1. Dynkin Systems
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Tutorial 1: Dynkin systems 1 1. Dynkin systems Definition 1 A Dynkin system on a set Ω is a subset D of the power set P(Ω), with the following properties: (i)Ω2D (ii) A; B 2D;A⊆ B ) B n A 2D +[1 (iii) An 2D;An ⊆ An+1;n≥ 1 ) An 2D n=1 Definition 2 A σ-algebra on a set Ω is a subset F of the power set P(Ω) with the following properties: (i)Ω2F 4 (ii) A 2F ) Ac =Ωn A 2F +[1 (iii) An 2F;n≥ 1 ) An 2F n=1 Exercise 1. Let F be a σ-algebra on Ω. Show that ;2F,thatifA; B 2F then A [ B 2F and also A \ B 2F. Recall that B n A = B \ Ac and conclude that F is also a Dynkin system on Ω. Exercise 2. Let (Di)i2I be an arbitrary family of Dynkin systems on Ω, with 4 I =6 ;. Show that D = \i2I Di is also a Dynkin system on Ω. Exercise 3. Let (Fi)i2I be an arbitrary family of σ-algebras on Ω, with I =6 ;. 4 Show that F = \i2I Fi is also a σ-algebra on Ω. Exercise 4. Let A be a subset of the power set P(Ω). Define: 4 D(A) = fD Dynkin system on Ω : A⊆Dg Show that P(Ω) is a Dynkin system on Ω, and that D(A) is not empty. Define: 4 \ D(A) = D D2D(A) Show that D(A) is a Dynkin system on Ω such that A⊆D(A), and that it is the smallest Dynkin system on Ω with such property, (i.e. if D is a Dynkin system on Ω with A⊆D,thenD(A) ⊆D). Definition 3 Let A⊆P(Ω).WecallDynkin system generated by A, the Dynkin system on Ω,denotedD(A), equal to the intersection of all Dynkin systems on Ω, which contain A. Exercise 5. Do exactly as before, but replacing Dynkin systems by σ-algebras. www.probability.net Tutorial 1: Dynkin systems 2 Definition 4 Let A⊆P(Ω).Wecallσ-algebra generated by A,the σ-algebra on Ω,denotedσ(A), equal to the intersection of all σ-algebras on Ω, which contain A. Definition 5 AsubsetA of the power set P(Ω) is called a π-system on Ω,if and only if it is closed under finite intersection, i.e. if it has the property: A; B 2A ) A \ B 2A Exercise 6. Let A be a π-system on Ω. For all A 2D(A), we define: 4 Γ(A) = fB 2D(A): A \ B 2D(A)g 1. If A 2A, show that A⊆Γ(A) 2. Show that for all A 2D(A), Γ(A) is a Dynkin system on Ω. 3. Show that if A 2A,thenD(A) ⊆ Γ(A). 4. Show that if B 2D(A), then A⊆Γ(B). 5. Show that for all B 2D(A), D(A) ⊆ Γ(B). 6. Conclude that D(A)isalsoaπ-system on Ω. Exercise 7. Let D be a Dynkin system on Ω which is also a π-system. 1. Show that if A; B 2Dthen A [ B 2D. 4 2D ≥ [n [+1 2. Let An ;n 1. Consider Bn = i=1Ai. Show that n=1An = [+1 n=1Bn. 3. Show that D is a σ-algebra on Ω. Exercise 8. Let A be a π-system on Ω. Explain why D(A)isaσ-algebra on Ω, and σ(A) is a Dynkin system on Ω. Conclude that D(A)=σ(A). Prove the theorem: Theorem 1 (Dynkin system) Let C be a collection of subsets of Ω which is closed under pairwise intersection. If D is a Dynkin system containing C then D also contains the σ-algebra σ(C) generated by C. www.probability.net Solutions to Exercises 3 Solutions to Exercises Exercise 1. 1. From (i) of definition (2), Ω 2F. Hence, from (ii), ; =Ωc 2F. 2. If A; B 2F, we can construct a sequence (An)n≥1 of elements of F by setting A1 = A, A2 = B and Ak = ; for all k ≥ 3. Then, using (iii)of definition (2): +[1 A [ B = An 2F n=1 3. From (ii), Ac and Bc are also elements of F.SoAc [ Bc 2F. Finally, again from (ii): A \ B =(Ac [ Bc)c 2F 4. B and Ac are both elements of F,so: 4 B n A = B \ Ac 2F 5. F being a σ-algebra, conditions (i)and(iii) of definition (1) are immedi- ately satisfied. But we have just proved that if A; B 2F,thenB n A 2F. Hence, condition (ii) of definition (1) is also satisfied, and F is therefore a Dynkin system. Exercise 1 Exercise 2. 1. Each Di is a Dynkin system. From (i) of definition (1), Ω 2Di.This being true for all i 2 I,Ω2\i2I Di = D. Hence (i)ofdefinition(1)is satisfied for D. 2. Let A; B 2Dwith A ⊆ B. Then for all i 2 I, A; B 2Di,withA ⊆ B. Since each Di is a Dynkin system, from (ii)ofdefinition(1)weseethat B n A 2Di. This being true for all i 2 I, B n A 2\i2I Di = D. Hence, (ii) of definition (1) is satisfied for D. 3. Let (An)n≥1 be a sequence of elements of D with An ⊆ An+1. Then, for all i 2 I,(An)n≥1 is a sequence of elements of Di with An ⊆ An+1. Since each Di is a Dynkin system, from (iii)ofdefinition(1)weseethat [+1 2D 2 [+1 2\ D D n=1An i. This being true for all i I, n=1An i2I i = . Hence, (iii) of definition (1) is satisfied for D. 4. Having checked (i); (ii); (iii) of definition (1), we conclude that D is indeed a Dynkin system on Ω. Exercise 2 Exercise 3. www.probability.net Solutions to Exercises 4 1. Each Fi is a σ-algebra. From (i) of definition (2), Ω 2Fi. This being true for all i 2 I,Ω2\i2I Fi = F. Hence (i) of definition (2) is satisfied for F. 2. Let A 2F. Then for all i 2 I, A 2Fi. Since each Fi is a σ-algebra, from c (ii) of definition (2) we see that A 2Fi. This being true for all i 2 I, c A 2\i2I Fi = F. Hence, (ii) of definition (2) is satisfied for F. 3. Let (An)n≥1 be a sequence of elements of F. Then, for all i 2 I,(An)n≥1 is a sequence of elements of Fi. Since each Fi is a σ-algebra, from (iii) [+1 2F 2 of definition (2) we see that n=1An i. This being true for all i I, [+1 2\ F F F n=1An i2I i = . Hence, (iii) of definition (2) is satisfied for . 4. Having checked (i); (ii); (iii) of definition (2), we conclude that F is indeed a σ-algebra on Ω. Exercise 3 Exercise 4. 1. Ω is obviously a subset of Ω, so Ω 2P(Ω), and (i) of definition (1) is satisfied for P(Ω). If A; B 2P(Ω), whether or not A ⊆ B, B n A is still a subset of Ω, i.e. B n A 2P(Ω). So (ii) of definition (1) is also satisfied for P(Ω). If (An)n≥1 is a sequence of subsets of Ω, whether or not this ⊆ [+1 sequence is increasing (i.e. An An+1), n=1An is still a subset of Ω, i.e. belongs to P(Ω). So (iii) of definition (1) is satisfied for P(Ω), and finally, P(Ω) is a Dynkin system on Ω. 2. By assumption, A⊆P(Ω). Since P(Ω) is also a Dynkin system on Ω, we see that P(Ω) 2 D(A). In particular, D(A)isnotempty. 3. Take I = D(A), and for all i 2 I, define Di = i.Then(Di)i2I is a family of Dynkin systems on Ω (where I =6 ;) and since: 4 \ \ D(A) = D = Di D2D(A) i2I using exercise (2), we conclude that D(A) is a Dynkin system on Ω. 4. Let A 2A. For all D2D(A), we have A⊆D. Hence, for all D2D(A), A 2D.So: \ 4 A 2 D = D(A) D2D(A) It follows that A⊆D(A). 5. Suppose D is another Dynkin system on Ω such that A⊆D.Then D2D(A), from which we conclude that: 4 \ D(A) = D0 ⊆D D02D(A) www.probability.net Solutions to Exercises 5 Exercise 4 Exercise 5. 4 1. We define similarly: F (A) = fF σ-algebra on Ω : A⊆Fgand: 4 \ σ(A) = F F2F (A) 2. Ω 2P(Ω), and (i) of definition (2) is satisfied for P(Ω). If A 2P(Ω), then c A 2P(Ω) and (ii) of definition (2) is also satisfied for P(Ω). If (An)n≥1 [+1 is a sequence of subsets of Ω, n=1An is still a subset of Ω, and (iii)of definition (2) is satisfied for P(Ω). Finally, P(Ω) is a σ-algebra on Ω. 3. By assumption, A⊆P(Ω). Since P(Ω) is also a σ-algebra on Ω, we see that P(Ω) 2 F (A). In particular, F (A)isnotempty. 4. Take I = F (A), and for all i 2 I, define Fi = i.Then(Fi)i2I is a family of σ-algebra on Ω (where I =6 ;) and since: 4 \ \ σ(A) = F = Fi F2F (A) i2I using exercise (3), we conclude that σ(A)isaσ-algebra on Ω.