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MIT15 070JF13 Lec14.Pdf MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Fall 2013 Lecture 14 10/28/2013 Introduction to Ito calculus. Content. 1. Spaces L2; M2; M2;c. 2. Quadratic variation property of continuous martingales. 1 Doob-Kolmogorov inequality. Continuous time version Let us establish the following continuous time version of the Doob-Kolmogorov inequality. We use RCLL as abbreviation for right-continuous function with left limits. Proposition 1. Suppose Xt ≥ 0 is a RCLL sub-martingale. Then for every T; x ≥ 0 2 E[XT ] P( sup Xt ≥ x) ≤ 2 : 0≤t≤T x Proof. Consider any sequence of partitions Π = f0 = tn < tn < ::: < tn = n 0 1 Nn T g such that Δ(Π ) = max jtn − tnj ! 0. Additionally, suppose that the n j j+1 j sequence Πn is nested, in the sense the for every n1 ≤ n2, every point in Πn1 is n n also a point in Π . Let X = X n where j = maxfi : t ≤ tg. Then X is a n2 t tj i t sub-martingale adopted to the same filtration (notice that this would not be the case if we instead chose right ends of the intervals). By the discrete version of the D-K inequality (see previous lectures), we have [X2 ] n n E T (max X ≥ x) = (sup X ≥ x) ≤ : P tj P t 2 j≤Nn t≤T x n By RCLL, we have supt≤T Xt ! supt≤T Xt a.s. Indeed, fix E > 0 and find t0 = t0(!) such that Xt0 ≥ supt≤T Xt − E. Find n large enough and 1 n j = j(n) such that tj(n)−1 ≤ t0 ≤ tj(n). Then tj(n) ! t0 as n ! 1. X X ! X By right-continuity of , tj(n) t0 . This implies that for sufficiently n sup Xn ≥ X ≥ X − 2E large , t≤T t tj(n) t0 , and the a.s. convergence is es­ tablished. On the other hand, since the sequence Πn is nested, then the se­ n quence supt≤T Xt is non-decreasing. By continuity of probabilities, we obtain n P(supt≤T Xt ≥ x) ! P(supt≤T Xt ≥ x). 2 Stochastic processes and martingales Consider a probability space (Ω; F; P) and a filtration (Ft; t 2 R+). We assume that all zero-measure events are ”added” to F0. Namely, for every A ⊂ Ω, such 0 0 0 that for some A 2 F with P(A ) = 0 we have A ⊂ A 2 F, then A also belongs to F0. A filtration is called right-continuous if Ft = \E>0Ft+E. From now on we consider exclusively right-continuous filtrations. A stochastic process Xt adopted to this filtration is a measurable function X :Ω × [0; 1) ! R, such that Xt 2 Ft for every t. Denote by L2 the space of processes s.t. the R T X (!)dt [R T X2dt] < 1 Riemann integral 0 t exists a.s. and moreover E 0 t for T > 0 (! : R T jX (!)jdt < 1; 8T ) = 1 every . This implies P 0 t . Let M2 consist of square integrable right-continuous martingales with left 2 limits (RCLL). Namely E[Xt ] < 1 for every X 2 M2 and t ≥ 0. Finally M2;c ⊂ M2 is a further subset of processes consisting of a.s. continuous processes. For each T > 0 we define a norm on M2 by kXk = kXkT = ( [X2 ])1=2. Applying sub-martingale property of X2 we have [X2 ] ≤ [X2 ] E T t E T1 E T2 for every T1 ≤ T2. A stochastic process Yt is called a version of Xt if for every t 2 R+; P(Xt = Yt) = 1. Notice, this is weaker than saying P(Xt = Yt; 8t) = 1. Proposition 2. Suppose (Xt; Ft) is a submartingale and t ! E[Xt] is a con­ tinuous function. Then there exists a version Yt of Xt which is RCLL. We skip the proof of this fact. Proposition 3. M2 is a complete metric space and (w.r.t. k ·k) M2;c is a closed subspace of M2. (n) Proof. We need to show that if X 2 M2 is Cauchy, then there exists X 2 (n) M2 with kX − Xk ! 0. (n) (n) (m) Assume X is Cauchy. Fix t ≤ T Since X − X is a martingale (n)(m) 2 (n)(m) 2 (n) as well, E[(Xt − Xt ) ] ≤ E[(XT − XT ) ]. Thus Xt is Cauchy as well. We know that the space L2 of random variables with finite second moment 2 (n) 2 is closed. Thus for each t there exists a r.v. Xt s.t. E[(Xt − Xt) ] ! 0 as (n) (n) n ! 1. We claim that since Xt 2 Ft and X is RCLL, then (Xt; t ≥ 0) is adopted to Ft as well (exercise). Let us show it is a martingale. First E[jXtj] < 2 (n) 1 since in fact E[Xt ] < 1. Fix s < t and A 2 Fs. Since each Xt is a (n) (n) martingale, then E[Xt 1(A)] = E[Xs 1(A)]. We have (n) (n) E[Xt1(A)] − E[Xs1(A)] = E[(Xt − Xt )1(A)] − E[(Xs − Xs )1(A)] (n) (n) (n) 2 1=2 We have E[jXt − Xt j1(A)] ≤ E[jXt − Xt j] ≤ (E[(Xt − Xt ) ]) ! 0 as n ! 1. A similar statement holds for s. Since the left-hand side does not depend on n, we conclude E[Xt1(A)] = E[Xs1(A)] implying E[XtjFs] = Xs, namely Xt is a martingale. Since E[Xt] = E[X0] is constant and therefore continuous as a function of t, then there exists version of Xt which is RCLL. For simplicity we denote it by Xt as well. We constructed a process Xt 2 M2 (n) 2 s.t. E[(Xt − Xt) ] ! 0 for all t ≤ T . This proves completeness of M2. (n) Now we deal with closeness of M2;c. Since Xt − Xt is a martingale, (n) 2 (n) 2 (Xt − Xt) is a submartingale. Since Xt 2 M2, then (Xt − Xt) is RCLL. Then submartingale inequality applies. Fix E > 0. By submartingale inequality we have 1 (sup jX(n) − X j > E) ≤ [(X(n) − X )2] ! 0; P t t 2 E T T t≤T E as n ! 1. Then we can choose subsequence nk such that 1 (sup jX(nk) − X j > 1=k) ≤ : P t t k t≤T 2 1=2k sup jX(nk) − Since is summable, by Borel-Cantelli Lemma we have t≤T t (nk) Xtj ! 0 almost surely: P(f! 2 Ω : supt≤T jXt (!) − Xt(!)j ! 0g) = 1. Recall that a uniform limit of continuous functions is continuous as well (first lecture). Thus Xt is continuous a.s. As a result Xt 2 M2;c and M2;c is closed. 3 Doob-Meyer decomposition and quadratic variation of processes in M2;c Consider a Brownian motion Bt adopted to a filtration Ft. Suppose this filtration makes Bt a strong Markov process (for example Ft is generated by B itself). 2 Recall that both Bt and Bt − t are martingales and also B 2 M2;c. Finally recall that the quadratic variation of B over any interval [0; t] is t. There is a 3 generalization of these observations to processes in M2;c. For this we need to recall the following result. Theorem 1 (Doob-Meyer decomposition). Suppose (Xt; Ft) is a continuous non-negative sub-martingale. Then there exist a continuous martingale Mt and a.s. non-decreasing continuous process At with A0 = 0, both adopted go Ft such that Xt = At + Mt. The decomposition is unique in the almost sure sense. The proof of this theorem is skipped. It is obtained by appropriate discretiza­ tion and passing to limits. The discrete version of this result we did earlier. See [1] for details. 2 Now suppose Xt 2 M2;c. Then Xt is a continuous non-negative submartin­ gale and thus DM theorem applies. The part At in the unique decomposition of 2 Xt is called quadratic variation of Xt (we will shortly justify this) and denoted hXti. Theorem 2. Suppose Xt 2 M2;c. Then for every t > 0 the following conver­ gence in probability takes place X 2 lim (Xtj+1 − Xtj ) ! hXti; Πn:Δ(Πn)!0 0≤j≤n−1 where the limit is over all partitions Πn = f0 = t0 < t1 < ··· < tn = tg and Δ(Πn) = maxj jtj − tj−1j. Proof. Fix s < t. Let X 2 M2;c. We have 2 2 2 E[(Xt − Xs) − (hXti − hXsi)jFs] = E[Xt − 2XtXs + Xs − (hXti − hXsi)jFs] 2 2 = E[Xt jFs] − 2XsE[XtjFs] + Xs − E[hXtijFs] + hXsi 2 2 = E[Xt − hXtijFs] − Xs + hXsi = 0: Thus for every s < t ≤ u < v by conditioning first on Fu and using tower property we obtain 2 2 E (Xt − Xs) − (hXti − hXsi)(Xu − Xv) − (hXui − hXvi) = 0 (1) The proof of the following lemma is application of various ”carefully placed” tower properties and is omitted. See [1] Lemma 1.5.9 for details. Lemma 1. Suppose X 2 M2 satisfies jXsj ≤ M a.s. for all s ≤ t. Then for every partition 0 = t0 ≤ · ·· ≤ tn = t 2 X 2 4 E (Xtj+1 − Xtj ) ≤ 6M : j 4 Lemma 2. Suppose X 2 M2 satisfies jXsj ≤ M a.s. for all s ≤ t. Then X 4 lim E[(Xtj+1 − Xtj ) ] = 0; Δ(Πn)!0 j where Πn = f0 = t0 < ··· < tn = tg; Δ(Πn) = maxj jtj+1 − tj j. Proof. We have X 4 X 2 2 (Xtj+1 − Xtj ) ≤ (Xtj+1 − Xtj ) supfjXr − Xsj : jr − sj ≤ Δ(Πn)g: j j Applying Cauchy-Schwartz inequality and Lemma 1 we obtain 2 2 X 4 X 2 4 E[(Xtj+1 − Xtj ) ] ≤ E (Xtj+1 − Xtj ) E[supfjXr − Xsj : jr − sj ≤ Δ(Πn)g] j j 4 4 ≤ 6M E[supfjXr − Xsj : jr − sj ≤ Δ(Πn)g]: Now X(!) is a.s.
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