Transcription of CONDITIONAL EXPECTATION AND MARTINGALES
1 CONDITIONAL EXPECTATION AND MARTINGALES1. IN T RO D U CT I O NMartingalesplay a role in stochastic processes roughly similar to that played byconservedquantitiesin dynamical systems. Unlike a conserved quantity in dynamics, which remainsconstant in time, a martingale s value can change; however, itsexpectationremains constantin time. More important, the EXPECTATION of a martingale is unaffected byoptional fact, this can be used as a provisional definition: A discrete-timemartingaleis a sequence{Xn}n 0of integrable real (or complex) random variables with the property that for every boundedstopping time , theOptional Sampling Formula(1)E X =E X0is have seen the Optional Sampling Formula before, in various guises.
2 In particular, theWald Identities I,II, and III are all instances of (1). Let 0, 1, .. be independent, identicallydistributed random variables, and letSn= 1+ 2+ nbe thenth partial sum. Denote by , 2, and ( ) the mean, variance, and moment generating function of 1, that is, =E 1, 2=E( 1 )2,and ( )=Eexp{ 1}.Corresponding to each of these scalar quantities is amartingale:Mn:=Sn n ,(2)Vn:=(Sn n )2 n 2,and(3)Zn( ) :=exp{ Sn}/ ( )n.(4)Observe that there is a separate martingaleZn( ) for every real value of such that ( )< .The Optional Sampling Formula could be taken as the definition of a martingale, but usuallyisn t. The standard approach, which we will follow, uses the notion ofconditional CO N D I T I O N ALEX PE CTATI O of CONDITIONAL random variables defined on discrete proba-bility spaces, CONDITIONAL EXPECTATION can be defined in an elementary manner: In particular,the CONDITIONAL EXPECTATION of a discrete random variableXgiven the valueyof another dis-crete random variableYmay be defined by(5)E(X|Y=y)= xx P(X=x|Y=y),where the sum is over the set of all possible valuesxofX.
3 Note that this expression depends onthe valuey. For discrete random variables that take values in finite sets there are no difficulties1regarding possible divergence of the sum, nor is there any difficulty regarding the meaning ofthe CONDITIONAL probabilityP(X=x|Y=y).For continuous random variables, or, worse, random variables that are neither discrete norhave probability densities, the definition (5) is problematic. There are two main difficulties: (a)IfXis not discrete, then the sum must be replaced by an integral of some sort; and (b) IfYisnot discrete then it is no longer clear how to define the condtional probabilitiesP(X=x|Y=y).Fortunately, there is an alternative way of defining CONDITIONAL EXPECTATION that works in boththe discrete and the indiscrete cases, and additionally allows for conditioning not only on thevalue of a single random variable, but for conditioning simultaneously on the values of finitelyor even countably many random variables or random vectors:Definition a real-valued random variable such that eitherE|X|< orX 0, and letYbe a random variable, random vector, or other random object taking values in a measurablespace (Y,H).
4 Theconditional EXPECTATION E(X|Y) is the essentially unique measurable real-valued function ofYsuch that for every bounded, measurable, real-valued functiong(Y),(6)E X g(Y)=E(E(X|Y)g(Y)).Definition generally1, ifXis defined on a probability space ( ,F,P) andG Fis a algebra contained inF, thenE(X|G) is the essentially uniqueG-measurable random vari-able such that(7)E(X Z)=E(E(X|G)Z)for every bounded,G measurable random is by no means cleara priorithat such functionsE(X|Y) orE(X|G) should always exist, noris it obvious that they should be unique. In fact, theexistenceof such a function is an importanttheorem of measure theory, theRadon-Nikodymtheorem, which I will take as known. Theuniquenessof the function is not difficult to prove:Proof of that there are distinct functionsh1(y) andh2(y) such that, forevery bounded functiong(y),E X g(Y)=E h1(Y)g(Y)andE X g(Y)=E h2(Y)g(Y).
5 Then by the linearity of ordinary EXPECTATION (taking the difference of the two equations) it mustbe the case that for every bounded functiong(y),0=E(h1(Y) h2(Y))g(Y);in particular, this equation must hold for the functiong(y) that is 1 ifh1(y)>h2(y) and 0 oth-erwise. But this implies thatP{h1(Y)>h2(Y)}=0. A similar argument shows thatP{h2(Y)=h1(Y)}=0. It follows thatP{h1(Y)=h2(Y)}=1. 1 Nearly all of the algebras that one encounters in typical situations are generated, either explicitly or im-plicitly, by random variablesYvalued in a Polish space (a complete, separable metric space). In particular, the algebras that arise naturally in the study of stochastic processes are usually generated bysample paths.
6 IfYgeneratesG, that is, ifGconsists of all events of the form {Y B} whereBis a Borel subset of the range ofY, theneveryG measurable random variable is a measurable function ofY, and conversely (Exercise!). Hence, Defini-tions 1 and 2 of the Naive and Modern is not difficult to show that the naivedefinition (6) coincides with the modern Definition 1 when the random variableXand therandom vectorY=(Y1,Y2, .. ,Ym) are discrete and assume only finitely many possible valueswith positive probability. Defineh(y)= xx P(X=x|Y=y)= xxP{X=xandY=y}P{Y=y}for those values ofysuch thatP{Y=y}>0. To show thatE(X|Y)=h(Y), it suffices, by Defini-tion 1, to show that, for any bounded functiong(y),E X g(Y)=E h(Y)g(Y).
7 ButE X g(Y)= y xx g(y)P{X=xandY=y}= yg(y)P{Y=y} xx P(X=x|Y=y)= yg(y)P{Y=y}h(y)=E h(Y)g(Y). of CONDITIONAL raw definition given above can be clumsy towork with directly. In this section we present a short list of important rules for manipulating andcalculating CONDITIONAL expectations . The bottom line will be that, in many important respects, CONDITIONAL expectations behave like ordinary expectations , with random quantities that arefunctions of the conditioning random variable being treated as a random variable, vector, or object valued in a measurable space, and letXbe anintegrablerandom variable (that is, a random variable withE|X|< ).Summary: Basic Properties of CONDITIONAL EXPECTATION .
8 (1)Definition:E X g(Y)=E E(X|Y)g(Y) for all bounded measure functionsg(y).(2)Linearity:E(aU+bV|Y)=aE (U|Y)+bE(V|Y) for all scalarsa,b R.(3)Positivity:IfX 0 thenE(X|Y) 0.(4)Stability:IfXis a (measurable) function ofY, thenE(X Z|Y)=X E(Z|Y).(5)Independence Law:IfXis independent ofYthenE(X|Y)=E Xis constant (6)Tower Property:IfZis a function ofYthenE(E(X|Y)|Z)=E(X|Z).(7) EXPECTATION Law:E(E(X|Y))=E X.(8)Constants:For any scalara,E(a|Y)= we ll prove a theorem to the effect that CONDITIONAL expectationsareordinary expectations in a certainsense. For those of you who already know the terminology, this theorem is the assertion that every real randomvariableY or more generally, every random variable valued in a Polish space has aregular CONDITIONAL distri-butiongiven any (9)Jensen Inequalities:If :R Ris convex andE|X|< thenE( (X)) (E X) andE( (X)|Y) (E(X|Y).)
9 In all of these statements, the relations=and are meant to holdalmost surely. Similar state-ments can be formulated for CONDITIONAL expectationE(X|G) on a algebra. Also, properties(3) (7) extend to nonnegative random variablesXwith infinite EXPECTATION . All of the propertiescan be proved easily, using only Definition 1 and elementary properties give an idea of how these arguments go, we shall outline the proofs of the Linearity, Positivity,and Independence properties below. You should try to check the Stability and Tower Propertiesyourself. The Jensen inequality is of a somewhat different character, but it is not difficult toprove see :The definition (1) requires only that the equationE X g(Y)=E E(X|Y)g(Y) be valid forboundedfunctionsg.
10 A standard argument from measure theory then implies that it holds forall functions such that the productX g(Y) has finite first moment. Similarly, Property (4) holdsprovided the product has finite first of the Positivity idea is to exploit the defining property (6) of conditionalexpectation. First, suppose thatX 0. DefineBto be the set of possible values ofYfor whichthe CONDITIONAL expectationE(X|Y)<0, so that the event {E(X|Y)<0} coincides with the event{Y B}. Then by equation (6),E X1B(Y)=E(E(X|Y)1B(y)).SinceX 0, the left side of this equality is nonnegative; but by definition ofB, the right side isnegative unlessP{Y B}=0. It follows thatP{Y B}=0, that is,E(X|Y) 0 with probabilityone. Proof of the Linearity each of the CONDITIONAL expectationsE(U|Y) andE(V|Y)is a function ofY, so is the linear combinationaE(U|Y)+bE(V|Y).
