Stochastics II - Ulm€¦ · Deﬁnition1.1.7 LetX X t ,t> T bearandomfunctionwithvaluesin S,B...

Stochastics II

Lecture notes

Prof. Dr. Evgeny Spodarev

Ulm2015

Contents

1 General theory of random functions 11.1 Random functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Elementary examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Regularity properties of trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Differentiability of trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.5 Moments und covariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.6 Stationarity and Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.7 Processes with independent increments . . . . . . . . . . . . . . . . . . . . . . . . . 151.8 Additional exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2 Counting processes 192.1 Renewal processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 Poisson processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.2.1 Poisson processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.2.2 Compound Poisson process . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.2.3 Cox process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.3 Additional exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3 Wiener process 383.1 Elementary properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.2 Explicit construction of the Wiener process . . . . . . . . . . . . . . . . . . . . . . 39

3.2.1 Haar- and Schauder-functions . . . . . . . . . . . . . . . . . . . . . . . . . . 393.2.2 Wiener process with a.s. continuous paths . . . . . . . . . . . . . . . . . . 42

3.3 Distribution and path properties of Wiener processes . . . . . . . . . . . . . . . . 453.3.1 Distribution of the maximum . . . . . . . . . . . . . . . . . . . . . . . . . . 453.3.2 Invariance properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.4 Additional exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4 Lévy Processes 524.1 Lévy Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.1.1 Infinitely Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524.1.2 Lévy-Khintchine Representation . . . . . . . . . . . . . . . . . . . . . . . . 554.1.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594.1.4 Subordinators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

4.2 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

5 Martingales 675.1 Basic Ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 675.2 (Sub-, Super-)Martingales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 695.3 Uniform Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

i

ii Contents

5.4 Stopped Martingales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 745.5 Lévy processes and Martingales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 785.6 Martingales and Wiener Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805.7 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

Bibliography 85

1 General theory of random functions

1.1 Random functionsLet Ω,A,P be a probability space and S,B a measurable space, Ω,S x g.Definition 1.1.1A random element X Ω S is a ASB-measurable mapping (Notation: X > ASB), i.e.,

X1B ω > Ω Xω > B > A, B > B.

If X is a random element, then Xω is a realization of X for arbitrary ω > Ω.We say that the σ-algebra B of subsets of S is induced by the set system M (Elements of M

are also subsets of S), ifB

FaMF-σ-algebra on S

F

(Notation: B σM).If S is a toplological or metric space, then M is often chosen as a class of all open sets of S

and σM is called the Borel σ-algebra (Notation: B BS).Example 1.1.1 1. If S R, B BR, then a random element X is called a random

variable.

2. If S Rm, B BRm, m A 1, then X is called a random vector. Random variables andrandom vectors are considered in the lectures „Elementare Wahrscheinlichkeitsrechnungund Statistik“ and „Stochastik I“.

3. Let S be the class of all closed sets of Rm. Let

M A > S A 9B x g , B – arbitrary compactum of Rm .Then X Ω S is a random closed set.

As an example we consider n independent uniformly distributed points Y1, . . . , Yn > 0,1mand R1, . . . ,Rn A 0 (almost surely) independent random variables, which are defined on thesame probability space Ω,A,P as Y1, . . . , Yn. Consider X 8ni1BRiYi, where Brx y >Rm SSyxSS B r. Obviously, this is a random closed set. An example of a realization is providedin Figure 1.1.Exercise 1.1.1Let Ω,A and S,B be measurable spaces, B σM, where M is a class of subsets of S.Prove that X Ω S is ASB-measurable if and only if X1C > A, C >M.Definition 1.1.2Let T be an arbitrary index set and St,Btt>T a family of measurable spaces. A familyX Xt, t > T of random elements Xt Ω St defined on Ω,A,P and ASBt-measurablefor all t > T is called a random function (associated with St,Btt>T ).

1

2 1 General theory of random functions

Abb. 1.1: Example of a random set X 86i1BRiYi

Therefore it holds X Ω T St, t > T , i.e. Xω, t > St for all ω > Ω, t > T andX, t > ASBt, t > T . We often omit ω in the notation and write Xt instead of Xω, t.Sometimes St,Bt does not depend on t > T as well: St,Bt S,B for all t > T .

Special cases of random functions:

1. T b Z X is called a random sequence or stochastic process in discrete time.Example: T Z, N.

2. T b R X is called a stochastic process in continuous time.Example: T R, a, b, ª @ a @ b @ª, R.

3. T b Rd, d C 2 X is called a random field.Example: T Zd, Rd

, Rd, a, bd.

4. T b BRd X is called set-indexed process.If X is almost surely non-negative and σ-additive on the σ-algebra T , then X is calleda random measure.

The tradition of denoting the index set with T comes from the interpretation of t > T for thecases 1 and 2 as time parameter.For every ω > Ω, Xω, t, t > T is called a trajectory or path of the random function X.We would like to prove that the random function X Xt, t > T is a random element

within the corresponding function space, which is equipped with a σ-algebra that now is spec-ified.Let ST Lt>T St be the cartesian product of St, t > T , i.e., x > ST if xt > St, t > T . The

elementary cylindric set in ST is defined as

CT B, t x > ST xt > B ,where t > T is a selected point from T and B > Bt a subset of St. CT B, t therefore containsall trajectories x, which go through the „gate“ B, see Figure 1.2.Definition 1.1.3The cylindric σ-algebra BT is introduced as a σ-algebra induced in ST by the family of all

1 General theory of random functions 3

Abb. 1.2: Trajectories which pass a „gate“ Bt.

elementary cylinders. It is denoted by BT at>TBt. If Bt B for all t > T , then BT is writteninstead of BT .Lemma 1.1.1The family X Xt, t > T is a random function on Ω,A,P with phase spaces St,Btt>Tif and only if for every ω > Ω the mapping ω (Xω, is ASBT -measurable.Exercise 1.1.2Proof Lemma 1.1.1.Definition 1.1.4Let X be a random element X Ω S, i.e. X be ASB-measurable. The distribution of X isthe probability measure PX on S,B such that PXB PX1B, B > B.Lemma 1.1.2An arbitrary probability measure µ on S,B can be considered as the distribution of a randomelement X.

Proof Take Ω S, A B, P µ and Xω ω, ω > Ω.

When does a random function with given properties exist? A random function, which consistsof independent random elements always exists. This assertion is known asTheorem 1.1.1 (Lomnicki, Ulam):Let St,Bt, µtt>T be a sequence of probability spaces. It exists a random sequence X Xt, t > T on a probability space Ω,A,P (associated with St,Btt>T ) such that

1. Xt, t > T are independent random elements.

2. PXt µt on St,Bt, t > T .A lot of important classes of random processes is built on the basis of independent randomelements; cf. examples in Section 1.2.Definition 1.1.5Let X Xt, t > T be a random function on Ω,A,P with phase space St,Btt>T . Thefinite-dimensional distributions of X are defined as the distribution law Pt1,...,tn of Xt1, . . . ,XtnTon St1,...,tn ,Bt1,...,tn, for arbitrary n > N, t1, . . . , tn > T , where St1,...,tn St1 . . . Stn and


Bt1,...,tn Bt1 a . . .aBtn is the σ-algebra in St1,...,tn , which is induced by all sets Bt1 . . .Btn ,Bti > Bti , i 1, . . . , n, i.e., Pt1,...,tnC PXt1, . . . ,XtnT > C, C > Bt1,...,tn . In particular,for C B1 . . . Bn, Bk > Btk one has

Pt1,...,tnB1 . . . Bn PXt1 > B1, . . . ,Xtn > Bn.Exercise 1.1.3Prove that Xt1,...,tn Xt1, . . . ,XtnT is a ASBt1,...,tn-measurable random element.Definition 1.1.6Let St R for all t > T . The random function X Xt, t > T is called symmetric, if allof its finite-dimensional distributions are symmetric probability measures, i.e., Pt1,...,tnA

Pt1,...,tnA for A > Bt1,...,tn and all n > N, t1, . . . , tn > T , whereby

Pt1,...,tnA PXt1, . . . ,XtnT > A.Exercise 1.1.4Prove that the finite-dimensional distributions of a random function X have the followingproperties: for arbitrary n > N, n C 2, t1, . . . , tn ` T , Bk > Stk , k 1, . . . , n and an arbitrarypermutation i1, . . . , in of 1, . . . , n it holds:

1. Symmetry: Pt1,...,tnB1 . . . Bn Pti1 ,...,tin Bi1 . . . Bin2. Consistency: Pt1,...,tnB1 . . . Bn1 Stn Pt1,...,tn1B1 . . . Bn1The following theorem evidences that these properties are sufficient to prove the existence of

a random function X with given finite-dimensional distributions.Theorem 1.1.2 (Kolmogorov):Let Pt1,...,tn , n > N, t1, . . . , tn ` T be a family of probability measures on

Rm . . . Rm,BRma . . .a BRm,which fulfill conditions 1 and 2 of Exercise 1.1.4. Then there exists a random function X Xt, t > T defined on a probability space Ω,A,P with finite-dimensional distributionsPt1,...,tn .

Proof See [13], Section II.9.

This theorem also holds on more general (however not arbitrary!) spaces than Rm, on so-called Borel spaces, which are in a sense isomorphic to 0,1 ,B 0,1 or a subspace of that.Definition 1.1.7Let X Xt, t > T be a random function with values in S,B, i.e., Xt > S almost surelyfor arbitrary t > T . Let (T,C) be itself a measurable space. X is called measurable if themapping X ω, t(Xω, t > S, ω, t > Ω T , is AaC SB-measurable.Thus, Definition 1.1.7 not only provides the measurability of X with respect to ω > Ω:

X, t > ASB for all t > T , but X, > AaC SB as a function of ω, t. The measurability of Xis of significance if Xω, t is considered at random moments τ Ω T , i.e., Xω, τω. Thisis in particular the case in the theory of martingales if τ is a so-called stopping time for X. Thedistribution of Xω, τω might differ considerably from the distribution of Xω, t, t > T .


1.2 Elementary examplesThe theorem of Kolmogorov can be used directly for the explicit construction of random pro-cesses only in few cases, since for a lot of random functions their finite-dimensional distributionsare not given explicitly. In these cases a new random function X Xt, t > T is built asXt gt, Y1, Y2, . . ., t > T , where g is a measurable function and Yn a sequence of randomelements (also random functions), whose existence has already been ensured. For that we giveseveral examples.Let X Xt, t > T be a real-valued random function on a probability space Ω,A,P.1. White noise:

Definition 1.2.1The random function X Xt, t > T is called white noise, if all Xt, t > T , areindependent and identically distributed (i.i.d.) random variables.

White noise exists according to the Theorem 1.1.1. It is used to model the noise in(electromagnetic or acoustical) signals. If Xt Berp, p > 0,1, t > T , one meansSalt-and-pepper noise, the binary noise, which occurs at the transfer of binary data incomputer-networks. If Xt N 0, σ2, σ2 A 0, t > T , then X is called Gaussian whitenoise. It occures e.g. in acoustical signals.

2. Gaussian random function:Definition 1.2.2The random function X Xt, t > T is called Gaussian, if all of its finite-dimensionaldistributions are Gaussian, i.e. for all n > N, t1, . . . , tn ` T it holds

Xt1,...,tn Xt1, . . . ,Xtn N µt1,...,tn , Qt1,...,tn

,where the mean is given by µt1,...,tn EXt1, . . . ,EXtn and the covariance matrixis given by Pt1,...,tn covXti,Xtjni,j1.Exercise 1.2.1Proof that the distribution of an Gaussian random function X is uniquely determined byits mean value function µtEXt, t > T , and covariance function Cs, tEXsXt,s, t > T , respectively.

An example for a Gaussian process is the so-called Wiener process (or Brownian motion)X Xt, t C 0, which has the expected value zero (µt 0, t C 0) and the covariancefunction Cs, t min s, t, s, t C 0. Usually it is addionally required that the paths ofX are continuous functions.We shall investigate the regularity properties of the paths of random functions in moredetail in Section 1.3. Now we can say that such a process exists with probability one(with almost surely continuous trajectories).Exercise 1.2.2Prove that the Gaussian white noise is a Gaussian random function.

3. Lognormal- and χ2-functions:The random function X Xt, t > T is called lognormal, if Xt eY t, where Y


Y t, t > T is a Gaussian random function. X is called χ2-function, if Xt YY tY2,where Y Y t, t > T is a Gaussian random function with values in Rn, for whichY t N 0, I, t > T ; here I is the n n-unit matrix. Then it holds that Xt χ2

n,t > T .

4. Cosine wave:X Xt, t > R is defined by Xt º2 cos2πY tZ, where Y U0,1 and Z is arandom variable, which is independent of Y .Exercise 1.2.3Let X1,X2, . . . be i.i.d. cosine waves. Determine the weak limit of the finite-dimensionaldistributions of the random function 1º

n Pnk1Xkt, t > R for nª.

5. Poisson process:Let Ynn>N be a sequence of i.i.d. random variables Yn Expλ, λ A 0. The stochasticprocess X Xt, t C 0 defined as Xt max n > N Pnk1 Yk B t is called Poissonprocess with intensity λ A 0. Xt counts the number of certain events until the timet A 0, where the typical interval between two of these events is Expλ-distributed. Theseevents can be claim arrivals of an insurance portfolio the records of elementary particlesin the Geiger counter, etc. Then Xt represents the number of claims or particles withinthe time interval 0, t.

1.3 Regularity properties of trajectories

The theorem of Kolmogorov provides the existence of the distribution of a random functionwith given finite-dimensional distributions. However, it does not provide a statement aboutthe properties of the paths of X. This is understandable since all random objects are definedin the almost surely sense (a.s.) in probability theory, with the exception of a set A ` Ω withPA 0.Example 1.3.1Let Ω,A,P 0,1 ,B0,1, ν1, where ν1 is the Lebesgue measure on 0,1. We defineX Xt, t > 0,1 by Xt 0, t > 0,1 and Y Y t, t > 0,1 by

Y t 1, t U,0, sonst,

where Uω ω, ω > 0,1, is a U0,1-distributed random variable defined on Ω,A,P.Since PY t 0 1, t > T , because of PU t 0, t > T , it is clear that X d

Y . Nevertheless,X and Y have different path properties since X has continuous and Y has discontinuoustrajectories, and PXt 0, ¦t > T 1, where PY t 0, ¦t > T 0.It may well be that the „set of exceptions“ A (see above) is very different for Xt for every

t > T . Therefore, we require that all Xt, t > T , are defined simultaneously on a subset Ω0 b Ωwith PΩ0 1. The so defined random function X Ω0 T R is called modification ofX ΩT R. X and X differ on a set Ω~Ω0 with probability zero. Therefore we indicate laterwhen stating that „random function X possesses a property C“ that it exists a modificationof X with this property C. Let us hold it in the following definition:


Definition 1.3.1The random functions X Xt, t > T and Y Y t, t > T defined on the same prob-ability space Ω,A,P associated with St,Btt>T have equivalent trajectories (or are calledstochastically indistinguishable) if

A ω > Ω Xω, t x Y ω, t for a t > T > Aand PA 0.This term implies that X and Y have paths, which coincide with probability one.

Definition 1.3.2The random functionsX Xt, t > T and Y Y t, t > T defined on the same probabilityspace Ω,A,P are called (stochastically) equivalent, if

Bt ω > Ω Xω, t x Y ω, t > A, t > T,and PBt 0, t > T .We also can say that X and Y are versions or modifications of one and the same random

function. If the space Ω,A,P is complete (i.e. the implication of A > A PA 0 is for allB ` A: B > A (and then PB 0)), then the indistinguishable processes are stochasticallyequivalent, but vice versa is not always true (it is true for so-called separable processes. Thisis the case if T is countable).Exercise 1.3.1Prove that the random functions X and Y in Example 1.3.1 are stochastically equivalent.Definition 1.3.3The random functions X Xt, t > T and Y Y t, t > T (not necessarily defined onthe same probability space) are called equivalent in distribution, if PX PY on St,Btt>T .Notation: X d

Y .According to Theorem 1.1.2 it is sufficient for the equivalence in distribution of X and Y that

they possess the same finite-dimensional distributions. It is clear that stochastic equivalenceimplies equivalence in distribution, but not the other way around.Now, let T and S be Banach spaces with norms S ST and S SS , respectively. The random

function X Xt, t > T is now defined on Ω,A,P with values in S,B.Definition 1.3.4The random function X Xt, t > T is called

a) stochastically continuous on T , if Xs PÐÐst

Xt, for arbitrary t > T , i.e.PSXs XtSS A εÐÐ

st0, for all ε A 0.

b) Lp-continuous on T , p C 1, if Xs LpÐÐst

Xt, t > T , i.e. ESXsXtSp ÐÐst

0. For p 2the specific notation „continuity in the square mean “is used.

c) a.s. continuous on T , if Xs f.s.ÐÐst

Xt, t > T , i.e., PXsÐÐst

Xt 1, t > T .

d) continuous, if all trajectories of X are continuous functions.


In applications one is interested in the cases c) and d), although the weakest form of continuityis the stochastic continuity.

Lp-continuity Ô stochastic continuity Ô a.s. continuity Ô continuity of all paths

Why are cases c) and d) important? Let us consider an example.Example 1.3.2Let T 0,1 and Ω,A,P be the canonical probability space with Ω R0,1, i.e. Ω Lt>0,1R.Let X Xt, t > 0,1 be a stochastic process on Ω,A,P. Not all events are elements ofA, like e.g. A ω > Ω Xω, t 0 for all t > 0,1 9t>0,1 Xω, t 0, since this is anintersection of measurable events from A in uncountable number. If however X is continuous,then all of its paths are continuous functions and one can write A 9t>D Xω, t 0, whereD is a dense countable subset of 0,1, e.g., D Q 9 0,1. Then it holds that A > A.However, in many applications (like e.g. in financial mathematics) it is not realistic to

consider stochastic processes with continuous paths as models for real phenomena. Therefore,a bigger class of possible trajectories of X is allowed: the so-called càdlàg-class (càdlàg =continue à droite, limitée à gauche (fr.)).Definition 1.3.5A stochastic process X Xt, t > R is called càdlàg, if all of its trajectories are right-sidecontinuous functions, which have left-side limits.Now, we would like to consider the properties of the notion of continuity (introduced above)

in more detail. One can note e.g. that the stochastic continuity is a property of the two-dimensional distribution Ps,t of X. This is shown by the following lemma.Lemma 1.3.1Let X Xt, t > T be a random function associated with S,B, where S and T are Banachspaces. The following statements are equivalent:

a) Xs PÐÐst0

Y ,

b) Ps,td

ÐÐÐs,tt0

PY,Y ,

where t0 > T and Y is a S-valued ASB-random element. For the stochastic continuity of X, oneshould choose t0 > T arbitrarily and Y Xt0.Proof a bXs P

ÐÐst0

Y means Xs,Xt PÐÐÐs,tt0

Y,Y .PSXs,Xt Y,Y S2´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶SXsY S2SSXtY S2S1~2

A ε D PSXs Y SS A ε~2 PSXt Y SS A ε~2ÐÐÐs,tt0

0

This results in Ps,td PY,Y , since P

-convergence is stronger than d-convergence.

b aFor arbitrary ε A 0 we consider a continuous function gε R 0,1 with gε0 0, gεx 1,


x ¶ Bε0. It holds for all s, t > T that

EgεSXs XtSS PSXs XtSS A ε EgεSXs XtSS1SXs XtSS B ε,hence PSXs XtSS A ε B EgεSXs XtSS RS RS gεSx ySSPs,tdx, y ÐÐ

st0tt0

RS RS gεSx ySSPY,Y dx, y 0, since PY,Y is concentrated on x, y > S2 x y and

gε0 0. Thus Xsst0 is a fundamental sequence (in probability), therefore Xs PÐÐst0

Y.

It may be that X is stochastically continuous, although all of the paths of X have jumps,i.e. X cannot possess any a.s. continuous modification. The descriptive explanation for thatis that such X may have a jump at concrete t > T with probability zero. Therefore jumps ofthe paths of X always occur at different locations.Exercise 1.3.2Prove that the Poisson process is stochastically continuous, although it does not possess anya.s. continuous modification.Exercise 1.3.3Let T be compact. Prove that if X is stochastically continuous on T , then it also is uniformlystochastically continuous, i.e., for all ε, η A 0 §δ A 0, such that for all s, t > T with Ss tST @ δ itholds that PSXs XtSS A ε @ η.Now let S R, EX2t @ ª, t > T , EXt 0, t > T . Let Cs, t E XsXt be the

covariance function of X.Lemma 1.3.2For all t0 > T and a random variable Y with EY 2 @ª the following assertions are equivalent:

a) Xs L2ÐÐst0

Y

b) Cs, tÐÐÐs,tt0

EY 2

Proof a bThe assertion results from the Cauchy-Schwarz inequality:SCs, t EY 2S SEXsXt EY 2S SE Xs Y Y Xt Y Y EY 2S

B ESXs Y Xt Y S ESXs Y Y S ESXt Y Y SB

¿ÁÁÁÀEXs Y 2 EXt Y 2´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶SSXsY SS2L2 SSXtY SS2

L2

¿ÁÁÁÀEY 2 EXs Y 2´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶SSXsY SS2L2

¿ÁÁÁÀEY 2 EXt Y 2´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶SSXtY SS2L2

ÐÐÐs,tt0

0

with assumption a).b a

EXs Xt2 EXs2

2EXsXt EXt2

Cs, s Ct, t 2Cs, tÐÐÐs,tt0

2EY 2 2EY 2

0.


Thus, Xs, s t0 is a fundamental sequence in the L2-sense, and we get Xs L2ÐÐst0

Y .

A random function X, which is continuous in the mean-square sense, may still have uncon-tinuous trajectories. In most of the cases which are practically relevant, X however has an a.s.continuous modification. Later on this will become more precise by stating a correspondingtheorem.Corollary 1.3.1The random function X, which satisfies the conditions of Lemma 1.3.2, is continuous on T inthe mean-square sense if and only if its covariance function C T 2 R is continuous on thediagonal diagT 2 s, t > T 2

s t, i.e., lims,tt0 Cs, t Ct0, t0 VarXt0 for all t0 > T.

Proof Choose Y Xt0 in Lemma 1.3.2.

Remark 1.3.1If X is not centered, then the continuity of µ together with the continuity of C on diagT 2

is required to ensure the L2-continuity of X on T .Exercise 1.3.4Give an example of a stochastic process with a.s. discontinuous trajectories, which is L2-continuous.Now we consider the property of (a.s.) continuity in more detail. As mentioned before,

we can merely talk about continuous modification or version of a process. The possibility topossess such a version also depends on the properties of the two-dimensional distributions ofthe process. This is proven by the following theorem (originally proven by A. Kolmogorov).Theorem 1.3.1Let X Xt, t > a, b, ª @ a @ b B ª be a real-valued stochastic process. X has acontinuous version, if there exist constants α, c, δ A 0 such that

ESXt h XtSα @ cShS1δ, t > a, b, (1.3.1)

for sufficiently small ShS.Proof See, e.g. [7], Theorem 2.23.

Now we turn to processes with càdlàg-trajectories. Let Ω,A,P be a complete probabilityspace.Theorem 1.3.2Let X Xt, t C 0 be a real-valued stochastic process and D a countable dense subset of0,ª. If

a) X is stochastically right-handside continuous, i.e., Xt h PÐÐÐh0

Xt, t > 0,ª,b) the trajectories of X at every t > D have finite right- and left-handside limits, i.e.,S limh0Xt hS @ª, t >D a.s.,

then X has a version with a.s. càdlàg-paths.Without proof.


Lemma 1.3.3Let X Xt, t C 0 and Y Y t, t C 0 be two versions of a random function, both definedon the probability space Ω,A,P, with property that X and Y have a.s. right-handsidecontinuous trajectories. Then X and Y are indistinguishable.

Proof Let ΩX ,ΩY be „sets of exception“, for which the trajectories of X and Y , respec-tively are not right-sided continuous. It holds that PΩX PΩY 0. Consider At ω > Ω Xω, t x Y ω, t, t > 0,ª and A 8t>QAt, where Q Q 9 0,ª. Since X andY are stochastically equivalent, it holds that PA 0 and therefore

P A B PA PΩX PΩY 0,

where A A 8 ΩX 8 ΩY . Therefore Xω, t Y ω, t holds for t > Q and ω > Ω A. Now,we prove this for all t C 0. For arbitrary t C 0 a sequence tn ` Q exists, such that tn t.Since Xω, tn Y ω, tn for all n > N and ω > Ω A, it holds that Xω, t limnªXω, tn limnª Y ω, tn Y ω, t for t C 0 and ω > Ω A. Therefore X and Y are indistinguishable.

Corollary 1.3.2If càdlàg-processes X Xt, t C 0 and Y Y t, t C 0 are versions of the same randomfunction then they are indistinguishable.

1.4 Differentiability of trajectoriesLet T be a linear normed space.Definition 1.4.1A real-valued random function X Xt, t > T is differentiable on T in direction h > Tstochastically, in the Lp-sense, p C 1, or a.s., if

liml0

Xt hl Xtl

X

ht, t > Texists in the corresponding sense, namely stochastically, in the Lp-space or a.s..The Lemmas 1.3.1 - 1.3.2 show that the stochastic differentiability is a property that is deter-

mined by three-dimensional distributions ofX (because the common distribution of XthlXtl

and XthlXtl

should converge weakly), whereas the differentiability in the mean-squaresense is determined by the smoothness of the covariance function Cs, t.Exercise 1.4.1Show that

1. the Wiener process is not stochastically differentiable on 0,ª.2. the Poisson process is stochastically differentiable on 0,ª, however not in the Lp-mean,p A 1.

Lemma 1.4.1A centered random function X Xt, t > T (i.e., EXt 0, t > T ) with EX2t @ª, t > Tis L2-differentiable in t > T in direction h > T if its covariance function C is differentiabletwice in t, t in direction h, i.e., if § C

hht, t ∂2Cs,t∂sh∂th

Ust. X

ht is L2-continuous in t > T


if C

hhs, t ∂2Cs,t∂sh∂th

is continuous in s t. Moreover, C

hhs, t is the covariance function ofX

h X

ht, t > T.Proof According to Lemma 1.3.2 it is enough to show that

I liml,l0

EXt lh Xtl

Xs lh Xs

l

exists for s t. Indeed we get

I 1llCt lh, s lh Ct lh, s Ct, s lh Ct, s

1lCt lh, s lh Ct lh, s

lCt, s lh Ct, s

lÐÐÐ

l,l0C

hh s, t .All other statements of the lemma result from this relation.

Remark 1.4.1The properties of the L2-differentiability and a.s. differentiability of random functions aredisjoint in the following sense: there are stochastic processes that have L2-differentiable paths,although they are a.s. discontinuous, and vice versa, processes with a.s. differentiable pathsare not always L2-differentiable, since e.g. the first derivative of their covariance function isnot continuous.Exercise 1.4.2Give appropriate examples!

1.5 Moments und covarianceLet X Xt, t > T be a random function that is real-valued, and let T be an arbitraryindex space.Definition 1.5.1The mixed moment µj1,...,jnt1, . . . , tn of X of order j1, . . . , jn > Nn, t1, . . . , tn > T is givenby µj1,...,jnt1, . . . , tn E Xj1t1 . . . Xjntn, where it is required that the expected valueexists and is finite. Then it is sufficient to assume that ESXtSj @ ª for all t > T and j

j1 . . . jn.Important special cases:

1. µ t µ1t EXt, t > T – mean value function of X.

2. µ1,1 s, t E XsXt Cs, t – (non-centered) covariance function of X. Whereasthe centered covariance function is: Ks, t covXs,Xt µ1,1s, t µsµt,s, t > T .

Exercise 1.5.1Show that the centered covariance function of a real-valued random function X

1. is symmetric, i.e., Ks, t Kt, s, s, t > T .


2. is positive semidefinite, i.e., for n > N, t1, . . . , tn > T , z1, . . . , zn > R it holds thatn

Qi,j1

Kti, tjzizj C 0.

3. satisfies Kt, t VarXt, t > T .Property 2) also holds for the non-centered covariance function Cs, t.The mean value function µt shows a (non random) trend. If µt is known, the random

function X can be centered by considering a random function Y Y t, t > T with Y t Xt µt, t > T .The covariance function Ks, t (Cs, t, respectively) contains information about the depen-

dence structure of X. Sometimes the correlation function Rs, t Ks,t»Ks,sKt,t for all s, t > T :

Ks, s VarXs A 0, Kt, t VarXt A 0 is used instead of K and C, respectively. Becauseof the Cauchy-Schwarz inequality it holds that SRs, tS B 1, s, t > T . The set of all mixedmoments in general does not (uniquely) determine the distribution of a random function.Exercise 1.5.2Give an example of different random functions X Xt, t > T und Y Y t, t > T, forwhich it holds that EXt EY t, t > T and EXsXt EY sY t, s, t > T .Exercise 1.5.3Let µ T R be a measurable function and K T T R be a positive semidefinite sym-metric function. Prove that a random function X Xt, t > T exists with EXt µt,covXs,Xt Ks, t, s, t > T .Let now X Xt, t > T be a real-valued random function with E SXtSk @ª, t > T , for a

k > N.Definition 1.5.2The mean increment of order k of X is given by γks, t EXs Xtk, s, t > T .Special attention is paid to the function γs, t 1

2γ2s, t 12EXsXt2, s, t > T , which

is called variogram of X. In geostatistics the variogram is often used instead of the covariancefunction. A lot of times we discard the condition EX2t @ ª, t > T , instead we assume thatγs, t @ª for all s, t > T .Exercise 1.5.4Prove that there exist random functions without finite second moments with γs, t @ ª,s, t > T .Exercise 1.5.5Show that for a random function X Xt, t > T with mean value function µ and covariancefunction K it holds that:

γs, t Ks, s Kt, t2

Ks, t 12µs µt2, s, t > T.

If the random function X is complex-valued, i.e., X Ω T C, with E SXtS2 @ ª, t > T ,then the covariance function of X is introduced as Ks, t EXs EXsXt EXt,s, t > T , where z is the complex conjugate of z > C. Then it holds that Ks, t Kt, s, s, t > T ,and K is positive semidefinite, i.e, for all n > N, t1, . . . , tn > T , z1, . . . , zn > C it holds thatPni,j1Kti, tjzizj C 0.


1.6 Stationarity and Independence

T be a subset of the linear vector space with operations , over space R.Definition 1.6.1The random function X Xt, t > T is called stationary (strict sense stationary) if for alln > N, h, t1, . . . , tn > T with t1 h, . . . , tn h > T it holds that:

PXt1,...,Xtn PXt1h,...,Xtnh,i.e., all finite-dimensional distributions of X are invariant with repsect to translations in T .Definition 1.6.2A (complex-valued) random function X Xt, t > T is called second-order stationary (orwide sense stationary) if ESXtS2 @ ª, t > T , and µt EXt µ, t > T , Ks, t

covXs,Xt Ks h, t h for all h, s, t > T s h, t h > T .If X is second-order stationary, it is convenient to introduce a function Kt K0, t, t > T

whereby 0 > T .Strict sense stationarity and wide sense stationarity do not result from each other. However

it is clear that if a complex-valued random function is strict sense stationary and possessesfinite second-order moments, then the function is also second-order stationary.Definition 1.6.3A real-valued random function X Xt, t > T is intrinsic second-order stationary if γks, t,s, t > T exist for k B 2, and for all s, t, h > T , s h, t h > T it holds that γ1s, t 0,γ2s, t γ2s h, t h.For real-valued random functions, intrinsic second-order stationarity is more general than

second-order stationarity since the existence of ESXtS2, t > T is not required.The analogue of the stationarity of increments of X also exists in strict sense.

Definition 1.6.4Let X Xt, t > T be a real-valued stochastic process, T ` R. It is said that X

1. possesses stationary increments if for all n > N, h, t0, t1, t2, . . . , tn > T , witht0 @ t1 @ t2 @ . . . @ tn, ti h > T , i 0, . . . , n the distribution ofXt1 h Xt0 h, . . . ,Xtn h Xtn1 hdoes not depend on h.

2. possesses independent increments if for all n > N, t0, t1, . . . , tn > T with t0 @ t1 @ . . . @ tnthe random variables Xt0,Xt1Xt0, . . . ,XtnXtn1 are pairwise independent.

Let S1,B1 and S2,B2 be measurable spaces. In general it is said that two randomelements X Ω S1 and X Ω S2 are independent on the same probability space Ω,A,Pif PX > A1, Y > A2 PX > A1PY > A2 for all A1 > B1, A2 > B2.This definition can be applied to the independence of random functions X and Y with phase

space ST ,BT , since they can be considered as random elements with S1 S2 ST , B1 B2 BT (cf. Lemma 1.1.1). The same holds for the independence of a random element (or a randomfunction) X and of a sub-σ-algebra G > A: this is the case if PX > A 9G PX > APG,for all A > B1, G > G (or A > BT , G > G).


1.7 Processes with independent incrementsIn this section we concentrate on the properties and existence of processes with independentincrements.Let ϕs,t, s, t C 0 be a family of characteristic functions of probability measures Qs,t,

s, t C 0 on BR, i.e., for z > R, s, t C 0 it holds that ϕs,tz RR eizxQs,tdx.Theorem 1.7.1There exists a stochastic process X Xt, t C 0 with independent increments with theproperty that for all s, t C 0 the characteristic function of Xt Xs is equal to ϕs,t if andonly if

ϕs,t ϕs,uϕu,t (1.7.1)

for all 0 B s @ u @ t @ª. Thereby the distribution of X0 can be chosen arbitrarily.

Proof The necessity of the condition (1.7.1) is clear since for all s > 0,ª s @ u @ t it holdsXtXs Xt Xu´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

Y1

Xu Xs´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶Y2

, and XtXu and XuXs are independent.Then it holds ϕs,t ϕY1Y2 ϕY1ϕY2 ϕs,uϕu,t.Now we prove the sufficiency.If the existence of a process X with independent increments and property ϕXtXs ϕs,ton a probability space Ω,A,P had already been proven, one could define the characteristicfunctions of all of its finite-dimensional distributions with the help of ϕs,t as follows.Let n > N, 0 t0 @ t1 @ . . . @ tn @ª and Y Xt0,Xt1Xt0, . . . ,XtnXtn1. Theindependence of increments results in

ϕY z0, z1, . . . , zn´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶z

Eei`z,Y e ϕXt0z0ϕt0,t1z1 . . . ϕtn1,tnzn, z > Rn1,

where the distribution of Xt0 is an arbitrary probability measure Q0 on BR. For Xt0,...,tn Xt0,Xt1, . . . ,Xtn however it holds that Xt0,...,tn AY , where

A

1 0 0 . . . 01 1 0 . . . 01 1 1 . . . 0. . . . . . . . . . . . . . .1 1 1 . . . 1

.

Then ϕXt0,...,tn z ϕAY z Eei`z,AY e Eei`Az,Y e ϕY Az holds. Therefore the distribu-tion ofXt0,...,tn possesses the characteristic function ϕXt0,...,tn z ϕQ0l0ϕt0,t1l1 . . . ϕtn1,tnln,where l l1, l1, . . . , ln Az, thus¢

¦¤l0 z0 . . . znl1 z1 . . . zn

ln zn

Thereby ϕXt0 ϕQ0 and ϕXt1,...,tn z1, . . . , zn ϕXt0,...,tn 0, z1, . . . , zn holds for all zi > R.Now we prove the existence of such a process X.


For that we construct the family of characteristic functions

ϕt0 , ϕt0,t1,...,tn , ϕt1,...,tn , 0 t0 @ t1 @ . . . @ tn @ª, n > Nfrom ϕQ0 and ϕs,t, 0 B s @ t as above, thus

ϕt0 ϕQ0 , ϕt1,...,tn0, z1, . . . , zn ϕt0,t1,...,tn0, z1, . . . , zn, zi > R,ϕt0,...,tnz ϕt0z1 . . . znϕt0,t1z1 . . . zn . . . ϕtn1,tnzn.

Now we have to check whether the corresponding probability measures of these characteristicfunctions fulfill the conditions of Theorem 1.1.2. We will do that in equivalent form since byExercise 1.8.1 the conditions of symmetry and consistency in Theorem 1.1.2 are equivalent to:

a) ϕti0 ,...,tin zi0 , . . . , zin ϕt0,...,tnz0, . . . , zn for an arbitrary permutation 0,1, . . . , n (i0, i1, . . . , in,b) ϕt0,...,tm1,tm1,...,tnz0, . . . , zm1, zm1, . . . , zn ϕt0,...,tnz0, . . . ,0, . . . , zn, for all

z0, . . . , zn > R, m > 1, . . . , n.Condition a) is obvious. Conditon b) holds since

ϕtm1,tm0 zm1 . . . znϕtm,tm1zm1 . . . zn ϕtm1,tm1zm1, . . . , znfor all m > 1, . . . , n. Thus, the existence of X is proven.

Example 1.7.1 1. If T N0 N80, thenX Xt, t > N0 has independent incrementsif and only if Xn d

Pni0 Yi, where Yi are independent random variables and Ynd

Xn Xn 1, n > N. Such a process X is called random walk. It also may be definedfor Yi with values in Rm.

2. The Poisson process with intensity λ has independent increments (we will show thatlater).

3. The Wiener process possesses independent increments.Exercise 1.7.1Proof that!Exercise 1.7.2LetX Xt, t C 0 be a process with independent increments and g 0,ª R an arbitrary(deterministic) function. Show that the process Y Y t, t C 0 with Y t Xtgt, t C 0,also possesses independent increments.

1.8 Additional exercisesExercise 1.8.1Prove the following assertion: The family of probability measures Pt1,...,tn on Rn,BRn,n C 1, t t1, . . . , tn > Tn fulfills the conditions of the theorem of Kolmogorov if and only iffor n C 2 and for all s s1, . . . , sn > Rn the following conditions are fulfilled:

a) ϕPt1,...,tn s1, . . . , sn ϕPtπ1,...,tπn sπ1, . . . , sπn for all π > Sn.


b) ϕPt1,...,tn1s1, . . . , sn1 ϕPt1,...,tn s1, . . . , sn1,0.

Remark: ϕ denotes the characteristic function of the corresponding measure. Sn denotes thegroup of all permutations π 1, . . . , n 1, . . . , n.Exercise 1.8.2Show the existence of a random function whose finite-dimensional distributions are multivariate-normally distributed and explicitly give the measurable spaces Et1,...,tn ,Et1,...,tn.Exercise 1.8.3Give an example of a family of probability measures Pt1,...,tn , which do not fulfill the conditionsof the theorem of Kolmogorov.Exercise 1.8.4Let X Xt, t > T and Y Y t, t > T be two stochastic processes which are defined onthe same complete probability space Ω,F ,P and which take values in the measurable spaceS,B.

a) Proof that: X and Y are stochastically equivalent Ô PX PY .

b) Give an example of two processes X and Y for which holds: PX PY , but X and Y arenot stochastically equivalent.

c) Proof that: X and Y are stochastically indistinguishable ÔX and Y are stochasticallyequivalent.

d) Proof in the case of countability of T : X and Y are stochastically equivalent ÔX andY are stochastically indistinguishable.

e) Give in the case of uncountability of T an example of two processes X and Y for whichholds: X and Y are stochastically equivalent but not stochastically indistinguishable.

Exercise 1.8.5Let W W t, t > R be a Wiener Process. Which of the following processes are Wienerprocesses as well?

a) W1 W1t W t, t > R,b) W2 W2t ºtW 1, t > R,c) W3 W3t W 2t W t, t > R.

Exercise 1.8.6Given a stochastic process X Xt, t > 0,1 which consists of idependent and identicallydistributed random variables with density fx, x > R. Show that such a process can not becontinuous in t > 0,1.Exercise 1.8.7Give an example of a stochastic process X Xt, t > T which is stochastically continuouson T , and prove why this is the case.Exercise 1.8.8In connection with the continuity of stochastic processes the so-called criterion of Kolmogorov


plays a central role. (see also theorem 1.3.1 in the lecture notes): Let X Xt, t > a, b bea real-valued stochastic process. If constants α, ε A 0 and C Cα, ε A 0 exist such that

ESXt h XtSα B C ShS1ε (1.8.1)

for sufficient small h, then the process X possesses a continuous modification. Show that:

a) If you fix the variable ε 0 in condition (1.8.1), then in general the condition is notsufficient for the existence of a continuous modification. Hint: Consider the Poissonprocess.

b) The Wiener process W W t, t > 0,ª possesses a continuous modification. Hint:Consider the case α 4.

Exercise 1.8.9Show that the Wiener process W is not stochastically differentiable at any point t > 0,ª.Exercise 1.8.10Show that the covariance function Cs, t of a complex-valued stochastic process X Xt, t >T

a) is symmetric, i.e. Cs, t Ct, s, s, t > T ,b) fulfills the identity Ct, t VarXt, t > T ,c) is positive semidefinite, i.e. for all n > N, t1, . . . , tn > T , z1, . . . , zn > C it holds that:

n

Qi1

n

Qj1

Cti, tjzizj C 0.

Exercise 1.8.11Show that it exists a random function X Xt, t > T which simultaneously fulfills theconditions:

• The second moment EX2 does not exist.

• The variogram γs, t is finite for all s, t > T .

Exercise 1.8.12Give an example of a stochastic process X Xt, t > T whose paths are simultaneouslyL2-differentiable but not almost surely differentiable, and prove why this is the case.Exercise 1.8.13Give an example of a stochastic process X Xt, t > T whose paths are simultaneouslyalmost surely differentiable but not L1-differentiable, and prove why this is the case.Exercise 1.8.14Proof that the Wiener process possesses independent increments.Exercise 1.8.15Proof: A (real-valued) stochastic process X Xt, t > 0,ª with independent incrementsalready has stationary increments if the distibution of the random variable Xt h Xh isindependent of h.

2 Counting processes

In this chapter we consider several examples of stochastic processes which model the countingof events and thus possess piecewise constant paths.Let Ω,A,P be a probability space and Snn>N a non-decreasing sequence of a.s. non-

negative random variables, i.e. 0 B S1 B S2 B . . . B Sn B . . ..Definition 2.0.1The stochastic process N Nt, t C 0 is called counting process if

Nt ª

Qn1

1Sn B t,where 1A is the indicator function of the event A > A.Nt counts the events which occur at Sn until time t. Sn e.g. may be the time of occurence

of

1. the n-th elementary particle in the Geiger counter, or

2. a damage in the insurance of material damage, or

3. a data paket at a server within a computer network, etc.

A special case of the counting processes are the so-called renewal processes.

2.1 Renewal processesDefinition 2.1.1Let Tnn>N be a sequence of i.i.d. non-negative random variables with PT1 A 0 A 0. Acounting process N Nt, t C 0 with N0 0 a.s., Sn Pnk1 Tk, n > N, is called renewalprocess. Thereby Sn is called the time of the n-th renewal, n > N.The name „renewal process“ is given by the following interpretation. The „interarrival times“

Tn are interpreted as the lifetime of a technical spare part or mechanism within a system, thusSn is the time of the n-th break down of the system. The defective part is immediately replacedby a new part (comparable with the exchange of a lightbulb). Thus, Nt is the number ofrepairs (the so-called „renewals“) of the system until time t.Remark 2.1.1 1. It is Nt ª if Sn B t for all n > N.

2. Often it is assumed that only T2, T3, . . . are identically distributed with ETn @ ª. Thedistribution of T1 is freely selectable. Such a process N Nt, t C 0 is called delayedrenewal process (with delay T1).

3. Sometimes the requirement Tn C 0 is omitted.

19

20 2 Counting processes

Abb. 2.1: Konstruktion und Trajektorien eines Erneuerungsprozesses

4. It is clear that Snn>N0 with S0 0 a.s., Sn Pnk1 Tk, n > N is a random walk.

5. If one requires that the n-th exchange of a defective part in the system takes a time T

n,then by Tn Tn T

n, n > N a different renewal process is given. Its stochastic propertydoes not differ from the process which is given in definition 2.1.1.

In the following we assume that µ ETn > 0,ª, n > N.

Theorem 2.1.1 (Individual ergodic theorem):Let N Nt, t C 0 be a renewal process. Then it holds that:

limtª

Ntt

1µ

a.s..

Proof For all t C 0 and n > N it holds that Nt n Sn B t @ Sn1, therefore SNt B t @SNt1 and

SNtNt B

t

Nt B SNt1

Nt 1Nt 1Nt .

If we can show that SNtNt a.s

ÐÐtª

µ and Nt a.s.ÐÐtª

ª, then tNt a.s

ÐÐtª

µ holds and therefore theassertion of the theorem.According to the strong law of large numbers of Kolmogorov (cf. lecture notes „Wahrschein-lichkeitsrechnung“ (WR), theorem 7.4) it holds that Sn

n

a.s.ÐÐÐnª

µ, thus Sna.s.ÐÐÐnª

ª and thereforePNt @ª 1 since PNt ª P Sn B t,¦n 1 P§n ¦m > N0 Snm A t´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

1, if Sna.sÐÐÐnª

ª

1 1 0.

Then Nt, t C 0, is a real random variable.We show that Nt a.s.

ÐÐtª

ª. All trajectories of Nt are monotonously non-decreasing in t C 0,

2 Counting processes 21

thus § limtªNω, t for all ω > Ω. Moreover it holds that

P limtª

Nt @ª limnª

P limtª

Nt @ n lim

nª

limtª

PNt @ n lim

nª

limtª

PSn A t limnª

limtª

P n

Qk1

Tk A tB lim

nª

limtª

n

Qk1

PTk A t

n´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

ÐÐtª

0

0.

The equality () holds since limtªNt @ n §t0 > Q ¦t C t0 Nt @ n 8t0>Q 9t>QtCt0Nt @ n lim inf t>Q

tª

Nt @ n, then the continuity of the probability measure is used,

where Q Q9R q > Q q C 0. Since for every ω > Ω it holds that limnª

Snn limtª

SNtNt

(the codomain of a realization of N is a subsequence of N), it holds that limtª

SNtNt a.s

µ.

Remark 2.1.2One can generalize the ergodic theorem to the case of non-identically distributed Tn. Therebywe require that µn ETn, Tn µnn>N are uniformly integrable and 1

n Pnk1 µk ÐÐÐnª

µ A 0.

Then we can prove that Ntt

PÐÐtª

1µ (cf. [2], page 276).

Theorem 2.1.2 (Central limit theorem):If µ > 0,ª, σ2 VarT1 > 0,ª, it holds that

µ32 Nt t

µ

σºt

dÐÐtª

Y,

where Y N 0,1.Proof According to the central limit theorem for sums of i.i.d. random variables (cf. theorem7.5, WR) it holds that

Sn nµºnσ2

dÐÐÐnª

Y. (2.1.1)

Let x be the whole part of x > R. It holds for a σ2

µ3 that

PNt

tµº

atB x

PNt B xºat t

µ P Smt A t ,

where mt xºat tµ 1, t C 0, and limtªmt ª. Therefore we get that

RRRRRRRRRRRPNt t

µºat

B x ϕxRRRRRRRRRRR TP Smt A t ϕxT

RRRRRRRRRRRPSmt µmtσ»mt A

t µmtσ»mt ϕx

RRRRRRRRRRR Itx


for arbitrary t C 0 and x > R, where ϕ is the distribution function of the N 0,1-distribution.For fixed x > R we introduce Zt tµmt

σ»mt x, t C 0. Then it holds that

Itx RRRRRRRRRRRPSmt µmtσ»mt Zt A x

ϕxRRRRRRRRRRR .If we can prove that Zt ÐÐ

tª

0, then applying (2.1.1) and the theorem of Slutsky (theorem

6.4.1, WR) would result in Smtµmtσ»mt Zt

dÐÐtª

Y N 0,1 since Zt ÐÐtª

0 a.s. results

in Ztd

ÐÐtª

0. Therefore we could write Itx ÐÐtª

Sϕx ϕxS Sϕx ϕxS 0, whereϕx 1 ϕx is the tail function of the N 0,1-distribution, and the property of symmetryof N 0,1 ϕx ϕx, x > R was used.Now we show that Zt ÐÐ

tª

0, thus tµmtσ»mt ÐÐtª

x. It holds that mt xºat tµ εt,

where εt > 0,1. Then it holds thatt µmtσ»mt

t µxºat t µεt

σ»mt x

ºat µ

σ¼xºat t

µ εt µεt

σ»mt

xµ

σ

½xºat

1µa

εtat

µ εtσ»mt

xµσ½

µ2

σ2 xºatεtat´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

ÐÐtª

x

µεt

σ»mt´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

ÐÐtª

0

ÐÐtª

x.

Remark 2.1.3In Lineberg form, the central limit theorem can also be proven for non-identically distributedTn, cf. [2], pages 276 - 277.Definition 2.1.2The function Ht ENt, t C 0 is called renewal function of the process N (or of the sequenceSnn>N).Let FT x PT1 B x, x > R be the distribution function of T1. For arbitrary distribution

functions F,G R 0,1 the convolution F G is defined as F Gx R xª F x ydGy.The k-fold convolution F k of the distribution F with itself, k > N0, is defined inductive:

F 0x 1x > 0,ª, x > R,F 1x F x, x > R,

F k1x F k F x, x > R.

Lemma 2.1.1The renewal function H of a renewal process N is monotonously non-decreasing and right-sidedcontinuous on R. Moreover it holds that

Ht ª

Qn1

PSn B t ª

Qn1

F nT t, t C 0. (2.1.2)


Proof The monotony and right-sided continuity of H are consequences from the almost surelymonotony and right-sided continuity of the trajectories of N . Now we prove (2.1.2):

Ht ENt Eª

Qn1

1Sn B t

ª

Qn1

E1Sn B t ª

Qn1

PSn B t ª

Qn1

F nT t,

since PSn B t PT1 . . .Tn B t F nT t, t C 0. The equality holds for all partial sums

on both sides, therefore in the limit as well.

Except for exceptional cases it is impossible to calculate the renewal function H by theformula (2.1.2) analytically. Therefore the Laplace transform of H is often used in calulations.For a monotonously (e.g. monotonously non-decreasing) right-sided continuous function G 0,ª R the Laplace transform is defined as lGs R ª0 esxdGx, s C 0. Here the integral isto be understood as the Lebesgue-Stieltjes integral, thus as a Lebesgue integral with respect tothe measure µG on BR defined by µGx, y GyGx, 0 B x @ y @ª, if G is monotonouslynon-decreasing.Just to remind you: the Laplace transform lX of a random variable X C 0 is defined bylXs R ª0 esxdFXx, s C 0.Lemma 2.1.2For s A 0 it holds that:

lHs lT1s1 lT1s .

Proof It holds that:

lHs Sª

0esxdHx (2.1.2)

Sª

0esxd ª

Qn1

F nT x ª

Qn1S

ª

0esxdF nx

ª

Qn1

lT1...Tns ª

Qn1

lT1sn lT1s1 lT1s ,

where for s A 0 it holds that lT1s @ 1 and thus the geometric series Pª

n1 lT1sn converges.

Remark 2.1.4If N Nt, t C 0 is a delayed renewal process (with delay T1), the statements of lemmas2.1.1 - 2.1.2 hold in the following form:

1.Ht ª

Qn0

FT1 FnT2 t, t C 0,

where FT1 and FT2 , respectively are the distribution functions of T1 and Tn, n C 2,respectively.

2.lHs lT1s

1 lT2s , s C 0, (2.1.3)

where lT1 and lT2 are the Laplace transforms of the distribution of T1 and Tn, n C 2.


For further observations we need a theorem (of Wald) about the expected value of a sum(with random number) of independent random variables.Definition 2.1.3Let ν be a N-valued random variable and be Xnn>N a sequence of random variables definedon the same probability space. ν is called independent of the future, if for all n > N the eventν B n does not depend on the σ-algebra σXk, k A n.Theorem 2.1.3 (Wald’s identity):Let Xnn>N be a sequence of random variables with sup ESXnS @ª, EXn a, n > N and be ν aN-valued random variable which is independent of the future, with Eν @ª. Then it holds that

E ν

Qn1

Xn a Eν.Proof Calculate Sn Pnk1Xk, n > N. Since Eν Pª

n1 Pν C n, the theorem follows fromLemma 2.1.3.

Lemma 2.1.3 (Kolmogorov-Prokhorov):Let ν be a N-valued random variable which is independent of the future and it holds that

ª

Qn1

Pν C nESXnS @ª. (2.1.4)

Then ESν Pª

n1 Pν C nEXn holds. If Xn C 0 a.s., then condition (2.1.4) is not required.

Proof It holds that Sν Pνn1Xn Pª

n1Xn1ν C n. We introduce the notation Sν,n

Pnk1Xk1ν C k, n > N. First, we prove the lemma for Xn C 0 f.s., n > N. It holds Sν,n Sν ,nª for every ω > Ω, and thus according to the monotone convergence theorem it holds that:ESν limnª ESν,n limPnk1 EXk1ν C k. Since ν C k ν B k 1c does not dependon σXk ` σXn, n C k it holds that EXk1ν C k EXkPν C k, k > N, and thusESν Pª

n1 Pν C nEXn.Now, let Xn be arbitrary. Take Yn SXnS, Zn Pnn1 Yn, Zν,n Pnk1 Yk1ν C k, n > N.Since Yn C 0, n > N, it holds that EZν Pª

n1 EXn S Pν C k @ ª from (2.1.4). SinceSSν,nS B Zν,n B Zν , n > N, according to the dominated convergence theorem of Lebesgue it holdsthat ESν limnª ESν,n Pª

n1 EXnPν C n, where this series converges absolutely.

Conclusion 2.1.1 1. Ht @ª, t C 0.

2. For an arbitrary Borel measurable function g R R and the renewal process N Nt, t C 0 with interarrival times Tn, Tn i.i.d., µ ETn > 0,ª it holds that

ENt1Qk1

gTn 1 HtEgT1, t C 0.

Proof 1. For every t C 0 it is obvious that ν 1 Ht does not depend on the future ofTnn>N, the rest follows from theorem 2.1.3 with Xn gTn, n > N.

2. For s A 0 consider T sn minTn, s, n > N. Choose s A 0 such that for freely selected

(but fixed) ε A 0 µs ET s1 C µ ε A 0. Let N s be the renewal process which is based


on the sequence T sn n>N of interarrival times: N st Pª

n1 1T sn B t, t C 0. It holds

Nt B N st, t C 0, a.s., according to conclusion 2.1.1:

µ εEN st 1 B µsEN st 1 ESsNst1 ESs

Nst´¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¶Bt

TsNst1´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

Bs

B t s,t C 0, where Ss

n Ts1 . . . T

sn , n > N. Thus Ht ENt B EN st B ts

µε , t C 0.Since ε A 0 is arbitrary, it holds that lim suptª

Htt B

1µ , and also our assertionHt @ª,

t C 0.

Conclusion 2.1.2 (Elementary renewal theorem):For a renewal process N as defined in conclusion 2.1.1, 1) it holds:

limtª

Htt

1µ.

Proof In conclusion 2.1.1, part 2) we already proved that lim suptª

Htt B

1µ . If we show

lim inftª

Htt C

1µ , our assertion would be proven. According to theorem 2.1.1 it holds that

Ntt ÐÐ

tª

1µ a.s., therefore according to Fatou’s lemma

1µ E lim inf

tª

Ntt

B lim inftª

ENtt

lim inftª

Htt

.

Remark 2.1.5 1. We can prove that in the case of the finite second moment of Tn (µ2

ET 21 @ª) we can derive a more exact asymptotics for Ht, tª:

Ht t

µµ22µ2 o1, tª.

2. The elementary renewal theorem also holds for delayed renewal processes, where µ ET2.We define the renewal measure H on BR by HB Pª

n1 RB dF nT x, B > BR. It

holds Hª, t Ht, Hs, t Ht Hs, s, t C 0, if H is the renewal functionas well as the renewal measure.

Theorem 2.1.4 (Fundamental theorem of the renewal theory):Let N Nt, t C 0 be a (delayed) renewal process associated with the sequence Tnn>N,where Tn, n > N are independent, Tn, n C 2 identically distributed, and the distributionof T2 is not arithmetic, thus not concentrated on a regular lattice with probability 1. Thedistribution of T1 is arbitrary. Let ET2 µ > 0,ª. Then it holds that

St

0gt xdHxÐÐ

tª

1µS

ª

0gxdx,

where g R R is Riemann integrable 0, n, for all n > N, and Pª

n0 maxnBxBn1 SgxS @ª.


Abb. 2.2:

Without proof.

In particular Ht u, t ÐÐtª

uµ holds for an arbitrary u > R, thus H asymptotically (for

tª) behaves as the Lebesgue measure.

Definition 2.1.4The random variable χt SNt1 t is called excess of N at time t C 0.

Obviously χ0 T1 holds. We now give an example of a renewal process with stationaryincrements.Let N Nt, t C 0 be a delayed renewal process associated with the sequence of interarrivaltimes Tnn>N. Let FT1 and FT2 be the distribution functions of the delays T1 and Tn, n C 2.We assume that µ ET2 > 0,ª, FT20 0, thus T2 A 0 a.s. and

FT1x 1µS

x

0FT2ydy, x C 0. (2.1.5)

In this case FT1 is called the integrated tail distribution function of T2.

Theorem 2.1.5Under the conditions we mentioned above, N is a process with stationary increments.

Abb. 2.3:

Proof Let n > N, 0 B t0 @ t1 @ . . . @ tn @ ª. Because N does not depend on Tn, n > N thecommon distribution of Nt1 tNt0 t, . . . ,Ntn tNtn1 t does not depend ont, if the distribution of χt does not depend on t, thus χt d χ0 T1, t C 0, see Figure ....


We show that FT1 Fχt, t C 0.

Fχtx Pχt B x ª

Qn0

PSn B t, t @ Sn1 B t x PS0 0 B t, t @ S1 T1 B t x

ª

Qn1

EE1Sn B t, t @ Sn Tn1 B t x S Sn FT1t x FT1t ª

Qn1S

t

0Pt y @ Tn1 B t x ydFSny

FT1t x FT1t S t

0Pt y @ T2 B t x yd ª

Qn1

FSny´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶Hy

.If we can prove that Hy y

µ , y C 0, then we would get

Fχtx FT1t x FT1t 1µS

0

tFT2z x 1 1 FT2zdz

FT1t x FT1t 1µS

t

0FT2z FT2z xdz

FT1t x FT1t FT1t 1µS

tx

xFT2ydy

FT1t x FT1t x FT1x FT1x, x C 0,

according to the form (2.1.5) of the distribution of T1.Now we like to show that Ht t

µ , t C 0. For that we use the formula (2.1.4): it holds that

lT1s 1µS

ª

0est1 FT2tdt 1

µS

ª

0estdt´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶1s

1µS

ª

0estFT2tdt

1µs

1 Sª

0FT2tdest 1

µs1 estFT2t´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

FT200

Tª0 Sª

0estdFT2t´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶lT2s

1µs1 lT2s, s C 0.

Using the formula (2.1.4) we get

lHs lT1s1 lT2s 1

µs

1µS

ª

0estdt l t

µs, s C 0.

Since the Laplace transform of a function uniquely determines this function, it holds thatHt t

µ , t C 0.

Remark 2.1.6In the proof of Theorem 2.1.5 we showed that for the renewal process with delay which possessesthe distribution (2.1.5), Ht t

µ not only asymptotical for tª (as in the elementary renewal


theorem) but it holds Ht tµ , for all t C 0. This means, per unit of the time interval we

get an average of 1µ renewals. For that reason such a process N is called homogeneous renewal

process.We can prove the following theorem:

Theorem 2.1.6If N Nt, t C 0 is a delayed renewal process with arbitrary delay T1 and non-arithmeticdistribution of Tn, n C 2, µ ET2 > 0,ª, then it holds that

limtª

Fχtx 1µS

x

0FT2ydy, x C 0.

This means, the limit distribution of excess χt, tª is taken as the distribution of T1 whendefining a homogeneous renewal process.

2.2 Poisson processes

2.2.1 Poisson processes

In this section we generalize the definition of a homogeneous Poisson process (see section 1.2,example 5)Definition 2.2.1The counting process N Nt, t C 0 is called Poisson process with intensity measure Λ if

1. N0 0 a.s.

2. Λ is a locally finite measure R,i.e., Λ BR R possesses the property ΛB @ª forevery bounded set B > BR.

3. N possesses independent increments.

4. Nt Ns PoisΛs, t for all 0 B s @ t @ª.

Sometimes the Poisson process N Nt, t C 0 is defined by the corresponding randomPoisson counting measure N NB, B > BR, i.e., N 0, t, t C 0, where a countingmeasure is a locally finite measure with values in N0.Definition 2.2.2A random counting measure N NB, B > BR is called Poissonsh with locally finiteintensity measure Λ if

1. For arbitrary n > N and for arbitrary pairwise disjoint bounded sets B1,B2, . . . ,Bn >

BR the random variables NB1,NB2, . . . ,NBn are independent.

2. NB PoisΛB, B > BR, B-bounded.

It is obvious that properties 3 and 4 of definition 2.2.1 follow from properties 1 and 2 ofdefinition 2.2.2. Property 1 of definition 2.2.1 however is an autonomous assumption. NB,B > BR is interpreted as the number of points of N within the set B.


Remark 2.2.1As stated in definition 2.2.2, a Poisson counting measure can also be defined on an arbitrarytopological space E equipped with the Borel-σ-algebra BE. Very often E Rd, d C 1 ischosen in applications.Lemma 2.2.1For every locally finite measure Λ on R there exists a Poisson process with intensity measureΛ.

Proof If such a Poisson process had existed, the characteristic function ϕNtNs of theincrement Nt Ns, 0 B s @ t @ ª would have been equal to ϕs,tz ϕPoisΛs,tz eΛs,teiz1, z > R according to property 4 of definition 2.2.1. We show that the family ofcharacteristic functions ϕs,t, 0 B s @ t @ ª possesses property 1.7.1: for all n 0 B s @ u @ t,ϕs,uzϕu,tz eΛs,ueiz1eΛu,teiz1 eΛs,uΛu,teiz1 eΛs,teiz1 ϕs,tz,z > R since the measure Λ is additive. Thus, the existence of the Poisson process N followsfrom theorem 1.7.1.

Remark 2.2.2The existence of a Poisson counting measure can be proven with the help of the theorem ofKolmogorov, yet in a more general form than in theorem 1.1.2.From the properties of the Poisson distribution it follows that ENB VarNB ΛB,

B > BR. Thus ΛB is interpreted as the mean number of points of N within the set B,B > BR.We get an important special case if Λdx λdx for λ > 0,ª, i.e., Λ is proportional to theLebesgue measure ν1 on R. Then we call λ EN1 the intensity of N .Soon we will prove that in this case N is a homogeneous Poisson process with intensity λ. Toremind you: In section 1.2 the homogeneous Poisson process was defined as a renewal processwith interarrival times TN Expλ: Nt supn > N Sn B t, Sn T1 . . . Tn, n > N, t C 0.Exercise 2.2.1Show that the homogeneous Poisson process is a homogeneous renewal process with T1

d T2

Expλ. Hint: you have to show that for an arbitrary exponential distributed random variableX the integrated tail distribution function of X is equal to FX .Theorem 2.2.1Let N Nt, t C 0 be a counting process. The following statements are equivalent.

1. N is a homogeneous Poisson process with intensity λ A 0.

2. a) Nt Poisλt, t C 0b) for an arbitrary n > N, t C 0, it holds that the random vector S1, . . . , Sn under

condition Nt n possesses the same distribution as the order statistics of i.i.d.random variables Ui > U0, t, i 1, . . . , n.

3. a) N has independent increments,b) EN1 λ, andc) property 2b) holds.

4. a) N has stationary and independent increments, and


b) PNt 0 1 λt ot, PNt 1 λt ot, t 0 holds.

5. a) N hast stationary and independent increments,

b) property 2a) holds.

Remark 2.2.3 1. It is obvious that Definition 2.2.1 with Λdx λdx, λ > 0,ª is anequivalent definition of the homogeneous Poisson process according to Theorem 2.2.1.

2. The homogeneous Poisson process N was introduced in the beginning of the 20th centuryfrom the physicists A. Einstein and M. Smoluchovsky to be able to model the countingprocess of elementary particle in the Geiger counter.

3. From 4b) it follows PNt A 1 ot, t 0.

4. The intensity of N has the following interpretation: λ EN1 1

ETn , thus the meannumber of renewals of N within a time interval with length 1.

5. The renewal function of the homogeneous Poisson process is Ht λt, t C 0. TherebyHt Λ0, t, t A 0 holds.

Proof Structure of the proof: 1 2 3 4 5 11 2:From 1) follows Sn Pnk1 Tk Erln,λ since Tk Poisλ, n > N, thus PNt 0 PT1 At eλt, t C 0, and for n > N

PNt n PNt C n Nt C n 1 PNt C n PNt C n 1 PSn B t PSn1 B t S t

0

λnxn1n 1!eλxdx S t

0

λn1xn

n!eλxdx

St

0

d

dxλxn

n!eλxdx λtn

n!eλt, t C 0.

Thus 2a) is proven.Now let’s prove 2b). According to the transformation theorem of random variables (cf. theorem3.6.1, WR), it follows from ¢

¦¤S1 T1S2 T1 T2

Sn1 T1 . . . Tn1

that the density fS1,...,Sn of S1, . . . , Sn1 can be expressed by the density of T1, . . . , Tn1,Ti Expλ, i.i.d.:

fS1,...,Sn1t1, . . . , tn1 n1Mk1

fTktk tk1 n1Mk1

λeλtktk1 λn1eλtn1

for arbitrary 0 B t1 B . . . B tn1, t0 0.For all other t1, . . . , tn1 it holds fS1,...,Sn1t1, . . . , tn1 0.


Therefore

fS1,...,Snt1, . . . , tnSNt n fS1,...,Snt1, . . . , tnSSk B t, k B n, Sn1 A t

R ªt fS1,...,Sn1t1, . . . , tn1dtn1

R t0 R tt1 . . . R ttn1 R ªt fS1,...,Sn1t1, . . . , tn1dtn1dtn . . . dt1

R ªt λn1eλtn1dtn1

R t0 R tt1 . . . R ttn1 R ªt λn1eλtn1dtn1dtn . . . dt1

I0 B t1 B t2 B . . . B tn B t

n!tn

I0 B t1 B t2 B . . . B tn B t.This is exactly the density of n i.i.d. U0, t-random variables.Exercise 2.2.2Proof this.

2 3From 2a) obviously follows 3b). Now we just have to prove the independence of the incrementsof N . For an arbitrary n > N, x1, . . . , xn > N, t0 0 @ t1 @ . . . @ tn for x x1 . . . xn it holdsthat

P9nk1Ntk Ntk1 xk P9nk1Ntk Ntk1 xkSNtn x´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶x!

x1!...xn! Lnk1 tktk1

tnxk according to 2b)

PNtn x´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶eλtn

λtnxx! according to 2a)

n

Mk1

λtk tk1xkxk!

eλtktk1,

since the probability of () belongs to the polynomial distribution with parameters n, tktk1tn

nk1

.Because the event ()is that at the independent uniformly distributed toss of x points on 0, t,exactly xk points occur within the basket of length tk tk1, k 1, . . . , n:

Abb. 2.4:

Thus 3a) is proven since P9nk1Ntk Ntk1 xk Lnk1 PNtk Ntk1 xk.


3 4We prove that N possesses stationary increments. For an arbitrary n > N0, x1, . . . , xn > N,t0 0 @ t1 @ . . . @ tn and h A 0 we consider Ih P9nk1Ntk h Ntk1 h xk andshow that Ih does not depend on h > R. According to the formula of the total probability itholds that

Ih

ª

Qm0

P9nk1Ntk h Ntk1 h xk S Ntn h m PNtn h m

ª

Qm0

m!x1! . . . xn!

n

Mk1

tk h tn1 h

tn h hxk eλtnh λtn hm

m!

ª

Qm0

P9nk1Ntk Ntk1 xk S Ntn h m PNtn h m I0.We now show property 4b) for h > 0,1:

PNh 0

ª

Qk0

PNh 0,N1 k ª

Qk0

PNh 0,N1 Nh k

ª

Qk0

PN1 Nh k,N1 k

ª

Qk0

PN1 kPN1 Nh k S N1 k

ª

Qk0

PN1 k1 hk.We have to show that PNh 0 1λh oh, i.e., limhª

1h1PNh 0 λ. Indeed

it holds that

1h1 PNh 0

1h1

ª

Qk0

PN1 k1 hk ª

Qk1

PN1 k 1 1 hkh

ÐÐh0

ª

Qk1

PN1 k limh0

1 1 hkh´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

k

ª

Qk0

PN1 kk EN1 λ,since the series uniformly converges in h because it is dominated by Pª

k0 PN1 kk λ @ª

because of the inequality 1 hk C 1 kh, h > 0,1, k > N.Similarly one can show that limh0

PNh1h limh0Pª

k1 PN1 kk1hk1 λ. 4 5We have to show that for an arbitrary n > N and t C 0

pnt PNt n eλt λtnn!

(2.2.1)

holds. We will prove that by induction with respect to n. First we show that p0t eλt, h 0.For that we consider p0th PNth 0 PNt 0,NthNt 0 p0tp0h p0t1λh oh, h 0. Similarly one can show that p0t p0th1λh oh, h 0.


Thus p0t limh0p0thp0t

h λp0t, t A 0 holds. Since p00 PN0 0 1, itfollows from p0t λp0t

p00 1,

that it exists an unique solution p0t eλt, t C 0. Now for n the formular (2.2.1) be approved.Let’s prove it for n 1.

pn1t h PNt h n 1 PNt n,Nt h Nt 1 PNt n 1,Nt h Nt 0 pnt p1h pn1t p0h pntλh oh pn1t1 λh oh, h 0, h A 0.

Thus pn1t λpn1t λpnt, t A 0pn10 0 (2.2.2)

Since pnt eλt λtnn! , we obtain pn1t eλt λtn1

n1! as solution of (2.2.2). (Indeed pn1t Cteλt C teλt λCteλt........... λpntC t λn1tn

n! Ct λn1tn1n1! , C0 0)5 1Let N be a counting process Nt maxn Sn B t, t C 0, which fulfills conditions 5a) and5b). We show that Sn Pnk1 Tk, where Tk i.i.d. with Tk Expλ, k > N. Since Tk Sk Sk1,k > N, S0 0, we consider for b0 0 B a1 @ b1 B . . . B an @ bn

P 9nk1ak @ Sk B bk P9n1

k1Nak Nbk1 0,Nbk Nak 19Nan Nbn1 0,Nbn Nan C 1

n1Mk1

PNak bk1 0´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶eλakbk1

PNbk ak 1´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶λbkakeλbkak

PNan bn1 0´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

eλanbn1

PNbn an C 1´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶1eλbnan eλanbn11 eλbnan n1

Mk1

λbk akeλbkbk1

λn1eλan eλbn n1Mk1

bk ak S b1

a1. . .S

bn

anλneλyndyn . . . y1.

The common density of S1, . . . , Sn therefore is given by λneλyn1y1 B y2 B . . . B yn.2.2.2 Compound Poisson processDefinition 2.2.3Let N Nt, t C 0 be a homogeneous Poisson process with intensity λ A 0, build by meansof the sequence Tnn>N of interarrival times. Let Unn>N be a sequence of i.i.d. randomvariables, independent of Tnn>N. Let FU be the distribution function of U1. For an arbitrary


t C 0 let Xt PNtk1 Uk. The stochastic process X Xt, t C 0 is called compound Poisson

process with parameters λ, FU . The distribution of Xt thereby is called compound Poissondistribution with parameters λt, FU .The compound Poisson process Xt, t C 0 can be interpreted as the sum of „marks“ Un of

a homogeneous marked Poisson process N,U until time t.In queueing theory Xt is interpreted as the overall workload of a server until time t if therequests to the service occur at times Sn Pnk1 Tk, n > N and represent the amount of workUn, n > N.In actuarial mathematics Xt, t C 0 is the total damage in a portfolio until time t C 0 withnumber of damages Nt and amount of loss Un, n > N.Theorem 2.2.2Let X Xt, t C 0 be a compound Poisson process with parameters λ, FU . The followingproperties hold:

1. X has independent and stationary increments.

2. If mUs EesU1 , s > R, is the moment generating function of U1, such that mUs @ª,s > R, then it holds that

mXts eλtmU s1, s > R, t C 0, EXt λtEU1, VarXt λtEU21 , t C 0.

Proof 1. We have to show that for arbitrary n > N, 0 B t0 @ t1 @ . . . @ tn and h

P Nt1h

Qi1Nt0h1

Ui1 B x1, . . . ,NtnhQ

inNtn1h1Uin B xn

n

Mk1

P Ntk

QikNtk11

Uik B xk

for arbitrary x1, . . . , xn > R. Indeed it holds that

P Nt1h

Qi1Nt0h1

Ui1 B x1, . . . ,NtnhQ

inNtn1h1Uin B xn

ª

Qk1,...,kn0

n

Mj1

Fkjn xjP 9nm1 Ntm h Ntm1 h km

ª

Qk1,...,kn0

n

Mj1

Fkjn xj n

Mm1

PNtm Ntm1 km

n

Mm1

ª

Qkm0

F kmn xmPNtm Ntm1 km

n

Mm1

P Ntm

QkmNtm11

B xm

2.Exercise 2.2.3


2.2.3 Cox process

A Cox process is a (in general inhomogeneous) Poisson process with intensity measure Λ whichas such is a random measure. The intuitive idea is stated in the following definition.Definition 2.2.4Let Λ ΛB, B > BR be a random a.s. locally finite measure. The random countingmeasure N NB, B > BR is called Cox counting measure (or doubly stochastic Poissonmeasure) with random intensity measure Λ if for arbitrary n > N, k1, . . . , kn > N0 and 0 B a1 @ b1 B

a2 @ b2 B . . . B an @ bn it holds that P9ni1Nai, bi ki E Lni1 e

Λai,bi Λkiai,biki! .

The process Nt, t C 0 with Nt N0, t is called Cox process (or doubly stochasticPoisson process) with random intensity measure Λ.Example 2.2.1 1. If the random measure Λ is a.s. absolutely continuous with respect to

the Lebesgue measure, i.e., ΛB RB λtdt, B - bounded, B > BR, where λt, t C 0is a stochastic process with a.s. Borel-measurable Borel-integrable trajectories, thenλt C 0 a.s. for all t C 0 is called the intensity process of N .

2. In particular, it can be that λt Y where Y is a non-negative random variable. Thenit holds that ΛB Y ν1B, thus N has a random intensity Y . Such Cox processes arecalled mixed Poisson processes.

A Cox process N Nt, t C 0 with intensity process λt, t C 0 can be build explicitlyas the following. Let N Nt, t C 0 be a homogeneous Poisson process with intensity 1,which is independent of λt, t C 0. Then N

d N1, where the process N1 N1t, t C 0

is given by N1t NR t0 λydy, t C 0. The assertion Nd N1 of course has to be proven.

However, we shall assume it without proof. It is also the basis for the simulation of the Coxprocess N .

2.3 Additional exercisesExercise 2.3.1Let NttC0 be a renewal process with interarrival times Ti, which are exponentially dis-tributed, i.e. Ti Expλ.

a) Prove that: Nt is Poisson distributed for every t A 0.

b) Determine the parameter of this Poisson distribution.

c) Determine the renewal function Ht ENt.Exercise 2.3.2Prove that a (real-valued) stochastic process X Xt, t > 0,ª with independent incre-ments already has stationary increments if the distribution of the random variable Xt h Xh does not depend on h.Exercise 2.3.3Let N Nt, t > 0,ª be a Poisson process with intensity λ. Calculate the probabilitiesthat within the interval 0, s exactly i events occur under the condition that within the interval0, t exactly n events occur, i.e. PNs i S Nt n for s @ t, i 0,1, . . . , n.


Exercise 2.3.4Let N 1 N 1t, t > 0,ª and N 2 N 2t, t > 0,ª be independent Poissonprocesses with intensities λ1 and λ2. In this case the independence indicates that the sequencesT11 , T

12 , . . . and T 2

1 , T22 , . . . are independent. Show that N Nt N 1tN 2t, t >0,ª is a Poisson process with intensity λ1 λ2.

Exercise 2.3.5 (Queuing paradox):Let N Nt, t > 0,ª be a renewal process. Then T t SNt1 t is called the time ofexcess, Ct t SNt the current lifetime and Dt T t Ct the lifetime at time t A 0.Now let N Nt, t > 0,ª be a Poisson process with intensity λ.

a) Calculate the distribution of the time of excess T t.b) Show that the distribution of the current lifetime is given by PCt t eλt and the

density is given by fCtSNtA0s λeλs1s B t.c) Show that PDt B x 1 1 λmint, xeλx1x C 0.d) To determine ET t, one could argue like this: On average t lies in the middle of the

surrounding interval of interarriving time SNt, SNt1, i.e. ET t 12ESNt1

SNt 12ETNt1

12λ . Considering the result from part (a) this reasoning is false.

Where is the mistake in the reasoning?

Exercise 2.3.6Let X Xt PNt

i1 Ui, t C 0 be a compound Poisson process. Let MNts EsNt,s > 0,1, be the generating function of the Poisson processes Nt, LUs E expsU theLaplace Transform of Ui, i > N, and LXts the Laplace Transform of Xt. Prove that

LXts MNtLUs, s C 0.

Exercise 2.3.7Let X Xt, t > 0,ª be a compound Poisson process with Ui i.i.d., U1 Expγ, wherethe intensity of Nt is given by λ. Show that for the Laplace transform LXts of Xtit holds:

LXts exp λts

γ s .

Exercise 2.3.8Write a function in R (alternatively: Java) to which we pass time t, intensity λ and a value γas parameters. The return of the function is a random value of the compound Poisson processwith characteristics λ,Expγ at time t.Exercise 2.3.9Let the stochastic process N Nt, t > 0,ª be a Cox process with intensity functionλt Z, where Z is a discrete random variable which takes values λ1 and λ2 with probabilities1~2. Determine the moment generating function as well as the expected value and the varianceof Nt.Exercise 2.3.10Let N 1 N 1t, t > 0,ª and N 2 N 2t, t C 0 be two independent homogeneousPoisson processes with intensities λ1 and λ2. Moreover, let X C 0 be an arbitrary non-negative


random variable which is independent of N 1 and N 2. Show that the process N Nt, t C0 with

Nt ¢¦¤N1t, t BX,

N 1X N 2t X, t AX

is a Cox process whose intensity process λ λt, t C 0 is given by

λt ¢¦¤λ1, t BX,

λ2, t AX.

3 Wiener process

3.1 Elementary properties

In Example 2) of Section 1.2 we defined the Brownian motion (or Wiener process)W W t, t C 0 as an Gaussian process with EW t 0 and covW s,W t mins, t,s, t C 0. The Wiener process is called after the mathematician Norbert Wiener (1894 - 1964).Why does the Brownian motion exist? According to theorem of Kolmogorov (Theorem 1.1.2)it exists a real-valued Gaussian process X Xt, t C 0 with mean value EXt µt, t C 0,and covariance function covXs,Xt Cs, t, s, t C 0 for every function µ R R andevery positive semidefinite function C R R R. We just have to show that Cs, t

mins, t, s, t C 0 is positive semidefinite.Exercise 3.1.1Prove this!We now give a new (equivalent) definition.

Definition 3.1.1A stochastic process W W t, t C 0 is called Wiener process (or Brownian motion) if

1. W 0 0 a.s.

2. W possesses independent increments

3. W t W s N 0, t s, 0 B s @ t

The existence of W according to Definition 3.1.1 follows from Theorem 1.7.1 since ϕs,tz EeizW tW s e

tsz22 , z > R, and e

tuz22 e

usz22 e

tsz22 for 0 B s @ u @ t, thus

ϕs,uzϕu,tz ϕs,tz, z > R. From Theorem 1.3.1 the existence of a version with continuoustrajectories follows.Exercise 3.1.2Show that Theorem 1.3.1 holds for α 3, σ

12 .

Therefore, it is often assumed that the Wiener process possesses continuous paths (just takeits corresponding version).Theorem 3.1.1Both definitions of the Wiener process are equivalent.

Proof 1. From definition in Section 1.2 follows Definition 3.1.1.W 0 0 a.s. follows from VarW 0 min0,0 0. Now we prove that the incrementsof W are independent. If Y N µ,K is a n-dimensional Gaussian random vectorand A a n n-matrix, then AY N Aµ,AKA holds, this follows from the explicitform of the characteristic function of Y . Now let n > N, 0 t0 B t1 @ . . . @ tn, Y

38

3 Wiener process 39

W t0,W t1, . . . ,W tn. For Z W t0,W t1 W t0, . . . ,W tn W tn1 itholds that Z AY , where

A

1 0 0 . . . . . . 0

1 1 0 . . . . . . 00 1 1 0 . . . 0. . . . . . . . . . . . . . . . . . . . . .0 0 0 . . . 1 1

.

Thus Z is also Gaussian with a covariance matrix which is diagonal. Indeed, it holdscovW ti1 W ti,W tj1 W tj minti1, tj1 minti1, tj minti, tj1 minti, tj 0 for i x j. Thus the coordinates of Z are uncorrelated, which means inde-pendence in case of a multivariate Gaussian distribution. Thus the increments of W areindependent. Moreover, for arbitrary 0 B s @ t it holds that W t W s N 0, t s.The normal distribution follows since Z AY is Gaussian, obviously it holds thatEW tEW s 0 and VarW tW s VarW t2 covW s,W tVarW s t 2 mins, t s t s.

2. From Definition 3.1.1 the definition in Section 1.2 follows.Since W t W s N 0, t s for 0 B s @ t, it holds

covW s,W t EW sW tW sW s EW sEW tW sVarW s s,thus it holds covW s,W t mins, t. FromW tW s N 0, ts andW 0 0it also follows that EW t 0, t C 0. The fact that W is a Gaussian process, follows frompoint 1) of the proof, relation Y A1Z.

Definition 3.1.2The process W t, t C 0,W t W1t, . . . ,Wdt, t C 0, is called d-dimensional Brownianmotion if Wi Wit, t C 0 are independent Wiener processes, i 1, . . . , d.The definitions above and Exercise 3.1.2 ensure the existence of a Wiener process with

continuous paths. How do we find an explicit way of building these paths? We will showthat in the next section.

3.2 Explicit construction of the Wiener processFirst we construct the Wiener process on the interval 0,1. The main idea of the constructionis to introduce a stochastic process X Xt, t > 0,1 which is defined on a probabilitysubspace of Ω,A,P with X d

W , where Xt Pª

n1 cntYn, t > 0,1, Ynn>N is a sequenceof i.i.d. N 0,1-random variables and cnt R t0 Hnsds, t > 0,1, n > N. Here, Hnn>N isthe orthonormed Haar basis in L20,1 which is introduced shortly now.

3.2.1 Haar- and Schauder-functionsDefinition 3.2.1The functions Hn 0,1 R, n > N, are called Haar functions if H1t 1, t > 0,1,H2t 10, 1

2 t 1 12 ,1t, Hkt 2n2 1In,kt 1Jn,kt, t > 0,1, 2n @ k B 2n1, where

In,k an,k, an,k 2n1, Jn,k an,k 2n1, an,k 2n, an,k 2nk 2n 1, n > N.

40 3 Wiener process

Abb. 3.1: Haar functions

Lemma 3.2.1The function system Hnn>N is an orthonormal basis in L20,1 with scalar product@ f, g A R 1

0 ftgtdt, f, g > L20,1.Proof The orthonormality of the system `Hk,Hne δkn, k,n > N directly follows from def-inition 3.2.1. Now we prove the completeness of Hnn>N. It is sufficient to show that forarbitrary function g > L20,1 with `g,Hne 0, n > N, it holds g 0 almost everywhere on0,1. In fact, we always can write the indicator function of an interval 1an,k,an,k2n1 as alinear combination of Hn, n > N.

10, 12

H1 H22

,

1 12 ,1

H1 H22

,

10, 14

10, 12 1º

2H22

,

1 14 ,

12

10, 12 1º

2H22

,

1an,k,an,k2n1 1an,k,an,k2n 2n2Hk

2, 2n @ k B 2n1.

Therefore it holds Rk1

2nk

2ngtdt 0, n > N0, k 1, . . . ,2n 1, and thus Gt R t0 gsds 0 for

t k2n , n > N0, k 1, . . . ,2n 1. Since G is continuous on 0,1, it follows Gt 0, t > 0,1,

and thus gs 0 for almost every s > 0,1.From lemma 3.2.1 it follows that two arbitrary functions f, g > L20,1 have expan-

sions f Pª

n1`f,HneHn and g Pª

n1`g,HneHn (these series converge in L20,1) and`f, ge Pª

n1`f,Hne`g,Hne (Parseval identity).Definition 3.2.2The functions Snt R t0 Hnsds `10,t,Hne, t > 0,1, n > N are called Schauder functions.

3 Wiener process 41

Abb. 3.2: Schauder functions

Lemma 3.2.2It holds:

1. Snt C 0, t > 0,1, n > N 1,2. P2n

k1 S2nkt B 122n2 , t > 0,1, n > N,

3. Let ann>N be a sequence of real numbers with an Onε, ε @ 12 , nª. Then the series

Pª

n1 anSnt converges absolutly and uniformly in t > 0,1 and therefore is a continuousfunction on 0,1.

Proof 1. follows directly from definition 3.2.2.

2. follows since functions S2nk for k 1, . . . ,2n have disjoint supports and S2nkt B

S2nk2k12n1 2n2 1, t > 0,1.

3. It suffices to show that Rn supt>0,1PkA2n SakSSkt ÐÐÐnª

0. For every k > N and c A 0it holds SakS B ckε. Therefore it holds for all t > 0,1, n > N

Q2n@kB2n1

SakSSkt B c 2n1ε Q2n@kB2n1

Skt B c 2n1ε 2

n2 1

B c 2εn 12ε.

Since ε @ 12 , it holds Rm B c 2εPnCm 2n 1

2ε ÐÐÐmª

0.

Lemma 3.2.3Let Ynn>N be a sequence of (not necessarily independent) random variables defined on Ω,A,P,Yn N 0,1, n > N. Then it holds SYnS Ologn 1

2 , nª, a.s.

Proof We have to show that for c Aº

2 and almost all ω > Ω it exists a n0 n0ω, c > N suchthat SYnS B clogn 1

2 for n C n0. If Y N 0,1, x A 0, it holds

PY A x 1º2π S

ª

xe

y22 dy

1º2π S

ª

x1yde y2

2

1º2π

1xe

y22 S

ª

xe

y22

1y2dy B 1º

2π1xe

x22 .

42 3 Wiener process

(We also can show that Φx 1º2π

1xe

x22 , xª.) Thus for c A

º2 it holds

QnC2

PSYnS A clogn 12 B c1 2º

2πQnC2

logn 12 e

c22 logn

c1º2º

πQnC2

logn 12n

c22 @ª.

According to the Lemma of Borel-Cantelli (cf. WR, Lemma 2.2.1) it holds P9n 8kCn Ak 0if Pk PAk @ª with Ak SYkS A e log k 1

2 , k > N. Thus Ak occurs in infinite number onlywith probability 0, with SYnS B clogn 1

2 for n C n0.

3.2.2 Wiener process with a.s. continuous paths

Lemma 3.2.4Let Ynn>N be a sequence of independent N 0,1-distributed random variables. Let ann>Nand bnn>N be sequences of numbers with P2m

k1 Sa2mkS B 2m2 , P2mk1 Sb2mkS B 2m2 , m > N. Then

the limits U Pª

n1 anYn and V Pª

n1 bnYn, U N 0,Pª

n1 a2n, V N 0,Pª

n1 b2n exist a.s.,

where covU,V Pª

n1 anbn. U and V are independent if and only if covU,V 0.

Proof Lemma 3.2.2 and 3.2.3 reveal the a.s. existence of the limits U and V (replace an by Ynand Sn by e.g. bn in Lemma 3.2.2). From the stability under convolution of the normal distri-bution it follows for U m Pmn1 anYn, V m Pmn1 bnYn, that U m N 0,Pmn1 a

2n, V m

N 0,Pmn1 b2n. Since U m d

Ð U , V m dÐ V it follows U N 0,Pª

n1 a2n, V N 0,Pª

n1 b2n.

Moreover, it holds

covU,V limmª

covU m, V m lim

mª

m

Qi,j1

aibj covYi, Yj lim

mª

m

Qi1aibi

ª

Qi1aibi,

according to the dominated convergence theorem of Lebesgue, since according to Lemma 3.2.3it holds SYnS B c logn 1

2´¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¶Bcnε, ε@ 1

2

, for n C N0, and the dominated series converges according to Lemma

3.2.2:

2m1

Qn,k2m

anbkYnYka.s.B

2m1

Qn,k2m

anbkc2nεkε B 22εm1

2m2 2

m2 212εm, 1 2ε A 0.

For sufficient large m it holds Pª

n,km anbkYnYk B Pª

jm 212εj @ª, and this series convergesa.s.Now we show

covU,V 0 U and V are independent

Independence always results in the uncorrelation of random variables. We prove the other

3 Wiener process 43

direction. From U m, V m dÐÐÐmª

U,V it follows ϕUm,V m ÐÐÐmª

ϕU,V , thus

ϕUm,V ms, t limmª

E expit mQk1

akYk sm

Qn1

bnYn lim

mª

E expi mQk1

tak sbkYk limmª

m

Mk1

E expitak sbkYk lim

mª

m

Mk1

exptak sbk2

2 exp ª

Qk1

tak sbk2

2

exp t22

ª

Qk1

a2k¡ exp

¢¦¨¤ts

ª

Qk1

akbk´¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¶covU,V 0

£§¨¥

exps2

2ª

Qk1

b2k¡ ϕUtϕV s,

s, t > R. Thus, U and V are independent if covU,V 0.

Theorem 3.2.1Let Yn, n > N be a sequence of i.i.d. random variables that are N 0,1-distributed, definedon a probability space Ω,A,P. Then there exists a probability space Ω0,A0,P of Ω,A,Pand a stochastic process X Xt, t > 0,1 on it such that Xt, ω Pª

n1 YnωSnt,t > 0,1, ω > Ω0 and X d

W . Here, Snn>N is the family of Schauder functions.

Proof According to Lemma 3.2.2, 2) the coefficients Snt fulfill the conditions of Lemma 3.2.4for every t > 0,1. In addition to that it exists according to Lemma 3.2.3 a subset Ω0 ` Ω,Ω0 > A with PΩ0 1, such that for every ω > Ω0 the relation SYnωS Oºlogn, n ª,holds. Let A0 A 9 Ω0. We restrict the probability space to Ω0,A0,P. Then conditionan Ynω Onε, ε @ 1

2 , is fulfilled sinceº

logn @ nε for sufficient large n, and accordingto Lemma 3.2.2, 3) the series Pª

n1 YnωSnt converges absolutely and uniformly in t > 0,1to the function Xω, t, ω > Ω0, which is a continuous function in t for every ω > Ω0. X, tis a random variable since in Lemma 3.2.4 the convergence of this series holds almost surely.Moreover it holds Xt N 0,Pª

n1 S2nt, t > 0,1.

We show that this stochastic process, defined on Ω0,A0,P, is a Wiener process. For that wecheck the conditions of Definition 3.1.1. We consider arbitrary times 0 B t1 @ t2, t3 @ t4 B 1 and

44 3 Wiener process

evaluate

covXt2 Xt1,Xt4 Xt3 cov ª

Qn1

YnSnt2 Snt1, ª

Qn1

YnSnt4 Snt3

ª

Qn1

Snt2 Snt1Snt4 Snt3

ª

Qn1

`Hn,10,t2e `Hn,10,t1e `Hn,10,t4e `Hn,10,t3e

ª

Qn1

`Hn,10,t2 10,t1e`Hn,10,t4 10,t3e `10,t2 10,t1,10,t4 10,t3e `10,t2,10,t4e `10,t1,10,t4e

`10,t2,10,t3e `10,t1,10,t3e mint2, t4 mint1, t4 mint2, t3 mint1, t3,

by Parseval inequality and since @ 10,s,10,t A R mins,t0 du mins, t, s, t > 0,1. If 0 B

t1 @ t2 B t3 @ t4 @ 1, it holds covXt2 Xt1,Xt4 Xt3 t2 t1 t2 t1 0, thusthe increments of X (according to Lemma 3.2.4) are uncorrelated. Moreover it holds X0 N 0,Pª

n1 S2n0 N 0,0, therefore X0 a.s. 0. For t1 0, t2 t, t3 0, t4 t it follows that

VarXt t, t > 0,1, and for t1 t3 s, t2 t4 t, that VarXtXs tsss ts,0 B s @ t B 1. Thus it holds XtXs N 0, t s, and according to Definition 3.1.1 it holdsX

dW .

Remark 3.2.1 1. Theorem 3.2.1 is the basis for an approximative simulation of the pathsof a Brownian motion through the partial sums Xnt Pnk1 YkSkt, t > 0,1, forsufficient large n > N.

2. The construction in Theorem 3.2.1 can be used to construct the Wiener process withcontinuous paths on the interval 0, t0 for arbitrary t0 A 0. If W W t, t > 0,1 is aWiener process on 0,1 then Y Y t, t > 0, t0 with Y t ºt0W tt0 , t > 0, t0, isa Wiener process on 0, t0.Exercise 3.2.1Prove that.

3. The Wiener process W with continuous paths on R can be constructed as follows. LetW n W nt, t > 0,1 be independent copies of the Wiener process as in Theorem3.2.1. Define W t Pª

n1 1t > n 1, nPn1k1 W

k1 W nt n 1, t C 0, thus,

W t ¢¦¤W 1t, t > 0,1,W 11 W 2t 1, t > 1,2,W 11 W 21 W 3t 2, t > 2,3,etc.

Exercise 3.2.2Show that the introduced stochastic process W W t, t C 0 is a Wiener process on R.

3 Wiener process 45

Abb. 3.3:

3.3 Distribution and path properties of Wiener processes

3.3.1 Distribution of the maximum

Theorem 3.3.1Let W W t, t > 0,1 be the Wiener process defined on a probability space Ω,F ,P.Then it holds:

Pmaxt>0,1W t A x ¾ 2

πS

ª

xe

y22 dy (3.3.1)

for all x C 0.

The mapping maxt>0,1W t Ω 0,ª given in relation (3.3.1) is a well-defined randomvariable since it holds: maxt>0,1W t, ω limnª maxi1,...,kW ik , ω for all ω > Ω since thetrajectories of W t, t > 0,1 are continuous. From 3.3.1 it follows that maxt>0,1W t hasan exponential bounded tail: thus maxt>0,1W t has finite k-th moments.Useful ideas for the proof of Theorem 3.3.1

Let W t, t > 0,1 be a Wiener process and Z1, Z2, . . . a sequence of independent randomvariables with PZi 1 PZi 1 1

2 for all i C 1. For every n > N we define Wnt, t >0,1 by Wnt Sntºn nt ntZnt1º

n, where Si Z1 . . . Zi, i C 1, S0 0.

Lemma 3.3.1For every k C 1 and arbitrary t1, . . . , tk > 0,1 it holds:

W nt1, . . . , W ntk d W t1, . . . ,W tk .

Proof Consider the special case k 2 (for k A 2 the proof is analogous). Let t1 @ t2. For all

46 3 Wiener process

s1, s2 > R it holds:

s1Wnt1 s2W

nt2 s1 s2Snt1ºn

s2Snt2 Snt11º

n

Znt11 nt1 nt1 s1ºns2ºn

Znt21nt2 nt2 s2ºn,

since Snt2 Snt1 Snt2 Snt11 Snt11.Now observe that the 4 summands on the right-hand-side of the previous equation are inde-pendent and moreover that the latter two summands converge (a.s. and therefor particularlyin distribution) to zero.Consequently, it holds

limnª

Eeis1W nt1s2W nt2 limnª

Eeis1s2º

nSnt1Eei

s2ºnSnt2Snt11

limnª

Eeis1s2¼ nt1

n

Snt1»nt1 Eeis2ºnSnt2nt11

CLT, CMT e

t12 s1s22

et2t1

2 s22

e12 s2

1t12s1s2t1s22t2

e12 s2

1t12s1s2 mint1,t2s22t2

ϕW t1,W t2s1, s2,where ϕW t1,W t2 is the characteristic function of W t1,W t2.Lemma 3.3.2Let W n maxt>0,1 W nt. Then it holds:

W n d

1ºn

maxk1,...,n

Sk, for all n > N

and

limnª

PW nB x ¾ 2

πS

x

0e

y22 dy, for all x C 0.

Without proof

Proof of Theorem 3.3.1. We shall prove only the upper bound in Theorem 3.3.1.From Lemma 3.3.1 and the continuous mapping theorem it follows for x C 0, k C 1 and t1, . . . , tk >0,1 that

limnª

P maxt>t1,...,tk W

nt A x P maxt>t1,...,tkW t A x ,

since x1, ..., xk(maxx1, ..., xk is continuous.Consequently, it holds

lim infnª

Pmaxt>0,1 W

nt A x C P maxt>t1,...,tkW t A x ,

3 Wiener process 47

since maxt>t1,...,tk W

nt A x b maxt>0,1 W

nt A x.With t1, . . . , tk 1

k , . . . ,kk and max

t>0,1W t limkª

maxi1,...,k

W ik a.s. (and therefore partic-

ularly in distribution) it holds

lim infnª

Pmaxt>0,1 W

nt A x C limkª

P maxi1,...,k

W ik A x Pmax

t>0,1W t A x .Conclusively, the assertion follows from lemma 3.3.2.

Corollary 3.3.1Let W t, t C 0 be a Wiener process. Then it holds:

P limtª

W tt

0 1.

Proof Exercise.

3.3.2 Invariance propertiesSpecific transformations of the Wiener process again reveal the Wiener process.Theorem 3.3.2Let W t, t C 0 be a Wiener process. Then the stochastic processes Y it, t C 0,i 1, . . . ,4, with

Y 1t W t, (Symmetry)Y 2t W t t0 W t0 for a t0 A 0, Translation of the origin)Y 3t

ºcW tc for a c A 0, (Scaling)

Y 4t tW 1t , t A 0,0, t 0. (Reflection at t=0)

are Wiener processes as well.

Proof 1. Y i, i 1, . . . ,4, have independent increments with Y it2Y it1 N 0, t2t1.

2. Y i0 0, i 1, . . . ,4.

3. Y i, i 1, . . . ,3, have continuous trajectories. Y it, t C 0 has continuous trajectoriesfor t A 0.

4. We have to prove that Y 4t is continuous at t 0, i.e. that limt0 tW 1t 0.

limt0 tW 1t limtª

W tt

a.s. 0 because of Corollary 3.3.1.

Corollary 3.3.2Let W t, t C 0 be the Wiener process. Then it holds:

PsuptC0

W t ª PinftC0

W t ª 1.

48 3 Wiener process

and consequently

PsuptC0

W t ª, inftC0

W t ª .Proof For x, c A 0 it holds:

PsuptC0

W t A x PsuptC0

W tc A xº

c Psup

tC0W t A xº

c

PsuptC0

W t 0¡ 8 suptC0

W t ª¡ PsuptC0

W t 0 PsuptC0

W t ª 1.

Moreover it holds

PsuptC0

W t 0 PsuptC0

W t B 0 B PW t B 0, suptC1

W t B 0 PW 1 B 0, sup

tC1W t W 1 B W 1

S0

ª

PsuptC1

W t W 1 B W t SW 1 xP W 1 > dx S

0

ª

PsuptC0

W t W 1 B x SW 1 xP W 1 > dx S

0

ª

PsuptC0

W t 0P W 1 > dx Psup

tC0W t 0 1

2,

thus P suptC0W t 0 0 and thus P suptC0W t ª 1.Analogously one can show that P inftC0W t ª 1.

The remaining part of the claim follows from P A 9 B 1 for any A,B > F with P A

P B 1.

Remark 3.3.1P suptC0Xt ª, inftC0Xt ª 1 implies that the trajectories of W oscillate betweenpositive and negative values on 0,ª an infinite number of times.Corollary 3.3.3Let W t, t C 0 be a Wiener process. Then it holds

P ω > Ω W ω is nowhere differentiable in 0,ª 1.

Proof ω > Ω W ω is nowhere differentiable in 0,ª 9

ª

n0ω > Ω W ω is nowhere differentiable in n,n 1.It is sufficient to show that Pω > Ω W ω is differentiable for a t0 t0ω > 0,1 0. Definethe set

Anm ω > Ω it exists a t0 t0ω > 0,1 with SW t0ω h,ω W t0ω, ωS Bmh, ¦h > 0, 4n .

3 Wiener process 49

Then it holds

ω > Ω W ω differentiable for a t0 t0ω mC1

nC1

Anm.

We still have to show P8mC1 8nC1 Anm 0.Let k0ω mink1,2,... kn C t0ω. Then it holds for ω > Anm and j 0,1,2

WW k0ω j 1n

,ω W k0ω jn

,ωW B WW k0ω j 1n

,ω W t0ω, ωW WW k0ω j

n,ω W t1ω, ωW

B8mn.

Let ∆nk W k1n W kn. Then it holds

PAnm B P n

k0

2j0

S∆nk jS B 8mn

B

n

Qk0

P 2j0

S∆nk jS B 8mn n

Qk0

PS∆n0S B 8mn3

B n 1 16mº2πn

3 0, nª,

by the independence of the increments of the Wiener Process.Since Anm ` An1,m, it follows P Anm 0.

Corollary 3.3.4With probability 1 it holds:

supnC1

sup0Bt0@...@tnB1

n

Qi1

SW ti W ti1S ª,i.e. W t, t > 0,1 possesses a.s. trajectories with unbounded variation.

Proof Since every continuous function g 0,1 R with bounded variation is differentiablealmost everywhere, the assertion follows from Corollary 3.3.3.

Alternative proofIt is sufficient to show that limnªP2n

i1 UW it2n W i1t2n U ª.

Let Zn P2ni1 W it2n W i1t

2n 2t. Hence EZn 0 and EZ2

n t22n1 and with Tchebysh-

eff’s inequality

P SZnS @ ε B EZ2n

ε2 tε2

2n1, i.e.ª

Qi1

P SZnS A ε a.s. 0.

50 3 Wiener process

From lemma of Borel-Cantelli it follows that limnªZn 0 almost surely and thus

0 B t B

2n

Qi1

W it2n W i 1t

2n2

B lim infnª

max1BkB2n

WW kt2n W k 1t

2nW 2n

Qi1

WW it2n W i 1t

2nW .

Hence the assertion follows since W has continuous trajectories and therefore

limnª

max1BkB2n

WW kt2n W k 1t

2nW 0.

3.4 Additional exercisesExercise 3.4.1Give an intuitive (exact!) method to realize trajectories of a Wiener process W W t, t >0,1. Thereby use the independence and the distribution of the increments of W . Addition-ally, write a program in R for the simulation of paths of W . Draw three paths t(W t, ω fort > 0,1 in a common diagram.Exercise 3.4.2Given are the Wiener process W W t, t > 0,1 and L argmaxt>0,1W t. Show that itholds:

PL B x 2π

arcsinºx, x > 0,1.

Hint: Use relation maxr>0,tW r d SW tS.Exercise 3.4.3For the simulation of a Wiener processW W t, t > 0,1 we also can use the approximation

Wnt n

Qk1

Sktzkwhere Skt, t > 0,1, k C 1 are the Schauder functions, and zk N 0,1 i.i.d. random variablesand the series converges almost surely for all t > 0,1 (nª).

a) Show that for all t > 0,1 the approximation Wnt also converges in the L2-sense toW t.

b) Write a program in R (alternative: C) for the simulation of a Wiener process W W t, t > 0,1.c) Simulate three paths t ( W t, ω for t > 0,1 and draw these paths into a common

diagram. Hereby consider the sampling points tk kn , k 0, . . . , n with n 28

1.

Exercise 3.4.4For the Wiener process W W t, t C 0 we define the process of the maximum that is givenby M Mt maxs>0,tW s, t C 0. Show that it holds:

3 Wiener process 51

a) The density fMt of the maximum Mt is given by

fMtx ¾ 2πt

expx2

2t¡1x C 0.

Hint: Use property PMt A x 2PW t A x.b) Expected value and variance of Mt are given by

EMt ¾2tπ, VarMt t1 2~π.

Now we define τx argmin s>RW s x as the first point in time for which the Wienerprocess takes value x.

c) Determine the density of τx and show that: Eτx ª.

Exercise 3.4.5Let W W t, t C 0 be a Wiener process. Show that the following processes are Wienerprocesses as well:

W1t ¢¦¤0, t 0,tW 1~t, t A 0,

W2t ºcW t~c, c A 0.

Exercise 3.4.6The Wiener process W W t, t C 0 is given. Size Qa, b denotes the probability that theprocess exceeds the half line y at b, t C 0, a, b A 0. Proof that:

a) Qa, b Qb, a and Qa, b1 b2 Qa, b1Qa, b2,b) Qa, b is given by Qa, b exp2ab.

4 Lévy Processes

4.1 Lévy Processes

Definition 4.1.1A stochastic process Xt, t C 0 is called Lévy process, if

1. X0 0,

2. Xt has stationary and independent increments,

3. Xt is stochastically continuous, i.e for an arbitrary ε A 0, t0 C 0:

limtt0

PSXt Xt0S A ε 0.

Remark 4.1.1One can easily see, that compound Poisson processes fulfill the 3 conditions, since for arb. ε A 0it holds

P SXt Xt0S @ ε C P SXt Xt0S A 0 B 1 eλStt0S ÐÐtt0

0.

Further holds for the Wiener process for arb. ε A 0

P SXt Xt0S A ε

¾2

πt t0 S ª

texp y2

2t t0dyx y»

tt0

2πS

ª

t»tt0

ex22 dxÐÐ

tt00.

4.1.1 Infinitely Divisibility

Definition 4.1.2Let X Ω R be an arbitrary random variable. Then X is called infinitely divisible, if forarbitrary n > N there exist i.i.d. random variables Y n

1 , . . . , Ynn with X d

Yn

1 . . . Ynn .

Lemma 4.1.1The random variable X Ω R is infinitely divisible if and only if the characteristic functionϕX of X can be expressed for every n C 1 in the form

ϕXs ϕnsn for all s > R,

where ϕn are characteristic functions of random variables.

52

4 Lévy Processes 53

Proof „ “Y

n1 , . . . , Y

nn i.i.d., X d

Yn

1 . . . Ynn . Hence, it follows that ϕXs Ln

i1ϕY ni

s ϕnsn.“ “

ϕXs ϕnsn there exist Y n1 , . . . , Y

nn i.i.d. with characteristic function ϕn and

ϕY1,...,Yns ϕnsn ϕXs. With the uniqueness theorem for characteristic functionsit follows that X d

Yn

1 . . . Ynn .

Theorem 4.1.1Let Xt, t C 0 be a Lévy process. Then the random variable Xt is infinitely divisible forevery t C 0.

Proof For arbitrary t C 0 and n > N it obviously holds that

X t X tn X 2t

n X t

n . . . X nt

n X n 1t

n .

Since Xt has independent and stationary increments, the summands are obviously inde-pendent and identically distributed random variables.

Lemma 4.1.2Let X1,X2, . . . Ω R be a sequence of random variables. If there exists a function ϕ R C,such that ϕs is continuous in s 0 and limnªϕXns ϕs for all s > R, then ϕ is thecharacteristic function of a random variable X and it holds that Xn

dÐX.

Definition 4.1.3Let ν be a measure on the measure space R,BR. Then ν is called a Lévy measure, ifν0 0 and

SR

min y2,1νdy @ª. (4.1.1)

Abb. 4.1: y (miny2,1Note

• Apparently every Lévy measure is σ-finite and

ν ε, εc @ ε, for all ε A 0, (4.1.2)

where ε, εc R ε, ε.

54 4 Lévy Processes

• In particular every finite measure ν is a Lévy measure, if ν0 0.

• If νdy gydy then gy 1SySδ for y 0, where δ > 0,3.• An equivalent condition to (4.1.2) is

SR

y2

1 y2 νdy @ª, since y2

1 y2 B min y2,1 B 2 y2

1 y2 . (4.1.3)

Theorem 4.1.2Let a > R, b C 0 be arbitrary and let ν be an arbitrary Lévy measure. Let the characteristicfunction of a random variable X Ω R be given through the function ϕ R C with

ϕs expias bs2

2 S

Reisy 1 isy1y > 1,1νdy¡ for all s > R. (4.1.4)

Then X is infinitely divisible.Remark 4.1.2 • The formula (4.1.4) is also called Lévy-Khintchine formula.

• The inversion of Theorem 4.1.2 also holds, hence every infinitely divisible random variablehas such a representation. Therefore the characteristic triplet a, b, ν is also called Lévycharacteristic of an infinitely divisible random variable.

Proof of Theorem 4.1.2 1st stepShow that ϕ is a characteristic function.For y > 1,1 it holds

• Teisy 1 isyT W ªQk0

isykk!

1 isyW W ªQk2

isykk!

W B y2 W ªQk2

sk

k!W´¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¶

c

B y2c

Hence it follows from (4.1.1) that the integral in (4.1.4) exists and therefore it is well-defined.

• Let now cn be an arbitrary sequence of numbers with cn A cn1 A . . . A 0 and limnª cn 0. Then the function ϕn R C with

ϕns expisa Scn,cnc91,1 yνdy bs2

2¡ expScn,cnc eisy 1νdy

is the characteristic function of the sum of independent random variables Zn1 and Zn

2 ,since– the first factor is the characteristic function of the normal distribution with expec-

tation a Rcn,cnc91,1 yνdy and variance b.– the second factor is the characteristic function of a compound Poisson process with

parameters

λ νcn, cnc and PU ν 9 cn, cnc~νcn, cnc.


• Furthermore limnªϕns ϕs for all s > R, where ϕ is obviously continuous in 0,since it holds for the function ψ R C in the exponent of (4.1.4)

ψs SReisy 1 isy1 y > 1,1νdy for all s > R

that SψsS cs2 R1,1 y2νdy R1,1c Teisy 1Tνdy. Out of this and from (4.1.3) itfollows by Lebesgue’s theorem that lim

s0ψs 0.

• Lemma 4.1.2 yields that the function ϕ given in (4.1.4) is the characteristic function of arandom variable.

2nd stepThe infinite divisibility of this random variable follows from Lemma 4.1.1 and out of the fact,that for arbitrary n > N ν

n is also a Lévy measure and that

ϕs exp¢¦¤i ans

bns

2

2 S

Reisy 1 isy1y > 1,1 ν

n dy£§¥ for all s > R.

Remark 4.1.3The map η R C with

ηs ias bs2

2 S

Reisy 1 isy1y > 1,1νdy

from (4.1.4) is called Lévy exponent of this infinitely divisible distribution.

4.1.2 Lévy-Khintchine Representation

Xt, t C 0 – Lévy process. We want to represent the characteristic function of Xt, t C 0,through the Lévy-Khintchine formula.Lemma 4.1.3Let Xt, t C 0 be a stochastically continuous process, i.e. for all ε A 0 and t0 C 0 it holdsthat limtt0 PSXtXt0S A ε 0. Then for every s > R, tz ϕXts is a continuous mapfrom 0,ª to C.

Proof • y z eisy continuous in 0, i.e. for all ε A 0 there exists a δ1 A 0, such that

supy>δ1,δ1 Teisy 1T @ ε

2.

• Xt, t C 0 is stochastically continuous, i.e. for all t0 C 0 there exists a δ2 A 0, such that

suptC0, Stt0S@δ2

P SXt Xt0S A δ1 @ ε4 .


Hence, it follows that for s > R, t C 0 and St t0S @ δ2 it holds

TϕXts ϕXt0sT UE eisXt eisXt0U B E UeisXt0 eisXtXt0 1U

B E UeisXtXt0 1U SRTeisy 1TPXtXt0dy

B Sδ1,δ1 Teisy 1TPXtXt0dySδ1,δ1c Teisy 1T´¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¶

B2

PXtXt0dyB sup

y>δ1,δ1 Teisy 1T 2P SXt Xt0S A δ1 B ε.Theorem 4.1.3Let Xt, t C 0 be a Lévy process. Then for all t C 0 it holds

ϕXts etηs, s > R,

, where η R C is a continuous function. In particular it holds that

ϕXts etηs eηst ϕX1st , for all s > R, t C 0.

Proof Due to stationarity and independence of increments we have

ϕXtts EeisXtt E eisXteisXttXt ϕXtsϕXts, s > R.

Let gs 0,ª C be defined by gst ϕXts, s > R, gst t gstgst, t, t C 0.X0 0. ¢¦¤

gst t gstgst, t, t C 0,gs0 1,gs 0,ª C continuous.

Hence there exists η R C, such that gst eηst for all s > R, t C 0. ϕX1s eηs and itfollows that η is continuous.

Lemma 4.1.4 (Prokhorov):Let µ1, µ2, . . . be a sequence of finite measures (on BR) with

1. supnC1 µnR @ c, c const @ª (uniformly bounded)

2. for all ε A 0 there exists Bε > BR compact, such that fulfills the tightness conditionsupnC1 µnBc

ε B ε. Hence follows that there exists a subsequence µn1 , µn2 , . . . and a finitemeasure over BR, such that for all f R C, bounded, continous, it holds that

limkª

SRfyµnkdy SR fyµdy

Proof See [14], page 122 - 123.


Theorem 4.1.4Let Xt, t C 0 be a Lévy process. Then there exist a > R, b C 0 and a Lévy measure ν, suchthat

ϕX1s eias bs22 S

Reisy 1 iy1y > 1,1νdy, for all s > R.

Proof For all sequences tnn>N b 0,ª, with limnª

tn 0, it holds

ηs etηsVt0

limnª

etnηs 1tn

limnª

ϕXtns 1tn

, (4.1.5)

since η R C is continuous. The latter convergence is even uniform in s > s0, s0 for anys0 A 0, since Taylor’s theorem yields

limnª

Wηs etnηs 1tn

W limnª

Wηs 1tn

ª

Qk1

tnηskk!

W lim

nª

W 1tn

ª

Qk2

tnηskk!

W lim

nª

Wηs ª

Qk1

tnηskk 1! W lim

nª

Wη2stn ª

Qk1

tnηsk1k 1! WB lim

nª

M2tnª

Qk1

StnM Sk1k 1! , where M sups>s0,s0 SηsS @ª

limnª

M2tnª

Qk1

StnM Sk1k 1! 1kk 1

B limnª

M2tnª

Qk1

StnM Sk1k 1! lim

nª

M2tneStnM S

0.

Now let tn 1n and Pn be the distribution of X 1

n. Hence it follows that

limnª

nSReisy 1Pnds lim

nª

nϕX 1

ns 11n

ηslimnª

SRnS

s0

s0eisy 1Pndyds S s0

s0ηsds

and consequently

limnª

nSR1 sins0y

s0yPndy lim

nª

nSR

12s0S

s0

s0eisy 1dsPndy 1

2s0S

s0

s0ηsds.

Since η R C is continuous with η0 0 and it follows from the mean value theorem, thatfor all ε A 0 it exists δ0 A 0, such that U 1

2s0 Rs0s0

ηsdsU @ ε. Since 1 sins0ys0y

C12 , for Ss0yS C 2,


it holds: for all ε A 0 there exist s0 A 0, n0 A 0, such that

lim supnª

n

2 SySySC 2s0

Pndy B lim supnª

nSR1 sins0y

s0yPndy @ ε.

Hence for all ε A 0 there exist s0 A 0, n0 A 0, such that

nSySySC 2s0

Pndy B 4ε, for all n C n0.

Decreasing s0 givesnSySySC 2

s0 Pndy B 4ε, for all n C 1.

y2

1 y2 B c1 sin yy

, for all y x 0 and a c A 0.

Hence, it follows that

supnC1

nSR

y2

1 y2 Pndy B c for a c @ª.

Let now µn BR 0,ª be defined as

µnB nSB

y2

1 y2 Pndy for all B > BR.It follows that µnn>N is uniformly bounded, supnC1 µnR @ c. Furthermore holds y2

1y2 B 1,supnC1 µn y SyS A 2

s0 B 4ε and µnn>N relatively compact. After lemma 4.1.3 it holds: there

exists µnkk>N, such that

limkª

SRfyµnkdy SR fyµdy

for a measure µ and f continuous and bounded. Let for s > R the function fs R C be definedas

fsy ¢¦¤ eisy 1 is siny 1y2

y2 , y x 0,s2

2 , otherwise.

Hence follows that fs is bounded and continuous and

ηs limnª

nSReisy 1Pndy

limnª

SRfsyµndy isnS

Rsin yPndy

limkª

SRfsyµnkdy isnk SR sin yPnkdy

SRfsyµdy lim

kª

isnk SR

sin yPnkdyηs ias

bs2

2 S

Reisy 1 is sin yνdy,


for all s > R with a limkª isnk RR sin yPnkdy @ª, b µ 0, ν BR 0,ª,νdy ¢¦¤

1y2

y2 µdy, y x 0,0 , y 0.

SRSy1y > 1,1 sin ySνdy @ª.

Sy1y > 1,1 sin yS 1 y2

y2 @ c, for all y x 0 and a c A 0.

Hence follows that

ηs ias bs2

2 S

Reisy 1 isy1 y > 1,1νdy, for all s > R.

a a SRy1y > 1,1 sin y νdy.

4.1.3 Examples

1. Wiener process (it is enough to look at X1)X1 N 0,1, ϕX1s e s2

2 and hence follows

a, b, ν 0,1,0.Let X Xt, t C 0 be a Wiener proess with drift µ, i.e. Xt µt σW t, W W t, t C 0 – Brownian motion. It follows

a, b, ν µ,σ2,0.ϕX1s EeisX1

EeµσW 1is eµisϕW 1σs eisµσ2 s2

2 , s > R.

2. Compound Poisson process with parameters λ,PnXt PNt

i1 Ui, Nt Poisλt, Ui i.i.d. PU .

ϕX1s expλSReisx 1PUdx

expλisSRx1x > 1,1PUdx λS

Reisx 1 isx1x > 1,1PUdx

expλisS 1

1xPUdx λS

Reisx 1 isx1x > 1,1PUdx , s > R.

Hence follows

a, b, ν λS 1

1xPUdx,0, λPU , PU – finite on R.


3. Process of Gauss-Poisson typeX Xt, t C 0, Xt X1t X2t, t C 0.X1 X1t, t C 0 and X2 X2t, t C 0 independent.X1 – Wiener process with drift µ and variance σ2,X2 – Compound Poisson process with parameters λ,PU .

ϕXts ϕX1tsϕX2ts exp

isµ σ2s2

2 λS

R

eisx 1PUdx expisµ λS 1

1xPUdx σ2s2

2

SRλ eisx 1 isx1x > 1,1PUdx , s > R.

Hence follows a, b, ν µ λS 1

1xPUdx, σ2, λPU .

4. Stable Lévy processX Xt, t C 0 – Lévy process with Xt α stable distribution, α > 0,2. Tointroduce α-stable laws ν, let us begin with an example.If X W (Wiener process), then X1 N 0,1. Let Y,Y1, . . . , Yn be i.i.d. N µ,σ2-variables. Since the normal distribution is stable w.r.t. convolution it holds

Y1 . . . Yn N nµ,nσ2 d

ºnY nµ

ºnµ

ºnY µ n ºn

n12Y µ n 2

2 n12

n1αY µ n n 1

α , α 2.

Definition 4.1.4The distribution of a random variable Y is called α-stable, if for all n > N independent copiesY1, . . . , Yn of Y exist, such that

Y1 . . . Ynd n

1αY dn,

where dn is deterministic. The constant α > 0,2 is called index of stability.Moreover, one can show that

dn

¢¦¤ µ n n 1α , α x 1,

µn logn , α 1.

Example 4.1.1 • α 2: Normal distribution, with any mean and any variance.

• α 1: Cauchy distribution with parameters µ,σ2. The density:

fY x σ

π x µ2 σ2 , x > R.

It holds EY 2 ª, EY does not exist.


• α 12 : Lévy distribution with parameters µ,σ2. The density:

fY x ¢¦¤ σ2π 1

2 1xµ 3

2exp σ

2xµ, x A µ,

0 , otherwise.

These examples are the only examples of α-stable distribution, where an explicit form ofthe density is available. For other α > 0,2, α x

12 , 1, the α-stable distribution is introduced

through its characteristic function. In general holds: If Y α-stable, α > 0,2, then ESY Sp @ ª,0 @ p @ α.Definition 4.1.5The distribution of a random variable is called symmetric, if Y d

Y .If Y has a symmetric α-stable distribution, α > 0,2, then

ϕY s expc SsSα , s > R.Indeed, it follows from the stability of Y that

ϕY sn eidnsϕY n 1α s , s > R.

It follows that dn 0, since ϕY s ϕY s varphiY s. It holds: eidns eidns, s > R anddn 0. The rest is left as an exercise.Lemma 4.1.5Lévy-Khintchine representation of the characteristic function of a stable distribution. Anystable law is infinitely divisible with the Lévy triplet a, b, ν, where a > R arbitrary,

b σ2, α 2,0 , α @ 2.

andνdx 0 , α 2,

c1x1α 1x C 0dx c2SxS1α 1x @ 0dx, α @ 2, c1, c2 C 0 c1 c2 A 0

Without proofExercise: Prove that

P SY S C x

xª

¢¦¤ ex2

2σ2 , α 2,cxα , α @ 2.

Definition 4.1.6The Lévy process X Xt, t C 0 is called stable, if X1 has an α-stable distribution,α > 0,2 (α 2: Brownian motion (with drift)).

4.1.4 Subordinators

Definition 4.1.7A Lévy process X Xt, t C 0 is called subordinator, if for all 0 @ t1 @ t2, Xt1 BXt2 a.s.Since

X0 0 a.s. Xt C 0, t C 0, a.s.


This class of Subordinators is important since you can easily introduce R ba gtdXt as aLebesgue-Stieltjes-integral.Theorem 4.1.5The Lévy process X Xt, t C 0 is a subordinator if and only if the Lévy-Khintchine repre-sentation can be expressed in the form

ϕX1s expias SReisx 1νdx , s > R, (4.1.6)

where a > 0,ª and ν is the Lévy measure, with

ν ª,0 0, Sª

0min 1, y2νdy @ª.

Proof SufficiencyIt has to be shown that Xt2 CXt1 a.s., if t2 C t1 C 0.First of all we show that X1 C 0 a.s.. If ν 0, then X1 a C 0 a.s., hence

ϕXts ϕX1st eiats, s > R.

Xt at a.s. and therefore it follows that Xt and X is a subordinator.If ν0,ª A 0, then there exists N A 0 such that for all n C N it holds 0 @ ν 1

n ,ª @ª. Itfollows

ϕX1s expias limnª

Sª

1n

eisx 1νdx¡ eias limnª

ϕns, s > R,

where ϕns R ª1neisx 1νdx is the characteristic function of a compound Poisson process

distribution with parametersν 1

n ,ª , ν 9 1n,ª

ν 1n,ª

for all n > N. Let Zn be the random

variable with characteristic function ϕn. It holds: Zn PNni1Ui, Nn Pois ν 1n ,ª,

Ui ν9 1

n,ª

ν 1n,ª ; hence follows Zn C 0 a.s. and X1 d

a®C0

limZn´¹¹¹¹¹¸¹¹¹¹¹¹¶C0

C 0 a.s.. Since X is a Lévy

process, it holds

X 1 X 1n X 2

n X 1

n . . . X n

n X n 1

n ,

where, because of stationarity and independence of the increments, X kn X k1

n a.s.C 0 for

1 B k B n for all n. Hence Xq2 Xq1 C 0 a.s. for all q1, q2 > Q, q2 C q1 C 0. Now let t1, t2 > Rsuch that 0 B t1 B t2. Let qn1 , q

n2 be sequences of numbers from Q with q

n1 B q

n2 such

that qn1 t1, qn2 t2, nª. For ε A 0

P Xt2 Xt1 @ ε PXt2 X qn2 X qn2 X qn1 ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

C0

X qn1 X t1 @ εB P Xt2 X qn2 X qn1 X t1 @ εB PXt2 X qn2 @ ε

2 PX qn1 Xt1 B ε2ÐÐÐnª

0,


since X is stochastically continuous. Then

P Xt2 Xt1 @ ε 0 for all ε A 0 andP Xt2 Xt1 @ 0 lim

ε0P Xt2 Xt1 @ ε 0

Xt2 CXt1 a.s.

NecessityLet X be a Lévy process, which is a subordinator. It has to be shown that ϕX1 has theform (4.1.6).After the Lévy-Khintchine representation for X1 it holds that

ϕX1s expias b2s2

2 S

ª

0eisx 1 isx1x > 1,1νdx¡ , s > R.

The measure ν is concentrated on 0,ª, since Xt a.s.C 0 for all t C 0 and from the proof ofTheorem 4.1.4 ν ª,0 0 can be chosen.

ϕX1s expias b2s2

2¡´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

ϕY1s

expS ª

0eisx 1 isx1 x > 1,1νdx ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

ϕY2s

Hence it follows that X1 Y1 Y2, where Y1 and Y2 are independent, Y1 N a, b2 andtherefore b 0. (Otherwise Y1 could attain negative values and consequently X1 could attainnegative values as well.)For all ε > 0,1

ϕX1s expisa S 1

εxνdx S ε

0eisx 1 isxνdx S ª

0eisx 1νdx .

It has to be shown that for ε 0 it holds R ªε eisx 1νdx R ª0 eisx 1νdx @ ª withR 1

0 min x,1νdx @ ª. ϕX1s expis a R 1ε xνdxϕZ1sϕZ2s, where Z1 and Z2

are independent, ϕZ1s exp ε

R0eisx 1 isxνdx¡, ϕZ2s expR ªε eisx 1νdx,

s > R. Then X1 d a R 1ε xνdxZ1 Z2. There exist ϕ2

Z10 EZ2

12 @ª, ϕ1

Z10 0 iEZ1

and it therefore follows that EZ1 0 and PZ1 B 0 A 0. On the other hand, Z2 has a compoundPoisson distribution with parameters ν ε,ª , ν9ε,ª

νε,ª , ε > 0,1. P Z2 B 0 A 0, since PZ2 0 A 0. P Z1 Z2 B 0 C P Z1 B 0, Z2 B 0 P Z1 B 0P Z2 B 0 A 0

For X1 to be positive it follows that a R 1ε xνdx C 0 for all ε > 0,1. Hence a C 0 and

Sª

0min x,1dx @ª.

Moreover, for ε 0 it holds Z1d 0 and consequently

ϕX1s expisa S 1

0xνdx S ª

0eisx 1νdx , s > R.


Example 4.1.2 (α-stable subordinator):Let X Xt, t C 0 be a subordinator, with a 0 and the Lévy measure

νdx αΓ1α 1

x1αdx, x A 0,0 1

x1αdx 0 , x B 0.

By Lemma 4.1.5 it follows that X is an α-stable Lévy process.We show that lXts EesXt etsα for all s, t C 0.

ϕXts ϕX1st exptS ª

0eisx 1 α

Γ1 α 1x1αdx¡ , s > R.

It has to be shown that

uα α

Γ1 α S ª

01 eux dx

x1α , u C 0.

This is enough since ϕXt can be continued analytically to z > C Imz C 0, i.e. ϕXtiu lXtu, u C 0. In fact, it holds that

Sª

01 eux dx

x1d Sª

0uS

x

0euydyx1αdx

Fubini S

ª

0 Sª

yueuyx1αdxdy

Sª

0 Sª

yx1αdxueuydy

u

αS

ª

0euyyαdy

Subst.

u

αS

ª

0ezzα

1uα

d zu

uα

αS

ª

0ezz1α1dz

uα

αΓ1 α

and hence follows lXts etsα , t, s C 0.

4.2 Additional Exercises

Exercise 4.2.1Let X eb a random variable with distribution function F and characteristic function ϕ. Showthat the following statements hold:

a) If X is infinitely divisible, then it holds ϕt x 0 for all t > R. Hint: Show thatlimnª SϕnsS2 1 for all s > R, if ϕs ϕnsn. Note further that SϕnsS2 isagain a characteristic function and limnª x

1n 1 holds for x A 0.

b) Give an example (with explanation) for a distribution, which is not infinitely divisible.


Exercise 4.2.2Let X Xt, t C 0 be a Lévy process. Show that the random variable Xt is then infinitelydivisible for every t C 0.Exercise 4.2.3Show that the sum of two independent Lévy processes is again a Lévy process, and state thecorresponding Lévy characteristic.Exercise 4.2.4Look at the following function ϕ R C with

ϕt eψt, where ψt 2ª

Qkª

2kcos2kt 1.Show that ϕt is the characteristic function of an infinitely divisible distribution. Hint: Lookat the Lévy-Khintchine representation with measure ν2k 2k, k > Z.Exercise 4.2.5Let the Lévy process Xt, t C 0 be a Gamma process with parameters b, p A 0, that is, forevery t C 0 it holds Xt Γb, pt. Show that Xt, t C 0 is a subordinator with the Laplaceexponent ξu R ª0 1 euyνdy with νdy py1ebydy, y A 0. (The Laplace exponentof Xt, t C 0 is the function ξ 0,ª 0,ª, for which holds that EeuXt etξu forarbitrary t, u C 0)Exercise 4.2.6Let Xt, t C 0 be a Lévy process with charactersistic Lévy exponent η and τs, s C 0 aindependent subordinator witch characteristic Lévy exponent γ. The stochastic process Y bedefined as Y Xτs, s C 0.(a) Show that

E eiθY s eγiηθs, θ > R,

where Imz describes the imaginary part of z.Hint: Since τ is a process with non-negative values, it holds Eeiθτs eγθs for allθ > z > C Imz C 0 through the analytical continuation of Theorem 4.1.3.

(b) Show that Y is a Lévy process with characteristic Lévy exponent γiη.Exercise 4.2.7Let Xt, t C 0 be a compound Poisson process with Lévy measure

νdx λº

2σºπe

x22σ2 dx, x > R,

where λ,σ A 0. Show that σW Nt, t C 0 has the same finite-dimensional distributions asX, where Ns, s C 0 is a Poisson process with intensity 2λ and W is a standard Wienerprocess independent from N .Hint to exercise 4.2.6 a) and exercise 4.2.7

• In order to calculate the expectation for the characteristic function, the identity EX EEX SY RR EX SY yFY dy for two random variables X and Y can be used. Indoing so, it should be conditioned on τs.


• R ªª cossye y22a dy

º2πa eas

22 for a A 0 and s > R.

Exercise 4.2.8Let W be a standard Wiener process and τ an independent α

2 -stable subordinator, whereα > 0,2. Show that W τs, s C 0 is a α-stable Lévy process.Exercise 4.2.9Show that the subordinator T with marginal density

fT ts t

2ºπs

32 e

t24s 1s A 0

is a 12 -stable subordinator. (Hint: Differentiate the Laplace transform of T t and solve the

differential equation)

5 Martingales

5.1 Basic Ideas

Let Ω,F ,P be a complete probability space.Definition 5.1.1Let Ft, t C 0 be a family of σ-algebras Ft ` F . It is called

1. a filtration if Fs b Ft, 0 B s @ t.

2. a complete filtration if it is a filtration such that F0 (and therefore all Fs, s A 0) containsall sets of zero probability.Later on we will always assume, that we have a complete filtration.

3. a right-continuous filtration if for all t C 0 Ft 9sAtFs.

4. a natural filtration for a stochastic process Xt, t C 0, if it is generated by the past ofthe process until time t C 0, i.e. for all t C 0 Ft is the smallest σ-algebra which containsthe sets ω > Ω Xt1, . . . ,Xtn ` B, for all n > N, 0 B t1, . . . , tn B t, B > BRn.

A random variable τ Ω R is called stopping time (w.r.t. the filtration Ft, t C 0), if forall t C 0 ω > Ω τω B t > Ft.If Ft, t C 0 is the natural filtration of a stochastic process Xt, t C 0, then τ being astopping time means that by looking at the past of the process X you can tell, whether themoment τ occurred.Lemma 5.1.1Let Ft, t C 0 be a right-continuous filtration. τ is a stopping time w.r.t. Ft, t C 0 if andonly if τ @ t > Ft´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ω>Ωτω@t>Ft

for all t C 0.

Proof „ “Let τ @ t > Ft, t C 0. To show: τ B t > Ft.τ B t 9s>t,tετ @ s for all ε A 0 τ B t > 9sAtFs Ft

„ “To show: τ B t > Ft, t C 0 τ @ t > Ft, t C 0.τ @ t 8s>0,tτ B t s > 8s>0,tFts ` Ft

Definition 5.1.2Let Ω,F ,P be a probability space, Ft, t C 0 a filtration (Ft ` F , t C 0) and X Xt, t C0 a stochastic process on Ω,F ,P. X is adapted w.r.t. the filtration Ft, t C 0, if Xt isFt-measurable for all t C 0, i.e. for all B > BR Xt > B > Ft.

67

68 5 Martingales

Definition 5.1.3The time τBω inft C 0 Xt > B, ω > Ω, is called first hitting time of the set B > BR bythe stochastic process X Xt, t C 0 (also called: first passage time, first entrance time).Theorem 5.1.1Let Ft, t C 0 be a right-continuous filtration and X Xt, t C 0 an adapted (w.r.t.Ft, t C 0) càdlàg process. For open B ` R, τB is a stopping time. If B is closed, thenτBω inft C 0 Xt > B or Xt > B is a stopping time, where Xt limstXs.Proof 1. Let B > BR be open. Because of Lemma 5.1.1 it is enough to show that τB @

t > Ft, t C 0. Because of right-continuity of the trajectories of X it holds:τB @ t 8s>Q90,tXs > B > 8s>Q90,tFs b Ft, since Fs b Ft, s @ t.

2. Let B > BR be closed. For all ε A 0 let Bε x > R dx,B @ ε be a parallel set of B,where dx,B infy>B Sx yS. Bε is open for all ε A 0.

τB B t s>Q90,tXs > B 8 nC1

s>Q90,tXs > B 1

n > Ft,

since X is adapted w.r.t. Ft, t C 0.Lemma 5.1.2Let τ1, τ2 be stopping times w.r.t. the filtration Ft, t C 0. Then minτ1, τ2, maxτ1, τ2,τ1 τ2 and ατ1, α C 1, are stopping times (w.r.t. Ft, t C 0).Proof For all t C 0 holds:minτ1, τ2 B t τ1 B t´¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¶

>Ft

8τ2 B t´¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¶>Ft

> Ft,

maxτ1, τ2 B t τ1 B t 9 τ2 B t > Ft,ατ1 B t τ1 Btα > F t

α` Ft, since t

α B t,τ1 τ2 B tc τ1 τ2 A t τ1 A t´¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¶>Ft

8τ2 A t´¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¶>Ft

8τ1 C t, τ2 A 0´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶>Ft

8τ2 C t, τ1 A 0´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶>Ft

8

0 @ τ2 @ t, τ1 τ2 A t 8 0 @ τ1 @ t, τ2 τ1 A t,To show: 0 @ τ2 @ t, τ1 τ2 A t > Ft. (That also 0 @ τ1 @ t, τ2 τ1 A t > Ft works analogously.)0 @ τ2 @ t, τ1 τ2 A t

s>Q90,ts @ τ1 @ t, τ2 A t s > FtTheorem 5.1.2Let τ be an a.s. finite stopping time w.r.t. the filtration Ft, t C 0 on the probability spaceΩ,F ,P, i.e. Pτ @ ª 1. Then there exists a sequence of discrete stopping times τnn>N,τ1 C τ2 C τ3 C . . ., such that τn τ , nª a.s.

Proof For all n > N let

τn 0, if τω 0k12n , if k

2n @ τω B k12n , for a k > N0

For all t C 0 and for all n > N §k > N0 k

2n B t B k12n , i.e. it holds τn B t τn B

k2n τ B k

2n > F k2n

` Ft τn is a stopping time. Obviously τn τ , nª a.s.

5 Martingales 69

Corollary 5.1.1Let τ be an a.s. finite stopping time w.r.t. the filtration Ft, t C 0 and X Xt, t C 0 acàdlàg process on Ω,F ,P, Ft ` F for all t C 0. Then Xω, τω, ω > Ω is a random variableon Ω,F ,P.Proof To show: Xτ Ω R measurable, i.e. for all B > BR Xτ > B > F . Let τn τ ,n ª be as in Theorem 5.1.2. Since X is càdlàg, it holds that Xτn ÐÐÐ

nª

Xτ a.s.. ThenXτ is F-measurable as the limit of Xτn, which are themselves F-measurable. Indeed, forall B > BR it holds

Xτn > B ª

k0

τn k

2n ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

>F

9X k2n > B ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

>F

> F

5.2 (Sub-, Super-)Martingales

Definition 5.2.1Let X Xt, t C 0 be a stochastic process adapted w.r.t. to a filtration Ft, t C 0, Ft ` F ,t C 0, on the probability space Ω,F ,P, E SXtS @ª, t C 0.X is called martingale, resp. sub- or supermartingale, if EXt S Fs Xs a.s., resp.EXt S Fs C Xs a.s. or EXt S Fs B Xs a.s. for all s, t C 0 with t C s: EXt EXs const for all s, t.Examples

Very often martingales are constructed on the basis of a stochastic process Y Y t, t C 0as follows: Xt gY t EgY t for some measurable function g R R, or byXt eiuY t

ϕY tu , for any fixed u > R.

1. Poisson processLet Y Y t, t C 0 be the homogeneous Poisson process with intensity λ A 0. EY t VarY t λt since Y t Poisλt, t C 0.a) Xt Y t λt, t C 0 Xt is a martingale w.r.t. the natural filtrationFs, s C 0.

EXt S FssBt EY t λt Y s λs Y s λs S Fs Y s λs EY t Y s λt s S Fs

ind.incr. Y s λs EY t Y s λt s Y s λs EY t s´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

λtsλt s

Y s λs a.s. Xs

70 5 Martingales

b) Xt X2t λt, t C 0 Xt is a martingale w.r.t. Fs, s C 0.EXt S Fs EX2t λt S Fs EXt Xs Xs2

λt S Fs EXt Xs2

2Xt XsXs X2s λs λt s S Fs Xs EXt Xs2´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

VarY tY sλts2XsEXt Xs´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

0

λt sa.s. X s, s B t.

2. Compound Poisson processY t PNt

i1 Ui, t C 0, N – homogeneous Poisson process with intensity λ A 0, Ui –independent identically distributed random variables, ESUiS @ª, Ui independent of N .Let Xt Y t EY t Y t λtEU1, t C 0.Exercise 5.2.1Show that X Xt, t C 0 is a martingale w.r.t. its natural filtration.

3. Wiener processLet W W t, t C 0 be a Wiener process, Fs, s C 0 be the natural filtration.a) Y Y t, t C 0, where Y t W 2t EW 2t W 2t t, t C 0, is a martingale

w.r.t. Fs, s C 0.Indeed, it holds

EY t S Fs EW t W s W s2

s t s S Fs EW t W s2

2W sW t W s W s2 S Fs s t s EW t W s2 2W sEW t W s EW 2s S Fs s t s t s W 2s s t s W 2s s a.s. Y s, s B t.

b) Y t euW tu2 t2 , t C 0 and a fixed u > R.

ESY tS eu2 t2 EeuW t eu2 t

2 eu2 t

2 1 @ ª. We show that Y Y t, t C 0 is amartingale w.r.t. Fs, s C 0.

EY t S Fs EeuW tW sW su2 s2u

2 ts2 S Fs

eu2 s

2 euW s´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶Y s

eu2 ts

2 EeuW tW s S Fs´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶EeuW tseu2 ts

2

Y seu2 ts2 eu

2 ts2 Y s, s B t.

4. Regular martingaleLet X be a random variable (on Ω,F ,P) with ESX S @ª. Let Fs, s C 0 be a filtrationon Ω,F ,P.Construct Y t EX S Ft, t C 0. Y Y t, t C 0 is a martingale.Indeed, ESY tS ESEX S FtS B EESX S S Ft ESX S @ª, t C 0.

5 Martingales 71

EY t S Fs EEX S Ft S Fs a.s. EX S Fs Y s, s B t since Fs b Ft.If Y nS n 1, ...,N, where N > N, is a martingale, then it is regular since Y n

EY N S Fn for all n 1, ...,N , i.e. X Y N.However, the latter is not possible for processes of the form Y nS n > N orY tS t C 0.

5. Lévy processesLet X Xt, t C 0 be a Lévy process with Lévy exponent η and natural filtrationFs, s C 0.a) If ESX1S @ ª, define Y t Xt tEX1´¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¶

EXt, t C 0. As in the previous cases it can

be shown that Y Y t, t C 0 is martingale w.r.t. the filtration Fs, s C 0.(Compare Example 1 and 2)

b) In the general case one can use the combination from example 3b – normalize eiuXtby the characteristic function of Xt, i.e. let Y t eiuXt

ϕuXt

eiuXtetηu eiuXttηu,

t C 0, u > R.To show: Y Y t, t C 0 is a complex-valued martingale.ESY tS SetηuS @ª, since η R C. EY t 1, t C 0. Furthermore, it holds

EY t S Fs EeiuXtXstsηueiuXssηu S Fs eiuXssηuetsηuEeiuXtXs Y setsηuetsηu a.s. Y s

6. Monotone Submartingales/SupermartingalesEvery integrable stochastich process X Xt, t C 0, which is adapted w.r.t. toa filtration Fs, s C 0 and has a.s. monotone nondecreasing (resp. non-increasing)trajectories, is a sub- (resp. a super-)martingale.In fact, it holds e.g. Xt a.s.C Xs, t C s EXt S Fs a.s.C EXs S Fs a.s. Xs. Inparticular, every subordinator is a submartingale.

Lemma 5.2.1Let X Xt, t C 0 be a stochastic process, which is adapted w.r.t. a filtration Ft, t C 0and let f R R be convex such that ESfXtS @ ª, t C 0. Then Y fXt, t C 0 is asub-martingale, if

a) X is a martingale, or

b) X is a sub-martingale and f is monotone nondecreasing.

Proof Use Jensen’s inequality for conditional expectations.

EfXt S Fs a.s.C fEXt S Fs´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

b)CXs, a)

Xs

a.s.C fXs.

72 5 Martingales

5.3 Uniform Integrability

It is known, that in general Xna.s.ÐÐÐnª

X does not yield XnL1ÐÐÐnª

X. Here X,X1,X2, . . .

are random variables, defined on the probability space Ω,F ,P. When does „Xna.s.ÐÐÐnª

X“

„XnL1ÐÐÐnª

X“hold? The answer for this provides the notion of uniformly integrability ofXn, n > N.Definition 5.3.1The sequence Xn, n > N of random variables is called uniformly integrable, if ESXnS @ ª,n > N, and sup

n>NESXnS1SXnS A εÐÐÐ

εª

0.

Lemma 5.3.1The sequence Xn, n > N of random variables is uniformly integrable if and only if

1. supn>N

ESXnS @ª (uniformly bounded) and

2. for every ε A 0 there is a δ A 0 such that ESXnS1A @ ε for all n > N and all A > F withPA @ δ.

Proof Let Xn be a sequence of random variables.It has to be shown that

supn>N

ESXnS1SXnS A xÐÐÐxª

01 sup

n>NESXnS @ª

2 ¦ε A 0 §δ A 0 ESXnS1A @ ε ¦A > F PA @ δ„“Set An SXnS A x for all n > N and x A 0. It holds PAn B 1

xESXnS by virtue of Markov’s in-equality and consequently supn PAn B 1

x supn ESXnS B cx ÐÐÐxª

0 §N A 0 ¦x A N PAn @ δ2 supn ESXnS1An B ε.Since ε A 0 can be chosen arbitrarily small supn ESXnS1SXnS A xÐÐÐ

xª

0.

„“

1.

supn

ESXnS B supnESXnS1SXnS A x ESXnS1SXnS B x

B supnESXnS1SXnS A x xPSXnS B x´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

B1

B ε x @ª

2. For all ε A 0 §x A 0 such that ESXnS1SXnS A x @ ε2 because of uniform integrability.

Choose δ A 0 such that xδ @ ε2 . Then it holds

ESXnS1A E SXnS±Bx

1SXnS B x´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶B1

1A ESXnS1SXnS A x 1A²B1

B xPA´¹¹¹¹¹¹¸¹¹¹¹¹¹¹¶

Bε2

ESXnS1SXnS A x´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶Bε2

,

5 Martingales 73

Lemma 5.3.2Let Xnn>N be a sequence of random variables with ESXnS @ ª, n > N, Xn

PÐÐÐnª

X. Then

XnL1ÐÐÐnª

X if and only if Xnn>N is uniformly integrable.

Particularly, XnP

ÐÐÐnª

X implies EXn ÐÐÐnª

EX.

Proof 1) Let Xnn>N be uniformly integrable. It has to be shown that ESXn X SÐÐÐnª

0.

Since XnP

ÐÐÐnª

X one obtains that Xnn>N has a subsequence Xnkk>N converging almostsurely to X. Consequently, Fatou’s Lemma yields

ESX S B lim infkª

ESXnk S B supn>N

ESXnS,and therefore ESX S @ª by Lemma 5.3.1.1).Moreover, one infers from lim

nª

P SXn X S A ε 0 and Lemma 5.3.1.2) that

limnª

ESXn X S1 SXn X S A ε 0

for all ε A 0. (It is an easy exercise to verify that the uniform integrability of Xnn>N impliesthe uniform integrability of Xn Xn>N.)Conclusively it follows

limnª

ESXn X S limnª

ESXn X S1 SXn X S A ε ESXn X S1 SXn X S B ε B εfor all ε A 0, i.e. Xn

L1ÐÐÐnª

X.

2) Now let ESXn X SÐÐÐnª

0. The properties 1 and 2 of Lemma 5.3.1 have to be shown.

1. supn

ESXnS B supn

ESXn X S ESX S @ª, since XnL1ÐX.

2. For all A ` F , PA B δ:ESXnS1A B ESXn X S 1A²

B1

ESX S1A B ESXn X S´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶@ε2

ε

2 ε

with an appropriate choice of δ, since ESX S @ª and since for all ε A 0 §N , such that forall n A N ESXn X S @ ε

2 .

74 5 Martingales

5.4 Stopped MartingalesNotation: x x maxx,0, x > R.Theorem 5.4.1 (Doob’s inequality):Let X Xt, t C 0 be a càdlàg process, adapted w.r.t. the filtration F , t C 0. Let X bea submartingale. Then for arbitary t A 0 and arbitrary x A 0 it holds:

P sup0BsBt

Xx A x B EXtx

Proof W.l.o.g. assume Xt C 0, t C 0 a.s..Psup0BsBtXs A x Psup0BsBtXs A x, for all t C 0, x A 0. A supt1,...,tnXs A x,0 B t1 @ t2 @ . . . @ tn B t – arbitrary times. A 8nk1Ak,

A1 Xt1 A xA2 Xt2 B x,Xt2 A x

Ak Xt1 B x,Xt1 B x, . . . ,Xtk1 B x,Xtk A x,k 2, . . . , n, Ai 9Aj g, i x j.It has to be shown that PA B EXtn

x .EXtn C EXtn1A Pnk1 EXtn1Ak C xPnk1 PAk xPA, k 1, . . . , n 1,since X is a martingale and thus follows that EXtn1Ak C EXtk1Ak C Ex1Ak xPAk, k 1, . . . , n 1, tn A tk.Let B ` 0, t be a finite subset, 0 > B, t > B it is proven similarly that Pmaxs>BXs Ax B EXt

x .Q is dense in R 0, t 9Q 8 t 8ªk1Bk, Bk ` 0, t 9Q 8 t finite, Bk ` Bn, k @ n. By themonotonicity of the probability measure it holds:

limnª

Pmaxs>B

Xs C x P8nmaxs>Bn

Xs A x P sups>8nBn

Xs A x B EXtx

By the right-continuity of the paths of X it holds Psup0BsBtXs A x B EXtx .

Corollary 5.4.1For the Wiener process W W t, t C 0 we are looking at the Wiener process with negativedrift: Y t W t µt, µ A 0, t C 0. From example nr.3 of section 5.3 Xt expuY t tµ u2t

2 , t C 0 is a martingale w.r.t. the natural filtration of W . For u 2µ it holds

Xt exp2µY t, t C 0.

Psup0BsBt Y s A x Psup0BsBt e2µY s A e2µx B Ee2µY t

e2µx e2µx, x A 0and consequently

P supsC0

Y s A x limtª

P sup0BsBt

Y s A x B e2µx.

Theorem 5.4.2Let X Xt, t C 0 be a martingale w.r.t. the filtration Ft, t C 0 with càdlàg paths. Ifτ Ω 0,ª is a finite stopping time w.r.t. the filtration Ft, t C 0, then, the stochasticprocess Xτ,tt C 0 is called a stopped martingale. Here a , b mina, b.

5 Martingales 75

Lemma 5.4.1Let X Xt, t C 0 be a martingale with càdlàg-trajectories w.r.t. the filtration Ft, t C 0.Let τ be a finite stopping time and let τnn>N be the sequence of discrete stopping times outof Theorem 5.1.2, for which τn τ , nª, holds. Then Xτn , tn>N is uniformly integrablefor every t C 0.

Proof

τn 0 , if τ 0k12n , if k

2n @ τ B k12n , for a k > N0

1. It is to be shown: ESXτn , tS @ª for all n.ESXτn,tS B Pk k2n @t ESX k2n SESXtS @ª, sinceX is a martingale, therefore integrable.

2. It is to be shown: supn ESXτn , tS1SXτn , tS A x´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶An

ÐÐÐxª

0.

supn

ESXτn , tS1An sup

n

Qk k2n @tEVX k2nV1τn k

2n 9An E SXtS1 τn A t1 An

subm.B sup

n

Qk k2n @tESXtS1τn k

2n 9An E SXtS1 τn A t 9An

supn

ESXtS Qk k2n @t1τn

k

2n 9An E SXtS1 τn A t 9An

supn

E SXtS1 An B supn

E SXtS1 Y A x E SXtS1 Y A x ,

where 1An B 1supn SXτn , tS´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶Y

A x. It remains to prove that PY A x ÐÐÐxª

0. (The

latter obviously implies limxª

E SXtS1 Y A x 0, since ESXtS @ª for all t C 0.)Doob’s inequality yields

PY A x B P sup0BsBt

SXsS A x B ESXtSx

ÐÐÐxª

0.

Proof of Theorem 5.4.2It is to be shown that Xτ , t, t C 0 is a martingale.

1. ESXτ,tS @ª for all t C 0. As in conclusion 5.1.1 τn τ , nªXτn,t a.s.ÐÐÐnª

Xτ,tis approximated, but since ESXτn , tS @ª for all n it follows ESXτ , tS @ª because ofLemma 5.4.1, since uniform integrability gives L1-convergence.

76 5 Martingales

2. Martingale propertyIt is to be shown:

EXτ , t S Fs a.s. Xτ , s, s B t

EXτ , t1A a.s. EXτ , s1A, A > Fs

First of all, we show that ESXτn , tS1A ESXτn , sS1A, A > Fs, n > N. Lett1, . . . , tk > s, t be discrete values, which τn takes with positive probability in s, t.

EXτn , t S Fs EEXτn , t S Ftk S Fs EEXτn , t´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

Xtk1τn B tk S Ftk S Fs

EEXτn , t´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶Xt

1τn A tk S Ftk S Fs EXtk1τn B tk S Fs E1τn A tkEXt S Ftk S Fs EXtk , τn S Fs . . . EXtk1 , τn S Fs . . . EXt1 , τn S Fs . . . EXτn , s S Fsa.s. Xτn , s

Since X is càdlàg and τn τ , n ª, it holds Xτn , t a.s.ÐÐÐnª

Xτn , t. FurthermoreXτn , tn>N is uniformly integrable because of L1-convergence. Therefore follows that

EXτn , t1A EXτn , s1A for all A > Fs

EXτ , t1A EXτ , s1A Xτ , t, t C 0 is a martingale.

Definition 5.4.1Let τ Ω R be a stopping time w.r.t. the filtration Ft, t C 0, Ft ` F , t C 0. The „stopped“σ-algebra Fτ is defined by A > Fτ A 9 τ B t > Ft for all t C 0.

Lemma 5.4.2 1. Let η, τ be stopping times w.r.t. the filtration Ft, t C 0, η a.s.B τ . Thenit holds Fη ` Fτ .

2. Let X Xt, t C 0 be a martingale with càdlàg-trajectories w.r.t. the filtrationFt, t C 0 and let τ be a stopping time w.r.t. Ft, t C 0. Then Xτ is Fτ -measurable.

Proof 1. A > Fs A9 η B t > Ft, t C 0. A9 τ B t A 9 η B t´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶>Ft

9τ B t´¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¶>Ft

> Ft for all t C 0

A > Fτ .

2. Xτ g X f , f Ω Ω R, fω ω, τω, g Ω R R, gω, s Xs,ω.It has to be shown: f -F S F BR-measurable, g-F BR S Fτ -measurable g Xf -F S Fτ -measurable.

5 Martingales 77

f -F S F BR-measurable is obvious, since τ is a random variable. If we’re looking at therestriction of X Xs, s C 0 on s > 0, t, t C 0.It has to be shown: Xτ > B 9 τ B t > Ft for all t C 0, B > BR.X – càdlàg Xs,ω X0, ω1s 0 limnªP2n

k1Xt k2n , ω1k12n t @ s B

k2n t

Xs,ω is B0,t Ft-measurable Xτ is F S Fτ -measurable.

Theorem 5.4.3 (Optional sampling theorem):Let X Xt, t C 0 be a martingale with càdlàg-trajectories w.r.t. a filtration Ft, t C 0,and let τ be a finite stopping time w.r.t. Ft, t C 0. Then EXt S Fτ a.s. Xτ , t, t A 0.

Proof First of all we show that EXt S Fτn a.s. Xτn , t, t C 0, n > N, where τn τ , nª,is the discrete approximation of τ , cf. Theorem 5.1.2. Let t1 B t2 B . . . B tk t be the values,attained by τn , t with positive probability. It is to be shown that for all A > Fτn it holds:EXt1A EXτn , t1A.

Xt Xτn , t1A tkt

k1Qi1Xtk Xti1τn , t ti 9A

k

Qi2Xtk Xti11τn , t ti1 9A

k

Qi2

k

Qji

Xtj Xtj11τn , t ti1 9A

k

Qj2

j

Qi2Xtj Xtj11τn , t ti1 9A

k

Qj2Xtj Xtj11τn , t B tj1 9A

EXt Xτn , t1A

k

Qj2

EXtj Xtj11τn , t B tj1 9A

k

Qj2

EEXtj Xtj1´¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¶>Ftj1

1τn , t B tj1 9A´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶>Ftj1

S Ftj1

k

Qj2

E1τn , t B tj1 9AEXtj S Ftj1´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶Xtj1

Xtj1 0

by Definition 5.4.1 and martingale property. Hence it holds EXt S Fτn a.s. Xτn , t, sinceXτn is Fτn-measurable. Since τ B τn, ¦n, by Lemma 5.4.2, Fτ b Fτn . Then

EXt S Fτ a.s. EXt S Fτn Xτn , t a.s. Xτ , t, nª,since X is càdlàg.

78 5 Martingales

Corollary 5.4.2Let X Xt, t C 0 be a càdlàg-martingale and let η, τ be finite stopping times, suchthat Pη B τ 1. Then it holds EXτ , t S Fη a.s.

EXη , t, t C 0. In particularEXτ , t EX0 holds.

Proof Since X is a martingale, by Theorem 5.4.2 Xτ , t, t A 0 is also a martingale. ApplyTheorem 5.4.3 to it:

EXτ , t S Fη a.s. Xτ , η , t a.s. Xη , t,since η

a.s.B τ . Set η 0, then EEXτ , t S F0 EX0 , t EX0.

5.5 Lévy processes and Martingales

Theorem 5.5.1Let X Xt, t C 0 be a Lévy process with characteristics a, b, ν.

1. There exists a càdlàg-modification of X Xt, t C 0 ofX with the same characteristicsa, b, ν.2. The natural filtration of càdlàg-Lévy processes is right-continuous.

Without proof

Theorem 5.5.2 (Regeneration theorem for Lévy processes):Let X Xt, t A 0 be a càdlàg-Lévy process with natural filtration FX

t , t C 0 and letτ be a finite stopping time w.r.t. FX

t , t C 0. The process Y Y t, t C 0, given byY t Xτ tXτ, t C 0, is also a Lévy process, adapted w.r.t. the filtration FX

τt, t C 0,which is independent of FX

τ and has the same characteristics as X. τ is called regenerationtime.

Abb. 5.1:

5 Martingales 79

Proof For any n > N, take arbitrary 0 B t0 @ @ tn and u1, . . . , un > R. We state that allassertions of the theorem follow from the relation

E1A exp

¢¦¤n

Qj1

iujY tj Y tj1£§¥ PAEexp

¢¦¤n

Qj1

iujXtj Xtj1£§¥ (5.5.1)

for all A > FXτ .

1. By Lemma 5.4.2, since τ and τ t, t C 0 are stopping times with τ B τ t a.s., we haveFXτs b FX

τt, i.e. FXτttC0 is a filtration, and Y t Xτ t Xτ is FX

τt-adapted:Xτ is FX

τ -measurable, Xτ t is FXτt-measurable, FX

τ b FXτt.

2. It follows from (5.5.1) for A Ω that X d Y , i.e., Y is a Lévy process with the same Lévy

exponent η as X.

3. It also follows from (5.5.1) that Y and FXτ are independent, since arbitrary increments

of Y do not depend on FXτ .

4. Now let us prove (5.5.1). We begin with the case ofa.) § c A 0: Pτ B c 1. By example 5, b) of section 5.2., Yj YjttC0, j 1, . . . , n with

Yjt expiujXt tηuj, t C 0 are complex-valued martingales. Furthermore,it holds

E1A exp n

Qj1

iujY tj Y tj1 E1A exp n

Qj1

iujXτ tj Xτ Xτ tj1 Xτ E

1A n

Mj1

Yjτ tjYjτ tj1 expηujτ tj

expηujτ tj1 E

E1A n

Mj1

Yjτ tjYjτ tj1 exptj tj1ηuj S FX

τtj1

E

1A n1Mj1

Yjτ tjYjτ tj1etjtj1ηuj etntn1ηun

Ynτ tn1 EYnτ tn S FXτtn1

E1A <@@@@>

n1Mj1

Yjτ tjYjτ tj1etjtj1ηuj=AAAA? etntn1ηun

. . . E1A n

Mj1

etjtj1ηuj PA n

Mj1

etjtj1ηuj

PAEexpi n

Qj1ujXtj Xtj1

b.) Prove (5.5.1) for arbitrary finite stopping times τ : Pτ @ ª 1. For any k > N itholds Ak A9 τ B k > FX

τ,k, if A > FXτ . Then it follows from 4. a.) and (5.5.1) that

E1Ak exp

¢¦¤n

Qj1

iujYk,tj Yk,tj1£§¥ PAkEexp

¢¦¤n

Qj1

iujXtj Xtj1£§¥

(5.5.2)

80 5 Martingales

for Yk,t Xτ , k t Xτ , k. Now let n ª on both sides of (5.5.2). ByLebesgue’s theorem on majorized convergence, (5.5.1) follows for any a.s. finite τ ,since τ , k a.s.

τ , nª.

5.6 Martingales and Wiener ProcessOur goal: We would like to show that if W W t, t C 0 is a Wiener process, then it holds

Pmaxs>0,tW s A x ¾ 2

πtS

ª

xe

y22t dy, x C 0.

Theorem 5.6.1 (Reflection principle):Let τ be an arbitrary finite stopping time w.r.t. the natural filtration FW

t , t C 0. Let X Xt, t C 0 be the reflected Wiener process at time τ , i.e. Xt W τ ,tW tW τ ,t,t C 0. Then X d

W holds.

Abb. 5.2: Reflection principle.

Proof It holds

Xt W τ , t W t W τ , t ¢¦¤W t , t @ τ

2W τ W t , t C τ

Let X1t W τ , t, X2t W τ t W τ, t C 0. From Theorem 5.5.2 it follows thatX2 is independent of τ,X1 (W – Lévy process and τ – regeneration time). It holds W t X1t X2t τ, Xt X1t X2t τ, t C 0. Indeed,

X1t X2t τ ¢¦¤W t X20 W t , t @ τ

W τ W τ t τ W τ W t , t C τ

5 Martingales 81

W t X1t X2t τ, t C 0. Furthermore

X1t X2t τ ¢¦¤W t X20 W t , t @ τ

W τ W t W τ 2W τ W t , t C τ

Xt X1t X2t τ, t C 0. From theorems 5.5.2 and 3.3.2 it follows that X2d W ,

WdW and hence τ1,X1,X2 d

τ,X1,X2+ +

Wd X

Let W W t, t C 0 be the Wiener process on Ω,F ,P, let FWt , t C 0 be the natural

filtration w.r.t. W . For z > R let τWz inft C 0 W t z. τWz τWz is an a.s. finitestopping time w.r.t. FW

t , t C 0, z A 0, since it obviously holds FWz B t > FW

t . Since W hascontinuous paths (a.s.), FW

t , t C 0 is right-continuous.Corollary 5.6.1Let Mt maxs>0,tW s, t C 0. Then it follows for all z A 0, y C 0, that PMt C z,W t Bz y PW t A y z.Proof Mt is a random variable, sinceW has continuous paths. Let τ τWz . By Theorem 5.6.1,it holds for Y t W τ , tW tW τ , t, t C 0, that Y d

W . Hence, τWz ,W d τYz , Y ,

since W τ z, τWz τYz . Therefore

Pτ B t,W t @ z y PτYz B t, Y t @ z y,whereas τYz B t 9 Y t @ z y τYz B t 9 2z W t @ z y. If τ τYz B t thenY t 2W τ W t 2z W t, and hence

Pτ B t,W t @ z y Pτ B t,2z W t @ z y Pτ B t,W t A z y PW t A z y.Per definition of τ τWz it holds:

Pτ B t,W t @ z y PMt C z,W t @ z y PW t A y z,since τWz B t maxs>0,tW s C z.Theorem 5.6.2 (Distribution of the maximum of W ):For t A 0 and x C 0 it holds

PMt A x ¾ 2πtS

ª

xe

y22t dy

Proof In Corollary 5.6.1, set y 0 PMt C z,W t @ z PW t A z. It holds PW t Az PW t C z for all t and all z, since W t N 0, t, thus continuously distributed PMt C z,W t @ z PW t C z PW t A z PW t A z PMt C z,W t @ z PMt C z,W t C z PMt C z 2PW t A z PMt A z 2PW t A z 2 1º

2πt Rª

z ey22z dy

¼2πt R ªz e

y22t dy

82 5 Martingales

Let Xt W t tµ, t C 0, µ A 0, be the Wiener process with negative drift. ConsiderPsuptC0Xt A x, x C 0.

Motivation Calculation of the following ruin probability in risk theory.

Assumptions Initial capital x C 0. Let µ be the volume of premiums per time unit. µt– earned premiums at time t C 0. Let W t be the loss process (price development). Y t x tµ W t – remaining capital at time t. The ruin probability is

PinftC0

Y t @ 0 Px suptC0

Xt @ 0 PsuptC0

Xt A xTheorem 5.6.3It holds

PsuptC0

Xt A x e2µx, x C 0, µ A 0.

Proof Let τ τXx inft C 0 Xt x. It holdsPsup

tC0Xt A x Pτ @ª lim

tª

Pτ @ t.Compute this limit. For that, introduce the process Y Y t, t C 0,

Y t expuXt tu2

2 µu¡ , t C 0,

u C 0 fixed. It can be easily shown that Y is a martingale. τ τ , t is a finite stopping timew.r.t. filtration FX

t , t C 0. By Corollary 5.4.1, EY τ EY 0 e0 1. On other hand,

1 EY τ EY τ 1τ @ t EY τ 1τ C t EY τ1τ @ t EY τ 1τ C t.

If we can show thatlimtª

EY τ 1τ C t 0, (5.6.1)

then limtª EY τ1τ @ t 1. Since Y τ expux τ u2

2 µu, it followslimtª

E expτ u2

2 uµ¡1τ @ t eux,

and for u 2µ it holds

limtª

E e01τ @ t limtª

Pτ @ t e2µx

PsuptC0

Xt A x e2µx. Now let us prove (5.6.1). By Corollary 3.3.2, it is known that

W tt

a.s. 0 t ª.

5 Martingales 83

Hence,

limtª

Xtt

limtª

W tt

µ µXt a.s. ª, t ª. Then,

Y τ 1τ C t exp2µXt1τ C t a.s. 0 t ªBy Lebesgue’s theorem, it holds E Y τ 1τ C t 0, t ª.Theorem 5.6.4Let X Xt, t C 0 be a Lévy process and let τ τt, t C 0 be a subordinator, whichare both defined on a probability space Ω,F ,P. Let X and τ be independent. Then Y Y t, t C 0 definded by Y t Xτt, t C 0, is a Lévy process.Without proof

5.7 Additional ExercisesExercise 5.7.1Let X,Y Ω R be arbitrary random variables on Ω,F ,P with

ESX S @ª, ESY S @ª, ESXY S @ª,and let G ` F be an arbitrary sub-σ-Algebra of F . Then it holds a.s. that

(a) EX Sg,Ω EX,EX SF X,

(b) EaX bY SG aEX SG bEY SG for arbitrary a, b > R,

(c) EX SG B EY SG, if X B Y ,

(d) EXY SG Y EX SG, if Y is a G,BR-measurable random variable,

(e) EEX SG2SG1 EX SG1, if G1 and G2 are sub-σ-algebras of F with G1 ` G2,

(f) EX SG EX, if the σ-algebra G and σX X1BR are independent, i.e.,if PA 9A PAPA for arbitrary A > G and A > σX.

(g) EfXSG C fEX SG, if f R R is a convex function such that ESfXS @ª.

Exercise 5.7.2Look at the two random variables X and Y on the probability space 1,1,B1,1, 1

2νwith ESX S @ª, where ν is the Lebesgue measure on 1,1. Determine σY and a version ofthe conditional expectation EX SY for the following random variables.

(a) Y ω ω5 (Hint: Show first that σY B1,1)(b) Y ω 1k for ω > k3

2 , k22 , k 1, . . . ,4 and Y 1 1

(Hint: It holds EX SB EX1BPB for B > σY with PB A 0)

(c) Calculate the distribution of EX SY in (a) and (b), if X U1,1.

84 5 Martingales

Exercise 5.7.3Let X and Y be random variables on a probability space Ω,F ,P. The conditional varianceVarY SX is defined by

VarY SX EY EY SX2SX.Show that

VarY EVarY SX VarEY SX.Exercise 5.7.4Let S and τ be stopping times w.r.t. the filtration Ft, t C 0. Show:(a) A 9 S B τ > Fτ , A > FS

(b) FminS,τ FS 9Fτ

Exercise 5.7.5 (a) Let Xt, t C 0 be a martingale. Show that EXt EX0 holds forall t C 0.

(b) Let Xt, t C 0 be a sub- resp. supermartingale. Show that EXt C EX0 (resp.,EXt B EX0) holds for all t C 0.

Exercise 5.7.6The stochastic process X Xt, t C 0 is adapted and càdlàg. Show that

P sup0BvBt

Xv A x B EXt2

x2 EXt2

holds for arbitrary x A 0 and t C 0, if X is a submartingale with EXt 0 and EXt2 @ª.Exercise 5.7.7 (a) Let g 0,ª 0,ª be a monotone increasing function with

gxx

ª, xª.

Show that the sequence X1,X2, . . . of random variables is uniformly integrable, ifsupn>N EgSXnS @ª.

(b) Let X Xn, n > N be a martingale. Show that the sequence of random variablesXτ , 1,Xτ , 2, . . . is uniformly integrable for every finite stopping time τ , ifESXτS @ª and ESXnS1τAn 0 for nª.

Exercise 5.7.8Let S Sn a Pni1Xi, n > N be a symmetric random walk with a A 0 and PXi 1 PXi 1 1~2 for i > N. The random walk is stopped at the time τ , when it exceeds or fallsbelow one of the two values 0 and K A a for the first time, i.e.

τ minkC0

Sk B 0 or Sk CK.Show that Mn Pni0 Si

13S

3n is a martingale and EPτi0 Si 1

3K2 a2a a holds.

Hint: To calculate EMnSFMm , n A m, you can use EPlikXi3 0, 1 B k B l, Mn

Pmr0 Sr Pnrm1 Sr 13S

3n and Sn Sn Sm Sm.

5 Martingales 85

A discrete martingale w.r.t. a filtration Fnn>N is a sequence of random variables Xnn>N ona probability space Ω,F ,P, such that Xn is measurable w.r.t. Fnn>N and EXn1SXn Xn

a.s. for all n > N. A discrete stopping time w.r.t. Fnn>N is a random variableτ Ω N 8 ª such that τ B n > Fn for all n > N 8 ª, where Fª σª

n1 Fn.Exercise 5.7.9Let Xnn>N be a discrete martingale and τ a discrete stopping time w.r.t. Fnn>N. ShowthatXτ,nn>N is also a martingale w.r.t. Fnn>N.Exercise 5.7.10Let Snn>N be a symmetric random walk with Sn Pni1Xi for a sequence of independent andidentically distributed random variables X1,X2, . . ., such that PX1 1 PX1 1 1

2 . Letτ infn SSnS Aºn and Fn σX1, . . . ,Xn, n > N.

(a) Show that τ is a stopping time w.r.t. Fnn>N.(b) Show that Gnn>N with Gn S2

τ,n τ , n is a martingale w.r.t. Fnn>N. (Hint: UseExercise 5.7.9)

(c) Show that SGnS B 4τ holds for all n > N.(Hint: It holds SGnS B SS2

τ,nS Sτ , nS B S2τ,n τ)

Exercise 5.7.11Let X1,X2, . . . be a sequence of independent and identically distributed random variables withESX1S @ ª. Let Fn σX1, . . . ,Xn, n > N, and let τ be a stopping time w.r.t. Fnn>N withEτ @ª.

(a) Let τ be independent of X1,X2, . . .. Derive a formula for the characteristic function ofSτ Pτi1Xi and verify Wald’s identity ESτ EτEX1.

(b) Let additionally EX1 0 and τ infn Sn @ 0. Use Theorem 2.1.3, to show thatEτ ª. (Hint: Proof by contradiction)

Bibliography

[1] D. Applebaum. Lévy processes and stochastic calculus. Cambridge University Press, Cam-bridge, 2004.

[2] A. A. Borovkov. Wahrscheinlichkeitstheorie. Eine Einfürung. Birkhäuser, Basel, 1976.

[3] W. Feller. An introduction to probability theory and its applications. Vol. I/II. John Wiley& Sons Inc., New York, 1970/71.

[4] P. Gänssler and W. Stute. Wahrscheinlichkeitstheorie. Springer, Berlin, 1977.

[5] G. Grimmett and D. Stirzaker. Probability and random processes. Oxford University Press,New York, 2001.

[6] C. Hesse. Angewandte Wahrscheinlichkeitstheorie. Vieweg, Braunschweig, 2003.

[7] O. Kallenberg. Foundations of modern probability. Springer, New York, 2002.

[8] J. F. C. Kingman. Poisson processes. Oxford University Press, New York, 1993.

[9] N. Krylov. Introduction to the theory of random processes. American Mathematical Society,Providence, RI, 2002.

[10] S. Resnick. Adventures in stochastic processes. Birkhäuser, Boston, MA, 1992.

[11] G. Samorodnitsky and M. Taqqu. Stable non-Gaussian random processes. Chapman &Hall, New York, 1994.

[12] K.-I. Sato. Lévy Processes and Infinite Divisibility. Cambridge University Press, Cam-bridge, 1999.

[13] A. N. Shiryaev. Probability. Springer, New York, 1996.

[14] A. V. Skorokhod. Basic principles and applications of probability theory. Springer, Berlin,2005.

[15] J. Stoyanov. Counterexamples in probability. John Wiley & Sons Ltd., Chichester, 1987.

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