Brownian Motion An Introduction to Stochastic Processes de Gruyter Graduate, Berlin 2012 ISBN:

Transcription

1 Browia Moio A Iroducio o Sochasic Processes de Gruyer Graduae, Berli 22 ISBN: Soluio Maual Reé L. Schillig & Lohar Parzsch Dresde, May 23

2 R.L. Schillig, L. Parzsch: Browia Moio Ackowledgeme. We are graeful o Björ Böcher, Kaharia Fischer, Fraziska Küh, Julia Holleder, Felix Lider ad Michael Schwarzeberger who suppored us i he preparaio of his soluio maual. Daiel Tillich poied ou quie a few mispris ad mior errors. Dresde, May 23 Reé Schillig Lohar Parzsch 2

3 Coes Rober Brow s New Thig 5 2 Browia moio as a Gaussia process 3 3 Cosrucios of Browia Moio 25 4 The Caoical Model 3 5 Browia Moio as a Marigale 35 6 Browia Moio as a Markov Process 45 7 Browia Moio ad Trasiio Semigroups 55 8 The PDE Coecio 7 9 The Variaio of Browia Pahs 79 Regulariy of Browia Pahs 89 The Growh of Browia Pahs 95 2 Srasse s Fucioal Law of he Ieraed Logarihm 99 3 Skorokhod Represeaio 7 4 Sochasic Iegrals: L 2 Theory 9 5 Sochasic Iegrals: Beyod L 2 T 9 6 Iô s Formula 2 7 Applicaios of Iô s Formula 29 8 Sochasic Differeial Equaios 39 9 O Diffusios 53 3

4

5 Rober Brow s New Thig Problem. (Soluio) a) We show he resul for R d -valued radom variables. Le ξ, η R d. By assumpio, lim E exp [i (ξ η ), (X ) ] E exp [i ( ξ Y η ), (X Y ) ] lim E exp [i ξ, X + i η, Y ] E exp [i ξ, X + i η, Y ] If we ake ξ ad η, respecively, we see ha Sice X Y we fid lim E exp [i η, Y d ] E exp [i η, Y ] or Y Y lim E exp [i ξ, X ] E exp [i ξ, X ] or X d X. E exp [i ξ, X + i η, Y ] lim E exp [i ξ, X + i η, Y ] ad his shows ha X Y. b) We have Y X X X + lim E exp [i ξ, X ] E exp [i η, Y ] lim E exp [i ξ, X ] lim E exp [i η, Y ] E exp [i ξ, X ] E exp [i η, Y ] almos surely d X X X almos surely d X X Y X X + Y almos surely d X + Y. A simple direc calculaio shows ha X 2 (δ + δ ) Y. Thus, X d X, Y d Y X, X + Y d. Assume ha (X, Y ) d (X, Y ). Sice X Y, we fid for he disribuio of X + Y : X + Y 2 (δ + δ ) 2 (δ + δ ) 4 (δ δ + 2δ δ + δ δ ) 4 (δ + 2δ + δ 2 ). Thus, X + Y / δ lim (X + Y ) ad his shows ha we cao have ha (X, Y ) d (X, Y ). 5

6 R.L. Schillig, L. Parzsch: Browia Moio c) If X Y ad X Y, he we have X + Y d X + Y : his follows sice we have for all ξ R: lim E eiξ(x+y) lim E e iξx E e iξy lim E e iξx E e iξx E e iξy a) E [e iξx e iξy ] E e iξ(x+y ). lim E e iξy A similar (eve easier) argume works if (X, Y ) d (X, Y ). The we have f(x, y) e iξ(x+y) is bouded ad coiuous, i.e. we ge direcly lim E eiξ(x+y) lim E f(x, Y ) E f(x, Y ) E e iξ(x+y ). For a couerexample (if X ad Y are o idepede), see par b). Noice ha he idepedece ad d-covergece of he sequeces X, Y already implies X Y ad he d-covergece of he bivariae sequece (X, Y ). This is a cosequece of he followig Lemma. Le (X ) ad (Y ) be sequeces of radom variables (or radom vecors) o he same probabiliy space (Ω, A, P). If he (X, Y ) d X Y for all ad X X ad Y d Y, d (X, Y ) ad X Y. Proof. Wrie φ X, φ Y, φ X,Y (X, Y ). By assumpio for he characerisic fucios of X, Y ad he pair lim φ X (ξ) lim E e iξx E e iξx φ X (ξ). A similar saeme is rue for Y ad Y. For he pair we ge, because of idepedece lim φ X,Y (ξ, η) lim E e iξx+iηy lim E e iξx E e iηy lim E e iξx lim E e iηy E e iξx E e iηy φ X (ξ)φ Y (η). 6

7 Soluio Maual. Las updae Jue 2, 27 Thus, φ X,Y (ξ, η) h(ξ, η) φ X (ξ)φ Y (η). Sice h is coiuous a he origi (ξ, η) ad h(, ), we coclude from Lévy s coiuiy heorem ha h is a (bivariae) characerisic fucio ad ha (X, Y ) d (X, Y ). Moreover, h(ξ, η) φ X,Y (ξ, η) φ X (ξ)φ Y (η) which shows ha X Y. Problem.2 (Soluio) Usig he elemeary esimae e iz iz e ζ dζ sup e iy z z (*) y z we see ha he fucio e i ξ,, ξ, R d is locally Lipschiz coiuous: Thus, e i ξ, e i ξ,s e i ξ, s ξ, s ξ s for all ξ,, s R d, E e i ξ,y E [e i ξ,y X e i ξ,x ] E [(e i ξ,y X )e i ξ,x ] + E e i ξ,x. Sice lim E e i ξ,x E e i ξ,x, we are doe if we ca show ha he firs erm i he las lie of he displayed formula eds o zero. To see his, we use he Lipschiz coiuiy of he expoeial fucio. Fix ξ R d. E [(e i ξ,y X )e i ξ,x ] E (e i ξ,y X )e i ξ,x E e i ξ,y X Y X δ ei ξ,y X d P + Y X >δ ei ξ,y X d P (*) δ ξ + Y X >δ 2 d P δ ξ + 2 P ( Y X > δ) δ ξ, δ where we used i he las sep he fac ha X Y P. Problem.3 (Soluio) Recall ha Y d Y wih Y c a.s., i. e. where Y δc for some cosa c R. Sice he d-limi is rivial, his implies Y P Y. This meas ha boh is his sill rue -quesios ca be aswered i he affirmaive. We will show ha (X, Y ) d (X, c) holds wihou assumig ayhig o he joi disribuio of he radom vecor (X, Y ), i.e. we do o make assumpio o he correlaio srucure of X ad Y. Sice he maps x x + y ad x x y are coiuous, we see ha lim E f(x, Y ) E f(x, c) f C b (R R) 7

8 R.L. Schillig, L. Parzsch: Browia Moio implies boh ad This proves (a) ad (b). lim E g(x Y ) E g(xc) lim E h(x + Y ) E h(x + c) g C b (R) h C b (R). I order o show ha (X, Y ) coverges i disribuio, we use Lévy s characerizaio of disribuioal covergece, i.e. he poiwise covergece of he characerisic fucios. This meas ha we ake f(x, y) e i(ξx+ηy) for ay ξ, η R: E e i(ξx+ηy) E e i(ξx+ηc) E e i(ξx+ηy) E e i(ξx+ηc) + E e i(ξx+ηc) E e i(ξx+ηc) E e i(ξx+ηy) E e i(ξx+ηc) + E e i(ξx+ηc) E e i(ξx+ηc) E e iηy e iηc + E e iξx E e iξx. d The secod expressio o he righ-had side coverges o zero as X X. For fixed η we have ha y e iηy is uiformly coiuous. Therefore, he firs expressio o he righ-had side becomes, wih ay ɛ > ad a suiable choice of δ δ(ɛ) > E e iηy e iηc E [ e iηy e iηc { Y c >δ}] + E [ e iηy e iηc { Y c δ}] 2 E [ { Y c >δ}] + E [ɛ { Y c δ}] 2 P( Y c > δ) + ɛ P -covergece as δ,ɛ are fixed ɛ. ɛ Remark. The direc approach o (a) is possible bu relaively ugly. Par (b) has a relaively simple direc proof: Fix ξ R. E e iξ(x+y) E e iξx ( E e iξ(x+y) E e iξx ) + ( E e iξx E e iξx ) by d-covergece For he firs erm o he righ we fid wih he uiform-coiuiy argume from Problem..2 ad ay ɛ > ad suiable δ δ(ɛ, ξ) ha E e iξ(x+y) E e iξx E e iξx (e iξy ) E e iξy where we use P-covergece i he peulimae sep. ɛ + P ( Y > δ) ɛ fixed ɛ ɛ. 8

9 Soluio Maual. Las updae Jue 2, 27 Problem.4 (Soluio) Le ξ, η R ad oe ha f(x) e iξx ad g(y) e iηy are bouded ad coiuous fucios. Thus we ge ad we see ha (X, Y ) d (X, Y ). E e i (ξ η ), (X Y ) E e iξx e iηy E f(x)g(y ) lim E f(x )g(y ) lim E e iξx e iηy lim E e i (ξ ), (X η Y ) Assume ow ha X φ(y ) for some Borel fucio φ. Le f C b ad pick g f φ. Clearly, f φ B b ad we ge E f(x )f(x) E f(x )f(φ(y )) E f(x )g(y ) E f(x)g(y ) E f(x)f(x) E f 2 (X). Now observe ha f C b f 2 C b ad g B b. By assumpio Thus, i.e. f(x ) L2 f(x). E f 2 (X ) E f 2 (X). E ( f(x) f(x ) 2 ) E f 2 (X ) 2 E f(x )f(x) + E f 2 (X) E f 2 (X) 2 E f(x)f(x) + E f 2 (X), Now fix ɛ > ad R > ad se f(x) R x R. Clearly, f C b. The P( X X > ɛ) P( X X > ɛ, X R, X R) + P( X R) + P( X R) P( f(x ) f(x) > ɛ, X R, X R) + P( X R) + P( f(x ) R) P( f(x ) f(x) > ɛ) + P( X R) + P( f(x ) R) P( f(x ) f(x) > ɛ) + P( X R) + P( f(x) R/2) + P( f(x ) f(x) R/2) where we used ha { f(x ) R} { f(x) R/2} { f(x ) f(x) R/2} because of he riagle iequaliy: f(x ) f(x) + f(x) f(x ) P( f(x ) f(x) > ɛ) + P( X R/2) + P( X R/2) + P( f(x ) f(x) R/2) 9

10 R.L. Schillig, L. Parzsch: Browia Moio P( f(x ) f(x) > ɛ) + 2 P( X R/2) + P( f(x ) f(x) R/2) ( ɛ R 2 ) E ( f(x) f(x ) 2 ) + 2 P( X R/2) ɛ,r fixed ad ff R C b X is a.s. R-valued 2 P( X R/2). R Problem.5 (Soluio) Noe ha E δ j ad V δ j E δ 2 j. Thus, E S ad V S. a) We have, by he ceral limi heorem (CLT) S S CLT G where G N(, ), hece G G N(, ). b) Le s <. Sice he δ j are iid, we have, S S s S s, ad by he ceral limi heorem (CLT) S s s S s CLT s G G s. s If we kow ha he bivariae radom variable (S s, S S s ) coverges i disribuio, we do ge G G s +G s because of Problem.. Bu his follows agai from he lemma which we prove i par d). This lemma shows ha he limi has idepede coordiaes, see also par c). This is as close as we ca come o G G s G s, uless we have a realizaio of ALL he G o a good space. I is Browia moio which will achieve jus his. c) We kow ha he eries of he vecor (X m X m,..., X 2 X, X ) are idepede (hey deped o differe blocks of he δ j ad he δ j are iid) ad, by he oe-dimesioal argume of b) we see ha X k X k k k G k G k k k d for all k,..., m where he G k, k,..., m are sadard ormal radom vecors. By he lemma i par d) we eve see ha (X m X m,..., X 2 X, X ) d ( G,..., m m G m ) ad he G k, k,..., m are idepede. Thus, by he secod asserio of par b) ( G,..., m m G m ) (G,..., G m m m ) (G,..., G m G m ). d) We have he followig Lemma. Le (X ) ad (Y ) be sequeces of radom variables (or radom vecors) o he same probabiliy space (Ω, A, P). If he (X, Y ) d X Y for all ad X X ad Y d Y, d (X, Y ) ad X Y (for suiable versios of he rv s).

11 Soluio Maual. Las updae Jue 2, 27 Proof. Wrie φ X, φ Y, φ X,Y (X, Y ). By assumpio for he characerisic fucios of X, Y ad he pair lim φ X (ξ) lim E e iξx E e iξx φ X (ξ). A similar saeme is rue for Y ad Y. For he pair we ge, because of idepedece lim φ X,Y (ξ, η) lim E e iξx+iηy lim E e iξx E e iηy lim E e iξx lim E e iηy E e iξx E e iηy φ X (ξ)φ Y (η). Thus, φ X,Y (ξ, η) h(ξ, η) φ X (ξ)φ Y (η). Sice h is coiuous a he origi (ξ, η) ad h(, ), we coclude from Lévy s coiuiy heorem ha h is a (bivariae) characerisic fucio ad ha (X, Y ) d (X, Y ). Moreover, which shows ha X Y. h(ξ, η) φ X,Y (ξ, η) φ X (ξ)φ Y (η) Problem.6 (Soluio) Necessiy is clear. For sufficiecy wrie B() B(s) B() B( s+ 2 ) + s 2 s 2 B( s+ 2 ) B(s) s 2 2 (X + Y ). By assumpio X Y, X Y ad X 2 (X + Y ). This is already eough o guaraee ha X N(, ), cf. Réyi [8, Chaper VI.5, Theorem 2, pp ]. Aleraive Soluio: Fix s < ad defie j s + j ( s) for j,...,. The B B s B j B j s j j j j B j B j j j G j By assumpio, he radom variables (G j ) j, are ideically disribued (for all j, ) ad idepede (i j). Moreover, E(G j ) ad V(G j ). Applyig he ceral limi heorem (for riagular arrays) we obai G j where G N(, ). Thus, B B s N(, s). d G

12

13 2 Browia moio as a Gaussia process Problem 2. (Soluio) Le us check firs ha f(u, v) g(u)g(v)( si u si v) is ideed a probabiliy desiy. Clearly, f(u, v). Sice g(u) (2π) /2 e u2 /2 is eve ad si u is odd, we ge f(u, v) du dv g(u) du g(v) dv g(u) si u du g(v) si v dv. Moreover, he desiy f U (u) of U is f U (u) f(u, v) dv g(u) g(v) dv g(u) si u g(v) si v dv g(u). This, ad a aalogous argume show ha U, V N(, ). Le us show ha (U, V ) is o a ormal radom variable. Assume ha (U, V ) is ormal, he U + V N(, σ 2 ), i.e. O he oher had we calculae wih f(u, v) ha ad his coradics (*). E e iξ(u+v ) e 2 ξ2 σ 2. (*) E e iξ(u+v ) e iξu+iξv f(u, v) du dv 2 2 ( e iξu g(u) du) ( e iξu g(u) si u du) e ξ2 ( 2i e iξu (e iu e iu 2 )g(u) du) e ξ2 ( 2i (e i(ξ+)u e i(ξ )u 2 )g(u) du) e ξ2 ( 2i (e 2 (ξ+)2 e 2 2 (ξ )2 )) e ξ2 + 4 (e 2 (ξ+)2 e 2 (ξ )2 ) 2 e ξ2 + 4 e e ξ2 (e ξ e ξ ) 2, Problem 2.2 (Soluio) Le (ξ,..., ξ ) (,..., ) ad se. The we fid from (2.2) ( j k ) ξ j ξ k ( j j )(ξ j + + ξ ) 2. (2.) k > Equaliy ( ) occurs if, ad oly if, (ξ j + + ξ ) 2 for all j,...,. This implies ha ξ... ξ. 3

14 R.L. Schillig, L. Parzsch: Browia Moio Absrac aleraive: Le (X ) I be a real-valued sochasic process which has a secod mome (such ha he covariace is defied!), se µ E X. For ay fiie se S I we pick λ s C, s S. The Cov(X s, X )λ s λ E ((X s µ s )(X µ ))λ s λ s, S s, S E (X s µ s )λ s (X µ )λ s, S E ((X s µ s )λ s(x µ )λ ) s S S E s S(X s µ s )λ s 2. Remark: Noe ha his aleraive does o prove ha he covariace is sricly posiive defiie. A sadard couerexample is o ake X s X. Problem 2.3 (Soluio) These are direc & sraighforward calculaios. Problem 2.4 (Soluio) Le e i (,...,,,...) R be he ih sadard ui vecor. The i a ii Ae i, e i Be i, e i b ii. Moreover, for i j, we ge by he symmery of A ad B ad which shows ha a ij b ij. Thus, A B. We have A(e i + e j ), e i + e j a ii + a jj + 2b ij B(e i + e j ), e i + e j b ii + b jj + 2b ij Le A, B R be symmeric marices. If Ax, x Bx, x for all x R, he A B. Problem 2.5 (Soluio) a) X 2B /4 is a BM : scalig propery wih c /4, cf b) Y B 2 B is o a BM, he idepede icremes is clearly violaed: E(Y 2 Y )Y E(Y 2 Y ) E Y 2 E(B 4 B 2 )(B 2 B ) E(B 2 B ) 2 (B) E(B 4 B 2 ) E(B 2 B ) E(B 2 B ) 2 (B) E(B 2 ). c) Z B is o a BM, he idepede icremes propery is violaed: E(Z Z s )Z s ( s) s E B 2 ( s) s. 4

15 Soluio Maual. Las updae Jue 2, 27 Problem 2.6 (Soluio) We use formula (2.b). a) f B(s),B() (x, y) 2π s( s) exp [ (y 2 (x2 x)2 + s s )]. b) Deoe by f B() he desiy of B(). The we have f B(s),B() B() (x, y B() z) f B(s),B(),B()(x, y, z) f B() (z) Thus, (2π) 3/2 s( s)( ) exp [ (y 2 (x2 x)2 + + s s f B(s),B() B() (x, y B() ) Noe ha x 2 s + (y x)2 s Therefore, (z y)2 )] (2π) /2 exp [ z2 2 ]. 2π s( s)( ) exp [ (y 2 (x2 x)2 + + y2 s s )]. + y2 s( s) (x s 2 y) + y2 + y2 s( s) (x s 2 y) y 2 + ( ). E(B(s)B() B() ) c) I aalogy o par b) we ge xyf B(s),B() B() (x, y B() ) dx dy 2π s( s)( ) y y exp [ 2 y 2 ( ) ] x exp [ x 2 s( s) (x s 2 y) ] dx dy 2π ( ) s ( ) s( ). s( s) s y y2 f B(2 ),B( 3 ) B( ),B( 4 )(x, y B( ) u, B( 4 ) z) f B( ),B( 2 ),B( 3 ),B( 4 )(u, x, y, z) f B( ),B( 4 )(u, z) 2π s y exp [ 2 y 2 ( ) ] dy 2π [ ( 4 ) ( 2 )( 3 2 )( 4 3 ) ] 2 (x exp [ 2 (u2 u)2 (y x)2 (z y) )] Thus, exp [ (z 2 (u2 u)2 + )]. 4 f B(2 ),B( 3 ) B( ),B( 4 )(x, y B( ) B( 4 ) ) 5

16 R.L. Schillig, L. Parzsch: Browia Moio 2π [ ( 4 ) ( 2 )( 3 2 )( 4 3 ) ] 2 exp [ 2 ( x2 (y x)2 + + y2 )] Observe ha x 2 (y x)2 + + y ( 2 )( 3 2 ) (x 2 2 y) ( 3 )( 4 3 ) y2. Therefore, we ge (usig physiciss oaio: dy h(y) h(y) dy for easier readabiliy) xy f B(2 ),B( 3 ) B( ),B( 4 )(x, y B( ) B( 4 ) ) dx dy 2π( 4 3 ) y dy exp [ 2 4 ( 3 )( 4 3 ) y2 ] y 2π(2 )( 3 2 ) x exp [ x 2 (x y) 3 ( 2 )( 3 2 ) ] dx y ( 4 3 )( 3 ) 4 ( 2 )( 4 3 ) 4. Problem 2.7 (Soluio) Le s. The C(s, ) E(X s X ) E(B 2 s s)(b 2 ) E(B 2 s s)([b B s + B s ] 2 ) E(B 2 s s)(b B s ) E(B 2 s s)b s (B B s ) + E(B 2 s s)b 2 s E(B 2 s s) (B) E(B 2 s s) E(B B s ) E(B 2 s s)b s E(B B s ) + E(B 2 s s)b 2 s E(B 2 s s) ( s) + 2 E(B 2 s s)b s + E B 4 s s E B 2 s 2s 2 2(s 2 2 ) 2(s ) 2. Problem 2.8 (Soluio) a) We have for s, m() E X e α/2 E B e α. C(s, ) E(X s X ) e α 2 (s+) E B e αsb e α e α 2 (s+) (e αs e α ) e α 2 s. b) We have P(X( ) x,..., X( ) x ) P (B(e α ) e α /2 x,..., B(e α ) e α/2 x ) Thus, he desiy is f X( ),...,X( )(x,..., x ) k e α k/2 f B(e α ),...,B(e α)(e α /2 x,..., e α/2 x ) 6

17 k e αk/2 (2π) /2 ( k (2π) /2 ( ( e α( k k ) )) k /2 (e α k e α k )) (we use he coveio ad x ). Soluio Maual. Las updae Jue 2, 27 e 2 k (eα k /2 x k e α k /2 x k ) 2 /(e α k e α k ) /2 e 2 k (x k e α( k k )/2 x k ) 2 /( e α( k k ) ) Remark: he form of he desiy shows ha he Orsei Uhlebeck is sricly saioary, i.e. (X( + h),..., X( + h) (X( ),..., X( )) h >. Problem 2.9 (Soluio) Assume ha we have (B). Observe ha he family of ses σ(b u,..., B u ) u u s, is a -sable family. This meas ha i is eough o show ha By (B) we kow ha B B s (B u,..., B u ) for all s. B B s (B u, B u2 B u,..., B u B u ) ad so... B u... B u2 B u B B s... B u3 B u2... B u B u B u B u2 B u3 B u Le < 2 <... < <,. The we fid for all ξ,..., ξ R d E (e i k ξ k, B( k ) B( k ) ) E (e i ξ, B() B( ) e i ) F mble., hece B( ) B( ) k ξ k, B( k ) B( k ) E (e i ξ, B() B( ) ) E (e i k ξ k, B( k ) B( k ) ) This shows (B). k E (e i ξ k, B( k ) B( k ) ). Problem 2. (Soluio) Reflecio ivariace of BM, cf. 2.8, shows τ a if{s B s a} if{s B s a} if{s B s a} τ a. The scalig propery 2.2 of BM shows for c /a 2 τ a if{s B s a} if{s ab s/a 2 a} if{a 2 r ab r a} a 2 if{r B r } a 2 τ. 7

18 R.L. Schillig, L. Parzsch: Browia Moio Problem 2. (Soluio) a) No saioary: E W 2 C(, ) E(B 2 ) 2 E(B 4 2B ) cos. b) Saioary. We have E X ad E X s X e α(+s)/2 E B e αsb e α e α(+s)/2 (e αs e α ) e α s /2, i.e. i is saioary wih g(r) e α r /2. c) Saioary. We have E Y. Le s. The we use E B s B s o ge E Y s Y E(B s+h B s )(B +h B ) E B s+h B +h E B s+h B E B s B +h + E B s B (s + h) ( + h) (s + h) s ( + h) + s, if > s + h h < s (s + h) (s + h) h ( s), if s + h h s. Swappig he roles of s ad fially gives: he process is saioary wih g() (h ) + (h ). d) No saioary. Noe ha E Z 2 E Be 2 e cos. Problem 2.2 (Soluio) Clearly, W is coiuous for. If we ge ad his proves coiuiy for. lim W (ω) W (ω) B (ω) lim W (ω) B (ω) lim β / (ω) β (ω) B (ω); Le us check ha W is a Gaussia process wih E W ad E W s W s. Corollary 2.7, W is a BM. By Pick ad < <... <. Case : If, here is ohig o show sice (B ) [,] is a BM. Case 2: Assume ha >. The we have W W 2 W 2 3 B β / β / β 8

19 Soluio Maual. Las updae Jue 2, 27 ad sice B (β /,..., β /, β ) are boh Gaussia, we see ha (W,..., W ) is Gaussia. Furher, le ad i < j : E W E B + E β / E β E W i W j E(B + i β /i β )(B + j β /j β ) + i j j i i j j + i i j. Case 3: Assume ha < <... < k < k+ <... <. The we have W W 2 W k W k+ k+2 B B k B β /k+ β / β. Sice (B,..., B k, B ) (β /k+,..., β /, β ) are Gaussia vecors, (W,..., W ) is also Gaussia ad we fid E W E W i W j E B i (B + j β /j β ) i i j for i k < j. Problem 2.3 (Soluio) The process X() B(e ) has o memory sice (cf. Problem 2.9) σ(b(s) s e a ) σ(b(s) B(e a ) s e a ) ad, herefore, σ(x() a) σ(b(s) s e a ) σ(b(e a+s ) B(e a ) s ) σ(x( + a) X(a) ). The process X() e /2 B(e ) is o memoryless. For example, X(a + a) X(a) is o idepede of X(a): E(X(2a) X(a))X(a) E (e a B(e 2a ) e a/2 B(e a ))e a/2 B(e a ) e 3a/2 e a e a e a. 9

20 R.L. Schillig, L. Parzsch: Browia Moio Problem 2.4 (Soluio) The process W B a B a, a clearly saisfies (B) ad (B4). For s a we fid W W s B a B a s B a s B a B s N(, ( s) id) ad his shows (B2) ad (B3). For < <... < a we have W j W j B a j B a j B a j B a j j ad his proves ha W iheris (B) from B. Problem 2.5 (Soluio) We kow from Paragraph 2.3 ha Moreover, B(s) lim B(/) lim a.s. s s E ( B(s) 2 ) s s s 2 s i.e. we ge also covergece i mea square. s Remark: a direc proof of he SLLN is a bi more ricky. Of course we have by he classical SLLN ha B (B j B j ) SLLN a.s. Bu he we have o make sure ha B s /s coverges. This ca be doe i he followig way: fix s >. The here is a uique ierval (, + ] such ha s (, + ]. Thus, B s s B s B + + B + s + + sup s + B s B s B ad we have o show ha he expressio wih he sup eds o zero. This ca be doe by showig, e.g., ha he L 2 -limi of his expressio goes o zero (usig he reflecio priciple) ad wih a subsequece argume. Problem 2.6 (Soluio) Se Clearly, Σ σ(b() J) J [,), J couable σ(b ) Σ σ(b ) def F B (*) The firs iclusio follows from he fac ha each B is measurable wih respec o Σ. Le us show ha Σ is a σ-algebra. Obviously, Σ ad F Σ F c Σ. 2

21 Soluio Maual. Las updae Jue 2, 27 Le (A ) Σ. The, for every here is a couable se J such ha A σ(b() J ). Sice J J is sill couable we see ha A σ(b() J) for all. Sice he laer family is a σ-algebra, we fid A σ(b() J) Σ. Sice σ(b ) Σ, we ge oe: F B is, by defiiio, he smalles σ-algebra for which all B are measurable ha This shows ha Σ F B. F B Σ. Problem 2.7 (Soluio) Assume ha he idices,..., m ad s,..., s are give. Le {u,..., u p } {s,..., s } {,..., m }. By assumpio, (X(u ),..., X(u p )) (Y (u ),..., Y (u p )). Thus, we may hi ou he idices o each side wihou edagerig idepedece: {s,..., s } {u,..., u p } ad {,..., m } {u,..., u p }, ad so (X(s ),..., X(s )) (Y ( ),..., Y ( m )). Problem 2.8 (Soluio) Sice F F ad G G i is clear ha F G F G. Coversely, sice (F ) ad (G ) are filraios we fid F F, G G, F F, G G. By assumpio: P(F G) P(F ) P(G). Thus, F G. Sice he families F ad G are -sable (use agai he argume ha we have filraios o fid for F, F F some wih F, F F ec.), he σ-algebras geeraed by hese families are idepede: F σ ( F ) σ ( G ) G. Problem 2.9 (Soluio) Le U R d d be a orhogoal marix: UU id ad se X UB for a BM d (B ). The E exp i ξ j, X( j ) X( j ) E exp i ξ j, UB( j ) UB( j ) 2

22 R.L. Schillig, L. Parzsch: Browia Moio E exp i U ξ j, B( j ) B( j ) exp 2 ( j j ) U ξ j, U ξ j exp 2 ( j j ) ξ j 2. (Observe U ξ j, U ξ j UU ξ j, ξ j ξ j, ξ j ξ j 2 ). The claim follows. Problem 2.2 (Soluio) Noe ha he coordiae processes b ad β are idepede BM. a) Sice b β, he process W (b + β )/ 2 is a Gaussia process wih coiuous sample pahs. We deermie is mea ad covariace fucios: E W 2 (E b + E β ) ; Cov(W s, W ) E(W s W ) 2 E(b s + β s )(b + β ) 2 ( E b sb + E β s b + E b s β + E β s β ) (s s ) s 2 where we used ha, by idepedece, E b u β v E b u E β v. Now he claim follows from Corollary 2.7. b) The process X (W, β ) has he followig properies W ad β are BM E(W b ) 2 /2 E(b + β )β 2 /2 (E b E β + E β 2 ) / 2, i.e. W ad β are NOT idepede. This meas ha X is o a BM 2, as is coordiaes are o idepede. The process Y ca be wrie as b + β 2 b β U b β b 2 β. Clearly, UU id, i.e. Problem 2.9 shows ha (Y ) is a BM 2. Problem 2.2 (Soluio) Observe ha b β sice B is a BM 2. Sice E X Cov(X, X s ) E X X s E(λb s + µβ s )(λb + µβ ) λ 2 E b s b + λµ E b s β + λµ E b β s + µ 2 β s β 22

23 Soluio Maual. Las updae Jue 2, 27 λ 2 E b s b + λµ E b s E β + λµ E b E β s + µ 2 E β s β λ 2 (s ) µ 2 s (λ 2 + µ 2 )(s ). Thus, by Corollary 2.7, X is a BM if, ad oly if, λ 2 + µ 2. Problem 2.22 (Soluio) X (b, β s β ), s, is NOT a Browia moio: X (, β s ) (, ). O he oher had, Y (b, β s β s ), s, IS a Browia moio, sice b ad β s β s are idepede BM, cf. Time iversio 2. ad Theorem 2.6. Problem 2.23 (Soluio) We have W UB cos α si α si α b cos α β. The marix U is a roaio, hece orhogoal ad we see from Problem 2.9 ha W is a Browia moio. Geeralizaio: ake U orhogoal. Problem 2.24 (Soluio) If G N(, Q) he Q is he covariace marix, i.e. Cov(G j, G k ) q jk. Thus, we ge for s < The characerisic fucio is Cov(X j s, X k ) E(X j sx k ) E X j s(x k X k s ) + E(X j sx k s ) E X j s E(X k X k s ) + sq jk (s )q jk. E e i ξ,x E e i Σ ξ,b e 2 Σ ξ 2 e 2 ξ,σσ ξ, ad he rasiio probabiliy is, if Q is o-degeerae, f Q (x) (2π) de Q exp ( x, Qx ). 2 If Q is degeerae, here is a orhogoal marix U R such ha UX (Y,..., Y k,,..., k ) where k < is he rak of Q. disribuio i R k. The k-dimesioal vecor has a odegeerae ormal 23

24

25 3 Cosrucios of Browia Moio Problem 3. (Soluio) The parial sums W N (, ω) N G (ω)s (), [, ], coverge as N P-a.s. uiformly for owards B(, ω), [, ] cf. Problem 3.2. Therefore, he radom variables W N() d N G S () d P -a.s. X N B() d. This shows ha W N() d is he sum of idepede N(, )-radom variables, hece iself ormal ad so is is limi X. From he defiiio of he Schauder fucios (cf. Figure 3.2) we fid ad his shows S () d 2 S 2 j +k() d j, k,,..., 2 j, j. W 2 +() d 2 G + 4 j 2 j j G 2 j +l. l Cosequely, sice he G j are iid N(, ) radom variables, This meas ha E W 2 +() d, V W 2 +() d X 2 G + j j 2 3j l j 2 2j j (+) j 2 j l G 2 j +l N(,2 j ) where he series coverges P-a.s. ad i mea square, ad X N(, 3 ). 25

26 R.L. Schillig, L. Parzsch: Browia Moio Problem 3.2 (Soluio) [, ], we fid a) From he defiiio of he Schauder fucios S (),, S (), S 2 j +k() S 2 j +k((2k + )/2 j+ ) 2 j/2 /2 j+ 2 2 j/2 j, k, 2 j S 2 j +k() k 2 2 j/2 (disjoi suppors!) By assumpio, C >, ɛ (, 2 ), a C ɛ. Thus, we fid a S () a + j a + j a + (The series is coverge sice ɛ < /2). j 2 j a 2 j +k S 2 j +k() k 2 j C (2 j+ ) ɛ S 2 j +k() k C 2 (j+)ɛ 2 2 j <. This shows ha a S () coverges absoluely ad uiformly for [, ]. b) For C > 2 we fid from P ( G > log ) 2 π C log e 2 C2 log 2 π C C2 /2 3 ha he followig series coverges: P ( G > log ) <. By he Borel Caelli Lemma we fid ha G (ω) O( log ) for almos all ω, hus G (ω) O( ɛ ) for ay ɛ (, /2). From par a) we kow ha he series G (ω)s () coverges a.s. uiformly for [, ]. Problem 3.3 (Soluio) Se f p ( E f p ) /p Soluio : We observe ha he space L p (Ω, A, P; S) {X X S, d(x, ) p < } is complee ad ha he codiio saed i he problem jus says ha (X ) is a Cauchy sequece i he space L p (Ω, A, P; S). A good referece for his is, for example, he moograph by F. Trèves [3, Chaper 46]. You will fid he pedesria approach as Soluio 2 below. 26

27 Soluio Maual. Las updae Jue 2, 27 Soluio 2: By assumpio k N k sup m N k d(x Nk, X m ) p 2 k. Wihou loss of geeraliy we ca assume ha N k N k+. I paricular d(x Nk, X Nk+ ) p 2 k l l>k d(x Nk, X Nl ) p jk 2 j 2 2 k. Fix m. The we see ha d(x Nk, X m ) d(x Nl, X m ) p d(x Nk, X Nl ) p k,l. This meas ha ha (d(x Nk, X m )) k is a Cauchy sequece i L p (P; R). By he compleeess of he space L p (P; R) here is some f m L p (P; R) such ha d(x Nk, X m ) ad, for a subsequece ( k ) (N k ) k we fid d(x k, X m ) The subsequece k may also deped o m. i Lp f m k almos surely f m. k Sice ( k (m)) k is sill a subsequece of (N k ), we sill have d(x k (m), X m+ ) f m+ i L p, hece we ca fid a subsequece ( k (m + )) k ( k (m)) k such ha d(x k (m+), X m+ ) f m+ a.s. Ieraig his we see ha we ca assume ha ( k ) k does o deped o m. I paricular, we have almos surely Moreover, ɛ > L L(ɛ) k L d(x k, X m ) f m ɛ. lim f m p lim lim d(x k, X m ) p lim lim d(x k, X m ) p m m k m k lim sup d(x k, X m ) p. m k k Thus, f m i L p ad, for a subsequece m k we ge Therefore, ɛ > K K(ɛ) r K f mr ɛ. d(x k, X l ) d(x k, X mr ) + d(x k, X mr ) d(x k, X mr ) f mr + d(x k, X mr ) f mr + 2 f mr. Fix ɛ > ad pick r > K. The le k, l. This gives d(x k, X l ) d(x k, X mr ) f mr + d(x k, X mr ) f mr + 2ɛ 4ɛ k, l L(ɛ). 27

28 R.L. Schillig, L. Parzsch: Browia Moio Sice S is complee, his proves ha (X k ) k coverges o some X S almos surely. Remark: If we replace he codiio of he Problem by higs become MUCH simpler: lim E (sup d p (X, X m )) m This codiio says ha he sequece d sup m d p (X, X m ) coverges i L p (P; R) o zero. Hece here is a subsequece ( k ) k such ha lim sup d(x k, X m ) k m k almos surely. This shows ha d(x k, X l ) as k, l, i.e. we fid by he compleeess of he space S ha X k X. Problem 3.4 (Soluio) Fix,... ad Borel ses A,..., A. By assumpio, we kow ha P(X Y ) P(X j Y j j,..., ) P {X j Y j }. Thus, P {X j A j } P {X j A j } {X j Y j } P {X j A j } {X j Y j } P {Y j A j } {X j Y j } P {Y j A j }. Problem 3.5 (Soluio) idisiguishable modificaio: P(X Y ) P(X Y ). modificaio equivale: see he previous Problem 3.4 Now assume ha I is couable or X, Y are (lef- or righ-)coiuous. modificaio idisiquishable: By assumpio, P(X Y ) for ay I. Le D I be ay couable dese subse. The P q D{X q Y q } P(X q Y q ) q D 28

29 Soluio Maual. Las updae Jue 2, 27 which meas ha P(X q Y q q D). If I is couable, we are doe. I he oher case we have, by he desiy of D, P(X Y I) P (lim D q X q lim D q Y q I) P (X q Y q q D). equivale / modificaio: To see his le (B ) ad (W ) be wo idepede oe-dimesioal Browia moios defied o he same probabiliy space. Clearly, hese processes have he same fiie-dimesioal disribuios, i.e. hey are equivale. O he oher had, for ay > P(B W ) P(B y) P(W dy) P(W dy). Problem 3.6 (Soluio) Sice (B q ) q Q [,) is uiformly coiuous, here exiss a uique process (B ) such ha B lim q B q ad B is coiuous. We use he characerizaio from Lemma 2.4. Is proof shows ha we ca derive (2.7) E exp i j, X qj X qj + i ξ, X q ξ exp 2 o he basis of (B) (B3) for (B q ) q Q [,) ad q,..., q Q [, ). ξ j 2 (q j q j ) Now se q ad pick,..., R ad approximae each j by a raioal sequece, k. Sice (2.7) holds for q (k), j,..., ad every k, we ca easily perform q (k) j j he limi k o boh sides (o he lef we use domiaed covergece!) sice B is coiuous. This proves (2.7) for (B ), ad sice (B ) has coiuous pahs, Lemma 2.4 proves ha (B ) is a BM. Problem 3.7 (Soluio) The joi desiy of (W ( ), W (), W ( )) is f,, (x, x, x ) (2π) 3/2 exp ( ( )( ) 2 [(x x) 2 + (x x ) 2 + x2 ]) while he joi desiy of (W ( ), W ( )) is f, (x, x ) (2π) exp ( ( ) 2 [(x x ) 2 + x2 ]). The codiioal desiy of W () give (W ( ), W ( )) is f, (x x, x 2 ) f,, (x, x, x ) f, (x, x ) (2π) 3/2 exp ( ( )( ) 2 [ (x x) 2 + (x x ) 2 + x2 ]) (2π) exp ( ( ) 2 [ (x x ) 2 + x2 ]) 29

30 R.L. Schillig, L. Parzsch: Browia Moio ( ) 2π ( )( ) exp ( 2 [(x x) 2 + (x x ) 2 (x x ) 2 ]) ( ) 2π ( )( ) exp ( 2 [( )(x x) 2 + ( )(x x ) 2 (x x ) 2 ]) ( )( ) Now cosider he argume i he square brackes [ ] of he exp-fucio [ ( )(x x) 2 + ( )(x x ) 2 ( )( ) (x x ) 2 ] ( ) ( )( ) [ (x x) 2 + (x x ) 2 ( )( ) ( ) 2 (x x ) 2 ] ( ) ( )( ) [ ( + ) x 2 + ( ( )( ) ( ) 2 ) x 2 + ( ( )( ) ( ) 2 ) x 2 2 x x 2 xx + 2 ( )( ) ( ) 2 x x ] ( ) ( )( ) [x2 + ( ) 2 ( ) 2 x2 + ( ) 2 ( ) 2 x2 2 x x 2 xx + 2 ( )( ) ( ) 2 x x ] ( ) ( )( ) [x x 2 x ] ( ) ( )( ) [x ( x + 2 x )]. Se σ 2 ( )( ) ( ) ad m x + x he our calculaio shows ha f, (x x, x 2 ) (x m)2 exp ( 2π σ 2σ 2 ). 3

31 4 The Caoical Model Problem 4. (Soluio) Le F R [, ] be a disribuio fucio. We begi wih a geeral lemma: F has a uique geeralized moooe icreasig righ-coiuous iverse: F (u) G(u) if{x F (x) > u} [ sup{x F (x) u}]. (4.) We have F (G(u)) u if F () is coiuous i G(u), oherwise, F (G(u)) u. Ideed: For hose where F is sricly icreasig ad coiuous, here is ohig o show. Le us look a he wo problem cases: F jumps ad F is fla. F () G(u) w + w w v u G(u) G(v ) G(v) G(w) u Figure 4.: A illusraio of he problem cases If F () jumps, we have G(w) G(w + ) G(w ) ad if F () is fla, we ake he righ edpoi of he flaess ierval [G(v ), G(v)] o defie G (his leads o righ-coiuiy of G) a) Le (Ω, A, P) ([, ], B[, ], du) (du sads for Lebesgue measure) ad defie X G (G F as before). The P({ω Ω X(ω) x}) λ({u [, ] G(u) x}) (he discoiuiies of F are couable, i.e. a Lebesgue ull se) 3

32 R.L. Schillig, L. Parzsch: Browia Moio λ({ [, ] F (x)}) λ([, F (x)]) F (x). Measurabiliy is clear because of moooiciy. b) Use he produc cosrucio ad par a). To be precise, we do he cosrucio for wo radom variables. Le X Ω R ad Y Ω R be wo iid copies. We defie o he produc space (Ω Ω, A A, P P ) he ew radom variables ξ(ω, ω ) X(ω) ad η(ω, ω ) Y (ω ). The we have ξ, η live o he same probabiliy space ξ X, η Y P P (ξ A) P P ({(ω, ω ) Ω Ω ξ(ω, ω ) A}) P P ({(ω, ω ) Ω Ω X(ω) A}) P P ({ω Ω X(ω) A} Ω ) P({ω Ω X(ω) A}) P(X A). ad a similar argume works for η. ξ η P P (ξ A, η B) P P ({(ω, ω ) Ω Ω ξ(ω, ω ) A, η(ω, ω ) B}) P P ({(ω, ω ) Ω Ω X(ω) A, Y (ω ) B}) P P ({ω Ω X(ω) A} {ω Ω Y (ω ) B}) P({ω Ω X(ω) A}) P ({ω Ω Y (ω ) B}) P(X A) P(Y B) P P (ξ A) P P (η B) The same ype of argume works for arbirary producs, sice idepedece is always defied for ay fiie-dimesioal subfamily. I he ifiie case, we have o ivoke he heorem o he exisece of ifiie produc measures (which are cosruced via heir fiie margials) ad which ca be see as a paricular case of Kolmogorov s heorem, cf. Theorem 4.8 ad Theorem A.2 i he appedix. c) The saemes are he same if oe uses he same cosrucio as above. A difficuly is o ideify a mulidimesioal disribuio fucio F (x). Roughly speakig, hese are fucios of he form F (x) P (X (, x ] (, x ]) where X (X,..., X ) ad x (x,..., x ), i.e. x is he upper righ edpoi of a ifiie recacle. A absrac characerisaio is he followig 32

33 Soluio Maual. Las updae Jue 2, 27 F R [, ] x j F (x) is moooe icreasig x j F (x) is righ coiuous F (x) if a leas oe ery x j F (x) if all eries x j + ( ) k ɛ k F (ɛ a +( ɛ )b,..., ɛ a +( ɛ )b ) where < a j < b j < ad where he ouer sum rus over all uples (ɛ,..., ɛ ) {, } The las propery is equivale o () h () h F (x) h,..., h where (k) h F (x) F (x + he k) F (x) ad e k is he kh sadard ui vecor of R. I priciple we ca cosruc such a mulidimesioal F from is margials usig he heory of copulas, i paricular, Sklar s heorem ec. ec. ec. Aoher way would be o ake (Ω, A, P) (R, B(R ), µ) where µ is he probabiliy measure iduced by F (x). The he radom variables X are jus he ideiy maps! The idepede copies are he obaied by he usual produc cosrucio. Problem 4.2 (Soluio) Sep : Le us firs show ha P(lim s X s exiss) <. Sice X r X s ad X s X s we ge Thus, X r X s X r + X s N(, s + r) s + r N(, ). P( X r X s > ɛ) P ( X > ɛ ) P ( X > ɛ ). s + r r,s 2 This proves ha X s is o a Cauchy sequece i probabiliy, i. e. i does o eve coverge i probabiliy owards a limi, so a.e. covergece is impossible. I fac we have {ω lim X s (ω) does o exis} { sup X s X r > } s k s,r [ /k,+/k] ad so we fid wih he above calculaio P ( lim X s does o exis) lim P ( sup X s X r > ) P ( X > ɛ ) s k s,r [ /k,+/k] 2 This shows, i paricular ha for ay sequece we have P ( lim X exiss) < q <. where q q() (bu idepede of he sequece). Sep 2: Fix >, fix a sequece ( ) wih, ad se A {ω Ω lim s X s (ω) exiss} ad A( ) {ω Ω lim X (ω) exiss}. 33

34 R.L. Schillig, L. Parzsch: Browia Moio Clearly, A A( ) for ay such sequece. Moreover, ake wo sequeces (s ), ( ) such ha s ad ad which have o pois i commo; he we ge by idepedece ad sep (X s, X s2, X s3...) (X, X 2, X 3...) A( ) A(s ) ad so, P(A) P(A(s ) A( )) P(A(s )) P(A( )) q 2. By Sep, q <. Sice here are ifiiely may sequeces havig all o pois i commo, we ge P(A) lim k q k. 34

35 5 Browia Moio as a Marigale Problem 5. (Soluio) a) We have F B σ(σ(x), F B ) σ(x, B s s ) F. Le s. The σ(b B s ), F B s ad σ(x) are idepede, hus σ(b B s ) is idepede of σ(σ(x), F B ) F. This shows ha F B (B ). b) Se N {N M A such ha N M, P(M) }. The we have F B σ(f B, N) F B. is a admissible filraio for From measure heory we kow ha (Ω, A, P) ca be compleed o (Ω, A, P ) where We fid for all A B(R d ), F F s, N N A {A N A A, N N}, P (A ) P(A) for A A N A. P ({B B s A} (F N)) P (({B B s A} F ) ({B B s A} N) ) A N P({B B s A} F ) P(B B s A) P(F ) P (B B s A) P (F N). Therefore F B is admissible. Problem 5.2 (Soluio) Le <... <, ad cosider he radom variables B( ) B( ),..., B( ) B( ). Usig he argume of Problem 2.9 we see for ay F F E (e i k ξ k, B( k ) B( k ) F ) E (e i ξ, B() B( ) e i k ξ k, B( k ) B( k ) F ) F mble., hece B( ) B( ) E (e i ξ, B() B( ) ) E (e i k ξ k, B( k ) B( k ) F ) 35

36 R.L. Schillig, L. Parzsch: Browia Moio k E (e i ξ k, B( k ) B( k ) ) E F. This shows ha he icremes are idepede amog hemselves (use F Ω) ad ha hey are all ogeher idepede of F (use he above calculaio ad he fac ha he icremes are amog hemselves idepede o combie agai he expeced value) uder he Thus, F σ(b( k ) B( k ) k,..., ) Therefore he saeme is implied by F < <...< σ(b( k ) B() k,..., ). Problem 5.3 (Soluio) a) i) E X <, sice he expecaio does o deped o he filraio. ii) X is F measurable ad F F. Thus X is F measurable. iii) Le N deoe he se of all ses which are subses of P-ull ses. Deoe by P he measure of he compleio of (Ω, A, P) (compare wih he soluio o Exercise 5..b)). Le s. For all F F s here exis F F s, N N such ha F F N ad F X s d P F X s d P F X d P F X d P. Sice F is arbirary his implies ha E(X F s ) X s. b) i) E Y E X <. ii) Noe ha {X Y }, is compleme ad ay of is subses is i F. Le B B(R d ). The we ge {Y B} ({X B} {X Y } ) {Y B, X Y }. F F F iii) Similar o par a-iii). For each F F s we ge i.e. E(Y F s ) Y s. F Y s d P F X s d P a) F X d P F Y d P, Problem 5.4 (Soluio) Le s < ad pick s s such ha s < s <. The E(X F s+ ) sub-mg E(X() F s ) X(s ) s s c a.e. coiuous X(s+) X(s). pahs The covergece o he lef side follows from he (sub-)marigale covergece heorem (Lévy s dowward heorem). 36

37 Soluio Maual. Las updae Jue 2, 27 Problem 5.5 (Soluio) Here is a direc proof wihou usig he hi. We sar wih calculaig he codiioal expecaios E(B 4 F s) E ((B B s + B s ) 4 F s ) Bs 4 + 4Bs 3 E(B B s ) + 6Bs 2 E((B B s ) 2 ) + 4B s E((B B s ) 3 ) + E((B B s ) 4 ) Bs 4 + 6Bs 2 ( s) + 3( s) 2 Bs 4 6Bs 2 s + 6Bs 2 + 3( s) 2, ad E(B 2 F s) E ((B B s + B s ) 2 F s ) s + 2B s E(B B s ) + Bs 2 Bs 2 + s. Combiig hese calculaios, such ha he erm 6Bs 2 vaishes from he firs formula, we ge E (B 4 6B 2 F s) Bs 4 6sBs s s + 3s 2 Bs 4 6sB s + 3s Therefore π(, B ) B 4 6B is a marigale. Problem 5.6 (Soluio) For ad all c we have E e c B E e c B 2. ad for c E e c B ad E e c B 2. Now le > ad c >. There exiss some R > such ha c x < 4 x 2 for all x > R. Thus E e c B c e c x e 2 x 2 dx c e c x e 2 x 2 dx + c e 4 x 2 e 2 x 2 dx x R x >R e cr + c e 4 x 2 dx <, x >R i.e., E e c B < for all c,. Furhermore E e c B 2 c e c x 2 2 x 2 dx c e x 2 (c 2 ) dx ad his iegral is fiie if, ad oly if, c 2 < or equivalely c < 2. 37

38 R.L. Schillig, L. Parzsch: Browia Moio Problem 5.7 (Soluio) a) We have p(, x) (2π) d 2 e x 2 2. By he chai rule we ge p(, x) d d 2 2 (2π) d x 2 e (2π) d 2 ( ) 2 ( ) x 2 x 2 2 e 2 ad for all j,..., d p(, x) (2π) d 2x j 2 ( x j 2 ) x 2 e 2, 2 x 2 j p(, x) (2π) d 2 ( ) x 2 e 2 + (2π) d 2 x 2 j x 2 2 e 2. Addig hese erms ad oig ha x 2 d x2 j we ge 2 d 2 x 2 j b) A formal calculaio yields p(, x) d 2 (2π) d 2 e x (2π) d 2 p(, x) 2 2 x 2 j f(, x) dx p(, x) f(, x) 2 x j x j p(, x) 2 f(, x) 2 x 2 p(, x) f(, x) dx. j 2 2 x 2 2 x 2 e 2 p(, x). p(, x) f(, x) dx x j 2 x j + 2 x 2 p(, x) f(, x) dx j 2 By he same argumes as i Exercise 5.6 we fid ha all erms are iegrable ad vaish as x. This jusifies he above calculaio. Furhermore summig over j,... d we obai he saeme. Problem 5.8 (Soluio) Measurabiliy (i.e. adapedess o he Filraio F ) ad iegrabiliy is o issue, see also Problem 5.6. a) U is oly a marigale for c. Soluio : see Exercise 5.9. Soluio 2: if c, E U is o cosa, i.e. cao be a marigale. If c, U is rivially a marigale. b) V is a marigale sice E (V F s ) E(B B s ) + B s E ( B r dr F s ) E ( B r dr F s ) s s B s B r dr E ( (B r B s ) + B s dr F s ) s B s B r dr ( s)b s V s. s s 38

39 Soluio Maual. Las updae Jue 2, 27 c) ad e) Le a R. The we ge E (ab 3 B F s ) E (a(b B s + B s ) 3 (B B s ) B s F s ) abs 3 + 3aBs 2 E B s + 3aB s E B s 2 + a E B s 3 B s abs 3 + (3a( s) )B s. This is a marigale if, ad oly if, s 3a( s), i.e., a 3. marigale ad W is o a marigale. Thus Y is a d) We have see i par c) ad b) ha ad Thus, X is a marigale. E (B 3 F s) B 3 s + 3( s)b s 3 E ( B r dr F s ) 3 B r dr + 3( s)b s. f) Z is oly a marigale for c 2, see Exercise 5.9. Problem 5.9 (Soluio) Noe ha E X < for all a, b, cf. Problem 5.6. We have E (e ab+b F s ) E (e a(b Bs) e abs+b F s ) e abs+b E e ab s e abs+b+( s)a2 /2. Thus, X is a marigale if, ad oly if, bs b + ( s) a2 2, i.e., b 2 a2. Problem 5. (Soluio) We have E ( d B 2 F s ) + d d E ((B (j) ) 2 Pr. 5.5 F s ) + d s d ((B (j) s ) 2 + s) d B s 2 s. Problem 5. (Soluio) For a) c) we prove oly he saemes for τ, he saemes for τ are proved aalogously. a) The followig implicaios hold: A C { X A} { X C} τ A τ C. b) By par a) we have τ A C τ A ad τ A C τ C. Thus, τ A C a) mi{τ A, τ C}. To see he coverse, mi{τa, τ C } τ A C, i is eough o show ha X (ω) A C mi{τ A(ω), τ C(ω)} sice his implicaio shows ha τa C (ω) mi{τ A (ω), τ C (ω)} holds. 39

40 R.L. Schillig, L. Parzsch: Browia Moio Now observe ha X (ω) A C X (ω) A or X (ω) C τ A(ω) or τ C(ω) mi{τ A(ω), τ C(ω)}. c) Par a) implies max{τa, τ C } τ A C. Remark: we cao expec. To see his cosider a BM sarig a B ad he se A [4, 6] ad C [, 2] [5, 7]. The B has o reach firs C ad A before i his A C. d) as i b) i is clear ha τa τ A for all, hece τ A if τ A. I order o show he coverse, τ A if τ A, i is eough o check ha X (ω) A if τ A (ω) sice, if his is rue, his implies ha τ A (ω) if τ A (ω). Now observe ha X (ω) A A X (ω) A for some N τ A (ω) for some N if τ A (ω). e) Noe ha if {s X s+. Thus we ge A} if {s X s A} is moooe decreasig as if ( + if{s X s A}) + if if{s X s A} if{s > X s A} τ A. f) Le X x +. The τ{x } ad τ {x }. More geerally, a similar siuaio may happe if we cosider a process wih coiuous pahs, a closed se F, ad if we le he process sar o he boudary F. The τf a.s. (sice he process is i he se) while τ F > is possible wih posiive probabiliy. 4

41 Soluio Maual. Las updae Jue 2, 27 Problem 5.2 (Soluio) We have τ U τ U. Le x U. The τ U ad, sice U is ope ad X is coiuous, here exiss a N > such ha Thus τ U. X U for all N. If x U, he X (ω) U ca oly happe if >. Thus, τ U τ U. Problem 5.3 (Soluio) Suppose d(x, A) d(z, A). The d(x, A) d(z, A) if x y if z y y A y A if ( x z + z y ) if z y y A y A x z ad, wih a aalogous argume for d(x, A) d(z, A), we coclude d(x, A) d(z, A) x z. Thus x d(x, A) is globally Lipschiz coiuous, hece uiformly coiuous. Problem 5.4 (Soluio) We rea he wo cases simulaeously ad check he hree properies of a sigma algebra: i) We have Ω F ad Ω {τ } {τ } F F +. ii) Le A F τ(+). Thus A F, A c F ad A c {τ } Ω A {τ } (Ω {τ } ) (A {τ } ) F (+). F F + iii) Le A F τ(+). The A, A F ad Therefore F τ ad F τ+ are σ-algebras. A {τ } F (+) sice A F τ(+) (A {τ } ) F (+). F (+) Problem 5.5 (Soluio) a) Le F F τ+, i.e., F F ad for all s we have F {τ s} F s+. Le >. The F {τ < } (F {τ s}) F s+ F. s< s< For he coverse: If τ < a.s. he F > (F {τ }) F ad F {τ s} (F {τ < }) F F s+. >s >s If τ occurs wih sricly posiive probabiliy, he we have o assume ha F F. 4

42 R.L. Schillig, L. Parzsch: Browia Moio b) We have {τ } F F ad {τ } F if r ; {τ } {τ r} {τ r} F r F if r <. Problem 5.6 (Soluio) a) e iξb+ 2 ξ 2 is a marigale for all ξ R by Example 5.2 d). By opioal soppig Sice he lef-had side is real, we ge E e 2 (τ )c2 +icb τ. E (e 2 (τ )c2 cos(cb τ )). Se m a b. Sice B τ m, we see ha for mc < 2 π he cosie is posiive. By Faou s lemma we ge for all mc < 2 π lim E (e 2 (τ )c2 cos(cb τ )) E ( lim e 2 (τ )c2 cos(cb τ )) E (e 2 τc2 cos(cb τ )) cos(mc) E e 2 τc2. Thus, E e γτ < for ay γ < 2 c2 ad all c < π/(2m). Sice e j j j!, j e j j! we see ha E τ j j! γ j E e γτ < for ay j. b) By Exercise 5.8 d) he process B 3 3 B s ds is a marigale. By opioal soppig we ge E (B 3 τ 3 τ B s ds) for all. (*) Se m max{a, b}. By he defiiio of τ we see ha B τ m; sice τ is iegrable we ge B 3 τ m 3 ad τ B s ds τ m. Therefore, we ca use i (*) he domiaed covergece heorem ad le : E ( τ B s ds) 3 E(B3 τ ) 3 ( a)3 P(B τ a) + 3 b3 P(B τ b) (5.2) a 3 b + b 3 a 3 a + b ab(b a). 3 42

43 Soluio Maual. Las updae Jue 2, 27 Problem 5.7 (Soluio) By Example 5.2 c) B 2 d is a marigale. Thus we ge by opioal soppig E( τ R ) d E B τ R 2 for all. Sice B τr R, we ca use moooe covergece o he lef ad domiaed covergece o he righ-had side o ge E τ R sup E( τ R ) lim d E B τ R 2 d E B τ R 2 d R2. Problem 5.8 (Soluio) a) For all we have {σ τ } {σ } {τ } F. F F b) For all we have {σ < τ} {σ τ } ({σ r < τ} {σ τ }) r Q (({σ r} {τ r} c ) {σ τ }) F. r Q [,] This shows ha {σ < τ}, {σ τ} {σ < τ} c F σ τ. Sice σ ad τ play symmeric roles, we ge wih a similar argume ha {σ > τ}, {σ τ} {σ > τ} c F σ τ, ad he claim follows. c) Sice τ σ is a iegrable soppig ime, we ge from Wald s ideiies, Theorem 5., ha Followig he hi we ge E B 2 τ σ E(τ σ) <. E(B σ B τ {σ τ} ) E(B σ τ B τ {σ τ} ) E ( E(B σ τ B τ {σ τ} F τ σ )) b) E (B σ τ {σ τ} E (B τ F τ σ )) (*) E(B 2 σ τ {σ τ} ). (We will discuss he sep marked by (*) below.) Wih a aalogous calculaio for τ σ we coclude E(B σ B τ ) E(B σ τ B τ {σ<τ} ) + E(B σ τ B τ {τ σ} ) E(B 2 σ τ ) E σ τ. I he sep marked wih (*) we used ha for iegrable soppig imes σ, τ we have E(B τ F σ τ ) B σ τ. To see his we use opioal soppig which gives E(B τ k F σ τ k ) B σ τ k for all k. 43

44 R.L. Schillig, L. Parzsch: Browia Moio This is he same as o say ha F B τ k d P F B σ τ k d P for all k, F F σ τ k. Sice B τ k B τ i L 2 (P), see he proof of Theorem 5., we ge for some fixed k i < k because of F σ τ i F σ τ k ha F B τ d P lim k F B τ k d P lim k F B σ τ k d P F B σ τ d P for all F F σ τ i. Le ρ σ τ (or ay oher soppig ime). Sice F ρ k F ρ F k we see ha F ρ is geeraed by he -sable geeraor i F ρ i, ad (*) follows. d) From he above ad Wald s ideiy we ge E( B τ B σ 2 ) E(B 2 τ 2B τ B σ + B 2 σ) E τ 2 E τ σ + E σ E(τ 2(τ σ) + σ) E τ σ. I he las sep we used he elemeary relaio (a + b) 2(a b) a b + a b 2(a b) a b a b a b. 44

45 6 Browia Moio as a Markov Process Problem 6. (Soluio) We wrie g (x) (2π) /2 e x2 /(2) for he oe-dimesioal ormal desiy. a) This follows immediaely from our proof of b). b) Le u B b (R) ad s,. The, by he idepede ad saioary icremes propery of a Browia moio E u( B +s F s ) E u( (B +s B s ) + B s F s ) Sice B B we also ge E u( (B +s B s ) + y ) ybs E u( B + y ) ybs. ad, herefore, E u( B +s F s ) E u( B + y ) y Bs E u( B y ) ybs E u( B +s F s ) 2 [E u( B + y ) + E u( B y )] ybs 2 [ (u( z + y ) + u( z y )) g (z) dz]yb s 2 [ u( z ) (g (z + y) + g (z y)) dz]yb s here we use ha he iegrad is eve i z u( z ) (g (z + y) + g (z y)) dz yb s u( z ) (g (z + y ) + g (z y )) dz g u,s,+s(y) i is idepede of s! sice he iegrad is also eve i y! This shows ha E u( B +s F s ) is a fucio of B s, i.e. Markoviaiy. P y ( B dz) g (z y) + g (z + y) for z, y, i.e. he form of he rasiio fucio. Remark: B is called reflecig (also: refleced) Browia moio. yb s 45

46 R.L. Schillig, L. Parzsch: Browia Moio c) Se M sup s B s for he ruig maximum, i.e. Y M B. From he reflecio priciple, Theorem 6.9 we kow ha Y B. So he guess is ha Y ad B are wo Markov processes wih he same rasiio fucio! Le s, ad u B b (R). We have by he idepede ad saioary icremes propery of Browia moio E (u(y +s ) F s ) E (u(m +s B +s ) F s ) E (u( max { sup B r, sup B s+u } B +s ) F s ) u s u E (u( max { sup(b r B s ) + (B s B +s ), sup (B s+u B s+ )}) F s ) u s u ad, as sup u s (B r B s ) is F s measurable ad (B s B +s ), sup u (B s+u B s+ ) F s, we ge E (u( max {y + (B s B +s ), sup (B s+u B s+ )})) u ysup u s (B r B s) E (u( max {y B, sup (B u B )})) u yy s Usig ime iversio (cf. 2.) we see ha (B u B ) u [,] is agai a BM, ad we ge (B, sup u (B u B )) ( B, sup u B u )) E (u( max {y + B, sup B u )})). u yy s Usig Soluio 2 of Problem 6.8 we kow he joi disribuio of (B, sup u B u ): E (u( max {y + B, sup B u )})) z z u u(max{y + x, z}) 2 2z x e (2z x)2 /2 dx dz. x 2π Spliig he iegral z x io wo pars z x,y+x z + z x,y+x>z we ge I u(z) 2 z y 2z x e (2z x)2 /2 2 dx dz z 2π x u(z) /2 2π z e (z+y)2 dz e (2z x)2 /2 z y ad 2 z II 2z x u(y + x) e (2z x)2 /2 dx dz 2π z x z y 2 u(y + x) x+y 2z x e (2z x)2 /2 dz 2π x y zx /2 x+y 2 e (2z x)2 dx u(y + x) /2 2π x y [e x2 e (x+2y)2 /2 ] dx zx 46

47 Soluio Maual. Las updae Jue 2, 27 u(ξ) /2 2π x y [e (ξ y)2 e (ξ+y)2 /2 ] dξ. Fially, addig I ad II we ed up wih E (u( max {y + B, sup B u )})) u which is he same rasiio fucio as i par b). d) See par c). u(z)(g (z + y) + g (z y)) dz, y Problem 6.2 (Soluio) Le s,. We use he followig abbreviaios: s I s B r dr ad M s sup B u ad F s Fs B. u s a) Le f R 2 R measurable ad bouded. The E (f(m s+, B s+ ) F s ) E (f( sup B u M s, (B s+ B s ) + B s ) F s ) s u s+ E (f([b s + sup (B u B s )] M s, (B s+ B s ) + B s ) F s ). s u s+ By he idepede icremes propery of BM we ge ha he radom variables sup s u s+ (B u B s ), B s+ B s F s while M s ad B s are F s measurable. Thus, we ca rea hese groups of radom variables separaely (see, e.g., Lemma A.3: where E (f(m s+, B s+ ) F s ) E (f([z + sup (B u B s )] y, (B s+ B s ) + z) F s ) s u s+ ym s,zb s φ(m s, B s ) φ(y, z) E (f([z + sup s u s+ b) Le f R 2 R measurable ad bouded. The E (f(i s+, B s+ ) F s ) E (f( E (f( s+ s s+ s (B u B s )] y, (B s+ B s ) + z) F s ). B u du + I s, (B s+ B s ) + B s ) F s ) (B u B s ) du + I s + B s, (B s+ B s ) + B s ) F s ). By he idepede icremes propery of BM we ge ha he radom variables s+ s (B u B s ) du, B s+ B s F s while I s + B s ad B s are F s measurable. Thus, we ca rea hese groups of radom variables separaely (see, e.g., Lemma A.3: E (f(i s+, B s+ ) F s ) 47

48 R.L. Schillig, L. Parzsch: Browia Moio for he fucio E (f( s φ(i s, B s ) s+ φ(y, z) E (f( (B u B s ) du + y + z, (B s+ B s ) + z)) yi s,zb s s s+ (B u B s ) du + y + z, (B s+ B s ) + z)). c) No! If we use he calculaio of a) ad b) for he fucio f(y, z) g(y), i.e. oly depedig o M or I, respecively, we see ha we sill ge E (g(i +s ) F s ) ψ(b s, I s ), i.e. (I, F ) cao be a Markov process. The same argume applies o (M, F ). Problem 6.3 (Soluio) We follow he hi. Firs, if f R d R, f f(x,..., x ), x,..., x R d, we see ha E x f(b( )),..., B( )) E f(b( )) + x,..., B( ) + x) f(y R d R d + x,..., y + x) P(B( ) dy,..., B( ) dy ) imes ad he las expressio is clearly measurable. This applies, i paricular, o f A j where G {B( j) A j }, i.e. E x G is Borel measurable. Se Γ {B( j ) A j }, <, A,... A B b (R d ). Le us see ha Σ is a Dyki sysem. Clearly, Σ. If A Σ, he x E x A c E x ( A ) E x A B b (R d ) A c Σ. Fially, if (A j ) j Σ are disjoi ad A j A j we ge A j Aj. Thus, x E x A E x Aj B b (R d ). j This shows ha Σ is a Dyki Sysem. Deoe by δ( ) he Dyki sysem geeraed by he argume. The Γ Σ F B δ(γ) δ(σ) Σ F B. Bu δ(γ) σ(γ) sice Γ is sable uder fiie iersecios ad σ(γ) F B. This proves, i paricular, ha Σ F B. Sice we ca approximae every bouded F B measurable fucio Z by sep fucios wih seps from F B, he claim follows. 48

49 Soluio Maual. Las updae Jue 2, 27 Problem 6.4 (Soluio) Followig he hi we se u (x) ( ) x. The u (x) u(x) x. Usig (6.7) we see Now ake o ge ad we ge E [u (B +τ ) F τ+ ](ω) E Bτ (ω) u (B ). E [u (B τ ) F τ+ ](ω) u (B τ )(ω) lim E [u (B τ ) F τ+ ](ω) lim u (B τ )(ω) B τ (ω). Sice he l.h.s. is F τ+ measurable (as limi of such measurable fucios!), he claim follows. Problem 6.5 (Soluio) By he reflecio priciple, Theorem 6.9, P (sup s B s x) P (sup B s x) + P (if B s x) P( B x) + P( B x). s s Problem 6.6 (Soluio) a) Sice B( ) B( ), we ge τ b if{s B s b} if{s B s b} if{s B s b} τ b. b) Sice B(c 2 ) c B( ), we ge τ cb if{s B s cb} if{s c B s b} if{s B s/c 2 b} if{rc 2 B r b} c 2 if{r B r b} c 2 τ b. c) We have τ b τ a if{s B s+τa b} if{s B s+τa B τa b a} which shows ha τ b τ a is idepede of F τa Browia moio. by he srog Markov propery of Now we fid for all s, ad c [, a] {τ c s} {τ a } τ c τ a {τ c s } {τ a } F s F F. This shows ha {τ c s} F τa, i.e. τ c is F τa measurable. Sice c is arbirary, {τ c } c [,a] is F τa measurable, ad he claim follows. Problem 6.7 (Soluio) We begi wih a simpler siuaio. As usual, we wrie τ b for he firs passage ime of he level b: τ b if{ sup s B s b} where b >. From Example 5.2 d) we kow ha (M ξ exp(ξb 2 ξ2 )) is a marigale. By opioal soppig we ge 49

50 R.L. Schillig, L. Parzsch: Browia Moio ha (M ξ ) τ is also a marigale ad has, herefore, cosa expecaio. Thus, for b ξ > (ad wih E E ) E M ξ E ( exp(ξb τ b 2 ( τ b)ξ 2 )) Sice he RV exp(ξb τb ) is bouded (mid: ξ ad B τb b), we ca le ad ge E ( exp(ξb τb 2 τ bξ 2 )) E ( exp(ξb 2 τ bξ 2 )) or, if we ake ξ 2λ, E e λτ b e 2λb. As B B, τ b τ b, ad he above calculaio yields E e λτ b e 2λ b b R. Now le us ur o he siuaio of he problem. Se τ τ (a,b) c. Here, B τ is bouded (i is i he ierval (a, b), ad his makes higs easier whe i comes o opioal soppig. As before, we ge by soppig he marigale (M ξ ) ha e ξx lim E x ( exp(ξb τ 2 ( τ)ξ2 )) E x ( exp(ξb τ 2 τξ2 )) (ad o, as before, for posiive ξ! Mid also he sarig poi x, bu his does o chage higs dramaically.) by, e.g., domiaed covergece. The problem is ow ha B τ does o aai a paricular value as i may be a or b. We ge, herefore, for all ξ R e ξx E x ( exp(ξb τ 2 τξ2 ) {Bτ a}) + E x ( exp(ξb τ 2 τξ2 ) {Bτ b}) E x ( exp(ξa 2 τξ2 ) {Bτ a}) + E x ( exp(ξb 2 τξ2 ) {Bτ b}) Now pick ξ ± 2λ. This yields 2 equaios i wo ukows: e 2λ x e 2λ a E x (e λτ {Bτ a}) + e 2λ b E x (e λτ {Bτ b}) e 2λ x e 2λ a E x (e λτ {Bτ a}) + e 2λ b E x (e λτ {Bτ b}) Solvig his sysem of equaios gives e 2λ (x a) E x (e λτ {Bτ a}) + e 2λ (b a) E x (e λτ {Bτ b}) e 2λ (x a) E x (e λτ {Bτ a}) + e 2λ (b a) E x (e λτ {Bτ b}) ξ ad so E x (e λτ {Bτ b}) sih ( 2λ (x a)) sih ( 2λ (b a)) This aswers Problem b). ad E x (e λτ {Bτ a}) sih ( 2λ (b x)) sih ( 2λ (b a)). For he soluio of Problem a) we oly have o add hese wo expressios: E e λτ E (e λτ {Bτ a}) + E (e λτ {Bτ b}) sih ( 2λ (b x)) + sih ( 2λ (x a)) sih (. 2λ (b a)) 5