Vaiance and Covaiance Pocesses Pakash Balachandan Depamen of Mahemaics Duke Univesiy May 26, 2008 These noes ae based on Due s Sochasic Calculus, Revuz and Yo s Coninuous Maingales and Bownian Moion, Kaazas and Sheve s Bownian Moion and Sochasic Calculus, and Kuo s Inoducion o Sochasic Calculus 1 Moivaion In his secion, we moivae he consucion of vaiance and covaiance pocesses fo coninuous local maingales, which is cucial in he consucion of sochasic inegals w... coninuous local maingales as we shall see. In his secion, unless ohewise specified, we fix a Bownian moion B and a filaion {F } such ha: 1. Fo each, B is F -measuable. 2. Fo and s, he andom vaiable B B s is independen of he σ-field F s. Recall ha fo any Bownian moion, B immediaely implies (2) in he following = whee B is he quadaic vaiaion of B. This Definiion: Define L 2 ad (a, b Ω) o be he space of all sochasic pocesses f(, ω), a b, ω Ω such ha: 1. f(, ω) is adaped o he filaion {F }. 2. b a E f() 2 d = b a E f() 2 d B <. 1
lso ecall when consucing a heoy of inegaion w... a Bownian moion, we begin wih consucing he sochasic inegal fo f L 2 ad (a, b Ω). b a f()db Now, we wan a moe geneal fomalism of inegaing a class of pocesses w... a genealized maingale ha in he case Bownian moion will educe o he above. Definiion: Le G be a igh-coninuous filaion. We define L denoe he collecion of all joinly measuable sochasic pocesses X(, ω) such ha: 1. X is adaped w... G. 2. lmos all sample pahs of X ae lef coninuous. Fuhemoe, we define P o be he smalles σ-field of subses of a, b Ω wih espec o which all he sochasic pocesses in L ae measuable. sochasic pocesses Y (, ω) ha is P measuable is said o be pedicible. The moivaion fo he definiion of a pedicable pocess comes fom he following agumen: If Y is a pedicable pocess, hen almos all is values a ime can be deemined wih ceainy wih he infomaion available sicly befoe ime, since lef coninuiy of he pocess Y implies ha fo almos evey ω Ω and any sequence n as n : lim n Y n (ω) = Y (ω). Now, we have he following heoem which we shall pove a vesion of in he nex secion fo coninuous local mainagles: Theoem 1 (Doob-Meye) Le M, a b be a igh coninuous, squae inegable maingale wih lef hand limis. Then, hee exiss a unique decomposiion (M ) 2 = L +, a b whee L is a igh-coninuous maingale wih lef-hand limis, and is a pedicable, igh coninuous, inceasing pocess such ha a 0 and E < fo all a b. The above heoem ceainly applies o he squae inegable pocess B. 2
Claim 1 In he case M = B in Doob-Meye, = B =. Poof of Claim 1: WLOG, we may ake a = 0 and b = 0. Define P 0 s 0 : = (B ) 2. Then, fo E(B ) 2 F s = E(B B s + B s ) 2 F s = E(B B s ) 2 + 2B s (B B s ) + (B s ) 2 F s = E(B B s ) 2 + 2B s EB B s + (B s ) 2 = s + (B s ) 2 EP F s = E(B ) 2 F s = (B s ) 2 s = P s. Thus, P = (B ) 2 is a maingale, so ha (B ) 2 = P +. Clealy, saisfies all he condiions ha mus saisfy in Doob-Meye, so ha by uniqueness of, = = B. So, anohe way of viewing he inegal w... he maingale M w... he filaion G is he following: Fis, we look fo he unique pocess (guaaneed by Doob-Meye) M such ha L = (M ) 2 M is a maingale. Then, we make he Definiion: Define L 2 ped (a, b M Ω) o be he space of all sochasic pocesses f(, ω), a b, ω Ω such ha: 1. f(, ω) is pedicable w... {G }. 2. b a E f() 2 d M <. Then, we poceed o consuc he inegal b a f()dm fo f L 2 ped (a, b M Ω). I s clea ha in he case M = B and G = F ha he above fomulaion coincides wih he oiginal consucion of he sochasic inegal w... B eviewed a he beginning of his secion. Fo igh coninuous, squae inegable maingales M wih lef hand limis, a leas, his pocess woks. In he case whee M is a coninuous local maingale, we do he same hing. Howeve, i s no immediaely clea: 1. ha we have a vesion of Doob-Meye fo coninuous local maingales. 2. how he consucion of he inegal is affeced by he sopping imes T n ha educe M, if a all. In he nex secion, we deal wih he fis poblem. Then, we poceed o emedy he second. 3
2 Vaiance and Covaiance Pocesses We ake L and P as defined in secion 1. Theoem 2 If X is a coninuous local maingale, hen we define he vaiance pocess X o be he unique coninuous pedicable inceasing pocesses ha has 0 0 and makes X 2 a local maingale. Definiion: If X and Y ae wo coninuous local maingales, we le X, Y = 1 4 ( X + Y X Y ). We call X, Y he covaiance of X and Y. Based on he discussion in he fis secion, i s clea why we e ineesed in vaiance pocesses. I is convenien o define covaiance pocesses since hey ae vey useful and have quie nice popeies, such as: Theoem 3, is a symmeic bilinea fom on he class of coninuous local mainagles. We migh pove i his ime aound. If no, hopefully nex ime. Two quesions I m sill pondeing is 1. Can you un his ino an inne poduc? 2. If so, how you can chaaceize he class of pocesses ha is he compleion of his space? The poof of heoem 2 is long, bu i is insucive o go hough i, since i develops echniques ha will be useful lae. In ode o poceed, ecall ha any pedicable discee ime maingale is consan why?. Thee is a esul analogous o his in coninuous ime, and we use i o pove he uniqueness saemen in heoem 2: Theoem 4 ny coninuous local maingale X ha is pedicable and locally of bounded vaiaion is consan (in ime). Poof: By subacing X 0, WM ha X 0 0. Thus, we wish o show ha X 0 fo all > 0 almos suely. Le V (ω) = sup π Π T π (ω) be he vaiaion of X s (ω) on 0,, whee Π denoes he se of all (finie) paiions of 0,, π = {0 = 0 < 1 < < N = }, and whee fo a given paiion of his so, N T π (ω) = X m (ω) X m 1 (ω). 4
Lemma 1 Fo almos all ω Ω, V (ω) is coninuous Poof of Lemma: Fis, noice ha fo any ω Ω, V (ω) is inceasing: Fo s <, 0, s 0,, so ha any finie paiion π = {0 = 0 < 1 < < N = s} of 0, s gives a finie paiion π = {0 = 1 < < N = s < N+1 = } of 0,. Thus, fo any finie paiion π of 0, s, T π (ω) T π (ω), whee π is a finie paiion of 0,, so ha Since ω was abiay, his is ue fo all ω Ω. T π (ω) T π (ω) sup π Π T π (ω) = V (ω) V s (ω) = sup π Π s T π (ω) sup π Π T π (ω) = V (ω). Thus, o show ha V is coninuous a.s., i suffices o show ha fo almos all ω Ω, V (ω) has no disconinuiies (of he fis kind). Claim 2 Fo any ω Ω, V u (ω) = V s (ω) + V u s (ω) whee V u s (ω) is he vaiaion of X (ω) on s, u. Poof of Claim: Take any wo paiions {s = 0 < 1 < < N = u}, {0 = N < N 1 < < 0 = s}. Then: 0 m= N +1 X m (ω) X m 1 (ω) + N X m (ω) X m 1 (ω) V s (ω) + V u s (ω). Now, he LHS is T π (ω) fo π = {0 = N < < N = u}. Thus, V u (ω) V s (ω) + V u s (ω). Fo he ohe inequaliy, noe ha {0 = N Thus: < < 0 = s < < N = u} is a paiion of 0, u. V u (ω) N m= N +1 X m (ω) X m 1 (ω) = 0 m= N +1 X m (ω) X m 1 (ω) + N X m (ω) X m 1 (ω). Now, fixing one of he paiions on he RHS, we may ake he supemum of he emaining, and hen poceed o ake he supemum of he final em. Thus: V u (ω) V s (ω) + V u s (ω), so ha V u (ω) = V s (ω) + V u s (ω). Now, by hypohesis, X s is of locally bounded vaiaion. So, hee exiss a sequence of sopping imes T n a.s. such ha Xs Tn (ω) is of bounded vaiaion in ime. Le = {ω Ω : T n (ω) }. By definiion, P = 1. Now, le ω be fixed, and suppose ha s V s (ω) has a disconinuiy a. Choosing n lage enough so ha T n (ω) >, hee exiss s 0 < u 0 such ha X s (ω) is of bounded vaiaion on s 0, u 0. 5
Since s V s (ω) has a disconinuiy a, hee exiss ɛ > 0 such ha fo evey δ > 0, u s < δ implies V u (ω) V s (ω) > 3ɛ whee s < < u. By Claim 2 hen, fo evey δ > 0, u s < δ implies V u s (ω) > 3ɛ whee s < < u. Pick δ > 0 so ha if s < δ hen X s X < ɛ (using unifom coninuiy of X s (ω) on s 0, u 0 ). ssuming s n and u n have been defined, pick a paiion of s n, u n no conaining wih mesh less han δ and vaiaion geae han 2ɛ (his is possible since fo evey δ > 0, u s < δ implies V u s (ω) > 3ɛ whee s < < u). Le s n+1 be he lages poin in he paiion less han, and u n+1 be he smalles poin in he paiion lage han. Then u n+1 s n+1 < δ X sn+1 (ω) X un+1 (ω) < ɛ. Thus: N X m (ω) X m 1 (ω) > 2ɛ N, m s n+1,u n+1 X m (ω) X m 1 (ω) > 2ɛ X un+1 (ω) X sn+1 (ω) > ɛ. By omiing he poins s n+1 and u n+1 fom he paiion, we obain a paiion fo s n, u n s n+1, u n+1. Thus, afe aking supemums: V sn,u n s n+1,u n+1 (ω) = V un s n (ω) V u n+1 s n+1 (ω) > ɛ. Thus, V u 0 s 0 (ω) > Mɛ fo abiaily lage inege values M. Thus, i mus be infiniy, conadicing ha X s (ω) has bounded vaiaion on s 0, u 0. Thus, V (ω) mus be coninuous fo evey ω Ω, since ω was abiay. Now, we needed Lemma 1 in ode o guaanee ha he funcions ae sopping imes (why?). Lemma 2 {S n } educe X. Poof of Lemma 2: We poved las ime ha S n (ω) = inf{s : V s (ω) n} If X is a coninuous local maingale, we can always ake he sequence which educes X o be T n = inf{ : X > n} o any ohe sequence T n T n ha has T n as n. Now, suppose ha saisfies n < X = X X 0 V. Then, V > n, so ha { : X > n} { : V n} S n = inf{ : V n} inf{ : X n} = T n. 6
Ceainly, V s n + 1 n V s n so ha { : V s n + 1} { : V s n} S n = inf{{ : V s n} { : V s n + 1} = S n+1. Finally, since V is coninuous a.s., i s clea ha lim n S n (ω) = almos suely. Thus, S n educe X. Now, fix some n > 0. Then, S n implies X n. By Lemma 2, M = X Sn maingale. is a bounded Now, if s < : E(M M s ) 2 F s = EM 2 F s 2M s EM F s + M 2 s = EM 2 F s M 2 s = EM 2 M 2 s F s. (we efe o his elaionship as ohogonaliy of maingale incemens). If 0 = 0 < 1 < < N = is a paiion of 0,, we have: N N EM 2 = E M 2 m M 2 m 1 = E (M m M m 1 ) 2 E ne sup M m M m 1 m V Sn sup M m M m 1 m Taking a sequence of paiions n = {0 = n 0 < n 1 < < n k(n) = } in which he mesh n = sup m n m n m 1 0 coninuiy of sample pahs imply sup m M n m M n m 1 0 a.s. Since sup m M m M m 1 2n, he bounded convegence heoem implies E sup M n m M n m 1 0. m Thus, EM 2 = 0 so ha M = 0 a.s. Le = {ω Ω : M (ω) 0}. Then, since above was abiay, P = 0 fo any, so ha P = 0. Q, 0 Thus, wih pobabiliy 1, M = 0 fo all aional. By coninuiy of sample pahs, we have ha M = 0 wih pobabiliy 1 fo all. Uniqueness in heoem 2: Suppose ha and ae wo coninuous, pedicable, inceasing pocesses ha have 0 = 0 0, and make X2, X 2 local maingales. If T n educe X 2 and T n educe X 2 i s clea ha T n T n educe X 2 (X 2 ) =, so ha is a coninuous local maingale. I s clea ha is pedicable, since each and ae pedicable. 7
Finally, is locally of bounded vaiaion. To see his, ake he sopping imes S n = T n T n. Clealy, T n T n, and he sopped pocesses T n T n T n T n ae of bounded vaiaion fo each ω, being he diffeence of wo inceasing pocesses on he andom ineval 0, T n (ω) T n(ω). Thus, by heoem 4, mus be consan, so ha since 0 = 0 = 0, = 0 fo all. Thus, = fo all. The exisence poof is a lile moe difficul, bu uses some gea analysis. Exisence in heoem 2: We poceed in seps: Sep 1: Poof of exisence in heoem 2 when X is a bounded maingale: (noe ha uniqueness follows fom he pevious agumen) Given a paiion = {0 = 0 < 1 < } wih lim n n =, le k() = sup{k : k < } be he index of he las poin befoe ; noe ha k() is no a andom vaiable, bu a numbe. Define k() Q (X) = (X k X k 1 ) 2 + k=1 ( X X k() ) 2. Lemma 3 If X is a bounded coninuous maingale, hen X 2 Q (X) is a maingale. Poof of Lemma 3: Fis, noice ha k() Q Q ( ) ( ) k(s) 2 2 ( ) ) 2 2 s = Xk X k 1 + X X k() Xk X k 1 (X s X k(s) = k=1 (X k(s)+1 X k(s) ) 2 (X s X k(s) ) 2 + k() k=k(s)+2 k=1 ( Xk X k 1 ) 2 + ( X X k() ) 2. Define u i = i fo k(s) i k() and u k()+1 =. Then, wiing Q = Q s + (Q Q s ): = EX 2 F s Q s (X) E = EX 2 F s Q s (X) E k()+1 i=k(s)+2 k()+1 i=k(s)+2 EX 2 Q (X) F s ( ) 2 Xui X Fs ui 1 E ( 2 ( ) 2 X X k(s)+1 k(s)) Fs +E X s X k(s) Fs Xu 2 i Xu 2 Fs i 1 E X 2 k(s)+1 2X k(s)+1 X k(s) + X 2 k(s) F s +E X s 2 2X s X k(s) + X 2 k(s) F s 8
= Q s (X) + E X 2 k(s)+1 F s E X 2 k(s)+1 F s + 2X s X k(s) X 2 k(s) + Xs 2 2X s X k(s) + X 2 k(s) = X 2 s Q s (X) whee in he fis equaliy, we have used he fac ha Q s (X) is F s measuable, and in he second equaliy, we have used he ohogonaliy of maingale incemens. Lemma 4 Le X be a bounded coninuous maingale. Fix > 0 and le n be a sequence of paiions 0 = n 0 < < n k n = of 0, wih mesh n = sup k n k n k 1 0. Then, Q n (X) conveges o a limi in L 2 (Ω, F, P ). Poof of Lemma 4: Fis, we begin wih some noaion. If and ae wo paiions of 0,, we le denoe he paiion obained by aking all he poins in and. Now, by lemma 3, fo fixed paiions and of 0,, we have ha fo a bounded coninuous maingale X : Y = (X 2 Q ) (X 2 Q ) = Q Q is again a bounded maingale (Since X M fo all 0 implies Q Q KM since he paions and ae fixed). Thus, again by lemma 3: Z = (Y ) 2 Q (Y ) is a maingale wih Z 0 = 0, so ha ( ) 2 EZ = 0 E Q Q = E (Y ) 2 = E Q (Y ). Now, 2a 2 + 2b 2 (a + b) 2 = (a b) 2 0 fo any eal numbes a and b, so ha (a + b) 2 2(a 2 + b 2 ) fo any eal numbes a, b. Thus: Q (Y ) = k=1 k=1 k() ( ) ( ) k() 2 2 Yk Y k 1 + Y Y k() = k=1 ( + Q Q k=1 ) ( ) 2 Q k() Q k() ( ) ( ) 2 Q k Q k Q k 1 Q k 1 k() ) ( ) 2 ( ) ( ) 2 = (Q k Q k 1 Q k Q k 1 + Q Q k() Q Q k() k() ) 2+ ( ) 2+ ( ) 2+ ( ) 2 2 (Q k Q k 1 Q k Q k 1 Q Q ( k() Q Q k() = 2 Q Q ) ( + Q Q ). 9
Puing i all ogehe hen, we have: ( ) 2 E Q ( Q = E Q (Y ) 2 Q Q ) ( + Q Q ). Thus, o show ha {Q n (X)} is Cauchy in L 2 (Ω, F, P ), and hence conveges in his space since i is complee, i is sufficien o show ha + ( 0 E Q Q ) 0. To do his, le {s k } n k=1 = and { j } =. Le s k and j such ha j s k < s k+1 j+1. Then: Q s k+1 Q s k = (X sk+1 X j ) 2 (X sk X j ) 2 = (X sk+1 X sk ) 2 + 2(X sk+1 X sk )(X sk X j ) Q = (X sk+1 X sk )(X sk+1 + X sk 2X j ) (Q ) Q (X) sup(x sk+1 + X sk 2X j(k) ) 2 k whee j(k) = sup{j : j s k }. By he Cauchy-Schwaz inequaliy: ( E Q Q ) E Q (X) 2 1 2 E sup k (X sk+1 + X sk 2X j(k) ) 4 1 2. Since he sample pahs of X ae coninuous almos suely, sup k (X sk+1 + X sk 2X j(k) ) 4 0 almos suely as + 0. Since sup k (X sk+1 + X sk 2X j(k) ) 4 (4M) 4, he bounded convegence heoem implies ha as + 0. Thus, i emains o show ha E Q (X) 2 = E Q = E sup k ( n ) 2 (X sm X sm 1 ) 2 = (X sk+1 + X sk 2X j(k) ) 4 1 2 0 Q (X) 2 is bounded. To do his, noe ha: n n 1 n (X sm X sm 1 ) 4 +2 (X sm X sm 1 ) 2 n n 1 ( ) (X sm X sm 1 ) 4 + 2 (X sm X sm 1 ) 2 Q (X) Q s m (X) n (X) 2 = E (X sm X sm 1 ) +2E 4 n 1 To bound he fis em on he RHS, noe ha X M fo all implies: n n n E (X sm X sm 1 ) 4 (2M) 2 E (X sm X sm 1 ) 2 = 4M 2 E 10 j=m+1 (X sj X sj 1 ) 2 ( ) (X sm X sm 1 ) 2 Q (X) Q s m (X) Xs 2 m Xs 2 m 1 4M 2 E X 2 4M 4
whee in he fis equaliy, we have used ha ohogonaliy of maingale incemens: E(X sm X sm 1 ) 2 F sm 1 = EX 2 s m X 2 s m 1 F sm 1 implies: E(X sm X sm 1 ) 2 = EX 2 s m X 2 s m 1. Fo he second em on he RHS, noe ha (X sm X sm 1 ) 2 F sm. By lemma 3, and ohogonaliy of maingale incemens: So: E E E E Q (X) Q s (X) F s = E X 2 X 2 s F s = E (X X ) 2 F s. (X sm X sm 1 ) 2 ( Q n 1 n 1 E ) ( (X) Q s m (X) F sm = (X sm X sm 1 ) 2 E Q = (X sm X sm 1 ) 2 E (X X sm ) 2 F sm (2M) 2 (X sm X sm 1 ) 2 (X sm X sm 1 ) 2 ( Q ( (X sm X sm 1 ) 2 Q ) (X) Q s m (X) F sm 4M 2 E (X) Q s m (X)) 4M 2 E n 1 n 1 ) (X) Q s m (X) F sm (X sm X sm 1 ) 2 X 2 s m X 2 s m 1 4M 2 E X 2 4M 4. Thus, E Q (X) 2 4M 4 + 2 4M 4 = 12M 4. { Lemma 5 Le { n } be as in Lemma 4. Then, hee exiss a subsequence { nk } such ha Q n k conveges unifomly a.s. on 0,. Poof of Lemma 5: Since { } Q n conveges in L 2 (Ω, F, P ), i is Cauchy in his space. So, choose a { } subsequence such ha fo m n k, Le Q n k By Chebyshev s inequaliy: P k = P E Q m { k = ω Ω : sup sup Q n k+1 Q n k Q n k Q n k+1 2 < 2 k. (ω) Q n k (ω) > 1 } k 2. 1 Q k 2 k 4 nk+1 E Q n k 2 < k4 2 k. } 11
Since he RHS is summable, Boel-Canelli implies ha P lim sup k k = 0. So, fo almos all ω Ω, hee exiss N ω such ha k > N ω implies So, fo m > m > N ω : sup Q n k+1 (ω) Q n k (ω) < 1 k 2. sup (ω) Q n m (ω) Q nm Since he seies k=1 1 k 2 implies hus, fo m, m > max{n, N ω }: { Thus, Q n k m 1 k=m sup Q n k+1 (ω) Q n k (ω) < m 1 1 k 2. k=m conveges, we have ha given ɛ > 0, hee exiss N such ha m, m > N sup Q nm m 1 1 k 2 < ɛ. k=m (ω) Q n m (ω) < ɛ. } conveges unifomly almos suely on 0,. In wha follows, call he limiing funcion in Lemma 5, and define i o be zeo ouside 0,. Now, fo each nk in lemma 5, we can exend i o a paiion n k of 0, + 1, such ha n k 0. Then, fo his sequence of meshes, lemma 4 implies ha Q n k +1 conveges { o } a limi in L2 (Ω, F, P ). Repeaing he pocedue in lemma 5 hen, we can selec a subsequence Q n kj such ha i conveges unifomly almos suely on 0, + 1. Call he limiing funcion +1, and similaly define i o be zeo ouside 0, + 1. I s clea ha fo, Q n kj = Q n k, so ha = +1 fo. Repeaing he pocedue above, we obain a sequence of funcions { +j } j=0 such ha +j is coninuous on 0, + j a.s., and +j = +k fo min{ + j, + k}. So, we can unambiguously define (ω) = lim j +j (ω). If D j = {ω Ω : +j (ω) is no coninuous on 0, + j}, hen clealy D = j=0 D j has measue zeo. Thus, fo ω D c, i s clea ha (ω) will be coninuous, so ha is coninuous a.s. I s clea fom he consucion ha is pedicable. To show ha is inceasing, i is sufficien o show ha each +j is inceasing. To his end, le n be he paiion of 0, + j wih poins a k2 n ( + j) fo 0 k 2 n ; clealy, aking his sequence of paiions doesn ale he above agumens, so ha Q n +j unifomly a.e. on 0, + j. 12
Clealy, n+1 is a efinemen of n and n=1 n is dense in 0, + j. Thus, fo any pai s,, s <, in n=1 n hee exiss n 0 such ha s and belong o n fo n n 0. Thus, Q n s ha +j s Q n fo n n 0, so +j. Since his is ue fo any s <, s, n=1 n, by coninuiy of he pocess, i mus hold eveywhee on 0, + j. Thus, is coninuous, pedicable, and inceasing. ll we need o veify now is ha (X ) 2 is a maingale. Now, fo each j, +j is he limi of pocesses of he fom Q n k suely o +j. Thus, we have convegence in pobabiliy. which convege unifomly almos Similaly, since Q n k was obained as a subsequence of a sequence conveging in L 2 (Ω, F, P ) o say Q, he subsequence Q n k conveges o Q in L 2 (Ω, F, P ), so ha we also have convegence in pobabiliy. Thus, Q = R+j wih pobabiliy 1, so ha Q n k conveges o +j in L 2 (Ω, F, P ). Claim 3 Suppose ha fo each n, Z n is a maingale w... F, and ha fo each, Z n Z in L p (Ω, F, P ) whee p 1. Then, Z is a maingale. Poof of Claim 3: Recall ha since we e woking ove he finie measue space (Ω, F, P ), convegence in L p (Ω, F, P ) implies convegence in L 1 (Ω, F, P ) (why?). 2 Now, he maingale popey implies ha fo s <, EZ n F s = Zs n, so ha fo any F s EZ n F s = Since Zs n Z s in L p (and hence in L 1 ) we have ha lim EZ n n F s = lim n Now, Z n s. Zs n = E EZ n F s EZ F s p = E EZ n Z F s p EE Z n Z p F s = E Z n Z p whee we have used he condiional Jensen inequaliy, and lineaiy of condiional expecaion. Thus, EZ n F s EZ F s in L p (Ω, F, P ), so ha EZ n F s EZ F s in L 1 (Ω, F, P ). Thus: EZ F s = lim EZ n n F s = Z s so ha since F s was abiay, EZ F s = Z s. Z s. Thus, by lemma 3, since (X ) 2 Q n k is a maingale, and Q n k is a maingale. Thus, (X ) 2 is obviously a maingale. +j in L 2 (Ω, F, P ), (X ) 2 +j 13
Sep 2: Poof of exisence in heoem 2 when X is a local maingale Lemma 6 Le X be a bounded maingale, and T be a sopping ime. Then, X T = X T. Poof of Lemma 6: By he consucion in sep 1, M = (X ) 2 X is a maingale. Then, M T = (X T ) 2 X T is a maingale, so ha by uniqueness of he pocess X T, X T = X T. Now, le X be a coninuous local maingale, wih a sequence of sopping imes {T n } ha educe i. WLOG, we may ake he sopping imes o be he canonical imes: T n = inf{ : X > n}. Then, Y n = X Tn 1 Tn>0 is a bounded maingale. By he esuls in sep 1, hee is a unique, coninuous pedicable, inceasing pocess n such ha (Y n ) 2 n is a maingale. By lemma 6, fo T n, n = n+1, so ha we may unambiguously define X = n fo T n. Clealy, X is coninuous, pedicable, and inceasing. By definiion: XT 2 n 1 Tn>0 X Tn is a maingale, so ha (X ) 2 X is a local maingale. We poceed now o pove he analogue above fo he covaiance pocess. In paicula, i is vey useful in compuing X, Y. Theoem 5 Suppose ha X and Y ae coninuous local maingales. X, Y is he unique coninuous pedicable pocess ha is locally of bounded vaiaion, has 0 = 0, and makes X Y a local maingale. Poof of Theoem 5: By definiion: X Y X, Y = 1 (X + Y ) 2 X + Y 4 { (X Y ) 2 } X Y is obviously a coninuous local maingale. To pove uniqueness, noice ha if and ae wo pocesses wih he desied popeies, hen = (X Y ) (X Y ) is a coninuous local maingale ha is of locally bounded vaiaion. Hence, by heoem 4, his mus be idenically zeo, so ha = fo all. 14