Notes 18 : Optional Sampling Theorem

Transcription

1 Notes 18 : Optional Sampling Theorem Math : Theory of Probability Lecturer: Sebastien Roch References: [Wil91, Chapter 14], [Dur10, Section 5.7]. Recall: DEF 18.1 (Uniform Integrability) A collection C of RVs on (Ω, F, P) is uniformly integrable (UI) if: ε > 0, K > + s.t. E[ X ; X > K] < ε, X C. THM 18.2 (Necessary and Sufficient Condition for L 1 Convergence) Let {X n } L 1 and X L 1. Then X n X in L 1 if and only if the following two conditions hold: X n X in probability {X n } is UI THM 18.3 (Convergence of UI MGs) Let {M n } be UI MG. Then a.s. and in L 1. Moreover, M n M F = σ ( n F n ), M n = E[M F n ], n. THM 18.4 (Lévy s upward theorem) Let Z L 1 and define M n = E[Z F n ]. Then {M n } is a UI MG and a.s. and in L 1. M n M = E[Z F ], 1

2 Lecture 18: Optional Sampling Theorem 2 1 Optional Sampling Theorem 1.1 Review: Stopping times Recall: DEF 18.5 A random variable T : Ω Z + {0, 1,..., + } is called a stopping time if {T = n} F n, n Z +. EX 18.6 Let {A n } be an adapted process and B B. Then is a stopping time. T = inf{n 0 : A n B}, LEM 18.7 (Stopping Time Lemma) Let {M n } be a MG and T be a stopping time. Then the stopped process {M T n } is a MG and in particular E[M T n ] = E[M 0 ]. THM 18.8 Let {M n } be a MG and T be a stopping time. Then M T L 1 and if any of the following conditions holds: 1. T is bounded E[M T ] = E[M 0 ]. 2. {M n } is bounded and T is a.s. finite 3. E[T ] < + and {M n } has bounded increments 4. {M n } is UI. (This one is new. The proof follows from the Optional Sampling Theorem below.) Proof: From the previous theorem, we have ( ) E[M T n M 0 ] = Take n = N in ( ) where T N a.s. 2. Take n to + and use (DOM).

3 Lecture 18: Optional Sampling Theorem 3 3. Note that M T n M 0 = (M i M i 1 ) KT, i T n where M n M n 1 K a.s. Use (DOM). DEF 18.9 (F T ) Let T be a stopping time. Denote by F T the set of all events F such that n Z + F {T = n} F n. 1.2 More on the σ-field F T The following two lemmas help clarify the definition of F T : LEM F T = F n if T n, F T = F if T and F T F for any T. Proof: In the first case, note F {T = k} is empty if k n and is F if k = n. So if F F T then F = F {T = n} F n and if F F n then F = F {T = n} F n. Moreover F n so we have proved both inclusions. This works also for n =. For the third claim note F = k Z+ F {T = n} F. LEM If {X n } is adapted and T is a stopping time then X T F T (where we assume that X F, e.g., X = lim inf n X n ). Proof: For B B {X T B} {T = n} = {X n B} {T = n} F n. LEM If S, T are stopping times, then S T is a stopping time and F S T F T. Proof: We first show that S T is a stopping time. Note that {S T = k} = [{S = k} {T k}] [{S k} {T = k}] F k, since all event above are in F k by the fact that S and T are themselves stopping times. For the second claim, let F F S T. Note that F {T = n} = k n [(F {S T = k}) {T = n}] F n. Indeed, the expression in parenthesis is in F k F n and {T = n} F n.

4 Lecture 18: Optional Sampling Theorem Optional Sampling Theorem (OST) We show that the MG property extends to stopping times under UI MGs. THM (Optional Sampling Theorem) If {M n } is a UI MG and S, T are stopping times with S T a.s. then E M T < + and E[M T F S ] = M S. Proof: Since {M n } is UI, M L 1 s.t. M n M a.s. and in L 1. We prove a more general claim: LEM E[M F T ] = M T. Indeed, we then get the theorem by (TOWER) (and (JENSEN) for the integrability claim). Proof:(of the lemma) We divide M = M + M X Y into positive and negative parts and write M n = E[M F n ] = E[X F n ] E[Y F n ] X n Y n, by linearity. We show that E[X F T ] = X T. The same argument holds for {Y n }, which then implies E[M F T ] = X T Y T = M T, as claimed. We have of course that X 0, and hence X n = E[X F n ] 0 n. Let F F T. Then E[X ; F {T = }] = E[X T ; F {T = }], since X n = E[X F n ] X a.s. by Lévy s Upward Theorem. Therefore it suffices to show E[X ; F {T < + }] = E[X T ; F {T < + }]. In fact, by (MON), it suffices to show i 0 E[X ; F {T = i}] = i 0 But note that F {T = i} F i so that E[X T ; F {T = i}]. E[X T ; F {T = i}] = E[X i ; F {T = i}] = E[X ; F {T = i}], since X i = E[X F i ]. That concludes the proof of the stronger claim.

5 Lecture 18: Optional Sampling Theorem 5 2 Wald s identities Often additional properties of T hold (typically E[T ] < + ), which can be taken advantage of by considering instead the MG {M T n } in the OST above (noticing of course that M T S = M S and M T T = M T whenever S T a.s.). In that context, the following is useful. LEM Suppose {M n } is a MG such that E[ M n+1 M n F n ] B a.s. for all n. Suppose T is a stopping time with E[T ] < +. Then the stoppped MG {M T n } is UI. Proof: Assume WLOG that M 0 = 0, to simplify. Observe first that M T n M m+1 M m 1 {T >m}, n. Taking expectations on the RHS (which does not depend on n), we get E [ M m+1 M m 1 {T >m} ] = = = B E [ M m+1 M m 1 {T >m} ] E [ E [ M m+1 M m 1 {T >m} F m ]] E [ E [ M m+1 M m F m ] 1 {T >m} ] E [ B1 {T >m} ] P [T > m] B E[T ] < +, where we used that {T > m} F m. As an application, we recover Wald s first identity. For X 1, X 2,... R, let S n = n i=1 X i. THM (Wald s first identity) Let X 1, X 2,... L 1 be i.i.d. with µ = E[X 1 ] and let T L 1 be a stopping time. Then E[S T ] = µe[t ].

6 Lecture 18: Optional Sampling Theorem 6 Proof: Recall that M n = S n nµ is a MG. By LEM and the assumption T L 1, the MG {M T n } is UI. Indeed E [ M n+1 M n F n ] = E [ X n+1 µ F n ] µ + E X 1 B < +, by the triangle inequality and the Role of independence lemma. Apply THM 18.8 to {M T n }. We also recall Wald s second identity. We give a MG-based proof (but argue about convergence directly rather than using THM 18.8). THM (Wald s second identity) Let X 1, X 2,... L 2 be i.i.d. with E[X 1 ] = 0 and σ 2 = Var[X 1 ] and let T L 1 be a stopping time. Then E[S 2 T ] = σ 2 E[T ]. Proof: Recall that M n = S 2 n nσ 2 is a MG. Hence so is M T n and 0 = E[M T n ] = E[S 2 T n (T n)σ 2 ] = E[S 2 T n] σ 2 E[T n]. (1) We have that E[T n] E[T ] as n + by (MON). To argue about the convergence of E[ST 2 n ] we note that, by the assumption E[X 1 ] = 0, it follows that {S n } is a MG and hence so is {S T n }. The latter is bounded in L 2 since, by (1), we have E[S 2 T n] = σ 2 E[T n] σ 2 E[T ] < +, for all n. Hence S T n converges a.s. and in L 2 to S T (since T < + a.s. by assumption). Convergence in L 2 also implies convergence of the second moment. Indeed, by the triangle inequality, Hence, S T n 2 S T 2 S T n S T = E[S 2 T n] σ 2 E[T n] E[S 2 T ] σ 2 E[T ], which concludes the proof. To establish E[T ] < +, the following lemma can be used. LEM (Waiting for the inevitable) Let T be a stopping time. Assume there is N Z + and ε > 0 such that for every n P[T n + N F n ] > ε a.s. then E[T ] < +.

7 Lecture 18: Optional Sampling Theorem 7 Proof: For any integer m 1, P[T > mn T > (m 1)N] 1 ε, by assumption. (Indeed, by definition of Z = P[T > n + N F n ] 1 ε with n = (m 1)N, we have for F = {T > (m 1)N} F n P[T > mn] = E[1{T > n + N}; F ] = E[Z; F ] (1 ε)p[f ], and apply the definition of the conditional probability.) By the multiplication rule (i.e., the undergraduate rule P[A 1 A n ] = n i=1 P[A i A 1 A i 1 ]) and the monotonicity of the events {T > mn}, we have P[T > mn] (1 ε) m. We conclude using E[T ] = k 1 P[T k]. 3 Application I: Simple RW DEF Simple RW on Z is the process {S n } n 0 with S 0 = 0 and S n = k n X k where the X k s are iid in { 1, +1} s.t. P[X 1 = 1] = 1/2. THM Let {S n } as above. Let a < 0 < b. Define T x = inf{n 0 : S n = x} and T = T a T b. Then we have T < + a.s. P[T a < T b ] = E[T ] = ab b b a T a < + a.s. but E[T a ] = + Proof: 1) Apply the Waiting for the inevitable lemma with N = b a and ε = (1/2) b a (corresponding to moving right b a times in a row which takes you to b, no matter where you are within the {a,..., b} interval). That shows E[T ] < +, from which the claim holds.

8 Lecture 18: Optional Sampling Theorem 8 2) By Wald s first identity, E[S T ] = 0 or a P[S T = a] + b P[S T = b] = 0, that is (taking b in the second expression) P[T a < T b ] = b b a and P[T a < + ] P[T a < T b ] 1. 3) Wald s second identity says that E[S 2 T ] = E[T ] (by σ2 = 1). Also so that E[T ] = ab. E[S 2 T ] = b b a a2 + a b a b2 = ab, 4) Taking b + above shows that E[T a ] = + by monotone convergence. (Note that this case shows that the L 1 condition on the stopping time is necessary in Wald s second identity.) 4 Application II: Biased RW DEF Biased simple RW on Z with parameter 1/2 < p < 1 is the process {S n } n 0 with S 0 = 0 and S n = k n X k where the X k s are iid in { 1, +1} s.t. P[X 1 = 1] = p. Let q = 1 p. Let φ(x) = (q/p) x and ψ n (x) = x (p q)n. THM Let {S n } as above. Let a < 0 < b. Define T x = inf{n 0 : S n = x} and T = T a T b. Then we have T < + a.s. P[T a < T b ] = φ(0) φ(b) φ(a) φ(b) P[T a < + ] = 1/φ(a) < 1 and P[T b = + ] = 0 E[T b ] = b 2p 1

9 Lecture 18: Optional Sampling Theorem 9 Proof: There are two MGs here: E[φ(S n ) F n 1 ] = p(q/p) S n q(q/p) S n 1 1 = φ(s n 1 ), (noting that φ(s n ) (p/q) n a.s.) and E[ψ n (S n ) F n 1 ] = p[s n 1 +1 (p q)(n)]+q[s n 1 1 (p q)(n)] = ψ n 1 (S n 1 ), (noting that ψ n (S n ) (1 + p)n a.s.) 1) Follows by the same argument as in the unbiased case. 2) Now note that {φ(s T n )} is a bounded MG and, therefore, by THM 18.8, we get φ(0) = E[φ(S T )] = P[T a < T b ]φ(a) + P[T a > T b ]φ(b), or P[T a < T b ] = φ(b) φ(0) φ(b) φ(a) (where we used 1)). 3) By 2), taking b +, by monotonicity P[T a < + ] = 1 T a = + with positive probability. Similarly take a. φ(a) < 1 so 4) By LEM 18.7 applied to {Ψ n (S n )}, 0 = E[S Tb n (p q)(t b n)]. (We cannot use Wald s first identity directly because it is not immediately clear whether T b is integrable.) By (MON) and the fact that T b < + a.s. from 3), E[T b n] E[T b ]. Finally, inf n S n 0 a.s. and for x 0, P[ inf n S n x] = P[T x < + ] = ( ) q x, p so that E[ inf n S n ] = x 1 P[ inf t S t x] < +. Hence, we can use (DOM) with S Tb n max{b, inf n S n } to deduce that E[T b ] = E[S T b ] p q = b 2p 1.

10 Lecture 18: Optional Sampling Theorem 10 References [Dur10] Rick Durrett. Probability: theory and examples. Cambridge Series in Statistical and Probabilistic Mathematics. Cambridge University Press, Cambridge, [Wil91] David Williams. Probability with martingales. Cambridge Mathematical Textbooks. Cambridge University Press, Cambridge, 1991.