202 STOCHASTIC INTEGRATION9.4. Extensions of Itˆo’s formula. First I explained the idea of “co-variation” and the product rule. Then I used this to derive the secondand third version of Itˆo’s formula.9.4.1. covariation.Definition 9.25. The covariation process of Atand Btis defined by!A, B"t:= limδt→0!δAδB = limδt→0!(Ati− Ati−1)(Bti− Bti−1)I used the word “pro cess” to emphasize the fact that this is randomand it varies with time.Mathematically, this is the symmetric bilinear form which is asso-ciated to the quadratic variation. Two prop erties follow immediatelyfrom the definition:Properties!A, A"t= !A"t(quadratic variation)d !A, B"t= dAtdBt(by definition)And we have another formula which can also be used to define !A, B"t(since the other three terms have already been defined):!A + B"t= !A"t+ !B"t+ 2 !A, B"t.This is “obvious” from the binomial formula:!(δiA + δiB)2=!(δiA)2+!(δiB)2+ 2!δiAδiB.9.4.2. product rule. One of the really nice uses of the new definition isthe following product formula which holds without error! (See picture.)δ(AB) = AδB + BδA + δAδBFigure 1. The term δAδB becomes the covariation.MATH 56A SPRING 2008 STOCHASTIC PROCESSES 203The infinitesmal version isd(AB) = AdB + BdA + d !A, B"t.9.4.3. bounded variation. Last time (Exercise 9.23) I showed that:Lemma 9.26 (Lemma 1). If f is continuous with bounded variation(e.g. if f is C1) then!f"t= 0(The quadratic variation of f is zero.)Using the same argument as in the proof of the Schwarz inequality,you get the following.Lemma 9.27 (Lemma 2). If !f"t= 0 and !X"t< ∞ then !f, X"t= 0a.s. for all t.Proof. Suppose that, at some time t, there is a chance (nonzero prob-ability) that !f, X"t%= 0. Say, it could be positive. Then there is apossibility that !f, X"t> c > 0. Then!X − af"t= !X"t− 2a !f, X"t" #$ %>c+ a2!f"t" #$ %0If we make a really big then we get !X − af"t< 0 with nonzero prob-ability. But this is impossible because quadratic variations are sums ofsquares! This contradiction proves the lemma. !204 STOCHASTIC INTEGRATION9.4.4. review. During the review I used the notationd !A, B"t= dAtdBt.Then the properties of covariation become:(1) (L´evy) d !W "t= (dWt)2= dt.(2) d !f, X"t= dfdXt= 0 if f has bdd variation.(3) (product rule) d(AB) = AdB + BdA + dA dB.Before doing Itˆo’s second and third formulas, I went over Itˆo’s firstformula which was:df(Wt) = f"(Wt)dWt+12f""(Wt)dt.Exercise 9.28. Take f(x) = eσx. Then f"(x) = σeσx, f""(x) = σ2eσx.So, the formula gives:deσx= σeσxdWt+12σ2eσxdtdeσx= eσx&σdWt+σ22dt'.This is very similar to the model for stock prices that we will beusing:St:= value of one share of stock at time t.Our model assumes thatdSt= St(σdWt+ µdt)whereµ = driftσ = volatility(assumed constantWe are notassuming that µ = σ2/2. However, it would be convenientif this were true.9.4.5. quadratic variation of Zt. Here I calculated the quadratic vari-ation of Zt.Theorem 9.29. Suppose that(1) dZt= XtdWt+ Ytdt where(2) Ytis integrable (L1), i.e.,)t0|Ys|ds < ∞ and(3) Xtis square summable (L2), i.e.,)t0X2sds < ∞.Thend !Z"t= X2tdt.MATH 56A SPRING 2008 STOCHASTIC PROCESSES 205First, I rephrased the statement: Condition (1) says thatZt− Z0=*t0XsdWs" #$ %=Btstochastic integral+*t0Ysds" #$ %f(t)The conclusion is:!Z"t= Bt+ f(t).Proof. f(t) has bounded variation sincevariation of f =*t0|f"(s)|ds =*t0|Ys|ds < ∞since Ytis L1. But thenZt= Bt+ f(t) + Z0" #$ %g(t)= Bt+ g(t)where g(t) has bounded variation because it has the same derivative asf(t). So,!Z"t= !B"t+ !g"t"#$%=0 by Lem 1+2 !Bt, g"t" #$ %=0 by Lem 2= !B"t= dBtdBt= X2t(dWt)2= X2tdt.!9.4.6. Itˆo’s second formula. We want to replace WtwithZt=*t0XsdWs+*t0Ysdsin Itˆo’s first formula. The question is: What isdf(Zt) =?Taylor’s formula says:δf = f(x + δx) − f(x) = f"(x)δx +12f""(x)(δx)2+ O+(δx)3,Substituting x = Ztwe get:δf(Zt) = f(Zt+δt) − f(Zt) = f"(Zt)δZt+12f""(Zt)(δZt)2+ #Taking the limit as δt → 0 we get:df(Zt) = f"(Zt)dZt+12f""(Zt)d !Z"tNow we can insert expressions for dZtand d !Z"t:dZt= XtdWt+ Ytdt206 STOCHASTIC INTEGRATIONd !Z"t= X2tdtand we get:Theorem 9.30 (Itˆo II).df(Zt) = f"(Zt)XtdWt+ f"(Zt)Ytdt +12f""(Zt)X2tdt.Exercise 9.31. Calculate d(S2t) if µ = −5 and σ = 3. [Use f(x) = x2.]Answer: f(x) = x2. So, f"(x) = 2x, f""(x) = 2 and Itˆo’s equationgives:dS2t= 2StXtdWt+ 2StYtdt + X2tdt.The formula for dStis:dSt= σSt"#$%XtdWt+ µSt"#$%YtdtSo, Xt= 3Stand Yt= −5StmakingdS2t= 6S2tdWt− 10S2tdt + 9S2tdt = 6S2tdWt− S2tdt.9.4.7. Itˆo’s third formula. We need the next formula which gives thedifferential of a function of t and Zt:df(t, Zt) =?The function we really want to know about isV (t, x) = the value at time t of a stock option if St= xThe kind of stock option we are talking about is the option to buy oneshare of stock at some future time T at a fixed price K. We want toknow how much the option is worth:V (t, St) =?A general function of this kind is f(t, Zt). This is a function of thevectorx =&tZt', δx =&δtδZt'The derivative of f is given by the gradient∇f = (˙f, f") =&∂f∂t,∂f∂z'Here I use the physicist notation that a dot indicates time derivativeand prime indicates space derivative:˙f =∂f∂t, f"=∂f∂zMATH 56A SPRING 2008 STOCHASTIC PROCESSES 207The second derivative of f is given by the Hessian of f:D2f =-∂2f∂t2∂2f∂tdz∂2f∂tdz∂2f∂z2.=&¨f˙f"˙f"f""'This is a symmetric matrix which tells you if you have a local max ormin or a saddle point (from multivariable calculus). Taylor’s formulain two variables is:δf = ∇f · δx +12D2f(δx)2+ #The first term on the right is the dot product of vectors (which is amatrix product when ∇f is a written as a row vector):∇f · δx = (˙f, f")&δtδZt'=˙fδt + f"δZtThe second term is a product of three matrices:12D2f(δx)2=12(δt, δZt)&¨f˙f"˙f"f""'&δtδZt'So,δf =˙fδt + f"δZt+12(δt, δZt)&¨f˙f"˙f"f""'&δtδZt'+ #Take the limit as δt → 0 and plug in dtdt = d !t" = 0, dtdZt=d !t, Z"t= 0 and (dZt)2= X2tdt:df(t, Zt) =˙fdt + f"dZt+12f""X2tdtSince dZt= XtdWt+ Ytdt we get:Theorem 9.32 (Itˆo III).df(t, Zt) =˙fdt + f"XtdWt+ f"Ytdt +12f""X2tdt.Exercise 9.33. If a, b, c are constants then find dZtwhenZt= at2+ btWt+ cW2t[Use f(t, x) = at2+ btx + cx2]208 STOCHASTIC INTEGRATIONAnswer: We need:˙f = 2at + bx = 2at + bWtf"= bt + 2cx = bt + 2cWtf""= 2cSo,df(t, Wt) = (2at + bWt)dt+ (bt + 2cWt)(XtdWt+ Ytdt)+12(2c)X2tdtSince we are using dWt= 1dWt+ 0 instead of dZt=