FICHE DE TRAVAUX DIRIGÉS NO 1

Transcription

1 Universié de Poiiers Déparemen de Mahémaiques 3M21 Processus à emps coninu Maser MMAS 2, année FICHE DE TAVAUX DIIGÉS NO 1 Il a éé suggéré qu une parie des enseignemens de maser s effecue en anglais. Sans sombrer dans le ridicule, on peu inroduire un cerain mulilinguisme à ravers la documenaion ou le maériel pédagogique. À ire d essai, cerains énoncés d exercices, peu-êre ous, de ces fiches de ravaux dirigés seron écris en anglais. Un bénéfice immédia e indiscuable sera l acquisiion des ermes usuels de la discipline dans la langue de Shakespeare. Oherwise saed,, n, and more generally opological spaces, are equipped wih heir Borel measurable srucure. So, he Borel σ-algebras B(), B( n ), B(E), will be usually omied in noaions. General saes spaces will be denoed by (E, E), (F, F), ec. Exercice 1. Le (, A, P) be a probabiliy space. Suppose ha X : is a funcion which assumes only counably many values a 1, a 2,.... (i) Show ha X is a random variable if and only if X 1 {a k } A for all k = 1, 2,... ( ) (ii) Suppose ha he former propery ( ) holds. Show ha E[ X ] = a k P{X = a k }. (iii) If ( ) holds and E[ X ] <, show ha E[X] = a k P{X = a k }. (iv) If ( ) holds and f : is a bounded Borel measurable funcion, show ha E[f(X)] = f(a k ) P{X = a k }. Answer. (i) The condiion ( ) is necessary: since X 1 (B) A for every Borel measurable se B B(), i is also he case for every B = {a k }. Conversely, if ( ) holds, hen for every B, especially B B(), one has X 1 (B) = X 1 {a k : a k B} = X 1 ( a k B {a k}) = a k B X 1 {a k } A since i is a counable union of A-measurable ses. (ii) The ses ({X = a k }) form a counable measurable pariion of, hus 1 = 1 = 1 {X=a k} = k 1 {X=a k }. Then i is a simple use of Fubini Tonelli s heorem o he nonnegaive measurable funcion (ω, k) N 1 {X=ak }(ω) X(ω) inegraed wih respec o he σ-finie produc measure P ( δ {k}) : E[ X ] = X(ω) P(dω) = 1 X(ω) P(dω) = 1 {X=ak } X(ω) P(dω) = 1 {X=ak } a k P(dω) = k a k 1 {X=ak } P(dω) = a k P{X = a k }.

2 2 Coninuous Time Processes. Fiche de ravaux dirigés n o 1 This resul is a such usual corollary of Fubini Tonelli s heorem, i is invoked by saying ha expecacion is addiive wih respec o counable sums of non-negaive random variables. (iii) When one of he hree (double or muliple) inegrals is finie, all of hem are so and one can remove safely absolue values: ha s Fubini s heorem. Here, E[ X ] < if and only if a k P{X = a k } < and hen E[X] = a k P{X = a k }. (iv) Consider f : a bounded Borel measurable funcion, or more generally any bounded funcion (since we consider in fac f {ak :1 k< } : {a k : 1 k < } here is no real measubiliy condiion). Since f is bounded E[ f(x) ] is finie, hen Once again, by Fubini s heorem applied o he non-negaive measurable funcion (ω, k) N 1 {X=ak }(ω)f(x(ω)) inegraed wih respec o he σ-finie produc measure P ( δ {k}), we have n E[f(X)] = f(a k ) P{X = a k }. emark. The resuls of he previous exercice are obvious when X akes only finiely many values. Exercice 2. Le X : (, A, P) be a random variable. The disribuion funcion F X of he law or disribuion of X is defined by F X (x) = P{X x}, x. (i) Prove ha F X has he following properies: a) 0 F X 1, lim x F X (x) = 0, lim x + F X (x) = 1 ; b) F X is increasing, i.e. non decreasing ; c) F X is righ coninuous, i.e. F X (x) = lim h 0 F X (x + h). (ii) Le g : be a Borel mesurable funcion such ha E[ g(x) ] <. Prove ha E[g(X)] = + g(x) df X (x), where he inegral on he righ is inerpreed in he Lebesgue Sieljes sense. (iii) Le p : + be a non negaive measurable funcion on. We say ha he law of X admis he densiy p if F X (x) = x p(y) dy for all x. We know ha he law a ime of a 1-dimensional Brownian moion sared from 0, B, has he densiy p(x) = 1 2π e x2 /2, x. Find he densiy of B 2. Answer. (i) a) Obvious, use coninuiy properies of probabliy measures along monoone sequences of measurable ses. b) Obvious, probabiliy measures are monoone. c) Once again, his is he coninuiy propery of probabiliy measures along decreasing measurable ses.

3 (ii) By definiion + Maser MMAS 2 3 g(x) df X (x) = + g(x)µ X (dx) which is equal o E[g(X)] by ransfer heorem. More precisely, df X mus be seen as he unique probabiliy measure whose F X is he disribuion funcion, ha is o say µ X. (iii) The random variable B 2 is non-negaive, i disribuion funcion is 0 over and for x 0, F B 2 (x) = P{B 2 x} = P 0 {B [ x, x]} = 2 P 0 {B x} 1 = 2F B ( x) 1. This shows ha F B 2 is quie regular over + wih derivaive p(x) = 1 x F B ( x) = 1 2πx e x/2 Thus, p(x) = 1 + (x) 2πx e x/2, x is a densiy of he law of B 2 when he iniial value is 0 (i is a bi more complicaed when B 0 0 since he symmery argumen used in compuaion is no longer rue). Exercice 3. Le (H i ) i I be a family of σ-algebras on. Prove ha H = H i = { A : A H i for all i I } i I is again a σ-algebra on. Answer. This is absoluely obvious. I is always funny o look a he case where I is an empy se. Wha should be H in his case? Since he infimum of he empy se in an ordered se should he he naural supremum, if any, of his se, one should ake H = P(). Exercice 4. (i) Le X : (, A, P) be a random variable such ha Prove he Markov( Chebychev) inequaliy E [ X p] < for some 0 < p <. P{ X λ} 1 λ p E[ X p] for all λ 0. (Hin. We have X p dp A X p dp, where A = { X λ}.) (ii) Suppose ha here exiss k > 0 such ha M = E[exp(k X )] <. Prove ha P{ X λ} M e kλ for all λ 0. Answer. (i) For λ 0, we have P{ X > λ} = P{ X p > λ p } = { X p >λ p } { X p >λ p } 1 P(dω) X p λ p P(dω) Noe ha we allow λ o be 0 since here is no 0/0 case in his compuaion. (ii) One has P{ X > λ} = P{exp(k X ) > exp(kλ)} X p λ p P(dω) = 1 λ p E[ X p]. 1 E[exp(k X )] M exp( kλ), exp(kλ)

4 4 Coninuous Time Processes. Fiche de ravaux dirigés n o 1 bry he former inequaliy. Exercice 5. Le X, Y : (, A, P) be wo independen random variables and assume for simpliciy ha X and Y are bounded. Prove ha E[XY ] = E[X] E[Y ]. (Hin. Assume ha X M, Y N. Approximae X and Y by simple random variables X m (ω) = m i=1 x i1 Ai (ω), Y m (ω) = n j=1 y j1 Bj (ω) respecively, where A i = X 1 ([x i, x i+1 [), B j = Y 1 ([y j, y j+1 [), M = x 0 < x 1 < < x m 1 < x m = M, N = y 0 < y 1 < < y n 1 < y n = N. Then E[X] E[X m ] = x i P(A i ), E[Y ] E[Y n ] = y i P(B i ) and and so on...) 0 i m E[XY ] E[X m Y n ] = 0 i m 0 j n x i y j P(A i B j ) 0 j n Answer. We skip he reducion of he from inegrable variables o bounded ones: replace X by 1 { X M} X and Y by 1 { Y N} Y which are sill independen variables and whose expecaions are close o X and Y ones for M and N sufficienly large. The case of he producs may be a lile more delicae, so, forge abou i. Discreizaion allows ha X X m ε and Y Y m ε. Then E[X] E[X m ] ε, E[Y ] E[Y n ] ε, and E[X] E[Y ] E[X m ] E[Y n ] = E[X] E[Y ] E[X] E[Y n ] + E[X] E[Y n ] E[X m ] E[Y n ] Moreover, E[X] E[Y ] E[X] E[Y n ] + E[X] E[Y n ] E[X m ] E[Y n ] Mε + Nε = (M + N)ε. XY X m Y m = XY XY m + XY m X m Y m XY XY m + XY m X m Y m Nε + Mε = (M + N)ε and E[XY ] E[X m Y n ] (M + N)ε. Finally, since E[X m Y n ] = E[X m ] E[Y n ], E[XY ] E[X] E[Y ] = E[XY ] E[X m Y n ] + E[X m Y n ] E[X] E[Y ] E[XY ] E[X m Y n ] + E[X m Y n ] E[X] E[Y ] E[XY ] E[X m Y n ] + E[X m ] E[Y n ] E[X] E[Y ] (M + N)ε + (M + N)ε = 2(M + N)ε. Thus E[XY ] E[X] E[Y ] is arbirarily close oo zero, so E[XY ] = E[X] E[Y ]. Exercice 6. Le (, A, P) be a probabiliy space and le A 1, A 2,... be ses in A such ha P(A n ) <. Prove he weak sense of he Borel Canelli lemma: ( P A k ), = 0, n=1 k=n ha is o say: he se of all ω such ha ω belongs o infiniely many A k s has probabiliy zero.

5 n=1 k=n k=n Maser MMAS 2 5 Answer. This is obvious: for every N 1, on has ( ) ( ) A k A k and P A k P A k n=1 k=n k=n P(A k ). The erm on he righ being he remainder of a convergen series, i ends o zero as N. This also proves ha if P ( n=1 k=n A k) > 0, hen n=1 P(A n) =. Le us see he srong sens of he Borel Canelli lemma. Suppose moreover ha (A n ) n 1 are independen evens. Then, if n P(A n) =, P ( n=1 k=n A k) = 1 (i is also a sricly posiive probabiliy). For 1 n N, ( ) ( ) P A m = 1 P = 1 P(A c m) = 1 ( 1 P(Am ) ). n m N n m N Since 1 x e x for every x 0, ( P n m N A c m n m N ) ( A m 1 exp N m=n ) P(A m ). k=n n m N By hypohesis, when N goes o infiniy, N m=n P(A m) ends, for every n, o infiniy and hen ( ) P A m = 1 n m for every n, hen he conclusion follows by decreasing limi. Exercice 7. Le be non empy se. (i) Suppose ha A 1,..., A m are disjoins subse of such ha = m A k. Prove ha he family A consising of all unions of some, none, or all A 1,..., A m is a σ-algebra on. (ii) Prove ha any finie σ-algebra A on is of he ype described in he former quesion. (iii) Le A be a finie σ-algebra on and le X : be a A-measurable funcion. Prove ha X assumes only finiely many possible values. More precisely, here exiss a disjoin family of subses B 1,..., B n A and real numbers x 1,..., x m such ha m X = x i 1 Bi. i=1 Answer. (i) Le (A k ) m be a finie pariion of he se and le A = { k K A k : K {1,..., m}. We shall prove ha A is a σ-algebra on : a) aking K = {1,..., m}, one has A (and aking K =, one has ); b) if A = k K A k A, hen ( A c = \ A = which is in A; k {1,...,m} ) ( ) A k \ A k = k K k {1,...,m}\K A k = c) if B l = k K l A k A, l = 1, 2,..., hen B l = ( ) A k = A k, wih K = l K l, l l k K l k K which is in A. k K c A k

6 6 Coninuous Time Processes. Fiche de ravaux dirigés n o 1 (ii) Le A a finie σ-algebra on. For ω, ω we se ω ω if and only if, for every A A, ω A implies ω A. I is easy o prove ha : {rue, false} is an equivalence relaion (reflexiviy, symmery, ransiiviy). Is equivalence classes A 1,..., A m form a pariion of and, moreover, each A k is in A: for any ω A k, ω = A k = A A,ω A A. Furhermore, if A A and ω A hen ω = A k A, hus A is a (finie) union of A k. (iii) Le P X = {X 1 {x} : x }. Since X is A/B()-measurable, P X A and is finie. This shows ha here is only finiely many x such ha X 1 {x} =, i.e. X assumes only finiely many possible values. Le x 1,..., x n be hese values, B i = X 1 {x i }, and hen X = n x i 1 Bi. i=1 One should noice ha, since X is consan over each A k, m X = x k1 Ak where x k is he value aken by X on A k. emark. The las quesion can be seed wih X : (, A) (E, E) where (E, E) is a measurable space such ha all singleons belong o E, bu one would have o wrie anew he las formula since he sum may have no meaning if E is no a vecor space. Moreover, one can ake B i = A i if here is no need for x 1,..., x n o be disincs. Exercice 8. Le B = (B ) 0 be a Brownian moion on wih B 0 = 0, and pu E = E 0. (i) Show ha E [ e iθx ] = exp( θ 2 /2) for all θ. (Hin. Compue he characerisic funcion of he sandard normal disribuion N (0, 1) by solving an ordinary differenial equaion and deduce he resul from his compuaion.) (ii) Use he power series expansion of he exponenial funcion on boh sides, compare he erms wih he same power of θ and deduce ha E [ B 4 ] = 3 2 and more generally ha E [ B 2k ] (2k)! = 2 k k! k, k N. (iii) If you feel uneasy abou he lack of rigour in he mehod in (ii), you can proceed as follows: we have E 0 [f(b )] = 1 f(x) e x2 /2 dx 2π for all funcion f such ha he inegral on he righ converges ; apply his o f(x) = x 2k and use inegraion by pars and inducion on k. (iv) Suppose ha B is a n-dimensional Brownian moion. Prove ha E x[ B B s 4] = n(n + 2) s 2 by using (ii) and inducion on n. Answer. (i) Le Z : (, A, P) a random variable wih sandard normal disribuion. Is characerisic funcion is defined by φ(θ) = E [ e iθz] = e iθx e x2 /2 dx, θ. 2π

7 We can differeniae φ o obain φ (θ) = ix e iθx e x2 /2 dx = 2π = i Maser MMAS 2 7 ix e iθx e x2 /2 dx = i 2π e iθx d ( e x 2 /2 ) [ dx e iθx ] e x2 /2 + = i dx 2π 2π θ e iθx( x e x2 /2 ) dx 2π e iθx e x2 /2 dx = θφ(θ) 2π The differenial equaion φ (θ) = θφ(θ), φ(0) = 1 admis a unique soluion on. We check ha θ exp( θ 2 /2) is a soluion, hence he soluion: φ(θ) = exp( θ 2 /2) for all θ. Le s consider B. We know ha B / has sandard normal disribuion, hen (ii) On one side ϕ B (θ) = E [ e iθb ] = E [ e i(θ )(B / ) ] = φ ( θ ) = exp( θ 2 /2), θ. on he oher side, E [ e iθb ] [ = E n=0 exp( θ 2 /2) = (iθb ) n ] = n! n=0 (iθ) n E[B n ], n! n=0 ( θ 2 ) n 2 n. n! By idenifying he differen powers, one has for every n 0 E[B 2n+1 ] = 0 and E[B 2n in paricular, for n = 2, E[B 4 ] = 24/4/2 2 = 3 2. ] = (2n)! 2 n n! n, (iii) Noe ha B has N (0, ) disribuion when he iniial value is 0. Then call M n () = E[B n ]. One has, for n 1, M n () = x n e x2 /2 dx = x n 1( x/ e x2 /2 ) dx 2π 2π [ x n 1 ] e x2 /2 = + (n 1) x n 2 e x2 /2 dx 2π 2π Then, and = 0 + (n 1)M n 2 (). M 2n () = (2n 1)(2n 3) 3 1 n M 0 () = 2n(2n 1)(2n 2)(2n 3) n n(n 1) 2 1 n M 0 () = (2n)! 2 n n! n M 2n+1 () = (2n)(2n 2) 4 2 n M 1 () = 2 n n! n M 1 () = 0 since M 0 () = 1 and M 1 () = 0. (iv) Se τ = s. Whaever is he iniial value, we know ha Z = B B s = (Z 1,..., Z n ) has N (0, τ Id n ) n-dimensional normal disribuion: Z i are independen and each Z i has N (0, τ) normal disribuion. Then Z 2 = Z Z 2 n and Z 4 = n Zi i=1 1 i<j n Z 2 i Z 2 j,

8 8 Coninuous Time Processes. Fiche de ravaux dirigés n o 1 hen, by lineariy and independance, n E[ Z 4 ] = E [ Zi 4 ] + 2 E [ Zi 2 ] [ ] E Z 2 j i=1 1 i<j n = n E [ Z1 4 ] n(n 1) + 2 E [ Z ] = 3nτ 2 + n(n 1)τ 2 = n(n + 2)τ 2. Exercice 9. To illusrae ha he (finie-dimensional) disribuions alone do no give all he informaion regarding he coninuiy properies of a process, consider he following example: Le (, A, P) = ( +, B( + ), µ) where µ is a probabiliy measure on + wih no mass on single poins. Define and X (ω) = { 1 if = ω 0 oherwise Y (ω) = 0 for all (, ω) + +. Prove ha x = (X ) 0 and y = (Y ) 0 have he same (finie-dimensional) disribuions and ha X is a version of Y. And ye we have ha Y (ω) is coninuous for all ω, while X (ω) is disconinuous for all ω. Answer. For every 0, P{X = 1} = P{ω : X (ω) = 1} = P{ω : ω = } = µ{} = 0. Thus, since all possible values of X are {0, 1}, for every 0 1 < 2 < < n, P{X 1 = 0,..., X n = 0} = 1 which is he same as Y. This shows ha X and Y have he same finie dimensional disribuions. We recall ha X and Y are versions of each ohers if and only if P{ω : X (ω) = Y (ω)} = 1 for all 0. Here for a given 0, we have P{ω : X (ω) Y (ω)} = µ{} = 0. Thus X and Y are also versions of each ohers. Exercice 10. A sochasic process X = (X ) 0 is called saionary if X has he same disribuion as (X +h ) 0 for all h 0. Prove ha he Brownian moion B = (B ) 0 has saionary incremens, i.e. ha he process (B +h B ) h 0 has he same disribuion for all 0. Answer. Immediae. Use ( ) for insance. More generally, if X is a homogeneous Markov process wih values in whose ransiion funcion P saisfies P (x, B) = µ (B x) where µ is a probabiliy measure on, hen X as saionary incremens. Exercice 11. Prove ha, if B = (B (1),..., B (n) ) is a n-dimensional Brownian moion wih independen iniial coordinaes (B (1) 0,..., B(n) 0 ), hen he 1-dimensional processes B (i) = ) 0, 1 i n, are independen 1-dimensional Brownian moions. (B (i) Answer. Il suffi de regarder la caracérisaion en erme de processus à accroissemens indépendans gaussiens e de voir que oues les variables réelles obenues son indépendanes e on la loi qui convien. (C es rop lourd pour ere ecri. Passer par les foncions caracérisiques? Ça laisse les éudians rès scepiques. Exercice 12. Le B = (B ) 0 be a Brownian moion and fix 0 0. Prove ha is a Brownian moion. B = B 0 + B 0, 0, Answer. Il suffi de regarder la caracérisaion en erme de processus à accroissemens indépendans gaussiens. Exercice 13. Le B = (B ) 0 be a 2-dimensional Brownian moion and pu D ϱ = {x 2 : x < ϱ} for ϱ > 0.

9 Compue P 0 {B D ϱ }. Answer. We have P 0 {B D ϱ } = D ϱ e ((x 2 +y 2) /2 dx dy 2π Maser MMAS 2 9 = ϱ 2π 0 0 r 2 e r /2 dr = [ e r2 /2 ] ϱ 0 = 1 /2. e ϱ2 Exercice 14. Le B = (B ) 0 be a n-dimensional Brownian moion and le K n have zero n-dimensional Lebesgue measure. Prove ha he expeced oal ime spen in K by B is zero. (Poenial heoriical remark. This implies ha he Green measure associaed wih B is absoluely coninuous wih respec o he Lebesgue measure.) Answer. By Fubini s heorem, we have [ ] E x 1 {B K} d = E x[ 1 {B K}] d 0 0 = 1 {y K} e y x 0 n 2 /2 dy d = (2π) n/2 0 0 d = 0. This resul simply use he fac fac he law of each B, > 0, is absoluely coninuous wih respec o he Lebesgue measure. Many oher processes share he same propery (uniform moion on, compensaed Poisson process,...). Exercice 15. Le B = (B ) 0 be a n-dimensional Brownian moion sared from 0 and le O n n be a (consan) orhogonal marix, i.e. OO = Id n. Prove ha is also a Brownian moion. B = OB, 0, Answer. C es clairemen un processus à accroissemens indépendans. Ces accroissemens son des images par une applicaion linéaire de veceurs gaussiens, ce son donc des veceurs gaussiens. Leur veceur moyen es nul comme il se doi e leur marice de covariance es inchangée (il suffi de l écrire). Exercice 16 (Brownian scaling propery). Le B = (B ) 0 be a 1-dimensional Brownian moion and le c > 0 be a consan. Prove ha is also a Brownian moion. B = 1 c B c 2, 0, Answer. C es encore un processus à accroissemens indépendans. Il suffi de déerminer la loi des accroissemens, ce qui es presque immédia. Exercice 17 (à revoir!). If X : (, A, P), 0, is a coninuous sochasic process, hen, for p > 0, he p h variaion process of X, X, X (p) is defined by X, X (p) = lim Xk+1 X k p, k 0 0= 1 < 2 < < n = where k = k+1 k, as a limi for he convergence in probabiliy aken over all subdivisions of [0, ]. In paricular, if p = 1 his process is called he oal variaion process of X, and if p = 2 his is called he quadraic variaion process. For 1-dimensional brownian moion B we now prove ha he quadraic variaion process is simply B, B (ω) = B, B (2) (ω) = almos surely. Proceed as follows:

10 10 Coninuous Time Processes. Fiche de ravaux dirigés n o 1 (i) Define B k = B k+1 B k Show ha and pu Y (ω) = E [ (Y ) 2] = 2 0= 1 < 2 < < n = ( ) 2. Bk (ω) 0= 1 < 2 < < n = and deduce ha Y in L 2 (, A, P) as k 0. ( k ) 2 (ii) Use (i) o prove ha almos surely pahs of Brownian moions do no have a bounded variaion on [0, ], i.e. he oal variaion of Brownian moion is infinie almos surely. Exercice 18. Le = {1, 2, 3, 4, 5, } and U be he collecion U = { {1, 2, 3, }, {3, 4, 5} } of subses of. Find he smalles σ-algebra conaining U (i.e. he σ-algebra σ(u) generaed by U). (i) Define X : by Is X measurable wih respec o σ(u)? (ii) Define Y : by X(1) = X(2) = 0, X(3) = 10, X(4) = X(5) = 1. Find he σ-algebra σ(y ) generaed by Y. Y (1) = 1, Y (2) = Y (3)Y (4) = Y (5) = 1. Exercice 19. Le (, A, P) be a probabiliy space and le p [1, ]. A sequence (f n ) n=1 of funcions in L p (P) = L p (, A, P; ) is called a Cauchy sequence if f n f m p as m, n. The sequence is called convergen if here exiss f L p (P) such ha f n f in L p (P). Prove ha every convergen sequence is a Cauchy sequence. A fundamenal heorem in measure heory saes ha he converse is also rue: every Cauchy sequence in L p (P) is convergen. A normed linear space wih his propery is called complee. Thus, he L p (P) spaces are complee. Exercice 20. Le B be a 1 dimensional Brownian moion, σ a consan and 0 s. Use ( ) o prove ha E [ exp ( σ(b s B ) )] = exp ( 1 2 σ2 (s ) ).