Lecture 16: Ito Integral

MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Lecture 16 Fall 2013 11/04/2013 Ito integral. Properties Content. 1. Definition of Ito integral 2. Properties of Ito integral 1 Ito integral. Existence We continue with the construction of Ito integral. Combining the results of Propositions 1-3 from the previous lecture we proved the following result. Proposition 1. Given any process X ∈ L2 there exists a sequence of simple processes Xn ∈ L02 such that T (Xn (t) − X(t))2 dt] = 0. lim E[ n→∞ (1) 0 Now, given a process X ∈ L2 , we fix any sequence of simple processes ∈ L02 which satisfies (1) for a given T . Recall, that we already have defined Ito integral for simple processes It (X n ). Xn Proposition 2. Suppose a sequence of simple processes X n satisfies (1). There exists a process Zt ∈ M2,c satisfying limn E[(Zt − It (X n ))2 ] = 0 for all 0 ≤ t ≤ T . This process is unique a.s. in the following sense: if X̂tn is another process satisfying (1) and Ẑ is the corresponding limit, then P(Ẑt = Zt , ∀t ∈ [0, T ]) = 1. Proof. Fix T > 0. Applying linearity of It (X) and Ito isometry E[(IT (Xm ) − IT (Xn ))2 ] = E[IT2 (Xm − Xn )] T (Xm (t) − Xn (t))2 dt] = E[ 0 T T (X(t) − Xm (t))2 dt] + 2E[ ≤ 2E[ 0 (X(t) − Xn (t))2 dt]. 0 1 But since the sequence Xn satisfies (1), it follows that the sequence IT (X n ) is Cauchy in L2 sense. Recall now from Theorem 2.2. previous lecture that each It (X n ) is a continuous square integrable martingale: It (X n ) ∈ M2,c Applying Proposition 2, Lecture 1, which states that M2,c is a closed space, there exists a limit Zt , t ∈ [0, T ] in M2,c satisfying E[(ZT − IT (X n ))2 ] → 0. The same applies to every t ≤ T since (Zt − It (X n ))2 is a submartingale. It remains to show that such a process Zt is unique. If Ẑt is a limit of some sequence X̂ n satisfying (1), then by submartingale inequality for every E > 0 we have P(supt≤T |Zt − Ẑt | ≥ E) ≤ E[(ZT − ẐT )2 ]/E2 . But E[(ZT − ẐT )2 ] ≤ 3E[(ZT − IT (X n ))2 ] + 3E[(IT (X n ) − IT (X̂ n ))2 ] + 3E[(IT (X̂ n ) − ẐT )2 ], and the right-hand side converges to zero. Thus E[(ZT − ẐT )2 ] = 0. It follows that Zt = Ẑt a.s. on [0, T ]. Since T was arbitrary we obtain an a.s. unique limit on R+ . Now we can formally state the definition of Ito integral. Definition 1 (Ito integral). Given a stochastic process Xt ∈ L2 and T > 0, its Ito integral It (X), t ∈ [0, T ] is defined to be the unique process Zt constructed in Proposition 2. We have defined Ito integral as a process which is defined only on a finite interval [0, T ]. With a little bit of extra work it can be extended to a process It (X) defined for all t ≥ 0, by taking T → ∞ and taking appropriate limits. Details can be found in [1] and are omitted, as we will deal exclusively with Ito integrals defined on a finite interval. 2 2.1 Ito integral. Properties Simple example Let us compute the Ito integral for a special case Xt = Bt . We will do this di­ rectly from the definition. Later on we will develop calculus rules for computing the Ito integral for many interesting cases. We fix a sequence of partitions Πn : 0 = t0 < · · · < tn = T and consider Btn = Btj , t ∈ [tj , tj+1 ). Assume that limn Δ(Πn ) = 0, where Δ(Πn ) = maxj |tj+1 − tj |. We first show that this is sufficient for having Z T lim E[ (Bt − Btn )2 dt] = 0. (2) n 0 2 Indeed T Z (Bt − Btn )2 dt = 0 n−1 n Z tj+1 (Bt − Btj )2 dt. tj j=0 We have Z E[ tj+1 tj+1 Z 2 (Bt − Btj ) dt = tj E[(Bt − Btj )2 ]dt tj tj+1 Z (t − tj )dt = tj = (tj+1 − tj )2 , 2 implying Z T n−1 n−1 n 1n E[ (Bt − Btn )2 dt] = (tj+1 − tj )2 ≤ Δ(Πn ) (tj+1 − tj ) = Δ(Πn )T → 0, 2 0 j=0 j=0 as n → ∞. Thus (2) holds. Thus we need to compute the L2 limit of n IT (Bn ) = Btj (Btj+1 − Btj ) j as n → ∞. We use the identity Bt2j+1 − Bt2j = (Btj+1 − Btj )2 + 2Btj (Btj+1 − Btj ), implying B 2 (T ) − B 2 (0) = n−1 n Bt2j+1 − Bt2j = j=0 n−1 n n−1 n j=0 j=0 (Btj+1 − Btj )2 + 2 Btj (Btj+1 − Btj ), But recall the quadratic variation property of the Brownian motion: lim n n−1 n (Btj+1 − Btj )2 = T j=0 in L2 (recall that the only requirement for this convergence was that Δ(Πn ) → 0). Therefore, also in L2 n−1 n j=0 1 T Btj (Btj+1 − Btj ) → B 2 (T ) − . 2 2 We conclude 3 Proposition 3. The following identity holds Z T 1 T IT (B) = Bt dBt = B 2 (T ) − . 2 2 0 Further, recall that since Bt ∈ M2,c then it admits a unique Doob-Meyer decomposition Bt2 = t + Mt , where t = (Bt ) is the quadratic variation of Bt and Mt is a continuous martingale. Thus we recognize Mt to be 2It (B). 2.2 Properties We already know that It (X) ∈ M2,c , in particular it a is continuous martin­ gale. Let us establish additional properties, some of which are generalizations of Theorem 2.2 from the previous lecture. Proposition 4. The following properties hold for It (X): It (αX + βY ) = αIt (X) + βIt (Y ), ∀α, β, Z t E[(It (X) − Is (X))2 |Fs ] = E[ Xu2 du|Fs ], ∀ 0 ≤ s < t ≤ T. (3) (4) s Furthermore, the quadratic variation of It (X) on [0, T ] is T 0 Xt2 dt. Proof. The proof of (3) is straightforward and is skipped. We now prove (4). Fix any set A ∈ Fs . We need to show that Z t 2 E[(It (X) − Is (X)) I(A)] = E[I(A) Xu2 du]. s Fix a sequence X n ∈ L02 satisfying (1). Then E[(It (X) − Is (X))2 I(A)] = E[(It (X) − It (X n ))2 I(A)] + E[(It (X n ) − Is (X n ))2 I(A)] + E[(Is (X n ) − Is (X))2 I(A)] + 2E(It (X) − It (X n ))(It (X n ) − Is (X n ))I(A) + 2E(It (X) − It (X n ))(Is (X n ) − Is (X))I(A) + 2E(It (X n ) − Is (X n ))(Is (X n ) − Is (X))I(A) But E[(It (X) − It (X n ))2 I(A)] → 0 since E[(It (X) − It (X n ))2 ] → 0 (def­ inition of Ito integral). Similarly E[(Is (X) − Is (X n ))2 I(A)] → 0. Applying Cauchy-Schwartz inequality |E(It (X) − It (X n ))(It (X n ) − Is (X n ))I(A)| ≤ (E[(It (X) − It (X n ))2 ])1/2 (E[(It (X n ) − Is (X n ))2 ])1/2 → 0 4 from the definition of It (X). Similarly we show that all the other terms with factor 2 in front converge to zero. By property (2.6) Theorem 2.2 previous lecture, we have Z t n n 2 E[(It (X ) − Is (X )) I(A)] = E[I(A) (Xun )2 du] s Now Z E[I(A) t (Xun )2 du] − E[I(A) Z s s t Z t Xu2 du] = E[I(A) (Xun − Xu )(Xun + Xu )du] s Z t ≤ E[ |(Xun − Xu )(Xun + Xu )|du] s Z t Z t 1 1 ≤ E 2 [ (Xun − Xu )2 du]E 2 [ (Xun + Xu )2 du] s s where Cauchy-Schwartz inequality was used in the last step. Now the first term in the product converges to zero by the assumption (1) and the second is uni­ formly bounded in n (exercise). The assertion then follows. Now we prove the last part. R t Applying Proposition 3 from Lecture 1, it suf­ fices to show that It2 (X) − 0 Xs2 ds is a martingale, since then by uniqueness Rt of the Doob-Meyer decomposition we must have that (It (X)) = 0 Xs2 ds. But note that (4) is equivalent to Z t 2 2 2 2 E[It (X) − Is (X)|Fs ] = E[It (X)|Fs ] − Is (X) = E[ Xu2 du|Fs ] s Z t Z s = E[ Xu2 du|Fs ] − E[ Xu2 du|Fs ] 0 0 Z t Z s = E[ Xu2 du|Fs ] − Xu2 du. 0 0 Namely, Z t Z E[It2 (X)|Fs ] − E[ Xu2 du|Fs ] = Is2 (X) − 0 namely, 3 It2 (X) − Rt 2 0 Xs ds s Xu2 du, 0 is indeed a martingale. Additional reading materials • Karatzas and Shreve [1]. • Øksendal [2], Chapter III. 5 References [1] I. Karatzas and S. E. Shreve, Brownian motion and stochastic calculus, Springer, 1991. [2] B. Øksendal, Stochastic differential equations, Springer, 1991. 6