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An Explicit Model of Default Time with given Survival Probability

Monique Jeanblanc Shiqi Song

22 juin 2010

Abstract For a given filtered probability space (Ω,F,P), whereF= (Ft)t≥0 is a filtration, anF-adapted continuous increasing processΛand a positive P-Flocal martingaleN such that Zt := Nte−Λt 1, t 0, we construct a model of default time, i.e., a probability measure QZ and a random time τ on an extension of (Ω,F,P), such that Q[τ > t|Ft] = Zt, t 0.

The probability QZ is linked with the well-known Cox model by an explicite density function.

There exist various properties, such as the proportionality or the invariance of conditional laws, which characterize QZ from others. We will make a particular attention on the complete separation property. As usual the model is equipped with an enlarged filtration G = (Gt)t≥0 withGt =Ft∨σ(τ ∧t). We will show that all P-F martingales areQZ-G semimartingale and give an explicite semimartingale decomposition formula.

1 Introduction

1.1 The problem-A

In this paper, we consider a filtered probability space (Ω,F,P), where the elements of the filtration Fare denoted by(Ft)0≤t≤∞withF=0≤t<∞Ft. We consider N a càdlàg positive P-F local martingale and Λ an F-adapted continuous increasing process.1 We assume that N0 = 1,Λ0 = 0and 0≤Nte−Λt 1 for all 0≤t <∞. Let Zt:=Nte−Λt, t≥0. We note that the process Z is a class (D) supermartingale whose Doob-Meyer decomposition, thanks to the continuity of Λ, is dZs =e−ΛsdNs−Zss. The random variableR

0 Zss is integrable and R·

0e−ΛsdNs is a BMO martingale.

Let ( ˆΩ,A,ˆ F,ˆ Q) be a second filtered probability space, where the elements of the filtration

This research benefited from the support of the “Chaire Risque de Crédit”, Fédération Bancaire Française.

Equipe Analyse et Probabilités, EA2172, Université d’Évry Val d’Essonne, 91025 Évry Cedex, France and Institut Europlace de Finance

Equipe Analyse et Probabilités, EA2172, Université d’Évry Val d’Essonne, 91025 Évry Cedex, France

1Notice that in this paper callingaa positive number means thata0and callingf(t), tR,an increasing function meansf(s)f(t)forst.

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Fˆ are denoted by ( ˆFt)t≥0. Let π be a map from Ωˆ into Ω. We say that ( ˆΩ,A,ˆ F,ˆ Q, π) is an extension of (Ω,F,P) ifFˆt =π−1(Ft) for all 0 t≤ ∞, and P= Q|Fˆ◦π−1. In this paper, if extension holds, we shall simply identify F,P with F,ˆ Q|Fˆ

and equally we shall consider a random variableY on(Ω,F)as the random variableY◦πon( ˆΩ,Fˆ). With this identification an F-adapted process onΩis aP-F martingale if and only if it is aQ-Fˆ martingale.

In this paper we study the following problem:

Problem-AConstruct an extension ( ˆΩ,A,ˆ F,ˆ Q, π)of (Ω,F,P) and a random timeτ on ( ˆΩ,A)ˆ such that Q[τ > t|Ft] =Nte−Λt for all 0≤t <∞.

The two processes N and Λ will be considered as the parameters of the problem-A. We solve the problem under the hypothesis:

Hy(N,Λ): Assume that 0 Zt < 1,0 Zt− < 1, for 0 < t < ∞, and almost surely, the random measure 1−Ztt is bounded on any closed interval contained in (0,∞).

We prove that for each pair of N and Λ satisfying Hy(N,Λ), there exists a solution to the problem-A (in a companion paper Jeanblanc and Song [6], we shall construct other solutions).

This particular solution is built entirely in terms of the two factorsN and Λ, whilst in [6] we show that the other solutions depend some other factors. We exhibit the different properties which characterize our solution from others. In particular, we study the relation with the Cox construction, the conditional law invariance property, the separation condition, the (H’) property for the enlargement of filtration problem with a decomposition formula of the same type as in the case of honest time, whilst our solutionτ is even notF-measurable.

1.2 Motivation

This question arises from credit risk modelling. Usually, in that setting, the starting point is a given increasing continuous processΛ, adapted with respect to some information filtration F= (Ft)t≥0. The processΛcontains the only parameters to be calibrated from the market data.

Then, one constructs a random timeτ on some probability space such that(Ht−Λt∧τ, t≥0)is a G-local martingale, whereHt= 11τ≤t, andG= (Gt)t≥0 withGt=Ft∨σ(τ∧t). The "canonical"

construction is the Cox model, where τ := inf{t : Λt Θ} and Θ is a r.v. independent of F, with unit exponential law. However, the Cox model is a fairly simple model, because its conditional survival probability Q[τ > t|Ft] is equal to e−Λt, whilst it is known that, for a general τ, for 11{τ≤t} Λt∧τ to be a G-local martingale, a necessary and sufficient condition is that Q[τ > t|Ft] has a multiplicative decomposition NteΛt, where N is a positive F-local martingale. Our aim here is to take into account this second parameter N and to study the properties of the models with conditional survival probability of the formN e−Λ.

The problem-A has been studied in Gapeev et al. [4] where, inspired by an example from filtering, a solution has been found by solving an integral equation (see Section 6). The result on the integral equation in [4] is important to well understand the problem-A. We shall give below a precise description on the relation between our approach and [4]. Another related work is Nikeghbali and Yor [11] where for a strictly positive martingale N with no positive jumps and limt→∞Nt= 0, the authors have established the multiplicative decomposition formula

P[g > t|Ft] = Nt

St =Nteln(St), 0≤t <∞,

whereSt= sup0≤s≤tNs andg= sup{t0 :Nt=St}. It is a particular case of the problem-A

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withΛt= ln(St). It is particular also because the solution(P, g)is living on the initial spaceΩ.

With the hypothesisHy(N,Λ) our situation is radically different. In fact none of our solutions can live in the initial spaceΩ.

1.3 Miscellaneous setting conditions

In this paper our response to the problem-A will be constructed on the product space [0,∞]×Ω. This space will exclusively be equipped with the productσ-fieldB[0,∞]⊗ F, with the mapπ :π(s, ω) =ω and the map τ:τ(s, ω) =s, with the filtrationFˆ=π−1(F) which will immediately be identified with F. In such a setting, for ([0,∞]×Ω,B[0,∞]⊗ F,ˆF,Q, π) to be an extension of(Ω,F,P), or for ([0,∞]×Ω,Q, τ)to be a solution model of the problem-A, the only element we have to determine is the probabilityQ. Therefore, in these cases, we shall simply say thatQis an extension, or thatQis a solution. We equip systematically the product space [0,∞]×Ω with the progressively enlarged filtration G= (Gt)t≥0 made from F with the random time τ, i.e.Gt=Ft∨σ(τ∧t), t≥0. This fix the framework for our discussion on the enlargement of filtration problem.

Notice that the notation τ can be used in a general situation for a general random time.

This will not cause confusion because, each time the product space is concerned,τ will be the projection map defined above.

We notice also that it is the filtration G which is considered instead of the filtration G+. The reason is that theQ-Gsemimartingale decomposition formula we will establish for theP-F local martingales exhibits only G-adapted càdlàg processes. This yields that in the filtration G+, we will have the same decomposition formula.

2 Basic Analysis

In this section, we collect diverse ideas we can have immediately on the problem-A.

2.1 Simple cases

The simplest form of the problem-A is the case where N 1 and Λ = ι, for which the solution on the product space is the probability measure making τ an exponential random variable independent ofF. More generally, ifN 1and Λ=∞, again a solution exists on the product space given by the probability measureνΛ:

νΛ[A∩ {s < τ ≤t}] =P[IA Z t

s

e−Λvv], A∈ F, 0< s < t <∞ (1) This construction2 coincides with the Cox construction. We will call νΛ the Cox measure.

2.2 Föllmer’s measure

The condition of the problem-Aon the conditional survival probability reminds us immedia- tely of the Föllmer’s measure (see Meyer [9] ; or the Doléans-Dade measure in our case of a class (D) supermartingale). We wonder if the solution of our problem is not simply the Föllmer’s

2In this paper, for a probability measureP, we use the notationP[X]forEP[X]

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measure. To simplify the formula, we suppose Z = 0. The Föllmer’s measure in this case is given by

QF[F] =P[

Z

0

F(s,·)Zss]

Under this measure, the Ft-conditional survival probability of τ is effectively Zt. In order to be a solution of the problem-A,QF must be an extension of(Ω,F,P). This is equivalent to the condition: R

0 Zss 1 (see Nikeghbali [10] for more discussion on such situation), because the restriction ofQF onto F is calculated by

QF[A] =P[IA Z

0

Zss], A∈ F.

We note that in such a case the Föllmer’s measure QF coincides with the Cox measure νΛ. (Although the Föllmer’s measure need not be a solution, we notice that any solution of the problem-A coincides with the Föllmer’s measure on theF-predictableσ-field.)

The above observation is completed by another observation. If a solution (Q, τ) exists, the tails conditional laws of τ are linked to the supermartingale Z by the identity:

Q[τ ∈du, t≤τ ≤ ∞|Ft] =Q[11{t≤u<∞}Zuu+Zδ(du)|Ft]

But few information on Q[τ du,0 τ t|Ft] can be drawn from the supermartingale Z.

That is the main difficulty of our problem.

2.3 Time change

We can believe that ifΛt=tfor all 0≤t <∞, the problem could be easier. The following result clarifies the situation. Set ι(t) =t,0≤t≤ ∞and

Ct= inf{s0 : Λs> t}, 0≤t <∞

Then, (Ct, t 0) is an increasing family of F-stopping times. Since Λ is a finite increasing process,C= limt↑∞Ct=∞. We consider the time change induced by (Ct, t≥0). We recall that ZCt = NCte−t∧Λ, t 0, is an FC-supermartingale with 0 ZCt 1 (we use the fact ΛCt = t∧Λ) and NC = (NCt, t 0) is a true FC-martingale. The following result can be checked straightforwardly :

Theorem 2.1 Assume the hypothesisHy(N,Λ). If(Q, τ)is a solution of the problem-Aon the space(Ω,F,P) with parameters N and Λ, (Q,Λτ) is a solution of the Problem-A on (Ω,FC,P) with parametersNC andι∧Λ. Conversely if(Q, τ)is a solution of the problem-Aon(Ω,FC,P) with parametersNC andι,(Q, Cτ−)is a solution of the problem-Aon(Ω,F,P)with parameters N andΛ.

2.4 Probability change

Suppose Λ = ∞. Let νΛ be Cox measure (see the formula (1)). A natural idea to solve the problem-Ais to look for solutions among the probability measuresQon the product space, which are absolutely continuous with respect toνΛ.

Let G be the progressively enlarged filtration from F by the random time τ. Let Q be a probability measure on the product space, absolutely continuous with respect to νΛ, with

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densityξ = dQΛ. Recall that, under the probability measure νΛ, the immersion property holds (see Bielecki et al. [2]). Then, it can be proved that Q is a solution of the problem-A, if and only if there exists a family L= (Lt, 0≤t <∞) of positiveB[0, t]⊗ Ftmeasurable functions, such that

1. Lsatisfies the equation :

(L) : Nte−Λt + Z t

0

Lt(s,·)e−Λss= 1, ∀t≥0.

2. for any 0< s <∞, the process Lt(s), t≥sis a càdlàg νΛ-Gmartingale 3. for 0≤t <∞,νΛ[ξ|Gt] =NtI{t<τ}+Lt(τ,·)I{τ≤t}.

The key point here is the integral equation(L). In [4] a sufficient condition (see Section 6.2 for a detailed discussion on that condition) has been found to solve the equation (L). We do not know how to solve the equation (L) in general.

2.5 No Solution Case

To solve the problem-A an extension of (Ω,F,P) is in general necessary. For example, no solution can exist onΩfor the problem-Awith parameterN 1because then random variable Λτ is independent ofF (see [2], p.136).

3 A solution of the problem-A

In this section, we assume HypothesisHy(N,Λ). We construct a solution of the problem-A which is absolutely continuous with respect to the measure defined by

P[IA Z

]s,t]

(dΛv+δ(dv))], A∈ F,0≤s < t <∞.

This looks like the probability change approach mentioned in the preceding section. But we will not pass by the equation(L).

3.1 Preliminary computations

Lemma 3.1 Let 0< θ <∞ be fixed and consider the process defined for θ≤t≤ ∞:

Jt=Jt(θ) := (1−Zt) exp

½

Z t

θ

Zs 1−Zss

¾

Then, the processJ = (Jt, θ≤t≤ ∞) is a P-F uniformly integrable martingale.

Proof : Notice thatHy(N,Λ) guarantees the well definiteness ofJ. Recall that the processZ is a supermartingale with the Doob-Meyer decomposition dZs =e−ΛsdNs−Zss. Applying integration by parts formula onJ, for θ≤t <∞,

dJt = exp

½

Z t

θ

Zs s 1−Zs

¾

e−ΛtdNt

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we see that J is aP-F local martingale on [θ,∞). But clearly positive and bounded by 1, it is a uniformly integrable martingale on[θ,∞].

As a corollary, we obtain

Lemma 3.2 Let 0< θ <∞ be fixed and define Et(θ) = Jt(θ)

1−Zθ = 1−Zt 1−Zθexp

½

Z t

θ

Zs s 1−Zs

¾

, θ≤t≤ ∞.

The processes (Et(θ), θ t ≤ ∞) is a P-F uniformly integrable martingale ; and, for t θ, E[Et(θ)|Fθ] = 1.

Note that, in different situation,Et(θ)may be denoted in different forms such asEt(θ, ω), or EtN,Λ(θ). We need to know the behaviour ofEt(θ)when tis fixed andθ→0.

Lemma 3.3 Let 0 < t ≤ ∞ be fixed. There is a sequencen) decreasing to 0 such that limn(1−Zθn)Etn) = 0 P-almost everywhere.

Proof.We begin with the expectation limit. Using the preceding lemma and the fact that Z is càdlàg bounded by 1 andZ0 = 1, we can write

limθ↓0P[(1−Zθ)Et(θ)] = lim

θ↓0P[(1−Zθ)] = 0.

But everything being positive, the sequence(1−Zθ)Et(θ)converges in L1 to zero. The lemma follows.

Lemma 3.4 For any 0< θ <∞, θ≤t <∞, we have Z t

θ

ZvEt(v)dΛv = (1−Zt)(1−Zθ)Et(θ),

and, forθ≤t≤ ∞, Z t

0

ZvEt(v)dΛv = 1−Zt Proof :We compute for 0< θ <∞, θ≤t <∞,

Z t

θ

ZvEt(v)dΛv = Z t

θ

Zv 1−Zt 1−Zv exp

½

Z t

v

Zs s 1−Zs

¾ v

= (1−Zt) exp

½

Z t

v

Zs s 1−Zs

¾¯¯

¯¯

v=t v=θ

= (1−Zt)(1−Zθ)Et(θ)

Letθ goes to zero along the sequence(θn) in the preceding lemma. We obtain Z t

0

ZvEt(v)dΛv = 1−Zt. Passing to the limit, this identity holds fort=∞.

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Remark 3.1 Suppose thatN is continuous. We have the identity

Et(θ) = exp{−

Z t

θ

e−sdNs 1−Zs 1

2 Z t

θ

e−2sdhNis (1−Zs)2 }

This explains the use of the notation of an exponential martingale. Let us write the identity:

1−Zt= (1−Zθ) exp{

Z t

θ

Zuu 1−Zu}Et(θ)

It is the multiplicative decomposition of the positive submartingale(1−Zt, t≥θ) whereEt(θ) is the martingale factor and(1−Zθ) exp{Rt

θ Zuu

1−Zu }is the bounded variation factor. Notice that we keep a θ >0 in the decomposition, because in a neighborhood ofθ= 0the integrability of

1−Z1 θ is troublesome.

3.2 The Construction Define

u(θ) =uZ(θ, ω) :=











0 θ= 0

Zθ(ω)E(θ, ω) 0< θ <∞

Z(ω) θ=

Notice that, for eachθ, the random functionu(θ)onθ∈[0,∞]is non negative and it is càdlàg inθ over(0,∞).

Lemma 3.5 The random function u(θ) is a probability density with respect to the random measure θ+δ(dθ) over [0,∞], i.e.,

Z

0

u(θ)dΛθ+u(∞) = 1. Proof.According to Lemma 3.4, we can write

Z

0

u(v)dΛv+u(∞) = Z

0

ZvE(v)dΛv+Z= (1−Z) +Z= 1

Now we construct a probability QZ on the product space by:

QZ[A∩ {t < τ ≤t0}] =P[IA Z

]t,t0]

u(θ)(dΛθ+δ(dθ))] (2)

for A∈ F,0≤t < t0 ≤ ∞. Applying Lemma 3.5, we see that QZ is an extension of(Ω,F,P), i.e. QZ[A] =P[A], for A ∈ F. Compute now the conditional survival probability under QZ. Let0< t <∞ andA∈ Ft. We have

QZ[A∩ {τ > t}] =P[IA( Z

t

u(v)dΛv+u(∞))] =P[IA( Z

t

ZvE(v)dΛv+u(∞))]

From Lemma 3.4, it follows

QZ[A∩ {τ > t}] =P[IA((1−Z)(1−Zt)E(t) +Z)] =P[IAIA(1−Zt)E(t)]

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According to Lemma 3.2, P[E(t)|Ft] = 1and so

QZ[A∩ {τ > t}] =P[IAIA(1−Zt)] =P[IAZt] =QZ[IAZt] i.e., QZ[τ > t|Ft] =Zt. We can now state the following theorem.

Theorem 3.1 Suppose Hy(N,Λ). Then, the probability measure QZ defined in (2) on the product space is a solution of the problem-A.

4 The density process with respect to Cox measure

Remark that, if N 1, it can be checked that Et(θ) 1,0 < θ t ≤ ∞. If moreover Λ = so that Z = 0, the random function ue−Λ(θ) is simply e−Λθ and the probability measureQZ becomes

Qe−Λ[A∩ {s < τ ≤t}] =P[IA Z t

s

e−Λθθ] =νΛ[A∩ {s < τ ≤t}]

whereνΛ is the Cox measure (see the formula (1)).

Return back to a general Z =N e−Λ. We know by the construction that QZ is absolutely continuous with respect toνΛ. Now we look at its density process.

Theorem 4.1 Suppose Hy(N,Λ) and Λ= so that Z= 0. Set L(s, ω) =

½ Ns(ω)E(s, ω), 0< s <∞, ω

0 s= 0 or s=

(in other words, L=NτE(τ) νΛ-a.s.). Then, L is the probability density of QZ with respect to νΛ: QZ = L·νΛ, and the G-density process Lt = νΛ[L|Gt], 0 t < ∞, is given by Lt = Nt∧τEt∨τ(τ), 0≤t≤ ∞.

Proof. The first part of the theorem is a direct consequence of the construction. For the assertion onG-density process(Lt,0≤t <∞), it is the consequence of the following lemma.

Lemma 4.1 Suppose Λ = ∞. Let 0 a < ∞. Let F(s, ω) be a positive B[0,∞]⊗ F measurable νΛ integrable function on [0,∞]×Ω. Then, we have

νΛ[F|Ga](s, ω) =G(s, ω)I{s≤a}+J(ω)I{a<s}

where

G(s, ω)I{s≤a}=P[F(s,·)|Fa](ω)I{s≤a} νΛ-almost surely J(ω) =eΛa(ω)P[R

a F(s,·)e−ΛsΛ(ds)|Fa](ω) P-almost surely In particular, we have

νΛ[NτE(τ)|Ga] =Na∧τEa∨τ(τ)

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Proof.The first part of the lemma is well-known. Applying this to F =NτE(τ), its G(s, ω) part is computed by

P[NsE(s)|Fa] =NsEa(s) =Na∧sEa∨s(s), s≤a.

ItsJ(ω) part is computed by,s > a, eΛaP[R

a NvE(v)e−ΛvΛ(dv)|Fa] = eΛaP[R

a ZvΛ(dv)|Fa]

= eΛaZa(recall thatZ= 0)

= Na∧sEa∨s(s) Combining these two results we obtain the lemma.

Remark 4.1 The density process (Nt∧τEt∨τ(τ),0 t < ∞) satisfies the equation (L) (see subsection 2.4). It is just a restatement of Lemma 3.4.

5 Characterizations of Q

Z

The probability measureQZ possesses many properties and can be characterized in different ways as we present now. We assume the hypothesisHy(N,Λ) andΛ=∞.

5.1 Definition of new Properties

We introduce the following properties. Let K to be the set of all families (χt,0 t < ∞) such that each χt=χt(s, ω)is a function defined on [0, t]×Ωand isB[0, t]⊗ Ft-measurable.

Partial SeparationAn elementχ= (χt,0≤t <∞)inKis said to be in a partially separable form, if, for any 0 < v < ∞, there exist two F-adapted càdlàg processes (Xt(v))t≥v,(Yt(v))t≥v (defined only fort≥v) such that

χt(s,·) =Xt(v)Ys(v),

for all v s t < ∞. The processes X(v), Y(v) will be called respectively the first and the second components in the partially separable form.

Complete Separation An element χ= (χt,0≤t <∞) inK is said to be in a completely separable form, if there exist twoF-adapted càdlàg processes (Xt)t≥0,(Yt)t≥0 such that

χt(s,·) =XtYs,

for all0≤s≤t <∞. The processes(X, Y) will be called respectively the first and the second components in the completely separable form.

Now consider a probability measure Qon the product space.

¦ Local Density Hypothesis3 We say that Qsatisfies the local density hypothesis, if there exists a positive element α= (αt,0≤t <∞)inK such that, for0≤t <∞ andB ∈ B[0, t],

Q[τ ∈B|Ft] = Z

B

αt(s,·)e−Λss

3To be distinguished from the density hypothesis introduced in Karoui et al. [3]

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The family α will be called a family of local density functions. To such a family α, we can ascribe different properties (We will see later the implications of these properties):

? Decreasing conditionRt

0αt(u,·)e−Λuu >0 for any 0< t <∞, and, for 0≤u <∞, the map

t→ αt(u,·) Rt

0 αt(s,·)e−Λss is a continuous decreasing function int∈[u,∞).

? Partial separation formα is in partially separable form.

? Complete separation formα is in a completely separable form.

? Canonical formThe familyα takes the formαt(s,·) =NsEt(s) for 0< s≤t <∞.

¦ Conditional law InvarianceWe say thatQ satisfies the conditional law invariance hypo- thesis, if, for any0< a≤b, for any A, B∈ B[0, a], we have

Q[τ ∈A|Fa]

Q[τ ∈B|Fa] = Q[τ ∈A|Fb] Q[τ ∈B|Fb] wheneverQ[τ ∈B|Fb]>0.

¦ Proportionality Hypothesis LetQ be a probability measure on (Ω,A). We say that the conditional proportionality hypothesis is satisfied, ifQ[τ ≤t|Ft]>0 for any0< t <∞, and, for any 0< u <∞, the map

t→ Q[τ ≤u|Ft] Q[τ ≤t|Ft]

has a version which is a continuous decreasing function int∈[u,∞).

Remark 5.1 Chronologically it was the proportionality hypothesis which leaded us to find our solution QZ to the problem-A.

5.2 QZ in canonical form

Theorem 5.1 The probabilityQZ constructed in Theorem 3.1 is the unique extension of(Ω,F,P) on the product space which satisfies the local density hypothesis with a local density in canonical form.

Proof.Let0≤t <∞, A∈ Ft, B∈ B[0, t], by construction ofQZ, we can write QZ[A∩ {τ ∈B}] =QZ[IA

Z

B

NsE(s)e−Λss] =QZ[IA Z

B

NsEt(s)e−Λss]

This proves, forQZ, the local density hypothesis with canonical form. The uniqueness is obvious.

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5.3 Computations of conditional laws

In this subsection we consider a probability measure Q on the product space which is a solution of the problem-A. Note that

Q[τ = 0] = lim

t↓0(1Q[τ > t]) = lim

t↓0(1Q[Zt]) = 0.

Lemma 5.1 For any 0< t <∞, for A∈ B[t,∞], Q[τ ∈A|Ft] =Q[

Z

A

Nue−Λuu+I{∞∈A}Z|Ft] Proof.Foru≥t,

Q[τ > u|Ft] =Q[Zu−Z+Z|Ft] =Q[

Z

u

Nse−Λss+Z|Ft] That is what should be proved.

Lemma 5.2 Suppose the proportionality hypothesis. Then, for any 0≤t <∞, forA∈ B[0, t], Q[τ ∈A|Ft] =

Z

A

NsEt(s)e−Λss i.e., the local density hypothesis holds in canonical form.

Proof.Let0< s <∞be fixed. Let rt:= Q[τ ≤s|Ft]

Q[τ ≤t|Ft] = Q[τ ≤s|Ft]

1−Zt , s≤t <∞.

By hypothesis (rt, t≥s) is a continuous decreasing function. It is clearlyF-adapted. Consider the identity

Q[τ ≤s|Ft] = (1−Zt)rt. (3)

The left-hand side of this identity is an F-martingale. Therefore, the right-hand side should have its drift null, i.e.,

rtZtt+ (1−Zt)drt= 0, s≤t <∞.

Solving this equation leads to

rt= exp{−

Z t

s

Zuu

1−Zu}, s≤t < η,

where η := inf{t ≥s :rt = 0}. But the above expression shows that necessarily η = ∞. We substitute this expression of rt into the identity (3) to get

Q[τ ≤s|Ft] = (1−Zt) exp{−

Z t

s

Zuu 1−Zu} Lets↓0. We see that

exp{−

Z t

0

Zuu 1−Zu}= 0

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Using these facts, we write finally

Q[τ ≤s|Ft] = (1−Zt) exp{−

Z t

s

Zuu 1−Zu}

= Z s

0

(1−Zt) exp{−

Z t

u

Zvv 1−Zv} Zu

1−Zuu = Z s

0

NuEt(u)e−Λuu That ends the proof.

Lemma 5.3 We assume that, for any 0< v < a <∞, on the set {Q[v < τ ≤a|Fa]>0} ∈ Fa the map

t→r(v)t = Q[v < τ ≤a|Ft] Q[v < τ ≤t|Ft]

has a version which is a strictly positive continuous decreasing random function in t∈[a,∞).

Then the local density hypothesis holds in canonical form.

Proof.The essential of the proof is the same as in the preceding lemma, but we have to take into account the parameterv >0. Notice

limv↓0Q[v < τ ≤a|Fa] =Q[τ ≤a|Fa] = 1−Za>0 Let0< v < a <∞ and consider

rt=r(v)t := Q[v < τ ≤a|Ft] Q[v < τ ≤t|Ft].

By hypothesis, on the set Av ={Q[v < τ a|Fa]>0} ∈ Fa, the function (rt, a ≤t <∞) is strictly positive, continuous and decreasing, and clearlyF-adapted. Rewrite this fact in another form

Q[v < τ ≤a|Ft] = (mt−Zt)rt, t≥a, (4) where mt = Q[v < τ|Ft] (a càdlàg version). On the set Av, the left-hand side of the identity (4) is an F-martingale on the interval [a,∞). Therefore the right-hand side must has its drift null, i.e.

rtZtt+ (mt−Zt)drt= 0, t≥a.

Let Sn= inf{t≥a:mt−Zt 1n}, ηv = supnSn,

Becausema−Za>0on the setAv, one has ηv > a. Leta≤t < ηv. We can write

−∞<ln(rt) = Z t

a

(mt−Zu) (mt−Zu)

dru ru =

Z t

a

Zuu (mt−Zu) where we use the fact thatln(ra) = 0. In other words,

rt= exp{−

Z t

a

Zuu

mu−Zu}, a≤t < ηv.

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We have now another equation forQ[v < τ ≤a|Fb]on the setAv: Q[v < τ ≤a|Ft] = (Q[v < τ|Ft]−Zt) exp{−

Z t

a

Zuu

Q[v < τ|Fu]−Zu}, a≤t < ηv. Lettingv↓0we getAv Ω,ηv ↑ ∞(because of Hy(N,Λ)) andQ[v < τ ≤a|Ft]Q[τ ≤a|Ft].

We obtain

Q[τ ≤a|Ft] = (1−Zt) exp{−

Z t

a

Zuu 1−Zu} Leta↓0. We see that

exp{−

Z t

0

Zuu 1−Zu}= 0 We have finally

Q[τ ≤a|Ft]

= (1−Zt) exp{−Rt

a Zuu

1−Zu}

= Ra

0(1−Zt) exp{−Rt

s Zuu

1−Zu }1−ZZsss

= Ra

0 NsEt(s)e−Λss The lemma is proved.

5.4 Equivalences

We now prove that, if Q is a solution of the problem-A, the various properties above are equivalent, leading to the fact thatQZ is the only solution that enjoys these properties.

Theorem 5.2 Let Q be a probability measure on the product space which is a solution of the problem-A. The following conditions are equivalent :

1. The proportionality hypothesis.

2. The local density hypothesis in canonical form, i.e. Q coincides with QZ.

3. The local density hypothesis in partially separable form with its first component strictly positive.

4. The local density hypothesis with decreasing condition.

5. The conditional law invariance and the existence of a version of f(a, b) =Q[τ a|Fb], such that, outside of a negligible set, for all 0≤a≤b <∞, we have the limits

lima0↑bf(a0, b) =f(b, b), lim

b0↓bf(a, b0) =f(a, b) (5) Proof.The implication1.2.has been proved in Lemma 5.2.

To see the implication 2.3.we need only to write, for 0< v < s < t, NsEt(s) = Ns1−Z1−Ztsexp{−Rt

s Zww

1−Zw }

= Ns1−Z1−Zt

sexp{−Rt

v Zww

1−Zw +Rs

v Zww

1−Zw }

= 1−Zt

exp{Rt

vZwdΛw 1−Zw }

Nsexp{Rs

v ZwdΛw 1−Zw } 1−Zs

This proves the partial separation form with

Xt(v) = 1−Zt exp{Rt

v Zww

1−Zw }, Ys(v)= Nsexp{Rs

v Zww

1−Zw } 1−Zs

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The condition 3. is proved.

Let us see the implication3.2.. Let0< v < a < b <∞. IfQ[v < τ ≤a|Fa]>0, we have Ra

v Ys(v)e−Λss >0. Since, for a≤t <∞,Xt(v)>0 by hypothesis, we can write r(v)t := Q[v < τ ≤a|Ft]

Q[v < τ ≤t|Ft] = Ra

v Ys(v)e−Λss Rt

vYs(v)e−Λss

This expression defines clearly a strictly positive continuous and decreasing function in t [a,∞). By the Lemma 5.3, we see that the condition 2 must be satisfied.

For the implication 2.4., we introduce the map t→ NsEt(s)

Rt

0NvEt(v)e−Λvv = NsEt(s)

1−Zt = Ns 1−Zsexp

½

Z t

s

Zv 1−Zvv

¾

on the interval[s,∞). It is clearly continuous and decreasing. The condition 4. is proved.

Consider the implication 4. 1.. Let 0 < s ≤t < ∞. First of all Q[τ ≤t|Ft] = 1−Zt is strictly positive. Notice also that

Q[τ ≤t|Ft] =Rt

0αt(v,·)e−Λvv= 1−Zt Q[τ ≤s|Ft] =Rs

0 αt(v,·)e−Λvv

So, Q[τ ≤s|Fs]

Q[τ ≤t|Ft] = 1 1−Zt

Z s

0

αt(v,·)e−Λvv. By hypothesis,

t→ αt(v,·) 1−Zt

is a continuous decreasing function in t [v,∞). The dominated convergence theorem yields

that Q[τ ≤s|Ft]

Q[τ ≤t|Ft]

has a version which is a continuous decreasing function int∈[s,∞). The condition 1. is proved.

Up to now, we have proved the equivalence between the conditions 1. to 4.. To see that these conditions are also equivalent to the condition 5., let us prove first that condition 3. implies the conditional law invariance. We begin with the fact that, for any 0 < v < a b, for any B∈ B[v, a],

Q[τ B|Fb] = Z

B

Xb(v)Yu(v)e−Λuu

IfQ[τ B|Fb]>0,R

BYu(v)e−Λuu >0. Recall that, by hypothesis, Xa(v) >0, Xb(v)>0. These facts yield, forA,B∈ B[v, a],

Q[τ A|Fa] Q[τ B|Fa] =

R

AYu(v)e−Λuu R

BYu(v)e−Λuu = Q[τ A|Fb] Q[τ B|Fb] Such an identity being true for all v >0, we have actually

Q[τ A|Fa]

Q[τ B|Fa] = Q[τ A|Fb] Q[τ B|Fb]

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for any A,B∈ B([0, a]). The invariance hypothesis is proved. Let us now prove that condition 2. implies the property (5). It is enough to set

f(a, b) = Z a

0

NuEb(u)e−Λuu

and we check directly thatf(a, b) is a version of Q[τ ≤a|Fb]which satisfies the property (5).

The implication from condition 1-4 to the condition 5. is proved.

We show finally the implication 5.1.. Letf(a, b),0≤a≤b <∞, be the function in the condition 5. According to the conditional law invariance hypothesis, we can write

f(a, b)

f(b, b) = f(a, b0)

f(b, b0), a≤b < b0 <∞

whenever f(b, b0)>0. The above identities hold outside of a commun exception set. Now the property (5) together with the conditional law invariance makes valid the two limits:

limb0↑b

f(a, b0) f(b0, b0) = lim

b0↑b

f(a, b)

f(b0, b) = f(a, b) f(b, b)

(notice that there is no problem of zero denominatorf(b0, b), because f(b, b) = 1−Zb >0 so that forb0 close to b,f(b0, b)>0.) and

limb0↓b

f(a, b0) f(b0, b0) = lim

b0↓b

f(a, b0)

1−Zb0 = f(a, b) f(b, b)

These two limits mean that f(a,b)f(b,b) is a continuous function in b∈ [a,∞). But this function is also decreasing as the following computation shows:

f(a, b)

f(b, b) = f(a, b0)

f(b, b0) f(a, b0)

f(b0, b0), a≤b≤b0 <∞ The condition 1. is proved

6 Complete Separation Condition

We assumeHy(N,Λ)andΛ=∞. We notice that the completely separable form is absent from Theorem 5.2. This will be the main subject in this section.

6.1 Local Solution

We shall need the notion of local solution.

Local Solution Let Q be a set function on the product space. We say that Q is a local solution to the problem-A, if there exists an increasing sequence(Tn)n≥1 of F-stopping times such that limn→∞Tn = and, for any n 1,Q is a probability measure on any of GTn and P|FTn =Q◦π−1|FTn, andQ[τ > Tn∧t|FTn∧t] =ZTn∧tfor all 0≤t <∞.

Notice that the above two conditions on Q are well defined because they concern only the behaviour ofQover σ-field GTn.

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