MATH 56A SPRING 2008 STOCHASTIC PROCESSES 1919.2. Integration wrt Brownian motion. We want to define the sto-chastic integralZt=!t0YsdWswhere Wsis standard Brownian motion and Ysis a predictable squaresummable process. Tomorrow, I will prove Itˆo’s formula so we cancalculate it. Today, I just want to define it. This is not that easy. I’lldo it in three steps.(1) First, we take Ysa step function.(2) Next, we have the theory: Ztis a martingale and Z2t− "Z#tisalso a martingale.(3) Finally, take the limit as δt → 0. I will use a theorem from realanalysis: The space of L2functions is complete. I will explainwhen we get there.9.2.1. simple processes. Here is the picture I drew in class.There are only finitely many steps. The last step is infinite. A simpleprocess is a step function where each step has a random height. Thenumbering is supposed to indicate that each step is determined at thebeginning of the time interval.Yt=0 if t = 0Y0if t ∈ (0, t1]Y1if t ∈ (t1, t2]· · ·Ynif t ∈ (tn, ∞)192 STOCHASTIC INTEGRATIONWe assume that Ykis square summable (L2). This means E(Y2k) < ∞.(This implies that the L2norm of any finite part of the function, suchas the shaded part, is finite.) We also assume that Ykis Ftk. Thismakes Ytinto a predictable L2process.Definition 9.6. The integral of Ytis given by!t0Ysds :=k&i=1Yi−1(ti− ti−1) + Yk(t − tk)'=&Y δt(if t ∈ (tk, tk+1]. This integral is the shaded area in the figure.The integral is the sum of areas of rectangles. Each rectangle hasa random height. Note that at t one of these rectangles is cut in halfvertically by a colored line. Each piece is again a rectangle with randomheight.Definition 9.7. The stochastic integral of Ytis given by!t0YsdWs:=k&i=1Yi−1(Wti− Wti−1) + Yk(Wt− Wtk)'=&Y δW(if t ∈ (tk, tk+1].The first thing I pointed out is thatE(Zt| F0) = 0This is because each term is a product Y δW where, at some time s ≥ 0,Y is Fs-measurable and E(δW | Fs) = 0. So, using the law of iteratedexpectation,E(Y δW ) = E(E(Y δW | Fs)) = E(0) = 0.Since this is the third time I used this argument, I decided to make alemma, so that, next time, I can just say “by Lemma A.”Lemma 9.8 (Lemma A). If there will be some time in the future (∃s ≥t), at which you will know the value of A (A is Fs-measurable) andB will still be random with zero expectation (E(B | Fs) = 0) then thecurrent expected value of the product is zero:E(AB | Ft) = 0.Theorem 9.9. Ztis a martingale.In other words, you cannot make a profit on a stock with µ = 0.MATH 56A SPRING 2008 STOCHASTIC PROCESSES 193Proof. All we did was to separate past from future and notice that thefuture stuff has expectation zero by Lemma A.If s > t thenZs=m&i=1Yi−1)Wti− Wti−1*+ Ym(Ws− Wtm)=k&i=1Yi−1δW + Yk)Wtk+1− Wtk*+ later terms+,- .Yk(Wt− Wtk) + Yk)Wtk+1− Wt*ZtZs− ZtThis is in the past (before t) This is in the future (after t)Zs− Ztis a sum of Y δW ’s with E(Y δW | Ft) = 0 by Lemma A. So,E(Zs| Ft) = Zt+ E(Zs− Zt| Ft)- .+ ,=0= Zt.So, Ztis a martingale. !9.2.2. square summability. The next question is:E(Z2t) =?If we use the shorthand: Zt=/Y δW thenZ2t=&Y2δW2+ cross termsLemma 9.10 (Lemma B). The expected value of the cross terms iszero.Proof. This follows from Lemma A. In each cross terms, one part is inthe past and one part is in the future. The future part is δW whichhas expected value 0:E(YiδiW YjδjW ) = 0by Lemma A since YiδWiYjis Ftj-measurable and E(δjW | Ftj) = 0. !Since the cross terms don’t count, you just get the square terms:E(Z2t) =&E(Y2δW2)This is equal to:=&E(Y2)E(δW2)194 STOCHASTIC INTEGRATIONsince Y, δW are independent. In fact, we know thatδW = Wti− Wti−1is independent of Fti−1which includes everything that happened up totime ti−1. To find the expected value recall thatδW ∼ N(0, δt).So, E(δW2) = δt. This means thatE(Z2t) =&E(Y2)δt =!t0E(Y2s)dsby definition of the integral of a step function.Definition 9.11. The quadratic variation of Ztis defined by"Z#t:= limδt→0&(δZ)2.In this case this is:"Z#t= limδt→0&(Y δW )2This is Z2twithout its cross terms. So, its expected value is:E("Z#t) = limδt→0E'&(Y δW )2(= E(Z2t) =!t0E(Y2s)ds.Theorem 9.12. Z2t− "Z#tis a martingale.Proof. If s > t then Zsis equal to Ztplus a sum of increments of Z:Zs= Zt+&δZSo,Z2s= (Zt+ ΣδZ)2= Z2t+&(δZ)2+ cross termsAnd, by definition,"Z#s= "Z#t+&(δZ)2If we subtract, the/(δZ)2terms cancel and we haveZ2s− "Z#s= Z2t− "Z#t+ cross termsBut the cross terms have expectation zero:E(cross terms | Ft) = 0So,E(Z22− "Z#s| Ft) = Z2t− "Z#tmaking Z2t− "Z#tinto a martingale. !MATH 56A SPRING 2008 STOCHASTIC PROCESSES 1959.2.3. general stochastic integral. Suppose now that Ytis any L2pre-dictable process. Then, we make it into a simple process by dividingthe line [0, t] into n pieces at points tk= kt/n and lettingY(n)k= average of Yton (tk−1, tk]=1tk− tk−1!tktk−1YtdtThen you define:Z(n)t=!t0Y(n)sdWs.Since Y(n)sis a predictable L2process, Z(n)tis a square summable mar-tingale.Definition 9.13. The general stochastic integral Zt=0t0YtdWsisdefined to be the limit:Zt= limn→∞Z(n)n.This limit exists and is L2by the following theorem.Theorem 9.14. The space of L2functions is complete in the L2norm.The word complete means that any Cauchy sequence converges inthe L2norm. So, the theorem is saying that, if we know that Z(n)tis aCauchy sequence, we know it will converge.Lemma 9.15. {Z(n)t} is a Cauchy sequence. In other words, ∀" > 0∃N so thatE((Z(n)t− Z(m)t)2) < "if n, m > N.Remark 9.16. The usual definition of the L2norm is the integral of thesquare of a function with respect to the given measure:||g||2:=!g2dµBut, we have a probability space and our measure is the probabilitymeasure dµ = f(x)dx. So,||g||2= E(g2).I ran of out time. But there were a couple more things:Theorem 9.17. Ztis a martingale.196 STOCHASTIC INTEGRATIONProof. This follows from the fact that E(− | Fs) is continuous in theL2-norm. In other words, it commutes with limits. SoE(Zt| Fs) = E'limn→∞Z(n)t| Fs(= lim E(Z(n)t| Fs)= lim Z(n)s= Zsmaking Zta martingale. !Since the cross terms for the product of a sum of increments of amartingale have expectation zero (by Lemma A), we have:Corollary 9.18. Z2t− "Z#tis a