STOCHASTIC ANALYSIS, CONTROLLED DYNAMICAL SYSTEMS AN APPLICATIONS

Transcription

1 STOCHASTIC ANALYSIS, CONTROLLED DYNAMICAL SYSTEMS AN APPLICATIONS Jena, March 2015 Monique Jeanblanc, LaMME, Université d Évry-Val-D Essonne Enlargement of filtration in discrete time

2

3 Page 1 sur 1 09/03/2015

4 Jena Fuchsturm Sieben Wunder (Jena) Wikipedia Page 1 sur 2 10/03/2015 Der Fuchsturm um 1900 Weitere Einzelheiten Unbekannt - Original image: Photochrom print (color photo lithograph) Reproduction number: LC-DIG-

5

6 The general problem of enlargement of filtration is the following one : let X be an F-martingale and G a filtration larger that F. Find conditions such that X is a G semimartingale and then, give the G semimartingale decomposition of the process X, i.e. write X as a G martingale plus a predictable bounded variation process. This is important in finance to exclude arbitrages (for example studying insider trading) and to study the impact of new information in optimizing wealth processes. Few results are known from the 70 s in continuous time, however, in that setting the proofs are not trivial.

7 Many references and books in continuous time (Jeulin, Jacod, Mansuy and Yor, Yor; Protter) Recent thesis with applications to finance: Amendinger, Ankirchner, Aksamit, Deng, Kreher and works of these authors. Some results (presented as particular cases of continuous time) in a discrete time setting can be found in Deng s thesis and the related paper Tahir Choulli and Jun Deng : Non-arbitrage for Informational Discrete Time Market Models, available on Arxiv.

8 In this talk, based on work in progress with Ankirchner and Blanchet-Scaillet, we study enlargement of filtration in discrete time. Our goal is to compute more explicitly the semimartingale decomposition, and to show, with elementary computation, that we recover the classical general formula established in the literature in continuous time. The interest is mainly from a pedagogical point of view.

9 Doob s decomposition: Any discrete time process is a semimartingale in any filtration for which it is adapted: X = M F + V F where M F is an F-martingale and V F is F-predictable, V F 0 = 0 and V F n := V F n V F n 1 = E(X n X n 1 F n 1 ) = E( X n F n 1 ). Setting V F as above, it remains to check that M F is a martingale. Note that M F n := M F n M F n 1 = X n E(X n F n 1 ). If X is an F-martingale, and G any filtration such that F G, it is a G-semimartingale with decomposition X = M G + V G where M G is a G-martingale and V G is G-predictable, and V G n := V G n V G n 1 = E(X n X n 1 G n 1 ) = E( X n G n 1 )

10 Predictable bracket of two martingales If X and Y are two F martingales, there exists an F predictable process K such that XY K is a martingale and K n = E(Y n X n F n 1 ) = E( Y n X n F n 1 ) Indeed, K n = E(X n Y n X n 1 Y n 1 F n 1.

11 Predictable bracket of two martingales If X and Y are two F martingales, there exists an F predictable process K such that XY K is a martingale and K n = E(Y n X n F n 1 ) = E( Y n X n F n 1 ) Indeed, K n = E(X n Y n X n 1 Y n 1 F n 1 ).

12 Predictable bracket of two semi martingales The predictable bracket of two semimartingales X, Y is defined as the dual predictable projection of the covariation process X, Y = [X, Y ] p. For discrete time semimartingales, [X, Y ] n = n k=1 X k Y k, and [X, Y ] p is the only predictable (bounded variation) process such that [X, Y ] [X, Y ] p is a martingale, i.e. [X, Y ] p is the predictable part of [X, Y ]. From Doob s decomposition ( [X, Y ] p ) n = E([X, Y ] n [X, Y ] n 1 F n 1 ) = E( X n Y n F n 1 ) Then, X, Y F n = E( X n Y n F n 1 ).

13 There are, in continuous time, mainly two kinds of enlargement Initial enlargement, where L is a random variable and G t = F t σ(l). Progressive enlargement, where τ is a positive random variable G t = F t σ(t τ).

14 Initial enlargement: an example (Bridge) Let X n = n k=1 Y i, where Y i, i 1 are i.i.d. centered, be a martingale and let G n = Fn X σ(y N ), for n N Then, X = X G + V G where X G is a G martingale and V G n = E(X n X n 1 G n 1 ) = 1 N (n 1) (X N X n 1 ) Indeed, for n k N, E(Y k G n 1 ) = E(Y n G n 1 ), hence E(Y k G n 1 ) = E(Y n G n 1 ) = 1 N N (n 1) E( n Y k G n 1 ) In continuous time: Brownian bridge. For G = F σ(b 1 ), B t = B G t + t 0 B 1 B s 1 s ds, t 1

15 Initial enlargement: an example (Bridge) Let X n = n k=1 Y i, where Y i, i 1 are i.i.d. centered, be a martingale and let G n = Fn X σ(y N ), for n N Then, X = X G + V G where X G is a G martingale and V G n = E(X n X n 1 G n 1 ) = 1 N (n 1) (X N X n 1 ) Indeed, for n k N, E(Y k G n 1 ) = E(Y n G n 1 ), hence E(Y k G n 1 ) = E(Y n G n 1 ) = 1 N N (n 1) E( n Y k G n 1 ) In continuous time: Brownian bridge. For G = F σ(b 1 ), B t = B G t + t 0 B 1 B s 1 s ds, t 1

16 Initial enlargement: another example Let X be a martingale, L be a r.v. taking values in Z and p n (j) = P(L = j Fn X ) and let G n = Fn X σ(l). Then V G n = E( X n F n σ(l)) and V G n 1 {L=j} = 1 {L=j} E( X n 1 {L=j} F n 1 ) P(L = j F n 1 ) = 1 {L=j} E( X, p(j) n F n 1 ) p n 1 (j). = 1 {L=j} E( X n ( F n ) F n 1 ) p n 1 (j) = In continuous time, under Jacod s hypothesis P(τ du F t ) << p t (u)p(τ du) X t = X G t + d X, p(u) s L=u p s (L)

17 Initial enlargement: another example Let X be a martingale, L be a r.v. taking values in Z and p n (j) = P(L = j F X n ) and let G n = F X n Then V G n σ(l). = E( X n F n σ(l)) and V G n 1 {L=j} = 1 {L=j} E( X n 1 {L=j} F n 1 ) P(L = j F n 1 ) = 1 {L=j} E(p n (j) X n F n ) p n 1 (j) = 1 {L=j} E( X, p(j) n F n 1 ) p n 1 (j). In continuous time, under Jacod s hypothesis P(τ du F t ) << p t (u)p(τ du) X t = X G t + t 0 d X, p(u) s L=u p s (L)

18 Progressive enlargement We assume here that τ is a random variable valued in N {+ }, and introduce G n = F n σ(τ n). If Y n G n, then there exists y n F n such that Y n 1 {n<τ} = y n 1 {n<τ}. Any G-predictable process can be written as V n = Vn b 1 {n τ} + Vn a (τ) 1 {τ<n} where, for any u, V a (u) is F-predictable. We introduce two supermartingales Z n = P(τ > n F n ), Zn = P(τ n F n ) and the Doob-Meyer decomposition of Z = M à where M is an F-martingale and à an F-predictable increasing process.

19 We shall use the trivial equalities Z n = P(τ > n 1 F n ), Z n = P(τ n + 1 F n ). On the set {n τ}, Z n and Z n 1 are (strictly) positive. The proof follows from simple arguments E( 1 {n τ} 1 {Zn 1 =0}) = E(P(n τ F n 1 ) 1 {Zn 1 =0}) = E(Z n 1 1 {Zn 1 =0}) = 0 On the set {n > τ}, Z n and Z n 1 are strictly smaller than 1.

20 We shall use the trivial equalities Z n = P(τ > n 1 F n ), Z n = P(τ n + 1 F n ). On the set {n τ}, Z n and Z n 1 are (strictly) positive. The proof follows from simple arguments E( 1 {n τ} 1 {Zn 1 =0}) = E(P(n τ F n 1 ) 1 {Zn 1 =0}) = E(Z n 1 1 {Zn 1 =0}) = 0 On the set {n > τ}, Z n and Z n 1 are strictly smaller than 1.

21 We shall use the trivial equalities Z n = P(τ > n 1 F n ), Z n = P(τ n + 1 F n ). On the set {n τ}, Z n and Z n 1 are (strictly) positive. The proof follows from simple arguments E( 1 {n τ} 1 {Zn 1 =0}) = E(P(n τ F n 1 ) 1 {Zn 1 =0}) = E(Z n 1 1 {Zn 1 =0}) = 0 On the set {n τ}, Z n and Z n 1 are (strictly) positive. On the set {n > τ}, Z n and Z n 1 are strictly smaller than 1.

22 One has, for any random time τ, a) if Y is integrable b) for Y n F n E(Y G n ) 1 {n<τ} = 1 {n<τ} E(Y 1 {n<τ} F n ) Z n E(Y n G n 1 ) 1 {τ n} = 1 {τ n} 1 Z n 1 E(Y n Zn F n 1 ) E( Y n Z n 1 {τ n} G n 1 ) = 1 {τ n} 1 Z n 1 E(Y n 1 { Zn >0} F n 1).

23 Credit Risk Let Z n = P(τ > n F n ) and Z n = M n A n its F-supermartingale decomposition. The process H n Λ n τ is a G martingale, where Λ n = A n Z n 1. Assume now that τ = inf{n : Γ n Θ} where Γ is increasing. Then Z n = e Γ n. If Γ is predictable, Z n = 1 A n = 1 e Γ n, and Λ n = A n Z n 1. If Γ is not predictable, Λ n = e Γ n 1 1 E(e Γ n F n 1 ) Z n 1

24 Credit Risk Let Z n = P(τ > n F n ) and Z n = M n A n its F-supermartingale decomposition. The process H n Λ n τ is a G martingale, where Λ n = A n Z n 1. Assume now that τ = inf{n : Γ n Θ} where Γ is increasing. Then Z n = e Γ n. If Γ is predictable, Z n = 1 A n = 1 e Γ n, and Λ n = A n Z n 1. If Γ is not predictable, Λ n = e Γ n 1 1 E(e Γ n F n 1 ) Z n 1

25 Credit Risk Let Z n = P(τ > n F n ) and Z n = M n A n its F-supermartingale decomposition. The process H n Λ n τ is a G martingale, where Λ n = Z n Z n 1. Assume now that τ = inf{n : Γ n Θ} where Γ is increasing. Then Z n = e Γ n. If Γ is predictable, Z n = 1 A n = 1 e Γ n, and Λ n = A n Z n 1. If Γ is not predictable, Λ n = e Γ n 1 1 E(e Γ n F n 1 ) Z n 1

26 Credit Risk Let Z n = P(τ > n F n ) and Z n = M n A n its F-supermartingale decomposition. The process H n Λ n τ is a G martingale, where Λ n = A n Z n 1. Assume now that τ = inf{n : Γ n Θ} where Γ is increasing. Then Z n = e Γ n. If Γ is predictable, Z n = 1 A n = 1 e Γ n, and Λ n = A n Z n 1 = 1 e Γ n. If Γ is not predictable, Λ n = 1 E(e Γ n F n 1 )

27 Immersion in progressive enlargement F is immersed in G iff any F martingale is a G martingale, or equivalently if Z n = P(τ > n F ) = P(τ > n F k ) for any k n. F is immersed in G if and only if Z is predictable and Z n = P(τ n F ).

28 Assume that F is immersed in G. Then, Z n = P(τ n F n ) = P(τ > n 1 F n ) = P(τ > n 1 F n 1 ) = P(τ > n 1 F ) = P(τ n F ) where the second and the next to last equality follow from immersion assumption. The second equality establishes the predictability of Z. Note that one has Z n = Z n 1.

29 Assume now that Z is predictable and Z n = P(τ n F ). Then, Z n = P(τ n F n 1 ) and P(τ > n F n ) = P(τ n + 1 F n ) = Z n+1 = P(τ > n F )

30 Semi martingale decomposition, Before τ Any F-martingale X stopped at τ is a G-semimartingale with decomposition where Z = M Ã. X τ = X G + τ k=0 1 Z k 1 M, X k

31 To compute the quantity E(X n X n 1 G n 1 ) on the set n 1 < τ, we apply previous results 1 {τ>n 1} (A G n A G n 1) = 1 {τ>n 1} E(X n X n 1 G n 1 ) = 1 {τ>n 1} 1 Z n 1 E( 1 {τ>n 1} (X n X n 1 ) F n 1 ) = 1 {τ>n 1} 1 Z n 1 E(E( 1 {τ>n 1} F n )(X n X n 1 ) F n 1 ) = 1 {τ>n 1} 1 Z n 1 E(E( 1 {τ n} F n )(X n X n 1 ) F n 1 ) = 1 {τ>n 1} 1 Z n 1 E( Z n (X n X n 1 ) F n 1 ).

32 Using now the Doob-Meyer decomposition of Z, and the martingale property of X, we obtain E( Z n (X n X n 1 ) F n 1 ) = E(( M n Ãn)(X n X n 1 ) F n 1 ) = E( M n (X n X n 1 ) F n 1 ) = M, X n and finally 1 {τ>n 1} (A G n A G n 1) = 1 {τ>n 1} 1 Z n 1 M, X n. In continuous time X τ = X G + τ 0 1 Z s d X, M where M = Z A 0, and A0 is the dual optional projection of 1 τ t.

33 Using now the Doob-Meyer decomposition of Z, and the martingale property of X, we obtain E( Z n (X n X n 1 ) F n 1 ) = E(( M n Ãn)(X n X n 1 ) F n 1 ) = E( M n (X n X n 1 ) F n 1 ) = M, X n and finally 1 {τ>n 1} (A G n A G n 1) = 1 {τ>n 1} 1 Z n 1 M, X n. In continuous time X τ = X G + τ 0 1 Z s d X, M s where M = Z A 0, with A0 being the F-dual optional projection of 1 τ t and Z t = P(τ t F t ).

34 After τ, Honest times We now consider the case where τ is honest (and valued in N). We recall the definition and some of the main properties. A random time is honest, if, for any n N, there exists an F n -measurable random variable τ(n) such that or equivalently if there exists τ(n) such that 1 {τ n} τ = 1 {τ n} τ(n) 1 {τ<n} τ = 1 {τ<n} τ(n) It follows that any G-predictable process V can be written as V n = V b n 1 {n τ} + V a n 1 {τ<n} where V a, V b are F-predictable processes.

35 If τ is honest, Z n = Z n on the set n > τ. Furthermore, τ is honest if and only if Z τ = 1 For any n, P(τ = n F n ) 1 {n>τ} = P(τ = n F n ) 1 {n>τ;n>τ(n)} = E( 1 {τ=n} 1 {n>τ(n)} F n ) 1 {n>τ} = E( 1 {τ=n} 1 {n>τ(n)} 1 {n>τ} F n ) 1 {n>τ} = 0 It follows that Z n 1 {τ<n} = Z n 1 {τ<n}. Furthermore, Z n 1 {τ=n} = 1 {τ=n} P(τ n F n ) = 1 {τ=n} 1 {τ(n)=n} P(τ n F n ) = 1 {τ=n} E( 1 {τ(n)=n} 1 {τ n} F n ) = 1 {τ=n} which implies Z τ = 1. Let l(n) = sup{k n : Zk = 1}. Then, on τ n one has τ = l(n).

36 Decomposition in the enlarged filtration. Let X be an F-martingale. Then, X = X + τ k=0 1 Z k 1 M, X k k=τ 1 1 Z k 1 M, X k where X is a G-martingale. One has 1 τ n (A G n+1 A G n) = E( 1 τ n (X n+1 X n ) G n ). We now take the conditional expectation w.r.t. F n. Taking into account that A F is predictable, and that, from the property of honest times, there exists A F, an F-predictable process, such that A G n 1 τ n = A F n 1 τ n one has E( 1 τ n F n )(A G n+1 A G n) = E( 1 τ n (X n+1 X n ) F n ) = E(E( 1 τ n F n+1 )(X n+1 X n ) F n )

37 E( 1 τ n F n )(A F n+1 A F n) = E(E( 1 τ n F n+1 )(X n+1 X n ) F n ) Now, using the fact that E( 1 τ n F n ) = 1 E( 1 τ>n F n ) = 1 Z n E( 1 τ n F n+1 ) = 1 E( 1 τ>n F n+1 ) = 1 E( 1 τ n+1 F n+1 ) = 1 Z n+1 and that X is an F-martingale, we obtain (1 Z n )(A G n+1 A G n) = E( Z n+1 (X n+1 X n ) F n ) = M, X n. It seems important to note that the Doob-Meyer decomposition of Z is not needed.

38 Brackets in F and G. Let X and Y be F adapted processes (hence, semi martingales) Let τ be an honest time. Then X, Y G n 1 n τ = 1 n τ 1 Z n 1 E( Z n X n Y n F n 1 ) X, Y G n 1 τ<n = 1 τ<n 1 1 Z n 1 E((1 Z n ) X n Y n F n 1 )

39 Enlargement with a process For n 0 let U n (dy) be a regular conditional distribution of the random vector Ȳ n 1 = (Y 0,..., Y n 1 ) with respect to F n. Moreover, for n 1 let V n (dy) be a regular conditional distribution of Ȳn 1 with respect to F n 1. Assume that U n (dy) is absolutely continuous wrt V n (dy) for all n 1 and d n (y) := U n(dy) V n (dy). Then, the information drift of X wrt to (G n ) is given by A n = n M, d(z) z=(y0 k,...,y k 1 ). k=1

40 THANK YOU FOR YOUR ATTENTION