov chain with countable state space S and transition kernel P , and let S0 ⊂ S be ﬁnite. Assume 0 that P := P S0 is irreducible and aperiodic. Let α and η, h denote its Perron- Frobenius eigenvalue and left and right eigenvectors, respectively, normalized such P P that x∈S0 η(x) = 1 and x∈S0 η(x)h(x) = 1. Then η is the unique quasi- stationary law of P on S0 and the process started in any initial law satisﬁes lim Xn ∈ · n < τ] = η, n→∞ P

0 where τ := inf{k ≥ 0 : Xk 6∈ S }.

0 Proof Deﬁne ﬁnite measures µn on S by

0 µn(x) := P[Xn = x, n < τ](x ∈ S , n ≥ 0). Then

X 0 µn+1(x) = P[Xn+1 = x, n + 1 < τ] = P[Xn = y, n < τ]P (y, x) = µnP (x) y0∈S0

0 0 −1 (x ∈ S ), i.e., µn+1 = µnP . As before, we let h and h denote the operators that correspond to multiplication with these functions. Then P h = α−1h−1P 0h and therefore

0n h −1n n h n −1 µn = µ0P = µ0 αhP h = α µ0h(P ) h . (2.7)

Let us deﬁne a probability measure ν on S0 by

−1 0 ν(x) := (µ0h) µ(x)h(x)(x ∈ S ).

3Usually, the terms ‘stationary law’ and ‘invariant law’ can be used exchangeably. In this case, this may lead to confusion, however, since the term ‘quasi-invariant measure’ is normally used in ergodic theory for a measure that is mapped into an equivalent measure by some transformation. 2.9. CONDITIONING TO STAY INSIDE A SET 59

We observe that n h n n h n α µ0h(P ) = α (µ0h)ν(P ) , where ν(P h)n is just the law at time n of a Markov chain with transition kernel P h and initial law ν. Since P h is irreducible and aperiodic, ν(P h)n converges as n → ∞ to the invariant law π given by π(x) = η(x)h(x). Inserting this into (2.7), we see that −n h n −1 −1 α µn = µ0h(P ) h −→ (µ0h)ηhh = (µ0h)η. n→∞ It follows that

−n µn α µn (µ0h)η P Xn ∈ · n < τ] = = −n =⇒ = η, µn1 α µn1 n→∞ (µ0h)η1 where we have used that η1 = 1 by the choice of our normalization. 60 CHAPTER 2. EIGENVALUES Chapter 3

Intertwining

3.1 Intertwining of Markov chains

In linear algebra, an intertwining relation is a relation between linear operators of the form AB = BA.˜ In particular, if B is invertible, then this says that A = BAB˜ −1, or, in other words, that the matrices A and A˜ are similar. In this case, we may associate B with a change of basis, and A˜ corresponds to the matrix A written in terms of a diﬀerent basis. In general, however, B need not be invertible. It is especially in this case that the word intertwining is used. We will be especially interested in the case that A, A˜ and B are (linear operators corresponding to) probability kernels. So let us assume that we have probability kernels P, P˜ on countable spaces S, S˜, respectively, and a probability kernel K from S to S˜ (i.e., S × S˜ 3 (x, y) 7→ K(x, y)), such that

PK = KP.˜ (3.1)

˜ ˜ Note that P, P˜ and K correspond to linear operators P : CS → CS, P˜ : CS → CS, ˜ and K : CS → CS, so both sides of the equation (3.1) correspond to a linear ˜ operator from CS into CS. We need some examples to see this sort of relations between probability kernels can really happen.

Exclusion process. Fix n ≥ 2 and let Cn := Z/n, i.e., Cn = {0, . . . , n − 1} with addition modulo n. (I.e., Cn is the cyclic group with n elements.) Let

61 62 CHAPTER 3. INTERTWINING

S := {0, 1}Cn , i.e., S consists of all ﬁnite sequences x = (x(0), . . . , x(n − 1)) indexed by Cn. Let (Ik)k≥1 be i.i.d. and uniformly distributed on Cn. For each x ∈ S, we may deﬁne a Markov chain X = (Xk)k≥0, started in X0 = x and with values in S, by setting  Xk(I + 1) if i = I,  Xk+1(i) := Xk(I) if i = I + 1, (3.2)  Xk(i) otherwise.

In words, this says that in each time step, we choose a uniformly distributed position I ∈ Cn and exchange the values of X in I and I + 1 (where we calculate modulo n). Note that we have described our Markov chain in terms of a random mapping representation. In particular, it is clear from this construction that X is a Markov chain. Thinning Let C be any countable set and let S := {0, 1}C be the set of all x = (x(i))i∈C with x(i) ∈ {0, 1} for all i ∈ C, i.e., S is the set of all sequences of zeros and ones indexed by C. Fix 0 ≤ p ≤ 1, and let (χi)i∈C be i.i.d. Bernoulli (i.e., {0, 1}-valued) random variables with P[χi = 1] = p. We deﬁne a probability kernel Kp from S to S by Kp(x, · ) = P χix(i) i∈C ∈ · (x ∈ S). (3.3) Note that χix(i) i∈C is obtained from x by setting some coordinates of x to zero, independently for each i ∈ C, where each coordinate x(i) that is one has probability p to remain one and probability 1 − p to become a zero. We describe this procedure as thinning the ones with parameter p.

Exercise 3.1 (Thinning of exclusion processes) Let P be the transition ker- Cn nel of the exclusion process on Cn described above, with state space S = {0, 1} , and for 0 ≤ p ≤ 1, let Kp be the kernel from S to S corresponding to thinning with parameter p. Show that

PKp = KpP (0 ≤ p ≤ 1).

Sum of two processes Let P be a transition kernel on a countable space S and let X(1) = (Xk(1))k≥0 and X(2) = (Xk(2))k≥0 be two independent Markov chains with transition kernel P and possibly diﬀerent deterministic initial states X0(i) = x(i)(i = 1, 2). Then (X(1),X(2)) = (Xk(1),Xk(2))k≥0 is a Markov chain with values in the product space S × S. We may view (X(1),X(2)) as two particles, 3.1. INTERTWINING OF MARKOV CHAINS 63 walking around in the space S. Now maybe we are not interested in which particle is where, but only in how many particles are on which place. In that case, we may look at the process

Yk(i) := 1{Xk(1)=i} + 1{Xk(2)=i} (i ∈ S, k ≥ 0), which takes values in the space S˜ consisting of all functions y : S → {0, 1, 2} such P that i∈S y(i) = 2. Note that Y just counts how many particles are present on each site i ∈ S. It is not hard to see that Y = (Yk)k≥0 is an autonomous Markov chain.

Exercise 3.2 (Counting process) Let (X(1),X(2)) and Y be the Markov chains ˜ with state spaces S × S and S described above. Let P2 be the transition kernel of (X(1),X(2)) and let P˜ be the transition kernel of Y . For each y ∈ S˜, let K(y, · ) be the uniform distribution on the set Uy := x(1), x(2) ∈ S × S : y = 1x(1) + 1x(2) ,

˜ where 1x(i) := 1{x=i}. Note that K is a probability kernel from S to S × S. From y, we can see that there are two particles and where these are, but not which is the ﬁrst and which is the second particle. All K does is arbitrarily ordering the particles in y. Show that ˜ PK = KP2. This example can easily be generalized to any number of independent Markov chains (all with the same transition kernel).

Conditioning on the future As a third example, we look at a Markov chain X with ﬁnite state space S and transition kernel P . We assume that P is such that all states in S are transient, except for two states z0 and z1, which are traps. We set

x hi(x) := P Xk = zi for some k ≥ 0 (i = 0, 1).

By Lemma 1.2, we know that h0 and h1 are harmonic functions. Since all states except z0 and z1 are transient and our state space is ﬁnite, we have h0 + h1 = 1. By Proposition 1.5, the process X conditioned to be eventually trapped in zi is itself a Markov chain, with state space {x ∈ S : hi(x) > 0} and Doob transformed transition kernel P hi . Set

˜ S := (x, i): x ∈ S, i ∈ {0, 1}, hi(x) > 0 . 64 CHAPTER 3. INTERTWINING

We deﬁne a probability kernel P˜ on S˜ by

˜ hi P (x, i), (y, j) := 1{i=j}P (x, y)(x, y ∈ S, i, j ∈ {0, 1}). ˜ ˜ ˜ Note that if (X,I) = (Xk,Ik)k≥0 is a Markov chain with transition kernel P , then I never changes its value, and depending on whether Ik = 0 for all k ≥ 0 or Ik = 1 for all k ≥ 0, the process X˜ is our original Markov chain X conditioned to be trapped in either z0 or z1. Exercise 3.3 (Conditioning on the future) In the example above, deﬁne a probability kernel K from S to S˜ by K x, (y, i) := 1{x=y}hi(x)(x, y ∈ S, i ∈ {0, 1}). Show that PK = KP.˜

Returning to our general set-up, we observe that the intertwining relation (3.1) implies that for any probability measure µ on S µP nK = µKP˜n (n ≥ 0). This function has the following interpretation. If we start the Markov chain with transition kernel P in the initial law µ, run it till time n, and then apply the kernel K to its law, then the result is the same as if we start the the Markov chain with transition kernel P˜ in the initial law µK, run it till time n, and look at its law. More concretely, in our three examples, this says:

• If we start an exclusion process in some initial law, run it till time n, and then thin it with the parameter p, then the result is the same as when we thin the initial state with p, and then run the process till time n. • If we start the counting process Y in any initial law, run it to time n, and then arbitrarily order the particles, then the result is the same as when we ﬁrst arbitrarily order the particles, and then run the process (X(1),X(2)) till time n. • If we run the two-trap Markov chain X till time n, and then assign it a value 0 or 1 according to the probability, given its present state Xn, that it will eventually get trapped in z0 or z1, respectively, then the result is the same as when we assign such values at time zero, and then run the appropriate Doob transformed Markov chain till time n. 3.2. MARKOV FUNCTIONALS 65

None of these statement comes as a big surprise, but we will later see less trivial examples of this phenomenon.

3.2 Markov functionals

We already know that functions of Markov chains usually do not have the Markov property. An exception, as we have seen in Lemma 0.12, is the case when a function of a Markov chain is autonomous. Let us quickly recall what this means. Let X be a Markov chain with countable state space S and transition kernel P , and let ψ : S → S˜ be a function from S into some other countable set S˜. Then we say that (Yk)k≥0 := (ψ(Xk))k≥0 is an autonomous Markov chain if P ψ(Xk+1) = y Xk = x depends on x only through ψ(x). Equivalently, this says that there exists a transition kernel P˜ on S˜ such that X P˜(y, y0) = P (x, x0) ∀x ∈ S s.t. ψ(x) = y. (3.4) x0: ψ(x0)=y0 If (3.4) holds, then, regardless of the initial law of X, one has that the process ˜ ψ(X) = (ψ(Xk))k≥0 is a Markov chain with transition kernel P . The next theorem shows that sometimes, a function ψ(X) of a Markov chain X can be a Markov chain itself, even when it is not autonomous. In this case, however, this is usually only true for certain special initial laws of X. It seems this result is due to Rogers and Pitman [RP81]. For better comparison with other results, in the theorem below, it will be convenient to interchange the roles of X and Y , i.e., X will be a function of Y and not the other way around. Note that if X and Y are random variables such that P[Y ∈ · |X] = K(X, · ), then formula (3.5) below says that ψ(Y ) = X a.s. Thus, the probability kernel K from S to S˜ is in a sense the ‘inverse’ of the function ψ : S˜ → S. Theorem 3.4 (Markov functionals) Let Y be a Markov chain with countable state space S˜ and transition kernel P , and let ψ : S˜ → S be a function from S˜ into some other countable set S. Let K be a probability kernel from S to S˜ such that y ∈ S˜ : K(x, y) > 0 ⊂ y ∈ S˜ : ψ(y) = x (x ∈ S), (3.5) and let P˜ be a transition kernel on S such that PK˜ = KP. (3.6) 66 CHAPTER 3. INTERTWINING

Then ˜ P Y0 = y ψ(Y0) = K(ψ(Y0), y) a.s. (y ∈ S) (3.7) implies that ˜ P Yn = y ψ(Y0), . . . , ψ(Yn) = K(ψ(Yn), y) a.s. (y ∈ S, n ≥ 0). (3.8)

Moreover, (3.7) implies that the process ψ(Y ) = (ψ(Yk))k≥0, on its own, is a Markov chain with transition kernel P˜.

Proof Formula (3.8) says that P Yn = y ψ(Y0), . . . , ψ(Yn) = (x0, . . . , xn) = K(xn, y) a.s.

n n ≥ 0, (x0, . . . , xn) ∈ S such that P[ ψ(Y0), . . . , ψ(Yn) = (x0, . . . , xn) > 0, ˜ and y ∈ S such that ψ(y) = xn. We will prove this by induction. Assume that the statement holds for some n ≥ 0. We wish to prove the statement for n + 1. Let (x0, . . . , xn+1) be such that P[ ψ(Y0), . . . , ψ(Yn+1) = (x0, . . . , xn+1) > 0, ˜ and let y ∈ S be such that ψ(y) = xn+1. Then P Yn+1 = y ψ(Y0), . . . , ψ(Yn+1) = (x0, . . . , xn+1) P Yn+1 = y, ψ(Yn+1) = xn+1 ψ(Y0), . . . , ψ(Yn) = (x0, . . . , xn) = P ψ(Yn+1) = xn+1 ψ(Y0), . . . , ψ(Yn) = (x0, . . . , xn) (3.9) P Yn+1 = y ψ(Y0), . . . , ψ(Yn) = (x0, . . . , xn) = . P ψ(Yn+1) = xn+1 ψ(Y0), . . . , ψ(Yn) = (x0, . . . , xn) We will treat the nominator and denominator separately. To shorten notation, let us denote the conditional law given ψ(Y0), . . . , ψ(Yn) = (x0, . . . , xn) by P(x0,...,xn) · := P · ψ(Y0), . . . , ψ(Yn) = (x0, . . . , xn) . Then P(x0,...,xn) Yn+1 = y X 0 0 = P(x0,...,xn) Yn+1 = y Yn = y P(x0,...,xn) Yn = y . 0 0 y :ψ(y )=xn 3.2. MARKOV FUNCTIONALS 67

Here, by our induction assumption,

0 0 0 ˜ 0 P(x0,...,xn) Yn = y = K(xn, y ) ∀y ∈ S s.t. ψ(y ) = xn, while by the Markov property of Y ,

0 P(x0,...,xn) Yn+1 = y Yn = y 0 = P Yn+1 = y Yn = y , ψ(Y0), . . . , ψ(Yn) = (x0, . . . , xn) 0 0 = P Yn+1 = y Yn = y = P (y , y). Thus, we see that

X 0 0 P(x0,...,xn) Yn+1 = y = K(xn, y )P (y , y) 0 0 y : ψ(y )=xn X 0 0 0 ˜ = K(xn, y )P (y , y) = KP (xn, y ) = PK(xn, y), y0∈S˜

0 0 where we have used that K(xn, · ) is concentrated on {y : ψ(y ) = xn} and in the last step we have applied our intertwining assumption.

We observe that by (3.5) and the fact that ψ(y) = xn+1, ˜ X PK(xn, y) = P (xn, x)K(x, y) x∈S˜ X = P (xn, x)K(x, y)1{ψ(y)=x} = P (xn, xn+1)K(xn+1, y), x∈S˜ and therefore P(x0,...,xn) Yn+1 = y = P (xn, xn+1)K(xn+1, y). It follows that X P(x0,...,xn) ψ(Yn+1) = xn+1 = P(x0,...,xn) Yn+1 = y y: ψ(y)=xn+1 (3.10) X ˜ = P (xn, xn+1)K(xn+1, y) = P (xn, xn+1).

y: ψ(y)=xn+1

Inserting our last two formulas into (3.9), we obtain that P Yn+1 = y ψ(Y0), . . . , ψ(Yn+1) = (x0, . . . , xn+1) P (x , x )K(x , y) = n n+1 n+1 = K(x , y). ˜ n+1 P (xn, xn+1) 68 CHAPTER 3. INTERTWINING

This completes our proof of (3.8). Moreover, formula (3.10) shows that ψ(Y ) is a Markov chain with transition kernel P˜.

Let us see how Theorem 3.4 relates to the examples developed in the previous section. Counting process Let (X(1),X(2)) be two independent Markov processes, where each process takes values in S and has transition kernel P , as in Excercise 3.2. Let Y and S˜ be as deﬁned there and let ψ : S × S → S˜ be the function

ψ(x(1), x(2)) := 1x(1) + 1x(2), so that Y = ψ(X(1),X(2)). Then the kernel K defined in Excercise 3.2 satisfies (x(1), x(2)) : Ky, (x(1), x(2)) > 0 ⊂ {(x(1), x(2)) : ψ(x(1), x(2)) = y . ˜ Since moreover PK = KP2, all assumptions of Theorem 3.4 are fulfilled, so we find that P (X0(1),X0(2)) = (x(1), x(2)) Y0 = K Y0, (x(1), x(2)) a.s. (3.11) for all (x(1), x(2)) ∈ S × S implies that P (Xn(1),Xn(2)) = (x(1), x(2)) (Y0,...,Yn) = K Yn, (x(1), x(2)) a.s. (3.12) for all (x(1), x(2)) ∈ S × S and n ≥ 0, and under the same assumption, Y = ψ(X(1),X(2)) is a Markov chain with transition kernel P˜. Note that this latter conclusion holds in fact even without the assumption (3.11), since Y is autonomous. Formula (3.12) tells us that if initially we do not know the order of the particles, then by observing the process Y up to time n, we obtain no information about the order of the particles. Conditioning on the future Let X and (X,I˜ ) be Markov chains with state spaces S and S˜ and transition kernels P and P˜ as in Excercise 3.3. Thus, X is a Markov ˜ chain that eventually gets trapped in one of two traps z0 and z1, and (X,I) is the same process where the second coordinate I tells us from the beginning in which of the two traps we will end up. Define ψ : S˜ → S by ψ(x, i) := x (x, i) ∈ S˜, i.e., ψ is just projection on the first coordinate. Then the kernel K from Excer- cise 3.3 satisfies (˜x, i) ∈ S˜ : Kx, (˜x, i) > 0 ⊂ (˜x, i) ∈ S˜ : ψ(˜x, i) = x 3.3. INTERTWINING AND COUPLING 69 and PK = KP˜, so all assumptions of Theorem 3.4 are fulfilled. We therefore conclude that ˜ ˜ ˜ P (X0,I0) = (˜x, i) X0] = K X0, (˜x, i) a.s. (˜x, i) ∈ S implies that ˜ ˜ ˜ P (Xn,In) = (˜x, i) (X0,..., Xn)] = K Xn, (˜x, i) a.s. for all n ≥ 0 and (˜x, i) ∈ S˜. More simply formulated, this says that ˜ P I0 = i X0 = hi(X0) a.s. (i = 0, 1) implies that ˜ ˜ ˜ P In = i (X0,..., Xn) = hi(Xn) a.s. (n ≥ 0, i = 1, 2). (3.13) Moreover, under the same assumption, the process X˜, on its own, is a Markov chain with transition kernel P . Note that this is more surprising than in the previous example, since in the present set-up, X˜ is not autonomous as part of the joint Markov chain (X,I˜ ). Indeed, X˜ evolves according to the transition kernel P h0 , resp. P h1 , depending on whether I = 0 or = 1. To understand how it is possible that X˜ has the Markov property even though it is not autonomous, we observe that by (3.13), if we observe the whole path of the process X˜ up to time n, then we do not gain more information about the present state of I than we would get from knowing Xn. As a result, ˜ ˜ ˜ P Xn+1 = xn+1 (X0,..., Xn) = (x0, . . . , xn) ˜ ˜ = P Xn+1 = xn+1 (Xn,In) = (xn, 0) h0(xn) ˜ ˜ + P Xn+1 = xn+1 (Xn,I) = (xn, 1) h1(xn), which depends only on xn and not on (x0, . . . , xn−1). Thinning of exclusion processes This example does not satisfy condition (3.5), hence Theorem 3.4 is not applicable. In view of this and other examples, we will prove a more general theorem in the next section.

3.3 Intertwining and coupling

In Theorem 3.4 we have seen how intertwining is related to the problem of Markov functionals, i.e., the question whether certain functions of a Markov chain themselves have the Markov property. The intertwinings in Theorem 3.4 are of a special 70 CHAPTER 3. INTERTWINING kind, because of the condition (3.5). In this section, we will see that intertwining relations between two Markov chains X and Y in general give rise to couplings between X and Y such that (3.8) holds. The following result is due to Diaconis and Fill [DF90, Thm 2.17]; the continuous-time analogue has been proved in [Fil92, Thm. 2].

Theorem 3.5 (Intertwining coupling) Let P and P˜ be probability kernels on countable state spaces S and S˜, respectively. Assume that K is a probability kernel from S to S˜ such that PK = KP.˜ Let 0 0 0 0 P (x, x )K(x , y ) 0 0 0 P 0 (x, x ) := x, x ∈ S, y ∈ S,PK˜ (x, y ) > 0 , (3.14) y PK(x, y0)

0 and choose for Py0 (x, · ) any probability law on S if PK(x, y ) = 0. Then Py0 is a probability kernel on S for each y0 ∈ S˜ and

ˆ 0 0 ˜ 0 0 0 0 ˆ P (x, y; x , y ) := P (y, y )Py0 (x, x ) (x, y), (x , y ) ∈ S) (3.15) deﬁnes a probability kernel on Sˆ := {(x, y) ∈ S × S˜ : K(x, y) > 0}, where Pˆ does not depend on the freedom in the choice of the Py0 . If (X,Y ) is the Markov chain with transition kernel Pˆ started in an initial law such that ˜ P Y0 = y X0 = K(X0, · )(y ∈ S), then ˜ P Yn = y (Xk)0≤k≤n = K(Xn, · )(y ∈ S, n ≥ 0), and X, on its own, is a Markov chain with transition kernel P .

Remark It is clear from (3.15) that in the joint Markov chain (X,Y ), the second component Y is autonomous with transition kernel P˜, but X is in general not 0 autonomous, unless Py0 can be chosen so that it does not depend on y . Proof of Theorem 3.5 We observe that 0 X 0 PK(x, y ) 0 0 P 0 (x, x ) = = P˜(y, y )(PK(x, y ) > 0), y PK(x, y0) x0∈S ˆ which shows that Py0 is a probability kernel. To show that the deﬁnition of P 0 does not depend on how we deﬁne Py0 (x, · ) when PK(x, y ) = 0, we observe that 3.3. INTERTWINING AND COUPLING 71

(x, y) ∈ S and P˜(y, y0) > 0 imply that K(x, y)P˜(y, y0) > 0 and hence PK(x, y0) = KP˜ (x, y0) > 0. To prove the remaining statements, let Kˆ be the probability kernel from S to Sˆ deﬁned by ˆ 0 0 0 0 0 ˆ K x, (x , y ) := 1{x=x0}K(x, y ) x ∈ S, (x , y ) ∈ S , and deﬁne ψ : Sˆ → S by ψ(x, y) := x. If we show that P Kˆ = Kˆ P,ˆ then the remaining claims follow from Theorem 3.4. We observe that ˆ 0 0 X 00 00 0 0 0 0 P K(x; x , y ) = P (x, x )1{x00=x0}K(x , y ) = P (x, x )K(x , y ) x00∈S and ˆ ˆ 0 0 X 00 ˆ 00 00 0 0 KP (x; x , y ) = 1{x=x00}K(x, y )P (x , y ; x , y ) (x00,y00)∈Sˆ X = K(x, y00)Pˆ(x, y00; x0, y0), y00∈S˜ so that P Kˆ = Kˆ Pˆ can be written coordinatewise as X P (x, x0)K(x0, y0) = K(x, y00)Pˆ(x, y00; x0, y0) x ∈ S, (x0, y0) ∈ Sˆ. y00∈S˜ To check this, we write X X P (x, x0)K(x0, y0) K(x, y00)Pˆ(x, y00; x0, y0) = K(x, y00)P˜(y00, y0) PK(x, y0) y00∈S˜ y00∈S˜ KP˜(x, y0) = P (x, x0)K(x0, y0) = P (x, x0)K(x0, y0), PK(x, y0) where we have used our assumption that KP˜ = PK.

Thinning of exclusion processes In this example, Sˆ = (x, y): x, y ∈ {0, 1}Cn , y ≤ x , and the joint evolution of (X,Y ) can be described in the same way as in (3.2), using the same random pair I for both X and Y . While this example is rather trivial (we could have invented this coupling without knowing anything about intertwining!), we will see in the next section that there exist less trivial examples of intertwinings where the coupling of Theorem 3.5 is much more surprising. 72 CHAPTER 3. INTERTWINING

3.4 First exit problems revisited

In this section we return to the set-up of Section 2.9. We will be interested at the behavior of a Markov chain X until it leaves some finite set S0. For simplicity, we will assume that the whole state space of X just consists of S0 plus one additional point z, which is a trap. More generally, the results we derive can be used to study Markov chains stopped at the first time they leave a finite set. Thus, we let X be a Markov chain with finite state space S = S0 ∪ {z}, where z is a trap. We let P denote the transition kernel of X, we write P 0 for the restriction of P to S0 and assume that P 0 is irreducible. It follows that P 0 has a unique Perron-Frobenius eigenvalue α and associated positive right eigenvector h, which is unique up to a multiplicative constant. If we extend h to S by putting h(z) := 0, then h is an eigenvector of P . Indeed, since h(z) = 0, we have X X P h(x) = P (x, y)h(y) = P 0(x, y)h(y) = P 0h(x) = αh(x)(x ∈ S0), y∈S y∈S0 P while P h(z) = y∈S P (z, y)h(y) = h(z) = 0 by the fact that z is a trap. Proposition 3.6 (Intertwining for chain with one trap) Let X be a Markov chain with finite state space S = S0 ∪ {z} and transition kernel P . Assume that z is a trap and the restriction of P to S0 is irreducible. Let h be the, up to multiple constants unique, right eigenvector of P such that h(z) = 0 and h > 0 on S0, and assume that h is normalized such that supx∈S0 h(x) ≤ 1. Let α be the eigenvalue of h and let Y be a Markov process with state space S˜ := {1, 2} and transition kernel P (1, 1) P (1, 2) α 1 − α P˜ = = . P (2, 1) P (2, 2) 0 1

Let K be the probability kernel from S to S˜ deﬁned by h(x) if y = 1, K(x, y) := 1 − h(x) if y = 2. Then PK = KP.˜

Proof This follows by writing X P h(x) = αh(x) if y = 1, PK(x, y) = P (x, x0)K(x0, y) = P (1 − h)(x) = 1 − αh(x) if y = 1, x0∈S 3.4. FIRST EXIT PROBLEMS REVISITED 73 while X KP˜(x, y) = K(x, y0)P (y0, y) y0∈S˜ K(x, 1)α = αh(x) if y = 1, = K(x, 2) + K(x, 1)(1 − α) = 1 − h(x) + (1 − α)h(x) if y = 2, which shows that PK = KP˜.

By Theorem 3.5 and the remark following the latter, we see that the processes X and Y from Proposition 3.6 can be coupled in such a way that their joint transition probabilities are given by

ˆ 0 0 ˜ 0 0 0 0 ˆ P (x, y; x , y ) = P (y, y )Py0 (x, x ) (x, y), (x , y ) ∈ S , ˆ ˜ where S := {(x, y) ∈ S × S : K(x, y) > 0} and Py0 is given by

0 0 0 0 P (x, x )K(x , y ) 0 P 0 (x, x ) := (PK(x, y ) > 0). y PK(x, y0) Filling in the definition of K and using our formulas for PK, we see that P (x, x0)K(x0, 1) P (x, x0)h(x0) P (x, x0) = = = P h(x, x0)(x, x0 ∈ S0) 1 PK(x, 1) αh(x0) is the generalized Doob transform of P 0 as defined in Lemma 2.7 and more specif- ically in Theorem 2.17. Thus, as long as the autonomous process Y stays in 1, the process X jumps according to the generalized Doob transformed transition kernel h 0 P . Recall that Ph is concentrated on S , hence X cannot get trapped as long as Y = 1. Starting with the time step when Y jumps to the state 2, and from thereon, the process X jumps according to the transition kernel P (x, x0)K(x0, 2) P (x, x0)(2 − h(x0)) P (x, x0) = = , 2 PK(x, 2) 2 − αh(x) which is well-defined for all x ∈ S since α < 1 and h ≤ 1. Since 1 − h is not an eigenvector of P , this is not a kind of transform we have seen so far. Remark The coupling of X and Y immediately gives us the estimate

x 0 1 n P [Xn ∈ S ] ≥ K(x, 1)P [Yn = 1] = α h(x). (3.16) In view of (2.6), this estimate quite good, but it is not completely sharp since the function h in Section 2.9 is normalized in a diﬀerent way than in the present 74 CHAPTER 3. INTERTWINING

P section. Indeed, in Section 2.9 we chose h such that x∈S0 η(x)h(x) = 1, where η is a probability measure, while at present we need supx∈S0 h(x) ≤ 1. Thus (unless h is constant on S0), the h in our present section is always smaller than the one in Section 2.9. Interesting examples of intertwining relations for birth-and-death processes can be found in [DM09, Swa11]. Lower estimates in the spirit of (3.16) but in the more complicated set-up of hierarchical contact processes have been derived in [AS10]. The ‘evolving set process’ in [LPW09, Thm 17.23] (which can be deﬁned for quite general Markov chains) provides another nontrivial example of an intertwining relation. Chapter 4

Branching processes

4.1 The branching property

Let S be a countable set and let N (S) be the set of all functions x : S → N such P that i∈S x(i) < ∞. It is easy to check that N (S) is a countable set (even though the set of all functions x : S → N is uncountable). We interpret x ∈ N (S) as a collection of ﬁnitely many particles or individuals, where x(i) is the number of individuals of type i, where S is the type space, or sometimes also the number of particles at the position i, where S represents physical space. Each x ∈ N (S) can be written as |x| X x = δiβ , (4.1) β=1 P where |x| := i x(i), i1, . . . , i|x| ∈ S, and δi ∈ N (S) is deﬁned as

δi(j) := 1{i=j} (i, j ∈ S).

This way of writing x is of course not unique but depends on the way we order the individuals. We will be interested in Markov processes with state space N (S), where in each time step, each individual, independently of the others, is replaced by a ﬁnite number of new individuals (its oﬀspring). Let Q be a probability kernel from S to N (S). For a given x ∈ N (S) of the form (4.1), we can construct independent N (S)-valued random variables V 1,...,V |x| such that

β P[V ∈ · ] = Q(iβ, · )(β = 1,..., |x|).

75 76 CHAPTER 4. BRANCHING PROCESSES

Then |x| X β P (x, · ) := P V ∈ · β=1 defines a probability law on N (S). Doing this for each x ∈ N (S) defines a probability kernel P on N (S). By definition, the Markov chain X with state space N (S) and transition kernel P is called the multitype branching process with offspring distribution Q.

Lemma 4.1 (Branching property) Let X = (Xk)k≥0 and Y = (Yk)k≥0 be independent branching processes with the same type space N (S) and oﬀspring distribution Q. Then Z = (Zk)k≥0, deﬁned by

Zk(i) := Xk(i) + Yk(i)(k ≥ 0, i ∈ S) (4.2) is distributed as a branching processes with type space N (S) and oﬀspring distribution Q.

Proof We need to check that

Z P Zk+1 = z Fk = P (Zk, z) a.s. k ≥ 0, z ∈ N (S) ,

Z where (Fk )k≥0 is the ﬁltration generated by Z. We will show that actually

(X,Y ) P Zk+1 = z Fk = P (Zk, z) a.s. k ≥ 0, z ∈ N (S) ,

(X,Y ) Z where (Fk )k≥0 is the ﬁltration generated by (X,Y ) = (Xk, yk)k≥0. Since Fk ⊂ (X,Y ) Z Fk and P (Zk, z) is Fk -measurable, this then implies that

Z (X,Y ) Z Z P Zk+1 = z Fk = E P Zk+1 = z Fk Fk = E P (Zk, z) Fk = P (Zk, z).

Thus, by the Markov property of (X,Y ), it suﬃces to show that P Zk+1 = z Xk = x, Yk = y = P (x + y, z) k ≥ 0, x, y, z ∈ N (S) .

Here P Zk+1 = z Xk = x, Yk = y X 0 0 X 0 0 = P Xk+1 = x ,Yk+1 = z − x Xk = x, Yk = y = P (x, x )P (y, z − x ), x0≤z x0≤z 4.1. THE BRANCHING PROPERTY 77 where we have used that by the independence of X and Y , the process (X,Y ) is 0 0 0 0 a Markov process with transition kernel P2(x, y; x , y ) := P (x, x )P (y, y ). Thus, we are left with the task of showing that X P (x + y, z) = P (x, x0)P (y, z − x0) x, y, z ∈ N (S). (4.3) x0≤z Let us write |x| |y| X X x = δiβ and y = δjγ , β=1 γ=1 and let V 1,...,V |x| and W 1,...,W |y| be all independent of each other such that

β P[V ∈ · ] = Q(iβ, · )(β = 1,..., |x|), γ P[W ∈ · ] = Q(jγ, · )(γ = 1,..., |y|). Then

|x| |y| X β X γ P (x + y, z) = P V + W = z β=1 γ=1 |x| |y| X X β 0 X γ 0 X 0 0 = P V = x , W = z − x = P (x, x )P (y, z − x ), x0≤z β=1 γ=1 x0≤z as required.

Remark We may deﬁne the convolution of two probability laws µ, ν on N (S) by X µ ∗ ν(z) := µ(x0)ν(z − x0). x0≤z Then (4.3) says that P (x + y, · ) = P (x, · ) ∗ P (y, · ) x, y ∈ N (S). (4.4) In general, any probability kernel on N (S) with this property is said to have the branching property. It is not hard to see that a Markov process with state space N (S) has the branching property (4.2) if and only if its transition kernel has the branching property (4.4). In particular (4.4) implies that if

|x| X x = δiβ , β=1 78 CHAPTER 4. BRANCHING PROCESSES then

P (x, · ) = P (δi1 , · ) ∗ · · · ∗ P (δi|x| , · ), where P (δi, · ) = Q(i, · ). Thus, each Markov process that has the branching property is a branching process, and its transition probabilities are uniquely characterized by the oﬀspring distribution.

4.2 Generating functions

For each function φ : S → R and x ∈ N (S), let us write

|x| |x| x Y x(i) Y X φ := φ(i) = φ(iβ) where x = δiβ , i∈S β=1 β=1 where φ0 := 1. It is easy to see that

x+y x y φ = φ φ x, y ∈ N (S), φ : S → R .

Because of all the independence coming from the branching property, the linear operator P associated with the transition kernel of a branching process maps such ‘multiplicative functions’ into multiplicative functions. We will especially be interested in the case that φ takes values in [0, 1]. We let [0, 1]S denote the space of all functions φ : S → [0, 1].

Lemma 4.2 (Generating operator) Let P denote the transition kernel of a multitype branching process with type space S and oﬀspring distribution Q. Let U be the nonlinear operator deﬁned by X 1 − Uφ(i) := Q(i, x)(1 − φ)x i ∈ S, φ ∈ [0, 1]S. x∈N (S)

Then S P fφ = fUφ φ ∈ [0, 1] , S x where for any φ ∈ [0, 1] , we deﬁne fφ : N (S) → [0, 1] by fφ(x) := (1 − φ) .

P|x| Proof If X is started in X0 = x with x = β=1 δiβ , then X1 is equal in distribution P|x| β β to β=1 V where the V ’s are independent with distribution Q(iβ, · ). It follows 4.2. GENERATING FUNCTIONS 79 that |x| P|x| V β Y V β P fφ(x) = E (1 − φ) β=1 = E (1 − φ) β=1 |x| |x| Y V β Y = E (1 − φ) = (1 − Uφ)(iβ) = fUφ(x). β=1 β=1

Remark 1 It would seem that the formulation of the lemma is simpler if we replace φ by 1 − φ everywhere, but as we will see later there are good reasons to formulate things in terms of 1 − φ. P x Remark 2 Our assumption that 0 ≤ φ ≤ 1 guarantees that the sum x Q(i, x)φ in the definition of Uφ(i) is finite. Under more restrictive assumptions on Q, we can define Uφ also for more general real-valued φ. We call the nonlinear operator U from Lemma 4.2 the generating operator of the branching process with offspring distribution Q. By induction, Lemma 4.2 shows n that P fφ = fU nφ, or, in other words

x Xn n x S E (1 − φ) = (1 − U φ) n ≥ 0, φ ∈ [0, 1] . (4.5) The advantage of the operator U is that it acts on functions ‘living’ on the space S, while P acts on functions on the much larger space N (S). The price we pay for this is that U, unlike P , is not linear. The next lemma shows that U contains, in a sense ‘all information we need’.

Lemma 4.3 (Generating functions are distribution determining) Let µ, ν be probability measures on N (S) such that X X µ(x)(1 − φ)x = ν(x)(1 − φ)x φ ∈ [0, 1]S. x∈N (S) x∈N (S)

Then µ = ν.

Proof We will prove the statement first under the assumtion that S is finite. Let N (S) ∪ {∞} be the one-point compactification of N (S). For each ψ ∈ [0, 1)S, x define gψ(x) := ψ and gψ(∞) := 0. Then the gψ’s are continuous functions on the compact space N (S)∪{∞}. It is not hard to see that they separate points, i.e., for 0 S 0 each x 6= x there exists a ψ ∈ [0, 1) such that gψ(x) 6= gψ(x ). Since gψgψ0 = gψψ0 , 80 CHAPTER 4. BRANCHING PROCESSES

S the class {gψ : ψ ∈ [0, 1) } is closed under multiplication. Let H be the space of all linear combinations of functions from this class and the identity function. Then H is an algebra that separates points, hence by the Stone-Weierstras theorem H is dense in the space of continuous functions on N (S) ∪ {∞}, equipped with the supremumnorm. By linearity and because µ and ν are probability measures, P P x µ(x)f(x) = x ν(x)f(x) for all f ∈ H. Since H is dense, it follows that µ = ν. If S is not finite, then by applying our argument to functions ψ that are zero outside a finite set, we see that the finite-dimensional marginals of µ and ν agree, which shows that µ = ν in general.

There is a nice suggestive way of writing the relation (4.5). Generalizing (3.3), for S any φ ∈ [0, 1] , we deﬁne a probability kernel Kφ from N (S) to N (S) by

|x| X Kφ(x, · ) := P χβδiβ ∈ · , (4.6) β=1

P|x| where x = β=1 δiβ and the χ1, . . . , χ|x| are independent Bernoulli random variables with P[χβ = 1] = φ(iβ). Thus, if Z is distributed according to the law Kφ(x, · ), then Z is obtained from x by independent thinning, where a particle of type i is kept with probability φ(i) and thrown away with the remaining probability. Note that Kφ has the branching property (4.4) and corresponds in fact to the oﬀspring distribution Qφ(i, y) := Kφ(δi, y) = ρ(i)1{y=δi} + (1 − ρ(i))1{y=0} i ∈ S, y ∈ N (S) .

Let Thinφ(x) denote a random variable with law Kφ(x, · ). Then

x S P Thinφ(x) = 0 = (1 − φ) x ∈ N (S), φ ∈ [0, 1] , where 0 denotes the conﬁguration in N (S) with no particles. In view of this,

X x 1 − Uφ(i) = Q(i, x)(1 − φ) = δiPKφ(0) x∈N (S) and hence Uφ(i) = δiQKφ(N (S)\{0}). (4.7) Note that this says that if we start with one particle of type i, let it produce oﬀspring, and then thin with φ, then Uφ(i) is the probability that we are left with at least one individual. Likewise, we may rewrite (4.5) in the form

n δxP Kφ(0) = δxKU nφ(0), (4.8) 4.3. THE SURVIVAL PROBABILITY 81

n where µP Kφ is the law obtained by starting the branching process in x, running it till time t, and then applying the kernel Kφ, while δxKU nφ is the law obtained by thinning x with U nφ.

S Exercise 4.4 (Repeated thinning) Show that KφKψ = Kφψ (φ, ψ ∈ [0, 1] ).

Exercise 4.5 (Thinning characterization) Let µ, ν be probability laws on S N (S) such that µKφ(0) = νKφ(0) for all φ ∈ [0, 1] . Show that µ = ν.

4.3 The survival probability

Let X be a branching process with type space S and generating operator U. We observe that

δi δi P Xn 6= 0] = 1 − P Thin1(Xn) = 0]

δi n n = 1 − P ThinU n1(δi) = 0] = 1 − 1 − U 1(i) = U 1(i), where we use the symbol 1 also to denote the function that is constantly one. Since 0 is a trap for any branching process,

δi δi δi Xn 6= 0] = Xk 6= 0 ∀0 ≤ k ≤ n] −→ Xk 6= 0 ∀k ≥ 0], P P n→∞ P where we have used the continuity of our probability measure with respect to decreasing sequences of events. In view of this, let us write

n δi ρ(i) := lim U 1(i) = Xk 6= 0 ∀k ≥ 0] (4.9) n→∞ P for the probability to survive starting from a single particle of type i.

Lemma 4.6 (Survival probability) The function ρ in (4.9) is the largest solution (in [0, 1]S) of the equation Uρ = ρ, i.e., ρ solves this equation and any other solutions ρ0 of this equation, if they exist, satisfy ρ0 ≤ ρ.

Proof Since X Uφ(i) = Q(i, x)1 − (1 − φ)x, x 82 CHAPTER 4. BRANCHING PROCESSES and since φ 7→ 1 − (1 − φ)x is a nondecreasing function, we see that

φ ≤ ψ implies Uφ ≤ Uψ. (4.10)

By monotone convergence, we see moreover that

φn ↓ φ implies Uφn ↓ Uφ. (4.11)

Since 1 ≥ U1 we see by (4.10) and induction that U n1 ≥ U n+11 and U n1 ↓ ρ, which was in fact also clear from our probabilistic interpretation. By (4.11), it follows that Uρ = U lim U n1 = lim U n+1 = ρ. n→∞ n→∞ Now if ρ0 ∈ [0, 1]S is any other ﬁxed point of U, then by (4.10) and (4.11)

ρ0 ≤ 1 implies ρ0 = U nρ0 ≤ U n1 −→, ρ→∞ which shows that ρ0 ≤ ρ.

Exercise 4.7 (Galton-Watson process) Let X be a branching process whose type space S = {1} consists of a single point, and let Q be its offspring distribution. ∼ We identify N ({1}) = N. Since there is only one type of individual, we only need to now with which probability a single individual produces n offspring (n ≥ 0). Thus, we simply write Q(n) = Q(1, nδ1) which is a probability law on N. Assume that Q has a finite second moment and let ∞ X a := nQ(n) n=0 denote its mean. We identify [0, 1]S ∼= [0, 1] and let U : [0, 1] → [0, 1] be the generating operator of X, which is now just a (nonlinear) function from [0, 1] to [0, 1]. (a) Show that U is a concave function and that U 0(0) = a. (b) Assume that Q(1) < 1. Show that

1 P [Xk 6= 0 ∀k ≥ 0] > 0 if and only if a > 1. 4.3. THE SURVIVAL PROBABILITY 83

Remark Single-type branching processes as in Excercise 4.7 are called Galton- Watson processes after the seminal paper (from 1875) [WG75], where they proved that the survival probability ρ solves Uρ = ρ but incorrectly concluded from this that the process survives for all a ≥ 0, since ‘obviously the solution of this equation is ρ = 0’.

Exercise 4.8 (Spatial branching) Let (Ik)k≥0 be i.i.d. with P[Ik = −1] = 1/2 = P[Ik = 1], and let N be a Poisson distributed random variable with mean a, independent of (Ik)k≥0. Let X be the branching process with type space Z and oﬀspring distribution Q given by

N X Q(i, · ) := P δi+Ik ∈ · (i ∈ Z), k=1 which says that a particle at i produces Pois(a) oﬀspring which are independently placed on either i − 1 or i + 1, with equal probabilities. Show that

δ0 P Xk 6= 0 ∀k ≥ 0 > 0 if and only if a > 1.

Exercise 4.9 (Two-type process) Let X be a branching process with type space S = {1, 2} and the following offspring distribution. Individuals of both types produce a Poisson number of offspring, with mean a. If the parent is of type 1, then its offspring are, independently of each other, of type 1 or 2 with probability 1/2 each. All offspring of individuals of type 2 are again of type 2. Starting with a single individual of type 1, for what values of a is there a positive probability that there will be individuals of type 1 at all times?

Exercise 4.10 (Poisson oﬀspring) Let X be a Galton-Watson process where each individual produces a Poisson number of oﬀspring with mean a, and let

1 ρn := P [Xn = 0] (n ≥ 0) be the probability that the process started with a single individual is extinct after a(ρn−1) n steps. Prove that ρn+1 = e . 84 CHAPTER 4. BRANCHING PROCESSES

4.4 First moment formula

Proposition 4.11 (First moment formula) Let X be a branching process with type space S and oﬀspring distribution Q. Assume that the matrix X A(i, j) := Q(i, x)x(j)(i, j ∈ S) (4.12) x∈N (S) satisﬁes X sup A(i, j) < ∞. (4.13) i∈S j∈S Then x X n E [Xn(j)] = x(i)A (i, j) x ∈ N (S), j ∈ S, n ≥ 0 . i

P|x| Proof We ﬁrst prove the statement for n = 1. If x = β=1 δiβ , then X1 is equal in P|x| β β distribution to β=1 V where the V ’s are independent with distribution Q(iβ, · ). Therefore,

|x| |x| x X β X β E [X1(j)] = E V (j) = E V (j) β=1 β=1 |x| |x| X X X X = Q(iβ, y)y(j) = A(iβ, j) = x(i)A(i, j). β=1 y β=1 i By induction, it follows that

x X x 0 x 0 E [Xn+1(j)] = P [Xn = x ]E [Xn+1(j) | Xn = x ] x0 X x 0 X 0 X X x 0 0 = P [Xn = x ] x (i)A(i, j) = A(i, j) P [Xn = x ]x (i) x0 i i x0 X X x X X n n+1 = A(i, j) E [Xn(i)] = A(i, j) x(k)A (k, i) = A (i, k), i x0 i k where all expressions are ﬁnite by (4.13).

Lemma 4.12 (Subcritical processes) Let X be a branching process with finite type space S and offspring distribution Q. Assume that its first moment matrix A defined in (4.12) is irreducible and satisfies (4.13), and let α be its Perron- Frobenius eigenvalue. If α < 1, then x P Xk 6= 0 ∀k ≥ 0 = 0 x ∈ N (S) . 4.4. FIRST MOMENT FORMULA 85

Proof For any f : S → R, let lf : N (S) → R denote the ‘linear’ function X lf (x) := x(i)f(i) x ∈ N (S), f : S → R . i∈S

Then by Lemma 4.11,

n x X x P lf (x) = E lf (Xn) = f(i)E Xn(i) i X X n X n = f(i) x(j)A (j, i) = x(j)A f(j) = lAnf (x). i j j

In particular, if h is the (strictly positive) right eigenvector of A with eigenvalue α, then n P lh = lAnh = lαnh, which says that x n E h(Xn) = α h(x), which tends to zero by our assumption that α < 1. Since h is strictly positive, it x follows that P [Xn 6= 0] → 0.

Proposition 4.13 (Critical processes) Let X be a branching process with finite type space S and offspring distribution Q. Assume that its first moment matrix A defined in (4.12) is irreducible and satisfies (4.13), and let α be its Perron- Frobenius eigenvalue. If α = 1 and there exists some i ∈ S such that Q(i, 0) > 0, then x P Xk 6= 0 ∀k ≥ 0 = 0 x ∈ N (S) .

Proof Let A := {Xn 6= 0 ∀n ≥ 0} and let

δ ρ(i) := P i (A).

By the branching property,

x c x P (A ) = (1 − ρ) x ∈ N (S) .

By the principle ‘what can happen must happen’ (Proposition 0.14), we have

(1 − ρ)Xn −→ a.s. on the event A. 0→∞ 86 CHAPTER 4. BRANCHING PROCESSES

By irreducibility and the fact that Q(j, 0) > 0 for some j, it is not hard to see that

ρ(i) < 1 for all i ∈ S. By the ﬁniteness of S, it follows that supi∈S ρ(i) < 1 and hence inf (1 − ρ)x > 0 (N ≥ 0). x∈N (S), |x|≤N It follows that |Xn| −→ ∞ a.s. on the event A, (4.14) n→∞ i.e., the only way for the process to survive is to let the number of particles tend to inﬁnity. Let h be the (strictly positive) right eigenvector of A with eigenvalue α. Then

X E lh(Xn+1) Fn = (P lh)(Xn) = lαh(Xn) = αlh(Xn), which shows that (provided that E[h(X0)] < ∞, which is satisﬁed for processes started in deterministic initial states) the process

−n X Mn := α h(i)Xn(i)(n ≥ 0) (4.15) i∈S is a nonnegative martingale. By martingale convergence, it follows that there exists a random variable M∞ such that

Mn −→ M∞ a.s. (4.16) n→∞ In particular, if α = 1, this proves that

|Xn| 6−→ ∞ a.s., n→∞ which by (4.14) implies that P(A) = 0.

4.5 Second moment formula

We start with some general Markov chain theory. For any probability law µ on a countable set S and functions f, g : S → R, let us write

Covµ(f, g) := µ(fg) − (µf)(µg) for the covariace of f and g, whenever this is well-deﬁned. Note that if X is a random variable with law µ, then Covµ(f, g) = E[f(X)g(X)] − E[f(X)]E[g(X)], in accordance with the usual formula for the covariance of two functions. 4.5. SECOND MOMENT FORMULA 87

Lemma 4.14 (Covariance formula) Let X be a Markov chain with countable state space S and transition kernel P . Let µ be a probability law on S and let C be a class of functions f : S → R such that

(i) f ∈ C implies P f ∈ C. X (ii) µ(x)P (x, y)|f(y)g(y)| < ∞ for all x ∈ S and f, g ∈ C. x,y

Then, for any f, g ∈ C,

n n n X n−k k−1 k−1 CovµP n (f, g) = Covµ(P f, P g) + µP Γ(P f, P g), (4.17) k=1 where Γ(f, g) := P (fg) − (P f)(P g)(f, g ∈ C).

Remark 1 If X = (Xk)k≥0 is the Markov chain with initial law µ and transition kernel P , then CovµP n (f, g) = E f(Xn)g(Xn) − E f(Xn) E g(Xn) , =: Cov f(Xn), g(Xn) , and similarly

n n n n Covµ(P f, P g) = Cov (P f)(X0), (P g)(X0) .

Remark 2 The assumptions of the lemma are trivially fulfilled if we take for C the class of all bounded real functions on S. Often, we also need the lemma for certain unbounded functions, but in this case we need to find a class C satisfying the assumptions of the lemma to ensure that all second moments are finite. Proof of Lemma 4.14 The statement is trivial for n = 0. Fix n ≥ 1 and define a function H : {0, . . . , n} → R by H(k) := P k(P n−kf)(P n−kg) (0 ≤ k ≤ n). Then µH(n) − H(0) = µP n(fg) − µ(P nf)(P ng) = µP n(fg) − (µP hf)(µP ng) − µ(P nf)(P ng) − (µP hf)(µP ng) n n = CovµP n (f, g) − Covµ(P f, P g). 88 CHAPTER 4. BRANCHING PROCESSES

It follows that

n n n X CovµP n (f, g) − Covµ(P f, P g) = µ H(k) − H(k − 1) k=1 n X = µP k(P n−kf)(P n−kg) − P k−1(P n−k+1f)(P n−k+1g) k=1 n X = µP k−1Γ(P n−kf, P n−kg). k=1

Changing the summation order (setting k0 := n − k + 1), we arrive at (4.17).

We now apply this general formula to branching processes. To simplify matters, we will only look at ﬁnite type spaces.

Proposition 4.15 (Second moment formula) Let X be a branching process with ﬁnite type space S and oﬀspring distribution Q. Let V i denote a random variable with law Q(i, · ) and assume that

i A(i, j) := E[V (j)], (4.18) i i C(i; j, k) := E[V (j)V (k)] are ﬁnite for all i, j, k ∈ S. Let A be the linear operator with matrix A(i, j) and for functions f, g : S → R, let C(f, g): S → R be deﬁned by X C(f, g)(i) := C(i; j, k)f(j)g(k). j,k∈S P Fox x ∈ N (S) and f : S → R, let xf := i x(i)f(i). Then, for functions f, g : S → R, one has

x n E Xnf = xA f, n x X n−k k−1 k−1 (4.19) Cov Xnf, Xng = xA C(A f, A g), k=1

x where Cov denotes covariance w.r.t. to the law Px.

Proof The ﬁrst formula in (4.19) has already been proved in Lemma 4.11. As in the proof of Lemma 4.12, for any real function f on S, let lf : N (S) → R denote 4.6. SUPERCRITICAL PROCESSES 89

P the ‘linear’ function lf (x) := i f(i)x(i). Then the ﬁrst formula in (4.19) says that n S P lf = lAnf (f ∈ R ), which motivates us to take for the class C in Lemma 4.14 the class of ‘linear’ functions lf with f : S → R any function. Using the fact that C(i; j, k) < ∞ for all i, j, k ∈ S, it is not hard to prove that C satisﬁes the assumptions of Lemma 4.14. P|x| 1 |x| β Let x = β=1 δiβ and let V ,...,V be independent such that V is distributed according to Q(iβ, · ). We calculate Γ(lf , lg)(x) = P (lf lg) − (P lf )(P lg) (x) X X X X = Cov f(j)V β(j) , g(k)V γ(k) = f(j)g(k) CovV β(j),V β(k) j,β k,γ jk β X = x(i)f(j)g(k)C(i; j, k) = lC(f, g)(x), ijk where we have used that CovV β(i),V γ(j) = 0 for β 6= γ by independence. Then Lemma 4.14 tells us that

n n X n−k k−1 k−1 X n−k n CovδxP (lf , lg) = δxP Γ(P lf ,P lg) = δxP Γ(lAk−1f , lAk−1g) k=1 k=1 n n X n−k X = δxP lC(Ak−1f, Ak−1g) = lAn−kC(Ak−1f, Ak−1g)(x), k=1 k=1 which proves the second formula in (4.19).

4.6 Supercritical processes

The aim of this section is to prove that supercritical branching processes survive.

Proposition 4.16 (Supercritical process) Let X be a branching process with finite type space S and offspring distribution Q. Assume that the first and second moments A(i, j) and C(i; j, k) of Q, defined in (4.18), are all finite and that A is irreducible. Assume that the Perron-Frobenius eigenvalue α of A satisfies α > 1. Then x P Xk 6= 0 ∀k ≥ 0 > 0 x ∈ N (S), x 6= 0 . 90 CHAPTER 4. BRANCHING PROCESSES

Proof Let h be the (strictly positive) right eigenvector of A with eigenvalue α. We have already seen in formulas (4.15)–(4.16) in the proof of Proposition 4.13 that −n Mn = α Xnh is a nonnegative martingale that converges to an a.s. limit M∞. By Proposition 0.8, if we can show that M is uniformly integrable, then

x E [M∞] = xh > 0 for all x 6= 0, which shows that the process survives with positive probability. Thus, it suﬃces to show that

x 2 sup E Mn] < ∞. n≥0 We write x 2 x −n 2 x −n E Mn] = E [α Xnh] + Var (α Xnh), where by the ﬁrst formula in (4.19)

x −n 2 −n n 2 2 E [α Xnh] = (α xA h) = (xh) is clearly bounded uniformly in n. We observe that since h is strictly positive and S is ﬁnite, we can ﬁnd a constant K < ∞ such that C(h, h) ≤ Kh. Therefore, applying the second formula in (4.19), we see that

n x −n −2n X n−k k−1 k−1 Var (α Xnh) = α xA C(A h, A h) k=1 n n X X = α−2n xAn−kα2(k−1)C(h, h) ≤ α−2nK xAn−kα2(k−1)h k=1 k=1 n n ∞ X X X = α−2nK αn−kα2(k−1)xh = Kxh αk−n−2 ≤ Kxhα−2 α−k < ∞, k=1 k=1 k=0 uniformly in n ≥ 1, where we have used that α > 1.

4.7 Trimmed processes

Let X be a multitype branching process with countable type space S and oﬀspring distribution Q. Recall from Lemma 4.6 that

δi ρ(i) := P Xk 6= 0 ∀k ≥ 0 (i ∈ S) 4.7. TRIMMED PROCESSES 91 is the largest solution of the equation Uρ = ρ, where U is the generating operator of X. Let ρ ∈ [0, 1]S be any solution of Uρ = ρ. The aim of the present section is to prove a result similar to conditioning on the future (as in Proposition 1.5) or intertwining of processes with one trap (Proposition 3.6), but now on the level of the individuals in a branching process. More precisely, we will divide the population into two ‘sorts’ of individuals, with probabilities (depending on the type) ρ(i) and 1 − ρ(i). In particular, if ρ is the survival probability, then we divide the population at each time k into those individuals which we know are going to survive (or, more precisely, that have living descendants at all times), and those whose descendants are going to die out completely. P|x| To this aim, let x = β=1 δiβ ∈ N (S) and let χ1, . . . , χ|x| be independent Bernoulli random variables with P[χβ = 1] = ρ(iβ). Doing this for any x ∈ N (S), we deﬁne a probability kernel Lρ from N (S) to N (S × {0, 1}) by

|x| X Lρ(x, · ) := P δ(iβ ,χβ ) ∈ · . β=1

Another way to describe Lρ is to note that Lρ has the branching property (4.4) and

ˆ Lρ(δi, y) = ρ(i)1{y=δ(i,1)} + (1 − ρ(i))1{y=δ(i,0)} i ∈ S, y ∈ N (S) .

Note that this is very similar to the thinning kernel Kρ deﬁned in (4.6). We set ˆ S := (i, σ): i ∈ S, σ ∈ {0, 1},K(δi, δ(i,σ)) > 0 , i.e., ˆ S := S0 ∪ S1, where

S0 := {(i, 0) : 1 − ρ(i) > 0} and S1 := {(i, 0) : 1 − ρ(i) > 0}. ˆ Then Lρ is in eﬀect a probability kernel from N (S) to N (S). For any x ∈ N (S) and S0 ⊂ S, let us write

0 0 x S0 := x(i))i∈S ∈ N (S ) for the restriction of x to S0 (and similarly for subsets of Sˆ). With this notation, if Y is a random variable with law Lρ(x, · ), then Y |S1 (resp. Y |S0 ) is a thinning of x with the function ρ (resp. 1 − ρ). 92 CHAPTER 4. BRANCHING PROCESSES

Recall that the oﬀspring distribution Q of X is a probability kernel from S to N (S). Let E := y ∈ N (Sˆ): y = 0 . S1 ˆ Observe that QLρ is a probability kernel from S to N (S). For given i ∈ S, the probability of E under QLρ(i, · ) is given by

X x QLρ(i, E) = Q(i, x)(1 − ρ) = 1 − Uρ(i) = 1 − ρ(i), (4.20) x∈N (S) where we have used that Uρ = ρ. We define a probability kernel Qˆ from Sˆ to N (Sˆ) by ( QLρ i, · E if σ = 0, Qˆ(i, σ; · ) := c QLρ i, · E if σ = 1, where we are conditioning the probability law QLρ i, · ) on the event E and its complement, respectively. By (4.20), these conditional probabilities are well-defined, c i.e., QLρ(i, E) > 0 for all (i, 0) ∈ S0 and QLρ(i, E ) > 0 for all (i, 1) ∈ S1. ˆ In words, our definition of Q says that an individual of type (i, 0) ∈ S0 produces offspring in the following manner. First, we produce offspring according to the law Q(i, · ) of our original branching process. Next, we assign to these individuals independent ‘signs’ 0, 1 with probabilities depending on the type of the individual through the function ρ. Finally, we condition on producing only offspring with sign 0. Individuals of type (i, 1) ∈ S1 reproduce similarly, but in this case we condition on producing at least one offspring with sign 1. Note that individuals of a type in S1 always produce at least one offspring in S1, and possibly also offspring in S0. Individuals in S0 produce only offspring in S0, and possibly no offspring at all.

Theorem 4.17 (Distinguishing surviving particles) Let X = (Xk)k≥0 be a branching process with finite type space S and irreducible offspring distribution Q. ˆ ˆ Assume that X survives (with positive probability). Let ρ, Lρ, S, and Q be as defined above. Then X can be coupled to a branching process Y with type space Sˆ and offspring distribution Qˆ, in such a way that ˆ P Yn = y (Xk)0≤k≤n = Lρ(Xn, · )(y ∈ S, n ≥ 0). (4.21)

Proof We apply Theorem 3.5. In fact, for our present purpose Theorem 3.4 is suﬃcient, where the function ψ occurring there is given by

ψ(y)(i) := y(i, 0) + y(i, 1) y ∈ N (Sˆ), i ∈ S. 4.7. TRIMMED PROCESSES 93

Let P and Pˆ denote the transition kernels of X and Y , respectively. We need to check that ˆ PLρ = LρP. ˆ Since P,Lρ, and P have the branching property (4.4), it suﬃces to check that ˆ δiPLρ = δiLρP (i ∈ S). Indeed, by our deﬁnition of Qˆ, ˆ δiLρP = (1 − ρ(i))Q(i, 0; · ) + ρ(i)Q(i, 1; · ) c = (1 − ρ(i))QLρ(i, · | E) + ρ(i)QLρ(i, · | E ) = QLρ(i, · ) = δiPLρ, where we have used (4.20).

Proposition 4.18 (Trimmed process) Let X and Y be the branching processes with type spaces S and Sˆ in Theorem 4.17, let ρ be as in that theorem and let U be the generating operator of X. Then Yk S1 k≥0 ρ is a branching process with type space S1 and generating operator U given by

ρ −1 S1 U φ(i) = ρ (i)U(ρφ)(i) i ∈ S1, φ ∈ [0, 1] , where ρφ denotes the pointwise product of ρ and φ, which is extended to a function on S by setting ρφ(j) = 0 for all j ∈ S\S1.

Proof Since individuals in S0 never produce offpring in S1, it is clear that the 0 restriction of Y to S1 is a branching process. Let Q be the offspring distribution 0 of the restricted process. Then Q (i, · ) is just QKρ(i, · ) conditioned on producing at least one offspring, where Kρ is the thinning kernel defined in (4.6). Then, setting G := N (S1)\{0}, we have by (4.7)

ρ 0 δiQKρKφ(G) δiQKρφ(G) U(ρφ)(i) −1 U φ(i) = δiQ Kφ(G) = = = = ρ(i) U(ρφ)(i), δiQKρ(G) δiQKρ(G) Uρ(i) where we have used that Uρ = ρ.

Remark In particular, if ρ is the survival probability, then the branching process Y from Proposition 4.18 has been called the trimmed tree of X in [FS04], which deals with continuous type spaces and continuous time. Similar constructions have been used in branching theory long before this paper but it is hard to ﬁnd a good reference. 94 CHAPTER 4. BRANCHING PROCESSES

Exercise 4.19 (Nonbranching process) Let S be a countable set, let P 0 be a probability kernel on S, and let Q be the oﬀspring distribution deﬁned by

0 Q(i, δj) := P (i, j)(i, j ∈ S), and Q(i, x) := 0 if |x|= 6 1. In other words: a particle at the position i produces exactly one oﬀspring on a position that is distributed according to P 0(i, · ). Let X be the branching process with type space S and ofspring distribution Q and let U be its generating operator. Show that U = P 0. In particular, this says that a function ρ ∈ [0, 1]S solves Uρ = ρ if and only if ρ is harmonic for P 0 and U ρ = (P 0)ρ is just the classical Doob transform of P 0.

If X is a multitype branching process with type space S and offspring distribution Q, then let us write i → j if Q(i, {x : x(j) > 0}) > 0 and i j if there exist i = i0 → · · · → in = j. We say that Q is irreducible if i j for all i, j ∈ S. In particular, if Q has finite first moments, then this is equivalent to irreducibility of the matrix A(i, j) in (4.18). We also define aperiodicity of Q in the obvious way, i.e., an irreducible Q is aperiodic if the greatest common divisior n of {n ≥ 1 : P (δi, {x : x ≥ δi}) > 0} is one for some, and hence for all i ∈ S. If Q is irreducible and

δi ρ(i) = P Xk 6= 0 ∀k ≥ 0 (i ∈ S), (4.22) then it is not hard to see that either ρ(i) > 0 for all i ∈ S, or ρ(i) = 0 for all i ∈ S. In the ﬁrst case, we say that X survives, while in the second case we say that X dies out.

Exercise 4.20 (Immortal process) Let X be a branching process with finite type space S, offspring distribution Q, and generating operator U. Assume that Q is irreducible and aperiodic and that Q(i, 0) = 0 for each i ∈ S, i.e., each individual always produces at least one offspring. Assume also that Q(i, {x : |x| ≥ 2}) > 0 for at least one i ∈ S. Then it is not hard to show that

δi [Xn(j) ≥ N] −→ 1 (i ∈ S, N < ∞). P n→∞ Use this to show that for any φ ∈ [0, 1]S that is not identically zero,

U nφ(i) −→ 1 (i ∈ S). n→∞ In particular, this shows that the equation Uρ = ρ has only two solutions: ρ = 0 and ρ = 1. 4.7. TRIMMED PROCESSES 95

Exercise 4.21 (Fixed points of generating operator) Let X be a branching process with ﬁnite type space S, oﬀspring distribution Q, and generating operator U. Assume that Q is irreducible and aperiodic and that Q(i, {x : |x| ≥ 2}) > 0 for at least one i ∈ S. Assume that the survival probability ρ in (4.22) is positive for some, and hence for all i ∈ S. Show that for any φ ∈ [0, 1]S that is not identically zero, U nφ(i) −→ ρ(i)(i ∈ S). n→∞ In particular, this shows that ρ is the only nonzero solution of the equation Uρ = ρ. Hint: use Proposition 4.18 to reduce the problem to the set-up of Excercise 4.20.

Exercise 4.22 (Exponential growth) In the set-up of Excercise 4.21, assume that Q has finite first moments. Let α be the Perron-Frobenis eigenvalue of the first moment matrix A defined in (4.12) and let h > 0 be the associated right eigenvector. Assume that α > 0, let M = (Mn)n≥0 be the martingale defined in (4.15), and let M∞ := limn→∞ Mn. Set

0 δi ρ (i) := P M∞ > 0 . Prove that ρ0 = ρ, where ρ is the survival probability deﬁned in (4.22). Hint: show that Uρ0 = ρ0. 96 CHAPTER 4. BRANCHING PROCESSES Appendix A

Supplementary material

A.1 The spectral radius

Proof of Lemma 2.10 Suppose that λ ∈ spec(A) and let f be an associated eigenvector. Then kAnfk = |λ|nkfk which shows that kAnk1/n ≥ |λ| and hence ρ(A) ≥ |λ|.

To complete the proof, it suﬃces to show that ρ(A) ≤ λ+, where λ+ := sup{|λ| : λ ∈ spec(A)}. We start with the case that A can be diagonalized, i.e., there exists a basis {e1, . . . , ed} of eigenvectors with associated eigenvalues λ1, . . . , λd. By Excercise 2.5 the choice of our norm on V is irrelevant. We choose the `1-norm with respect to the basis {e1, . . . , ed}, i.e., X kφk := |φ(i)|, i=1 where φ(1), . . . , φ(d) are the coordinates of φ w.r.t. this basis. Then

d d n X n X n n kA φk = k φ(i)λi eik = |φ(i)| |λi| ≤ λ+kφk, i=1 i=1

n n which proves that kA k ≤ λ+ for each n ≥ 1 and hence ρ(A) ≤ λ+. In general, A need not be diagonalizable, but we can choose a basis such that the matrix of A w.r.t. this basis has a Jordan normal form. Then we may write A = D + E where D is the diagonal part of A and E has ones only on some places just above the diagonal and zeroes elsewhere. One can check that E is nilpotent,

97 98 APPENDIX A. SUPPLEMENTARY MATERIAL i.e., Em = 0 for some m ≥ 1. Moreover E commutes with D and kEk ≤ 1 if we choose the `1-norm with respect to the basis {e1, . . . , ed}. Now

m−1 X n An = (D + E)n = Dn−kEk k k=0 and therefore

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absorbing state, 24 generalized eigenspace, 49 adaptedness, 6 generalized eigenvector, 49 algebraic multiplicity, 49 generating operator, 79 autonomous Markov chain, 13, 65 geometric multiplicity, 49

Bernoulli random variable, 62 h-transform, 25 branching process, 76 harmonic function, 23 branching property, 77 homogeneous Markov chain, 12 characteristic polynomial, 49 increments codimension, 52 independent, 31 compensator, 8 integer lattice, 34 convergence of σ-fields, 9 intertwining, 61 convolution, 77 invariant coupling, 34 law, 19 detailed balance, 21 measure, 19 Doob transform, 25 invariant subspace, 52 generalized, 46, 56 irreducibility, 17, 41, 94 eigenspace, 49 Jordan normal form, 49 eigenvector Markov left or right, 42 chain, 12 equivalence of states, 17 functional, 13 escape probability, 33 property, 10, 15 extinction, 94 strong property, 16 Fekete’s lemma, 43 martingale, 7 filtered probability space, 7 convergence, 9 filtration, 6 multiplicity first entrance time, 6 algebraic, 49 geometric, 49 Galton-Watson process, 83 Gelfand’s formula, 50 nonnegative definite, 49

101 102 INDEX nonnegative matrix, 41 convergence, 9 null recurrence, 18 superharmonic function, 23 supermartingale, 7 oﬀspring distribution, 76 convergence, 9 operator norm, 41 survival, 94 optional stopping, 8 thinning, 62, 80 period of a state, 18 total variation distance, 38 periodicity, 18 transience, 17 Perron-Frobenius theorem, 42 transition positive recurrence, 17, 19 kernel, 12 quasi-stationary law, 58 probabilities, 12 transposed matrix, 42 random mapping representation, 13 trap, 24 random walk type space, 75 on integer lattice, 34 on tree, 30 uniformly integrable, 10 real matrix, 41 recurrence, 17 null, 18 positive, 17, 19 reversibility, 21 sample path, 5 self-adjoint transition kernel, 21 similar matrices, 61 spectral gap absolute, 54 spectral radius, 41 spectrum, 41 stochastic process, 5 stopped Markov chain, 13 stochastic process, 7 submartingale, 8 stopping time, 6 strong Markov property, 16 subadditivity, 43 subharmonic function, 23 submartingale, 7