Lecture 3: Randomly Stopped Processes

Transcription

1 STAT 498: Selected Topcs Stochastc Processes Wter 2018 Lecture 3: Radomly Stopped Processes Lecturer: Fag Ha Ja 29 Dsclamer: These otes have ot bee subjected to the usual scruty reserved for formal publcatos. They may be dstrbuted outsde ths class oly wth the permsso of the Lecturer. Ths chapter s, aga, a rearragemet of results PG Wald s Equatos For troducg Wald s marvelous research o radomly stopped processes, we have to frst formally troduce the cocept of stoppg tme. Defto 1 Stoppg tme. Let F := σx 1,..., X represet the sgma-feld geerated by the jot dstrbuto of X 1,..., X or other words, all formato or varato geerated by X 1,..., X. We say a radom varable T takg values 1, 2,... to be a stoppg tme adapted to F f for every = 1, 2,..., T = F. Wth the defto of stoppg tme md, we are ow ready to preset Wald s equato, a corerstoe of the theory of sequetal aalyss. Theorem 2. Let X be a sequece of..d. radom varables adapted to F := σx 1,..., X, wth EX 1 = µ, µ <. Let T be a stoppg tme adapted to F. Set S = X. The Moreover, f EX 1 = 0 ad EX 2 1 <, the Wald s frst equato ES T = µet, wheever ET <. Wald s secod equato ES 2 T = EX 2 1 ET, wheever ET <. Proof. 1 We prove the frst equalty. By workg o the sequece X µ stead, we may assume µ = 0. Let X = X + X, wth the coveto that X + = X1X 0 ad X = X1X < 0. The EX 1 = 0 mples that EX + 1 = EX 1 <. Notcg that 1T = 1 1T 1 F 1, we have X s depedet of 1T, whch yelds T E X + = E X + 1T = EX + 1T = EX + P T = EX+ E 1T = EX 1 + ET. Smlarly we obta E T X = EX1 ET. Lettg go to fty ad usg the mootoe covergece theorem, we further obta T T E X + = ET EX 1 + ad E X = ET EX1. 3-1

2 3-2 Lecture 3: Radomly Stopped Processes Accordgly, T Wald s secod equato ES T = E X + T X = ET EX+ 1 X 1 = ET EX 1. 2 We the move o to prove the secod equalty. Realzg that S T = X 1T, we have We hece have ES 2 T = EX 2 1T = EX 2 P T = EX1 2 ET. lm ES2 T = EX1 2 ET < provded EX 2 1 ad ET s smaller tha fty. Some stadard argumets yeld lm ES 2 T = ES2 T ad hece fsh the proof. Remark 3. Do we really eed..d.-ess, or just some frst or up to secod order statoary requremet? Oe terestg observato stemmg from Wald s equato s the followg decouplg theorem, whch opes a gate to a ew set of techques that wll be the focus of the ext few sectos. Theorem 4. Let X be a sequece of..d. r.v. s wth EX 1 = µ wth µ <. Let T be a stoppg tme adapted to F. Let X be a depedet copy of X ad also depedet of T. Set S = X 1 + +X ad S = X X. The ES T = E S T, wheever ET <. Moreover, f EX 1 = 0 ad EX 2 1 <, the ES 2 T = E S 2 T, wheever ET <. Remark 5. Of ote, S T ad S T could be completely dfferet. For example, cosder the smplest radom walk wth P X = 1 = P X = 1 = 1/2 ad T := f : S = a or S = b for some postve tegers a, b. We the have S T must be ether a or b, whle S T s ot ecessarly to be so. Before dvg to more detals aroud S T ad ultmately T tself, t s better to frst uderstad the requremet ET <. The followg theorem gves such a aalyss for the frst passage crossg tme. Theorem 6. Let X be..d. wth P X 1 = 0 < 1. Defe T a,b = f 1 : S a, b wth T a,b = ff S a, b for all. The, there exsts a costat γ 0, 1 such that whch mmedately gves ET a,b <. P T a,b > γ, Proof. WLOG, let s assume there exst ɛ, δ such that δ > 0, ɛ 0, 1 ad P X 1 δ ɛ > 0. Take r > b a/δ. The P S r > b a P X 1 δ,..., X r δ, S r > b a P X 1 δ,..., X r δ, rδ > b a ɛ r.

3 Lecture 3: Radomly Stopped Processes 3-3 Accordgly, P T a,b > r = P T a,b > r, 1 ɛ r. mr P = m=1 =m 1r+1 mr P =m 1r+1 mr =m 1r+1 X b a, m = 1,..., X b a, m = 1,..., X b a Ths completes the proof va choosg γ = 1 ɛ r. 3.2 Trucato method The ultmate goal of ths secto s to prove the followg ear-weakest set of codtos o the stoppg tme T ad the dstrbuto X 1 uder whch Wald s frst equato holds. Theorem 7. Let X be a sequece of..d. r.v. s. Assume that EX 1 = 0 mplyg E X 1 <. The ES T = 0 f EK X T < Itroducto to K-fucto bouds Gve a radom varable X 0 symmetrc recallg that symmetrcty s usually ot a requremet such that E X <, the K-fucto K X y s mplctly defed as the soluto of the followg equato: K X y 2 = ye[x 2 X K X y]. K X y s exstece ad uqueess s guarateed by the fact that the fucto E[X 2 /t 2 X /t] s strcly decreasg ad cotuous at the rage t 0,, wth values ragg from to 0. The followg proposto hts oe tuto o troducg the K-fucto. Proposto 8. We have E S 2K X. Proof. By defto, we have K X y 2 = yex 2 1 X KX y + yk X ye X 1 X >KX y, yeldg ad We the obta K X 2 EXX1 2 X KX E X 1 X >KX K X. E S E X 1 X K X + E X 1 X >K X [EX 2 1 X KX ] 1/2 + E X 1 X >KX 2K X,

4 3-4 Lecture 3: Radomly Stopped Processes where the secod equalty uses the Holder s equalty ad symmetrcty of X, whch mples EX 1 X K X s cetered. The followg boud shows that K X s a sharp cotrol o E S. Proposto 9. If X, X are..d. symmetrc r.v. s wth L 1 orm exstg. The, for ay = 1, 2,..., 1 4e K X E S 2K X. Proof. We have already proved the upper boud, ad hece are focused o the lower boud ths proof. By defto of the K-fucto, we have ether EX 2 1 X KX 1 2 K X 2 or E X 1 X >KX 1 2 K X. 1 If the latter holds, by defto of the K fucto, whch mples P X > K X = P [X 2 X K X > K X 2 ] E[X2 X K X ] K X 2 = 1, E S E X 1#: X >K X =1 [ = E X 1 1 X1 >K X [ = E =2 1 X K X =2 1 X K X ] ] E X 1 X >KX 1 1 1E X 1 X >KX > e 1 E X 1 X >KX. 2 If EX 2 1 X KX 1 2 K X 2. Settg Y := X 1 X K X, remdg Y 2 ad depedece, E Y 4 = EY <j EY 2 EYj 2 K X 2 EY 2 + 3[ EY 2 ] 2 = K X 2 E Y 2 + 3[E Y 2 ] 2. Holder s equalty wth 1/p = 2/3 ad 1/q = 1/3 gves E [ Y 2 = E Y 2/3 Y 4/3] [ E ] 2/3 [ Y E Y 4] 1/3. Substtutg the above equalty to the prevous oe, we obta E Y Usg the fact that 1 2 K X 2 EY 2 E Y [E Y 2 ] 3/2 [K X 2 E Y 2 + 3[E Y 2 ] 2 ] 1/2. K X 2, we further have [K X 2 /2] 3/2 [K X 4 /2 + 3K X 4 /2] 1/2 = 1 10 K X, K X 2, by symmetry

5 Lecture 3: Radomly Stopped Processes 3-5 To fsh the proof, we employ the followg geometry equalty Lemma PG1999 whch wll be left as a HW: Lemma 10. If X are depedet symmetrc r.v. s. The for all measurable sets A symmetrc about the org ad for all t > 0, P X 1 X A > t 2P S > t, yeldg for all p > 0, E X 1 X A p 2E S p. Ths completes the proof. We ed ths secto wth the followg proposto, whose proof s left as HW. Proposto 11. We have, wheever E X <, as y, K X y wll mootocally crease to fty; K X y 2 /y wll mootocally crease to EX 2 ; ad K X y/y wll mootocally decrease to Proof of Theorem 7 We are ow ready to state the proof of the ma theorem ths chapter. Our am s to show It s because, f the, we have E max T S cek X T. 3.1 E sup S T cek X T <, < whch, combed wth the domated covergece theorem, gves ES T = E lm S T = lm ES T, where each term the last term, by the proof of Wald s frst equato, wll always equal to 0. Step 1. For provg Equato 3.1, we frst defe the followg two symbols: The followg observatos are mmedate: κ T := mk : 2 k > T ad T := 2 κ T 1. 2 κ T 1 T, T T, 1T 2 k = 1T 2 k+1 1, ad 1κ T 1 k = 1T 2 k, 3.2 where the last equalty s due to 1κ T 1 k = 12 κ T 2 k+1 = 12 κ T 1 2 k+1 1 = 1T 2 k+1 1 = 1T 2 k. Step 2. We are ow usg two famous martgale theorems, whose proofs are left to the readers. Theorem 12. If M s a martgale, the the stopped process M T s also a martgale wth regard to the same fltrato.

6 3-6 Lecture 3: Radomly Stopped Processes Theorem 13 Burkholder-Davs-Gudy equalty. Assume X to be a martgale dfferece sequece, the for ay p 1, E max S p/2. k p E X 2 k [] Usg the above two theorems, takg p = 1 BDG equalty, we ow have E max S = E sup S T = E sup T 1/2 T 1/2. X 1T ce X 2 1T = ce X 2 Step 3. Ths step cotas the ma body of the proof. We ow are usg a smlar sequece-splttg techque as what we have see the proof of Erdos equalty. T ce X 2 1/2 = ce 1/2 X 2 1T 2 k <2 k+1 1/2 c E X 2 1T 2 k <2 k+1 1/21T c E 2 k = c 2 k <2 k+1 X 2 E 2 k <2 k+1 X 2 1/2P T 2 k, where the last equalty s due to that T s a stoppg tme ad T 2 k F 2k 1 s depedet of 2 k <2 k+1 X2. To cotue the proof, we eed the followg verso of Khch s equalty amed as Marckewcz equalty, whose proof wll be left as a HW. Lemma 14 Marckewcz equalty. Gve p 1, there exst costats c p, C p oly depedg o p such that, f Y are cetered ad depedet ad L p, the c 1 p E Y 2 p/2 p E Y Cp E p/2. Y 2 By Marckewcz equalty wth p = 1, we ow have 1/2 E X 2 c1 E X = c 1 E S 2 k, 2 k <2 k+1 2 k <2 k+1

7 Lecture 3: Radomly Stopped Processes 3-7 whch yelds T ce X 2 1/2 c P T 2 k E S 2 k c ce ce ce P T 2 k K X 2 k 1T 2 k K X 2 k 1T 2 k K X 2 k 1κ T 1 kk X 2 k κ T 1 ce 2 k/2 K X 2 k 2 k/2. Usg Proposto 11, we have 2 k/2 K X 2 k s mootocally creasg, so that κ T 1 ce Ths the fshes the proof. κ T 1 2 k/2 K X 2 k 2 k/2 ce 2 κ T 1/2 K X 2 κ T 1 2 k/2 κ T 1 = ce2 κ T 1/2 K X 2 κ T 1 cek X 2 κ T 1 2 κ T 1 2 cek X 2 κ T 1 cek X T. 2 κ T k/2 3.3 Radomly stopped sums: the grad decouplg The proof of Theorem 7 rests o provg that E max T S cek X T, where the LHS s a expectato o both T ad X 1,..., X, whle the RHS volves a expectato oly o T thk about the dfferece. I other words, vew of Proposto 9, what we really proved s that E max S ce max S, T T where T s detcally dstrbuted as T whle depedet of X 1,..., X. observed studyg Wald s equatos. Ths s exactly what we have

8 3-8 Lecture 3: Radomly Stopped Processes The grad theorem of ths chapter s the followg oe, whch geeralzes the above argumet to a bgger famly of fuctos: A α := Φ : R + R +, Φ0 = 0, Φ odecreasg, cotuous, ad Φcx c α Φx for all c 2, x 0. Theorem 15. Fx ay α > 0. Let Φ A α. Let X 1, X 2,... be depedet ot ecessarly detcally dstrbuted radom varables. Set S = X ad let T be a stoppg tme adapted to F := σx 1,..., X. The there exst costats c α, C α oly depedg o α such that c α Ea T,Φ E max [T ] Φ S C α Ea T,Φ, 3.3 where I other words, we have a,φ := E max k [] Φ S k. c α E max Φ S E max Φ S C α E max Φ S. [ T ] [T ] [ T ] The proof of the lower boud Equato 3.3 s extremely volved, ad wo t be covered ths class. Our ma focus, stead, s o provg the upper boud. For ths, let s frst troduce some otato: S m,] := max S j S m = max X m X j ; m<j m<j S := S 0,] = max j S j ; X := max j [] X j ; y = := sup : a δy; a := E max j [] Φ S j. Note that ad a are fxed umbers. Fxg y > 0, we wrte T y = mm [T ] : ΦS m y wth the coveto that T y = f such a m does ot exst. Proof. The proof s separated to three steps. Step 1. We frst trasfer the orgal problem to somethg easer, volvg oly XT what we have see H-J equalty. Ths s due to the followg lemma. Lemma 16. We have P ΦS T βy, ΦX T a T < δy 6 α δ β 3 α 1 + δ P ΦS T y ad a T, smlar to wheever δ > 0 ad β > 3 α 1 + δ. Proof. We frst prove the frst probablty boud. The followg several observatos are mmedate. 1 For ay a, b, c 0, we have Φa + b + c Φ3 maxa, b, c Φ3a + Φ3b + Φ3c 3 α Φa + Φb + Φc. 2 O the set ΦS T βy, β > 1 mples that T y T.

9 Lecture 3: Radomly Stopped Processes O the set T y T, ST = max S j max S j j T y T y<j T S Ty max S j S Ty + S Ty T y<j T max S j S Ty + S T Ty 1 + X Ty. y<j T y 4 Puttg the above three observatos together, we have, o the set ΦS T βy, sce β > 3α 1 + δ > 1, we have T y T, ad hece whch yelds S T max S j S Ty + S T Ty 1 + X Ty = ST y,t ] + S T y<j T y y 1 + X Ty, ΦST 3 α ΦST y,t ] + Φ S T y 1 + Φ X Ty. 5 Usg the defto of T y, we have ΦS T y 1 y. 6 Usg the defto of, we have a T δy mples T. 7 Combg 4-6 yelds P ΦST βy, ΦXT a T < δy [ P 3 α ΦST y,t ] + Φ S T y 1 + Φ X Ty βy, ΦXT < δy, T ] P [3 α y + δy + ΦSTy,T ] βy, ΦX T < δy, T ] [ ] P 3 α ΦST y,t ] yβ 3α 1 + δ 3 α y 1 β 3 α 1 + δ 1 EΦS T y,t ]. 8 Gve 7, what s left s to upper boud EΦST y,t ]. For ths, we have 1 EΦST y,t ] = EΦSj,T ] 1T y = j, T > j = Combg 7 ad 8, we have EΦS j, ] 1T y = j, T > j EΦS j, ] P T y = j, T > j by defto of T ad T y EΦ2S P T y = j, T > j usg S j, ] 2S 2 α EΦS P T y < T 2 α δyp ΦS T y usg the defto of ad T y. P ΦS T βy, ΦX T a T < δy 6 α δy yβ 3 α 1 + δ P ΦS T y,

10 3-10 Lecture 3: Radomly Stopped Processes ad completes the proof. Step 2. For boudg EΦST usg Lemma 16, we eed the followg good-lambda equalty troduced by Burkholder ad Gudy, whose proof s left as a HW. Lemma 17 Good-lambda equalty. Let U ad V be two o-egatve radom varables, ad let ɛ be a postve umber. Suppose there exst postve reals β, γ, δ such that β 1 > ɛ ad The P U βy, V < δy ɛp U y for all y > 0. EU β 1 ɛ 1 δ 1 EV. Usg Lemma 17, settg ɛ = 6 α δ β 3 α 1 + δ wth βɛ < 1, we obta EΦST δ 1 β 1 ɛ 1 E ΦXT a T δ 1 β 1 ɛ 1 EΦXT + Ea T. It remas to boud EΦX T by Ea T. Step 3. We boud EΦXT va the followg lemma. Lemma 18. Let d, e be two sequeces of real radom varables adapted to the same creasg sequece of σ-felds F. Assume the two sequeces are taget, or other words, that for all, P d F 1 = P e F 1. The, for all t > 0, P sup d j > t 2P sup e j > t. Proof. Let := f : e > t. The P sup d j > t = P sup d j > t, = + P sup = E1 sup d j > t1 = + P E E = = sup 1d j > t1 = + P < 1d j > t1 j + P < d j > t, < d j > t, < E E1d j > t1 j F j 1 + P < E 1 je1d j > t F j 1 + P <

11 Lecture 3: Radomly Stopped Processes 3-11 where the last equalty uses the fact that 1 j = 1 1 < j 1 s adapted to F j 1. We ca the cotue to wrte P sup d j > t = E 1 je1e j > t F j 1 + P < = E 1e j > t + P < = E1e > t + P < E1 sup e j > t + P 2P. Ths completes the proof of the lemma. sup e j > t Let d = Φ X 1T, e = Φ X 1T, ad sup j< F = σx 1,..., X, X 1,..., X, e j > t where X s a depedet copy of X. It s the clear that d ad e are taget sce, gve F 1, 1T s a costat the oly addtoal radomess to d ad e comes through X ad X, whch are..d.. Furthermore, we have ΦX T = sup Φ X 1T We the apply Lemma 18 to obta, for ay t 0, Itegratg over t, we have EΦXT 2EΦ X T 2EΦ max S S 1 T 2 α+1 EΦ max S = 2 α+1 Ea T. T Ths completes the whole proof. ad P ΦX T > t 2P Φ X T > t. Φ X T = sup Φ X.1T 2EΦ max S + max S 1 2EΦ2 max S T T T 3.4 From dscrete to cotuous: Embeddg theorem Ths secto wll preset getly the embeddg theorem o how to coect dscrete radom processes to cotuous oes. The followg s a exteso to Theorem 15. Theorem 19. Fx α > 0 ad let Φ A α. Let N t t 0 be a real-valued process cotuous o the rght wth lmts from the left wth depedet cremets. Let T be a stoppg tme adapted to σn s, s t. The c α Ea T E sup Φ N s C α Ea T, 0 s T where a t := E sup Φ N s. 0 s t

12 3-12 Lecture 3: Radomly Stopped Processes Proof. For a fxed t > 0, we take a subdvso of [0, t], τ = s : 0 such that 0 = s 0 < s 1 < < s 1 < s = t, ad max s +1 s 0 as. For N we take N s j = j N s N s 1 + N s 0. By defto, N s N s 1 s a dscrete tme process wth depedet cremets. We choose τ N such that τ +1 cotas τ. Settg T ω = s +1 o s < T t s +1 makes T a stoppg tme wth respect to F s 0 where F s := σn u : u s. Also, T 0 s a decreasg sequece covergg to T t. Theorem 15 the yelds c α E sup Φ Ñs E sup Φ N s C α E sup Φ Ñs. s τ,s T s τ,s T s τ,s T The rght cotuty assumpto o N t, t 0 ad the Lebesgue DCT the mples Lastly, usg Lebesgue MCT fshes the proof. c α E sup Φ Ñs E sup Φ N s C α E sup Φ Ñs. s T t s T t s T t The ext s a exteso of Levy-type equalty to cotuous tme processes, whose proof follows the same techque as used above. Proposto 20 Levy-type equalty for cotuous tme processes. Let N t, t 0, N 0 = 0, be a realvalued procress cotuous the rght wth lmts from the left wth cremets that are statoary ad depedet. The, for all y 0, Proof. For a fxed t, let wth τ = s : 0. Set P sup N s > y s t 9P 30 N t > y. N s j = s = 2 t j N s N s 1 + N s 0. Use the Levy-type equalty, Theorem 6 Chapter 2, for the above sequece of..d. cremets we obta for fxed 1 ad all y > 0, P N s > y 9P 30 N t > y. sup s τ,s t The result follows usg the DCT by lettg. The followg s a combato of the prevous two theorems Theorem 21. Let N t, t 0, N 0 = 0, be a real-valued process cotuous o the rght wth lmts from the left wth statoary ad depedet cremets. Let T be a stoppg tme adapted to σn s, s t. Fx α > 0. The, for ay Φ A α, there exst costat c α, C α depedg oly o α s.t. c α EΦ ÑT E sup Φ N s C α EΦ ÑT. 0 s T

13 Lecture 3: Radomly Stopped Processes 3-13 Proof. The left s a trval boud. The left s by codtog argumet ad the above proposto: E sup t T Ths completes the proof. Φ N t C α E sup Φ Ñt 9C α EΦ30 ÑT 9C α 30 α EΦ ÑT. t T The above theorem mmedately mples the followg corollary, whch s kow as the Burkholder-Gudy equalty. Corollary 22. Let B t, t 0 be a stadard Browa moto ad T a stoppg tme adapted to σb s, s > 0. The for all p > 0, c p E B 1 p ET p/2 E sup B t p C 1 C p 2 E B 1 p ET p/2. t T Proof. Usg Theorem 21 ad otcg that fsh the proof. E B T p = E T 1/2 B1 p = ET p/2 E B 1 p