Stochastic Calculus (Lecture #3)

Transcription

1 Stochastic Calculus (Lecture #3) Siegfried Hörmann Université libre de Bruxelles (ULB) Spring 2014

2 Outline of the course 1. Stochastic processes in continuous time. 2. Brownian motion. 3. Itô integral: preliminaries. 4. Brownian motion calculus 5. Itô s formula. 6. Stochastic differential equations. 7. Strong and weak solutions of SDEs. 8. Change of measure.

3 Itô integral: preliminaries 3.1. Functions of bounded variation Lebesgue-Stieltjes/Riemann-Stieltjes integral The stochastic integral t 0 B(s)dB(s) Integration of simple previsible processes Some martingale results.

4 Functions of bounded variation Definition The variation of a real function g over the interval [a, b] is defined as V g ([a, b]) := sup g(t i ) g(t i 1 ), where the supremum is taken over partitions a = t 0 < t 1 < < t n = b. If V g (t) := V g ([0, t]) < for all t 0, then g is said to be a function of finite variation. i=1

5 Functions of bounded variation Example If g is increasing then V g (t) = g(t) g(0). If g is decreasing then V g (t) = g(0) g(t). (Exercise.) Example If g is differentiable with continuous derivative then (Exercise.) V g (t) = t 0 g (s) ds. Example The function g(t) = t sin(1/t) for t > 0 and g(0) = 0 is continuous and differentiable at all points except 0, but has infinite variation. (Exercise.)

6 Functions of bounded variation Lemma (Jordan Decomposition) Any function g of finite variation can be written as the difference of two monotone functions. If g is right-continuous, then also the two monotone functions can be chosen to be right-continuous. Proof. Take, for example, h 1 (t) = 1 2 (V g(t) + g(t)) and h 2 (t) = 1 2 (V g(t) g(t)). Remark (1) The representation is not unique. (2) If g(0) = 0 and if h i as above we have 0 h i (t) V g (t).

7 Lebesgue-Stieltjes/Riemann-Stieltjes integral Suppose that g is right-continuous and of bounded variation. Then with h 1, h 2 as before we can define the two (σ-finite) measures µ 1 ((a, b]) := h 1 (b) h 1 (a) and µ 2 ((a, b]) := h 2 (b) h 2 (a). Then for f : [0, ) R the Lebesgue-Stieltjes integral is defined as fdg := fdµ 1 fdµ 2, (0,t] (0,t] (0,t] provided (0,t] f dµ i < for i = 1, 2. One can show that this definition is independent of the choice of h 1 and h 2.

8 Lebesgue-Stieltjes/Riemann-Stieltjes integral If f is continuous, then the Lebesgue-Stieltjes integral coincides with the Riemann-Stieltjes integral t 0 fdg = lim f (t k )(g(t k) g(t k 1 )), t k 1 t k t k, where 0 = t 0 < t 1 < < t n = t, defines a partition, whose refinement tends to zero. Remark (1) The limit above is independent of the choice of t k. (2) One can show: if the Riemann-Stieltjes integral exists for all continuous f, then g is of finite variation.

9 Lebesgue-Stieltjes/Riemann-Stieltjes integral (3) One can show: the Riemann-Stieltjes integral exists, if one of the two functions f or g is continuous and the other one is of bounded variation (4) Due to (3): if f is of bounded variation and B s a BM, then is defined for all ω. t 0 f (s)db s (ω) = we could define the integral path-wise (5) Partial integration: suppose that g and f are continuous and of bounded variation. Then f (t)g(t) = f (0)g(0) + fdg + gdf. (0,t] (0,t]

10 The stochastic integral t 0 B sdb s One of the main targets of this course is to define t 0 X s db s. By the previous statement, if X s (ω) were of bounded variation for all ω, we could define the integral pathwise as a Riemann-Stieltjes integral. However, this would already exclude the case where X s = B s as is implied by the following result. Lemma A BM path is a.s. not of bounded variation. Proof. Fix ω and write B t for B t (ω). Then if V B( ) (t) were < (B tk B tk 1 ) 2 V B( ) (t) sup B tk B tk k n

11 The stochastic integral t 0 B sdb s Let us check what happens. B tk 1 (B tk B tk 1 ) = 1 2 = 1 2 (B tk + B tk 1 )(B tk B tk 1 ) 1 2 (Bt 2 k Bt 2 k 1 ) 1 2 = 1 2 B2 t 1 2 (B tk B tk 1 )(B tk B tk 1 ) (B tk B tk 1 ) 2 (B tk B tk 1 ) 2 P 1 2 (B2 t B t ) = 1 2 (B2 t t). Compared to partial integration, we get extra term t/2.

12 The stochastic integral t 0 B sdb s Now consider n B t k (B t k B tk 1 ), with Then B t k (B tk B tk 1 ) = t k = αt k 1 + (1 α)t k, 0 α 1. B tk 1 (B tk B tk 1 ) + (B t k B tk 1 )(B tk B tk 1 ). The first summand converges to 1 2 (B2 t t) as we have shown. The second, converges to (1 α)t (exercise).

13 The stochastic integral t 0 B sdb s Hence n B t k (B t k B tk 1 ), with t k = αt k 1 + (1 α)t k, 0 α 1. converges (in probability) to 1 2 B2 t + ( ) 1 2 α t. Unlike the usual Riemann-Stieltjes integral, this limit depends on the choice of the nodes t k.

14 The stochastic integral t 0 B sdb s A way out of this problem is to fix the nodes. Different choices of α yield different integrals. 1. The choice α = 0 gives rise to the Itô integral. 2. Another popular choice is α = 1/2, which corresponds to the Stratonovich integral. As we will see, the Itô integral has the important advantage, that the resulting integrated process becomes a (local) martingale, which is a fundamental property.

15 Integration of simple previsible processes Our previous investigations motivate to define the stochastic integral as a Riemann-Stieltjes integral, where the process is evaluated on the left limit of the intervals. Definition (Simple previsible) A process Y is called simple previsible with respect to a filtration (F t ) t 0, if it can be represented as Y t = ξ 0 I{t = 0} + ξ k I{t (t k 1, t k ]}, where 0 = t 0 t 1 t 2 t n <, ξ k are bounded r.v.s and F tk 1 -measurable, and ξ 0 is a bounded F 0 -measurable r.v. In the sequel (F t ) t 0 will be the filtration of a BM B(t), satisfying the usual conditions.

16 Integration of simple previsible processes Example Define Y t = ξ k := max{min{b tk 1, c} c} (c > 0), if t k 1 < t t k, k n, and Y 0 = 0. Then Y is simple previsible. We denote E the set of simple previsible processes (SPPs). Definition (Stochastic integral of SPPs) Let Y E. Then the stochastic integral of Y with respect to B is defined as I B (Y ) = 0 Y s db s := ξ k (B tk B tk 1 ).

17 Integration of simple previsible processes It is trivial to see that E ( I B (Y ) ) 2 <. Hence, I B : E L 2 = L 2 (Ω, A, P). Here are some other simple properties: Well def.: The definition is independent of the (non-unique) representation of Y. Linear: If X and Y are simple previsible and α and β real numbers, then I B (αx + βy ) = αi B (X) + βi B (Y ). Moments: EI B (Y ) = 0 and E ( I B (Y ) ) 2 = E 0 Y 2 s ds.

18 Integration of simple previsible processes The first two properties are simple and left as an exercise. We prove the moments properties. [ ] E Y s db s = E[ξ k E[B tk B tk 1 F tk 1 ]] = 0. k n 0 Furthermore [ ] 2 E Y s db s = E[ξ k ξ l (B tk B tk 1 )(B tl B tl 1 )]. 0 k n l n

19 Integration of simple previsible processes When k < l E[ξ k ξ l (B tk B tk 1 )(B tl B tl 1 )] = E[E[ξ k ξ l (B tk B tk 1 )(B tl B tl 1 ) F tl 1 ]] = E[ξ k ξ l (B tk B tk 1 )E[B tl B tl 1 F tl 1 ]] = 0. Same argument applies when l < k. Hence, [ ] 2 E Y s db s = = = 0 E[ξk 2 (B t k B tk 1 ) 2 ] E[E[ξk 2 (B t k B tk 1 ) 2 F tk 1 ]] = E[ξk 2 E[(B t k B tk 1 ) 2 F tk 1 ]] [ ] E[ξk 2 ](t k t k 1 ) = E ξk 2 (t k t k 1 ) = E 0 Y 2 s ds.

20 Integration of simple previsible processes The mapping I B is transforming a function (Y s ) into a (random) number. We now wish to obtain a random process instead, which should represent t 0 Y sdb s. To this end we define I B (Y ) t := For a simple function this gives I B (Y ) t = 0 Y s I{s t}db s. ξ k (B tk t B tk 1 t).

21 Integration of simple previsible processes We obtain the following further properties. Continuity The paths of I B (Y ) t = t 0 Y sdb s are continuous. Martingale I B (Y ) t an L 2 -bounded martingale, i.e. and for s t sup E ( ) 2 I B (Y ) t < t 0 E[I B (Y ) t F s ] = I B (Y ) s.

22 Integration of simple previsible processes The continuity follows easily from the continuity of the BM paths. The L 2 -boundedness follows from sup t 0 E ( ) 2 I B (Y ) t = sup E t 0 = sup E t 0 0 t 0 Y 2 s I{s t}ds Y 2 s ds = E 0 Y 2 s ds <. (In the last step we used the monotone convergence theorem.)

23 Integration of simple previsible processes Finally, we can add t = t j and s = t k, k j, to the partition and get j E[I B (Y ) t F s ] = E ξ l (B tl B tl 1 ) F s l=1 j = I B (Y ) s + E[ξ l (B tl B tl 1 ) F s ] l=k+1 j = I B (Y ) s + E[ξ l E[B tl B tl 1 F tl 1 ] F s ] = I B (Y ) s. l=k+1

24 Some martingale results Having developed the stochastic integral for the case where the integrand is a simple previsible function, is a first step towards a general definition. To extend the integral to a bigger class of integrals (next lecture), we need some background information about martingales. Recall again that X = (X t : t 0) is a martingale w.r.t. to F = (F t : t 0) if (i) E X t < for all t 0, and (ii) E[X t F s ] = X s a.s. for all 0 s t. Examples: Let (B t ) a Brownian motion and F t = σ(b s : s t). Then the following process are martingales w.r.t. F: 1. B t. 2. B 2 t t. 3. exp(θb t tθ 2 /2). 4. If E Y <, then E[Y F t ] is a martingale.

25 Some martingale results Many important properties of martingales can be shown under the assumption that the paths are càdlàg. Definition (càdlàg) A function f on [0, ) is called càdlàg (continue à droite, limite à gauche), if it is right-continuous and if left limits exist. Example Continuous functions are càdlàg. Distribution functions are càdlàg. f : [0, 2] R with f (t) = sin(1/(1 t))i{t < 1} isn t càdlàg. Since one can prove that if the filtration F fulfills the usual conditions, then every martingale X w.r.t. F possesses a càdlàg version, this is not really a restriction.

26 Some martingale results Theorem (Optional stopping theorem) If X is a martingale with right-continuous paths and if T is a bounded stopping time, then X T is integrable and F T measurable. In addition we have E[X T ] = E[X 0 ] Theorem (Optional sampling theorem) If X is a martingale with right-continuous paths and if S T are bounded stopping times, then E[X T F S ] = X S a.s.

27 Some martingale results Theorem (Doob s inequality) If X possesses right-continuous paths, then [ E ] sup X s 2 4E X t 2. 0 s t Remark Doob s inequality holds in a more general setting for p-th order moments, p > 1. We now provide some results when X is an L 2 -bounded martingale, i.e. a martingale X for which sup EXt 2 <. t 0

28 Some martingale results Theorem (Martingale convergence theorem) If X possesses right-continuous paths, then We have X t L 2 X. E[X F t ] = X t In addition, Doob s inequality holds: [ E a.s. ] sup X s 2 4E X 2. 0 s<

29 Exercises 1. Let (M t ) be a square integrable martingale. Show that for Y F s and E(Y 2 ) < we have E(Y (M t M s )) = 0 for t s. This shows that martingale increments are orthogonal. 2. Consider n B t k (B t k B tk 1 ), with t k = (1 α)t k 1 + αt k, 0 α 1. Show that if the mesh of the partition tends to zero, then the above term converges in probability to 1 2 B2 t + (α 1 2 )t.

30 Exercises 3. Show that the integral of a simple previsible process is independent of the representation of the integrand. 4. Show that the integral of a simple previsible process is linear. I.e. I B (αx + βy ) = αi B (X) + βi B (Y ). 5. Show that if E Y <, then E[Y F t ] is a martingale.