University:
Massachusetts Institute of Technology
Course:
15.501/516 | Accounting
Academic year:

2023
Views:

422

Pages:

5
Author:

Kyra Compton

Advanced Stochastic Processes. LECTURE 13 Ito integral. Properties Lecture outline • Deﬁnition of Ito integral • Properties of Ito integral 13.1. Ito integral. Existence We continue with the construction of Ito integral. We combine the results of Propositions 13 from Lecture 12 and prove the following result. Proposition 1. Given any process X ∈ H 2 there exists a sequence of simple processes Xn ∈ S 2 such that � T (13.1) lim E[ (Xn (t) − X(t))2 dt] = 0. n 0 Proof. Given any large n > 0, by Proposition 3 from Lecture 12, there exists a bounded process Xn1 ∈ H 2 satisfying � T 1 E[ (X(t) − Xn1 (t))2 dt] ≤ . n 0 Applying Proposition 2 from Lecture 12, for each n there exists a bounded continuous process Xn2 ∈ H 2 satisfying � T 1 E[ (Xn1 (t) − Xn2 (t))2 dt] ≤ . n 0 Finally, applying Proposition 1 from Lecture 12, for each n there exists a simple process Xn3 ∈ H 2 satisfying � T 1 E[ (Xn2 (t) − Xn3 (t))2 dt] ≤ . n 0 1 2 Let aj = Xnj (t) − Xnj+1 (t), j = 0, 1, 2, where Xn0 (t) stands for X(t). Then (X(t) − Xn3 (t))2 = (X(t) − Xn1 (t) + Xn1 (t) − Xn2 (t) + Xn2 (t) − Xn3 (t))2 � � = a2j + 2 a j ak j ≤ j 0, its Ito integral IT (X) is deﬁned to be the L2 limit of random variables IT (Xn ), where Xn ∈ S 2 is any sequence of simple processes satisfying (13.1). Namely, IT (X) is the unique random variable satisfying lim E[(IT (X) − IT (Xn ))2 ] = 0. n We write � IT (X) = T � X(t)dB(t) = 0 T X(t, ω)dB(t, ω). 0 Theorem 13.5. Ito integral is well deﬁned. That is the L2 limit of IT (Xn ) exists and does not depend on the choice of the sequence satisfying (13.1). Moreover, Ito integral satisﬁes Ito isometry � T 2 E[IT (X)] = E[ X 2 (t)dt]. 0 Proof. We need to prove the consistency of the Ito integral deﬁnition. Namely, that such a limiting random variable exists and does not depend on the choice of the sequence Xn . We have established in Proposition 1, the existence of a sequence Xn satisfying (13.1). We have established in Proposition 2 that the sequence IT (Xn ) is Cauchy. Applying Theorem 12.3 from Lecture 12 (Completeness of L2 ), there exists a unique (up to measure zero sets) random variable which is the L2 limit of IT (Xn ). Now we need to establish uniqueness. Namely, if Xn� is another sequence satisfying (13.1), then the unique limit of the sequence IT (Xn� ) is the same as that of the sequence IT (Xn ). Problem 1. Let IT (X) be a limit of IT (Xn ) along some sequence of simple processes Xn ∈ S 2 satisfying (13.1). Establish that IT (X) satisﬁes Ito isometry: � T 2 X 2 (t)dt]. E[IT (X)] = E[ 0 Using this ﬁnish the proof of the theorem, by establishing the uniqueness of the limit. � 4 13.2. Ito integral. Properties 13.2.1. Simple examples Let us compute the Ito integral for a special case X(t) = B(t). We will do this directly 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 Bn (t) = B(tj ), t ∈ [tj , tj+1 ). Assume that limn Δ(Πn ) = 0. We ﬁrst show that this is suﬃcient for having � T (13.6) lim E[ (B(t) − Bn (t))2 dt] = 0. n 0 Indeed T � 2 (B(t) − Bn (t)) dt = 0 n−1 � � tj+1 (B(t) − B(tj ))2 dt. tj j=0 We have � E[ tj+1 � 2 tj+1 (B(t) − B(tj )) dt = tj E[(B(t) − B(tj ))2 ]dt tj tj+1 � (t − tj )dt = tj = (tj+1 − tj )2 , 2 implying � T n−1 n−1 � 1� E[ (B(t) − Bn (t))2 dt] = (tj+1 − tj )2 ≤ Δ(Πn ) (tj+1 − tj ) = Δ(Πn )T. 2 j=0 0 j=0 By our assumption, though, the righthand side of this expression converges to zero implying that (13.6). Thus we need to compute the L2 limit of � IT (Bn ) = B(tj )(B(tj+1 ) − B(tj )) j as n → ∞. We use the identity B 2 (tj+1 ) − B 2 (tj ) = (B(tj+1 ) − B(tj ))2 + 2B(tj )(B(tj+1 ) − B(tj )), implying 2 2 B (T ) − B (0) = n−1 � 2 2 B (tj+1 ) − B (tj ) = j=0 n−1 � 2 (B(tj+1 ) − B(tj )) + 2 j=0 But recall the quadratic variation property of the Brownian motion: lim n n−1 � j=0 (B(tj+1 ) − B(tj ))2 = T n−1 � j=0 B(tj )(B(tj+1 ) − B(tj )), LECTURE 13. INTEGRATION 5 in L2 (recall that the only requirement for this convergence was that Δ(Πn ) → 0). Therefore, also in L2 n−1 � 1 T B(tj )(B(tj+1 ) − B(tj )) → B 2 (T ) − . 2 2 j=0 We conclude Proposition 3. The following identity holds � T 1 T B(t)dB(t) = B 2 (T ) − . IT (B) = 2 2 0 13.2.2. Simple properties The following properties follow in a straightforward way from the deﬁnition: �T �T �T (a) 0 (a1 X(t) + a2 Y (t))dB(t) = a1 0 X(t)dB(t) + a2 0 Y (t)dB(t). �T (b) E[ 0 X(t)dB(t)] = 0. �T (c) 0 X(t)dB(t) is FT measurable. 13.3. Additional reading materials • Course Packet. Chapter from Harrison’s book ”Brownian models and stochastic control”. • Øksendal [1], Chapter III.