Probability, Control and Finance In honor for Ioannis Karatzas - PowerPoint PPT Presentation

Probability, Control and Finance In honor for Ioannis Karatzas Columbia University, June 6, 2012 Monique Jeanblanc, Université d’Évry-Val-D’Essonne Random times and Azéma supermartingales Joint work with S. Song Financial support from Fédération Bancaire Française

Problem Problem Motivation: In credit risk, in mathematical finance, one works with a random time which represents the default time (in a single default context). Many studies are based on the intensity process: starting with a reference filtration F , the intensity process of τ is the F predictable increasing process Λ such that 1 1 τ ≤ t − Λ t ∧ τ is a G -martingale, where G t = ∩ ǫ> 0 F t + ǫ ∨ σ ( τ ∧ ( t + ǫ )) . Then, the problem is : given Λ , construct a random time τ which admits Λ as intensity . 2

Problem A classical construction is: extend the probability space (Ω , F , P ) so that there exists a random variable Θ , with exponential law, independent of F ∞ and define τ := inf { t : Λ t ≥ Θ } 3

Problem Our goal is to provide other constructions. One starts with noting that, in general, Z t = P ( τ > t |F t ) is a supermartingale (called the Azéma supermartingale) with multiplicative decomposition Z t = N t D t , where N is a local martingale and D a decreasing predictable process. Assuming that Z does not vanishes, we set D t = e − Λ t . We shall now assume that Λ is continuous , and that Z 0 = 1 . Then, one proves that Λ is the intensity of τ . 4

Problem Problem ( ⋆ ): let (Ω , F , P ) be a filtered probability space, Λ an increasing continuous process, N a non-negative local martingale such that 0 < N t e − Λ t ≤ 1 Construct, on the canonical extended space (Ω × [0 , ∞ ]) , a probability Q such that 1. restriction condition Q | F ∞ = P | F ∞ 2. projection condition Q [ τ > t |F t ] = N t e − Λ t Here, τ is the canonical map. We shall note P ( X ) := E P ( X ) . 5

Problem Particular case: Z = e − Λ . In that case a solution (the Cox solution) is τ = inf { t : Λ t ≥ Θ } where Θ is a random variable with unit exponential law, independent of F ∞ , or in other words Q = Q C where, for A ∈ F ∞ : � � � t e − Λ u d Λ u Q C ( A ∩ { s < τ ≤ t } ) = P 1 1 A s so that Q C ( τ > θ |F t ) = e − Λ θ , for t ≥ θ 6

Problem Outline of the talk • Increasing families of martingales • Semi-martingale decompositions • Predictable Representation Theorem • Exemple 7

Problem The link between the supermartingale Z and the conditional law Q ( τ ∈ du |F t ) for u ≤ t is: Let M u t = Q ( τ ≤ u |F t ) , then M is increasing w.r.t. u and M u = 1 − Z u u M u M t ≤ t = 1 − Z t t (Note that, for t < u , M u t = E (1 − Z u |F t ) ). Solving the problem ( ⋆ ) is equivalent to find a family M u 8

Family i M Z Family i M Z An increasing family of positive martingales bounded by 1 − Z (in short i M Z ) is a family of processes ( M u : 0 < u < ∞ ) satisfying the following conditions: 1. Each M u is a càdlàg P - F martingale on [ u, ∞ ] . 2. For any u , the martingale M u is positive and closed by M u ∞ = lim t →∞ M u t . 3. For each fixed t , 0 < t ≤ ∞ , u ∈ [0 , t ] → M u t is a right continuous increasing map . 4. M u u = 1 − Z u and M u t ≤ M t t = 1 − Z t for u ≤ t ≤ ∞ . 9

Family i M Z Given an i M Z , let d u M u ∞ be the random measure on (0 , ∞ ) associated with the increasing map u → M u ∞ . The following probability measure Q is a solution of the problem ( ⋆ ) �� � � � M 0 ∞ δ 0 ( du ) + d u M u ∞ + (1 − M ∞ F ( u, · ) Q ( F ) := P ∞ ) δ ∞ ( du ) [0 , ∞ ] The two properties for Q : • Restriction condition : For B ∈ F ∞ , � � � ( M 0 ∞ δ 0 ( du ) + d u M u ∞ + (1 − M ∞ Q ( B ) = P ∞ ) δ ∞ ( du )) = P [ B ] I B [0 , ∞ ] • Projection condition: For 0 ≤ t < ∞ , A ∈ F t , Q [ A ∩ { τ ≤ t } ] = P [ I A M t ∞ ] = P [ I A M t t ] = Q [ I A (1 − Z t )] are satisfied. 10

Constructions of i M Z Constructions of i M Z Hypothesis ( � ) For all 0 < t < ∞ , 0 ≤ Z t < 1 , 0 ≤ Z t − < 1 . The simplest i M Z Assume conditions ( � ). The family � � � t Z s M u t := (1 − Z t ) exp − d Λ s , 0 < u < ∞ , u ≤ t ≤ ∞ , 1 − Z s u defines an i M Z , called basic solution . We note that e − Λ t dM u t = − M u dN t , 0 < u ≤ t < ∞ . t − 1 − Z t − 11

Constructions of i M Z Other solutions To construct an i M Z , we have to check four constraints : i. M u u = (1 − Z u ) ii. 0 ≤ M u iii. M u ≤ 1 − Z iv. M u ≤ M v for u < v These constraints are easy to handle if M u are solutions of a SDE: The constraint i indicates the initial condition; the constraint ii means that we must take an exponential SDE; the constraint iv is a comparison theorem for one dimensional SDE, the constraint iii can be handled by local time as described in the following result : Let m be a ( P , F ) -local martingale such that m u ≤ 1 − Z u . Then, m t ≤ (1 − Z t ) on t ∈ [ u, ∞ ) if and only if the local time at zero of m − (1 − Z ) on [ u, ∞ ) is identically null. 12

Constructions of i M Z Other solutions To construct an i M Z , we have to check four constraints : i. M u u = (1 − Z u ) ii. 0 ≤ M u iii. M u ≤ 1 − Z iv. M u ≤ M v for u < v These constraints are easy to handle if M u are solutions of a SDE: The constraint i indicates the initial condition; the constraint ii means that we must take an exponential SDE; the constraint iv is a comparison theorem for one dimensional SDE, the constraint iii can be handled by local time as described in the following result : Let m be a ( P , F ) -local martingale such that m u ≤ 1 − Z u . Then, m t ≤ (1 − Z t ) on t ∈ [ u, ∞ ) if and only if the local time at zero of m − (1 − Z ) on [ u, ∞ ) is identically null. 13

Constructions of i M Z Other solutions To construct an i M Z , we have to check four constraints : i. M u u = (1 − Z u ) ii. 0 ≤ M u iii. M u ≤ 1 − Z iv. M u ≤ M v for u < v These constraints are easy to handle if M u are solutions of a SDE: The constraint i indicates the initial condition; the constraint ii means that we must take an exponential SDE; the constraint iv is a comparison theorem for one dimensional SDE, the constraint iii can be handled by local time as described in the following result : Let m be a ( P , F ) -local martingale such that m u ≤ 1 − Z u . Then, m t ≤ (1 − Z t ) on t ∈ [ u, ∞ ) if and only if the local time at zero of m − (1 − Z ) on [ u, ∞ ) is identically null. 14

Constructions of i M Z Generating equation when 1 − Z > 0 Hypothesis ( � � ): 1. For all 0 < t < ∞ , 0 ≤ Z t < 1 , 0 ≤ Z t − < 1 . 2. All P - F martingales are continuous. Assume ( � � ). Let Y be a ( P , F ) local martingale and f be a (bounded) Lipschitz function with f (0) = 0 . For any 0 ≤ u < ∞ , we consider the equation  � � − e − Λ t   dN t + f ( X t − (1 − Z t )) dY t , u ≤ t < ∞ dX t = X t 1 − Z t ( ⋆ u )   X u = x 15

Constructions of i M Z Generating equation when 1 − Z > 0 Hypothesis ( � � ): 1. For all 0 < t < ∞ , 0 ≤ Z t < 1 , 0 ≤ Z t − < 1 . 2. All P - F martingales are continuous. Assume ( � � ). Let Y be a ( P , F ) local martingale and f be a bounded Lipschitz function with f (0) = 0 . For any 0 ≤ u < ∞ , we consider the equation  � � − e − Λ t   dX t = X t dN t + f ( X t − (1 − Z t )) dY t , u ≤ t < ∞ 1 − Z t ( ⋆ u )   X u = x Let M u be the solution on [ u, ∞ ) of the equation ( ⋆ u ) with initial condition M u u = 1 − Z u . Then, ( M u , u ≤ t < ∞ ) defines an i M Z . 16

Constructions of i M Z Particular case: in the case of a Brownian filtration, for N = 1 (so that Z t = e − Λ t and f ( x ) = x ,   dM u M u t ( M u t − (1 − Z t )) dB t , u ≤ t < ∞ = t  M u = 1 − Z u u In that case, one can check that M u 1 τ ≤ u . The fact that Z is decreasing show ∞ = 1 that τ is a pseudo-stopping time (i.e., times such that, for any BOUNDED F martingale m , one has E ( m τ ) = m 0 hence, for any F martingale X , the stopped process x τ is a G martingale. 17

Constructions of i M Z Balayage formula when 1 − Z can reach zero We introduce Z = { s : 1 − Z s = 0 } and, for t ∈ (0 , ∞ ) , the random time g t := sup { 0 ≤ s ≤ t : s ∈ Z} Hypothesis ( Z ) The set Z is not empty and is closed. The measure d Λ has a decomposition d Λ s = dV s + dA s where V, A are continuous increasing processes such that dV charges only Z while dA charges its complementary Z c . Moreover, we suppose � t Z s dA s < ∞ I { g t ≤ u<t } 1 − Z s u for any 0 < u < t < ∞ . 18

Constructions of i M Z We suppose that Hy ( Z ). The family � � � t � s Z v M u e − Λ s dN s t = (1 − Z u ) − − I { g s ≤ u } exp dA v 1 − Z v u u defines an i M Z . Note that � � � t Z s M u = I { g t ≤ u } exp − dA s (1 − Z t ) , 0 < u < ∞ , u ≤ t ≤ ∞ . t 1 − Z s u 19