Fundamental Inequalities, Convergence and the Optional Stopping Theorem for Continuous-Time Martingales
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1 Fundamental Inequalities, Convergence and the Optional Stopping Theorem for Continuous-Time Martingales Prakash Balachandran Department of Mathematics Duke University April 2, Review of Discrete-Time Martingales We assume the reader is familiar with the basic results from probability theory, particularly those concerning discrete-time martingales Here, we briefly review the major theorems and definitions For more details, see Patrick Billingsley s Probability and Measure or Richard Durrett s Probability: Theory and Examples Throughout, Ω, F, P ) is a probability space, and BX) denotes the Borel σ-algebra generated by a topological space X Definition 1 A filtration is a collection of σ-algebras, F n n=1 such that F n F n+1 Definition 2 Let F n n=1 be a filtration A stopping time T is a random variable T : Ω, F) N, BN)) such that ω Ω : T ω) = n F n for all n < Definition 3 Let I be an index set, and X i i I be a collection of random variables The collection X i i I is said to be uniformly integrable if lim M sup E [ X i : X i > M] i I ) < Definition 4 If f, g are random variables, then f g is the random variable defined by f g)ω) = minfω), gω) Definition 5 If T is a stopping time and X is a random variable, then X T is the random variable defined by X T ω) = X T ω) ω) Definition 6 A collection of random variables X n n=0 discrete-time submartingale) if is called a discrete-time martingale resp 1
2 1 There exists a filtration F n n=0 2 X n is F n -adapted; ie X n is F n -measurable 3 X n L 1 Ω, F, P ), n 0 4 E[X n+1 F n ] = X n resp E[X n+1 F n ] X n ), n 0 Now we need to review the notion of an upcrossing Let [α, β] be an interval α < β), and let X 1,, X n be random variables Inductively define τ 1, τ 2, by: τ 1 ω) is the smallest j such that 1 j n and X j ω) α, and is n if there is no such j τ k ω) for even k is the smallest j such that τ k 1 ω) < j n and X j ω) β and is n if there is no such j τ k ω) for odd k exceeding 1 is the smallest j such that τ k 1 ω) < j n and X j ω) α and is n if there is no such j The number U n ω) of upcrossings of [α, β] by X 1,, X n is the largest j such that X τ2j 1 α < β X τ2j Now, recall the basic results from discrete time martingales: Theorem 1 Doob s Inequality): If X n n=0 is a submartingale and λ > 0, then, P ω Ω : max 0 m n X+ mω) λ E[X+ n ] λ Theorem 2 The Upcrossing Inequality): If X m m=0 is a submartingale, then the number of upcrossings U n on [a, b] satisfies E[U n ] E[X n a) + ] E[X 0 a) + ] b a Theorem 3 Doob s Maximal Inequality): If X n n=0 is a submartingale, then for 1 < p <, [ ) p ] ) p p E max 0 m n X+ m E[X n + ) p ] p 1 Theorem 4 The Martingale Convergence Theorem): If X n n=0 is a submartingale such that sup E[X+ n ] < for all n, then as n, X n converges as to a limit X with E[ X ] < Theorem 5 The Optional Stopping Theorem): If L M are stopping times and Y M n n=0 is a uniformly integrable submartingale, then E[Y L ] E[Y M ] and Y L E[Y M F L ] where F L = A F : A ω Ω : Lω) k F k, 1 k < 2
3 2 Continuous-Time Martingales We would like to extend the definitions and theorems in 1) to similar definitions and theorems involving continuous time martingales Definition 7 A filtration is a collection of σ-algebras, F t t 0 such that F s F t for s < t We recall the interpretation of σ-algebras as containing information: F t is the information available to the adapted process X t t 0 up to and including) time t Thus, it s reasonable to interpret ) F t = σ F s as the σ-algebra of events strictly prior to t > 0 and s<t F t+ = ɛ>0 F t+ɛ as the σ field of events immediately after t 0 We take the definition F 0 = F 0 and say that the filtration F t t 0 is right- resp left-) continuous if F t = F t+ resp F t = F t ) holds for every t 0 Definition 8 Let F t t 0 be a filtration A stopping time T is a random variable T : Ω, F) [0, ], B[0, ])) such that ω Ω : T ω) t F t for every t 0 A random variable T, T : Ω, F) [0, ], B[0, ])) is called an optional time of the filtration if ω Ω : T ω) < t F t for every t 0 We have the following immediate result: Proposition 1 Every stopping time is optional, and the two concepts coincide if the filtration is rightcontinuous Proof: If T is a stopping time, then [ T t n] 1 Ft 1 F t for all n 1, n Z So: n [ [T < t] = T t 1 ] F t n n=1 Now, suppose that T is an optional time, and F t t 0 is a right continuous filtration Fix ɛ > 0, and notice that for η < ɛ rational, F t+η F t+ɛ This implies [T < t + η] F t+η F t+ɛ since T is an optional time, so that [T t] = [T < t + η] F t+ɛ 0<η<ɛ rational 3
4 Since ɛ > 0 was arbitrary, this implies that [T t] F t+ɛ for every ɛ > 0, so that [T t] ɛ>0 F t+ɛ = F t+ = F t since the filtration is right continuous Definition 9 Let I be an index set, and X i i I be a collection of random variables The collection X i i I is said to be uniformly integrable if lim M sup E [ X i : X i > M] i I ) < Definition 10 If f, g are random variables, then f g is the random variable defined by f g)ω) = minfω), gω) Definition 11 If T is a stopping time and X t t 0 is a stochastic process, then X T variable defined by X T ω) = X T ω) ω) is the random Definition 12 Let T be a stopping time of the filtration F t t 0 The σ-algebra F T of events determined prior to the stopping time T consists of those events A F for which A ω Ω : T ω) t F t for every t 0 Clearly, if X t t 0 is adapted to F t t 0, then X T is F T measurable Definition 13 Let T be an optional time of the filtration F t t 0 The σ-field F T + of events determined immediately after the optional time T consists of those events A F for which A T t F t+ for every t 0 Definition 14 A collection of random variables X n n=0 is called a continuous-time martingale resp continuous-time submartingale) if 1 There exists a filtration F t t 0 2 X t is F t -adapted; ie X t is F t -measurable 3 X t L 1 Ω, F, P ) for each t 0 4 E[X s F t ] = X t resp E[X s F t ] X t ) for 0 t < s < Now we need to define the notion of an upcrossing Let X t t 0 be a real-valued stochastic process Consider two numbers α < β, and a finite subset F of [0, ) We define the number of upcrossings U F α, β; Xω)) of the interval [α, β] by the restricted sample path X t ω); t F as follows Set τ 1 ω) = mint F ; X t ω) α 4
5 and define recursively for j = 1, 2, σ j ω) = mint F ; t τ j ω), X t ω) > β τ j+1 ω) = mint F ; t σ j ω), X t ω) < α We define the minimum of the empty set to be and let U F α, β; Xω)) to be the largest number j such that σ j ω) < If I [0, ) is not necessarily finite, we define U I α, β; Xω)) = supu F α, β; Xω)); F I, F finite Theorem 6 Let X t t 0 be a submartingale wrt F t t 0, whose every path is right continuous Let [σ, τ] be a subinterval of [0, ) and let α < β, λ > 0 be real numbers We have the following results: 1 Doob s Inequality): P ω Ω : sup X t ω) λ σ t τ E[X+ τ ] λ 2 The Upcrossing Inequality): 3 Doob s Maximal Inequality): [ E Proof: E[U [σ,τ] α, β, Xω))] E [X+ τ ] + α β α ) p ] sup X t σ t τ ) p p E[X p t p 1 ] for p > 1, provided X t 0 as for every t 0 and X τ L p Ω, F, P ) 1) Consider the enumeration σ = t 0 < t 1 < of the numbers in σ, τ [σ, τ] Q) Define F j = j i=0 t j τ Clearly, F j j=0 is an increasing sequence of finite sets such that Let A j = F = F j = σ, τ [σ, τ] Q) j=0 ω Ω : sup t Fj X t ω) λ Then, F j F j+1 sup X t ω) sup X t ω) A j A j+1 t F j t F j+1 5
6 Using continuity from below on A = note that A j are measurable) we have ω Ω : sup X t ω) λ t F P A ) = lim n P A n) By the discrete-time Doob inequality: P A j E[X+ max F ] j λ = E[X+ τ ] λ where max F j denotes the largest element in the set F j, which is τ by construction of F j j=0, so that = j=0 P A ) = lim P A j ) E[X+ τ ] j λ A j Claim 1 sup t F X t ω) = sup t [σ,τ] X t ω) Proof of Claim 1: Since F [σ, τ], sup t F X t ω) sup t [σ,τ] X t ω) On the other hand, since X t t 0 is right continuous, X t ω) = lim j X qj ω) where q j j=1 is a sequence in F such that lim j q j = t, q j t So, since X qj ω) sup t F X t ω), Thus, sup t F X t ω) = sup t [σ,τ] X t ω) X t ω) sup X t ω) sup X t ω) sup X t ω) t F t [σ,τ] t F By Claim 1, we have that A = ω Ω : sup X t ω) λ t F = ω Ω : sup X t ω) λ t [σ,τ] so that P ω Ω : sup X t ω) λ σ t τ = P A ) E[X+ τ ] λ 2) Let F j j=0 be as in 1) Claim 2 U [σ,τ] α, β; Xω)) = sup j 0 U Fj α, β; Xω)) 6
7 Proof of Claim 2: Since each member of F j j=0 is a finite subset of [σ, τ], we have by the definition of U [σ,τ], U [σ,τ] α, β; Xω)) supu Fj α, β; Xω)) j 0 On the other hand, if F [σ, τ] is finite, and x F is its largest element, there exists F k such that if t is the largest element of F k x t Since U [σ,t] is increasing in t, we have U F U Fk so that U [σ,τ] α, β; Xω)) sup j 0 U Fj α, β; Xω)) Thus, U [σ,τ] α, β; Xω)) = sup j 0 U Fj α, β; Xω)) The reason we needed to prove this was that in general U [σ,τ] is not the supremum over a countable collection of functions, since there are uncountably many F [σ, τ] such that F is finite; thus, U [σ,τ] need not in general be measurable But, by Claim 2, U Fj U [σ,τ] since F j F j+1 Thus, it is measurable and since E[U Fj α, β, Xω))] ] E [X + max + α Fj β α = E [X+ τ ] + α β α for the finite sets F j by the discrete-time upcrossing inequality, we must have E[U [σ,τ] α, β, Xω))] E [X+ τ ] + α β α by the monotone convergence theorem In particular, this shows that U [σ,τ] L 1 Ω, F, P ) since X t t 0 a submartingale implies E[X + τ ] E[ X τ ] <, and U [σ,τ] 0 implies U [σ,τ] = U [σ,τ] Thus, U [σ,τ] is finite ae 3) Let F j j=0 and F be as in 1) Claim 3 sup σ t τ X t = sup j 0 sup t Fj X t ) Proof of Claim 3: Since F j [σ, τ] for all j 0, we must have sup X t sup X t, j 0 sup t F j σ t τ j 0 sup t F j X t ) sup X t σ t τ Conversely, consider X t Then, since X t t 0 is right continuous, X t ω) = lim j X qj ω) where q j j=1 is a sequence in F such that lim j q j = t and q j t for all j 1 7
8 Now, if G j is the set in F k k=0 such that q j G j, then: ) X qj sup t G j X t sup k 0 sup t F k X t sup X t sup σ t τ j 0 Thus, sup σ t τ X t = sup j 0 sup t Fj X t ) j 1 X t = lim j X qj sup j 0 sup t F j X t ) sup t F j X t ) Since Y j = sup t Fj X t is increasing recall F j F j+1 ), we have from Claim 3 that Y j sup σ t τ X t Thus, sup σ t τ X t is measurable, since obviously, each Y j is measurable So, the discrete-time Doob maximal inequality: ) p ] E[Y p j [max ] = E X t t F j implies p p 1 ) p ) p p E[X + max F j ) p ] = E[X τ ) p ] p 1 [ ) p ] ) p p E max X t E[X τ ) p ] σ t τ p 1 by the monotone convergence theorem, since X t 0 as for every t 0 and X τ L p Ω, F, P ) 8
9 Theorem 7 The Martingale Convergence Theorem): Let X t t 0 be a right continuous submartingale with respect to F t t 0 If C := sup t 0 E[X + t ] < then X ω) := lim t X t ω) exists for ae ω Ω and X L 1 Ω, F, P ) Proof: From Theorem 62), we have for any n 1 and real numbers α < β: E[U [0,n] α, β, Xω))] E [X+ n ] + α β α By hypothesis, E [X + n ] sup t 0 E[X + t ] = C <, so that the above is bounded for all n 1 So, since U [0,n] U [0,n+1], the pointwise limit lim n U [0,n] = U [0, ) exists, so that by the monotone convergence theorem and the above, E[ U [0, ) α, β; Xω)) ] = E[U [0, ) α, β, Xω))] = lim n E[U [0,n]α, β; Xω))] C + α β α < since U [0,n] 0, n 1 implies U [0, ) 0 Thus, U [0, ) α, β; Xω)) L 1 Ω, F, P ), and so U [0, ) α, β; Xω)) is finite ae Now, the events A α,β = ω Ω : U [0, ) α, β; Xω)) =, < α < β < must have measure zero since α, β in the above were arbitrary, and U [0, ) α, β; Xω)) is finite ae Thus, the event A = α<β,α,β Q A α,β must also have measure zero Claim 4 Let X and X denote the limits superior and inferior of X t t 0 If E = ω Ω : X ω) > X ω), then E is a measurable subset of A Proof of Claim 4: If ω E, then for some α, β > 0 So: and for all t 0 lim inf t β < inf X tω) < α < β < lim sup X t ω) t sup t 0 s t X s ω) β < sup X s ω) s t sup inf X sω) < α inf X sω) < α t 0 s t s t 9
10 Now, in the notation of the above discussion involving upcrossings in the continuous-time case, let τ 1 ω) be given With t = τ 1 ω), β < sup s τ1 ω) X s ω) implies that we can choose s 1 > τ 1 ω) such that 0 < sup X s ω) X s1 ω) < s τ 1 ω) sup X s ω) β β < X s1 ω) < s>τ 1 ω) sup X s ω) s τ 1 ω) So, either σ 1 ω) = s 1 or σ 1 ω) < s 1 Thus, σ 1 ω) s 1, so that there is at least one upcrossing on [0, s 1 ] Now, take t = s 1 Then, inf s s1 X s ω) < α implies that we can choose s 2 > s 1 such that 0 < X s2 ω) inf s s 1 X s ω) < α inf s s 1 X s ω) inf s s 1 X s ω) < X s2 ω) < α So, either τ 2 ω) = s 2 or τ 2 ω) < s 2 Thus, τ 2 ω) s 2, so that there is at least one downcrossing on [s 1, s 2 ] We can repeat the above procedure with t = τ 2 ω) and so on to obtain a sequence τ 1 ω) = s 0 < s 1 < s 2 < such that there is at least one upcrossing on [s 2k, s 2k+1 ] for k = 0, 1, Taking H j = j k=0 s k, we have that H j < j > 0 and U Hj α, β; Xω)) as j Since U [0, ) α, β; Xω)) is by definition the supremum of all such functions over all finite subsets of [0, ), this implies that U [0, ) α, β; Xω)) =, so that ω A α,β A Since ω E was arbitrary, E A Clearly, by the definition of E, it is measurable, so that the result follows Since a measurable subset of a set of measure zero has measure zero, P E) = 0 by Claim 4 Thus, for almost every ω Ω, X ω) = lim t X t ω) exists Since X t t 0 is a submartingale, E[ X t ] = 2E[X t + ] E[X t] 2C E[X 0 ] <, so that sup t 0 E[X t + ] < is equivalent to the stronger assumption sup t 0 E[ X t ] < Thus, by Fatou s lemma: E[ X ] lim inf E[ X n ] sup E[ X t ] < X L 1 Ω, F, P ) n t 0 10
11 Theorem 8 The Optional Stopping Theorem): Let X t t 0 be a right continuous submartingale with respect to F t t 0 with last element X, and let S T be two optional times of the filtration F t We have E[X T F S+ ] X S, as If S is a stopping time, then F S can replace F S+ above In particular, E[X T ] E[X 0 ], and for a martingale with last element, we have E[X T ] = E[X 0 ] Proof: Consider the sequence S n ω) = Sω) Sω) = + k k 1 2 n 2 n Sω) < k 2 n and T n ω) = T ω) T ω) = + k k 1 2 n 2 n T ω) < k 2 n defined for n 1 Clearly, these are random variables taking values in [0, ] Claim 5 S n n=1, T n n=1 are stopping times, and S n T n n Furthermore, each of the sequences are decreasing, and lim n S n = S, lim n T n = T Proof of Claim 5: Let n 1, n Z be fixed, and suppose t 0 is given Then, ω Ω : S n ω) t = ω Ω : k 1 2 n Sω) < k k 2 n 2 n t = u k=1 ω Ω : k 1 2 n Sω) < k 2 n = ω Ω : 0 Sω) < u 2 n F u F 2 n t where u denotes the greatest integer such that u t2 n Since t and n were arbitrary, this shows that S n n=1 are stopping times Clearly, the same argument holds for T n n=1, so that T n n=1 are stopping times Now, suppose that S n ω) > T n ω) for some ω Ω; WLOG, WMA that neither takes the value Then, if S n ω) = k 2 n and T n ω) = k 2 n we must have k 2 n > k 2 n, or in other words k > k 11
12 But, since and this implies that ω ω ω Ω : k 1 2 n Sω) < k 2 n ω Ω : k 1 2 n T ω) < k 2 n k 1 2 n T ω) < k 2 n k 1 2 n Sω) < k 2 n which implies that T ω) < Sω) which is a contradiction Thus, S n ω) T n ω), so that since ω Ω was arbitrary, S n ω) T n ω) ω Ω Thus, S n T n, so that since n 1 was arbitrary, this holds n 1 It s clear by the construction of S n and T n that the sequences are decreasing To show the final statement, note that there s nothing to show for T ω) = For T ω) <, we have by construction of T n that k 1 2 n T ω) < k 2 n = T nω) T n ω) T ω) 1 2 n lim n T nω) = T ω) Clearly, the same holds for S n So, by Claim 5, S n and T n are stopping times with S n T n Furthermore, each S n and T n take on countably many values, so that by the discrete-time optional sampling theorem, X Sn E[X Tn F Sn ] X Sn dp X Tn dp for every A F Sn Now, it s not too hard to show Claim 6 F S+ = n=1 F S n F Sn A A Thus, for every A F S+ X Sn dp A A X Tn dp 12
13 Now assume that S is a stopping time By construction of S n, S S n for all n Let A F S Then A S t F t t 0 and hence A S n t = A S n t S t F t t 0, n 1 Thus, F S F Sn for each n 1, so that F S F S+ by Claim 6 So, X Sn dp X Tn dp for every A F S A A Claim 7 Let F n n=1 be a decreasing sequence of sub σ-fields of F, and let X n be a backwards submartingale wrt F n [that is, E[ X n ] <, X n is F n measurable, and E[X n F n+1 ] X n+1 for ever n 1] Then, l := lim n EX n ) > implies that X n n=1 is uniformly integrable By construction, X Sn is a backward submartingale wrt F Sn Thus, E[X Sn ] is decreasing, and bounded below by E[X ] Thus, by Claim 7, X Sn is uniformly integrable Similarly, the same is true of T n Since the process X t t 0 is right-continuous, Claim 5 implies lim X T n n ω) = X T ω), lim X S n n ω) = X S ω) ae ω Ω [Note: we re assuming that S, T < ae] By uniform integrability, it follows that X T and X S are integrable and that the above ae convergence is actually L 1 convergence Thus, A X SdP A X T dp for every A F S+, and A F S when S is a stopping time 13
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