Brownian Motion and Stochastic Calculus

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1 Browia Motio ad tochastic Calculus Xiogzhi Che Uiversity of Hawaii at Maoa Departmet of Mathematics July 3, 28 Cotets Measurability of Radom Process 2 toppig imes 5 3 Martigales 3 Browia Motio ad tochastic Calculus Chapter : Martigales, toppig times, Filtratios Measurability of Radom Process Problem Let Y be a modi catio of X ad suppose that both processes have a.s. right-cotiuous sample paths. he X ad Y are idistiguishable. olutio. ice for ay t Choose Q fr i : i g. he P (X t Y t ) P r i (X ri 6 Y ri ) X P (X ri 6 Y ri ) X ( P (X ri Y ri )) r i r i that is, \ P (X ri Y ri ) r i ice for ay t ;there is a sequece fr i;t g i Q such that r i;t # t; the lim X ri;t X t ; lim Y ri;t Y t ; a:s:8! 2 i i by the hypothesis ad P (X t Y t ; 8t ) P t (X t Y t ) X ri;t limi Y ri;t limi P i that is, X ad Y are idistiguishable. P r i (X ri Y ri )

2 De itio 2 Let (X; ) ; (Y; ) be two measurable spaces. he productio measure space (X Y; ) is de ed to be ( ) (fa B 2 g) where each A B 2 is called a measurable rectagle. Lemma 3 Every sectio of a measurable set is a measurable set (Paul Halmos) Proof. Let he obviously ad E is a -rig. hus E E fe 2 X Y : E x 2 ; E y 2 g E Corollary 4 If X t (!) 2 B ([; )) F; the X t (!) 2 B ([; )) for xed! 2 Proof. By (3). he trajectory for a xed! 2 is a sectio of X t (!) ad hece is measurable w.r.t B ([; )); that is, sice (t;!) : X t (!) 2 A 2 B R do A A 2 2 B ([; )) F the ay xed! 2 ; t : X t (!) 2 A 2 B R do A 2 B ([; )) (Fubii s heorem is too much for this corollary). Problem 5 Let X be a process, every sample path of which is RCLL. Let A be the evet that X is cotiuous o [; t ): how that A 2 F X t Proof. Chose xed! 2 ad suppose X t (! ) is ot cotiuous at some poit s 2 (; t ) :he lim tj "s X t (! ) 6 X s (! ) for some sequece ft j : j g Q Q \ [; t );which implies 9;s.t. for 8m; 9t ; < s t < 3m ; but jx t (! ) X s (! )j > ice the sample path is RCLL, there exist q ; q 2 2 Q such that < q s; q 2 t < 3m s.t., jx q (! ) X s (! )j < 4 ; jx q 2 (! ) X t (! )j < 4 hus jq q 2 j < m ad jx q (! ) X q2 (! )j > > jjx t (! ) X s (! )j jx q (! ) X s (! )j jx q2 (! ) X t (! )jj > 2 2 2

3 Reversely, if there 9;s.t. 8m; 9q ; q 2 2 Q ; < jq q 2 j < m ; but jx q (! ) X q2 (! )j > he we ca take 2 m for each m ad pick q m; ; q m;2 2 Q t with jq m; q m;2 j < 2 m but Xqm; (! ) X qm; (! ) > Let p lim if m fq m; ; q m;2 g : he p 2 (; t ) ad X t (! ) is ot cotiuous at p. Cosequetly, by letttig A ;m! 2 : jx r (!) X s (!)j > (r;s)2q Q jr sj< m it s clear that i.e., A ( m A ;m) C A lim A C ;m ice A ;m 2 F X t ad Q Q is coutable, we have A C ;m 2 F X t ad A 2 F X t Problem 6 Let X be a process whose sample paths are RCLL almost surely, ad let A be the evet that X is cotiuous [; t ): how that A ca fail to be i F X t ; but if ff t ; t g is a ltratio satisfyig F X t F t ; t ad F t cotais all P -ull sets of F;the A 2 F t Problem 7 Let X be a process with sample paths are LCRL, ad let A be the evet that X is cotiuous [; t ] : Let X be adapted to a right-cotiuous itratio ff t g : how that A 2 F X t : Proof. ice the sample path is LCRL, we ca let B ;m (r)! 2 : the it s clear that m r k rm r ad! 2 A i for ay 2 R d ; t 2 [; t ] Cosequetly, A r2q X r (!) X r+ kr B ;m (r) f! 2 : X t (!) < g! 2 : lim X t+ (!) X t (!) < k k o! 2 : X t+ (!) < 2 k m t k tm t A 2 F t F t km t (!) F t+ k F t 3

4 Propositio 8 If X is adapted to the ltratio ff t g ad every sample path is right-cotiuous (or, left-cotiuous), the X is also progressively measurable with respect to ff t g Proof. Let P t B ([; t]) F t ad B B R d :ake right-cotiuity for example. For each ; k 2 ;de e X () X k+ t (u;!) 2 t;! if u 2 ( kt 2 ; (k+)t 2 ]; X (;!) if u he for ay A 2 R B B R d ; it s clear that (s;!) : X () 2 k 2 P t t o (u;!) 2 A kt (k + ) t ; 2 2 X [(k+)2 ]t (A) fg X (A) Moreover, from it s clear that lim X () t (u;!) X (u;!) ; 8u 2 [; t] X (u;!) 2 P t B; 8u 2 [; t] Problem 9 If X is measurable ad the r.t. is ite, the the fuctio X is a radom variable. Proof. De e :! R by! 7! ( (!) ;!) We ll show is measurable. ake A 2 B (R ) ; the f! 2 : ( (!) ;!) 2 Ag (A! ) 2 F where A! fx 2 R : (x;!) 2 Ag ;that is, A! is the x-sectio of A: ice is ite ad X (!) X where X t 2 F for ay t 2 [; );the X (!) (!) 2 F Problem Let X be measurable ad a r.t.. how that F ff! : X 2 Ag ; f! : X 2 Ag [ f! : g : A 2 B (R)g is a sub-- eld of F;which is deoted by (X ) 4

5 Proof. () 2 F : (b) Let B 2 F : he or B f! : X 2 Ag B f! : X 2 Ag [ f! : g for some A 2 B (R) : hus B C! : X 2 A C or B C! : X 2 A C \ f! : < g ; i both cases B C 2 F sice A C ; 2 B (R) :(c) uppose B 2 F ad B " B: he B (f! : X 2 A k g [ f! : g) f! : X 2 A j g k j! : X 2 ( A k [ f! : g! : X 2 ) A j k j for some A j ; A k 2 B (R) : From the fact that k A k; j A j 2 B (R) ; it s clear that Cosequetly, F (X ) is a sub-- eld of F 2 toppig imes B 2 F Propositio If E is a o-empty class of sets cotaiig?;write E E ad for ay ordial > ; write iductively E! E < where C deotes the class of all coutabe uios of di ereces of sets of C: If! is the rst ucoutable ordial the (E) E <! Lemma 2 Let z z (!) be such that E [jzj] < ad be a s.t. Problem 3 Let X (X t ; t ) be a,real-valued stochastic process ad a stoppig time for F X t W t s X (B (R)) ; t ; i.e., Ft X is a sub--algebra such that all X s ; s t are measurable. uppose that for some pair!;! 2 ;we have X t (!) X t (! ) for all t 2 [; (!)] \ [; ):how that (!) (! ) Proof. ake X t (!) ad [; ] ad (!)! 2 : he F X t B ([; ]) : But for! 6! 2 ; (!) 6 (! ) (Prof. Ramsey gives a proof) Problem 4 Let X fx t ; F t g be a adapted stochastic process with right-cotiuous paths. Cosider a subset 2 B R d of the state space of the process, ad de e the hittig time H (!) if ft : X t (!) 2 g with the stadard covetio that if? : how that if is ope, the H is a optioal time. 5

6 Proof. Note that ad f! : H (!) g s<t fx s (!) 2 g fx t (!) 2 g s<t o show the validity of () it su ces to show fx t (!) 2 r<t;r2q r<t;r2q fx t (!) 2 g () g fx t (!) 2 uppose X s (!) 2 for some s < t; the B (X s (!) ; ) for some > sice is ope. Cosequetly from the right-cotiuity of X t (!) ;there is some > such that s<t X ft:s ts + g (!) B (X s (!) ; ) hus for ay r 2 Q with r 2 [s ; s + mi f ; t s g] it s clear that X r (!) 2 ; cotradictig with fx t (!) 2 g : ad r<t;r2q Whece that is, H f! : H (!) g s<t fx s (!) 2 g f! : H (!) < g fx s (!) 2 g is a optioal time. s<t r<t;r2q r<t;r2q g fx r (!) 2 fx r (!) 2 g g 2 F t is closed ad the sample paths of the process X are coti- Problem 5 If i the above problem uous, the H is stoppig time. Proof. From x 2 R d : (x; ) < it s clear each is ope ad " : he the times : H (!) are optioal ad " ad H : Let : lim : Obviously H gives a dichotomy of as E f! : H g ad F f! : H > g. o for all o E, ad o F; there is some k k (!) such that if < k < + < H if k (else there exist a subsequece j which implies ) o show tha H ; it su ces to show j fh>; <g Hj fh>; <g By cotiuity of the sample path, X lim X ad X m m ; 8m > k: (else if X m 2 m for some m ; the the cotiutity of X t ad the opeess of m implies that there is some " > ad > such that X [m ; m +] B (X m ; "), which violates the miimality of m ) By lettig m! ; it s clear that X 2 ; 8 k ad thus X 2 : Hece H ad H :Fially from fh g f < g for all > ad fh g fx 2 g ; it s clear that the assertio is justi ed. 6

7 Problem 6 Let ; be optioal times, the + is optioal. It is a stoppig time, if oe of the followig coditios holds: (i) ; > ;(ii) >, is stoppig time. Proof. (i) From the decompositio f + > g f ; > g f ; > g f ; > g f < < ; + > g f g f < < ; + > g (2) ad f < < ; + > g fr < < ; r2q;<r< rg ad it s clear that f rg [ r ]+ < r + 2 F r+ F f + > g 2 F (ii) From the above decompositio (2) ad ote that f g [ ]+ F m 2 F + F it s clear that f + > g 2 F Problem 7 Give the is a stoppig time of the ltratio ff t g ; verify that F fa 2 F : A f tg 2 F t ; 8t g is a sub-- eld of F ad is F -measurable. how that if (!) t for some costat t ad every! 2 ; the F t F Proof. Clearly?; 2 F : Let A 2 F : he A f tg 2 F t implies A C f tg f tg (A f tg) C 2 F t that is, A C 2 F :If A "; A 2 F ; the lim A f tg (lim A ) f tg 2 F t ad lim A 2 F : hus F is a sub-- eld of F: For ay r 2 Q f < rg f < rg 2 Ft if r t f tg f tg 2 F t if r > t the is F -measurable. Firstly, F t F : o it su ces to show F t F ; which is justi ed by the fact that for costat ; A 2 F implies A f tg A f tg A 2 F t : 7

8 Lemma 8 how that for ay stoppig time ad positive costat t; ^ t is a F t -measurable radom variable. Proof. Pick ay r 2 Q fr 2 Q : r g. Case : t < r; the from it s clear that or from f ^ t < rg (f tg f > tg) f ^ t < rg (f tg f ^ t < rg) (f > tg f ^ t < rg) f ^ t < rg f tg f > tg 2 F t f ^ t tg it s that f ^ t < rg f ^ t t < rg. Case 2: r t; the from the decompositio, it s clear that f ^ t < rg Cosequetly, ^ t is F t -measurable (f tg f ^ t < rg) (f > tg f ^ t < rg) f < rg r F r F t Lemma 9 For ay two stoppig times ad ;ad for ay A 2 F ; we have I paritcular, if o ; the F F Proof. his is clear from ad the previous lemma. A f g 2 F A f g f tg (A f tg) f tg f ^ t ^ tg Problem 2 Let be a stoppig time ad a radom time such that o : If is F -measurable o ; the it is also a stoppig time Proof. ice for ay ; t 2 [; ] ad f g, the which meas is a stoppig time. f g f tg 2 F t f tg f tg f tg 2 F t Problem 2 Let ; be stoppig times ad Z a itegrable radom variable. We have (i) E [ZjF ] E [ZjF ^ ] ; P -a.s. o f g ;(ii) E [E [ZjF ] jf ] E [ZjF ^ ] ; P -a.s. 8

9 Proof. (see proof i textbook) the key is that for ay A 2 F Cosequetly Z A Z f g E (ZjF ^ ) dp For claim (ii), we coclude from (i) that A f g 2 F F F ^ A\f g Z ZdP A f g E (ZjF ) ZdP f g E [E (ZjF ) jf ] E f g E (ZjF ) jf E f g E (ZjF ^ ) jf f g E (ZjF ^ ) which proves the desired result o f g : Iterchagig the role of ad ad replace Z by E (ZjF ) ; we ca also coclude from (i) that f< g E [E (ZjF ) jf ] f< g E [E (ZjF ) jf ^ ] f< g E [ZjF ^ ] Lemma 22 Let be a stoppig time ad s 2 [; t]. he the mappig is B ([; )) F t -measurble (s;!) 7! ( (!) ^ s;!) Proof. his is part of the argumet i propositio 2.8. ake A 2 F t ad r 2 Q: f(s;!) : (!) ^ s < rg (fs 2 [; t] : s < rg f! 2 A : (!) < rg) (fs 2 [; t] : s rg f! 2 A : (!) < rg) (fs 2 [; t] : s < rg f! 2 A : (!) rg) which is obviously B ([; )) F t -measurble sice is a stoppig time. Problem 23 Uder the same assumptios as i propositio 2.8, ad with f (t; x) : [; )R d! R d a bouded, B ([; )) B R d -measurble fuctio, show tha the process Y t R t f (s; X s) ds; t is a progressively measurable with respect to ff t g ; ad Y is a F -measurable radom variable. Proof. Let L o f (t; x) : [; ) R d! R d ; kfk < ad L f 2 L : Y t f (s; X s ) ds ff t g ; t where deotes progressively measurable, ad A r;q [; r] ( ; q) : r 2 Q; q 2 Q do 9

10 Obviously, f 2 L i f + ; f 2 L ad is a -class, ad L is a L-classs i that (i) f (t; x) implies Y t (!) f (s; X s (!)) ds sice for ay A A A 2 2 B ([; t]) B R d ds t ff t g f(t;!) 2 Ag A o, 2 L: (ii) for ay f; g 2 L ad ; 2 R; it s clear that f + g 2 L (iii) for f 2 L; f " f with f beig bouded, it s clear from Levi s lemma Y t (!) f (s; X s (!)) ds lim f (s; X s (!)) ds lim f (s; X s (!)) ds lim Y ;t (!) < ; 8t ad Y t ff t g ; t sice each Y ;t (!) is: (iv) for ay A r;q 2 ; it s clear that Ar;q ds [;r] (s) ( ;q) (X s (!)) ds if Xs (!) 2 ( ; q) or s 2 [t ^ r; t] t ^ r if X s (!) 2 ( ; q) ; s 2 [; t ^ r] hus for ay 2 Q (t;!) : 8 < : Ar;q ds <? if fx s (!) 2 ( ; q)g ([t ^ r; t] ) if < t ^ r [; t] if > t ^ r i.e, Let Ar;q ds ff t g ; 8A r;q 2 ; t : D fa : A 2 Lg the D (). For ay f 2 L ad f 2 B ([; )) B R d there are simple fuctios f 2 B ([; )) B R d such that f Xm k k 2 f k 2 jfj< k 2 g " f ad f 2 L:hus L cotais all bouded, B ([; )) B R d -measurble fuctios, as desired. De itio 24 Let be a optioal time of the ltratio ff t g : he - eld F + of evets determied immediately after the optioal time is de ed to be F + fa 2 F : A f tg 2 F t+ ; 8t g

11 Problem 25 Verify that the class F + is ideed a - eld with respect to which is measurable, that it coicides with F fa 2 F : A f < tg 2 F t ; 8t g ad that if is a stoppig time (so that both F + ; F are de ed), the F F + Proof. Pick ay A 2 F +. he If A 2 F + ; A " A; the A C f tg f tg (A f tg) C 2 F + A f tg ( A ) f tg (A f tg) 2 F + Also, 2 F + : hus F + is a - eld. For ay r 2 Q f < rg f tg thus 2 F + : ice for ay A 2 F + ; A f tg 2 F t+ implies A f < tg A t f tg 2 F + if r > t f < rg 2 F r+ F + if r t 2 F t F ad F + F : O the other had, for ay B 2 F ; B f < tg 2 F t implies B f tg B < t + 2 F t+ F t+ ad F + F : hus F + F Lemma 26 For ay optioal time ad a positive costat t, ^ t is F + -measurable. Proof. Pick ay r 2 Q fr 2 Q : r g. Case : t r; the from it s clear that or from f ^ t rg (f tg f > tg) f ^ t rg (f tg f ^ t rg) (f > tg f ^ t rg) f ^ t rg f tg f > tg 2 F t F t+ f ^ t tg it s that f ^ t rg f ^ t tg. Case 2: r < t; the from the decompositio, it s clear that f ^ t rg (f tg f ^ t rg) (f > tg f ^ t rg) f rg 2 F t F t+ Cosequetly, ^ t is F t -measurable.

12 Problem 27 Verify that aalogues of Lemma 2.5 ad 2.6 hold if ad are assumed to be optioal ad F ; F ad F ^ are replaced by F + ; F + ad F ( ^)+ ;respectively. Prove that if is a optioal time ad is a positive stoppig time with ;ad < o f < g ;the F + F Proof. Aalogue of 2.5: For ay A 2 F + o, A f g f tg (A f tg) f tg f ^ t ^ tg 2 F t+ A f g 2 F + If f g, the F + F + : Particularly, take A ; it s clear that f g 2 F + Aalogue of 2.6: From ^ mi f; g ;it s clear that F ( ^)+ F + F+ : Now take A 2 F + F+ ; observe that A f ^ tg A (f tg ( t)) (A f tg) (A ( t)) 2 F t+ so, A 2 F ( ^)+ : Hece F ( ^)+ F + F+ ice f g 2 F + ; the f > g 2 F + : Let R ^ :he R is both F + - ad F + - measurable. herefore, f < g fr < g 2 F + (sice is F + -measurable). withig the role of ad to ge the last part. ake A 2 F + :he A r2q [A f < r < g]! [A f g] ( f g is decomposed i the above represetatio). ice A f < r < g A f < rg fr < g 2 F sice A f < rg 2 F r ad is stoppig time. O the other had hus A 2 F as desired. A f g (A f g) f g 2 F Remark 28 Let ; be stoppig times ad R ^ : he R is F -measurable. Just oberve that R is F R -measurable ad F R F F Problem 29 how that if f g is a sequece of optioal times ad if ; the F + F +: Besides, if each is a positive stoppig time ad < o f < g ;the we have F + F 2

13 Proof. is a optioal time ad so F + is de ed ad F + F +. For the other directio, pick A such that the hus A 2 F + (ecod claim omitted) A f < tg A A f < tg 2 F t ; 8 ; t f < tg! (A f < tg) 2 F t Problem 3 Give a optioal time of the ltratio ff t g ; cosider the sequece of f g of radom times give by (!) o f! : (!) g (!) k 2 o! : k 2 (!) < k 2 for ; k : Obviously + for every : how that each is a stoppig time, that lim! ;ad that for every A 2 F + we have A k 2 2 F k2 ; ; k Proof. imply from f tg f tg k k 2 < k! f (!) g 2 (f tg f g) f tg k k 2 < k 2 f tg k k 2 < k 2 f tg k k 2 [t2 ] k [t2 ] k 2 2 < k 2 2 F t k k it s clear that is a stoppig time for :, 3 Martigales Let X (x ; F ) ; ; ; N;be a submartigale ad let (a; b) be a oempty iterval. We eed to de e the "umber of crossigs of the iterval (a; b) by the submartigal X." For this purpose de e: mi f < N : x ag 2 mi f < N : x bg 2m mi f 2m 2 < N : x ag 2m mi f 2m < N : x bg 3

14 If l is the maximal such idex; the l+r : N + for all N r l: If if N x > a; the : N + : l for all l. If at least some j exists, the for each i > accordig to the values of x s ; s i ; there ca be decided a j such that j < i j+ : hus the key idea is to use j to seperate x i De itio 3 If 2 N; the the maximal m for which 2m N is de ed is called the umber of up-crossigs of the iterval (a; b) ad is deoted by (a; b) :Else if 2 > N;the this umber is de ed to be Lemma 32 For ay x 2 R the fuctio x 7! x + max fx; g is covex Proof. (i) (x) + x + for ay > : (2) x x + x x + implies x x + i x i x : o from x + y x + + y + ; it s clear that (x + y) + (x + + y + ) + x + + y + : Moreover, x+y + 2 x y+ 2 : hus this fuctio is covex. heorem 33 If X (x ; F ) ; ; ; N;is be a submartigale, the E [ (a; b)] E (x N a) + E x + N + jaj b a b a Proof. Let X + (x a) + ; F : he X + is still a submartigale ad X (a; b) X + (; b a) hus we ca suppose X ad a ad show that for b > ; E [ X (; b)] E [x N] b Assume x ad for each i ; ;let if i j < i j+ for some odd j if j < i j+ for some eve j he for 2m j we have NX i (x i x i ) b X (; b) i ( P m j 2j+ (x i x i ) m (b a) if 2m+ N P m j 2j+ (x i x i ) + N (x N x 2m ) m (b a) + (x N a) if 2m+ > N (Note that P 2j+ <i 2(j+) i (x i x i ) P 2j+ <i 2(j+) (x i x i ) x 2j+ x 2(j+) m (b a) for j m) ad Hece " X # be [ X (; b)] E i (x i x i ) f i g m odd [f m < ig f m+ < ig] 2 F i i NX Z i NX Z i E [x N ] f i g f i g (x i x i ) dp NX Z i f i g (E [x i jf i ] x i ) dp 4 E [(x i x i ) jf i ] dp NX Z i (E [x i jf i ] x i ) dp

15 Remark 34 he proof here is a combiatio of " heory of Radom Process" by Wag su-kwu ad "tatistics of Radom Process" by R.. Liptser. he dow-crossigs (a; b) of the iterval (a; b) satis es E [ (a; b)] E (x N a) + E x + N + jaj b a b a 5