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English Pages [51] Year 2018
Part III — Stochastic Calculus and Applications Based on lectures by R. Bauerschmidt Notes taken by Dexter Chua
Lent 2018 These notes are not endorsed by the lecturers, and I have modified them (often significantly) after lectures. They are nowhere near accurate representations of what was actually lectured, and in particular, all errors are almost surely mine.
– Brownian motion. Existence and sample path properties. – Stochastic calculus for continuous processes. Martingales, local martingales, semimartingales, quadratic variation and cross-variation, Itˆ o’s isometry, definition of the stochastic integral, Kunita–Watanabe theorem, and Itˆ o’s formula. – Applications to Brownian motion and martingales. L´evy characterization of Brownian motion, Dubins–Schwartz theorem, martingale representation, Girsanov theorem, conformal invariance of planar Brownian motion, and Dirichlet problems. – Stochastic differential equations. Strong and weak solutions, notions of existence and uniqueness, Yamada–Watanabe theorem, strong Markov property, and relation to second order partial differential equations.
Pre-requisites Knowledge of measure theoretic probability as taught in Part III Advanced Probability will be assumed, in particular familiarity with discrete-time martingales and Brownian motion.
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Contents
III Stochastic Calculus and Applications
Contents 0 Introduction
3
1 The Lebesgue–Stieltjes integral
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2 Semi-martingales 2.1 Finite variation processes . . 2.2 Local martingale . . . . . . . 2.3 Square integrable martingales 2.4 Quadratic variation . . . . . . 2.5 Covariation . . . . . . . . . . 2.6 Semi-martingale . . . . . . . 3 The 3.1 3.2 3.3 3.4 3.5 3.6 3.7
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stochastic integral Simple processes . . . . . . . . Itˆ o isometry . . . . . . . . . . . Extension to local martingales . Extension to semi-martingales . Itˆ o formula . . . . . . . . . . . The L´evy characterization . . . Girsanov’s theorem . . . . . . .
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25 25 26 28 30 32 35 37
4 Stochastic differential equations 41 4.1 Existence and uniqueness of solutions . . . . . . . . . . . . . . . 41 4.2 Examples of stochastic differential equations . . . . . . . . . . . . 45 4.3 Representations of solutions to PDEs . . . . . . . . . . . . . . . . 47 Index
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Introduction
III Stochastic Calculus and Applications
Introduction
Ordinary differential equations are central in analysis. The simplest class of equations tend to look like x(t) ˙ = F (x(t)). Stochastic differential equations are differential equations where we make the function F “random”. There are many ways of doing so, and the simplest way is to write it as x(t) ˙ = F (x(t)) + η(t), where η is a random function. For example, when modeling noisy physical systems, our physical bodies will be subject to random noise. What should we expect the function η to be like? We might expect that for |t − s| 0, the variables η(t) and η(s) are “essentially” independent. If we are interested in physical systems, then this is a rather reasonable assumption, since random noise is random! In practice, we work with the idealization, where we claim that η(t) and η(s) are independent for t 6= s. Such an η exists, and is known as white noise. However, it is not a function, but just a Schwartz distribution. To understand the simplest case, we set F = 0. We then have the equation x˙ = η. We can write this in integral form as Z x(t) = x(0) +
t
η(s) ds. 0
To make sense of this integral, the function η should at least be a signed measure. Unfortunately, white noise isn’t. This is bad news. We ignore this issue for a little bit, and proceed as if it made sense. If the equation held, then for any 0 = t0 < t1 < · · · , the increments Z ti x(ti ) − x(ti−1 ) = η(s) ds ti−1
should be independent, and moreover their variance should scale linearly with |ti − ti−1 |. So maybe this x should be a Brownian motion! Formalizing these ideas will take up a large portion of the course, and the work isn’t always pleasant. Then why should we be interested in this continuous problem, as opposed to what we obtain when we discretize time? It turns out in some sense the continuous problem is easier. When we learn measure theory, there is a lot of work put into constructing the Lebesgue measure, as opposed to the sum, which we can just define. However, P∞ what we end up is much easier — it’s easier to integrate x13 than to sum n=1 n13 . Similarly, once we have set up the machinery of stochastic calculus, we have a powerful tool to do explicit computations, which is usually harder in the discrete world. Another reason to study stochastic calculus is that a lot of continuous time processes can be described as solutions to stochastic differential equations. Compare this with the fact that functions such as trigonometric and Bessel functions are described as solutions to ordinary differential equations! 3
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There are two ways to approach stochastic calculus, namely via the Itˆ o integral and the Stratonovich integral. We will mostly focus on the Itˆo integral, which is more useful for our purposes. In particular, the Itˆo integral tends to give us martingales, which is useful. To give a flavour of the construction of the Itˆo integral, we consider a simpler scenario of the Wiener integral. Definition (Gaussian space). Let (Ω, F, P) be a probability space. Then a subspace S ⊆ L2 (Ω, F, P) is called a Gaussian space if it is a closed linear subspace and every X ∈ S is a centered Gaussian random variable. An important construction is Proposition. Let H be any separable Hilbert space. Then there is a probability space (Ω, F, P) with a Gaussian subspace S ⊆ L2 (Ω, F, P) and an isometry I : H → S. In other words, for any f ∈ H, there is a corresponding random variable I(f ) ∼ N (0, (f, f )H ). Moreover, I(αf + βg) = αI(f ) + βI(g) and (f, g)H = E[I(f )I(g)]. Proof. By separability, we can pick a Hilbert space basis (ei )∞ i=1 of H. Let (Ω, F, P) be any probability space that carries an infinite independent sequence of standard Gaussian random variables Xi ∼ N (0, 1). Then send ei to Xi , extend by linearity and continuity, and take S to be the image. In particular, we can take H = L2 (R+ ). Definition (Gaussian white noise). A Gaussian white noise on R+ is an isometry W N from L2 (R+ ) into some Gaussian space. For A ⊆ R+ , we write W N (A) = W N (1A ). Proposition. – For A ⊆ R+ with |A| < ∞, W N (A) ∼ N (0, |A|). – For disjoint A, B ⊆ R+ , the variables W N (A) and W N (B) are independent. S∞ – If A = i=1 Ai for disjoint sets Ai ⊆ R+ , with |A| < ∞, |Ai | < ∞, then W N (A) =
∞ X
W N (Ai ) in L2 and a.s.
i=1
Proof. Only the last point requires proof. Observe that the partial sum Mn =
n X
W N (A)
i=1
is a martingale, and is bounded in L2 as well, since EMn2
=
n X
2
EW N (Ai ) =
i=1
n X
|Ai | ≤ |A|.
i=1
So we are done by theP martingale convergence theorem. The limit is indeed ∞ W N (A) because 1A = n=1 1Ai . 4
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The point of the proposition is that W N really looks like a random measure on R+ , except it is not. We only have convergence almost surely above, which means we have convergence on a set of measure 1. However, the set depends on which A and Ai we pick. For things to actually work out well, we must have a fixed set of measure 1 for which convergence holds for all A and Ai . But perhaps we can ignore this problem, and try to proceed. We define Bt = W N ([0, t]) for t ≥ 0. Exercise. This Bt is a standard Brownian motion, except for the continuity requirement. In other words, for any t1 , t2 , . . . , tn , the vector (Bti )ni=1 is jointly Gaussian with E[Bs Bt ] = s ∧ t for s, t ≥ 0. Moreover, B0 = 0 a.s. and Bt − Bs is independent of σ(Br : r ≤ s). Moreover, Bt − Bs ∼ N (0, t − s) for t ≥ s. In fact, by picking a good basis of L2 (R+ ), we can make Bt continuous. We can now try to define some stochastic integral. If f ∈ L2 (R+ ) is a step function, n X f= fi 1[si ,ti ] i=1
with si < ti , then W N (f ) =
n X
fi (Bti − Bsi )
i=1
This motivates the notation Z W N (f ) =
f (s) dBS .
However, extending this to a function that is not a step function would be problematic.
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III Stochastic Calculus and Applications
The Lebesgue–Stieltjes integral
R1 In calculus, we are able to perform integrals more exciting than simply 0 h(x) dx. In particular, if h, a : [0, 1] → R are C 1 functions, we can perform integrals of the form Z 1
h(x) da(x). 0
For them, it is easy to make sense of what this means — it’s simply Z 1 Z 1 h(x) da = h(x)a0 (x) dx. 0
0
In our world, we wouldn’t expect our functions to be differentiable, so this is not a great definition. One reasonable strategy to make sense of this is to come up with a measure that should equal “da”. An immediate difficulty we encounter is that a0 (x) need not be positive all the R1 time. So for example, 0 1 da could be a negative number, which one wouldn’t expect for a usual measure! Thus, we are naturally lead to think about signed measures. From now on, we always use the Borel σ-algebra on [0, T ] unless otherwise specified. Definition (Signed measure). A signed measure on [0, T ] is a difference µ = µ+ −µ− of two positive measures on [0, T ] of disjoint support. The decomposition µ = µ+ − µ− is called the Hahn decomposition. In general, given two measures µ1 and µ2 with not necessarily disjoint supports, we may still want to talk about µ1 − µ2 . Theorem. For any two finite measures µ1 , µ2 , there is a signed measure µ with µ(A) = µ1 (A) − µ2 (A). If µ1 and µ2 are given by densities f1 , f2 , then we can simply decompose µ as (f1 − f2 )+ dt + (f1 − f2 )− dt, where + and − denote the positive and negative parts respectively. In general, they need not be given by densities with respect to dx, but they are always given by densities with respect to some other measure. Proof. Let ν = µ1 + µ2 . By Radon–Nikodym, there are positive functions f1 , f2 such that µi (dt) = fi (t)ν(dt). Then (µ1 − µ2 )(dt) = (f1 − f2 )+ (t) · ν(dt) + (f1 − f2 )− (t) · ν(dt). Definition (Total variation). The total variation of a signed measure µ = µ+ − µ− is |µ| = µ+ + µ− . We now want to figure out how we can go from a function to a signed measure. R1 Let’s think about how one would attempt to define 0 f (x) dg as a Riemann sum. A natural option would be to write something like Z
t
h(s) da(s) = lim 0
m→∞
nm X
(m) (m) (m) h(ti−1 ) a(ti ) − a(ti−1 )
i=1
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(m)
(m)
for any sequence of subdivisions 0 = t0 < · · · < tnm = t of [0, t] with (m) (m) maxi |ti − ti−1 | → 0. In particular, since we want the integral of h = 1 to be well-behaved, the P (m) (m) sum (a(ti ) − a(ti−1 )) must be well-behaved. This leads to the notion of Definition (Total variation). The total variation of a function a : [0, T ] → R is ( n ) X Va (t) = |a(0)| + sup |a(ti ) − a(ti−1 )| : 0 = t0 < t1 < · · · < tn = T . i=1
We say a has bounded variation if Va (T ) < ∞. In this case, we write a ∈ BV . We include the |a(0)| term because we want to pretend a is defined on all of R with a(t) = 0 for t < 0. We also define adl` ag if it is right-continuous Definition (C`adl`ag). A function a : [0, T ] → R is c` and has left-limits. The following theorem is then clear: Theorem. There is a bijection n
o
signed measures on [0, T ]
←→
c`adl`ag functions of bounded variation a : [0, T ] → R
that sends a signed measure µ to a(t) = µ([0, t]). To construct the inverse, given a, we define 1 a± = (Va ± a). 2 Then a± are both positive, and a = a+ − a− . We can then define µ± by µ± [0, t] = a± (t) − a± (0) µ = µ+ − µ− Moreover, Vµ[0,t] = |µ|[0, t]. Example. Let a : [0, 1] → R be given by ( 1 t< a(t) = 0 t≥
1 2 1 2
.
This is c`adl`ag, and it’s total variation is v0 (1) = 2. The associated signed measure is µ = δ0 − δ1/2 , and the total variation measure is |µ| = δ0 + δ1/2 . We are now ready to define the Lebesgue–Stieltjes integral.
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Definition (Lebesgue–Stieltjes integral). Let a : [0, T ] → R be c`adl`ag of bounded variation and let µ be the associated signed measure. Then for h ∈ L1 ([0, T ], |µ|), the Lebesgue–Stieltjes integral is defined by Z t Z h(r) da(r) = h(r)µ(dr), s
(s,t]
where 0 ≤ s ≤ t ≤ T , and Z t
Z h(r) |da(r)| =
h(r)|µ|(dr).
s
(s,t]
We also write
t
Z h · a(t) =
h(r) da(r). 0
To let T = ∞, we need the following notation: Definition (Finite variation). A c`adl`ag function a : [0, ∞) → R is of finite variation if a|[0,T ] ∈ BV [0, 1] for all T > 0. Fact. Let a : [0, T ] → R be c` adl`ag and BV, and h ∈ L1 ([0, T ], |da|), then Z Z T b h(s) da(s) ≤ |h(s)| |da(s)|, 0 a and the function h · a : [0, T ] → R is c`adl`ag and BV with associated signed measure h(s) da(s). Moreover, |h(s) da(s)| = |h(s)| |da(s)|. We can, unsurprisingly, characterize the Lebesgue–Stieltjes integral by a Riemann sum: ag and BV on [0, t], and h bounded and leftProposition. Let a be c`adl` continuous. Then Z t nm X (m) (m) (m) h(s) da(s) = lim h(ti−1 ) a(ti ) − a(ti−1 ) m→∞
0
Z
t
h(s) |da(s)| = lim
m→∞
0
i=1 nm X
(m) (m) (m) h(ti−1 ) a(ti ) − a(ti−1 )
i=1 (m)
(m)
< · · · < tnm = t of [0, t] with
for any sequence of subdivisions 0 = t0 (m) (m) maxi |ti − ti−1 | → 0.
Proof. We approximate h by hm defined by hm (0) = 0,
(m)
(m)
(m)
hm (s) = h(ti−1 ) for s ∈ (ti−1 , ti
].
Then by left continuity, we have h(s) = lim hm (s) n→∞
by left continuity, and moreover lim
m→∞
nm X i=1
(m)
(m)
h(ti−1 )(a(ti
Z
(m)
) − a(ti−1 )) = lim
m→∞
Z hm (s)µ( ds) =
(0,t]
h(s)µ(ds) (0,t]
by dominated convergence theorem. The statement about |da(s)| is left as an exercise.
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III Stochastic Calculus and Applications
Semi-martingales
The title of the chapter is “semi-martingales”, but we are not going even meet the definition of a semi-martingale till the end of the chapter. The reason is that a semi-martingale is essentially defined to be the sum of a (local) martingale and a finite variation process, and understanding semi-martingales mostly involves understanding the two parts separately. Thus, for most of the chapter, we will be studying local martingales (finite variation processes are rather more boring), and at the end we will put them together to say a word or two about semi-martingales. From now on, (Ω, F, (Ft )t≥0 , P) will be a filtered probability space. Recall the following definition: Definition (C`adl`ag adapted process). A c` adl` ag adapted process is a map X : Ω × [0, ∞) → R such that (i) X is c` adl` ag, i.e. X(ω, · ) : [0, ∞) → R is c`adl`ag for all ω ∈ Ω. (ii) X is adapted, i.e. Xt = X( · , t) is Ft -measurable for every t ≥ 0. Notation. We will write X ∈ G to denote that a random variable X is measurable with respect to a σ-algebra G.
2.1
Finite variation processes
The definition of a finite variation function extends immediately to a finite variation process. Definition (Finite variation process). A finite variation process is a c` adl`ag adapted process A such that A(ω, · ) : [0, ∞) → R has finite variation for all ω ∈ Ω. The total variation process V of a finite variation process A is Z T Vt = |dAs |. 0
adl`ag adapted process A is Proposition. The total variation process V of a c` also c` adl` ag, finite variation and adapted, and it is also increasing. Proof. We only have to check that it is adapted. But that follows directly from our previous expression of the integral as the limit of a sum. Indeed, let (m) (m) 0 = t0 < t1 < · · · < tnm = t be a (nested) sequence of subdivisions of [0, t] (m) (m) with maxi |ti − ti−1 | → 0. We have seen Vt = lim
nm X
m→∞
i=1
|At(m) − At(m) | + |A(0)| ∈ Ft . i
i−1
Definition ((H · A)t ). Let A be a finite variation process and H a process such that for all ω ∈ Ω and t ≥ 0, Z t Hs (ω)| |dAs (ω)| < ∞. 0
Then define a process ((H · A)t )t≥0 by Z (H · A)t =
t
Hs dAs . 0
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For the process H · A to be adapted, we need a condition. Definition (Previsible process). A process H : Ω × [0, ∞) → R is previsible if it is measurable with respect to the previsible σ-algebra P generated by the sets E × (s, t], where E ∈ Fs and s < t. We call the generating set Π. Very roughly, the idea is that a previsible event is one where whenever it happens, you know it a finite (though possibly arbitrarily small) before. Definition (Simple process). A process H : Ω × [0, ∞) → R is simple, written H ∈ E, if n X H(ω, t) = Hi−1 (ω)1(ti−1 ,ti ] (t) i=1
for random variables Hi−1 ∈ Fi−1 and 0 = t0 < · · · < tn . Fact. Simple processes and their limits are previsible. Fact. Let X be a c`adl`ag adapted process. Then Ht = Xt− defines a leftcontinuous process and is previsible. In particular, continuous processes are previsible. Proof. Since X is c`adl`ag adapted, it is clear that H is left-continuous and adapted. Since H is left-continuous, it is approximated by simple processes. Indeed, let 2n X n Ht = H(i−1)2−n 1((i−1)2−n ,i2−n ] (t) ∧ n ∈ E. i=1
Then
Htn
→ H for all t by left continuity, and previsibility follows.
Exercise. Let H be previsible. Then Ht ∈ Ft− = σ(Fs : s < t). Example. Brownian motion is previsible (since it is continuous). Example. A Poisson process (Nt ) is not previsible since Nt 6∈ Ft− . Proposition. Let A be a finite variation process, and H previsible such that Z t |H(ω, s)| |dA(ω, s)| < ∞ for all (ω, t) ∈ Ω × [0, ∞). 0
Then H · A is a finite variation process. Proof. The finite variation and c`adl`ag parts follow directly from the deterministic versions. We only have to check that H · A is adapted, i.e. (H · A)( · , t) ∈ Ft for all t ≥ 0. First, H ·A is adapted if H(ω, s) = 1(u,v] (s)1E (ω) for some u < v and E ∈ Fu , since (H · A)(ω, t) = 1E (ω)(A(ω, t ∧ v) − A(ω, t ∧ u)) ∈ Ft . Thus, H · A is adapted for H = 1F when F ∈ Π. Clearly, Π is a π system, i.e. it is closed under intersections and non-empty, and by definition it generates the 10
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previsible σ-algebra P. So to extend the adaptedness of H · A to all previsible H, we use the monotone class theorem. We let V = {H : Ω × [0, ∞) → R : H · A is adapted}. Then (i) 1 ∈ V (ii) 1F ∈ V for all F ∈ Π. (iii) V is closed under monotone limits. So V contains all bounded P-measurable functions. So the conclusion is that if A is a finite variation process, then as long as reasonable finiteness conditions are satisfied, we can integrate functions against dA. Moreover, this integral was easy to define, and it obeys all expected properties such as dominated convergence, since ultimately, it is just an integral in the usual measure-theoretic sense. This crucially depends on the fact that A is a finite variation process. However, in our motivating example, we wanted to take A to be Brownian motion, which is not of finite variation. The work we will do in this chapter and the next is to come up with a stochastic integral where we let A be a martingale instead. The heuristic idea is that while martingales can vary wildly, the martingale property implies there will be some large cancellation between the up and down movements, which leads to the possibility of a well-defined stochastic integral.
2.2
Local martingale
From now on, we assume that (Ω, F, (Ft )t , P) satisfies the usual conditions, namely that (i) F0 contains all P-null sets (ii) (Ft )t is right-continuous, i.e. Ft = (Ft+ =
T
s>t
Fs for all t ≥ 0.
We recall some of the properties of continuous martingales. Theorem (Optional stopping theorem). Let X be a c`adl`ag adapted integrable process. Then the following are equivalent: (i) X is a martingale, i.e. Xt ∈ L1 for every t, and E(Xt | Fs ) = Xs for all t > s. (ii) The stopped process X T = (XtT ) = (XT ∧t ) is a martingale for all stopping times T . (iii) For all stopping times T, S with T bounded, XT ∈ L1 and E(XT | FS ) = XT ∧S almost surely. (iv) For all bounded stopping times T , XT ∈ L1 and E(XT ) = E(X0 ).
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For X uniformly integrable, (iii) and (iv) hold for all stopping times. In practice, most of our results will be first proven for bounded martingales, or perhaps square integrable ones. The point is that the square-integrable martingales form a Hilbert space, and Hilbert space techniques can help us say something useful about these martingales. To get something about a general martingale M , we can apply a cutoff Tn = inf{t > 0 : Mt ≥ n}, and then M Tn will be a martingale for all n. We can then take the limit n → ∞ to recover something about the martingale itself. But if we are doing this, we might as well weaken the martingale condition a bit — we only need the M Tn to be martingales. Of course, we aren’t doing this just for fun. In general, martingales will not always be closed under the operations we are interested in, but local (or maybe semi-) martingales will be. In general, we define adl`ag adapted process X is a local martingale Definition (Local martingale). A c` if there exists a sequence of stopping times Tn such that Tn → ∞ almost surely, and X Tn is a martingale for every n. We say the sequence Tn reduces X. Example. (i) Every martingale is a local martingale, since by the optional stopping theorem, we can take Tn = n. (ii) Let (Bt ) to be a standard 3d Brownian motion on R3 . Then 1 (Xt )t≥1 = |Bt | t≥1 is a local martingale but not a martingale. To see this, first note that sup EXt2 < ∞,
EXt → 0.
t≥1
Since EXt → 0 and Xt ≥ 0, we know X cannot be a martingale. However, we can check that it is a local martingale. Recall that for any f ∈ Cb2 , M f = f (Bt ) − f (B1 ) −
1 2
Z
t
∆f (Bs ) ds 0
1 1 is a martingale. Moreover, ∆ |x| = 0 for all x 6= 0. Thus, if |x| didn’t have a singularity at 0, this would have told us Xt is a martingale. Thus, we are safe if we try to bound |Bs | away from zero.
Let
1 Tn = inf t ≥ 1 : |Bt | < n
and pick fn ∈ Cb2 such that fn (x) =
1 |x|
for |x| ≥
, 1 n.
Then XtT − X1Tn =
fn Mt∧T . So X Tn is a martingale. n
It remains to show that Tn → ∞, and this follows from the fact that EXt → 0. 12
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Proposition. Let X be a local martingale and Xt ≥ 0 for all t. Then X is a supermartingale. Proof. Let (Tn ) be a reducing sequence for X. Then E(Xt | Fs ) = E lim inf Xt∧Tn | Fs n→∞
≤ lim E(Xt∧Tn | Fs ) n→∞
= lim inf Xs∧Tn Tn →∞
= Xs . Recall the following result from Advanced Probability: Proposition. Let X ∈ L1 (Ω, F, P). Then the set χ = {E(X | G) : G ⊆ F a sub-σ-algebra} is uniformly integrable, i.e. sup E(|Y |1|Y |>λ ) → 0 as λ → ∞.
Y ∈χ
Recall also the following important result about uniformly integrable random variables: Theorem (Vitali theorem). Xn → X in L1 iff (Xn ) is uniformly integrable and Xn → X in probability. With these, we can state the following characterization of martingales in terms of local martingales: Proposition. The following are equivalent: (i) X is a martingale. (ii) X is a local martingale, and for all t ≥ 0, the set χt = {XT : T is a stopping time with T ≤ t} is uniformly integrable. Proof. – (a) ⇒ (b): Let X be a martingale. Then by the optional stopping theorem, XT = E(Xt | FT ) for any bounded stopping time T ≤ t. So χt is uniformly integrable. – (b) ⇒ (a): Let X be a local martingale with reducing sequence (Tn ), and assume that the sets χt are uniformly integrable for all t ≥ 0. By the optional stopping theorem, it suffices to show that E(XT ) = E(X0 ) for any bounded stopping time T . So let T be a bounded stopping time, say T ≤ t. Then E(X0 ) = E(X0Tn ) = E(XTTn ) = E(XT ∧Tn ) 13
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for all n. Now T ∧ Tn is a stopping time ≤ t, so {XT ∧Tn } is uniformly integrable by assumption. Moreover, Tn ∧ T → T almost surely as n → ∞, hence XT ∧Tn → XT in probability. Hence by Vitali, this converges in L1 . So E(XT ) = E(X0 ). Corollary. If Z ∈ L1 is such that |Xt | ≤ Z for all t, then X is a martingale. In particular, every bounded local martingale is a martingale. The definition of a local martingale does not give us control over what the reducing sequence {Tn } is. In particular, it is not necessarily true that X Tn will be bounded, which is a helpful property to have. Fortunately, we have the following proposition: Proposition. Let X be a continuous local martingale with X0 = 0. Define Sn = inf{t ≥ 0 : |Xt | = n}. Then Sn is a stopping time, Sn → ∞ and X Sn is a bounded martingale. In particular, (Sn ) reduces X. Proof. It is clear that Sn is a stopping time, since (if it is not clear) \ \ [ 1 1 {Sn ≤ t} = sup |Xs | > n − = |Xs | > n − ∈ Ft . k k s≤t k∈N s ε
→ 0 as n → ∞ for all t > 0, ε > 0.
s∈[0,t]
17
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Theorem. Let M be a continuous local martingale with M0 = 0. Then there exists a unique (up to indistinguishability) continuous adapted increasing process (hM it )t≥0 such that hM i0 = 0 and Mt2 − hM it is a continuous local martingale. Moreover, d2n te
hM it = lim
n→∞
(n) hM it ,
(n) hM it
=
X
(Mt2−n − M(i−1)2−n )2 ,
i=1
where the limit u.c.p. Definition (Quadratic variation). hM i is called the quadratic variation of M . It is probably more useful to understand hM it in terms of the explicit formula, and the fact that Mt2 − hM it is a continuous local martingale is a convenient property. Example. Let B be a standard Brownian motion. Then Bt2 − t is a martingale. Thus, hBit = t. The proof is long and mechanical, but not hard. All the magic happened when we used the magical Doob’s inequality to show that M2c and M2 are Hilbert spaces. Proof. To show uniqueness, we use that finite variation and local martingale are incompatible. Suppose (At ) and (A˜t ) obey the conditions for hM i. Then At − A˜t = (Mt2 − A˜t )−(Mt2 −At ) is a continuous adapted local martingale starting at 0. Moreover, both At and A˜t are increasing, hence have finite variation. So A − A˜ = 0 almost surely. To show existence, we need to show that the limit exists and has the right property. We do this in steps. Claim. The result holds if M is in fact bounded. Suppose |M (ω, t)| ≤ C for all (ω, t). Then M ∈ M2c . Fix T > 0 deterministic. Let
d2n T e
Xtn
X
=
M(i−1)2−n (Mi2−n ∧t − M(i−1)2−n ∧t ).
i=1
This is defined so that (n)
2 n hM ik2−n = Mk2 −n − 2Xk2−n . n This reduces the study of hM i(n) to that of Xk2 −n . n We check that (Xt ) is a Cauchy sequence in M2c . The fact that it is a martingale is an immediate computation. To show it is Cauchy, for n ≥ m, we calculate d2n T e n X∞
−
m X∞
=
X
(M(i−1)2−n − Mb(i−1)2m−n c2−m )(Mi2−n − M(i−1)2−n ).
i=1
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III Stochastic Calculus and Applications
We now take the expectation of the square to get n d2 T e X n m 2 E(X∞ − X∞ ) = E (M(i−1)2−n − Mb(i−1)2m−nc2−m )2 (Mi2−n − M(i−1)2−n )2 i=1
≤ E
d2n T e
|Mt − Ms |2
sup
|s−t|≤2−m
X
(Mi2−n − M(i−1)2−n )2
i=1
! =E
sup
|Mt − Ms |
2
|Mt − Ms |
4
|s−t|≤2−m
(n) hM iT
!1/2 ≤E
sup |s−t|≤2−m
1/2 (n) E (hM iT )2 (Cauchy–Schwarz)
We shall show that the second factor is bounded, while the first factor tends to zero as m → ∞. These are both not surprising — the first term vanishing in the limit corresponds to M being continuous, and the second term is bounded since M itself is bounded. To show that the first term tends to zero, we note that we have |Mt − Ms |4 ≤ 16C 4 , and moreover |Mt − Ms | → 0 as m → ∞ by uniform continuity.
sup |s−t|≤2−m
So we are done by the dominated convergence theorem. To show the second term is bounded, we do (writing N = d2n T e) !2 N X (n) E (hM iT )2 = E (Mi2−n − M(i−1)2−n )2 i=1
=
N X
E (Mi2−n − M(i−1)2−n )4
i=1
+2
N X
E (Mi2−n − M(i−1)2−n )2
i=1
N X
! (Mk2−n − M(k−1)2−n )2
k=i+1
We use the martingale property and orthogonal increments the rearrange the off-diagonal term as E (Mi2−n − M(i−1)2−n )(MN 2−n − Mi2−n )2 . Taking some sups, we get E
(n) (hM iT )2
N X ≤ 12C 2 E (Mi2−n − M(i−1)2−n )2 i=1
= 12C 2 E (MN 2−n − M0 )2 ≤ 12C 2 · 4C 2 . 19
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2
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III Stochastic Calculus and Applications
So done. So we now have X n → X in Mc2 for some X ∈ Mc2 . In particular, we have
sup |Xtn − Xt | → 0
t
L2
So we know that sup |Xtn − Xt | → 0 t
almost surely along a subsequence Λ. Let N ⊆ Ω be the events on which this convergence fails. We define ( Mt2 − 2Xt ω ∈ Ω \ N (T ) At = . 0 ω∈N (T )
2 Then A(T ) is continuous, adapted since M and X are, and (Mt∧T − At∧T )t is a (T ) 2 martingale since X is. Finally, A is increasing since Mt − Xtn is increasing on 2−n Z ∩ [0, T ] and the limit is uniform. So this A(T ) basically satisfies all the properties we want hM it to satisfy, except we have the stopping time T . (T ) (T +1) We next observe that for any T ≥ 1, At∧T = At∧T for all t almost surely. This essentially follows from the same uniqueness argument as we had at the beginning of the proof. Thus, there is a process (hM it )t≥0 such that (T )
hM it = At
for all t ∈ [0, T ] and T ∈ N, almost surely. Then this is the desired process. So we have constructed hM i in the case where M is bounded. Claim. hM i(n) → hM i u.c.p. Recall that (n)
hM it
= M22−n b2n tc − 2X2n−n b2n tc .
We also know that sup |Xtn − Xt | → 0
t≤T
in L2 , hence also in probability. So we have (n)
|hM it − hM it | ≤ sup |M22−n b2n tc − Mt2 | t≤T
+ sup |X2n−n b2n tc − X2−n b2n tc | + sup |X2−n b2n tc − Xt |. t≤T
t≤T
The first and last terms → 0 in probability since M and X are uniformly continuous on [0, T ]. The second term converges to zero by our previous assertion. So we are done. Claim. The theorem holds for M any continuous local martingale. We let Tn = inf{t ≥ 0 : |Mt | ≥ n}. Then (Tn ) reduces M and M Tn is a bounded martingale. So in particular M Tn is a bounded continuous martingale. We set An = hM Tn i. 20
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Then (Ant ) and (An+1 t∧Tn ) are indistinguishable for t < Tn by the uniqueness argument. Thus there is a process hM i such that hM it∧Tn = Ant are indistinguishable 2 for all n. Clearly, hM i is increasing since the An are, and Mt∧T − hM it∧Tn is a n martingale for every n, so Mt2 − hM it is a continuous local martingale. Claim. hM i(n) → hM i u.c.p. We have seen hM Tk i(n) → hM Tk i u.c.p. for every k. So (n) Tk (n) Tk P sup |hM it − hMt i| > ε ≤ P(Tk < T ) + P sup |hM it − hM it > ε . t≤T
t≤T
So we can fisrt pick k large enough such that the first term is small, then pick n large enough so that the second is small. There are a few easy consequence of this theorem. Fact. Let M be a continuous local martingale, and let T be a stopping time. Then alsmot surely for all t ≥ 0, hM T it = hM it∧T 2 Proof. Since Mt2 − hM it is a continuous local martingle, so is Mt∧T − hM it∧T = T 2 (M )t − hM it∧T . So we are done by uniqueness.
Fact. Let M be a continuous local martingale with M0 = 0. Then M = 0 iff hM i = 0. Proof. If M = 0, then hM i = 0. Conversely, if hM i = 0, then M 2 is a continuous local martingale and positive. Thus EMt2 ≤ EM02 = 0. Proposition. Let M ∈ M2c . Then M 2 − hM i is a uniformly integrable martingale, and kM − M0 kM2 = (EhM i∞ )1/2 . Proof. We will show that hM i∞ ∈ L1 . This then implies |Mt2 − hM it | ≤ sup Mt2 + hM i∞ . t≥0
Then the right hand side is in L1 . Since M 2 − hM i is a local martingale, this implies that it is in fact a uniformly integrable martingale. To show hM i∞ ∈ L1 , we let Sn = inf{t ≥ 0 : hM it ≥ n}. Then Sn → ∞, Sn is a stopping time and moreover hM it∧Sn ≤ n. So we have 2 Mt∧S − hM it∧Sn ≤ n + sup Mt2 , n t≥0
2 and the second term is in L1 . So Mt∧S − hM it∧Sn is a true martingale. n
21
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III Stochastic Calculus and Applications
So 2 EMt∧S − EM02 = EhM it∧Sn . n 2 Taking the limit t → ∞, we know EMt∧S → EMS2n by dominated convergence. n Since hM it∧Sn is increasing, we also have EhM it∧Sn → EhM iSn by monotone convergence. We can take n → ∞, and by the same justification, we have 2 EhM i ≤ EM∞ − EM02 = E(M∞ − M0 )2 < ∞.
2.5
Covariation
We know M2c not only has a norm, but also an inner product. This can also be reflected in the bracket by the polarization identity, and it is natural to define Definition (Covariation). Let M, N be two continuous local martingales. Define the covariation (or simply the bracket) between M and N to be process 1 (hM + N it − hM − N it ). 4
hM, N it =
Then if in fact M, N ∈ M2c , then putting t = ∞ gives the inner product. Proposition. (i) hM, N i is the unique (up to indistinguishability) finite variation process such that Mt Nt − hM, N it is a continuous local martingale. (ii) The mapping (M, N ) 7→ hM, N i is bilinear and symmetric. (iii) (n)
hM, N it = lim hM, N it n→∞
u.c.p.
n
d2 te (n) hM, N it
=
X
(Mi2−n −M(i−1)2−n )(Ni2−n − N(i−1) 2−n ).
i=1
(iv) For every stopping time T , hM T , N T it = hM T , N it = hM, N it∧T . (v) If M, N ∈ M2c , then Mt Nt − hM, N it is a uniformly integrable martingale, and hM − M0 , N − N0 iM2 = EhM, N i∞ . Example. Let B, B 0 be two independent Brownian motions (with respect to the same filtration). Then hB, B 0 i = 0. Proof. Assume B0 = B00 = 0. Then X± = √12 (B ± B 0 ) are Brownian motions, and so hX± i = t. So their difference vanishes. An important result about the covariation is the following Cauchy–Schwarz like inequality:
22
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III Stochastic Calculus and Applications
Proposition (Kunita–Watanabe). Let M, N be continuous local martingales and let H, K be two (previsible) processes. Then almost surely Z
∞
Z
∞
|Hs ||Ks ||dhM, N is | ≤ 0
Hs2 dhM is
1/2 Z
0
∞
Hs2 hN is
1/2 .
0
In fact, this is Cauchy–Schwarz. All we have to do is to take approximations and take limits and make sure everything works out well. Proof. For convenience, we write hM, N its = hM, N it − hM, N is . Claim. For all 0 ≤ s ≤ t, we have |hM, N its | ≤
p
hM, M its
p
hN, N its .
By continuity, we can assume that s, t are dyadic rationals. Then n 2 t X t |hM, N is | = lim (Mi2−n − M(i−1)2−n )(Ni2−n − N(i−1)2−n ) n→∞ n i=2 s+1
1/2 2n t X 2 ≤ lim (Mi2−n − M(i−1)2−n ) × n→∞ n i=2 s+1
1/2 2n t X (Ni2−n − N(i−1)2−n )2 n i=2 s+1 1/2 1/2 = hM, M its hN, N its ,
(Cauchy–Schwarz)
where all equalities are u.c.p. Claim. For all 0 ≤ s < t, we have Z t p p |dhM, N iu | ≤ hM, M its hN, N its . s
Indeed, for any subdivision s = t0 < t1 < · · · tn = t, we have n X
|hM, N ittii−1 | ≤
i=1
n q q X hM, M ittii−1 hN, N ittii−1 i=1
≤
n X
!1/2 hM, M ittii−1
i=1
n X hN, N ittii−1
!1/2 .
i=1
(Cauchy–Schwarz) Taking the supremum over all subdivisions, the claim follows. Claim. For all bounded Borel sets B ⊆ [0, ∞), we have sZ sZ Z |dhM, N iu | ≤ B
dhM iu B
23
dhN iu . B
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Semi-martingales
III Stochastic Calculus and Applications
We already know this is true if B is an interval. If B is a finite union of integrals, then we apply Cauchy–Schwarz. By a monotone class argument, we can extend to all Borel sets. Claim. The theorem holds for H=
k X
h` 1B` ,
K=
`=1
n X
k` 1B`
`=1
for B` ⊆ [0, ∞) bounded Borel sets with disjoint support. We have Z Z n X |Hs Ks | |dhM, N is | ≤ |h` k` | ≤
`=1 n X
¯ , N is | |dM
B`
|h` k` |
dhM is
n X
h2`
`=1
B`
!1/2
Z dhM is B`
1/2 dhN is
B`
`=1
≤
1/2 Z
Z
n X
k`2
`=1
!1/2
Z dhN is B`
To finish the proof, approximate general H and K by step functions and take the limit.
2.6
Semi-martingale
Definition (Semi-martingale). A (continuous) adapted process X is a (continuous) semi-martingale if X = X0 + M + A, where X0 ∈ F0 , M is a continuous local martingale with M0 = 0, and A is a continuous finite variation process with A0 = 0. This decomposition is unique up to indistinguishables. Definition (Quadratic variation). Let X = X0 + M + A and X 0 = X00 + M 0 + A0 be (continuous) semi-martingales. Set hXi = hM i,
hX, X 0 i = hM, M 0 i.
This definition makes sense, because continuous finite variation processes do not have quadratic variation. Exercise. We have d2n te
hX, Y
(n) it
=
X
(Xi2−n − X(i−1)2−n )(Yi2−n − Y(i−1)2−n ) → hX, Y i u.c.p.
i=1
24
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The stochastic integral
3 3.1
III Stochastic Calculus and Applications
The stochastic integral Simple processes
We now have all the background required to define the stochastic integral, and we can start constructing it. As in the case of the Lebesgue integral, we first define it for simple processes, and then extend to general processes by taking a limit. Recall that we have Definition (Simple process). The space of simple processes E consists of functions H : Ω × [0, ∞) → R that can be written as Ht (ω) =
n X
Hi−1 (ω)1(ti−1 ,ti ] (t)
i=1
for some 0 ≤ t0 ≤ t1 ≤ · · · ≤ tn and bounded random variables Hi ∈ Fti . Definition (H · M ). For M ∈ M2 and H ∈ E, we set Z
t
H dM = (H · M )t = 0
n X
Hi−1 (Mti ∧t − Mti−1 ∧t ).
i=1
If M were of finite variation, then this is the same as what we have previously seen. Recall that for the Lebesgue integral, extending this definition to general functions required results like monotone convergence. Here we need some similar results that put bounds on how large the integral can be. In fact, we get something better than a bound. Proposition. If M ∈ M2c and H ∈ E, then H · M ∈ M2c and Z ∞ kH · M k2M2 = E Hs2 dhM is .
(∗)
0
Proof. We first show that H · M ∈ M2c . By linearity, we only have to check it for Xti = Hi−1 (Mti ∧t − Mti−1 ∧t ) We have to check that E(Xti | Fs ) = 0 for all t > s, and the only non-trivial case is when t > ti−1 . E(Xti | Fs ) = Hi−1 E(Mti ∧t − Mti−1 ∧t | Fs ) = 0. We also check that kX i kM2 ≤ 2kHk∞ kM kM2 . So it is bounded. So H · M ∈ M2c . To prove (∗), we note that the X i are orthogonal and that 2 hX i it = Hi−1 (hM iti ∧t − hM iti−1 ∧t ).
So we have Z t X X i i 2 hH ·M, H ·M i = hX , X i = Hi−1 (hM iti ∧t −hM iti−1 ∧t ) = Hs2 dhM is . 0
25
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The stochastic integral
III Stochastic Calculus and Applications
In particular, kH · M k2M2 = EhH · M i∞ = E
Z
∞
Hs2 dhM is .
0
Proposition. Let M ∈
M2c
and H ∈ E. Then
hH · M, N i = H · hM, N i for all N ∈ M2 . In other words, the stochastic integral commutes with the bracket. P P Proof. Write H · M = X i = Hi−1 (Mti ∧t − Mti−1 ∧t ) as before. Then hX i , N it = Hi−1 hMti ∧t − Mti−1 ∧t , N i = Hi−1 (hM, N iti ∧t − hM, N iti−1 ∧t ).
3.2
Itˆ o isometry
We now try to extend the above definition to something more general than simple processes. Definition (L2 (M )). Let M ∈ M2c . Define L2 (M ) to be the space of (equivalence classes of) previsible H : Ω × [0, ∞) → R such that Z ∞ 1/2 2 kHkL2 (M ) = kHkM = E Hs dhM is < ∞. 0 2
For H, K ∈ L (M ), we set Z (H, K)L2 (M ) = E
∞
Hs Ks dhM is .
0 2
2
In fact, L (M ) is equal to L (Ω × [0, ∞), P, dP dhM i), where P is the previsible σ-algebra, and in particular is a Hilbert space. Proposition. Let M ∈ M2c . Then E is dense in L2 (M ). Proof. Since L2 (M ) is a Hilbert space, it suffices to show that if (K, H) = 0 for all H ∈ E, then K = 0. So assume that (K, H) = 0 for all H ∈ E and Z t Xt = Ks dhM is , 0
Then X is a well-defined finite variation process, and Xt ∈ L1 for all t. It suffices to show that Xt = 0 for all t, and we shall show that Xt is a continuous martingale. Let 0 ≤ s < t and F ∈ Fs bounded. We let H = F 1(s,t] ∈ E. By assumption, we know Z t 0 = (K, H) = E F Ku dhM iu = E(F (Xt − XS )). s
Since this holds for all Fs measurable F , we have shown that E(Xt | Fs ) = Xs . So X is a (continuous) martingale, and we are done. 26
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The stochastic integral
III Stochastic Calculus and Applications
Theorem. Let M ∈ M2c . Then (i) The map H ∈ E 7→ H ·M ∈ M2c extends uniquely to an isometry L2 (M ) → M2c , called the Itˆ o isometry. (ii) For H ∈ L2 (M ), H · M is the unique martingale in M2c such that hH · M, N i = H · hM, N i for all N ∈ M2c , where the integral on the LHS is the stochastic integral (as above) and the RHS is the finite variation integral. (iii) If T is a stopping time, then (1[0,T ] H) · M = (H · M )T = H · M T . Definition (Stochastic integral). H · M is the stochastic integral of H with respect to M and we also write Z t (H · M )t = Hs dMs . 0
It is important that the integral of martingale is still a martingale. After proving Itˆo’s formula, we will use this fact to show that a lot of things are in fact martingales in a rather systematic manner. For example, it will be rather effortless to show that Bt2 − t is a martingale when Bt is a standard Brownian motion. Proof. (i) We have already shown that this map is an isometry when restricted to E. So extend by completeness of M2c and denseness of E. (ii) Again the equation to show is known for simple H, and we want to show it is preserved under taking limits. Suppose H n → H in L2 (M ) with H n ∈ L2 (M ). Then H n · M → H · M in M2c . We want to show that hH · M, N i∞ = lim hH n · M, N i∞ in L1 . n→∞
H · hM, N i = lim H n · hM, N i in L1 . n→∞
for all N ∈
M2c .
To show the first holds, we use the Kunita–Watanabe inequality to get E|hH · M − H n · M, N i∞ | ≤ E (hH · M − H n · M i∞ )
1/2
1/2
(EhN i∞ )
,
and the first factor is kH · M − H n · M kM2 → 0, while the second is finite since N ∈ M2c . The second follows from E |((H − H n ) · hM, N i)∞ | ≤ kH − H n kL2 (M ) kN kM2 → 0. So we know that hH · M, N i∞ = (H · hM, N i)∞ . We can then replace N by the stopped process N t to get hH · M, N it = (H · hM, N i)t . To see uniqueness, suppose X ∈ M2c is another such martingale. Then we have hX − H · M, N i = 0 for all N . Take N = X − H · M , and then we are done. 27
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The stochastic integral
III Stochastic Calculus and Applications
(iii) For N ∈ M2c , we have h(H · M )T , N it = hH · M, N it∧T = H · hM, N it∧T = (H1[0,T ] · hM, N i)t for every N . So we have shown that (H · M )T = (1[0,T ] H · M ) by (ii). To prove the second equality, we have hH · M T , N it = H · hM T , N it = H · hM, N it∧T = ((H1[0,T ] · hM, N i)t . Note that (ii) can be written as + *Z (−)
Hs dhM, N is .
=
Hs dMs , N 0
t
Z 0
t
Corollary. hH · M, K · N i = H · (K · hM, N i) = (HK) · hM, N i. In other words, *Z (−)
Z Hs dMs ,
0
+
(−)
Ks dNs 0
t
Z
Hs Ks dhM, N is .
= 0
t
Corollary. Since H · M and (H · M )(K · N ) − hH · M, K · N i are martingales starting at 0, we have Z t E H dMs = 0 0 Z t Z t Z t E Hs dMs Ks dNs = Hs Ks dhM, N is . 0
0
0
2
2
Corollary. Let H ∈ L (M ), then HK ∈ L (M ) iff K ∈ L2 (H · M ), in which case (KH) · M = K · (H · M ). Proof. We have Z E
∞
Z Ks2 Hs2 dhMs i = E
0
∞
Ks2 hH · M is ,
0
so kKkL2 (H·M ) = kHKkL2 (M ) . For N ∈ M2c , we have h(KH) · M, N it = (KH · hM, N i)t = (K · (H · hM, N i))t = (K · hH · M, N i)t .
3.3
Extension to local martingales
We have now defined the stochastic integral for continuous martingales. We next go through some formalities to extend this to local martingales, and ultimately to semi-martingales. We are not doing this just for fun. Rather, when we later prove results like Itˆo’s formula, even when we put in continuous (local) martingales, we usually end up with some semi-martingales. So it is useful to be able to deal with semi-martingales in general. 28
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The stochastic integral
III Stochastic Calculus and Applications
Definition (L2bc (M )). Let L2bc (M ) be the space of previsible H such that Z t Hs2 dhM is < ∞ a.s. 0
for all finite t > 0. Theorem. Let M be a continuous local martingale. (i) For every H ∈ L2bc (M ), there is a unique continuous local martingale H · M with (H · M )0 = 0 and hH · M, N i = H · hM, N i for all N, M . (ii) If T is a stopping time, then (1[0,T ] H) · M = (H · M )T = H · M T . (iii) If H ∈ L2loc (M ), K is previsible, then K ∈ L2loc (H · M ) iff HK ∈ L2loc (M ), and then K · (H · M ) = (KH) · M. (iv) Finally, if M ∈ M2c and H ∈ L2 (M ), then the definition is the same as the previous one. Rt Proof. Assume M0 = 0, and that 0 Hs2 dhM is < ∞ for all ω ∈ Ω (by setting H = 0 when this fails). Set Z t 2 Sn = inf t ≥ 0 : (1 + Hs ) dhM is ≥ n . 0
These Sn are stopping times that tend to infinity. Then hM Sn , M Sn it = hM, M it∧Sn ≤ n. So M Sn ∈ M2c . Also, Z ∞
Hs dhM Sn is =
Z
0
Sn
Hs2 dhM is ≤ n.
0
So H ∈ L2 (M Sn ), and we have already defined what H · M Sn is. Now notice that H · M Sn = (H · M Sm )Sn for m ≥ n. So it makes sense to define H · M = lim H · M Sn . n→∞
This is the unique process such that (H · M )Sn = H · M Sn . We see that H · M is a continuous adapted local martingale with reducing sequence Sn . Claim. hH · M, N i = H · hM, N i. Indeed, assume that N0 = 0. Set Sn0 = inf{t ≥ 0 : |Nt | ≥ n}. Set Tn = 0 Sn ∧ Sn0 . Observe that N Sn ∈ M2c . Then 0
0
hH · M, N iTn = hH · M Sn , N Sn i = H · hM Sn , N Sn i = H · hM, N iTn . Taking the limit n → ∞ gives the desired result. The proofs of the other claims are the same as before, since they only use the characterizing property hH · M, N i = H · hM, N i. 29
3
The stochastic integral
3.4
III Stochastic Calculus and Applications
Extension to semi-martingales
Definition (Locally boounded previsible process). A previsible process H is locally bounded if for all t ≥ 0, we have sup |Hs | < ∞ a.s. s≤t
Fact. (i) Any adapted continuous process is locally bounded. (ii) If H is locally bounded and A is a finite variation process, then for all t ≥ 0, we have Z t |Hs | |dAs | < ∞ a.s. 0
Now if X = X0 + M + A is a semi-martingale, where X0 ∈ F0 , M is a continuous local martingale and A is a finite variation process, we want to define R Hs dXs . We already know what it means to define integration with respect to dMs and dAs , using the Itˆo integral and the finite variation integral respectively, and X0 doesn’t change, so we can ignore it. Definition (Stochastic integral). Let X = X0 + M + A be a continuous semimartingale, and H a locally bounded previsible process. Then the stochastic integral H · X is the continuous semi-martingale defined by H · X = H · M + H · A, and we write Z (H · X)t =
T
Hs dXs . 0
Proposition. (i) (H, X) 7→ H · X is bilinear. (ii) H · (K · X) = (HK) · X if H and K are locally bounded. (iii) (H · X)T = H1[0,T ] · X = H · X T for every stopping time T . (iv) If X is a continuous local martingale (resp. a finite variation process), then so is H · X. Pn (v) If H = i=1 Hi−1 1(ti−1 ,ti ] and Hi−1 ∈ Fti−1 (not necessarily bounded), then n X (H · X)t = Hi−1 (Xti ∧t − Xti−1 ∧t ). i=1
Proof. (i) to (iv) follow from analogous properties for H · M and H · A. The last part is also true by definition if the Hi are uniformly bounded. If Hi is not bounded, then the finite variation part is still fine, since for each fixed ω ∈ Ω, Hi (ω) is a fixed number. For the martingale part, set Tn = inf{t ≥ 0 : |Ht | ≥ n}. 30
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III Stochastic Calculus and Applications
Then Tn are stopping times, Tn → ∞, and H1[0,Tn ] ∈ E. Thus (H · M )t∧Tn =
n X
Hi−1 T[0,Tn ] (Xti ∧t − Xti−1 ∧t ).
i=1
Then take the limit n → ∞. Before we get to Itˆ o’s formula, we need a few more useful properties: Proposition (Stochastic dominated convergence theorem). Let X be a continuous semi-martingale. Let H, Hs be previsible and locally bounded, and let K be previsible and non-negative. Let t > 0. Suppose (i) Hsn → Hs as n → ∞ for all s ∈ [0, t]. (ii) |Hsn | ≤ Ks for all s ∈ [0, t] and n ∈ N. Rt Rt (iii) 0 Ks2 dhM is < ∞ and 0 Ks |dAs | < ∞ (note that both conditions are okay if K is locally bounded). Then
t
Z
Hsn dXs →
0
Z
t
Hs dXs in probability. 0
Proof. For the finite variation part, the convergence follows from the usual dominated convergence theorem. For the martingale part, we set Z t Tm = inf t ≥ 0 : Ks2 dhM is ≥ m . 0
So we have Z Z Tm ∧t E Hsn dMs − 0
!2
Tn ∧t
Hs dMs
Z
(Hsn
≤E
0
!
Tn ∧t 2
− Hs ) dhM is
→ 0.
0
R T ∧t using the usual dominated convergence theorem, since 0 n Ks2 dhM is ≤ m. Since Tn ∧ t = t eventually as n → ∞ almost surely, hence in probability, we are done. Proposition. Let X be a continuous semi-martingale, and let H be an adapted (m) (m) bounded left-continuous process. Then for every subdivision 0 < t0 < t1 < (m) (m) (m) · · · < tnm of [0, t] with maxi |ti − ti−1 | → 0, then Z
t
Hs dXs = lim 0
m→∞
nm X i=1
Ht(m) (Xt(m) − Xt(m) ) i−1
i
i−1
in probability. Proof. We have already proved this for the Lebesgue–Stieltjes integral, and all we used was dominated convergence. So the same proof works using stochastic dominated convergence theorem.
31
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The stochastic integral
3.5
III Stochastic Calculus and Applications
Itˆ o formula
We now prove the equivalent of the integration by parts and the chain rule, i.e. Itˆ o’s formula. Compared to the world of usual integrals, the difference is that the quadratic variation, i.e. “second order terms” will crop up quite a lot, since they are no longer negligible. Theorem (Integration by parts). Let X, Y be a continuous semi-martingale. Then almost surely, Z t Z t Xt Yt − X0 Y0 = Xs dYs + Ys dXs + hX, Y it 0
0
The last term is called the Itˆ o correction. Note that if X, Y are martingales, then the first two terms on the right are martingales, but the last is not. So we are forced to think about semi-martingales. Observe that in the case of finite variation integrals, we don’t have the correction. Proof. We have Xt Yt − Xs Ys = Xs (Yt − Ys ) + (Xt − Xs )Ys + (Xt − Xs )(Yt − Ys ). When doing usual calculus, we can drop the last term, because it is second order. However, the quadratic variation of martingales is in general non-zero, and so we must keep track of this. We have Xk2−n Yk2−n − X0 Y0 =
k X (Xi2−n Yi2−n − X(i−1)2−n Y(i−1)2−n ) i=1
=
n X
X(i−1)2−n (Yi2−n − Y(i−1)2−n )
i=1
+ Y(i−1)2−n (Xi2−n − X(i−1)2−n ) + (Xi2−n − X(i−1)2−n )(Yi2−n − Y(i−1)2−n ) Taking the limit n → ∞ with k2−n fixed, we see that the formula holds for t a dyadic rational. Then by continuity, it holds for all t. The really useful formula is the following: Theorem (Itˆo’s formula). Let X 1 , . . . , X p be continuous semi-martingales, and let f : Rp → R be C 2 . Then, writing X = (X 1 , . . . , X p ), we have, almost surely, f (Xt ) = f (X0 ) +
p Z X i=1
0
t
p Z ∂f 1 X t ∂2f (Xs ) dXsi + (Xs ) dhX i , X j is . ∂xi 2 i,j=1 0 ∂xi ∂xj
In particular, f (X) is a semi-martingale. The proof is long but not hard. We first do it for polynomials by explicit computation, and then use Weierstrass approximation to extend it to more general functions. 32
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III Stochastic Calculus and Applications
Proof. Claim. Itˆ o’s formula holds when f is a polynomial. It clearly does when f is a constant! We then proceed by induction. Suppose Itˆ o’s formula holds for some f . Then we apply integration by parts to g(x) = xk f (x). where xk denotes the kth component of x. Then we have Z t Z t g(Xt ) = g(X0 ) + Xsk df (Xs ) + f (Xs ) dXsk + hX k , f (X)it 0
0
We now apply Itˆ o’s formula for f to write Z
p Z X
t
Xsk
df (Xs ) =
0
0
i=1
t
Xsk
∂f (Xs ) dXsi ∂xi p Z 1 X t k ∂2f (Xs ) dhX i , X j is . X 2 i,j=1 0 s ∂xi ∂xj
+ We also have k
hX , f (X)it =
p Z X i=1
t
0
∂f (Xs ) dhX k , X i is . ∂xi
So we have g(Xt ) = g(X0 ) +
p Z X i=1
0
t
p Z ∂g 1 X t ∂2g i (Xs ) dhX i , X j is . (X ) dX + s s ∂xi 2 i,j=1 0 ∂xi ∂xj
So by induction, Itˆ o’s formula holds for all polynomials. Claim. Itˆo’s formula holds for all f ∈ C 2 if |Xt (ω)| ≤ n and all (t, ω).
Rt 0
|dAs | ≤ n for
By the Weierstrass approximation theorem, there are polynomials pk such that 2 ∂f ∂ f ∂p ∂pk 1 sup |f (x) − pk (x)| + max i − + max i j − ≤ . i i ∂xj i i,j ∂x ∂x ∂x ∂x ∂x k |x|≤k By taking limits, in probability, we have f (Xt ) − f (X0 ) = lim (pk (Xt ) − pk (X0 )) k→∞
Z 0
t
∂f ∂pk (Xs ) dXsi = lim (Xs ) dXsi k→∞ ∂xi ∂xi
by stochastic dominated convergence theorem, and by the regular dominated convergence, we have Z t 2 Z t ∂f ∂ pk i j dhX , X i = lim dhX i , X j i. s i j k→∞ 0 ∂xi ∂xj 0 ∂x ∂x 33
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The stochastic integral
III Stochastic Calculus and Applications
Claim. Itˆ o’s formula holds for all X. Let
Z t Tn = inf t ≥ 0 : |Xt | ≥ n or |dAs | ≥ n 0
Then by the previous claim, we have p Z X
f (XtTn ) = f (X0 ) +
i=1
+
∂f (X Tn ) d(Xi )Ts n ∂xi s
t
∂2f (X Tn ) dh(Xi )Tn , (X j )Tn is ∂xi ∂xj s
0
Z 1X 2
= f (X0 ) +
i,j 0 p Z t∧Tn X i=1
+
t
1 2
0
XZ i,j
∂f (Xs ) d(Xi )s ∂xi
t∧Tn
0
∂2f (Xs ) dh(Xi ), (X j )is . ∂xi ∂xj
Then take Tn → ∞. Example. Let B be a standard Brownian motion, B0 = 0 and f (x) = x2 . Then Bt2 = 2
Z
t
BS dBs + t. 0
In other words, Bt2
Z −t=2
t
Bs dBs . 0
In particular, this is a continuous local martingale. Example. Let B = (B 1 , . . . , B d ) be a d-dimensional Brownian motion. Then we apply Itˆ o’s formula to the semi-martingale X = (t, B 1 , . . . , B d ). Then we find that Z t f (t, Bt ) − f (0, B0 ) − 0
d Z t X ∂ 1 ∂ + ∆ f (s, Bs ) ds = f (s, Bs ) dBsi i ∂s 2 ∂x 0 i=1
is a continuous local martingale. There are some syntactic tricks that make stochastic integrals easier to manipulate, namely by working in differential form. We can state Itˆ o’s formula in differential form df (Xt ) =
p p X ∂f 1 X ∂2f i dX + dhX i , X j i, i i ∂xj ∂x 2 ∂x i=1 i,j=1
which we can think of as the chain rule. For example, in the case case of Brownian motion, we have 1 df (Bt ) = f 0 (Bt ) dBt + f 00 (Bt ) dt. 2
34
3
The stochastic integral
III Stochastic Calculus and Applications
Formally, one expands f using that that “(dt)2 = 0” but “(dB)2 = dt”. The following formal rules hold: Z t Zt − Z0 = Hs dXs ⇐⇒ dZt = Ht dXt 0 Z t Zt = hX, Y it = dhX, Y it ⇐⇒ dZt = dXt dYt . 0
Then we have rules such as Ht (Kt dXt ) = (Ht Kt ) dXt Ht (dXt dYt ) = (Ht dXt ) dYt d(Xt Yt ) = Xt dYt + Yt dXt + dXt dYt 1 df (Xt ) = f 0 (Xt ) dXt + f 00 (Xt ) dXt dXt . 2
3.6
The L´ evy characterization
A more major application of the stochastic integral is the following convenient characterization of Brownian motion: Theorem (L´evy’s characterization of Brownian motion). Let (X 1 , . . . , X d ) be continuous local martingales. Suppose that X0 = 0 and that hX i , X j it = δij t for all i, j = 1, . . . , d and t ≥ 0. Then (X 1 , . . . , X d ) is a standard d-dimensional Brownian motion. This might seem like a rather artificial condition, but it turns out to be quite useful in practice (though less so in this course). The point is that we know that hH · M it = Ht2 · hM it , and in particular if we are integrating things with respect to Brownian motions of some sort, we know hBt it = t, and so we are left with some explicit, familiar integral to do. Proof. Let 0 ≤ s < t. It suffices to check that Xt − Xs is independent of Fs and Xt − Xs ∼ N (0, (t − s)I). 2
1
Claim. E(eiθ·(Xt −Xs ) | Fs ) = e− 2 |θ|
(t−s)
for all θ ∈ Rd and s < t.
This is sufficient, since the right-hand side is independent of Fs , hence so is the left-hand side, and the Fourier transform characterizes the distribution. To check this, for θ ∈ Rd , we define Yt = θ · Xt =
d X
θi Xti .
i=1
Then Y is a continuous local martingale, and we have hY it = hY, Y it =
d X
θj θk hX j , X k it = |θ|2 t.
i,j=1
by assumption. Let 1
1
2
Zt = eiYt + 2 hY it = eiθ·Xt + 2 |θ| t . 35
3
The stochastic integral
III Stochastic Calculus and Applications
By Itˆ o’s formula, with X = iY + 12 hY it and f (x) = ex , we get 1 1 dZt = Zt idYt − dhY it + dhY it = iZt dYt . 2 2 So this implies Z is a continuous local martingale. Moreover, since Z is bounded on bounded intervals of t, we know Z is in fact a martingale, and Z0 = 1. Then by definition of a martingale, we have E(Zt | Fs ) = Zs , And unwrapping the definition of Zt shows that the result follows. In general, the quadratic variation of a process doesn’t have to be linear in t. It turns out if the quadratic variation increases to infinity, then the martingale is still a Brownian motion up to reparametrization. Theorem (Dubins–Schwarz). Let M be a continuous local martingale with M0 = 0 and hM i∞ = ∞. Let Ts = inf{t ≥ 0 : hM it > s}, the right-continuous inverse of hM it . Let Bs = MTs and Gs = FTs . Then Ts is a (Ft ) stopping time, hM iTs = s for all s ≥ 0, B is a (Gs )-Brownian motion, and Mt = BhM it . Proof. Since hM i is continuous and adapted, and hM i∞ = ∞, we know Ts is a stopping time and Ts < ∞ for all s ≥ 0. Claim. (Gs ) is a filtration obeying the usual conditions, and G∞ = F∞ Indeed, if A ∈ Gs and s < t, then A ∩ {Tt ≤ u} = A ∩ {Ts ≤ u} ∩ {Tt ≤ u} ∈ Fu , using that A ∩ {Ts ≤ u} ∈ Fu since A ∈ Gs . Then right-continuity follows from that of (Ft ) and the right-continuity of s 7→ Ts . Claim. B is adapted to (Gs ). In general, if X is c´adl´ ag and T is a stopping time, then XT 1{T Ts− . Then we have hM iTs = hM iTs− . Since hM it is increasing, it means hM iTs is constant in [Ts− , Ts ]. We will later prove that 36
3
The stochastic integral
III Stochastic Calculus and Applications
Lemma. M is constant on [a, b] iff hM i being constant on [a, b]. So we know that if Ts > Ts− , then MTs = MTs − . So B is left continuous. We then have to show that B is a martingale. Claim. (M 2 − hM i)Ts is a uniformly integrable martingale. To see this, observe that hM Ts i∞ = hM iTs = s, and so M Ts is bounded. So (M − hM i)Ts is a uniformly integrable martingale. We now apply the optional stopping theorem, which tells us 2
Ts E(Bs | Gr ) = E(M∞ | Gs ) = MTt = Bt .
So Bt is a martingale. Moreover, E(Bs2 − s | Gr ) = E((M 2 − hM i)Ts | FTr ) = MT2r − hM iTr = Br2 − r. So Bt2 − t is a martingale, so by the characterizing property of the quadratic variation, hBit = t. So by L´evy’s criterion, this is a Brownian motion in one dimension. The theorem is only true for martingales in one dimension. In two dimensions, this need not be true, because the time change needed for the horizontal and vertical may not agree. However, in the example sheet, we see that the holomorphic image of a Brownian motion is still a Brownian motion up to a time change. Lemma. M is constant on [a, b] iff hM i being constant on [a, b]. Proof. It is clear that if M is constant, then so is hM i. To prove the converse, by continuity, it suffices to prove that for any fixed a < b, {Mt = Ma for all t ∈ [a, b]} ⊇ {hM ib = hM ia } almost surely. We set Nt = Mt − Mt ∧ a. Then hN it = hM it − hM it∧a . Define Tε = inf{t ≥ 0 : hN it ≥ ε}. Then since N 2 − hN i is a local martingale, we know that 2 E(Nt∧T ) = E(hN it∧Tε ) ≤ ε. ε
Now observe that on the event {hM ib = hM ia }, we have hN ib = 0. So for t ∈ [a, b], we have 2 E(1{hM ib =hM ia } Nt2 ) = E(1{hM ib =hM ia Nt∧T ) = E(hN it∧Tε ) = 0. ε
3.7
Girsanov’s theorem
Girsanov’s theorem tells us what happens to our (semi)-martingales when we change the measure of our space. We first look at a simple example when we perform a shift.
37
3
The stochastic integral
III Stochastic Calculus and Applications
Example. Let X ∼ N (0, C) be an n-dimensional centered Gaussian with positive definite covariance C = (Cij )ni,j=1 . Put M = C −1 . Then for any function f , we have 1/2 Z 1 M Ef (X) = det f (x)e− 2 (x,M x) dx. 2π Rn Now fix an a ∈ Rn . The distribution of X + a then satisfies 1/2 Z 1 M f (x)e− 2 (x−a,M (x−a)) dx = E[Zf (X)], Ef (X + a) = det 2π Rn where
1
Z = Z(x) = e− 2 (a,M a)+(x,M a) . Thus, if P denotes the distribution of X, then the measure Q with dQ =Z dP is that of N (a, C) vector. Example. We can extend the above example to Brownian motion. Let B be a Brownian motion with B0 = 0, and h : [0, ∞) → R a deterministic function. We then want to understand the distribution of Bt + h. Fix a finite sequence of times 0 = t0 < t1 < · · · < tn . Then we know that (Bti )ni=1 is a centered Gaussian random variable. Thus, if f (B) = f (Bt1 , . . . , Btn ) is a function, then Z Pn (xi −xi−1 )2 −1 E(f (B)) = c · f (x)e 2 i=1 ti −ti−1 dx1 · · · dxn . Rn
Thus, after a shift, we get E(f (B + h)) = E(Zf (B)), n
n
1 X (hti − hti−1 )2 X (hti − hti−1 )(Bti − Bti−1 ) Z = exp − + 2 i=1 ti − ti−1 ti − ti−1 i=1
! .
In general, we are interested in what happens when we change the measure by an exponential: Definition (Stochastic exponential). Let M be a continuous local martingale. Then the stochastic exponential (or Dol´eans–Dade exponential ) of M is 1
E(M )t = eMt − 2 hM it The point of introducing that quadratic variation term is Proposition. Let M be a continuous local martingale with M0 = 0. Then E(M ) = Z satisfies dZt = Zt dM, i.e. Z
t
Zt = 1 +
Zs dMs . 0
In particular, E(M ) is a continuous local martingale. Moreover, if hM i is uniformly bounded, then E(M ) is a uniformly integrable martingale. 38
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The stochastic integral
III Stochastic Calculus and Applications
There is a more general condition for the final property, namely Novikov’s condition, but we will not go into that. Proof. By Itˆ o’s formula with X = M − 12 hM i, we have 1 1 dZt = Zt d Mt − dhM it + Zt dhM it = Zt dMt . 2 2 R Since M is a continuous local martingale, so is Zs dMs . So Z is a continuous local martingale. Now suppose hM i∞ ≤ b < ∞. Then 2 P sup Mt ≥ a = P sup Mt ≥ a, hM i∞ ≤ b ≤ e−a /2b , t≥0
t≥0
where the final equality is an exercise on the third example sheet, which is true for general continuous local martingales. So we get Z ∞ E exp sup Mt = P(exp(sup Mt ) ≥ λ) dλ t Z0 ∞ = P(sup Mt ≥ log λ) dλ 0 Z ∞ 2 ≤1+ e−(log λ) /2b dλ < ∞. 1
Since hM i ≥ 0, we know that sup E(M )t ≤ exp (sup Mt ) , t≥0
So E(M ) is a uniformly integrable martingale. Theorem (Girsanov’s theorem). Let M be a continuous local martingale with M0 = 0. Suppose that E(M ) is a uniformly integrable martingale. Define a new probability measure dQ = E(M )∞ dP Let X be a continuous local martingale with respect to P. Then X − hX, M i is a continuous local martingale with respect to Q. Proof. Define the stopping time Tn = inf{t ≥ 0 : |Xt − hX, M it | ≥ n}, and P(Tn → ∞) = 1 by continuity. Since Q is absolutely continuous with respect to P, we know that Q(Tn → ∞) = 1. Thus it suffices to show that X Tn − hX Tn , M i is a continuous martingale for any n. Let Y = X Tn − hX Tn , M i,
Z = E(M ).
Claim. ZY is a continuous local martingale with respect to P.
39
3
The stochastic integral
III Stochastic Calculus and Applications
We use the product rule to compute d(ZY ) = Yt dZt + Zt dYt + dhY, Zit = Y Zt dMt + Zt (dX Tn − dhX Tn , M it ) + Zt dhM, X Tn i = Y Zt dMt + Zt dX Tn So we see that ZY is a stochastic integral with respect to a continuous local martingale. Thus ZY is a continuous local martingale. Claim. ZY is uniformly integrable. Since Z is a uniformly integrable martingale, {ZT : T is a stopping time} is uniformly integrable. Since Y is bounded, {ZT YT : T is a stopping time} is also uniformly integrable. So ZY is a true martingale (with respect to P). Claim. Y is a martingale with respect to Q. We have EQ (Yt − Ys | Fs ) = EP (Z∞ Yt − Z∞ Ys | Fs ) = EP (Zt Yt − Zs Ys | Fs ) = 0. Note that the quadratic variation does not change since b2n tc
hX − hX, M ii = hXit = lim
n→∞
X i=1
along a subsequence.
40
(Xi2−n − X(i−1)2−n )2 a.s.
4
Stochastic differential equations
4
III Stochastic Calculus and Applications
Stochastic differential equations
4.1
Existence and uniqueness of solutions
After all this work, we can return to the problem we described in the introduction. We wanted to make sense of equations of the form x(t) ˙ = F (x(t)) + η(t), where η(t) is Gaussian white noise. We can now interpret this equation as saying dXt = F (Xt ) dt + dBt , or equivalently, in integral form, Z Xt − X0 =
T
F (Xs ) ds + Bt . 0
In general, we can make the following definition: Definition (Stochastic differential equation). Let d, m ∈ N, b : R+ × Rd → Rd , σ : R+ × Rd → Rd×m be locally bounded (and measurable). A solution to the stochastic differential equation E(σ, b) given by dXt = b(t, Xt ) dt + σ(t, Xt ) dBt consists of (i) a filtered probability space (Ω, F, (Ft ), P) obeying the usual conditions; (ii) an m-dimensional Brownian motion B with B0 = 0; and (iii) an (Ft )-adapted continuous process X with values in Rd such that Z Xt = X0 +
t
Z σ(s, Xs ) dBs +
0
t
b(s, Xs ) ds. 0
If X0 = x ∈ Rd , then we say X is a (weak) solution to Ex (σ, b). It is a strong solution if it is adapted with respect to the canonical filtration of B. Our goal is to prove existence and uniqueness of solutions to a general class of SDEs. We already know what it means for solutions to be unique, and in general there can be multiple notions of uniqueness: Definition (Uniqueness of solutions). For the stochastic differential equation E(σ, b), we say there is – uniqueness in law if for every x ∈ Rd , all solutions to Ex (σ, b) have the same distribution. – pathwise uniqueness if when (Ω, F, (Ft ), P) and B are fixed, any two solutions X, X 0 with X0 = X00 are indistinguishable. These two notions are not equivalent, as the following example shows:
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Stochastic differential equations
III Stochastic Calculus and Applications
Example (Tanaka). Consider the stochastic differential equation dXt = sgn(Xt ) dBt , where
( sgn(x) =
+1 −1
X0 = x, x>0 . x≤0
This has a weak solution which is unique in law, but pathwise uniqueness fails. To see the existence of solutions, let X be a one-dimensional Brownian motion with X0 = x, and set Z t
Bt =
sgn(Xs ) dXs , 0
which is well-defined because sgn(Xs ) is previsible and left-continuous. Then we have Z t Z t x+ sgn(Xs ) dBs = x + sgn(Xs )2 dXs = x + Xt − X0 = Xt . 0
0
So it remains to show that B is a Brownian motion. We already know that B is a continuous local martingale, so by L´evy’s characterization, it suffices to show its quadratic variation is t. We simply compute Z t hB, Bit = dhXs , Xs i = t. 0
So there is weak existence. The same argument shows that any solution is a Brownian motion, so we have uniqueness in law. Finally, observe that if x = 0 and X is a solution, then −X is also a solution with the same Brownian motion. Indeed, Z t Z t Z t −Xt = sgn(Xs ) dBs = sgn(−Xs ) dBs + 2 1Xs =0 dBs , 0
0
0
where the second term R t vanishes, since it is a continuous local martingale with quadratic variation 0 1Xs =0 ds = 0. So pathwise uniqueness does not hold. In the other direction, however, it turns out pathwise uniqueness implies uniqueness in law. Theorem (Yamada–Watanabe). Assume weak existence and pathwise uniqueness holds. Then (i) Uniqueness in law holds. (ii) For every (Ω, F, (Ft ), P) and B and any x ∈ Rd , there is a unique strong solution to Ex (a, b). We will not prove this, since we will not actually need it. The key, important theorem we are now heading for is the existence and uniqueness of solutions to SDEs, assuming reasonable conditions. As in the case of ODEs, we need the following Lipschitz conditions:
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Stochastic differential equations
III Stochastic Calculus and Applications
Definition (Lipschitz coefficients). The coefficients b : R+ × Rd → Rd , σ : R+ × Rd → Rd×m are Lipschitz in x if there exists a constant K > 0 such that for all t ≥ 0 and x, y ∈ Rd , we have |b(t, x) − b(t, y)| ≤ K|x − y| |σ(t, x) − σ(t, y)| ≤ |x − y| Theorem. Assume b, σ are Lipschitz in x. Then there is pathwise uniqueness for the E(σ, b) and for every (Ω, F, (Ft ), P) satisfying the usual conditions and every (Ft )-Brownian motion B, for every x ∈ Rd , there exists a unique strong solution to Ex (σ, b). Proof. To simplify notation, we assume m = d = 1. We first prove pathwise uniqueness. Suppose X, X 0 are two solutions with X0 = X00 . We will show that E[(Xt − Xt0 )2 ] = 0. We will actually put some bounds to control our variables. Define the stopping time S = inf{t ≥ 0 : |Xt | ≥ n or |Xt0 | ≥ n}. By continuity, S → ∞ as n → ∞. We also fix a deterministic time T > 0. Then whenever t ∈ [0, T ], we can bound, using the identity (a + b)2 ≤ 2a2 + 2b2 , 0 E((Xt∧S − Xt∧S )2 ) ≤ 2E
!2
t∧S
Z
(σ(s, Xs ) − σ(s, Xs0 )) dBs
0
!2
t∧S
Z
(b(s, Xs ) − b(s, Xs0 )) ds
+ 2E
.
0
We can apply the Lipschitz bound to the second term immediately, while we can simplify the first term using the (corollary of the) Itˆo isometry !2 ! Z t∧S Z t∧S E (σ(s, Xs ) − σ(s, Xs0 )) dBs = E (σ(s, Xs ) − σ(s, Xs0 ))2 ds . 0
0
So using the Lipschitz bound, we have E((Xt∧S −
0 Xt∧S )2 )
Z
2
!
t∧S
|Xs −
≤ 2K (1 + T )E
Xs0 |2
ds
0
≤ 2K 2 (1 + T )
Z
t 0 E(|Xs∧S − Xs∧S |2 ) ds.
0
We now use Gr¨ onwall’s lemma: Lemma. Let h(t) be a function such that Z t h(t) ≤ c h(s) ds 0
for some constant c. Then h(t) ≤ h(0)ect . 43
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Stochastic differential equations
III Stochastic Calculus and Applications
Applying this to 0 h(t) = E((Xt∧S − Xt∧S )2 ),
we deduce that h(t) ≤ h(0)ect = 0. So we know that 0 E(|Xt∧S − Xt∧S |2 ) = 0
for every t ∈ [0, T ]. Taking n → ∞ and T → ∞ gives pathwise uniqueness. We next prove existence of solutions. We fix (Ω, F, (Ft )t ) and B, and define Z
t
F (X)t = X0 +
t
Z σ(s, Xs ) dBs +
b(s, Xs ) ds.
0
0
Then X is a solution to Ex (a, b) iff F (X) = X and X0 = x. To find a fixed point, we use Picard iteration. We fix T > 0, and define the T -norm of a continuous adapted process X as 1/2 kXkT = E sup |Xt |2 . t≤T
In particular, if X is a martingale, then this is the same as the norm on the space of L2 -bounded martingales by Doob’s inequality. Then B = {X : Ω × [0, T ] → R : kXkT < ∞} is a Banach space. Claim. kF (0)kT < ∞, and kF (X) − F (Y )k2T ≤ (2T + 8)K 2
Z
T
kX − Y k2t dt.
0
We first see how this claim implies the theorem. First observe that the claim implies F indeed maps B into itself. We can then define a sequence of processes Xti by Xt0 = x, X i+1 = F (X i ). Then we have kX i+1 − X i k2T ≤ CT
Z
T
kX i − X i−1 k2t dt ≤ · · · ≤ kX 1 − X 0 k2T
0
So we find that
∞ X
CT i i!
.
kX i − X i−1 k2T < ∞
i=1
for all T . So X i converges to X almost surely and uniformly on [0, T ], and F (X) = X. We then take T → ∞ and we are done. To prove the claim, we write
Z t
Z t
+
kF (0)kT ≤ |X0 | + b(s, 0) ds σ(s, 0) dB s .
0
0
44
T
4
Stochastic differential equations
III Stochastic Calculus and Applications
The first two terms are constant, and we can bound the last by Doob’s inequality and the Itˆ o isometry: 2
Z t
Z T Z T
≤ 2E σ(s, 0) dB σ(s, 0) dB = 2 σ(s, 0)2 ds. s s
0 0 0 T To prove the second part, we use Z t 2 ! kF (X) − F (Y )k ≤ 2E sup b(s, X − s) − b(s, Ys ) ds t≤T 2
0
Z t 2 ! + 2E sup (σ(s, Xs ) − σ(s, Ys )) dBs . t≤T 0
We can bound the first term with Cauchy–Schwartz by ! Z T Z T 2 2 TE |b(s, Xs ) − b(s, Ys )| ≤ TK kX − Y k2t dt, 0
0
and the second term with Doob’s inequality by ! Z T Z 2 E |σ(s, Xs ) − σ(s, Ys )| ds ≤ 4K 2 0
4.2
T
kX − Y k2t dt.
0
Examples of stochastic differential equations
Example (The Ornstein–Uhlenbeck process). Let λ > 0. Then the Ornstein– Uhlenbeck process is the solution to dXt = −λXt dt + dBt . The solution exists by the previous theorem, but we can also explicitly find one. By Itˆ o’s formula applied to eλt Xt , we get d(eλt Xt ) = eλt dXt + λeλt Xt dt = dBt . So we find that Xt = e−λt X0 +
Z
t
e−λ(t−s) dBs .
0
Observe that the integrand is deterministic. So we can in fact interpret this as an Wiener integral. Fact. If X0 = x ∈ R is fixed, then (Xt ) is a Gaussian process, i.e. (Xti )ni=1 is jointly Gaussian for all t1 < · · · < tn . Any Gaussian process is determined by the mean and covariance, and in this case, we have 1 −λ|t−s| EXt = e−λt x, cov(Xt , Xs ) = e − e−λ|t+s| 2λ Proof. We only have to compute the covariance. By the Itˆo isometry, we have Z t Z s −λ(t−u) −λ(s−u) E((Xt − EXt )(Xs − EXs )) = E e dBu e dBu 0
= e−λ(t+s)
0
Z 0
45
t∧s
eλu du.
4
Stochastic differential equations
III Stochastic Calculus and Applications
In particular, Xt ∼ N
e−λt x,
1 − e−2λt 2λ
→N
0,
1 2λ
.
1 ), then (Xt ) is a centered Gaussian process with stationary Fact. If X0 ∼ N (0, 2λ covariance, i.e. the covariance depends only on time differences:
cov(Xt , Xs ) =
1 −λ|t−s| e . 2λ
The difference is that in the deterministic case, the EXt cancels the first e−λt X0 term, while in the non-deterministic case, it doesn’t. This is a very nice example where we can explicitly understand the long-time behaviour of the SDE. In general, this is non-trivial. Dyson Brownian motion Let HN be an inner product space of real symmetric N × N matrices with inner product N Tr(HK) for H, K ∈ HN . Let H 1 , . . . , H dim(HN ) be an orthonormal basis for HN . Definition (Gaussian orthogonal ensemble). The Gaussian Orthogonal Ensemble GOEN is the standard Gaussian measure on HN , i.e. H ∼ GOEN if H=
dim Hn X
H iX i
r=1
where each X i are iid standard normals. We now replace each X i by a Ornstein–Uhlenbeck process with λ = 12 . Then GOEN is invariant under the process. Theorem. The eigenvalues λ1 (t) ≤ · · · ≤ λN (t) satisfies r i X 1 λ 1 2 i dλt = − + dB i . dt + 2 N λi − λj Nβ j6=i
Here β = 1, but if we replace symmetric matrices by Hermitian ones, we get β = 2; if we replace symmetric matrices by symplectic ones, we get β = 4. This follows from Itˆ o’s formula and formulas for derivatives of eigenvalues. Example (Geometric Brownian motion). Fix σ > 0 and t ∈ R. Then geometric Brownian motion is given by dXt = σXt dBt + rXt dt. We apply Itˆ o’s formula to log Xt to find that σ2 Xt = X0 exp σBt + r − t . 2
46
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Stochastic differential equations
III Stochastic Calculus and Applications
Example (Bessel process). Let B = (B 1 , . . . , B d ) be a d-dimensional Brownian motion. Then Xt = |Bt | satisfies the stochastic differential equation dXt =
d−1 dt + dBt 2Xt
if t < inf{t ≥ 0 : Xt = 0}.
4.3
Representations of solutions to PDEs
Recall that in Advanced Probability, we learnt that we can represent the solution to Laplace’s equation via Brownian motion, namely if D is a suitably nice domain and g : ∂D → R is a function, then the solution to the Laplace’s equation on D with boundary conditions g is given by u(x) = Ex [g(BT )], where T is the first hitting time of the boundary ∂D. A similar statement we can make is that if we want to solve the heat equation ∂u = ∇2 u ∂t with initial conditions u(x, 0) = u0 (x), then we can write the solution as √ u(x, t) = Ex [u0 ( 2Bt )] This is just a fancy way to say that the Green’s function for the heat equation is a Gaussian, but is a good way to think about it nevertheless. In general, we would like to associate PDEs to certain stochastic processes. Recall that a stochastic PDE is generally of the form dXt = b(Xt ) dt + σ(Xt ) dBt for some b : Rd → R and σ : Rd → Rd×m which are measurable and locally bounded. Here we assume these functions do not have time dependence. We can then associate to this a differential operator L defined by X 1X aij ∂i ∂j + bi ∂i . L= 2 i,j i where a = σσ T . Example. If b = 0 and σ =
√
2I, then L = ∆ is the standard Laplacian.
The basic computation is the following result, which is a standard application of the Itˆ o formula: Proposition. Let x ∈ Rd , and X a solution to Ex (σ, b). Then for every f : R+ × Rd → R that is C 1 in R+ and C 2 in Rd , the process Z t ∂ f Mt = f (t, Xt ) − f (0, X0 ) − + L f (s, Xs ) ds ∂s 0 is a continuous local martingale. 47
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Stochastic differential equations
III Stochastic Calculus and Applications
We first apply this to the Dirichlet–Poisson problem, which is essentially to solve −Lu = f . To be precise, let U ⊆ Rd be non-empty, bounded and open; ¯ ) = C 2 (U ) ∩ C(U ¯) f ∈ Cb (U ) and g ∈ Cb (∂U ). We then want to find a u ∈ C 2 (U such that −Lu(x) = f (x) u(x) = g(x)
for x ∈ U for x ∈ ∂U.
If f = 0, this is called the Dirichlet problem; if g = 0, this is called the Poisson problem. We will have to impose the following technical condition on a: ¯ → Rd×d is uniformly elliptic if Definition (Uniformly elliptic). We say a : U ¯ , we have there is a constant c > 0 such that for all ξ ∈ Rd and x ∈ U ξ T a(x)ξ ≥ c|ξ|2 . If a is symmetric (which it is in our case), this is the same as asking for the smallest eigenvalue of a to be bounded away from 0. It would be very nice if we can write down a solution to the Dirichlet–Poisson problem using a solution to Ex (σ, b), and then simply check that it works. We can indeed do that, but it takes a bit more time than we have. Instead, we shall prove a slightly weaker result that if we happen to have a solution, it must be given by our formula involving the SDE. So we first note the following theorem without proof: Theorem. Assume U has a smooth boundary (or satisfies the exterior cone condition), a, b are H¨older continuous and a is uniformly elliptic. Then for ¯ → R and any continuous g : ∂U → R, the every H¨older continuous f : U Dirichlet–Poisson process has a solution. The main theorem is the following: Theorem. Let σ and b be bounded measurable and σσ T uniformly elliptic, U ⊆ Rd as above. Let u be a solution to the Dirichlet–Poisson problem and X a solution to Ex (σ, b) for some x ∈ Rd . Define the stopping time TU = inf{t ≥ 0 : Xt 6∈ U }. Then ETU < ∞ and Z u(x) = Ex
g(XTU ) +
!
TU
f (Xs ) ds . 0
In particular, the solution to the PDE is unique. Proof. Our previous proposition applies to functions defined on all of Rn , while u is just defined on U . So we set 1 Un = x ∈ U : dist(x, ∂U ) > , Tn = inf{t ≥ 0 : Xt 6∈ Un }, n and pick un ∈ Cb2 (Rd ) such that u|Un = un |Un . Recalling our previous notation, let Z t∧Tn n un Tn Mt = (M )t = un (Xt∧Tn ) − un (X0 ) − Lun (Xs ) ds. 0
48
4
Stochastic differential equations
III Stochastic Calculus and Applications
Then this is a continuous local martingale that is bounded by the proposition, and is bounded, hence a true martingale. Thus for x ∈ U and n large enough, the martingale property implies ! Z t∧Tn
u(x) = un (x) = E u(Xt∧Tn ) −
Lu(Xs ) ds 0
!
t∧Tn
Z = E u(Xt∧Tn ) +
f (Xs ) ds . 0
We would be done if we can take n → ∞. To do so, we first show that E[TU ] < ∞. Note that this does not depend on f and g. So we can take f = 1 and g = 0, and let v be a solution. Then we have ! Z t∧Tn E(t ∧ Tn ) = E − Lv(Xs ) ds = v(x) − E(v(Xt∧Tn )). 0
Since v is bounded, by dominated/monotone convergence, we can take the limit to get E(TU ) < ∞. Thus, we know that t ∧ Tn → TU as t → ∞ and n → ∞. Since ! Z TU E |f (Xs )| ds ≤ kf k∞ E[TU ] < ∞, 0
the dominated convergence theorem tells us ! Z t∧Tn Z E f (Xs ) ds → E 0
!
TU
f (Xs ) ds .
0
¯ , we also have Since u is continuous on U E(u(Xt∧Tn )) → E(u(Tu )) = E(g(Tu )). We can use SDE’s to solve the Cauchy problem for parabolic equations as well, just like the heat equation. The problem is as follows: for f ∈ Cb2 (Rd ), we want to find u : R+ × Rd → R that is C 1 in R+ and C 2 in Rd such that ∂u = Lu ∂t u(0, · ) = f
on R+ × Rd on Rd
Again we will need the following theorem: Theorem. For every f ∈ Cb2 (Rd ), there exists a solution to the Cauchy problem. Theorem. Let u be a solution to the Cauchy problem. Let X be a solution to Ex (σ, b) for x ∈ Rd and 0 ≤ s ≤ t. Then Ex (f (Xt ) | Fs ) = u(t − s, Xs ). In particular, u(t, x) = Ex (f (Xt )). 49
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In particular, this implies Xt is a continuous Markov process. ∂ ∂ Proof. The martingale has ∂t + L, but the heat equation has ∂t − L. So we set g(s, x) = u(t − s, x). Then ∂ ∂ + L g(s, x) = − u(t − s, x) + Lu(t − s, x) = 0. ∂s ∂t
So g(s, Xs ) − g(0, X0 ) is a martingale (boundedness is an exercise), and hence u(t − s, Xs ) = g(s, Xs ) = E(g(t, Xt ) | Fs ) = E(u(0, Xt ) | Fs ) = E(f (Xt ) | Xs ). There is a generalization to the Feynman–Kac formula. Theorem (Feynman–Kac formula). Let f ∈ Cb2 (Rd ) and V ∈ Cb (Rd ) and suppose that u : R+ × Rd → R satisfies ∂u = Lu + V u ∂t u(0, · ) = f
on R+ × Rd on Rd ,
where V u = V (x)u(x) is given by multiplication. Then for all t > 0 and x ∈ Rd , and X a solution to Ex (σ, b). Then Z t u(t, x) = Ex f (Xt ) exp V (Xs ) ds . 0
If L is the Laplacian, then this is Schr¨odinger equation, which is why Feynman was thinking about this.
50
Index
III Stochastic Calculus and Applications
Index (H · A)t , 9 H · X, 30 L2 (M ), 26 L2bc (M ), 29 X T , 11 E, 10 M2 , 16
Lebesgue–Stieltjes integral, 8 Lipschitz coefficients, 43 local martingale, 12 locally bounded previsible process, 30
Bessel process, 47 bounded variation, 7 bracket, 22 BV, 7
pathwise uniqueness, 41 Poisson problem, 48 previsible σ-algebra, 10 previsible process, 10
c` adl` ag, 7 c` adl` ag adapted process, 9 Cauchy problem, 49 covariation, 22
quadratic variation, 18 semi-martingale, 24
Dirichlet problem, 48 Dirichlet–Poisson problem, 48 Dol´eans–Dade exponential, 38 Doob’s inequality, 16 Dubins–Schwarz theorem, 36 Dyson Brownian motion, 46
semi-martingale, 24 signed measure, 6 simple process, 10, 25 stochastic differential equation, 41 stochastic dominated convergence theorem, 31 stochastic exponential, 38 stochastic integral, 27, 30 stopped process, 11 strong solution, 41
Ornstein–Uhlenbeck process, 45
reduces, 12
Feynman–Kac formula, 50 finite variation, 8 Finite variation process, 9 Gaussian orthogonal ensemble, 46 Gaussian space, 4 Gaussian white noise, 4 Geometric Brownian motion, 46 Girsanov’s theorem, 39
Tanaka equation, 42 total variation, 6, 7 total variation process, 9 u.c.p., 17 uniformly elliptic, 48 uniformly on compact sets in probability, 17 uniqueness in law, 41 usual conditions, 11
Hahn decomposition, 6 integration by parts, 32 Itˆ o correction, 32 Itˆ o isometry, 27 Itˆ o’s formula, 32
Vitali theorem, 13 Kunita–Watanabe, 23 weak solution, 41 L´evy’s characterization of Brownian motion, 35
Yamada–Watanabe, 42
51