Week 8 Diffusion Processes, Part 2

Week 8 Diffusion processes, part 2 Jonathan Goodman October 28, 2013 1 Integration and Ito's lemma for dXt sec:il Outline of this section: 1. Ways to define new processes using old ones: R t (a) An Ito integral with respect to an old process, Yt = 0 asdXs (b) A function of an old process, Yt = f(Xt; t) 2. Figure out stuff about the new processes (a) Ito's lemma 3. It is more useful when it is more general (a) General Ito process, not just diffusions (b) Multi-component processes 1.1 Technical overview A diffusion is a process that satisfies an SDE, which makes it a Markov process. The operations listed above produce processes that have Ito differentials but are not Markov. We call Xt an Ito process if (in these formulas, ∆t > 0 and ∆X = Xt+∆t − Xt): E[ ∆X j Ft] = at∆t + o(∆t) (1) eq:im h 2 i E (∆X) j Ft = µt∆t + o(∆t) (one component) (2) eq:iv h t i E (∆X) (∆X) j Ft = µt∆t + o(∆t) (multi-component) (3) eq:ivm h 4 i 2 E j∆Xj j Ft ≤ C∆t (4) eq:m4 The little oh notation A = o(∆t) means that A is smaller than ∆t. More precisely, A=∆t ! 0 as ∆t ! 0. It is usually correct to think O(∆t2) whenever 1 you see o(∆t). That's what it would be if it were a Taylor series expansion. But it is convenient to make do with an error estimate that is weaker than O(∆t2). That way you don't waste time proving that the error is very small, when it only matters that the error is o(∆t). If Xt is a Markov process (i.e., a diffusion), then the infinitesimal mean and variance are functions of Xt. We may write at = a(Xt; t), and µt = µ(Xt; t) in that case. A diffusion process must be Markov and satisfy an SDE. An Ito process just eq:imeq:iv eq:ivm eq:im needs to satisfy (1), (2) or (3), and (1). For example, suppose (X1;t;X2;t) is a two component diffusion process and Yt = X1;t +X2;t. Then Yt is an Ito process but not a diffusion, because the infinitesimal mean and variance of Yt are not functions of Yt alone, but they depend on X1;t and X2;t separately. There is one thing diffusions do but non-Markov Ito processes do not { satisfy backward equations. The backward equation of last week has a(x; t) and µ(x; t), not just at and µt. The Ito calculus for Ito process is almost the same as it is for Brownian motion, but there is one new technical thing. The general treatment here is a little more complicated, though not much harder, because general Ito processes are not martingales. A general Ito process may be separated into a martingale part, which looks like Brownian motion for our purposes here, and a \smoother" part that can be integrated in the ordinary way. Ito's lemma for general Ito process has a natural version of the informal 2 2 rule (dWt) \=" dt. This rule makes sense because E (dWt) j Ft = dt. The general rule is 2 2 (dXt) \=" E (dXt) j Ft = µ(Xt; t)dt : 2 This fake \=" is not a true equality because the standard deviation of (dXt) is also of order dt. We can substitute the right side for the left side in an Ito 2 integral because the total effect of the “fluctuations” in (dXt) vanishes in the limit ∆t ! 0. 1.2 Martingale diffusions By definition, a stochastic process is a martingale if, for any s > 0, E[ Xt+s j Ft] = Xt : (5) eq:m This definition applies to general processes, which do not have to have continuous paths or be Markov processes. But for Ito processes, the martingale eq:m eq:im property (5) is equivalent to a = 0, the zero drift property, in (1). This equiv- eq:m eq:m alence means that zero drift implies (5), and (5) implies zero drift. The second eq:m statement is \obvious", as (5) implies that E[ ∆X j Ft] = 0 for any ∆t > 0. The first statement is also \obvious", so much so that we defer the discussion to a later subsection. If an Ito process is not a martingale, we can separate out the martingale part using an ordinary integral Z t Yt = Xt − as ds : 0 2 The increment of Y is ∆Y = ∆X − a ∆t + (smaller), so Y satisfies the in- eq:im t t finitesimal drift equation (1) with drift coefficient equal to zero. Therefore, we have Xt = Yt + Zt ; (6) eq:md where Yt is a martingale and Z t Zt = as ds : (7) eq:cp 0 In this decomposition, the drift part of Xt is Zt, and the quadratic variation of Xt is in Yt. We can see this by calculating h 2 i h 2 i E (∆Y ) j Ft = E (∆X − ∆Z) j Ft h 2 i h 2 i = E (∆X) j Ft + 2E[ ∆X∆Z j Ft] + E (∆Z) j Ft 2 2 2 = µt∆t + 2E[ ∆X∆Z j Ft] + at ∆t + O(∆t ) : Cauchy Schwarz handles the middle term on the right: r r h 2 i h 2 i jE[ ∆X∆Z j Ft]j ≤ E (∆X) j Ft E (∆Z) j Ft p p ≤ C ∆t ∆t2 ≤ C∆t3=2 : The power 3=2 shows the convenience of writing o(∆t) instead of O(∆t2). With some more work, we could get C∆t2 instead of C∆t3=2. But this is unnecessary for the present purpose. To summarize, E[ ∆Y j Ft] = 0 2 E[ ∆Z j Ft] = at∆t + O(∆t ) h 2 i 2 E (∆Y ) j Ft = µt∆t + O(∆t ) h 2 i 2 E (∆Z) j Ft = O(∆t ) h 4 i 2 E j∆Y j j Ft ≤ C∆t h 4 i 4 E j∆Zj j Ft ≤ C∆t We divided X into two pieces Y and Z. The martingale piece is as irregular as X but is a martingale. The regular piece is continuous enough to be treated with ordinary calculus. The definition of the integral is Z t Z t Z t fsdXs = fsdYt + fsasds : (8) eq:ci 0 0 0 Only the first term on the right needs a definition. 3 1.3 Ito integral for martingale Ito processes sec:ii This subsection defines Z t Ut = fsdYs ; 0 where Ys is an Ito process and a martingale. In the notation from the past few weeks: (m) X Ut = lim Ut = lim ftj Ytj+1 − Ytj : m!1 m!1 tj <t −m Recall that ∆t = 2 and tj = j∆t. Our Borel Cantelli type lemma implies that the limit exists if 1 h i X (m+1) (m) E Ut − Ut < 1 : m=1 We show that in fact the expectations on the right form a geometric series: h i p p −m (m+1) (m) E Ut − Ut ≤ C ∆t = C 2 : (m+1) (m) We study Ut − Ut as we did when we did integration with respect (m+1) to Brownian motion. The finer approximation Ut is defined using values 1 f = f with t = j∆t, and the half points f 1 = f with t 1 = (j + )∆t. j tj j j+ tj+ 1 j+ 2 2 2 2 The Yj and Y 1 are defined analogously. The finer approximation is j+ 2 (m+1) X h i U = f 1 Yj+1 − Y 1 + fj Y 1 − Yj t j+ 2 j+ 2 j+ 2 tj <t (m) Subtracting Ut from this gives the thing we need to estimate: (m+1) (m) X Ut − Ut = Rj ; tj <t with Rj = f 1 − fj Yj+1 − Y 1 : j+ 2 j+ 2 The expected square is the sum of diagonal and off diagonal terms. For each off diagonal term with j > k there is an equal term with j < k. A factor of 2 in the off diagonal sum takes that into account. We have 2 (m+1) (m) X 2 X X E Ut − Ut = E Rj + 2 E[ RkRj] : tj <t tj <t tj <tk<t The off diagonal terms E[ RkRj] with tk > tj vanish because h i h i E R R j F = E Y − Y 1 j F f 1 − f R = 0 : k j tj+ 1 k+1 k+ tk+ 1 k+ k j 2 2 2 2 4 The first factor on the right vanishes because Y is a martingale. The diagonal sum is also as it was in the Brownian motion case. 2 2 E Yj+1 − Y 1 f 1 − fj j+ 2 j+ 2 2 2 = E E Yj+1 − Y 1 j F 1 f 1 − fj j+ 2 j+ 2 j+ 2 2 ≤ C∆t E f 1 − fj j+ 2 ≤ C∆t2 : Therefore the diagonal sum is X 2 X E Rj ≤ C∆t ∆t = Ct∆t : tj <t tj <t The Cauchy Schwarz inequality, or Jensen's inequality, then gives s h i 2 (m+1) (m) (m+1) (m) E Ut − Ut ≤ E Ut − Ut p ≤ Ct∆t : This proves the existence of the Ito integral with respect to a martingale diffusion. Now that the Ito integral is defined, what are its properties? If U = eq:imt R t f dX , then dU = f dX .

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