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Lecture 20 Itô's Formula

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Lecture 20 Itô's Formula Lecture 20:Itô’s formula 1 of 13 Course: Theory of Probability II Term: Spring 2015 Instructor: Gordan Zitkovic Lecture 20 Itô’s formula Itô’s formula Itô’s formula is for stochastic calculus what the Newton-Leibnitz for- mula is for (the classical) calculus. Not only does it relate differentia- tion and integration, it also provides a practical method for computa- tion of stochastic integrals. There is an added benefit in the stochastic case. It shows that the class of continuous semimartingales is closed under composition with C2 functions. We start with the simplest ver- sion: Theorem 20.1. Let X be a continous semimartingale taking values in a seg- ment [a, b] ⊆ R, and let f : [a, b] ! R be a twice continuously differentiable function ( f 2 C2[a, b]). Then the process f (X) is a continuous semimartin- gale and Z t Z t Z t 0 0 1 00 (20.1) f (Xt) − f (X0) = f (Xu) dMu + f (Xu) dAu + 2 f (Xu) dhMiu. 0 0 0 R t 0 Remark 20.2. The Leibnitz notation 0 f (Xu) dMu (as opposed to the “semimartingale notation” f 0(X) · M) is more common in the context of the Itô formula. We’ll continue to use both. Before we proceed with the proof, let us state and prove two useful related results. In the first one we compute a simple stochastic integral explicitly. You will immediately see how it differs from the classical Stieltjes integral of the same form. 2,c Lemma 20.3. For M 2 M0 , we have 1 2 1 M · M = 2 M − 2 hMi. Proof. By stopping, we may assume that M and hMi are bounded. D For a partition D = f0 = t0 < t1 < ... g, let M¯ denote the “left- continuous simple approximation” M¯ D = M , where kD(t) = sup t ≤ t. t tkD(t) k k2N0 Last Updated: April 30, 2015 Lecture 20:Itô’s formula 2 of 13 Nothing but rearrangement of terms yields that ¯ D 1 2 D (20.2) (M · M)t = 2 (Mt − hMit ) for all t ≥ 0, and all partitions D 2 P[0,¥). Indeed, assuming for notational simplic- ity that D is such that tn = t, we have n n ( ¯ D · ) = ( − ) = 1 ( + ) − 1 ( − ) ( − ) M M t ∑ Mtk−1 Mtk Mtk−1 ∑ 2 Mtk Mtk−1 2 Mtk Mtk−1 Mtk Mtk−1 k=1 k=1 (20.3) n n = 1 (M2 − M2 ) − 1 (M − M )2 = 1 M2 − 1 hMiD. ∑ 2 tk tk−1 ∑ 2 tk tk−1 2 t 2 t k=1 k=1 1 2 By Theorem 18.10, the right-hand side of (20.3) converges to 2 (Mt − 2 hMit) in L , as soon as D ! Id. To show that the limit of the left-hand side converges in M · M, it is enough to use the stochastic dominated- convergence theorem (Proposition 19.10). Indeed, the integrands are all, unifromly, bounded by a constant. The second preparatory result is the stochastic analogue of the integration-by-parts formula. We remind the reader that for two semi- martingales X = M + A and Y = N + C, we have hX, Yi = hM, Ni. Proposition 20.4. Let X = M + A, Y = N + C be semimartingale de- compositions of two continuous semimartingales. Then XY is a continuous semimartingale and Z t Z t (20.4) XtYt = X0Y0 + Yu dXu + Xu dYu + hX, Yit. 0 0 Proof. By stopping, we can assume that the processes M, N, A and C are bounded (say, by K ≥ 0). Moreover, we assume that X0 = Y0 = 0 - otherwise, just consider X − X0 and Y − Y0. We write XY as (M + A)(N + C) and analyze each term. Using the polarization identity, the 1 2 2 2 product MN can be written as MN = 2 (M + N) − M − N , and Lemma 20.3 implies that Z t Z t (20.5) Mt Nt = Mu dNu + NudMu + hM, Nit, 0 0 holds for all t ≥ 0. As far as the FV terms A and C are concerned, the equality Z t Z t (20.6) AtCt = Au dCu + Cu dAu 0 0 follows by a representation of both sides as a limit of Riemann-Stieltjes sums. Alternatively, you can view the left-hand side of the above equality as the area (under the product measure dA × dC) of the square [0, t] × [0, t] ⊆ R2. The right-hand side can also be interpreted as the Last Updated: April 30, 2015 Lecture 20:Itô’s formula 3 of 13 area of [0, t] × [0, t] - the two terms corresponding to the areas above and below the diagonal f(s, s) 2 R2 : s 2 [0, t]g (we leave it up to the reader to supply the details). Let us focus now on the mixed term MC. Take a sequence fDngn2N n in P[0,¥) with Dn = ftk gk2N and Dn ! Id and write ¥ 1 2 3 MtCt = (M ^ n C ^ n − M ^ n C ^ n ) = I + I + I , where ∑ t tk+1 t tk+1 t tk t tk n n n k=0 ¥ 1 I = M ^ n (C ^ n − C ^ n ) n ∑ t tk t tk+1 t tk k=0 ¥ 2 I = C ^ n (M ^ n − M ^ n ) n ∑ t tk t tk+1 t tk k=0 ¥ 3 I = (M ^ n − M ^ n )(C ^ n − C ^ n ) n ∑ t tk+1 t tk t tk+1 t tk k=0 By the properties of the Stieltjes integral (basically, the dominated con- 1 R t vergence theorem), we have In ! 0 Mu dCu, a.s. Also, by the uniform 3 continuity of the paths of M on compact intervals, we have In ! 0, a.s. Finally, we note that Z t 2 ¯Dn In = Cu dMu, 0 ¯Dn where Cu = CtDn (u). By uniform continuity of the paths of C on [0, t], ¯Dn ¯Dn ∗ C ! C, uniformly on [0, t], a.s., and Cu − Cu ≤ 2Cu, which is an adapted and continuous process. Therefore, we can use the stochastic dominated convergence theorem (Proposition 19.10) to conclude that 2 R t In ! 0 Cu dMu in probability, so that Z t Z t (20.7) MtCt = Mu dCu + Cu dMu 0 0 and, in the same way, Z t Z t (20.8) At Nt = Au dNu + Nu dAu 0 0 The required equality (20.4) is nothing but the sum of (20.5), (20.6), (20.7) and (20.8). Remark 20.5. By formally differentiating (20.4) with respect to t, we write d(XY)t = Xt dYt + Yt dXt + dXtdYt. While meaningless in the strict sense, the “differential” representa- tion above serves as a good mnemonic device for various formulas in stochastic analysis. One simply has to multiply out all the terms, dis- 3 2 regard all terms of order larger than 2 (such as (dXt) or (dXt) dYt), Last Updated: April 30, 2015 Lecture 20:Itô’s formula 4 of 13 and use the following multiplication table: dM dA dN dhM, Ni 0 dC 0 0 where X = M + A, Y = N + C are semimartingale decompositions of X and Y. When M = N = B, where B is the Brownian motion, the multiplication table simplifies to dB dA dB dt 0 dC 0 0 Proof of Theorem 20.1. Let A be the family of all functions f 2 C2[a, b] such that the formula (20.1) holds for all t ≥ 0 and all continuous semimartingales X which take values in [a, b]. It is clear that A is a linear space which contains all constant functions. Moreover, it is also closed under multiplication, i.e., f g 2 A if f , g 2 A. Indeed, we need to use the integration-by-parts formula (20.4) and the associativ- ity of stochastic integration (Problem 19.5,(2)) applied to f (X) and g(X). Next, the identity is clearly in A, and so, P 2 A for each poly- nomial P. It remains to show that A = C2[a, b]. For f 2 C2[a, b], 00 let fPn gn2N be a sequence of polynomials with the property that 00 00 Pn ! f , uniformly on [a, b]. This can be achieved by the Weierstrass- Stone theorem, because polynomials are dense in C[a, b]. It is easy 00 to show that the polynomials fPn gn2N can be taken to be second derivatives of a sequence fPngn2N of polynomials with the property 0 0 00 00 that Pn ! f , Pn ! f and Pn ! f , uniformly on [a, b]. Indeed, 0 0 R x 00 R x 0 just use Pn(x) = f (x) + a Pn (x) dx and Pn(x) = f (x) + a Pn(x) dx. Then, as the reader will easily check, the stochastic dominated con- vergence theorem 19.10 will imply that all terms in (20.1) for Pn con- verge in probability to the corresponding terms for f . This shows that C2[a, b] = A. Remark 20.6. The proof of the Itô’s formula given above is slick, but it does not help much in terms of intuition. One of the best ways of understanding the Itô formula is the following non-rigorous, heuristic, derivation, where 0 = t0 < t1 < ··· < tn < tn+1 = t is a partiton of [0, t]. The main insight is that the second-order term in the Taylor’s 0 1 00 2 2 expansion f (t) − f (s) = f (s)(t − s) + 2 f (s)(t − s) + o((t − s) ) has to be kept and cannot be discarded because - in contast to the classical Last Updated: April 30, 2015 Lecture 20:Itô’s formula 5 of 13 case - the second (quadratic) variation does not vanish: n ( ) − ( ) = ( ) − ( ) f Xt f X0 ∑ f Xtk+1 f Xtk k=0 n ≈ 0( )( − ) + 1 00( )( − )2 ∑ f Xtk Xtk+1 Xtk 2 f Xtk Xtk+1 Xtk−1 k=0 n n = 0( )( − ) + 0( )( − ) ∑ f Xtk Mtk+1 Mtk ∑ f Xtk Atk+1 Atk k=0 k=0 n + 1 00( )( − )2 2 ∑ f Xtk Xtk+1 Xtk k=0 Z t Z t Z t 0 0 1 00 ≈ f (Xu) dMu + f (Xu) dAu + 2 f (Xu) dhXiu.
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