# Chapter 3. Strong convergence. 3.1 Definition of almost sure convergence

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3 The coverse of Corollary 3.4 is ot true, as the followig example illustrates. Example 3.5 Take Ω to be the half-ope iterval (0, 1], ad for ay iterval J Ω, say J = (a, b], take P (J) = b a to be the legth of that iterval. Defie a sequece of itervals J 1, J 2,... as follows: J 1 = (0, 1] J 2 through J 4 = (0, 1 3 ], ( 1 3, 2 3 ], ( 2 3,1] J 5 through J 9 = (0, 1 5 ], ( 1 5, 2 5 ], ( 2 5, 3 5 ], ( 3 5, 4 5 ], ( 4 5,1] J m 2 +1 through J (m+1) 2 =.. ( ] ( ] 1 2m 0,,..., 2m + 1 2m + 1, 1 Note i particular that P (J ) = 1/(2m + 1), where m = 1 is the largest iteger ot greater tha 1. Now, defie X = I{J } ad take 0 < ɛ < 1. The P ( X 0 < ɛ) is the same as 1 P (J ). Sice P (J ) 0, we coclude X P 0 by defiitio. However, it is ot true that X 0. Sice every ω Ω is cotaied i ifiitely may J, the set A defied i Equatio (3.1) is empty for all. Alteratively, cosider the set S = {ω : X (ω) 0}. For ay ω, X (ω) has o limit because X (ω) = 1 ad X (ω) = 0 both occur for ifiitely may. Thus S is empty. This is ot covergece with probability oe; it is covergece with probability zero! Multivariate Extesios We may exted Defiitio 3.1 to the multivariate case i a completely straightforward way: Defiitio 3.6 X is said to coverge almost surely (or with probability oe) to X (X X) if P (X X as ) = 1. Alteratively, sice the proof of Theorem 3.3 applies to radom vectors as well as radom variables, we say X X if for ay ɛ > 0, P ( X k X < ɛ for all k ) 1 as. 52

5 It is possible to use fairly simple argumets to prove a versio of the Strog Law uder more restrictive assumptios tha those give above. See Exercise 3.4 for details of a proof of the uivariate Strog Law uder the additioal assumptio that X 4 <. To aid the proof of the Strog Law i its full geerality, we first establish a useful lemma. Lemma 3.9 If k=1 P ( X k X > ɛ) < for ay ɛ > 0, the X X. Proof: The proof relies o the coutable subadditivity of ay probability measure, a axiom statig that for ay sequece A 1, A 2,... of evets, ( ) P A k P (A k ). (3.3) k=1 To prove the lemma, we must demostrate that P ( X k X ɛ for all k ) 1 as, which (takig complemets) is equivalet to P ( X k X > ɛ for some k ) 0. Lettig A k deote the evet that X k X > ɛ, coutable subadditivity implies ( ) P (A k for some k ) = P A k P (A k ), ad the right had side teds to 0 as because it is the tail of a coverget series. Lemma 3.9 is early the same as a famous result called the First Borel-Catelli Lemma, or sometimes simply the Borel-Catelli Lemma; see Exercise 3.3. The utility of Lemma 3.9 is illustrated by the followig useful result, which allows us to relate almost sure covergece to covergece i probability (see Theorem 2.24, for istace). Theorem 3.10 X P X if ad oly if each subsequece X 1, X 2,... cotais a further subsequece that coverges almost surely to X. The proof of Theorem 3.10, which uses Lemma 3.9, is the subject of Exercise 3.7. i=k k= k= Idepedet but ot idetically distributed variables Here, we geeralize the uivariate versio of the Strog Law to a situatio i which the X are assumed to be idepedet ad satisfy a secod momet coditio: Theorem 3.11 Kolmogorov s Strog Law of Large Numbers: Suppose that X 1, X 2,... are idepedet with mea µ ad Var X i <. i 2 The X µ. i=1 54

6 Note that there is o reaso the X i i Theorem 3.11 must have the same meas: If E X i = µ i, the the theorem as writte implies that (1/) i (X i µ i ) 0. Theorem 3.11 may be proved usig Kolmogorov s iequality from Exercise 1.31; this proof is the focus of Exercise 3.6. I fact, Theorem 3.11 turs out to be very importat because it may be used to prove the Strog Law, Theorem 3.8. The key to completig this proof is to itroduce trucated versios of X 1, X 2,... as i the followig lemma. Lemma 3.12 Suppose that X 1, X 2,... are idepedet ad idetically distributed ad have fiite mea µ. Defie Xi = X i I{ X i i}. The ad X X 0. i=1 Var X i i 2 < (3.4) Uder the assumptios of Lemma 3.12, we see immediately that X = X +(X X ) µ, because Equatio (3.4) implies X µ by Theorem This proves the uivariate versio of Theorem 3.8; the full multivariate versio follows because X µ if ad oly if X j µ j for all j (Lemma 1.31). A proof of Lemma 3.12 is the subject of Exercise 3.5. Exercises for Sectio 3.2 Exercise 3.3 Let B 1, B 2,... deote a sequece of evets. Let B i.o., which stads for B ifiitely ofte, deote the set B i.o. def = {ω Ω : for every, there exists k such that ω B k }. Prove the first Borel-Catelli Lemma, which states that if =1 P (B ) <, the P (B i.o.) = 0. Hit: Argue that B i.o. = B k, =1 k= the adapt the proof of Lemma 3.9. Exercise 3.4 Use the hit below to prove that if X 1, X 2,... are idepedet ad idetically distributed ad E X1 4 <, the X E X 1. You may assume without loss of geerality that E X 1 = 0. 55

7 Hit: Use Markov s iequality (1.22) with r = 4 to put a upper boud o P ( X > ɛ ) ivolvig E (X X ) 4. Expad E (X X ) 4 ad the cout the ozero terms. Fially, argue that the coditios of Lemma 3.9 are satisfied. Exercise 3.5 Lemma 3.12 makes two assertios about the radom variables X i = X i I{ X i i}: (a) Prove that i=1 Var X i i 2 <. Hit: Use the fact that the X i are idepedet ad idetically distributed, the show that X1 2 1 k I{ X 1 k} 2 X 2 1, k=1 perhaps by boudig the sum o the left by a easy-to-evaluate itegral. (b) Prove that X X 0. Hit: Use Lemma 3.9 ad Exercise 1.32 to show that X X 0. The use Exercise 1.3. Exercise 3.6 Prove Theorem Use the followig steps: (a) For k = 1, 2,..., defie Y k = max X µ. 2 k 1 <2 k Use the Kolmogorov iequality from Exercise 1.31 to show that P (Y k ɛ) 4 2k i=1 Var X i 4 k ɛ 2. (b) Use Lemma 3.9 to show that Y k 0, the argue that this proves X µ. Hit: Lettig log 2 i deote the smallest iteger greater tha or equal to log 2 i (the base-2 logarithm of i), verify ad use the fact that k 3i. 2 k= log 2 i 56

8 Exercise 3.7 Prove Theorem Hit: To simplify otatio, let Y k = X k deote a arbitrary subseqece. If Y P k X, Show that there exist k 1, k 2,... such that the use Lemma 3.9. P ( Y kj X > ɛ) < 1 2 j, O the other had, if X does ot coverge i probability to X, argue that there exists a subsequece Y 1 = X 1, Y 2 = X 2,... ad ɛ > 0 such that P ( Y k X > ɛ) > ɛ for all k. The use Corollary 3.4 to argue that Y does ot have a subsequece that coverges almost surely. 3.3 The Domiated Covergece Theorem We ow cosider the questio of whe Y d Y implies E Y E Y. This is ot geerally the case: Cosider cotamiated ormal distributios with distributio fuctios ( F (x) = 1 1 ) Φ(x) + 1 Φ(x 37). (3.5) These distributios coverge i distributio to the stadard ormal Φ(x), yet each has mea 37. However, recall Theorem 2.25, which guaratees that Y d Y implies E Y Y if all of the Y ad Y are uiformly bouded say, Y < M ad Y < M sice i that case, there is a bouded, cotiuous fuctio g(y) for which g(y ) = Y ad g(y ) = Y : Simply defie g(y) = y for M < y < M, ad g(y) = My/ y otherwise. To say that the Y are uiformly bouded is a much stroger statemet tha sayig that each Y is bouded. The latter statemet implies that the boud we choose is allowed to deped o, whereas the uiform boud meas that the same boud must apply to all Y. Whe there are oly fiitely may Y, the boudedess implies uiform boudedess sice we may take as a uiform boud the maximum of the bouds of the idividual Y. However, i the case of a ifiite sequece of Y, the maximum of a ifiite set of idividual bouds might ot exist. 57

11 Because E Z <, subtractig E Z preserves the iequality, so we obtai lim sup Together, iequalities (3.8) ad (3.9) imply E lim if X lim if E X E lim sup X. (3.9) E X lim sup E X E lim sup X. Therefore, the proof would be complete if X X. This is where we ivoke the Skorohod Represetatio Theorem: Because there exists a sequece Y that does coverge almost surely to Y, havig the same distributios ad expectatios as X ad X, the above argumet shows that E Y E Y, hece E X E X, completig the proof. Exercises for Sectio 3.3 Exercise 3.8 This exercise proves two results used to establish theorem (a) Prove Lemma Hit: For ay δ > 0, let x be a cotiuity poit of F (t) i the iterval (Y (ω) δ, Y (ω)). Use the fact that F d F to argue that for large, Y (ω) δ < Y (ω). Take the limit iferior of each side ad ote that δ is arbitrary. Similarly, argue that for large, Y (ω) < Y (ω + ɛ) + δ. (b) Prove that ay odecreasig fuctio has at most coutably may poits of discotiuity. Hit: If x is a poit of discotiuity, cosider the ope iterval whose edpoits are the left- ad right-sided limits at x. Note that each such iterval cotais a ratioal umber, of which there are oly coutably may. Exercise 3.9 Prove Fatou s lemma: E lim if X lim if E X. (3.10) Hit: Argue that E X E if k X k, the take the limit iferior of each side. Use the mootoe covergece property o page 25. Exercise 3.10 If Y d Y, a sufficiet coditio for E Y E Y is the uiform itegrability of the Y. 60

12 Defiitio 3.18 The radom variables Y 1, Y 2,... are said to be uiformly itegrable if sup E ( Y I{ Y α}) 0 as α. Prove that if Y d Y ad the Y are uiformly itegrable, the E Y E Y. Exercise 3.11 Prove that if there exists ɛ > 0 such that sup E Y 1+ɛ <, the the Y are uiformly itegrable. Exercise 3.12 Prove that if there exists a radom variable Z such that E Z = µ < ad P ( Y t) P ( Z t) for all ad for all t > 0, the the Y are uiformly itegrable. You may use the fact (without proof) that for a oegative X, E (X) = 0 P (X t) dt. Hits: Cosider the radom variables Y I{ Y t} ad Z I{ Z t}. I additio, use the fact that E Z = E ( Z I{i 1 Z < i}) i=1 to argue that E ( Z I{ Z < α}) E Z as α. 61

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