Solutios to HW: 1 Course: Theory of Probability II Page: 1 of 6 Uiversity of Texas at Austi Solutios to HW Assigmet 1 Problem 1.1. Let Ω, F, {F } 0, P) be a filtered probability space ad T a stoppig time. 1) Show that F T := {A F A {T = } F ) } is a sigma-field. 2) If {X } is a adapted process, the the process {X T } is also adapted, ad the radom variable X T I {T < } is F T -measurable. Solutio: 1) this item is just a easy applicatio of the defiitio of sigma-algebra. 2) Let B ay Borel measurable set of R or, i geeral, ay measurable set of the state space). The we have {X T B} = 1 k=0 I additio for each, so by the defiitio of F T. ) {T = k} {X k B} ) {T } {X B} F. {X T I {T < } B} {T = } = {X B} {T = } F {X T I {T < } B} F T Problem 1.2. Let 0 S T two stoppig times with respect to the discrete-time filtratio F ) 0. Let A F S. Show that the radom time T defied by is also a stoppig time. Solutio: We have that sice {T = } = T = S 1 A + T 1 A c ) {S = } A {S } {T = } A c) F {S } A c F. Problem 1.3. 1) Doob s decompositio). Let {X, F } be a submartigale. Show that it ca be decomposed uiquely as X = M + A, = 0, 1,... where {M, F } is a martigale ad the process {A } is icreasig, predictable with respect to the filtratio {F } ad A 0 = 0. Fid the processes M ad A i terms of the process X. 2) Let X = k=1 I B for B F. What is the Doob decompositio for X? Solutio: 1. Use the decompositio relatio for + 1 ad ad substract term by term to get X +1 X = M +1 M + A +1 A ) Takig coditioal expectatio with respect to F ad usig the martigale property for M ad the fact that A is predictable we obtai Now E[X +1 X F ] = A +1 A. A = A 0 + A 1 A 0 ) + + A A 1 ) = E[X 1 X 0 F 0 ] + + E[X X 1 F 1 ].
Solutios to HW: 1 Course: Theory of Probability II Page: 2 of 6 We ca also see that ad sice M 0 = X 0 we have M +1 M = X +1 X A +1 A ) = X +1 E[X +1 F ] M = X 0 + X 1 E[X 1 F 0 ]) + + X E[X F 1 ]). Sow far we have oly showed uiqueess, but existece of this decompositio ca be easily prove takig these relatio as defiitios for M ad A. 2. Sice X X 1 = 1 B ad E[1 B F 1 ] = pb F 1 ), we coclude that A = pb i F i 1 ) ad i=1 M = X A = 1 Bi pb i F i 1 )) Problem 1.4. Let Y 1, Y 2,... be o-egative i.i.d. with E[Y ] = 1 ad PY = 1) < 1. 1) Show that X = k=1 Y k is a martigale 2) Use the martigale covergece theorem to coclude that X 0 a.s. 3) Use the SLLN to show that i=1 logx ) c < 0, a.s. Solutio: 1. E[X +1 F ] = E[X Y +1 F ] = X E[Y +1 F ] = X, sice Y +1 is idepedet o the sigma algebra F geerated by X 1,..., X 3. we have that logx ) = logy 1 ) + + logy ). The hypotheses esures that log Y ) + L 1 by Jese) ad E[log Y ] < 0 it CAN be ). If the expectatio is fiite, the we ca apply the SLLN. If the expectatio is, we ca TRUNCATE log Y at a egative level C ad the let C to coclude that logy 1 ) + + logy ) E[log Y ] [, 0). 2. Sice logx )/ c < 0 the X 0. Oe ca prove this part directly, usig martigale covergece theorem for X, cocludig that X X 0. Now, if X > 0 the Y = X /X 1 1, ad we ca obtai a cotradictio. Problem 1.5. A example of a Uiformly Itegrable Martigale which is ot i H 1 ). Cosider Ω = N, F = 2 N ad the probability measure P defied by P[k] = 2 k, k = 1, 2.... Cosider ow the radom variable K such that Kk) = k, k = 1, 2.... 1) fid a explicit example of a radom variable Y : Ω [1, ) such that E[Y ] < ad E[Y K] =. 2) cosider the filtratio F = σ {1}, {2},..., { 1}, {, + 1,... }). Fid a expressio for X = E[Y F ]. 3) Show that the martigale X ) is ot i H 1 ote that the martigale is UI by the way it is defied). Solutio: 1) We ca defie Y k) = 2 k 1 k 2.
Solutios to HW: 1 Course: Theory of Probability II Page: 3 of 6 2) It is clear the formal proof is ot very short though!) that X k) = Y k), k = 1, 2,... 1 ad X ) = X + 1) = = 2 1 Y ) + 2 2 Y + 1) +... 3) we see that X ) X ), so ) E[X )] 2 2 k 1 2 k 1 k 2 = 1 1 2 k 2 =. =1 k= =1 k= Problem 1.6. 1) Suppose X ) is a submartigale such that E[ X +1 X F ] B a.s for each where B is a costat. Show that if N is a stoppig time such that E[N] < the X N ) is u.i., so, accordig to the optioal samplig theorem E[X 0 ] E[X N ] 2) Wald I): Let ξ 1, ξ 2,... be i.i.d such that ξ i L 1. If S = ξ 1 +... ξ ad N is a stoppig time such that E[N] < the E[S N ] = E[N]E[ξ] Please ote that this implies that the first hittig time at level oe for the symmetric radom walk has INFINITE expectatio. Solutio: 1. sup X N X 0 + X 1 X 0 +... X N X N 1 = X 0 + =0 X +1 X 1 {N>}. Takig ito accout that {N > } F, we ca coditio each term o time to coclude that E[sup X N ] E[X 0 ] + E[E[ X +1 X F ]1 {N>} ] E[X 0 ] + B =0 PN > ) = E[X 0 ] + PN = 0) + E[N] <. =0 This meas that the sequece X N ) is domiated by a 2. Defie the process M = S E[ξ]. We have M +1 M = ξ +1 E[ξ +1 ], so E[M +1 M F ] = 0, ad E[ M +1 M F ] E[ ξ E[ξ] ] This meas that M is a martigale ad we ca apply Part 1 to get E[M N ] = 0 which meas E[S N ] = E[N]E[ξ]. Problem 1.7. Quadratic variatio of square itegrable martigales). Cosider a square itegrable martigale, by which we mea that X 0 = 0 ad X L 2 for each. We ca defie the square bracket or the quadratic variatio) of the martigale by [X, X] = X k X k 1 ) 2. k=1 At the same time, we defie the agle bracket, or the predictable quadratic variatio process X, X as the predictable icreasig process i the Doob decompositio X 2 = M + X, X. Show that 1) X, X = k=1 E[X k X k 1 ) 2 F k 1 ] ad the predictable process X, X is the compesator i the Doob decompositio of [X, X]. 2) let T be a stoppig time possibly udbouded ad ifiite). Show that E[ X, X T ] = E[[X, X] T ]. ote that the radom variables above are actually well defied)
Solutios to HW: 1 Course: Theory of Probability II Page: 4 of 6 3) show that E[ sup X k 2 ] 4E[ X, X T ] = 4E[[X, X] T ]. 0 k T 4) assume that E[ X, X T ] <. The the processes X T 2 [X, X] T, = 0, 1, 2..., X T 2 X, X T, = 0, 1, 2..., are uiformly itegrable martigales. I particular, this meas that X is well defied o the set {T = } for such a stoppig time T ad E[X 2 T ] = E[ X, X T ] = E[[X, X] ]. 5) show that X = lim X exists o ad is fiite o { X, X < }. Solutio: 1) Usig the fudametal property of martigales, we ca easily check that X 2 [X, X] is a martigale. Therefore, the predictable compesator of X 2 i the Doob decompositio) is the same as the predictable compesator of [X, X]. Writig ow the explicit represetatio for the predictable compesator of [X, X] we get the aswer. 2) accordig to the item above, [X, X] X, X is a martigale. I particular, for each fixed we have E[ X, X T ] = E[[X, X] T ]. We ca ow let ad use Mootoe Covergece. 3) We apply the Maximal Iequality for the stopped process X T up to time, ad use the martigale property of the process X 2 [X, X] stopped at T betwee time 0 ad. After that we let go to ifiity. 4) usig the previous item, we see that both radom variables i the differece are bouded by somethig itegrable as i class for cotiuous martigales) 5) let T k = if{ X, X +1 k}. Because X, X is predictable, the T k is a stoppig time. We also have that X, X Tk k, { X, X < k} {T k = }. Accordig to part 4, X = lim X is well defied o {T k = }, so the coclusio follows by lettig k. Problem 1.8. Wald II) Use the otatio of exercise 9.4.2 ad assume that E[ξ] = 0 ad E[ξ 2 ] = σ 2 <. Use exercise 9.5.3 to show that if N is a stoppig time such that E[N] < the E[S 2 N ] = σ2 E[N]. Solutio: Sice E[ξ] = 0 it is a easy exercise to show that S ) is a martigale. Deote Y = S σ 2. We have that so Y +1 Y = S 2 +1 S 2 σ 2 = S + ξ +1 ) 2 S 2 σ 2 = 2S ξ +1 + ξ 2 +1 σ 2, E[Y +1 Y F ] = 2S E[ξ +1 ] + E[ξ 2 +1] σ 2 = 0, which meas that Y is a martigale. This shows that the process A = σ 2 is the predictable quadratic variatio of the martigale S. We ca apply ow the previous problem to fiish the proof.
Solutios to HW: 1 Course: Theory of Probability II Page: 5 of 6 Problem 1.9. de la Vallée Poussi criterio) A set of radom variables X i ) ı I is u.i. if ad oly if there exists a fuctio ψ : [0, ) [0, ) such that lim x ψx)/x = ad sup E[ψ X i )] <. i I The fuctio ψ ca be chose to be covex. Solutio: Oe implicatio is quite easy. More precisely, if there exists such a fuctio ψ, the, for each ε there exists Mε) such that Now, for ay M Mε) we have ψx) x 1 ε, x Mε). E[ X i 1 { Xi M}] εe[ψ X i )1 { Xi M}] ε sup E[ψ X i )]. i I For the other implicatio, let us recall that, if ψ is piecewise) C 1 ad ψ0) = 0 the We also have that E[ X 1 { X M} ] = E[ψ X )] = 0 0 ψ x)p X x)dx. P X 1 { X M} x)dx = M P X x)dx. Usig the defiitio of uiform itegrability, we ca therefore choose a icreasig sequeces a, with a icreasig verly slowly), such that a sup P X i a )dx <. i We ca ow defie the piece-wise C 1 fuctio ψ by ψ0) = 0 ad ad fiish the proof. ψx) = a, x, + 1), Problem 1.10. backward martigales) Let Ω, F, P) be a probability space ad {F } 0 a decreasig filtratio, which meas that F +1 F. Let {X, F } be a bacward martigale, which meas that X L 1 Ω, F, P), 0 ad X +1 = E[X F +1 ]. Show that X coverges a.s. ad i L 1 to some variable X ad X = E[X F ], where F = F. Note: you kow already from Probability I that backward martigales coverge a.s., so the poit is to see the covergece is actually L 1 as well. Solutio: As already see i Probability I ad metioed above), we kow that X X F, a.s. Sice X = E[X 0 F ], we kow for example from Hw 1) that X is a UI sequece. This implies that X is itegrable, ad X X i L 1. Now, let A F. Let also k. Sice X k = E[X F k ], we have Lettig k we obtai or E[X 1 A ] = E[X k 1 A ]. E[X 1 A ] = E[X 1 A ], )A F, X = E[X F ].
Solutios to HW: 1 Course: Theory of Probability II Page: 6 of 6 Problem 1.11. backward submartigales) Let Ω, F, P) be a probability space ad {F } 0 a decreasig filtratio as above. Let {X, F } be a bacward submartigale, which meas that X L 1 Ω, F, P), 0 ad X +1 E[X F +1 ]. Show that {X } is u.i. if ad oly if if E[X ] >. Solutio: We ca use a backwards) upcrossig iequality to coclude that there exists a X L F ) such that X X a.s. The positive part X + is UI, so splittig each X ito X + X ) it is eough to show that E[X ] E[X ]. Actually, just usig Fatou for the egative part sice the positive part coverges i L 1 ) we have that E[X ] lim E[X ]. Now, for k we have X k E[X F k ]. Lettig k ad usig the previous problem, we have that X E[X F ], so E[X ] E[X ], fiishig the proof.