B8.1 Martingales Through Measure Theory. Concept of independence

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1 B8.1 Martingales Through Measure Theory Concet of indeendence Motivated by the notion of indeendent events in relims robability, we have generalized the concet of indeendence to families of σ-algebras. In these notes, we assume that (Ω, F, ) is a robability sace. If {F α : α Λ} (where Λ is a non-emty index set) is a family of some sub σ-algebras on the robability sace (Ω, F, ), then {F α : α Λ} are indeendent if (A 1 A n ) = (A 1 ) (A n ) (1) for any A i F αi where α i Λ (i = 1,, n) as long as α 1,, α n are different. By definition, if {F α : α Λ} are indeendent, then any its sub family of {F α : α Λ} are indeendent. Furthermore we don t need to test (1) for all A i F αi, and very often we only need to verify (1) for those A i in a π-system C αi as long as it generates the σ-algebra F αi = σ {C αi }. From definition, we can see immediately that a family {F α : α Λ} of σ-algebras are indeendent, if and only if any finite subfamily {F αi : i = 1,, n} (where α 1,, α n belong to Λ, for any n as long as it is not greater than the number of elements in the index set Λ) are indeendent. This is due to the required equality (1) involves only finite many indices, so only to do with finite many σ-algebras in the family. Another direct consequence from the definition of indeendence is that, if {F n : n = 1, 2, } is a sequence of sub σ-algebras, then {F n } are indeendent if and only if (A 1 A n ) = (A 1 ) (A n ) (2) for any n, and for any A 1 F 1,, A n F n. This consequence follows from the fact that Ω belongs to any σ-algebra, and (Ω) = 1, so that we can insert as many as you want the term Ω in the intersection on the left-hand side, and as many as you wish (Ω) on the right-hand side of (2), which will not alter the equality. From elementary course on robability, we have learned that, if {A α : α Λ} is a family of events, i.e. all A α F, then σ {A α } (where α Λ) are indeendent if and only if (A α1 A αn ) = (A α1 ) (A αn ) for every finite subset {α 1,, α n } Λ as long as α i are different. That is, events A α (where α Λ) are indeendent as defined in the relim robability. In articular, if A n (n = 1, 2, ) is a sequence of events, then A 1, A 2, are indeendent, equivalently their generated σ-algebras σ {A 1 }, σ {A 2 }, are indeendent. The discussion can be extended to random variables. A family {X α : α Λ} of random variables on (Ω, F, ) are indeendent, by definition, if the family σ {X α } (where α Λ) of sub σ-algebras are indeendent, and thus if and only if any finite sub family X α1,, X αn are indeendent. Since for any real random variable X, σ {X} = X 1 (B) = { X 1 (B) : B R Borel measurable }

2 where X 1 (B) = {ω Ω : X (ω) B} {X B}, therefore random variables X α1,, X αn are indeendent, if and only if X α1 B 1,, X αn B n = X α1 B 1 X αn B n (3) for any Borel subsets B 1,, B n. Examle 1. Real random variables X 1,, X n are indeendent, if and only if the joint distribution µ of (X 1,, X n ), defined to be the robability measure µ on (R n, B (R n )) by µ(e) = (X 1,, X n ) E for E B (R n ) coincides with the roduct measure µ 1 µ n on B(R n ), where µ i is the distribution of X i, that is µ i (B) = X i B for any B B(R). I roof. If the joint distribution µ = µ 1 µ n, then by taking E = B 1 B n, we obtain µ (B 1 B n ) = µ 1 (B 1 ) µ n (B n ) (4) for all B i B (R), which is equivalent to X 1 B 1,, X n B n = X 1 B 1 X n B n (5) so X 1,, X n are indeendent. Conversely, suose X 1,, X n are indeendent, then (5) holds for all B i B(R). Let C be the collection of all subsets of R n which have a form B 1 B n. Then C is a π-system on R n, and B (R n ) = σ {C }. (5) says exactly that the joint distribution µ of X 1,, X n coincides with µ 1 µ n on the π-system C, so they must equal on B(R n ) according to the uniqueness lemma for measures. This comletes the roof. Examle 2. Random variables (real valued) X α1,, X αn are indeendent if and only if for any Borel measurable functions f i (i = 1, ) such that E f 1 (X α1 ) f n (X αn ) = E f 1 (X α1 ) E f n (X αn ) (6) as long as integrals (exectations) exist. roof. If (6) holds, then since σ {X α } = Xα 1 (B) (where B is the Borel σ-algebra), so that if A j σ { } X αj, then Aj = Xα 1 j (B j ), where B j B. Thus (A 1 A n ) = ( X 1 α 1 (B 1 ) X 1 α n (B n ) ) = ({X α1 B 1 } {X αn B n }) = ({X α1 B 1,, X αn B n }) ( ) ( ) = E 1 {Xα1 B 1,,X αn B n} = E 1 {Xα1 B 1} 1 {X αn B n} = E (1 B1 (X α1 ) 1 Bn (X αn )) = E (1 B1 (X α1 )) E (1 Bn (X αn ))

3 where the last equality follows from (6) alying to f i = 1 Bi which are Borel measurable as B i B. Hence (A 1 A n ) = E (1 B1 (X α1 )) E (1 Bn (X αn )) = (A 1 ) (A n ). Now, we show that, if X α1,, X αn are indeendent, then (6) holds. In fact, let µ i denote the distribution of X αi, that is, µ i (E) = X αi E for E B, then, by the revious examle, the joint distribution µ of (X α1,, X αn ) is exactly the roduct measure µ 1 µ n, hence, by Fubini s theorem we have ˆ E f 1 (X α1 ) f n (X αn ) = f 1 (x 1 ) f n (x n )µ(dx 1,, dx n ) Ω 1 Ω ˆ ˆ n = f 1 (x 1 ) f n (x n )µ n (dx n ) µ 1 (dx 1 ) Ω n which roves (6). Ω 1 = E f 1 (X α1 ) E f n (X αn ) Examle 3. Suose X, Y and Z are three indeendent real valued random variables, then, it should be clear that X + Y and Z are indeendent, but how to rove this? While Dynkin s lemma and the uniqueness lemma for measures may hel for this kind of questions. roof. According to definition, we want to show σ {X + Y } and σ {Z} are indeendent, that is, want to show that for any D σ {X + Y } and C σ {Z}, D C = D Z C. (7) Since X, Y and Z are indeendent, so X and Y are indeendent too, hence A B C = A B C = A B C that is, (7) holds for D = A B as long as A σ {X} and B σ {Y }, all such sets consist of a π-system which generates the σ-algebra σ {X, Y }. Formally we let C = {A B : A σ {X}, B σ {B}}. Then C is a π-system, and σ {X} C, σ {Y } C, so that σ {X, Y } = σ {C }. Let C σ {Z} be fixed but arbitrary. Considers two measures µ 1 (D) = D C and µ 2 D = D Z C for D F. Both µ i are finite measures, µ 1 (Ω) = µ 2 (Ω) = C, and µ 1 = µ 2 on C, hence by the Uniqueness Lemma for measures, µ 1 = µ 2 on σ {C } = σ {X, Y }. It follows that (7) holds for any D σ {X, Y }. While we know that X + Y is measurable with resect to σ {X, Y }, so that σ {X + Y } σ {X, Y }, hence (7) holds for any D σ {X + Y } and C σ {Z}, by definition, X + Y and Z are indeendent. From the roof, we can see that f(x, Y ) and Z are indeendent for any Borel measurable function f. For examle X 3 + cos Y and Z are indeendent. You may extend this to any finite many indeendent random variables. For examle, if X 1,, X n, Y 1,, Y m are indeendent (real) random variables on (Ω, F, ), then f(x 1,, X n ) and g (Y 1,, Y m ) are indeendent as long as f and g are Borel measurable.

4 Examle 4. If X is a random variable, and G is a σ-algebra, then naturally we say X and G are indeendent (or say X is indeendent of the σ-algebra G) if σ {X} and G are indeendent. This notion can be generalized to a family of random variables {X α } and a family σ-algebras {F β } in a natural way leave for the reader as an exercise. By definition, X and G are indeendent, if and only if X and 1 A are indeendent, and if and only if E f(x) : A = E f(x) (A) for any A G, and for any Borel measurable function f such that f(x) is integrable. Examle 5. G F be a sub σ-algebra, and X be a random variable, non-negative or integrable, then E X G = E X. In articular, if X and Z are indeendent, then E X Z = E X. roof. Let Y = E X. Then Y is a constant so is G-measurable. For every A G, we have E X : A = E X1 A = E X E 1 A = E E X 1 A so that E X is the conditional exectation of X given G. Examle 6. Consider a sequence of indeendent Bernoulli trials {X n : n = 1, 2, }, which is an i.i.d. sequence, indeendent identically distributed, with the same distribution X n = 1 =, and X n = 0 = 1 where 0 < < 1. Let T be the waiting time after the first time until the first success occurs Then T = inf {j 0 : X j+1 = 1}. T = k = X i = 0 for i k and X k+1 = 1 = (1 ) i where k = 0, 1,. That is T has a geometric distribution. Similarly, for every n = 1, 2,, if denotes the number of the longest success run starting from n, that is, = j if X i = 1 for i = n,, n + j 1 but X n+j = 0, so that for j = 0, 1, 2,. Then = j = j (1 ) lim su n = 1 = 1. (8) The roof of this result is a tyical alication of Borel-Cantelli lemma. First we recall from relims Analysis, if {x n } is a sequence of reals, then l = lim su x n is real, then there is a sub-sequence x nk l. If there is no any sub-sequence such that x nl > a then lim su x n a. Let a > 1 be arbitrary but fixed and set { } A n = n > a for n = 1, 2,.

5 Then A n = > a n = = j j>a n = j>a n a n j (1 ) = (1 ) 1 a n = 1 n a so that An = By Borel-Cantelli lemma, A n i. o. = 0, so that which yields that Next we show that lim su lim su lim su n 1 n a <. n > a n 1 n 1 To this end, we aly the Borel-Cantelli lemma again. This time we assume that 0 < a < 1 is arbitrary but fixed. Although X n are indeendent, but events A n may be not indeendent, so it is not a good idea to aly the Borel-Cantelli (art 2) to A n. Instead, we aly B-C to a sub-sequence A nk where n k = k k for k = 1, 2,. Now, we show that n k+1 n k > a n k. for k large enough, so that {A nk } are indeendent. To rove this, we estimate the ga ( ) n k+1 n k + a n k = (k + 1) (k + 1) k k a n k ( ) (k + 1) (k + 1) 1 k k a k k = (k + 1) (k + 1) 1 k k a k a k (1 a) k a k = 0 = 1. = 1. as k (since a < 1). Therefore there is k 0 such that for k k 0 n k+1 n k > a n k + 1.

6 Hence {A nk } are indeendent for k k 0 and A nk = j>a n k j (1 ) = (1 ) ( k k ) a a n k 1 a n k +1 so that k k 0 A nk = as a < 1. Hence, by alying Borel-Cantelli to {A nk : k k 0 }, we conclude that A nk : i.o. = 1, so that for every a < 1, and therefore which comletes the roof. lim su lim su n a n 1 = 1. = 1