Ergodic Theorems. Samy Tindel. Purdue University. Probability Theory 2 - MA 539. Taken from Probability: Theory and examples by R.

Ergodic Theorems Samy Tindel Purdue University Probability Theory 2 - MA 539 Taken from Probability: Theory and examples by R. Durrett Samy T. Ergodic theorems Probability Theory 1 / 92

Outline 1 Definitions and examples Preliminaries on Markov chains Examples of stationary sequences Notion of ergodicity 2 Ergodic theorem 3 Recurrence Samy T. Ergodic theorems Probability Theory 2 / 92

Overview Stationary sequence: such that {X n+k ; n 0} (d) = {X n ; n 0}. Main result: law of large numbers of the type 1 n n f (X k ) k=1 exists a.s Samy T. Ergodic theorems Probability Theory 3 / 92

Markov chain Definition 1. Let X = {X n ; n 0} be a process. 1 X is a Markov chain if P (X n+1 = j X 0 = i 0,..., X n = i n ) = P (X n+1 = j X n = i n ) for all n 0, i 0,..., i n, j E. 2 The Markov chain is homogeneous whenever P (X n+1 = j X n = i) = P (X 1 = j X 0 = i) Samy T. Ergodic theorems Probability Theory 6 / 92

Law of the Markov chain Proposition 2. Let X be a homogeneous Markov chain with initial law ν and transition p. 1 For n N and i 0,..., i n E, we have P (X 0 = i 0,..., X n = i n ) = ν(i 0 ) p(i 0, i 1 ) p(i n 1, i n ) 2 The law of X is characterized by ν and p. Samy T. Ergodic theorems Probability Theory 7 / 92

Law of the Markov chain: general state space Proposition 3. Let X be a homogeneous Markov chain on (S, S) with initial law ν and transition p. 1 For n N and ϕ C b (S), we have E [ϕ(x 0,..., X n )] = ϕ(x 0,..., x n )ν(dx 0 ) p(x 0, x 1 ) p(x n 1, x n ) (1) E n+1 2 The law of X is characterized by ν and p. Samy T. Ergodic theorems Probability Theory 8 / 92

A criterion for Markovianity Proposition 4. Let Z : Ω E random variable. F countable set. {Y n ; n 1} i.i.d sequence, with Y Z and Y n F. f : E F E. We set X 0 = Z, and X n+1 = f (X n, Y n+1 ). Then X is a homogeneous Markov chain such that ν 0 = L(Z), and p(i, j) = P (f (i, Y 1 ) = j). Remark: A converse result exists, but it won t be used. Samy T. Ergodic theorems Probability Theory 9 / 92

Graph of a Markov chain Definition 5. Let X homogeneous Markov chain (initial law ν 0, transition p). We define a graph G(X) by G(X) is an oriented graph Vertices of G(X) are points of E. Edges of G(X) are defined by V {(i, j); i j, p(i, j) > 0}. Samy T. Ergodic theorems Probability Theory 10 / 92

Example Example 6. We consider E = {1, 2, 3, 4, 5} and 1/3 0 2/3 0 0 1/4 1/2 1/4 0 0 p = 1/2 0 1/2 0 0 0 0 0 0 1 0 0 0 2/3 1/3 Related graph: to be done in class Samy T. Ergodic theorems Probability Theory 11 / 92

Graph and accessibility Proposition 7. Let X homogeneous Markov chain (initial law ν 0, transition p). Then i j iff i = j or there exists an oriented path from i to j in G(X). Samy T. Ergodic theorems Probability Theory 12 / 92

Minimal class Definition 8. An equivalence class C is minimal if: For all i C and j C, we have i j. Example: For Example 6, C 1, C 3 are minimal, and C 2 is not minimal. Minimality criterions: (i) If there exists a unique class C, it is minimal. (ii) There exists a unique minimal class C! class C such that for all i E, we have i C. Application: Random walk Samy T. Ergodic theorems Probability Theory 13 / 92

Recurrence criterions Theorem 9. Let Then X homogeneous Markov chain (initial law ν 0, transition p). C class for the relation. 1 If C is not minimal then it is transient 2 If C is minimal and E <, then C is recurrent C is positive recurrent : for all i C, Ei [R i ] <. Samy T. Ergodic theorems Probability Theory 14 / 92

Example Recall: In Example 6 we had E = {1, 2, 3, 4, 5} (hence E < ) and 1/3 0 2/3 0 0 1/4 1/2 1/4 0 0 p = 1/2 0 1/2 0 0 0 0 0 0 1 0 0 0 2/3 1/3 Related classes: C 1 = {1, 3}, C 2 = {2} and C 3 = {4, 5}. C 1, C 3 minimal and C 2 not minimal. Conclusion: C 1, C 3 positive recurrent, C 2 transient. Samy T. Ergodic theorems Probability Theory 15 / 92

Stationary sequence Definition 10. Let {X n ; n 0} sequence of random variables We say that X is stationary if for all k 1 we have {X n+k ; n 0} (d) = {X n ; n 0}. Samy T. Ergodic theorems Probability Theory 17 / 92

iid random variables Example 11. Let X = {X n ; n 0} sequence of i.i.d random variables Then X is a stationary sequence. Samy T. Ergodic theorems Probability Theory 18 / 92

Markov chains Example 12. Let X = {X n ; n 0} Markov chain on state space S Transition probability: p(x, A) Hypothesis 1: unique stationary distribution π Hypothesis 2: L(X 0 ) = π Then X is a stationary sequence. Proof: Easy consequence of relation (1). Samy T. Ergodic theorems Probability Theory 19 / 92

Trivial example of Markov chain Example 13. On S = {0, 1} we take p(x, 1 x) = 1 π(0) = 1, π(1) = 1 2 2 Then P(X = (0, 1, 0, 1,...)) = P(X = (1, 0, 1, 0,...)) = 1 2 Samy T. Ergodic theorems Probability Theory 20 / 92

Rotation of the circle Example 14. We consider λ Lebesgue measure and: (Ω, F, P) = ([0, 1], Borel sets, λ) θ (0, 1) X = {X n ; n 0} with X n (ω) = (ω + nθ) mod 1 = ω + nθ [ω + nθ]. Then X is a stationary sequence. Samy T. Ergodic theorems Probability Theory 21 / 92

Proof Law of X 1 : For ϕ C b (R) we have E [ϕ(x 1 )] = = = 1 0 1 θ 0 1 0 ϕ (x + θ [x + θ]) dx ϕ (x + θ) dx + 1 1 θ ϕ (v) dv = E [ϕ(x 0 )] ϕ (x + θ 1) dx Thus L(X 1 ) = L(X 0 ). Samy T. Ergodic theorems Probability Theory 22 / 92

Proof (2) Altenative expression for X n : Due to the relation (a + b) mod 1 = [a mod 1 + b mod 1] mod 1, we have X n = (ω + nθ) mod 1 = (ω + (n 1)θ + θ) mod 1 = [(ω + (n 1)θ) mod 1 + θ mod 1] mod 1 = [X n 1 + θ] mod 1 Samy T. Ergodic theorems Probability Theory 23 / 92

Proof (3) Recall: We have seen that X n = [X n 1 + θ] mod 1. (2) Conclusion: It is readily checked that 1 From (2) and Proposition 4, X is a Markov chain 2 The relation L(X 1 ) = L(X 0 ) means that λ is an invariant distribution Therefore X is stationary thanks to Example 12. Samy T. Ergodic theorems Probability Theory 24 / 92

Transformation of a stationary sequence Theorem 15. Let X = {X n ; n 0} stationary sequence g : R N R measurable We consider a sequence Y defined by Y k = g ({X k+n ; n 0}). Then Y is a stationary sequence. Samy T. Ergodic theorems Probability Theory 25 / 92

σ-algebra on R N (1) Set R N : We define R N = {ω = (ω j ) j N ; ω j R}. Finite dimensional set: Of the form A = { ω R N ; ω j B j, for 1 j n }, with B j R, where R Borel σ-algebra in R. Samy T. Ergodic theorems Probability Theory 26 / 92

σ-algebra on R N (2) Definition 16. On R N we set: R N = σ (finite dimensional sets). Then R N is called Borel σ-algebra on R N. Remark: Kolmogorov s extension theorem is valid on (R N, R N ). Samy T. Ergodic theorems Probability Theory 27 / 92

Proof of Theorem 15 Notation: For x R N and k 0 we set g k (x) = g ({x k+n ; n 0}). Inverse image: Let B R N. Then { ω R N ; Y B } = { ω R N ; X A }, where A = { ω R N ; g(x) B }. Samy T. Ergodic theorems Probability Theory 28 / 92

Proof of Theorem 15 (2) Proof of stationarity: For the generic B R N we have P ( ω R N ; Y B ) = P ( ω R N ; X A ) This yields stationarity for Y. = P ( ω R N ; {X n ; n 0} A ) = P ( ω R N ; {X k+n ; n 0} A ) = P ( ω R N ; {Y k+n ; n 0} B ). Samy T. Ergodic theorems Probability Theory 29 / 92

Bernoulli shifts Example 17. We consider λ Lebesgue measure and: (Ω, F, P) = ([0, 1], Borel sets, λ) Y 0 = ω Y = {Y n ; n 0} where for n 1 we have Y n (ω) = 2Y n 1 mod 1. Then Y is a stationary sequence. Samy T. Ergodic theorems Probability Theory 30 / 92

Proof by Markov chains Law of Y 1 : For ϕ C b (R) we have E [ϕ(y 1 )] = = = 1 0 1 2 0 1 0 ϕ (2y ϕ (2y) dy + mod 1) dy 1 1 2 ϕ (v) dv = E [ϕ(y 0 )] ϕ (2y 1) dy Thus L(Y 1 ) = L(Y 0 ). Samy T. Ergodic theorems Probability Theory 31 / 92

Proof by Markov chains (2) Proof of stationarity: We have 1 Y n (ω) = 2Y n 1 mod 1 2 Thanks to proposition 4, Y is thus a Markov chain 3 The relation L(Y 1 ) = L(Y 0 ) means that λ is an invariant distribution for Y Therefore Y is a stationary sequence thanks to Example 12. Samy T. Ergodic theorems Probability Theory 32 / 92

Proof by Theorem 15 Another representation for Y : Let {X i ; i 1} i.i.d with common law B(1/2) g(x) = i 1 x i 2 (i+1) defined for x {0, 1} N g k (x) = g({x k+i ; i 1}) defined for x {0, 1} N Y k = g k (X) Then Y 0 λ, and Y n (ω) = 2Y n 1 mod 1. Stationarity of Y : We have X stationary Y k = g k (X) Therefore Y is a stationary sequence. Samy T. Ergodic theorems Probability Theory 33 / 92

Measure preserving map Definition 18. Let (Ω, F, P) a probability space A measurable map ϕ : Ω Ω We say that ϕ is measure preserving if P ( ϕ 1 (A) ) = P (A), for all A F. Otherwise stated, ϕ is measure preserving if E [ g ( X(ϕ(ω)) )] = E [g(x(ω))], for all g C b. Samy T. Ergodic theorems Probability Theory 34 / 92

Measure preserving map and stationarity Example 19. Let (Ω, F, P) a probability space A measure preserving map ϕ : Ω Ω X F a random variable For n 0 we set: X n = X (ϕ n (ω)) Then X is a stationary sequence. Samy T. Ergodic theorems Probability Theory 35 / 92

Proof Characterization with expected values: For g C(R n ) and k 1 we have: E [g (X k,..., X k+n )] = E [ g ( X 0 (ϕ k (ω)),..., X n (ϕ k (ω)) )] Thus X is a stationary sequence. = E [g (X 0 (ω),..., X n (ω))] = E [g (X 0,..., X n )]. Samy T. Ergodic theorems Probability Theory 36 / 92

Stationarity and measure preserving maps Example 20. Let Y stationary sequence with values in S = R n P probability measure on (S N, R(S N )) defined by: X n (ω) = ω n = L(X) = L(Y ) We define a shift operator ϕ by: Then ϕ is measure preserving. ϕ ({ω j ; j 0}) = {ω j+1 ; j 0}. Interpretation: Previous examples can be reduced to a measure preserving map Samy T. Ergodic theorems Probability Theory 37 / 92

Two sided stationary sequence Theorem 21. Let X stationary sequence Then X can be embedded into a two sided sequence {Y n ; n Z}. Samy T. Ergodic theorems Probability Theory 38 / 92

Proof Application of Kolmogorov s extension: It is enough to define a family of probability measures 1 On R A for any finite subset of Z 2 With consistent property Current situation: we consider Y defined by E [g (Y m,..., Y n )] = E [g (X 0,..., X m+n )], for all g C b (R m+n+1 ). Kolmogorov s extension applies Samy T. Ergodic theorems Probability Theory 39 / 92

Setup General setting: We are reduced to (Ω, F, P) probability space A map ϕ preserving P A random variable X A sequence X n (ω) = X(ω) Invariant set: Let A F. The set A is invariant if P ( A ϕ 1 (A) ) = 0 Notation abuse: For an invariant set A, we often write A = ϕ 1 (A) Samy T. Ergodic theorems Probability Theory 41 / 92

Ergodicity Definition 22. Let (Ω, F, P) probability space A map ϕ preserving P I σ-algebra of invariant events We say that ϕ is ergodic if I is trivial, i.e: A I = P(A) {0, 1}. Samy T. Ergodic theorems Probability Theory 42 / 92

Kolmogorov s 0-1 law Theorem 23. Let X = {X n ; n 0} sequence of i.i.d random variables For n 0, we set F n = σ({x k ; k n}) Tail σ-field: T n 0 F n Then T is trivial, i.e: A T = P(A) {0, 1}. Samy T. Ergodic theorems Probability Theory 43 / 92

Ergodicity for i.i.d sequences Example 24. Let X = {X n ; n 0} sequence of i.i.d random variables ϕ shift operator on Ω = R N Then ϕ is ergodic. Samy T. Ergodic theorems Probability Theory 44 / 92

Proof Measurability of an invariant set: Let A I. Then A = {ω Ω; ϕ(ω) A} = A F 1. Tail σ-field: Iterating the previous relation we get A I = A T Hence by Kolmogorov s 0-1 law we get that I is trivial Samy T. Ergodic theorems Probability Theory 45 / 92

Ergodicity for Markov chains Example 25. Let X = {X n ; n 0} MCH on countable state space S Transition probability: p(x, A) ϕ shift operator on Ω = S N Hypothesis 1: unique stationary distribution π Hypothesis 2: π(x) > 0 for all x S Hypothesis 3: L(X 0 ) = π Then 1 If X is not irreducible, θ is not ergodic 2 If X is irreducible, θ is ergodic Samy T. Ergodic theorems Probability Theory 46 / 92

Proof Basic Markov chains facts: Since π(x) > 0 for all x S, all states are recurrent State space decomposition: S = j J R j, where R j disjoint irreducible sets Non irreducible case: If J 2 we have X 0 R j X n R j for all n 0. Therefore for all j J we get: 1 (X0 R j ) = 1 (X0 R j ) 1 (X1 R j ) = 1 (X1 R j ) Iterating we get (X 0 R j ) I π(x 0 R j ) (0, 1) if J 2 Samy T. Ergodic theorems Probability Theory 47 / 92

Proof (2) General relation for 1 A : For A I, we have 1 A = 1 A θ n We set F n = σ(x 0,..., X n ) We define h(x) = E x [1 A ] Then E π [1 A F n ] invariance Markov prop = E π [1 A θ n F n ] = h(x n ) Levy s 0-1 law: Let F n F A F Then a.s and in L 1 (Ω) we have lim n E [1 A F n ] = 1 A Samy T. Ergodic theorems Probability Theory 48 / 92

Proof (3) Limit of h(x n ): Recall E π [1 A F n ] = h(x n ), and lim n E [1 A F n ] = 1 A Thus lim h(x n) = 1 A. (3) n Irreducible case: If X irreducible and π(y) > 0 for all y S, then h(x n ) = h(y) infinitely often for all y S According to (2) we thus have h = Cst We have thus found that whenever A I, 1 A (ω) = Cst {0, 1} Samy T. Ergodic theorems Probability Theory 49 / 92

Ergodicity for rotations of the circle Example 26. We consider λ Lebesgue measure and: (Ω, F, P) = ([0, 1], Borel sets, λ) θ (0, 1) X = {X n ; n 0} with X n (ω) = (ω + nθ) mod 1 = ω + nθ [ω + nθ]. ϕ shift transformation Then the following holds true: 1 If θ is rational, then ϕ is not ergodic 2 If θ is irrational, then ϕ is ergodic Samy T. Ergodic theorems Probability Theory 50 / 92

Transformation of an ergodic sequence Theorem 27. Let X = {X n ; n 0} ergodic sequence g : R N R measurable We consider a sequence Y defined by Y k = g ({X k+n ; n 0}). Then Y is an ergodic sequence. Samy T. Ergodic theorems Probability Theory 51 / 92

Proof Inverse image: Let B R N. Recall that { ω R N ; Y B } = { ω R N ; X A }, where A = { ω R N ; g(x) B }. Consequence for ergodicity: If B satisfies { ω R N ; (Y n ) n 0 B } = { ω R N ; (Y 1+n ) n 0 B } Then A satisfies { ω R N ; (X n ) n 0 A } = { ω R N ; (X 1+n ) n 0 A } Samy T. Ergodic theorems Probability Theory 52 / 92

Proof (2) Conclusion: Since A satisfies { ω R N ; (X n ) n 0 A } = { ω R N ; (X 1+n ) n 0 A }, we have Therefore: P (X A) {0, 1}. P (Y B) {0, 1}. Samy T. Ergodic theorems Probability Theory 53 / 92

Ergodicity for Bernoulli shifts Example 28. We consider λ Lebesgue measure and: (Ω, F, P) = ([0, 1], Borel sets, λ) Y 0 = ω Y = {Y n ; n 0} where for n 1 we have Y n (ω) = 2Y n 1 mod 1. ϕ shift transformation Then ϕ is ergodic. Samy T. Ergodic theorems Probability Theory 54 / 92

Proof Representation for Y : Recall that we have defined {X i ; i 1} i.i.d with common law B(1/2) g(x) = i 1 x i 2 (i+1) defined for x {0, 1} N g k (x) = g({x k+i ; i 1}) defined for x {0, 1} N Y k = g k (X) Then Y 0 λ, and Y n (ω) = 2Y n 1 mod 1. Ergodicity of Y : We have X ergodic Y k = g k (X) Hence owing to Theorem 27, Y is a stationary sequence. Samy T. Ergodic theorems Probability Theory 55 / 92

Birkhoff s ergodic theorem Theorem 29. Let X L 1 (Ω) ϕ a measure preserving map on Ω I the σ-field of invariant sets Then 1 lim n n n X(ϕ j (ω)) = E [X I], j=0 a.s and in L 1 (Ω) Samy T. Ergodic theorems Probability Theory 57 / 92

Ergodic theorem for ergodic maps Proposition 30. Let X L 1 (Ω) ϕ a measure preserving map on Ω Hypothesis: ϕ is ergodic Then 1 lim n n n X(ϕ j (ω)) = E [X], j=0 a.s and in L 1 (Ω) Samy T. Ergodic theorems Probability Theory 58 / 92

Maximal ergodic lemma Lemma 31. Let X L 1 (Ω) ϕ a measure preserving map on Ω X j = X(ϕ j (ω)) S n = n 1 j=0 X j M k = max{0, S 1,..., S k } Then E [ X 1 (Mk >0)] 0. Samy T. Ergodic theorems Probability Theory 59 / 92

Proof Lower bound for X: We will prove that for n = 1,..., k X(ω) S n (ω) M k (ϕ(ω)) (4) Relation (4) for n = 1: We have S 1 (ω) = X(ω) and M k (ϕ(ω)) 0. Thus X(ω) S 1 (ω) M k (ϕ(ω)) Relation (4) for n = 2,..., k: We have M k (ϕ(ω)) S j (ϕ(ω)) for j = 1,..., k. Thus X(ω) + M k (ϕ(ω)) X(ω) + S j (ϕ(ω)) = S j+1 (ω) and X(ω) S j+1 (ω) M k (ϕ(ω)) Samy T. Ergodic theorems Probability Theory 60 / 92

Proof (2) Consequence of relation (4): X(ω) max {S 1 (ω),..., S k (ω)} M k (ϕ(ω)) Integration of the previous relation: E [ ] X1 (Mk >0) = {M k >0} {M k >0} [max {S 1 (ω),..., S k (ω)} M k (ϕ(ω))] dp [M k (ω) M k (ϕ(ω))] dp Samy T. Ergodic theorems Probability Theory 61 / 92

Proof (3) Conclusion: We have seen E [ ] X1 (Mk >0) {M k >0} [M k (ω) M k (ϕ(ω))] dp In addition on {M k > 0} c we have 1 M k (ω) = 0 2 M k (ϕ(ω)) 0 3 Therefore {M k >0} c [M k(ω) M k (ϕ(ω))] dp 0 We thus get E [ X1 (Mk >0) ] [M k (ω) M k (ϕ(ω))] dp ϕ invariant = 0 Samy T. Ergodic theorems Probability Theory 62 / 92

Proof of Theorem 29 Reduction to E[X I] = 0: We Set ˆX = X E[X I] Recall that E[X I] is invariant Therefore we have 1 n X(ϕ j (ω)) E [X I] = 1 n ( X(ϕ j (ω)) E [X I] ) n j=0 n j=0 = 1 n [X E {X I}] (ϕ j (ω))= 1 n ˆX(ϕ j (ω)) n n j=0 We can thus prove Theorem 29 for X such that E[X I] = 0 j=0 Samy T. Ergodic theorems Probability Theory 63 / 92

Proof of Theorem 29 (2) Sufficient condition: We define X = lim sup n S n n D = {ω; X(ω) > ε} for ε > 0 We wish to prove that P(D) = 0 Invariance of D: Since X is invariant, we have D I Samy T. Ergodic theorems Probability Theory 64 / 92

Proof of Theorem 29 (3) Notation: We set X (ω) = (X(ω) ε)1 D (ω) X j = X (ϕ j (ω)) Sn = n 1 j=0 Xj Mk = max{0, S1,..., Sk } F n = {Mn > 0} (increasing sequence) F = n 0 F n Relation between F and D: We have F = {There exists n s.t X(ϕ n (ω)) > ε} D = D We thus wish to prove that P(D) = P(F ) = 0 Samy T. Ergodic theorems Probability Theory 65 / 92

Proof of Theorem 29 (4) Proof of E[X 1 D ] 0: Owing to Lemma 31 we have E [X 1 Fn ] 0 By dominated convergence we get: E [X 1 D ] = E [X 1 F ] 0 Proof of P(D) = 0: Since D I and E[X I] = 0 we get E [X 1 D ] = E [(X ε) 1 D ] = E {E[X I] 1 D } ε P(D) = ε P(D) We have seen that E[X 1 D ] 0, which yields P(D) = 0 Samy T. Ergodic theorems Probability Theory 66 / 92

Proof of Theorem 29 (5) Almost sure limit of Sn : For k 1 we have seen that n ({ P(D k ) P Taking limits as k we get: P ({ ω; lim sup n ω; lim sup n S n n > 1 }) = 0 k }) S n n > 0 = 0 Since the same result is true for the r.v X we end up with: P ({ ω; lim sup n }) S n n = 0 = 1 Samy T. Ergodic theorems Probability Theory 67 / 92

Proof of Theorem 29 (6) Truncation procedure: For the convergence in L 1 (Ω) we set and X 1 M = X 1 ( X M), and X 2 M = X 1 ( X >M) A n = 1 n Then for n 1 we have n 1 m=0 X(ϕ m (ω)) E[X I] E [ A n ] E [ A 1 n ] + E [ A 2 n ], where for j = 1, 2 we have defined: A j n = 1 n n 1 m=0 X j M (ϕm (ω)) E[X j M I] Samy T. Ergodic theorems Probability Theory 68 / 92

Proof of Theorem 29 (7) Limit for A 1 n: We have 1 a.s lim n A 1 n = 0 2 A 1 n 2M Therefore by dominated convergence we have: lim E [ ] A 1 n n = 0 Samy T. Ergodic theorems Probability Theory 69 / 92

Proof of Theorem 29 (8) Limit for A 2 n: We have E [ A 2 n ] 1 n 1 E [ X 2 n M (ϕ m (ω)) ] [ + E E[ X 2 M I] ] m=0 2 E [ XM ] 2 In addition, by dominated convergence we have: lim E [ X 2 n M ] = 0 Therefore lim n E [ A 2 n ] = 0 Samy T. Ergodic theorems Probability Theory 70 / 92

Proof of Theorem 29 (9) Conclusion for the L 1 (Ω) convergence: We have seen We thus get lim n E [ A 1 n ] = 0, and lim n E [ A 2 n ] = 0. L 1 1 (Ω) lim n n n 1 m=0 X j M (ϕm (ω)) = E[X j M I] Samy T. Ergodic theorems Probability Theory 71 / 92

LLN for iid random variables Example 32. Let X = {X n ; n 0} sequence of i.i.d random variables Hypothesis: X 0 L 1 (Ω) X n 1 nj=1 X n j Then lim X n = E[X 0 ], n a.s and in L 1 (Ω) Remark: W.r.t the usual LLN, we have obtained the L 1 (Ω) convergence Samy T. Ergodic theorems Probability Theory 72 / 92

Proof Previous results: We have seen that 1 X is an ergodic sequence 2 I T and T is trivial Conclusion: Theorem 29 applies and can be read as lim X n = E[X 0 I] = E[X 0 ], n a.s and in L 1 (Ω) Remark: The L 1 (Ω) convergence can also be obtained as follows Result: If P lim n Y n = Y, then L 1 (Ω) lim n Y n = Y iff lim n E[ Y n ] = E[ Y ] Apply this result successively to Y n = X + n and Y n = X n Samy T. Ergodic theorems Probability Theory 73 / 92

LLN for Markov chains Example 33. Let X = {X n ; n 0} MCH on countable state space S Hypothesis 1: unique stat. dist. π and L(X 0 ) = π Hypothesis 2: π(x) > 0 for all x S Hypothesis 3: X irreducible Hypothesis 4: f : S R satisfies f L 1 (π) Then 1 lim n n n f (X j ) = f (x) π(x), j=1 x S a.s and in L 1 (Ω) Samy T. Ergodic theorems Probability Theory 74 / 92

Proof Previous results: We have seen that 1 X is an ergodic sequence 2 Therefore f (X) is an ergodic sequence 3 I is trivial whenever X is irreducible Conclusion: Theorem 29 applies and can be read as 1 lim n n n f (X j ) = E[f (X 0 ) I] = E[f (X 0 )], j=1 a.s and in L 1 (Ω) Remark: W.r.t the usual LLN for Markov chains, we have obtained the L 1 (Ω) convergence Samy T. Ergodic theorems Probability Theory 75 / 92

LLN for rotation of the circle Example 34. We consider λ Lebesgue measure and: (Ω, F, P) = ([0, 1], Borel sets, λ) θ (0, 1) Q c A Borel subset of [0, 1] X = {X n ; n 0} with X n (ω) = (ω + nθ) mod 1 = ω + nθ [ω + nθ]. Then 1 lim n n n 1 A (X n ) = A, a.s and in L 1 (Ω) j=1 Samy T. Ergodic theorems Probability Theory 76 / 92

Deterministic LLN for rotation of the circle Theorem 35. In the context of Example 34, for 0 a < b < 1 we have for all x [0, 1) 1 lim n n n 1 [a,b) (X n (x)) = b a, j=1 Proof: Start from Example 34 Additional ingredient based on density arguments Samy T. Ergodic theorems Probability Theory 77 / 92

Benford s law Proposition 36. For k {1,..., 9} we have 1 lim n n n m=1 ( ) k + 1 1 (1 st digit of 2 m =k) = log 10 k Samy T. Ergodic theorems Probability Theory 78 / 92

Proof Notation: We set θ = log 10 (2) A k = [log 10 (k), log 10 (k + 1)) Expression 1 for log 10 (2 m ): we have log 10 (2 m ) = m θ Expression 2 for log 10 (2 m ): First digit of 2 m is k 0 iff 2 m = k 0 10 α + k 1, with α 0, k 1 < 10 α and ( log 10 (2 m ) = α + log 10 k 0 + k ) 1 10 α Samy T. Ergodic theorems Probability Theory 79 / 92

Proof (2) Ergodic sequence: We have seen that first digit of 2 m is k 0 iff which is equivalent to ( m θ = α + log 10 k 0 + k ) 1, 10 α m θ mod 1 A k Thus 1 lim n n n m=1 1 1 (1 st digit of 2 m =k) = lim n n n 1 (m θ mod 1 Ak ) m=1 Samy T. Ergodic theorems Probability Theory 80 / 92

Proof (3) Conclusion: We have θ = log 10 (2) Q c We can thus apply Theorem 35 with x = 0 and A = A k. We get 1 lim n n n m=1 ( ) k + 1 1 (m θ mod 1 Ak ) = A k = log 10 k Samy T. Ergodic theorems Probability Theory 81 / 92

Number of points visited Theorem 37. Let X stationary sequence with values in R d S n = n 1 j=0 X j A = {S k 0 for all k 1} R n = {S 1,..., S n } Card({S 1,..., S n }) Then R n lim n n = E [1 A I], almost surely Samy T. Ergodic theorems Probability Theory 83 / 92

Proof Lower bound: We have S j (ϕ m (ω)) = S j+m (ω) S m (ω) Thus and S j (ϕ m (ω)) 0 S j+m (ω) S m (ω) 1 A (ϕ m (ω)) = {S j+m (ω) S m (ω) for all j 1} We have thus obtained n R n 1 A (ϕ m (ω)). m=1 and owing to the ergodic theorem we get lim inf n R n n lim inf n 1 n n 1 A (ϕ m (ω)) = E [1 A I] m=1 Samy T. Ergodic theorems Probability Theory 84 / 92

Proof (2) Upper bound: For k 1 set Then and A k = {S j 0 for all 1 j k}. 1 Ak (ϕ m (ω)) = {S j (ω) S m (ω) for all m + 1 j m + k} R n k + n k m=1 1 Ak (ϕ m (ω)). Applying the ergodic theorem again we get lim inf n R n n lim sup n n k 1 1 Ak (ϕ m (ω)) = E [1 Ak I] (5) n m=1 Samy T. Ergodic theorems Probability Theory 85 / 92

Proof (3) Limit in k: We have k A k decreasing and A k = A. k=1 Therefore by monotone convergence we get lim k E [1 A k I] = E [1 A I]. Thanks to (5) we have thus proved that lim sup n R n n E [1 A I] Samy T. Ergodic theorems Probability Theory 86 / 92

Recurrence criterion in Z Theorem 38. Let X stationary sequence with values in Z Hypothesis: X 1 L 1 (Ω) S n = n 1 j=0 X j A = {S k 0 for all k 1} The following holds true: 1 If E[X 1 I] = 0 then P(A) = 0 2 If P(A) = 0 then P(S n = 0 i.o) = 1 Samy T. Ergodic theorems Probability Theory 87 / 92

Proof Standing assumption: E[X 1 I] = 0, which yields S n lim n n Relation for max S: For all K 1 we have = 0. (6) lim sup n max 1 k n S k n = lim sup n max K k n S k n max k K S k k Taking K and thanks to (6) we get lim max S k n 1 k n n = 0 (7) Samy T. Ergodic theorems Probability Theory 88 / 92

Proof (2) Consequence on R n : In d = 1 we have Hence and according to (7) we get lim max R n n 1 k n n R n = Range {X j ; 0 j n}. R n 1 + 2 max 1 k n S k, Theorem 37 Relation (7) = E [1 A I] = 0 Conclusion for item 1: We take expectations in previous relation, which yields P(A) = 0 Samy T. Ergodic theorems Probability Theory 89 / 92

Proof (3) Notation: We set F j = {S i 0 for i < j and S j = 0} G j,k = {S j+i S j 0 for i < k and S j+k S j = 0} Simple relations on F j, G j,k : 1 Since P(A) = 0 we have k 1 P(F k ) = 1 2 By stationarity, P(G j,k ) = P(F k ) 3 Hence we also have k 1 P(G j,k ) = 1 4 For fixed j, the sets G j,k are disjoint Samy T. Ergodic theorems Probability Theory 90 / 92

Proof (4) More relations on F j, G j,k : We have obtained P (F j G j,k ) = P (F j ), and P (F j G j,k ) = 1 k 1 j,k 1 Conclusion on recurrence: We have F j G j,k = {S n = 0 at least two times} j,k 1 Therefore P (S n = 0 at least two times) = 1 Generalization: With sets G j1,...,j m Taking m Samy T. Ergodic theorems Probability Theory 91 / 92

Extension by Kac Theorem 39. Let X stationary sequence with values in (S, S) A S and S n = n 1 j=0 X j T 0 = 0 and T n = inf{m > T n 1 ; X m A} t n = T n T n 1 Hypothesis: P (X n A at least once) = 1 Then the following holds true: 1 Under P( X 0 A) the sequence (t n ) is stationary 2 We have E [T 1 X 0 A] = 1 P (X 0 A) Samy T. Ergodic theorems Probability Theory 92 / 92