MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Lecture 17 Fall 2013 11/13/2013 Ito process. Ito formula Content. 1. Ito process and functions of Ito processes. Ito formula. 2. Multidimensional Ito formula. Integration by parts. 1 Ito process Observe that trivially t 0 dBs = Bt . In the previous lecture we computed t It (B) = 0 1 t Bs dBs = Bt2 − , 2 2 or t B 2 (t) = 2 B(s)dB(s) + t, (1) 0 t Observe also that we cannot have t = 0 Xs dBs for some process X ∈ L2 as Ito integral is a martingale, but t is not. Thus we see that applying a functional operation to a process which is an Ito integral we do not necessarily get another Ito integral. But there is a natural generalization of Ito integral to a broader family, which makes taking functional operations closed within the family. Deﬁnition 1. An Ito process or stochastic integral is a stochastic process on (Ω, F, P) adopted to Ft which can be written in the form t Xt = X0 + t Us ds + 0 Vs dBs , 0 where U, V ∈ L2 . As a shorthand notation, we will write (2) as dXt = Ut dt + Vt dBt . 1 (2) Rt Rt Thus Bt2 is an Ito process: Bt2 = 0 ds + 2 0 Bs dBs or d(Bt2 ) = dt + 2Bt dBt . Note the difference from the usual differentiation: dx2 = 2xdx. The additional term dt arises because Brownian motion B is not differentiable and instead has quadratic variation. Notation Given an Ito process dXt = Ut dt+Vt dBt , let us introduce the notation (dXt )2 which stands for Vt2 dt. Equivalently (dXt )2 is (dXt ) · (dXt ) which is computed using the rules dt · dt = dt · dBt = dBt · dt = 0, dBt · dBt = dt. 2 Ito formula We now introduce the most important formula of Ito calculus: Theorem 1 (Ito formula). Let Xt be an Ito process dXt = Ut dt + Vt dBt . Sup pose g(x) ∈ C 2 (R) is a twice continuously differentiable function (in particular all second partial derivatives are continuous functions). Suppose g(Xt ) ∈ L2 . Then Yt = g(Xt ) is again an Ito process and dYt = ∂g 1 ∂2g (Xt )dXt + (Xt )(dXt )2 ∂x 2 ∂x2 Using the notational convention for dXt = Ut dt + Vt dBt and (dXt )2 , we can rewrite the Ito formula as dYt = � ∂g ∂x (Xt )Ut + � 1 ∂2g ∂g 2 (X )V (Xt )Vt dBt . t t dt + 2 2 ∂x ∂x Thus, we see that the space of Ito processes is closed under twice-continuously differentiable transformations. Proof sketch of Theorem 1. We will do this for a very special case. We assume ∂g ∂ 2 g as well as U and V are all bounded simple processes. that the derivatives ∂x ∂x2 The general case is then obtained by approximating U and V by bounded simple processes in a way similar to how we deﬁned the Ito integral. Let Πn : 0 = t0 < t1 < · · · < tn = t be a sequence of partitions such that Δ(Πn ) → 0. We use the notation ΔB(tj ) = B(tj+1 ) − B(tj ), ΔX(tj ) = 2 X(tj+1 ) − X(tj ). Using Taylor expansion of g we obtain X g(X(t)) = g(X(0)) + g(X(tj+1 )) − g(X(tj )) j < n X ∂g = g(X(0)) + (X(tj ))ΔX(tj ) ∂x j < n X 1 X ∂2g 2 + (X(t ))Δ X(t ) + o(Δ2 X(tj )). j j 2 ∂x2 j < n j < n Now, we have ΔX(tj ) = X(tj+1 ) − X(tj ) = U (tj )(tj+1 − tj ) + V (tj )(B(tj+1 ) − B(tj ))). Thus, we obtain X ∂g X ∂g (X(tj )) U (tj )(tj+1 − tj ) + V (tj )(B(tj+1 ) − B(tj )) (X(tj ))ΔX(tj ) = ∂x ∂x j < n j < n We argue that in L2 the convergence Z t X ∂g ∂g (X(tj )) U (tj )(tj+1 − tj ) → (X(s))U (s)ds ∂x 0 ∂x j < n and Z t X ∂g ∂g (X(tj )) V (tj )(B(tj+1 ) − B(tj )) → (X(s))V (s)dB(s) ∂x ∂x 0 j 0. Therefore the ﬁrst sum converges to zero provided Δ(Πn ) → 0. To analyze the second sum, we square it and take expected value: X ∂2g 2 E[2 (X(t ))U (t )V (t )(t − t )ΔB(t ) ] j j j j+1 j j ∂x2 j < n =4 X j < n ∂2g E[ (X(tj ))U (tj )V (tj ) ∂x2 2 ](tj+1 − tj )3 , where to obtain this equality we ﬁrst condition on ﬁeld Ftj , note that the ex pected value of all cross products vanishes and use E[Δ2 B(tj )|Ftj ] = tj+1 −tj . Again, since the second partial derivative and U, V are bounded, then the entire term converges to zero provided that Δ(Πn ) → 0. In order to analyze the last sum the following result is needed. Problem 1 (Generalized Quadratic Variation). Suppose a(s) ∈ H 2 and Πn : 0 = t0 < · · · < tn = t is a sequence of partitions satisfying Δ(Πn ) → 0 as n → ∞. Then the following convergence occurs in L2 : Z t X 2 lim a(tj )(B(tj+1 ) − B(tj )) = a(s)ds. n 0 j HINT: use the same approach that we did in establishing the quadratic variation of B. Using this result we establish that the last sum converges in L2 to Z t 2 ∂ g (X(s))V 2 (s)ds. 2 0 ∂x 4 P It remains to analyze j o(Δ2 X(tj )) and using similar techniques, it can be shown that this term vanishes in L2 norm as Δ(Πn ) → 0. Putting all of this together, we conclude that g(X(t)) is approximated in L2 sense by Z t Z ∂g 1 t ∂2g g(X(0)) + (X(s))dX(s) + (X(s))V 2 (s)ds. 2 0 ∂x2 0 ∂x But recall that V 2 (s)ds = (dX(s))2 . Making this substitution, we complete the derivation of the Ito formula. Let us apply Theorem 1 to several examples. Exercise 1. Verify that in all of the examples below the underlying processes are in L2 . (1) using Ito formula. Since Bt = R t Example 1. Let us re-derive our formula 1 2 dB is an Ito process and g(x) = x is twice continuously differentiable, s 2 0 then by the Ito formula we have 1 ∂g 1 ∂2g d( Bt2 ) = dg(Bt ) = dBt + (dBt )2 2 ∂x 2 ∂x2 1 = Bt dBt + (dBt )2 2 dt = Bt dBt + , 2 which matches (1). Example 2. Let us apply Ito formula to Bt4 . We obtain 1 d(Bt4 ) = 4Bt3 dBt + 12Bt2 dt = 4Bt3 dBt + 6Bt2 dt, 2 namely, written in an integral form Z t Z t 4 3 Bt = 4 Bs dBs + 6 Bs2 ds. 0 0 Taking expectations of both sides and recalling that Ito integral is a martingale, we obtain Z t 1 E[ Bs2 ds] = E[Bt4 ] 6 0 5 which we find to be (1/6)3t2 as the 4-th moment of a normal zero mean dis4 tribution with R t std2 σ is 3σ . Recall an earlier exercise where you were asked to compute E[ 0 Bs ds] directly. We see that Ito calculus is useful even in computing conventional integrals. 3 Multidimensional Ito formula There is a very useful analogue of Ito formula in many dimensions. We state this result without proof. Before turning to the formula we need to extend our discussion to the case of Ito processes with respect to many dimensions, as so far we have we have considered Ito integrals and Ito processes with respect to just one Brownian motion. Thus suppose we have a vector of d independent Brownian motions Bt = (Bi,t , 1 ≤ i ≤ d, t ∈ R+ ). A stochastic process Xt is defined to be an Ito process with respect to BP t if there exists Ut ∈ L2 and Vi,t ∈ L2 , 1 ≤ i ≤ d such that Xt = Ut dt + i Vi,t dBi,t , in the sense explained above. The definition naturally extends to the case when Xt is a vector of processes. Theorem 2. Suppose dXt = Ut dt + Vt dBt , where vector U = (U1 , . . . , Ud ) and matrix V = (V11 , . . . , Vdd ) have L2 components and B is the vector of d independent Brownian motions. Let g(x) be twice continuously differentiable function from Rd into R. Then Yt = g(Xt ) is also an Ito process and d d X ∂g 1 X ∂2g dYt = (Xt )dXi,t + (Xt )dXi,t · dXj,t , ∂xi 2 ∂xi xj i=1 i,j=1 where dXi,t · dXj,t is computed using the rules dtdt = dtdBi = dBi dt = 0, dBi dBj = 0 for all i 6= j and (dBi )2 = dt. Let us now do a quick example illustrating the use of the Ito formula. Consider g(t, Bt ) = etBt . We will use Ito formula to find its derivative. Since both t and Bt are Ito processes and g(t, x) = etx is twice continuously differentiable ∂ ∂2 2 tx ∂ function g : R2 → R, the formula applies. ∂t g = xetx , ∂t 2 g = x e , ∂x g = 2 ∂ 2 tx tetx , ∂x 2 g = t e . Then we can find its Ito representation using the Ito formula as 1 1 d(etBt ) = etBt Bt dt + etBt Bt2 (dt)2 + tetBt dBt + t2 etBt (dBt )2 2 2 1 2 tBt tBt = e (Bt + t )dt + te dBt . 2 6 R Problem 2. Use the Ito formula to ﬁnd eBt dBt . In other words, you need to Rrepresent this integral in terms of expressions not involving dBt (as we did for Bt dBt ). Suppose R t f is a continuously differentiable function. Let us use Ito formula to ﬁnd 0 fs dBs and derive the integration by parts formula. In other words we look at a special simple case when X is a deterministic process i.e., Xs = fs a.s. First we observe that f ∈ L2 . Indeed, it is differentiable and therefore continuous. This that f is bounded on any ﬁnite interval and therefore Rt 2 R t implies 2 E[ 0 fs ds] = 0 fs ds < ∞. 2 2 ∂ g df ∂g d f Introduce g(t, x) = ft x. We ﬁnd that ∂g ∂t = x dt , ∂x = ft , ∂t2 = x dt2 and second order partial derivatives with respect to x disappear. Therefore, using Ito formula, we obtain d(g(t, Bt ) = df df d2 f Bt dt + 2 Bt (dt)2 + ft dBt + 0 = Bt dt + ft dBt dt dt dt This implies (since g(0, B(0)) = 0) Z t Z t fs dBs = ft Bt − 0 Bs 0 df ds. ds This does look like integration by parts. It turns out that a more general version of the integration by parts formula holds in Ito calculus. We start by recalling the deﬁnition of Stieltjes integral. We are given a function g which has bounded variation and another function fR, which t we assume for simplicity is continuous. P We deﬁne the Stieltjes integral 0 fs dgs as an appropriate limit of the sums j ftj (gtj+1 − gtj ) where Πn : 0 = t0 < · · · < tn = t is a sequence of partitions with Δ(Πn ) = maxj (tj+1 − tj ) → 0. We skip the formalities of the construction of such a limit. They are similar to (and simpler than) those of the Ito integral. Now let us state without proving the integration by parts theorem. Theorem 3. Suppose fs is a continuous function on [0, t] with bounded varia tion. Then Z t Z t fs dBs = ft Bt − Bs dfs , 0 0 where the second integral is the Stieltjes integral. 7 Is there a generalization of integration by parts when f is not necessarily deterministic? The answer is positive, but, unfortunately, it does not reduce an Ito process to a process not involving Brownian component dB. Theorem 4. Let Xt , Yt be Ito processes. Then Xt Yt is also an Ito process: Z t Z t Z t Xt Yt = X(0)Y (0) + Xs dYs + Ys dXs + dXs dYs 0 0 0 Proof. The proof is direct application of multidimensional Ito formula to the function g(Xt , Yt ) = Xt Yt . 4 Additional reading materials • Øksendal [1], Chapter IV. References [1] B. Øksendal, Stochastic differential equations, Springer, 1991. 8

Lecture 17: Ito Process and Formula

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