### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# THEORY OF PROBABILITY M 385C

UT

GPA 3.67

### View Full Document

## 54

## 0

## Popular in Course

## Popular in Mathematics (M)

This 29 page Class Notes was uploaded by Reyes Glover on Sunday September 6, 2015. The Class Notes belongs to M 385C at University of Texas at Austin taught by Gordan Zitkovic in Fall. Since its upload, it has received 54 views. For similar materials see /class/181456/m-385c-university-of-texas-at-austin in Mathematics (M) at University of Texas at Austin.

## Popular in Mathematics (M)

## Reviews for THEORY OF PROBABILITY

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/06/15

Lecture 7 STOCHASTIC INTEGRATION 1 of 10 Course MSSSD Theory of Probability ll Term Spring 2009 Instructor Gordan Zitkovic Lecture 7 STOCHASTIC INTEGRATION The importance of being of nite variation Let F QT A R be a continuous or cadlag if you want function and let us suppose that we can use it as an integrator In other words let 8 denote the set of all functions h QT A R of the form 71 ht a010t Zakimkmt n e N at 6 IR k 1 t k1 For h E S we naturally de ne T n 72gt hltugt dFltugt aoFltogt Zakwm 7 Fat 0 k1 For a continuous function h QT A R and a partition A E P0T 0 to lt t1 lt lt tn1 T we de ne the approximation hA to h in S by 73 hA t h010t Jr ihtk1tk tk1t TL 6 N 161 and note that hA A h uniformly We ask the following question Question 71 What properties does F need to have for the limit T A AlgrlidO h dFu to exist for each continuous h Remark 72 The reader should note that the question we are asking above really deals with a dominated convergencetype property for the eventual extension of the simple integral 72 to a larger class of integrands h Indeed if a functional h gt gt fh dF is to be called an integral it should be linear and it should allow for a version of a dominated convergence theorem The requirement that the uniform convergence of integrands hA A h implies convergence of the integrals is a particularly weak form of the dominated convergence theorem Nevertheless as we will see it restricts the class of functions F considerably The main analytic tool we are going to use is the celebrated BanachSteinhaus aka uniformboundedness theorem Remember that for a linear operator A z X A Y between two normed spaces X and Y Y we de ne its operator norm by 14 supfllAny I f 6 X7 llfllx S 1 Theorem 73 Banach Steinhaus Let X be a Banach space let Y be a normed space and let Agate be a family of continuous linear operators Au X A Y Then sup 1404 lt 00 and only Vf E X sup AafHY lt 00 046A 046A Last Updated April 19 2009 Lecture 7 STOCHASTIC INTEGRATION 2 of 10 Proposition 74 Let F be a continuous function on 0T A R the sequence 0T such that for each continuous function h z T hAquotudFun e N is bounded in R for each sequence AnneN in 13mm such that An A 1d Then F is a function of nite variation on 0 Proof Let X C0T be the Banach space of all continuous functions on 0T normed with suptqo w ht for h E C0 T and let Y R We x a sequence AnneN in 13mm with An A 1d and assume without loss of generality that T E An for all n E N A sequence of operators can be de ned by T A Xgt Y Anh hAquotudFu n E N 0 These operators are continuous why and by the assumption the sequence AnhneN is bounded for each h E Xi Therefore by Theorem 73 there exists a constant K 2 0 such that T 74 hwu dFu KHhH Vh 6 Own Vn e N 0 For each n E N one can construct a function hn E C0T such that hn0 sgnF0 and hntg sgnFtZ17FtZ for all h S ISAquot T where An tg eNi Moreover such a function can be constructed with 1 Using 714 we get T Mia K2 1 hquotudFul F0l Z lFtZ1 FOZNA 160 We have proved that the sequence of variations of F along any sequence of partitions converging to identity remains boundedi It follows why that F must be a function of nite variationi It is clear that any convergent sequence is bounded so the answer to Question 71 must be It is necessary that F be of nite vaiiationi On the other hand when F is of nite variation the dominated convergence theorem implies that the sequence in Question 71 indeed converges towards fOT hu dFui Squareintegrable martingales as integrators In spite of the impossibility result of the previous section we will still be able to construct a satisfactory integration theory when the integrand is not of nite variation In that case it will be important that some randomness is present and that the class of integrands be restricted to those that do not depend on the future Without knowing it we have constructed a very simple version of a stochastic integral when we proved the martingale convergence theoremi Indeed let XnneN be an iid sequence of cointosses 1PX1 1 lP X1 71 and let MJEQOM be de ned as follows 00 n M ZSk1172ik11727kt7 where 5k Z Xk for t lt 1 161 161 The martingale convergence theorem implies that the limit M1 limtnl Mt exists as The reader can check that the variation of the path M w on 01 is given by 221 00 so that no trajectory of Mtte01 is of nite variationli On the other hand let HQEQOJ be any bounded say by 1The reader will also note that Motdoy is not continuous a similar example with Mgtdoy continuous can be concocted but it would not be as transparent as the one we give here Also if one prefers to work with 000 instead of 01 one can de ne Mt 215k1k1ymt Last Updated April 19 2009 Lecture 7 STOCHASTIC INTEGRATION 3 of 10 K 2 0 and ftM qo adapted leftcontinuous stochastic process On each segment 0t t lt 1 M tqo is a process of nite variation and so we have the following expression 1 161 0 Eu dMu Z Xk 161 where hk H1727k1 and Mt is such that t E 17 2 k117 2 16 Moreover hk E 0X1X2111Xk1 so that the process n N Z 16 1 is a martingale it is really a martingale transform of themi Moreover it is a martingale bounded in L2 indeed by the orthogonality of martingale increments we have n 2 n 00 EM 2E1X l 2212K S K2252 lt 007 k1 k1 k1 so that supn lt 00 Consequently the martingale convergence theorem implies that the limit N limneN Nn exists as and in L21 Equivalently the random variable 1 t Hu dMu lim Hu dMu limNn N00 tl 0 n is well de ned even though no path of M is of nite variation It is important to note that the sequence NnneN only converges aisi There is no way to conclude that the limit exists for each wi lndeed take hn 1 for all n E N and note that for those w for which Xk w 1 for all 16 E N the sequence Z XMW 007 161 161 does not convergei Moreover the series 221 le Ad will never converge absolutelyi It is the fact that the sequence XkkeN uctuates so much and the fact that hk does not see Xk that guarantee convergence N7 A N00 and equivalently allow for the integral fol Hu dMu to be wellde ned We turn now to a general construction Let Mthqopw be a squareintegrable martingale we will use the notation and the properties from Problem 67 in Lecture 6 For a simple predictable H E H mp we have already de ned stochastic integral H M in the previous lecturer It follows directly from the de nition prove it as an exercise that for any local martingale Nthqopw and any H E H mp we have 75 H M N EH1 dM N The equality above has the nice property that it ties the stochastic integration which we are yet to de ne to the Stieltjes integral which we now how to andlei In fact we can use it to construct the general stochastic integrali We start with an application of the Kunita Watanabe inequalityi Let L2 denote the set of all progressively measurable processes Htte0oo such that HHHLW ElO H3dM7M1ultooi In order for HL2M to be a norm and not only a quasinorm we need to identify the processes whose H HL2M distance is zero More precisely let L2 denote the set of all equivalence classes of elements in 2 under the relation H N K if and only if H K for dMM almost all t 2 0 as It is not hard to see that L2M HL2M is a Hilbert space Last Updated April 19 2009 Lecture 7 STOCHASTIC INTEGRATION 4 of 10 Proposition 75 For Mthqopw E H2 and Htte0oo E L2M there exists a unique process in H3 denoted by Mtte0m such that L H MNt HudMNu for all t 2 0 as 0 for all Nthqopw 6 H2 Moreover the map Htte0oo gt gt Mtte0oo is a linear isometry called the Ito isometry from L2 into Hg Proof Suppose that M0 0 and set HM H MiMo otherwise By the Kunita Watanabe inequality or N E Hg we ave g Elm1dNNuElmH3dMMu HNHmHHme Therefore the mapping Ntte0m gt gt lEf0 o Hu dMNu is a continuous and linear functional on the E o Hu aw NM Hilbert space Hgl Hence by the Riesz representation theorem there exists an element in H3 which we denote by Mtte0m with the property that 76 EU Hamil EKH Mummy 0 Using the uniqueness of the Riesz representation we can imediately conclude that the map H gt gt H M from L2M to H3 is linear For a stopping time T we plug in NT instead of N in 76 and remember that MNT MNT to get T 00 E Ht dlMJviu E Hu WWW EKH Mum 0 0 Elma Wham EKH mm by uniform integrability of martingales in Hz Therefore the process MtNt 7 f Hu dM Nluhqopo is a martingale which means that L H MNt Hu dMNu for all t 2 0 as 0 Furthermore Htte0oo gt gt Mtte0m is an isometry because HH MHip EHH MH Mlm HudlMyH Mu HidlMMlu HHHiW To deal with N E H2 and not only in HE we simply need to remember that L N L N 7 Noll for all t 2 0 as and all martingales Lthqopoy D From this point on we drop the notation Mthqopw and simply write M whenever it is clear that M is a stochastic process Also M N is taken to mean Mt N for all t 2 0 as De nition 76 For H E L2M and M 6 H2 the process H M sometimes also denoted by Hu dMu is called the stochastic or Ito integral of H with respect to M The following properties of the stochastic integral follow directly from the de nition Problem 71 For M 6 H2 H E L2 and a stopping time T we have 1 H M coincides with the previously de ned elementary integral when H E H mp N L2 M 2 KH 6 L2M andK KHM for any K6 L2HM 3 H MT H10T M Last Updated April 19 2009 Lecture 7 STOCHASTIC INTEGRATION 5 of 10 The restriction H E L2 on the integrand and M E H2 on the integrator in the de nition of the stochastic integral H M can be relaxed For a continuous local martingale M we de ne the class LM which contains all progressively measurable processes H with the property H3 dMMu lt 00 as 0 We de ne the stochastic integral H M for H E LM as follows let TnneN be a sequence of stopping times de ned by z Tninft20 lMtl 2nAinft20 H5dMl u 2 0 This sequence clearly reduces M Moreover we know that MTquot 6 H2 because it is bounded and H10Tn E L2MTquot for all n E N Therefore the integral H10Tn MTquot is wellde ned and has the property that t t 77 H10TnMNt HudMTquotNu Hu10TnudMNu for alltZ 0 as 0 0 for all N 6 H2 The processes H10Tn MTquot and H10Tn1 MTquott1 coincide on 0117 so we can patch them together to obtain a stochastic process which we denote by H M Using 77 we get 2 HMNt HudMNu for alltZO as 0 for all N 6 H2 so that H M shares the fundamental property with its H2version At this point we know how to use 1 processes of nite variation and 2 local martingales as integrators One can show that linear combinations of those are in a sense all processes that can be used as integrators For this reason we give them a name De nition 77 A stochastic process Xthqopw is called a continuous semimartingale if there exists a continuous adapted process Athqopw of nite variation and a continuous local martingale Mthqopw such that M0 0 and X1 Mt Al for all t 2 0 as Using the fact that there are no nontrivial continuous local martingales of nite variation one can show that that for a 39 A T X into a continuous local mar tingale and a continuous adapted process of nite variation is unique This decomposition is called the semimartingale decomposition of X i 1 euumaiuu 415 t e Problem 72 Show that continuous semimartingales are of nite quadratic variation and that XXt M M where X M A is the semimartingale decomposition of X For a continuous semimartingale X with the semimartingale decomposition X X0 A M let LX denote the set of all progressivelymeasurable processes with the property that z z lHul dlAlu H3 dMMu lt 00 for all t 2 0 as 0 0 For H E LX we can de ne both the LebesgueStieltjes integral f Hu dAu which we also denoted by HA and the stochastic integral H M thus we de ne the stochastic integral H X of H with respect to X by HXt HALHML for all t 2 0 It is immediately clear that H A is an adapted process of nite variation and that HM is a local martingale so that H X is a continuous semimartingale and H X H AH M is its semimartingale decomposition As the reader can check the stochastic integral for semimartingales has the following properties Problem 73 Let X be a 39 emimaitiu ale with the emimaitiu ale A T X AM Then 1 Both maps H gt gt H X and X gt gt H X are linear on their natural domains Last Updated April 19 2009 Lecture 7 STOCHASTIC INTEGRATION 6 of 10 2 For H E LX and K E LH X we have KH 6 LX and KHXKHX 3 For H E LX and a stopping time T we have HT 6 LX and H XT H10T X H XT 4 For H E Hsimp with representation Ht 220 Knlamtn t we have H E LX and 00 H 39Xz ZKnXttn1 XtAtnl n0 Before we move on here is a very useful version of the dominated convergence theorem for stochastic integration A stochastic process H is said to be locally bounded if there exists a sequence TnneN of stopping times such that Tn S Tn1 as for all n E N Tn A 00 as as n A 00 and each stopped process HTquot is bounded The following two problems prepare the ground Problem 74 Let X be a continuous semimartingale Show that 1 each progressively measurable locally bounded process H is in LX and that 2 each leftcontinuous and adapted process is locally bounded Construct an example of an adapted and cadlag process which is not locally bounded Problem 75 Let TnneN be a sequence of stopping times such that Tn S Tn1 as for all n E N n A 00 as as n A 00 and let XkkeN be a sequence of adapted and continuous processes Suppose that XTquot 13 0 as h A 00 for each n E N Show that Xk LB 0 ln words if you want to prove ucp convergence you are allowed to localize rst Proposition 78 A Stochastic Dominated Convergence Theorem Let X M A be a continuous semimartingale and let H neN be a sequence of progressivelymeasurable processes with the property that there exists a caglad process H such that S H for all t 2 0 as for all n E N Then H E LX for all n 6 an limHt A 0 for all t 2 0 as implies H X Lg 0 as n A 00 n Proof All we need is in the two problems above and the result from homework problem 51 D Ito s formula ltols formula is for stochastic calculus what the NewtonLeibnitz formula is for classical calculus Not only does it relate differentiation and integration it also provides a practical method for computation of stochastic integrals There is an added bene t in the stochastic case It shows that the class of continuous semimartingales is closed under composition with C2 functions Theorem 79 Let X be a continous semimartingale taking values in a segment a b E R and let f z a b A R be a twice continuously di erentiable function E C2a Then the process is a continuous semimartingale and 78gt we 7 fXo I ma M ma dA g f Xu dlM Mu Before we proceed with the proof let us state and prove two useful related results In the rst one we compute a simple stochastic integral explicitly You will immediately see how it differs from the classical Stieltjes integral of the same form Lemma 710 Let Mthqopw be a local martingale with M0 0 Then M M 7 MMt for all t 2 0 Last Updated April 19 2009 Lecture 7 STOCHASTIC INTEGRATION 7 of 10 Proof By stopping we may assume that M and MM are bounded We have already shown in the proof of Lemma 635 of the previous lecture that 79 MA M M3 7 MM for all t 2 0 and all partitions A 6 Hope where 1WA denotes the leftcontinuous approximation 1WA MEA t to Mthqopoy Thanks to relation 66 in the proof of Lemma 637 the righthand side of 79 converges to 7 MMt in L2 as soon as A 7 Id To show that the limit of the lefthand side converges in L2 to M it suf ces to to use the Ito isometry in the form L 2 H W 7 MA M ng 7 M 7 M5gt2dMMu and use dominated convergence to conclude that the righthand side converges to 0 as A 7 Id D The second preparatory result is the stochastic analogue of the integrationbyparts formulai We remind the reader that for two semimartingales X M A and Y N C we have X Y M N Proposition 711 Let X M AY N C be semimartingale decompositions of two continuous semimartingales Then XY is a continuous semimartingale and t l me xx XOYO Yu qu X dYu KY 0 0 Proof By stopping we can assume that the processes M N A and C are bounded say by K 2 0 Moreover we assume that X0 Y0 0 otherwise just consider X 7 X0 and Y 7 Yo We write XY as M A N C and analyze each termi Using the polarization identity the product MN can be written as MN MJrN2 7 M2 7 N2 and Lemma 7 10 implies that z z 711 mm Mu dNu NudMu MNL 0 0 holds for all t 2 0 As far as the FV terms A and C are concerned the equality z z 712 AECE Au dCu Cu dAu 0 0 follows by a representation of both sides as a limit of RiemannStieltjes sumsi Alternatively you can view the lefthand side of the above equality as the area under the product measure dA gtlt dC of the square 0t gtlt 0t Q R The righthand side can also be interpreted as the area of 0t gtlt 0t the two terms corresponding to the areas above and below the diagonal ss E R2 z s E 0t we leave it up to the reader to supply the details Let us focus now on the mixed term MCi Take a sequence AnneN in 13mm with An tgkeN and An 7 Id and write 00 M20 2me Cmquot MzAngng 1711 172 13 Where 160 k1 k1 7 k1 1 ZMLMZ Ctwquot 02mg 160 k1 I 7 2 cm MW 7 MM k0 I 7 Ewing 7 Mmgxcmg 7 CM k0 By the properties of the Stieltjes integral see hw4 I 7gt f Mu dCu aisi Next we have 73 7gt 0 as because k1 12 thNM Zlcmw 7 szgl S WLAquotMlClt A 0 160 Last Updated April 19 2009 Lecture 7 STOCHASTIC INTEGRATION 8 of 10 because the paths of M are uniformly continuous Finally we note that z 172 Cfquot dMu 0 where C39squot CEAWW By uniform continuity of the paths of C on 0t CAquot 7gt C uniformly on 0t as and 05quot 7 Cul S lClu which is an adapted and cadlag process Therefore we can use the stochastic dominated convergence theorem 78 to conclude that If 7gt f Cu dMu in probability so that 713 MICE fMu do f Cu 4M and in the same way 0 0 714 ALNL t Au dNu t Nu dAu The required equality 710 is nothing but theosum of 7113 7 12 713 and 714 D Remark 712 By formally differentiating 710 with respect to t we write dXYt XE dK Y dXt dXtdY While meaningless in the strict sense the differential representation above serves as a good mnemonic device for various formulas in stochastic analysis One simply has to multiply out all the terms disregard all terms of order larger than 2 such as dXt2 or dXt2dYt and use the following multiplication table dM dA dN dM N 0 d0 0 0 where X M A Y N C are semimartingale decompositions of X and Y When M N B where B is the Brownian motion the multiplication table simpli es to dB dA dB at 0 d0 0 0 Proof of Theorem 79 Let A be the family of all functions f E C2ab such that the formula 78 holds for all t 2 0 and all continuous semimartingales X which take values in abi It is clear that A is a linear space which contains all constant functions Moreover it is also closed under multiplication iiei fg E A if fg E A Indeed we need to use the integrationbyparts formula 7 10 and the associativity of stochastic integration Problem 73 applied to and gXi Next the identity is clearly in A and so P E A for each polynomial P It remains to show that A C2abi For f E C2a 12 let P neN be a sequence of polynomials with the property that Pg 7 f uniformly on abi This can be achieved by the Weierstrass Stone theorem because polynomials are dense in Ca 12 It is easy to show that the polynomials Iihey can be taken to be second derivatives of a sequence PnneN of polynomials with the property that P7 7 f P 7gt f and P77 7 f uniformly on abi lndeed just use d5 and Pnz P72 d5 Then as the reader will easily check the stochastic dominated convergence theorem 78 will imply that all terms in 78 for Pn converge in probability to the corresponding terms for f This shows that C2ab A Remark 713 The proof of the lto7s formula given above is slick but it does not help much in terms of intuition One of the best ways of understanding the lto formula is the following nonrigorous heuristic derivation where 0 to lt t1 lt lt tn lt tn1 t is a partiton of 0tli The main insight is that the secondorder term in the Taylor s expansion 7 fs t 7 s f st 7 s2 0t 7 s2 has to Last Updated April 19 2009 Lecture 7 STOCHASTIC INTEGRATION 9 of 10 be kept and cannot be discarded because in contast to the classical case the second quadratic variation does not vanish n fXz fXo Z fsz1 fsz 160 W Z fthth1 1 bfthth1 X k712gt 160 Z fthMtk1 7 Mn 7 Z fthAtk1 7 Ant 7 Z fX kth1 7 X102 160 160 160 1 t t m f XudMu f XudAu f XudXXui o o o Finally7 one simply needs to remember that XX M7 The A 39 t that the emimaitiu ale is quot 39 or that it takes values in a compact set a7 b7 are not necessary in general Here is a general version of the lto formula for continuous semimartingales We do not give a proof as it is very similar to the onedimensional case The requirement that the process takes values in a compact set is relaxed by stopping Theorem 714 Let X1X27 X 1 be d continuous semimartingales and let D Q Rd be an open set such that XE thp X E D for allt 2 0 as Moreover let f z D A R be a twice continuously di erentiable function E C2 Then the process fXtte0oo is a continuous semimartingale an 715 lel7lel7i17 Wkal i fi X dXi le t 0 7k1 0 a u u Qua 0 azian v a Remark 715 1 When the local martingale parts of some of the components of the process X X1 Xd vanish7 one does not need the full C2 differentiability in those coordinates C1 will be enough 2 Using the differential notation and the multiplication table of Remark 7127 we can write the lto formula in the more compact form n d a a Z dfXz Z maxi dXE 2 mm qudXi k1 i39j1 Problem 76 Let f z 000 X R A R be a function in Clgt207 00 X R the space functions continuously differentiable in t and twice continuously differentiable in Such a function f is said to be spacetime harmonic if it satis es f 5f 0 For f E C13 000 X R and a Brownian motion B7 show that the process ftBt is a local martingale if and only if f is spacetime harmonic Show7 additionally7 that ft7 B1 is a martingale if f is spacetime harmonic and f7 is bounded on each domain of the form 07 t X R7 t 2 0 We conclude the lecture about stochastic integration with a useful application of lto7s formula7 known as Levy s Characterization of Brownian Motion Theorem 716 Let Mthqopw be a continuous local martingale with M0 0 and MMt t for all t as Then M is an ftte0mBrownian motion Proof Consider the complexvalued process Z3 eXpith tht 2 07 for u E R We can use ltols formula applied separately to its real and imaginary parts to show that Z is a local martingale for each u E R lndeed7 Z ftMt7 where ftz eiux 2 so that7 as you can Last Updated April 197 2009 Lecture 7 STOCHASTIC INTEGRATION 10 of 10 easily check fttz fm tz 0 Therefore ft7Mz 7 f07Mo szltu7Mugtdu tfmltu7MugtdMu tfmltu7MugtdlM7Miu f07 0 tltfzltu7Mugt gfmltuMugtgt du mu Mu dMu 1tfxuMudMui We can actually assert that Z is a true martingale because it is bounded by expeth on 0t and therefore of class DL Consequently we have lEZLl5 Zf for all u E R 0 S s S t lt 00 With a bit of rearranging we get EeiuM 7Ms ei u zis l 2 That implies that M 7 M5 is independent of f5 and that its characteristic function is given by e Qu 75 ie that M 7 M5 is normal With mean 0 and variance t 7 8 D Remark 7 17 Continuity is very important in the Levyls characterization The compensated Poisson process Mt NE 7 t With parameter A l is a martingale With MMt t but is certainly not a Brownian motion Problem 77 Remember that a stochastic process Xthqopw is said to be Gaussian if all of its nite dimensional distributions are multivariate normali Let Mthqopw be a continuous local martingale With MMt ft for some continuous deterministic nondecreasing function f 000 7 000 Show that Mthqopw is Gaussian Is it true that that each continuous Gaussian local martingale has deterministic quadratic variation Last Updated April 19 2009 Lecture 3 DISCRETE MARTINGALE THEORY 1 of 8 Course MSSD Theory of Probability ll Term Spring 2009 Instructor Gordan Zitkovic Lecture 3 DISCRETE MARTINGALE THEORY Throughout this lecture7 T N or T N0 At a certain point and the reader will be warned in time we will need T 7N0 Predictability and martingale transforms De nition 31 Let fn Tn1 for n E De nition 32 Let fn process Xn be a ltration A process Hnn6N is said to be Tn predictable if Hn E nENo nENo neNo be a ltration and let Xnn6ND be a process adapted to Tn The stochastic de ned by nENo neNo7 H X0 0 H X ZHkXk 7 X for n e N k1 is called the martingale transform of X by H Proposition 33 A martingale transform H X of a martingale X by a predictable process H is a martingale provided that HnXn 7 Xn1 6 1 for all n E N Proof Just check the de nition and use properties of conditional expectation D Remark 34 l The martingale transform is the discretetime analogue of the stochastic integral Note that it is crucial that H be predictable if we want a martingale transform of a martingale to be a martingale Otherwise7 we just take Hn sgnXn 7 Xn1 E f and obtain a process which is not a martingale unless X is constant 2 It is easy to show that martingale transforms preserve supermartingales and submartingales7 as well7 provided that the process Hnn6N is nonnegative Proposition 35 Bounded Optional Sampling Let Xnn6ND be a submartingale and let T be a stopping time Then the stopped process thleN is also a submartingale Moreover we have ElXol S ElXTAml S E XML for all m 6 N0 and the inequalities turn into equalities when X is a martingale Proof We note that for the processes Hn 1Tltn Kn l 7 Hn7 n E N are both predictable7 nonnegative and bounded7 so their martingale transforms H X and K S are submartingales Moreover7 H 39 Xn Xn XTwu K Xn XTAn X0 The submartingale property of K X is equivalent to the submartingale property of XT7 and implies that lEX0 S lEXTAm7 for m 6 No To prove the second inequality7 we use the submartingale property of H X to conclude that lEXm 7 XTAm 2 0 D Corollary 36 Doob s inequality Let Xnn6ND be a submartingale For A gt 0 and n E N we have mm Xm 2 A EanIpupwxmziy Elm Last Updated February 2 2009 Lecture 3 DISCRETE MARTINGALE THEORY 2 of 8 Proof For n 6 N0 and A gt 07 we de ne the stopping time T by Tinfm N0 XmZAAn and the event A Xm 2 A7 for some 1 S m S Proposition 35 implies that Ean 2 ElXTML lEXT7 so lEXn 2 lEXT1A lEXT1A4 lEXT1A lEXn1A4 2 lEA1A lEXn1Aa Therefore AMA S EanlAl S ElellA For a stochastic process Xtheqq the maximal process Xi g is de ned by X sup Xl 5675St Corollary 37 Let Xnn6ND be a martingale or a nonnegative submartingale Then PlXZ 2 Al S iEHanhszMl S iEHanl for all n 6 No A gt 0 Corollary 38 Kolmogorov s inequality Let Xn all n 6 N0 Then for A gt 0 neNo be a martingale with lEX0 0 and lt 00 for Msup lel 2 A S Va Xn mSn Proof The process X2 nnENO 1s a nonnegative submartingale D Convergence of martingales A judicious use of a predictable processes in a martingale transform yields the following important result Theorem 39 Martingale convergence theorem Let Xnn6ND be a submartingale such that sup lEX7 lt 00 6 0 Then there exists a random variable X 6 1107 such that I Xn AX as and 2 X 6 L1 One of the ways to remember this fact is the following submartingales are stochastic analogues of nondecreasing sequences and boundedfromabove nondecreasing sequences converge Corollary 310 I Uniformlyintegrable submartingales converge as and in L1 2 Nonnegative supermartingales and nonpositive submartingales converge as Corollary 311 If Xn is a martingale of the form Xn lElen nENo for some X E L1f then Xn A Elefoo as and in lLl7 where 700 0UneNofn Last Updated February 2 2009 Lecture 3 DISCRETE MARTINGALE THEORY 3 of 8 Proof The martingale XnneNO is uniformly integrable so the Martingale convergence theorem implies that there exists a random variable Y 6 1117 such that Xn 7gt Y as and in 11 For m E N and A 6 fm we have llEXn1A 7 YlAll S lEHXn 7 Y1 7gt 0 so lEXn1A 7gt lEY1A Since lEXn1A lElElen1A lEX1A for n 2 m we have lEY1A lEX1A for all A E Unfn The family Unfn is a 7rsystem which generated the sigma algebra foo 0Unfn and the family of all A E f such that ElYlA EleA is a A system Therefore by the 7r 7 A Theorem we have lEY1A lEX1A for all A E 700 Therefore since Y E 700 we conclude that Y Elefoo D Remark 312 The convergence in L1 or equivalently uniform integrability does not need to hold for a generic nonnegative supermartingale Let XnneNO be a simple random walk starting from 1 ie X0 1 and X 1 ELLIS16 where EnneN is an iid sequence with M5 1 M5 71 5 n E N Clearly XnneNO is a martingale and so is YnneNo where Yn X77 and T inf n E N Xn 0 By convention ianl 00 It is well known that a simple symmetric random walk hits any level eventually with probability 1 so MT lt 00 1 and since Yn 0 for n 2 T we have Y7 7gt 0 as as n 7gt 00 On the other hand YnneNO is a martingale so ElYn 1 for n E N Therefore ElYn 77 ElX which can happen only if Yn is not uniformly integrable nENo Martingale inequalities Doobls inequality of Corollary 36 is one of many important inequalities where the submartingale property enters in a crucial way T e rst one relates the U norm of the maximal process XneNo to the U norm of X We start with an easy but quite useless version Proposition 313 An almost useless maximal inequality Let Xnn6ND be a martingale or a nonnegative submartingale and p E 1 Then for n 6 N0 llXille S n11planlleA Proof Without loss of generality we assume that lEHanp lt 00 By Jensen s inequality and the submartingale property of we have Elanlplfkl 2 Elanllfle 21X for 0 S k S n so that lEHanp gt lEHXklp Therefore nENO7 n 131090171 Elsup lelpl S ZEllelpl S n1Elanlp17 k5 k0 and the statement follows E Proposition 314 Maximal inequalities Let Xnn6ND be a martingale or a nonnegative submartingale and p E 1 Then for n 6 N0 HXZHLP S lanlleA Before we give a proof here is a simple and useful result which in the special case Y 1 relates the tail of a distribution to its moments Lemma 315 For nonnegative random variables X and Y and p gt 0 we have lEYXp pAp llEY1XZAdAi o Last Updated February 2 2009 Lecture 3 DISCRETE MARTINGALE THEORY 4 of 8 Proof Since 11 DA pAp l dA Fubini s theorem implies that X 00 00 lEYXp lEY pAp l dA Y1SXpp 1d pAp llEY1XZd 0 0 0 D Proof of Proposition 314 The process anlneNo is a nonnegative submartingale so we can assume without loss of generality that Xnn6ND is a nonnegative submartingale and that lEX lt 00 By Corollary 37 PlXZ 2 Al 3 iElanthszl Lemma 315 implies that lEXp pAp IMX 2 A dA g p7 lp 2lEXn1XZ dA lEXnXp 1 0 0 Let L 17 be the conjugate exponent of p Holder s inequality applied to the righthand side above yields llXillfp S alanllelKXDp llqu S alanllellXlef qA One would like to divide by li q and nish the proof but it will not work before we show that lEXp lt 00 This estimate however follows directly from the not so useless maximal inequality of Proposition 313 D De nition 316 A stochastic process XnneNO is said to be bounded in L for p 6 100 if the family Xnn6ND of random variables is bounded in L5 ie if sup HXnHM lt oo nENO Corollary 317 LPbounded martingales Let Xnn6ND be an LIDbounded martingale or a nonnegative submartingale where p 6 100 Then there exists X E U such that X7 A X as and in L Proof Even without the Maximal inequality one can conclude that Xnn6ND is Ul simply by using I gt gt 11 as a test function The martingale convergence theorem then implies that Xn A X as for some X 6 1 In order to show that the convergence holds in U the full strength of the Maximal inequality needs to be used lndeed Proposition 314 implies that for any m E N we have llX HLp lt lanHLp lt M sgglanHLp lt 00 7 Moreover X Xgo supneNO anl and so by the Monotone convergence theorem we have llX olle S M Consequently sup Xm 6 U n N and we can use the Dominated convergence theorem to show that HXn 7 XHLP A 0 D Optional Sampling Theorems For a stopping time T and a process Xn Ler the following prescription XTM XTltwgt To lt oo limn Xnw Tw 00 limn Xn exists de nes a random variable why is it measurable on the set T lt 00 U limn Xnw exists There is really no nice way to extend it to the whole 9 if the intersection T 00 limn Xn does not exist in not empty We will have no need for such an extension anyway We start the discussion with a very general version of the optional sampling theorem Theorem 318 Let Xnn6ND be a submartingale and let T be a stopping time such that Last Updated February 2 2009 Lecture 3 DISCRETE MARTINGALE THEORY 5 of 8 I lEXE1Tltm lt 00 and 2 the family X1TgtnneNo is uniformly integrable Then I the random variable XT is well de ned ie limn X7 exists as on T 00 2 XT 6 L1 and 5 EXol S ElXTl Proof The process XT is a submartingale and EKXZW ElXAnl ElX1Tnl ElehTml S ElX1Tltool1 gEle1Tgtkl lt 00 Therefore by the Martingale Convergence Theorem there exists a random variable X E L1 such that XTM A X as This implies that X7 converges on T 00 so the random variable XT is well de ned and XT E lLll We use the following decomposition XTAn X1Tgtn X1Tgtn XT1T n By alsoconvergence and uniform integrability we have 117111 ElXTAn1Tgtnl 117111 ElX7l1Tgtnl 13le 1Tool Also X77195 A Xf1Too also so Fatouls lemma implies that liminflEXfAn1Tgtn liminflEX1Tgtn 2 1Tm and so limsuplElXTAn1Tgtnl S ElXT1Tool On the other hand since X1Tltoo 6 ll Monotone convergence yields 11511 ElXAn1Tnl 11511 ElX1Tnl ElX1Tltool Fatouls lemma implies that li infElXEthsml 2 13le 1Tltool7 so that limsuplEXTAn1TSn S lEXT1Tltooi Therefore limsuplEXTM S EXT and since lEXTM 2 lEX0 by Proposition 35 we have lEX0 S lEXTl D Since 7X is a submartingale with Xi 0 whenever X is a nonnegative supermartingale we have the following result Corollary 319 Let XnneNO be a nonnegative supermartingale and let T be a stopping time Then ElXol 2 mm Another common set of conditions under which the optional sampling theorem is used is the following Proposition 320 Let XnneNO be a submartingale and let T be a stopping time If X7JfneNo is uniformly integrable then the random variable XT is well de ned in L0 and XT 6 1 Moreover the process is also a uniformly integrable submartingale and nENo ElXol S ElXTl If additionally XnneNO is uniformly integrable then so is XneNo Last Updated February 2 2009 Lecture 3 DISCRETE MARTINGALE THEORY 6 of 8 Proof We note rst that the condition 1 of Theorem 318 is clearly satis ed In order to show 2 we note that the process X7JfneNo is a supermartingale so by Proposition 35 ElX ltAnl E Elel Therefore if we subtract lEX1nltT lt 00 from both sides we get that EX1T5n S lEX1TSn S sup lt 00 kEN and the Monotone Convergence Theorem implies that lEXE1Tltm lt 00 To show that the process is uniformly integrable we note that nENo XriAT S X X1t1Tltoo7 eriAT1X TgtM1Tgtnl EeriAT1XATgtM1TS l EeriAT1XATgtMl S E S ElX7l1X gtMl ElX1t1Tltoo1th1TltwgtMlA 0 uniformly in n as M A 00 Uniform integrability of the stopped process follows from the uniform integrability of Xn along 716 nENo the same lines we use the fact that XT1TltOO 6 1 In analogy with the 0algebra fn which models the information available at time n E N we denote by fT the 0algebra which models the information available at the moment the stopping time T happens More precisely fTAEfoo A TSnEfn VnENO where 700 0Unfn Proposition 321 For a stopping time T fT is the 0algebra generated by the family 8 XT X E X where X denotes the set of all fnneNOadapted stochastic processes for which XT is wellde ned Proof Suppose rst that A Xf1B for some B 6 EUR and X E X Then A TSnXT EBTSnUZ0Xk EBTk 677 Since the sets of the form Xf1B B 6 EUR generate 08 we conclude that 08 Q fT Conversely for A E fT the process Xn 1A4UnltT is adapted and XT lAa so A0 6 08 and consequently A 6 08 D Theorem 322 Let XnneNO be a uniformly integrable submartingale and let S and T be two stopping times with T S S as Then XTX5 E L1 and lEXsl T 2 XT 18 In particular Elel 2 ElXTl Proof We cover the case S 00 rst By Theorem 318 XT is wellde ned and in 1 Moreover the family XTML n 6 N0 is uniformly integrable by the last statement of Proposition 320 By Proposition 35 we ElXTvil S EanL and upon letting n A 00 we get that ElXTl S EleL 31 where XDo limn X which exists as thanks to the uniform integrability of XnneNo We now take A E fT and de ne the stopping time TA by TA TIA 001A why is TA a stopping time The inequality 31 with T replaced by TA implies that lEXT1A S lEXm1A which directly implies that EllElXoollelAl S ElXTlAlz Last Updated February 2 2009 Lecture 3 DISCRETE MARTINGALE THEORY 7 of 8 for all A E fT The only way something like this will work is if lEXoolfT S XT as why It remains to remove the assumption 5 00 We do this by considering the process Y7 X75 and using the last part of the statement of Proposition 320 Corollary 323 Let Xnn6ND be a uniformly integrable martingale and let S and T be two stopping times with T S S as Then XTX5 E L1 and lEXslfT XT as In particular lElXTl Elel The optional sampling theorem is sometimes stated using the notion of the last element Namely a martingale Xnn6ND is said to have a last element if there exists a random variable X 6 1117 such that X7 lElen Vn 6 N0 For a supermartingale the de nition is analogous It is said to have a last element if there exists X E L1 such that X7 S lElen Vn 6 No In either case the random variable plays the role of XDo in the sense that the process Xnn6ND U X00 is still a martingale submartingale on the extended time set N0 U Remark 324 Note that last elements of martingales if they exist are unique as while a submartingale can have many last elements In fact if X is a last element of a submartingale then any Y 2 X as with Y E L1 is also a last element Proposition 325 For a submartingale Xn I Xnn6ND admits a last element 2 the family Proof Suppose rst that X E L1 is a last element of XnneNo The process YnneNo given by Y lElen n 6 N0 is uniformly integrable and so is YneNo Since Xi S Y5 the family X7JfneNo must be uniformly integrable too Conversely suppose that neNo the following two statements are equivalent neNo is uniformly integrable neNo is uniformly integrable Then the limit XDo limn X7 exists and X00 6 1 by the Martingale Convergence Theorem Moreover X A X in l by the uniformintegrability assumption Using the submartingale property of Xnn6ND for m 6 No we have Eanfm 2 Xm for all n 2 m Since X A X in l we have lEXlfm A EXlfm in 1 Moreover there exists a subsequence erk have of XifhleNo such that lEXklm A EXlfm as On the other hand Fatouls lemma implies that EXolfm lElimianklm S liminflEXklm as k k It follows that lEXoolfm 2 limsuplEXnklfm 2 Xm as k which means that XDo is a last element of Xn D nENo Corollary 326 Let Xnn6ND be a submartingale which admits a last element X00 Then for any stopping times we ElXol S ElXTl If Xnn6ND is a martingale the inequality above turns into an equality Last Updated February 2 2009 Lecture 3 DISCRETE MARTINGALE THEORY 8 of 8 Backward Martingales In this section we use the set 7N0 0 7l 72 of nonpositive integers as the index set T De nition 327 A stochastic process XnneNo is said to be a backward submartingale with respect to the ltration fnneNo if 1 ann67No is fnheiNoadaptedy 2 Xn 6 1 for all n 6 N0 and 3 lEan n1 2 XnE1 for all n 6 7N0 lf in addition to l and 2 the inequality in 3 is in fact an equality we say that XnneNo is a backward martingale One of the most important facts about backward submartingales is that they almost always converge as and in 1 Proposition 328 Suppose that XnneNo is a backward submartingale such that lim lEXn gt 700 Then XnneNo is uniformly integrable and there exists a random variable XDo E L1 nfn such that X7 7 XDo as and in L1 32 and XDo S lEXml n fn as for all m 6 7N0 33 Proof For n 6 7N0 set AAn Ean 7 Xn1l n1 2 0 as and A 220 AAk for n 6 NO The backward submartingale property of thleNO implie that lEXn 2 L so lEAn lEX0 7 Xn S lEX0 7 L for all n E No The Monotone Convergence Theorem implies that lEA7m lt 00 where A00 220 An The process Mn neNo de ned by Mn Xn 7 An is a backward martingale lndeed lEMn 7 Mnalfna lEXn 7 Xn1 7 AAnlf nal 0 Since all backward martingales are uniformly integrable why and the sequence AnneNo is uniformly domi nated by A00 E L1 and therefore uniformly integrable we conclude that XnneNo is also uniformly integrable To prove convergence we note that the uniform integrability of XnneNo implies that supneiNO lt 00 and a slight modi cation of the proof of the Martingale Convergence Theorem left to a very diligent reader implies that X7 7 Xm as for some random variable X00 6 nfn Uniform integrability ensures that the convergence holds in L1 and that XDo 6 1 In order to show 33 it is enough to show that ElX7oltgt1Al S ElelA 34 for any A E nfn and any m 6 7N0 We rst note that since Xn S lEXmlfn for n S m S 0 we have lEXn1A S lElEXmlfn1A lEXm1A for any A E nfn It remains to use the fact the lLlconvergence of Xn neNo implies that lEXn1A 7gt lEXm1A for all A E f D Remark 329 Even if limlEXn 700 the convergence X7 7 XDo still holds but not in L1 and XDo may take the value 700 with positive probability Corollary 330 If XnneNo is a backward martingale then X7 7 XDo lEX0l n fn as and in L1 Last Updated February 2 2009 Lecture 8 REPRESENTATIONS OF MARTINGALES 1 of 11 Course MSSSD Theory of Probability ll Term Spring 2009 Instructor Gordan Zitkovic Lecture 8 REPRESENTATIONS OF MARTINGALES AND CHANGES OF MEASURE Rightcontinuous inverses Let be a cadlag and nondecreasing function de 127 ned on 000 and taking values in 0 Even though it does not need to be invertible we can still de ne a function 9 z 0 00 H 000 called the right 107 X continuous inverse of f which will play the role of 39 its inverse in most applications 08 I x 9s inft 2 0 z gt s y l To have complete symmetry we can de ne 900 06 l limsnm9s and limtH00 The picture on the right shows a portion of the graph of a typical function f black as well as its rightcontinuous inverse bluei Note how jumps cor respond to at stretches and vice versa and how the 027 noninjectivity is resolved to yield right continuityi As always ht limtHmG ht for t gt 0 and h07 0 for any cadlag function h on 000 02 04 as 08 10 12 Proposition 81 Let f 000 H 000 be a nondecreasing cadlagfunction andlet9 be its rightcontinuous inverse Then for all st E 0 00 we have I 9 is nondecreasing and cadlag 2 2 t with equality and only ift is a point of right increase of f meaning that ft5 gt ft for all 5 gt 0 5 infs 2 0 9s gt t for t 2 0 4 2 s with equality and only ifs is a point of right increase of9 meaning that 9s5 gt 9s for all 5 gt 0 5 9s7 inft 2 0 z 2 s for s 2 0 and ft7 infs 2 0 93 2 t for t 2 0 6 ff is continuous then s for all s 2 0 Proof 1 9 is nondecreasing since the sets over which the in ma are taken do not increase as 8 gets bigger Consequently 9 has both left and right limits at each s 2 0 It remains to show that 9 is right continuousi Suppose to the contrary that there exists 8 2 0 6 gt 0 and a sequence snneN such that 3 gt 8 sn s and 9sn 2 98 6 That means that for any t E 9s9s 6 we have s lt S sn which is impossible because 3 st How would this argument break down if we put S instead of lt in the de nition of 9 What properties of f were actually used in this argument 2 Fort 2 0 de ne the set Flatft t 2 0 ft ft and let tr sup Flatft t inf Flatf t so that Flatft corresponds to the at stretch of f at the height By rightcontinuity t E Flatf t but fr 6 Flatft if and only if f is continuous at tTi Note also that tr t if and only ift is a point of right increase of i Last Updated May 7 2009 Lecture 8 REPRESENTATIONS OF MARTINGALES 2 of 11 By noting that inft 2 0 ft gt is the in mum of the set of all points lying above Flatft we conclude that sup Flatf t tr 2 t and that t if and only ift is a point of increase of This takes care of 2 3 Our next task is to show that f and 9 play symmetric roles ie that ft where inf8 2 0 98 gt t Note that 98 gt t implies 8 gt otherwise we would have 98 S S t Therefore 2 ft for each t 2 0 To get the opposite inequality S ft t 2 0 assume to the contrary that there exists t gt 0 such gt That means that for each 8 E ft ft we have 98 S t ie that ft 5 gt ft for all 5 gt 0 By right continuity of f we have 2 a contradiction 4 This follows now from symmetry between f and 9 5 We only prove the rst equality and let symmetry take care of the other one We choose 8 gt 0 the claim is trivial for 8 0 de ne 98 inft 2 0 z 2 8 and pick a sequence SnneN in 0 00 with Sn 8 Sn lt S Then since tzo ftgts2t20 ft28 we have 9Sn S 98 for each n E N and consequently 987 S Conversely suppose that there exists 6 gt 0 such that 98 2 987 6 ie that 98 2 9Sn 6 for each n E N That means that for t E 9Sng8n 6 we have E Sn8 a contradiction since 8 A 8 By 4 it will be enough to show that that each point is a point of right increase for 9 The continuity of f implies that thanks to 5 for each t 2 0 we have inf8 2 0 98 2 t inf8 2 0 98 gt t Taking t 98 in the above equality implies immediately that 8 is a point of right increase A on V D Even though we will have no use for its result the following problem can come in handy from time to time Problem 81 Let f 000 A 0 00 be a nondecreasing and cadlag function 1 Let go be a nonnegative Borel function on 0 00 and let 9 be the rightcontinuous inverse of Then 0 W W m soltgltsgtgt1ltmds 2 For 0 S a lt b lt 00 let u ab H 000 be a nondecreasing and continuous function Then 12 ub soltultsgtgtdfltultsgtgt swam TimeChanges De nition 82 An adapted cadlag and nondecreasing stochastic process Ts560oo is called a Change of time time Change if the random variable TS is a stopping time for each 8 2 0 Given a measurable process Xtte0oo the mapping X75 de nes a random variable for each 8 2 0 the stochastic process XTS560OO is called the time Change of a Xthqopw by 75560m Similarly given a ltration ftte0m we may de ne the timeChanged ltration gshqmw by g 75 Note that if fthqopw is rightcontinuous then so is gshqmmy Moreover if additionally the process Xthqopw is progressively measurable then then the timechanged process XTS560OO is gshqopoymeasurable lf Xthqopw happens to be cadlag then so is X75 ls the same true for continuous caglad processes Even though we are asking only for TS to be a stopping time for deterministic 8 2 0 this property extends to the class of all gshqopoystopping times Proposition 83 Supposing that the ltration fthqopw is rightcontinuous let a be a gshqopoystopping time and let Ts560oo be a time change Then the random variable TU is an fthqmmystopping time Last Updated May 7 2009 Lecture 8 REPRESENTATIONS OF MARTINGALES 3 of 11 Proof It is enough to deal with countablevalued stopping times Indeed a general stopping time a can be approximated from the right by a countablevalued sequence anneN of stopping times such that an a so that by right contiuity of 739 we have TU TU as The right continuity of the ltration fthqopw takes care of the rest We turn to a countablevalued stopping time a of the form a 221 sklAk where Ag 6 95k fTSk for h E N and A1 A2 N form a partition of 9 For t 2 0 we have 00 n g t U m S tWAk 161 Since Ag 6 fTSk we have Ak 75k S t 6 ft for each h E N and so TU S t 6 f D De nition 84 Let 7395560m be a time change A process Xthqopw is said to be TContinuous if it is continuous and X75 X75 for all s 2 0 as It is clear that XTSLQOPQ is a continuous process if Ts560oo is a time change and Xthqopw is a Tcontinuous process A deeper property of Tcontinuity is that it preserves martingality Before we state the precise result let us show that can go wrong Example 85 Let Bthqopw be a Brownian motion and let Bhemow be the rightcontinuous augmen tation of its natural ltration We de ne the family Ts560oo of stopping times by Tsinft20 Btgtss20 To show that 7395560m is a time change we simply express its paths as rightcontinuous inverses of the continuous and nondecreasing process Mt 0oo where M supsStBs The continuity of Mthqopw and the part 6 of Proposition 81 we ave B75 s for all s 2 0 as Therefore we have managed to timechange a martingale B into a constant ntevariation process s Note that the time change 739 is by no means continuous as all at stretches in Mthqopw correspond to jumps in 739 In fact it can be shown that in some sense 739 grows by jumps only Proposition 86 Let Ts560oo be a time change and let Mthqopo be a Tcontinuous local martingale Then the timechanged process Mnhqopw is a continuous local martingale for the ltration gshqomw where g 775 Proof By Tcontinuity the process N5560m given by N M75 is a continuous process adapted to the ltration gshqmmy De ne T inft 2 0 ME 2 n and set S infs 2 0 7395 2 Tn so that SnneN is a nondecreasing sequence of gshqopoystopping times with 5 A 00 as It is therefore enough to show that N5quot is a gshqopoymartingale for each n E N For s gt 0 by Tcontinuity of M we have Nmsn MTSASW MUS S and since 75A5n S Tn we conclude that S n as For a gshqmmy stopping time a the random variable lms is according to Proposition 83 an ftte0oostopping time and so is lms why Since lms S Tn we can use the optional sampling theorem to get Elequotl ElesJ 13er ElM l Elel ENo ENo quotl7 ToSn which implies that N5quot is a martingale D ms l We have seen that the class of local martingales is closed under time changes only if additional conditions are fulfulled The situation is quite diferent with semimartingales We state two important results the second one is known as Monroe s theorem the proof of which are outside the scope of these notes Both results deal with cadlag semimartingales ie with processes which can be decomposed into sums of a cadlag local martingale and a cadlag and adapted process of nite variation Proposition 87 Let Xthqopw be a cadlag semimartingale and let Ts560oo be a timechange Then the timechanged process XTSLQOPQ is a cadlag semimartingale not necessarily continuous even in X is Proposition 88 Let X5560oo be a cadlag semimartingale Then there exist a ltered probability space Q f ftLEO lP which supports an fte0mBrownian motion Bthqopw and a timechange Ts560oo such that X5560oo and Bnhqopw have the same nitedimensional distributions Last Updated May 7 2009 Lecture 8 REPRESENTATIONS OF MARTINGALES 4 of 11 A theorem of Dambis Dubins and Schwarz As we have seen in Example 85 a rightcontinuous inverse computed wwise of an adapted cadlag and nondecreasing process is clearly a time change The most important example of such a timechange for our purposes is obtained by taking a rightcontinuous inverse of the quadratic variation process of a continuous local martingale Mthqopoy 7395 inft 2 0 MMt gt s We note that the process 739 will not take the value 00 if the local martingale Mthqopw is divergent ie if M loo 00 as Theorem 89 Dambis Dubins and Schwarz Let Mthqopw be a divergent continuous local martingale with M0 0 and MM00 00 as De ne 7395 inft 2 0 MMt gt s and g fTS fors 2 0 Then the timechanged process B5560m where B M75 is a QBrownian motion and the local martingale M is a timechange of B ie M BMMt for t 2 0 Proof We rst show that M is Tcontinuous ie that M is constant on the intervals where MM is constant By stopping we may assume that both Mthqopw and MJWthopo are bounded We pick a rational number r 2 0 and de ne the process NE Turn 7 T t 2 0 which is clearly a martingale with N N M MT 7 M M The random variable Tr inft 2 0 NNt gt 0 is a stopping time and therefore the stopped martingale NT is in H2 with NTmNTrlt 0 for all t 2 0 Therefore NETT 0 for all t 2 0 and so M is constant on 1qu It is not hard to see that any interval of constancy of M is a closure of a countable union of intervals of the form r t Tr our claim follows It now follows from 86 that B is a continuous local martingalei To compute its quadratic variation we start from the fact that M 7 M M is a Tcontinous local martingale and therefore so is its timechange M35 7 MM75i By continuity of MM and part 6 of 81 we have MMTS si Therefore B52 7 s is a local martingale and so BB5 si Levyls characterization implies that Bshqopw is a Brownian motion Finally we need to show that Mt BMMt for all t 2 0 as By the de nition of B we have Emirth MnMMh while TMMt does not need to equal to t it follows from the proof of Proposition 81 that TMMt is the the right edge of the interval t 2 0 MMt M Continuity of MM and Tcontinuity of M imply now that MnMMh M D Remark 810 The restriction that M be divergent is here mostly for convenience It can be shown try to prove it that in that case M can still be written as M BMMt t 2 0 where B is a Brownian motion This time however B may have to be de ned on an extension of the original probability space The reason is that we do not have enough of M77 to construct the whole path of B fromi Martingals as stochastic integrals Let Mthqopw be a quot local martin alei A local martin ale Zte0oo is called the stochastic Dol ans Dade exponential of M denoted by Z M if L 81 Zn l Zu dMu for all t 2 0 as 0 ltols formula implies readily that the following prescription Mt XPML M7Mlt7t Z 07 de nes a stochastic exponential of Mi It can be shown that the process 5M is the only such process ie that the stochastic integral equation 81 has a unique solution in a specific sense i Recall that a subset G of a Hilbert space H is called total in H if the set of nite linear combinations of elements in G is dense in Hi This is equivalent to the statement Gi 0 where denoted the orthogonal componenti Last Updated May 7 2009 Lecture 8 REPRESENTATIONS OF MARTINGALES 5 of 11 Proposition 811 Let Bthqopw be a Brownian motion7 let fthqopw be the usual augmentation of its natural ltration7 and let foo at 2 0 Let I denote the set of all deterministic functions f z 0700 7gt R with the representation n71 8 2 ZAk1tk7tk1t7 n 6 N7 160 for some A17 7A 6 R and 0 lt t1 lt lt tn1 lt 00 For f E I let 5f denote the stochastic exponential Mf7 where M 0 fudBu Then 5f f GI is total in Mme Proof Let K span5f f E I be the set of all nite linear combinations of elements in 5f f E I7 and let Y E L2foo be such that ligt 07 for all h E K where7 as usual7 ligt lEDhl Pick a nite partition 0 to lt t1 lt lt tn lt tn lt 007 and let Q 0Btl7 7B1 For complex numbers 21722772n7 we de ne n 1 1 go21772n lE exp szBtk1 7Btkgt Y 7 160 and note that it is an analytic why function on C 7 ie7 that each section 2 gt gt go217 227 21117 27 21117 7 2 is analytic on C for a i 17277n and all 21722772n E C When 21 A1722 272n n7 for some real numbers A17 7 A777 we have n71 n 0 lt5f7ygt E exp 2 AkBtk17 Bu 7 k0 k0 71 1 7 Z Atk1 16 Yl 675 its Atk itkboo lv 7 where f Ak1tktk1t7 so that 873 go177n 07 for all A17A277nER Consider now the analytic function 2 gt7 goz7 A27 7 An7 for some choice A27 7 An 6 R This function must vanish7 because it is analytic in 2 and vanishes on the real axis a set with a limit point For 21 E C and A37 7 E 7 we note that the function 2 gt gt go2172737 7An is also analytic and7 by what we have just shown7 vanishes on the real axis Therefore7 go217227 A37 7An 07 for all 21722 6 C7 A37 7A We continue in the same manner7 replacing real arguments by complex ones7 until we reach the conclusion that go21772n 07 for all 21772n E C Since Q is generated by the random variables B11 7 B107 B12 7 B11 7 and Btu 7 Btnil 7 there exists a Borel function 77 R 7gt R such that EYE 77Bt1 7 BEO7B12 7 B117 7B 7 Btnil It follows now from 83 that AMe371 nelwgu sl 7suds d5 o for all complex 217 7 2777 where f is thejoint density of B11 7Bto7 B12 7B11 7 H and Btu 7Btn71 ln particluar7 if the real parts 0 217 727 vanish7 we recover nothing else but the fact that the Fourier transform of the function 77f vanishes identically Therefore7 the function 77f must vanish as7 and7 by positivity of f7 we conclude that EYE 07 as D Proposition 812 Let Bthqopw be a Brownian motion7 let fthqopw be the usual augmentation of its natural ltration7 and let 7 a t 2 0 For any random variable X E L2 700 there exists a A X Ralmost everywhere unique progressivelymeasurable process Htte0oo7 such that H5 du lt 00 an 84 XEX H743 as 0 Proof Let us deal with uniqueness7 rst Suppose that 84 holds for two progressive processes H and K Then 00 Do Do 0E HudBu7 KudBu lE Hu7Ku2du7 0 0 0 and it follows that Hu 7 Ku2 07 for A almost all u 2 07 libas Last Updated May 77 2009 Lecture 8 REPRESENTATIONS OF MARTINGALES 6 of 11 Next we deal with existence and start by de ning the subset I of the Hilbert space L2 700 by I Hu dBu H is progressive and H3 du lt 0 0 Let YnneN be a sequence in I with Yn fem HS dBu which converges to some Y 6 L2 700 Therefore YnneN is Cauchy and thanks to ltols isometry the sequence H neN is Cauchy in L20oo X 9 L20 00 X 9 Prog A X P where Prog denotes the progressive aalgebra By completeness of L20 00 X 9 H7 A H for some H E L20 00 X Consequently the Hu dBu lim H dBu limYn Y as 0 n 0 Therefore I is a closed subspace of L2 foo Moreover the representation 81 of the random variables 5f de ned in the statement of Proposition 811 tells us that 5f 7 1 E I for all functions with representation 82 It remains now to use Proposition 811 to conclude that the subspace I1 of L2A X 1 generated by I and the constant 1 is dense in L2 A X 117 because it contains a total set On the other hand I1 is clearly closed and so 1 L2 A X P It follows that each X E L2foo can be written as X zo HudBu 0 for some progressive H with H5 du lt 00 and some constant 10 E R Taking expectations of both sides yields that 10 Corollary 813 Let fthqopw be the usual agumentation of the Brownian ltration and let Mthqopw be an L2bounded cadlag fthqmmylocal martingale Then there exists a progressive process H with lEUO00 H3 du lt 00 such that 2 Mt M0 Hu dBu for all t 2 0 as 0 Proof By Proposition 812 we can express the last element M00 of M as Mm Mo HudBu 0 for some progressive H such that H3 du Let the continuous martingale N be de ned as L N M0 H1 dB 0 so that NDo Moo It follows that M lEMool lENool NE as for all t 2 0 and right continuity implies that M and N are indistinguishable D Proposition 814 Let fthqog be the usual agumentation of the Brownian ltration and let Mthqopw be a cadlag fthqmmylocal martingale Then there exists a progressive process H in LB such that 2 Mt M0 Hu dBu for all t 2 0 as 0 In particular each local martingale in the Brownian ltration is continuous Proof By stopping we can reduce the statement to the case of an uniformlyintegrable martingale note that due to lack of continuity we cannot necessarily assume that M is a bounded martingale Given the last element M00 of the uniformlyintegrable cadlag martingales M we can nd a random variable ME 6 1112700 such that 7 MOOHLl lt 5 for each n E N We de ne the squareintegrable martingale by ME EM ol and take a cadlag modi cation remember we can do that for any martingale on a ltration satisfying the usual conditions By the maximal inequality for martingales we have 85 usup 1M 7 M512 613EHMSL 7 M501 3 a 20 Last Updated May 7 2009 Lecture 8 REPRESENTATIONS OF MARTINGALES 7 of 11 However Corollary 813 implies that ME is actually continuous and so the equation 85 can be understood as follows with probability at most 56 the trajectory of M is in a 6neighborhood of a continuous function That means in particular that 86 PKAM 2 5 S be where the process AM is de ned by AM ME 7 M1 The lefthand side of does not depend on 5 so we conclude that S 6 as for all 6 gt 0 The constant 6 gt 0 was however arbitrary so AME 0 for all t 2 0 as ie M is continuous Now that we know that M is continuous we can reduce it by stopping to a squareintegrable martingale and nish the proof Girsanov s theorem Let Q be a probability measure on 7 equivalent to P ie VA 6 7 MA 0 if and only if 0 lts RadonNikodym derivative Z y is a nonnegative random variable in L1 with lEZ l The uniformly integrable martingale Z EMU t t 2 0 is called the density of Q with respect to P note that we can and do assume that Zthqopw is cadlag We will often use the shortcut Q local7 semi etc martingale for a process which is a local semi etc martingale for Q 7 727 Proposition 815 Let Xthqopw be a cadlag and adapted process Then X is a Qlocal martingale and only the product ZtXtthpQ is a cadlag Plocal martingale Before we give a proof here is a simple and useful lemma Since the measures involved are equivalent we are free to use the phrase almost surely77 without explicit mention of the probability Lemma 816 Let 9 H P be a probability space and let Q E H be a subaalgebra of H Givan a probability measure Q on H equivalent to P let Z 3P be its RadonNikodym derivative with respect to P For a random variable X E L1f we have XZ E L1 P and EQleQ WEWM where EQHQ denotes the conditional expectation on QHQ Proof First of all note that the RadonNikodym theorem implies that XZ E L1 P and that the set ElZlQl 0 has Qprobability and therefore libprobability 0 Indeed QlElZlgl 0i EQl1EZlQOl ElZ1EZlg0l EllElZ1EZlg0lg lElEZlQ1EZ gO 0 Therefore the expression on the righthand side is wellde ned almost surely and is clearly Qmeasurable Next we pick A E Q observe that EQllA EZ1g lElXZlgll ElZIAMleg lElXZlgll EllElZlgllA EZ1g lElXZlgll lElEZX1AlQ EQX1A and remember the de nition of conditional expectation D Proof of Proposition 815 Suppose rst that X is a Qmartingale Then lEWthfs X5 Q as By the tower property of conditional expectation the random variable Z is the RadonNikodym derivative of the restriction of Q with respect to the restriction of P on the probability space 9 f P prove this yourself Therefore we can use Lemma 816 with 7 playing the role of H and 75 the role Q and rewrite the Q martingale property of X as ZiglEXtZtl5 X5 Q 7 as ie Q N P as 87 Exam ZSXS 1P 7 as l We leave the other direction as well as the case of a local martingale to the reader D Last Updated May 7 2009 Lecture 8 REPRESENTATIONS OF MARTINGALES 8 of 11 Proposition 817 Suppose that the density process Zthqopw is continuous Let X be a continuous semi martingale under P with decomposition X M A Then X is also a Qsemimartingale and its Q semimartingale decomposition is given by X N A where t NM7F NAFandFt ZitdMZl 0 Proof The process F is clearly wellde ned continuous adapted and of nite variation so it will be enough to show that MiF is a Q local martingale Using Proposition 815 we only need to show that Y ZM7 F is a P local martingalel By ltols formula integrationbyparts the nitevariation part of Y is given by z 7 mm ZME 0 and it is easily seen to vanish using the associative property of Stieltjes integrationl B One of the most important applications of the above result is to the case of a Brownian motion Theorem 818 Girsanov Cameron and Martin Suppose that the ltration fthqopw is the usual aug mentation of the natural ltration generated by a Brownian motion thqopo I Let Q N P be a probability measure on f and let Zthqopw be the corresponding cadlag density process ie Z El Then here exists a progressivelymeasurable process Othqop in LB such that Z SUD91L dBut and z B 7 Bu du is a QBrownian motion 0 3 Conversely let 9tqopo be a progressivelymeasurable process in LB with the property that the process Z SUD91L dBu is a uniformlyintegrable martingale with Z00 gt 0 as For any probability measure Q N P such that Elo Zoo 2 Bti Budu tZO 0 is a QBrownian motion Proof A H V We start with an quot 39 of the martin ale 1 39 theorem Proposition 814 It implies that there exists a process p E LB such that t Zt1 pudBul 0 Since Z is continuous and bounded away from zero on each segment the process 916 090 given by 6 ptZt is in LB and we have t Z 1 ZuOudBul 0 Hence Z SUD91L dBu Proposition 817 states that B is a Q semimartingale with decomposition B B 7 F F where the continuous FVprocess F is given by t t 1 FL iBZu iZuOudu Budd 0 0 0 In particular B is a Q local martingalel On the other hand its quadratic variation as a limit in P and therefor in Qprobability is that of B so by Lveyls characterization B is a QBrownian motion 2 We only need to realize that any measure Q N P with Elfm ZDo will have Z as its density process The rest follows from D Last Updated May 7 2009 Lecture 8 REPRESENTATIONS OF MARTINGALES 9 of 11 Even though we stated it on 000 most of applications of the Girsanovls theorem are on nite intervals 0T T lt 0 The reason is that the condition that SUD91L dBu be uniformly integrable is hard to check or not satis es for practically relevant 9 The simplest conceivable example 6 u for all t 2 0 and u E R 0 gives rise to the exponential martingale Z eHBti l t which is not uniformly integrable why On any nite horizon 0 T the deterministic process uthT satis es the conditions of Girsanovls theorem and there exists a probability measure lib T on fT with the property that B BE 7 Mt is a lib T Brownian motion on 0 T It is clear furthermore that for T1 lt T2 PWT1 coincides with the restriction of lib T2 onto lei Our life would be easier if this consistency property could be extended all the way up to fool As we have already observed Girsanovls theorem does not help but we can easily construct a nitelyadditive measure P on the algebra U120 so that the central question becomes can P be extended to 700 Without giving a proof which is based on Kolmogorov s extension theorem we state the following result Proposition 819 Suppose that It160oo be the canonical process on C0oo and let P be the Wiener measure on 517 C0ooBC0oolP De ne the ltration fthqopw by 7 5gttazuu S s Let Lthqopw be a local martingale for which 5L is a martingale Then there exists a unique probability measure lP L on BC0 700 such that L leL P t lP t and CUP 75Lt for all t 2 0 Since the process Lt MB is a local martingale for which 5L is a martingale we can apply Proposition 819 to our setting and construct the measure lib 3 usually denoted simply by P which extends all lib Ti Of course all of this works under the rather strict assumption that everything is constructed on the canonical spacer Proposition 819 is suf cient however for one of the most important applications of Girsanov s theorem the study of the distributions of hitting times for the Brownian motion with drifti Indeed since we are interested in a distributional property of a path of a Brownian motion there is no loss of generality in assuming that we live on a canonical space as described in Proposition 819 We start with a useful result to showcase the relevant techniques and leave the derivation of the distributional identities related to hitting times to the reader We assume in what follows that the probability space and the ltration are as described in Proposition 819 Not to introduce more confusion than necessary we write BE instead of 1 when referring to the canonical process on C0oo i De ne Ta inf t 2 0 z t a for a gt 0 so that Ta is a stopping time for the rightcontinuous ltration fthqopoy The stopping time Ta can be also be seen as the rst hitting time of the level a for the FltBrownian motion with drift BE BE uti We use the shortcut QX 6 dz dz for the statement that is the density of the distribution of X under the measure By the formula derived in a homework we have a2 MT 6 dt 1765 at For T 2 0 the martingale exp MB 7 gift is uniformly integrable on 0 T so the optional sampling theorem implies that EleXPWBT 7 M2TlfmTl exPltMBmAT n A T Since Ta lt T 6 THAT Q fT and P and PM are equivalent on fT with the RadonNikodym derivative expiBT 7 u2T we have lP WTa S T E 1HST EHgtT1RST lEexpiBT 7 u2T1mST lEexpuB7aAT 7 u27 a T1nST lEexpiB7a 7 u2739a1mST lEexpia 7 i ra hagn iTew wrh 6 dt few L 6 dt a r 0 0 27ft3 Last Updated May 7 2009 Lecture 8 REPRESENTATIONS OF MARTINGALES 10 of 11 It follows immediately that the density of Ta under lP is given by a 7017 2 lP Ta 6 dt We 2 dt Equivalently the expression above is the density of the rst hitting time of the level a for a Brownian motion with drift a We quote the word density because if one tries to integrate it over all t 2 0 one gets 1w 7 0 W NEW Ta lt 00 70 We Qt dt 7 expia 7 Wall ln words if u and a have the same sign the Brownian motion with drift u will hit a sooner or later On the other hand if they differ in sign the probability that it will never get there is strictly positive and equal to Kazamaki s and Novikov s criteria The message of the second part of Theorem 818 is that given a drift process 9tqopo we can turn a Brownian motion in to a Brownian motion with drift 9 provided essentially that a certain exponential martingale is a Ul martingale Even though useful sufficient conditions for martingality of stochastic integrals are known the situation is much less pleasant in the case of stochastic exponentialsi The most wellknown criterion is the one of Novikovi Novikov s criterion is in turn implied by a slightly stronger criterion of Kazamakii We start with an auxiliary integrability result In addition to the role it plays in the proof of Kazamaki s criterion it is useful when one needs 5M to be a little more than just a martingale Lemma 820 Let Mthqopw with M0 0 be a continuous local martingale and let M be its stochastic exponential Suppose that Wu suplE lt 00 7 for some constant a gt where the supremum is taken over all nitevalued stopping times 739 Then 5M is a2 an lebounded martingale for p 471 6 100 Proof We pick a nite stopping time 739 and start from the following identity which is valid for all constants p s gt 0 HM ltpsMgteltPWgtMv For 1 gt s gt 0 we can use Holder s inequality note that Us and ll 7 s are conjugate exponents to obtain 88 EWMW S 13WVin8M EleXPaIiEMJlV S The rst term of the product is the s th power of the expectation a positive local martingale and therefore supermartingale sampled at a nite stopping time By the optional sampling theorem it is always nite actually it is less then 1 As for the second term one can easily check that the expression attains its minimum in s over 01 for s 2p 7 1 7 2xp2 7 p and that this minimum value equals to fp where fp L If we pick p 4417 then a and both terms on the right hand side of 88 2177172 p are bounded uniformly in 739 so that M is in fact a martingale and bounded in L why did we have to consider all stopping times 739 and only deterministic times D Proposition 821 Kazamaki7s criterion Let M be a continuous uniformly integrable martingale with M0 1 sup lEe M lt 00 7 where the supremum is taken over all nitevalued stopping times then 5M is a uniformly integrable mar tingale Last Updated May 7 2009 Lecture 8 REPRESENTATIONS OF MARTINGALES 11 of 11 Proof Note rst that the function I gt gt expz is a test function of uniform integrability so that the local martingale M is a uniformly integrable martingale and admits the last element Moor For the continuous martingale CM Where 0 lt c lt 1 is an arbitrary constant Lemma 8 20 and the assumption imply that the local martingale 5CM is in fact a martingale bounded in U for p Hi In particular it is uniformly integrablei Therefore 89 5 CM 1 exp CM 7 152 MMt 5 M Czecuic w 2 2 By letting t A 00 in 89 we conclude that 5M has the last element 5Moo and that the equality in 89 holds at t 00 as well By Holder7s inequality With conjugate exponents 152 and ll 7 02 we have 1 Emma E5Mm 2Eexp Mml1 2c Jensen s inequality implies that lEeXp Moo S lEeXpMool1Trc and so 1 1wMM EiexpeMwWHX We let 5 A 1 to get EM 2 1 Which together With the nonnegative supermartingale property of M implies that 5M is a uniformlyintegrable martingalei D Theorem 822 Novikovls criterion Let M be a continuous uniformly integrable martingale with M0 0 If 1 lEe M Ml lt 00 then M is a uniformly integrable martingale l l 1 Proof Since e2M 5M 2 e4 M Ml the CauchySchwarz inequality implies that new WWW2E5WW2 Ei mwt and Kazamakils criterion can be applied D Last Updated May 7 2009

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "Selling my MCAT study guides and notes has been a great source of side revenue while I'm in school. Some months I'm making over $500! Plus, it makes me happy knowing that I'm helping future med students with their MCAT."

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.