1 Martingales. Martingales. (Ω, B, P ) is a probability space.

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1 Martingales January 8, 206 Debdee Pati Martingales (Ω, B, P ) is a robability sace. Definition. (Filtration) filtration F = {F n } n 0 is a collection of increasing sub-σfields such that for m n, we have F m F n. Definition 2. (datation) sequence of random variables X = {X n } is adated if for all n, X n is F n measurable. Definition 3. (Martingales) n adated air (X n, F n ) is called sub / suer / martingale if. X n (P ) for all n 2. EX n+ F n / / = X n a.e. [P ]. Remark. a. X n is n.n.g, 2 sub-martingale, then X 2 n is sub-martingale. b. X n martingale, φ is a convex function. Then φ(x n ) is a sub-martingale. c. X n is a sub-martingale, φ is convex, non-decreasing (say x ), then φ(x n ) is a submartingale. Examle:. {ξ n } i.i.d mean 0, S n = n i= ξ i, F n = σ(ξ,..., ξ n ) = σ(s 0,..., S n ), then (S n, F n ) is a martingale. (For a sequence of random variables {X n }, F n = σ(x 0, X,..., X n ) is called a natural filtration. ) Theorem. (X n, F n ) is a martingale. φ : R R is convex and φ(x n ) for all n. Then (φ(x n ), F n ) is a sub-martingale. Proof. (Use Jensen s Inequality). Proerties:. {X n } is a martingale. Then X n is a sub-martingale. 2. {X n } is a -martingale for >. Then { X n } is a sub-martingale.

2 3. {X n } is a sub-martingale. Then {X + n } is a sub-martingale. 4. {X n } is a non-negative submartingale, >. Then {X n} is a sub-martingale. Examle:. Suose (X n, F n ) is an adated sequence and E(X n+ F n ) = a n X n. How to convert it to a martingale? et Z n = c n X n be a martingale. Then E(Z n+ F n ) = c n+ a n X n which should be equal to c n X n. Hence we should have c n+ /c n = a n for all n. Taking c 0 =, we have c n = n i= a i. Hence X n / n i= a i is a martingale. (Discussed in class. The case of E(X n+ F n ) = a n X n + b n left as an exercise). 2. Suose ([0, ), B, P ) be a robability sace and Q be another robability measure such that Q P. et F 0 = {φ, [0, )} F n = σ({[r2 n, (r + )2 n ) : 0 r 2 n }) called a dyadic filtration. Clearly σ( F n ) = B and X n (ω) = Q()/P () where is an F n -atom containing ω. Clearly, X n is F n -measurable since X n is constant on F n atoms. It is easy to check that (X n, F n ) is a martingale. (Discussed in class).. Doob s maximal inequalities Theorem 2. et (X j, F j ) 0 j n is a sub-martingale and λ R. Then. λp [max 0 j n X j > λ] max X j >λ X ndp E X n. 2. λp [min 0 j n X j λ] min X j λ X ndp E(X n X 0 ) E(X 0 ) E X n. Proof. Proof of a.) et 0 = [X 0 > λ] F 0 = [X 0 λ, X > λ] F. n = [X 0 λ,..., X n λ, X n > λ] F n. Then [max X j > λ] = n j=0 j and n λp [max X j > λ] = λp ( j ) n j=0 j=0 λdp j 2 n j=0 j X j dp = n X n dp = j j=0 max X j >λ X n dp,

3 where the enultimate inequality follows from the fact that since E(X n F j ) X j E(X n F j )dp X jdp for all F j. Hence X ndp X jdp for all F j. Proof of b.) 0 = [X 0 λ] F 0 = [X 0 > λ, X λ] F. n = [X 0 > λ,..., X n > λ, X n λ] F n n+ = [X 0 > λ,..., X n > λ] F n. Then EX 0 = n+ X 0 dp = X 0 dp j=0 j = X 0 dp + X 0 dp 0 c 0 λp ( 0 ) + X dp c 0 λp ( 0 ) + X dp + X dp { 0 } c λp ( 0 ) + X 2 dp { 0 } c. λp (min X j λ) + X n dp n+ = λp (min X j λ) + X n dp c n+ X n dp. This imlies λp [min 0 j n X j λ] min X j λ X ndp E(X n X 0 ) E(X 0 ) E X n. 2 Convergence of Martingales Goal: What conditions do we need so that an bounded martingale converges in? Definition 4. subset S is called uniformly integrable if given ɛ > 0, there exists c > 0 such that su f S f >c f dp < ɛ. 3

4 Definition 5. subset S is called bounded if su f S E f <. Examles:. ny finite -subset is uniformly integrable. 2. If S is dominated by an function, then S is uniformly integrable. 3. ny uniformly integrable subset is bounded. (Converse is not true in general: construct an bounded subset which is not uniformly integrable (Discussed in class). Theorem 3. S is uniformly integrable subset iff:. S is bounded. 2. For all ɛ > 0, there exists δ > 0 such that P () < δ, then su S f < ɛ. Proof. if art: Suose su f S E f = M <. Fix ɛ > 0. Choose δ from 2. Observe that su f S P ( f > c) su f S E f /c M/c. Choose c = M/δ with = { f > c} to get the result. only if art: Fix ɛ > 0. Choose c such that su f S f >c f < ɛ/2. Then f dp f + f < cp () + ɛ/2. f c Choose δ = ɛ/(2c) to get the result. Theorem 4. f. et S = {E C f := E(f C), Ca sub-σ-field of F}. Then S is uniformly integrable. Proof. Observe that E C f >c E C f So. is satisfied. To verify 2., note that E C f >c f >c E C f = su P ( E C f E f > c) su C C c E C f >c = E f. c Choose c large enough such that su C P ( E C f > c) is sufficiently small to have E C f >c f < ɛ/2. Observe that for any set with P () < ɛ/(2c), f f + f < cp () + ɛ/2 < ɛ. E C f c E C f >c f. 4

5 Theorem 5. f n, f n a.s. f. Then {f n } is uniformly integrable iff f and f n f. Proof. only if art: Fix ɛ > 0. By Fatou s emma and boundedness of {f n }, f. Next we show that f n f. To that end, observe that E f n f = E f n fn c f f c + E f n fn >c + E f f >c. Choose c such that P ( f = c) = 0, E f n fn >c < ɛ/3, E f f >c < ɛ/2 for all n. Since {ω : f n (ω) > c, f n (ω) f(ω) = c} { f = c}, we have f n fn c f f c 0 Choose N large enough such that for all n N, E f n fn c f f c < ɛ/3 by DCT. almost surely. Hence for large enough n, E f n f ɛ/3. if art: ssume f n f. Then {fn } is bounded and f. Fix ɛ > 0. Then f n f n f + f. Choose N such that for all n N, f n f < ɛ/2 and choose δ 0 with P () < δ 0 imlies f < ɛ/2. Then for n N, P () δ0 imlies f n < ɛ, imlying {f n } uniformly integrable. Theorem 6. f n 0, f n a.s. f, f n, f in. Then {f n } is uniformly integrable if and only if Ef n Ef. Proof. The only if art follows from Theorem 5. if art: It is enough to show that f n f (This follows from Scheff s emma). Note that E f n f = Ef n + Ef 2E min{f n, f}. Since Ef n Ef by assumtion and E min{f n, f} Ef by DCT, imlying f n f. Theorem 7. (X n, F n ) n 0 is a martingale. et F = σ( n 0 F n ) and su E X n = K. (E X n K since X n is a submartingale). Then the following are equivalent. i. K <, X n X. ii. (X n, F n ) 0 n is a martingale. 5

6 iii. K <, E X = K. iv. {X n } n 0 is uniformly integrable. Proof. (Comlete verification is left as an exercise). (i) = (iv): Since K <, X n X a.e. (Since bounded martingale converges almost surely, using Doobs s ucrossing inequality). Since X n X, from Theorem 5, {X n } n 0 is uniformly integrable. (iv) = (i): Since {X n } n 0 is uniformly integrable, {X n } n 0 is bounded imlying a.s K <. Since X n X, X n X by Theorem 5. Hence (i) (iv). (ii) = (iv): This follows directly from Theorem 4. (iv) = (ii): We will infact show (iv), (i) = (ii). X is the almost sure limit of X n and hence F n measurable for all n < and hence F measurable. lso (i) imlies X. It is enough to check that E(X F n ) = X n a.e. We already know that for m > n, X n = E(X m F n ) a.s. It is enough to show that E(X m X F n ) 0 a.e. as m. We know that X m X. Then E E[X m X F n ] EE[ X m X F n ] = E[ X m X ] 0 as m, imlying E X n E[X F n ] = 0 imlying X n = EX F n a.e. Hence we have shown that (i), (ii) and (iv) are equivalent. (iii) = (iv): Since X n is a sub-martingale, E X n K imlying E X n E X. By Scheffe s theorem X n X. By Theorem 6 { X n } and hence {X n } is uniformly integrable. Now we will show that (i) and (iv) imlies (iii). Since X n X, by Scheffe s emma and Fatou s emma E X n E X <. lso E X n K. Hence K < and E X = K. Theorem 8. (X n, F n ) 0 n N is a non-negative martingale. et >. Then max X n 0 n N X N Proof. The roof follows from the following emma and Doob s maximal inequality. emma. U, V non-negative random variables. et λ > 0, P (U > λ) (/λ) U>λ V dp. Then, for >, U V. 6

7 Proof. EU = = = = = = λ P (U > λ)dλ λ 2 V dp dλ U>λ λ 2 V (ω) U>λ (ω)dp (ω)dλ Ω Ω V (ω) λ 2 dλdp (ω) Ω 0 V (ω)u dp = Ω {E(V )} / {EU ( )q } /q {E(V )} / {EU } /q E(V U ) imlying U V if U. Otherwise, work with min{u, n}. Corollary. (X n, F n ) n 0 non-negative -bounded sub-martingale. Then X = su X n (True for bounded martingale with X = su X n ). Proof. et XN = max 0 n N X n, XN X. By MCT, E(XN ) E(X ). Since {X n } are bounded, ( ) E(XN) E X N < M, imlying X. Theorem 9. et >. {X n } is -bounded martingale or a non-negative sub-martingale. Then. {X n } is uniformly integrable. 2. X n X. Proof. Proof is left as an Exercise. Remark 2. Counter examle to show that one cannot get rid of the non-negativity in case of sub-martingale. et ([0, ), B, λ) be the measure sace and F n is a dyadic filtration. et X n = 2 n/2 [0,2 n ) 0 a.s. Check that {X n, F n } is a sub-martingale. Clearly X n 2 0 as X n 2 =. Comlete verification is left as an Exercise. 7

8 3 Examle: Two color urn model Suose (X n, F n ) adated -sequence. EX n+ F n = a n X n + b n a n 0. Find the associated martingale. Start with (W 0, B 0 ) balls with 0 W 0, B 0 and W 0 + B 0 =. t n th stage, W n white, B n black balls are available with W n + B n = n. Draw a ball at random and R 2 2 is a stochastic matrix. If you see a white ball, add R white and R 2 black balls. If you see black ball, add according to second row. X n+ is a vector which is (, 0) if white is drawn in nth stage, (0, ) if black. Z n = (W n, B n ). Since Z n is bounded for each fixed n, Z n. Observe that Z n+ = Z n + X n+r. Find the associated martingale. (Discussed in class). 4 Stoing Time F = (F n ) n 0 is a filtration. τ : Ω {0,, 2,..., } is a measurable random time. σ- field on {0,,..., } is P({0,,..., }). random time is called a stoing time if [τ = n] F n for all n {0,, 2,..., }. Examle: Time at which one starts smoking is a stoing time. However, the time at which one stos smoking is not a stoing time. [τ = n] is equivalent to [τ n] F n for all n {0,, 2,..., }. This is easy to see for discrete dime. For continuous time [τ t] F t for all t 0. [τ = t] = [τ t] n [τ t /n : n N]. In the discrete time case [τ < n] F n, [τ n ] F n F n. Theorem 0. For discrete arameter saces [τ = n] F n [τ n] F n [τ < n] F n. Definition 6. (X n, F n ) adated sequence. τ is a random time, [τ < ] = Ω. Stoing random variable X τ (ω) = X τ(ω) (ω). Theorem. X τ is B measurable. Proof. For B B(R), Xτ (B) = n=0 [X τ B] [τ = n] = n=0 [X n B] [τ = n]. Definition 7. (Stoing σ-field) F τ = { B : [τ = n] F n for all n} Clearly X τ is F τ measurable. 8

9 4. Proerties. τ σ F τ = F σ 2. τ, σ are sto times, then [τ < σ], [τ > σ], [τ = σ] are all F τ F σ -measurable. ([τ < σ] [τ = n] = [σ > n] [τ = n] F n ) 3. τ is F τ measurable. Then [τ = k] [τ = n] F n for k n. 4. τ σ, then F τ F σ. 5. (X n, F n ) 0 n N is a (sub)-martingale. τ σ is a stoing time. (X τ, X σ ) is a (sub)- martingale corresonding to (F τ, F σ ). 4.2 Doob s Ucrossing Inequality {X n } 0 n N, a < b. Define τ 0 = 0 τ = inf{n : X n a}. τ 2k+ = inf{n τ 2k : X n a} τ 2k+2 = inf{n τ 2k+ : X n b} with the convention that inf{φ} = N. Define emma 2. τ i s are sto times. U({X n } 0 n N ; [a, b]) = su{l : X τ2l a < b X τ2l } U({X n } n 0 ; [a, b]) = lim N {l : X τ2l a < b X τ2l }. Proof. τ 0, τ are sto times. ssume τ 2k is a sto time. Then {τ 2k+ = i} = i j=0 {τ 2k+ = i, τ 2k = j} {τ 2k+ = i, τ 2k = j} = {τ 2k = j} {X j+ > a} {X i > a} {X i a} F i For an adated sequence, {τ k } forms a stoing time, which imlies su{l : X τ2l a < b X τ2l } is measurable imlying U({X n } n 0 ; [a, b]) is measurable. 9

10 Theorem 2. (Doob s Ucrossing Inequality). {X n, F n } is a submartingale and a < b. Then EU({X n } 0 n N ; [a, b]) E[(X N a) + ] E[(X 0 a) + ] b a E X N + a. b a Corollary 2. n -bounded martingale converges almost surely. 5 Problem branching rocess is defined as follows: We start with one member, namely the oulation size is Z 0 =. et ξi n, i =, 2,..., n denote the number of children of ith individual in the nth generation. ssume ξi n are indeendent with common mean µ > 0. et Z n denote the oulation size in the nth generation. et k = P (ξi n = k), k 0, µ = E(ξi n). Z n+ = Z n i= ξn i. It is easy to see that X n = Z n /µ n is a martingale.. Does {X n } has a limit? Since X n is non-negative, {X n } converges almost everywhere to an integrable random variable X. 2. If µ <, Z n = 0 almost surely for large n. Z n = X n µ n a.s.. Since X n X and µ n a.s. 0, Z n 0. There exists a P -null set N such that for all ω / N, Z n (ω) 0. Given ɛ = /2, there exists N 0 (ω) such that Z n (ω) < ɛ for all n N 0 (ω) imlying Z n (ω) = 0 for all n N 0 (ω). This imlies Z n converges to 0 with robability for all large n. 3. If µ <, X n = 0 almost surely for large n. n= P (X n > 0) = n= P (Z n > 0) = n= P (Z n ) n= E(Z n) = n= µn <. Hence by Borel Cantelli emma P (lim su n ) = 0 which means P ( n= k n k ) = 0 imlying P ( n occurs infinitely often) = 0 imlying P (X n = 0 for all large n) =. Hence X 0 = 0 eventually with robability. 4. If µ = and P (ξ > ) > 0, then Z n 0 a.s. Z n non-negative martingale, hence Z n Z a.e and Z. It is enough to show that P (Z = k) = 0 for all k, or in other words P (Z n = k eventually) = 0. Observe that P (Z n = k eventually) P (one of ξ n,..., ξn k, eventually). It is enough to show that P (ξ n,..., ξn k >, infinitely often) =. To that end, note that n= P (ξn i > ) k =. The result follows from second Borel Cantelli emma. 0