Mean variance hedging under defaults risk.
Sebastien Choukroun, St´
ephane Goutte, Armand Ngoupeyou
To cite this version:
Sebastien Choukroun, St´ephane Goutte, Armand Ngoupeyou. Mean variance hedging under defaults risk.. 2012. <hal-00720912>
HAL Id: hal-00720912
https://hal.archives-ouvertes.fr/hal-00720912
Submitted on 26 Jul 2012
HAL is a multi-disciplinary open access archive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.
L’archive ouverte pluridisciplinaire HAL, est destin´ee au d´epˆot et `a la diffusion de documents scientifiques de niveau recherche, publi´es ou non, ´emanant des ´etablissements d’enseignement et de recherche fran¸cais ou ´etrangers, des laboratoires publics ou priv´es.
Mean variance hedging under defaults risk
Sébastien CHOUKROUN∗†, Stéphane GOUTTE∗‡ AND Armand NGOUPEYOU∗§ July 20, 2012
Abstract
We solve a Mean Variance Hedging problem in an incomplete market where multiple defaults can appear. For this, we use a default-density modeling approach. The global market information is formulated as progressive enlargement of a default-free Brownian filtration and the dependence of default times is modeled by a conditional density hypothesis. We prove the quadratic form of each value process between consecutive defaults times and solve recursively systems of quadratic backward stochastic differential equations. Moreover, we obtain an explicit formula of the optimal trading strategy. We illustrate our results with some specific cases.
Keywords:Mean variance hedging; default-density modeling; Quadratic backward stochastic differential equation (BSDE); Dynamic programming.
MSC Classification (2010):60J75, 91B28, 93E20.
Introduction
In this paper, we study the problem of mean variance hedging in a financial market model subject to defaults and contagion risk. We consider multiple defaults events corresponding for example of a succession of crisis periods for a country or a succession of bad annual financial results for a firm. These defaults could induce loss or gain on the asset price. A classic approach to model this is to use an Itô process governed by some Brownian motionW for the asset price
S and jumps appearing at random default times, associated to a marked point processµ. Hence the mean variance hedging problem in this incomplete market framework may be then studied by stochastic control and dynamic programming methods in the global filtrationGgenerated by
W andµ. This leads in principle to Hamilton-Jacobi-Bellman integro-differential equations in a Markovian framework, and more generally to Backward Stochastic Differential Equations (BS-DEs) with jumps, and the derivation relies on a martingale representation underGwith respect toW andµ, which holds under intensity hypothesis on the defaults, and the so-called immersion
∗
Laboratoire de Probabilités et Modèles Aléatoires, CNRS, UMR 7599, Universités Paris 7 Diderot.
†
Mail: sebastien.choukroun@univ-paris-diderot.fr
‡
Supported by the FUI projectR=M C2. Mail: goutte@math.univ-paris-diderot.fr
§
property (or (H)-hypothesis). Such an approach was used in [4] for the multiple defaults case or in [3] for the mean variance hedging problem underGfor defaultable claims.
The mean variance hedging problem was introduced in [2] and many papers have followed and developed this approach. In most of these papers, this problem was solved with continuous filtration [11], [12]. The authors use the dual’s approach to show the existence of the variance optimal measure (VOM). Moreover, they can write the solution of the primal problem using Back-ward Stochastic Differential Equations (BSDEs) whose existence of solutions are deduced by the existence of the VOM. In the case of discontinuous filtration, the VOM is not always a probability measure (see [1] for conditions), so we cannot use the previous approach to solve our problem. That is why, in general, in the case of discontinuous filtration, the authors make the assumption the VOM is a true probability measure as [9] and then deduce the solution of the primal problem using BSDEs. They so prove the existence of the solution of each BSDE using the VOM. Indeed, without the fact that the VOM is a true probability, it is difficult to show the existence of solution of the corresponding BSDEs with jumps since these BSDEs coefficients are not standard.
In a general model with discontinuous filtration generated by a continuous process and a dis-continuous process, the author in [10] proved the existence of the solution of the BSDEs for the mean variance problem assuming that the coefficients of its asset are adapted with respect to the continuous filtrationF. This strong assumption allows him not to assume that the VOM is a true probability and leads him to solve directly the main BSDE without any assumption on the VOM.
In this paper we work also in the case of a discontinuous filtrationG. In our model, jumps are generated by default times. So, we cannot use the same technics as [10], since his strong assump-tion is not well satisfied in our framework. Indeed, our assets coefficients depend on the jumps (defaults) . Therefore, we use a different approach than the one mentioned previously. Indeed, we use an approach initiated and studied in [5]. By viewing the global filtrationGas a progres-sive enlargement of filtrations of the default-free filtrationFgenerated by the Brownian motion
W, with the default filtration generated by the random times, the basic idea is to split the global mean variance problem, into sub-control problems in the reference filtrationFand corresponding to mean variance problems in default-free markets between two default times. More precisely, we derive a backward recursive decomposition by starting from the mean variance problem when all defaults occurred, and then going back to the initial mean variance problem before any default. The main point is to connect this family of stochastic control problems in the F-filtration, and this is achieved by assuming the existence of a conditional density on the default times given the default-free informationF. So we will use the approach of [5] to show that between each default time, using dynamic programming method, we can first characterize each dynamic version of the mean variance hedging problem in a quadratic decomposition form. These decompositions will depend explicitly on the parameters and default times of our model. Secondly, we will express the three terms appearing in this quadratic decomposition form as solution of three explicits backward stochastic differential equations (BSDEs). Then, starting after the last default event and then going back to the initial mean variance problem we will obtain for this each subset a system of recursive BSDEs. We will prove explicitly the existence and uniqueness of the solution of theses systems of quadratic BSDEs which is not trivial and we will find the optimal mean variance hedging strategy. The paper is so structured as follows: In section 1, we will introduce our model and the corresponding mean variance hedging problem. We will give the systems of BSDEs. Then, in section 2, we will give the solution to the mean variance hedging problem. For this, firstly, we will begin by giving a proof of the existence of a solution of the recursive system of BSDEs. Secondly,
we will give the BSDEs characterization by a verification theorem. Finally, in section 3, we will give some numerical illustrations.
1
Multiple defaults model
1.1 Market information
We adopt in this paper the same model and notations as in [5]. Letτ = (τ1, ..., τn) be now a
vector of thenrandom times andL= (L1, ..., Ln)be a vector of thenmarks associated toτ,Li
being aG-measurable random variable taking values inE ⊂Rand representing for example the loss given default at timeτi. We denote, fork={1, . . . , n},Dk = (Dtk)t∈[0,T]whereDkt = ˜Dtk+ andD˜k
t = σ(1τk≤s, s ≤ t)∨σ(Lk1τk≤s, s ≤t)the filtrations generated by the associated jump
processes. ThenG = (Gt)t∈[0,T] will be the enlarged progressive filtrationF∨D1∨...∨Dn, representing the structure of the global information available for the investors over[0, T]. In other words,Gis the smallest right-continuous filtration containingFsuch that for any1≤k≤n,τkis
aG-stopping time andLkisGτk-measurable. We shall assume that the default times are ordered
(i.e.τ1 ≤...≤τn) and so valued in∆non{θn≤T}where, fork= 1, ..., n, we denote
∆k :=
n
(θ1, ..., θk)∈(R+)k :θ1 ≤...≤θk
o
.
This means that we do not distinguish specific credit names and only observe the successive default times. For any(θ1, ..., θn)∈∆n,(l1, ..., ln)∈En, we denote byθ= (θ1, ..., θn),l= (l1, ..., ln),
andθk = (θ1, ..., θk),lk = (l1, ..., lk)for0 ≤k≤nwith the conventionθ0 =l0 =∅. We also denoteτk= (τ1, ..., τk)andLk = (L1, ..., Lk). Moreover, for0≤t≤T, the setΩkt denotes the
event
Ωkt := {τk≤t < τk+1},
(withΩ0t = {t < τ1}andΩnt = {τn ≤ t}) and represents the scenario wherekdefaults occur
before timet. We callΩk
t thek-default scenario at time t. We define similarlyΩkt− ={τk < t≤
τk+1}. We denote byP(F)theσ-algebra ofF-predictable measurable subsets onR+×Ω, and by
PF(∆k, Ek)the set of indexedF-predictable processesZk(., .), i.e. s.t. the map(t, ω,θk,lk) →
Zk
t(ω,θk,lk) is P(F)⊗ B(∆k) ⊗ B(Ek)-measurable. We also denote by OF(∆k, Ek) the set
of indexed F-adapted processes Zk(., .), i.e. s.t. for all 0 ≤ t ≤ T, the map (ω,θk,lk) →
Ztk(ω,θk,lk)isFt⊗ B(∆k)⊗ B(Ek)-measurable. Hence we have that anyG-predictable process
Z= (Zt)0≤t≤T has a decomposition in the form
Zt = n X k=0 1Ωk t−Z k t(τk, Lk), 0≤t≤T
whereZklies inPF(∆k, Ek). We assume also thedensity hypothesiswhich is given in multiple defaults case by the following statement:
Assumption 1.1 (Density hypothesis). There exists α ∈ OF(∆n, En) such that for any Borel
functionf on∆n×Enand0≤t≤T :
E[f(τ, L)|Ft] =
Z
∆n×En
f(θ,l)αt(θ,l)dθη(dl) a.s., (1.1)
wheredθ = dθ1...dθnis the Lebesgue measure onRn, and η(dl) is a Borel measure onEn in
the formη(dl) = η1(dl1)Qkn=1−1ηk+1(lk, dlk+1), withη1 a nonnegative Borel measure onE and
Remark 1.1. The condition(1.1)implies that in the case thatαis separable in the formαt(θ,l) =
ατt(θ)αLt(l)that the random times and marks are independent givenFt.
1.2 Asset price model under default risk
The trading assetSis aG-adapted process which admits (as in [5]) the following decomposed form St= n X k=0 1Ωk tS k t(τk,Lk), (1.2)
where Sk(θk,lk), θk = (θ1, ..., θk) ∈ ∆k, lk = (l1, ..., lk) ∈ Ek, is an indexed process in
OF(∆k, Ek), valued inR+, representing the asset value in the k-default scenario, given the past default eventsτk = θk, and the marks at default Lk = lk. Notice thatStis equal to the value
Stkonly on the setΩkt, that is, only forτk ≤t < τk+1.The dynamic of the indexed processSkis given by
dStk(θk,lk) = Stk(θk,lk)(µkt(θk,lk)dt+σtk(θk,lk)dWt), θk≤t≤T (1.3)
where W is a one-dimensional (P,F)-Brownian motion, µk and σk are indexed processes in
PF(∆k, Ek), valued inR. We make, as in the one default case, the usual no-arbitrage assumption that there exists an indexed risk premium processλk∈ PF(∆k, Ek)s.t. for all(θk,lk)∈∆k×Ek,
σtk(θk,lk)λkt(θk,lk) =µkt(θk,lk), 0≤t≤T. (1.4)
Moreover, in this contagion risk model, each default time may induce a jump in the assets port-folio. This is formalized by considering a family of indexed processesγk, 0 ≤ k ≤ n−1, in
PF(∆k, Ek, E), and valued in[−1,∞). For(θk,lk)∈∆k×Ek, andlk+1∈E,γtk(θk,lk, lk+1) represents the relative vector jump size on the asset at timet=θk+1 ≥θkwith a marklk+1, given the past default events(τk,Lk) = (θk,lk). In other words, we have :
Sθk+1 k+1(θk+1,lk+1) =S k θ−k+1(θk,lk) 1 +γθkk+1(θk,lk, lk+1) (1.5)
1.3 Strategy and wealth process
The trading strategy is aG-predictable processπ, hence decomposed in the form
πt= n X k=0 1Ωk t−π k t(τk,Lk), 0≤t≤T (1.6)
whereπk is an indexed process inPF(∆k, Ek), and πk(θk,lk) is valued in closed setAk ofR containing the zero element, and representing the amount invested continuously in the asset in the
k-default scenario, given the past default eventsτk = θk and the marks at defaultLk = lk, for
(θk,lk) ∈∆k×Ek. We shall often identify the strategyπ with the family(πk)0≤k≤ngiven in
1.6, and we require the integrability conditions : for allθk∈∆k,lk∈Ek,
Z T 0 |πtk(θk,lk)µkt(θk,lk)|dt+ Z T 0 |πtk(θk,lk)σkt(θk,lk)|2dt <∞, a.s. (1.7)
Given a trading strategyπ= (πk)0≤k≤n, the corresponding wealth process is given by
Xt= n X k=0 1Ωk tX k t(τk,Lk), 0≤t≤T (1.8)
where Xk(τk,Lk), θk ∈ ∆k, lk ∈ Ek, is an indexed process in OF(∆k, Ek), representing
the wealth controlled byπk(θk,lk)in the price processSk(θk,lk), given the past default events
τk =θkand the marks at defaultLk=lk. From the dynamics (1.3) and under (1.7), it is governed
by
dXtk(θk,lk) =πkt(θk,lk)(µkt(θk,lk)dt+σtk(θk,lk)dWt), θk≤t≤T. (1.9)
Moreover, each default time induces a jump in the asset price process, and then also on the wealth process. From (1.5), it is given by
Xθk+1 k+1(θk+1,lk+1) =X k θ−k+1(θk,lk) +π k θk+1(θk,lk)γ k θk+1(θk,lk, lk+1).
Finally, the payoff is a boundedGT-measurable random variableHT which admits the
decom-position form given by
HT = n X k=0 1Ωk TH k t(τk,Lk), (1.10)
whereHTk(., .)isFT ⊗ B(∆k)⊗ B(Ek)-measurable and represents the payoff when kdefaults occurred before maturityT.
Remark 1.2. We have between each default time (i.e. in each time eventsΩkt := {τk ≤ t <
τk+1},t∈[0, T]) that the market is complete.
1.4 The mean variance problem
On our problem of mean variance hedging (MVH), the performance of an admissible trading strategyπ ∈ AGstarted with an initial capitalx∈Ris measured over the finite horizon T by
J0H(x, π) =E[(HT −XTx,π)2] (1.11)
and the MVH problem is formulated as
V0H(x) = inf
π∈AGJ
H
0 (x, π).
1.4.1 Value functions
We define, first, the corresponding multiple defaults admissible trading strategies set: Definition 1.1. For0≤k≤n,Ak
Fdenotes the set of indexed processesπ
kinP
F(∆k, Ek), valued
inAksatisfying(1.7), and such that
E " Z T θk |πsk(θk,lk)|2ds # <∞ (1.12) We then denote byAG= (Ak
F)0≤k≤nthe set of admissible trading strategiesπ = (π
k)
0≤k≤n.
Under the density hypothesis 1.1, let us define a family of auxiliary processesαk∈ OF(∆k, Ek),
0≤k≤n, which is related to the survival probability and is defined by recursive induction from
αn=α, αkt(θk,lk) = Z ∞ t Z E αkt+1(θk, θk+1,lk, lk+1)dθk+1ηk+1(lk, dlk+1), (1.13)
for0≤k≤n−1, so thatP[τk+1 > t|Ft] = R∆k×Ekαkt(θk,lk)dθkη(dlk)andP[τ1 > t|Ft] =
α0t, where dθk = dθ1...dθk, η(dlk) = η1(dl1)...ηk(lk−1, dlk). Given πk ∈ AkF, we denote by
Xk,x(θk,lk)the controlled process solution to (1.9) and starting fromxatθk. We now give our
model hypothesis:
Assumption 1.2. We assume for allt ∈ [θk, T]and0 ≤ k ≤ n thatµkt,σkt,γkt and the family
processesαk ∈ OF(∆k, Ek) are uniformly bounded. Moreover, we assume for0 ≤k ≤ nthat the measureηk(dlk)is uniformly bounded too.
1.4.2 The mean variance hedging problem
The value function to the global mean varianceG-problem (1.11) is then given, in the multiple defaults case, in a backward induction from theF-problems (see [5] for more details) :
Vn(x,θ,l) = ess inf πn∈An F E h (HTn−XTn,x(θ,l))2αT(θ,l)|Fθn i (1.14) Vk(x,θk,lk) = ess inf πk∈Ak F E[(HTk−X k,x T (θk,lk)) 2αk T(θk,lk) + (1.15) Z T θk Z E Vk+1(Xθk,x k+1(θk,lk) + π k θk+1(θk,lk).γ k θk+1(θk,lk, lk+1),θk+1,lk+1)ηk+1(lk, dlk+1)dθk+1|Fθk]
where we recall thatθn=θ,ln=l,θ0=θ0 =∅andl0 =l0=∅.
Remark 1.3. If there exists, for all0 ≤k ≤n, someπk,∗ ∈ Ak
Fattaining the essential infimum
in the previous equations, then the strategy π∗ = (πk,∗)0≤k≤n ∈ AG is optimal for the MVH
problem.
2
Solution to the mean variance hedging problem
We exploit the quadratic form of the mean variance hedging problem in order to characterize by dynamic programming methods the solutions to the stochastic optimization problems (1.14) and (1.15) in terms of a recursive system of indexed BSDEs with respect to the filtrationF. We use a verification approach in the following sense:
1. Firstly, we derive formally the system of BSDEs associated to theF-stochastic control prob-lems (1.14) and (1.15) using dynamic programming principle.
2. Secondly, we obtain the existence of the solutions of the corresponding system of BSDEs (see Theorem 2.1).
3. Finally, in a verification Theorem (see Theorem 2.2), we prove that these BSDEs solutions are unique and provide the solution to our mean variance hedging problem. We prove also that the strategy found in step 1 is optimal and admissible. Moreover, we prove that the quadratic representation form of our value function are true.
So let’s begin with point 1: Fort ∈ [θn, T], νn ∈ AnF, let us introduce the set of controls
coinciding with strategyνuntil timet:
An F(t, ν n) ={πn∈ An F :π n .∧t=ν.n∧t}
We can now define the dynamic version of (1.14) by considering the family ofF-adapted pro-cesses: Vtn(x,θ,l, νn) = ess inf πn∈An F(t,ν n)E h (HTn−XTn,x(θ,l))2αT(θ,l)|Ft i , t≥θn, (2.16) so thatVn θn(x,θ,l, ν n) = Vn(x,θ,l) for any νn ∈ An
F. From the dynamic programing
prin-ciple, one should have the submartingale property on {Vtn(x,θ,l, νn) , θn≤t≤T}, for any
ν ∈ An
F, and if an optimal strategy exists for (2.16), we should have the martingale property of
{Vn
t (x,θ,l, π∗,n), θn≤t≤T}for some π∗,n ∈ AnF. Moreover, since we work on a quadratic
minimization approach, the value processVtn(x,θ,l, νn)should admit the quadratic form decom-position given by
Vtn(x,θ,l, νn) =vtn,θ,l(Xtn,x(θ,l)−Ytn,θ,l)2+ξtn,θ,l, t∈[θn, T]
We search a triple(vn,θ,l, Yn,θ,l,, ξn,θ,l)in the form
(En) dvtn,θ,l vtn,θ,l =−g n,θ,l,(1) t (v n,θ,l t , β n,θ,l t )dt+β n,θ,l t dWt, dYtn,θ,l =−gtn,θ,l,(2)(Ytn,θ,l, Ztn,θ,l)dt+Ztn,θ,ldWt dξtn,θ,l =−gn,t θ,l,(3)(ξtn,θ,l, Rn,t θ,l)dt+Rn,t θ,ldWt. (2.17)
Then, by using the above submartingale and martingale property of the dynamic programming principle and sinceVTn(x,θ,l, νn) = (XTn,x(θ,l)−HTn(θ,l))2αT(θ,l) by (2.16), we see from
Itô calculus (see Proposition 3.5 of Goutte and Ngoupeyou [3] for more details) that the triple
(vn,θ,l, Yn,θ,l, ξn,θ,l)satisfies (2.17) for allt∈[θn, T]with terminal conditionsvTn,θ,l =αT(θ,l),
YTn,θ,l =HTn(θ,l)andξtn,θ,l = 0. The corresponding coefficients of the BSDEs are given by the following equations: gtn,θ,l,(1)=−(µ n(θ,l) +σn(θ,l)βn,θ,l t ) 2 (σn(θ,l))2 , g n,θ,l,(2) t =− µn(θ,l) σn(θ,l)Z n,θ,l t and g n,θ,l,(3) t = 0.
We have, also, that the optimal strategyπn,∗ (such thatVtn(x,θ,l, πn,∗) is a true martingale) is given for allt∈[θn, T]by
πtn,∗(θ,l) = ftn,θ,l,1Xtn,x(θ,l) +ftn,θ,l,2 (2.18) where ftn,θ,l,1:=− 1 σtn,θ,l2 µtn,θ,l+σtn,θ,lβtn,θ,l and ftn,θ,l,2:= 1 σtn,θ,l2 h σtn,θ,lZtn,θ,l+Ytn,θ,lµn,t θ,l+σtn,θ,lβtn,θ,li
Hence, the optimal strategy is linear inXwhich is the case in the no default model. We will refer in the sequel to this problem as the(En)problem.
Next, consider the problem (1.15) and define similarly the dynamic version by considering the value function process given by:
Vtk(x,θk,lk, νk) = ess inf πk∈Ak F(t,ν k)E [(HTk(θk,lk)−XTk,x(θk,lk))2αkT(θk,lk) + (2.19) Z T t Z E Vθk+1 k+1(X k,x θk+1(θk,lk) + π k θk+1(θk,lk).γ k θk+1(θk,lk, lk+1),θk+1,lk+1)ηk+1(lk, dlk+1)dθk+1|Ft]
for θk ≤ t ≤ T, where AkF(t, νk) = {πk ∈ AFk : πk.∧t = ν.k∧t}, for νk ∈ AkF so that
Vk
θk(x,θk,lk, ν
k) =Vk(x,θ
k,lk). Similarly, we will refer in the sequel to this problem as(Ek)
problem fork = 0, . . . , n−1. The dynamic programming principle for (2.19) formally implies that the process
Vtk(x,θk,lk, νk)+ Z t 0 Z E Vθk+1 k+1(X k,x θk+1(θk,lk)+π k θk+1(θk,lk).γ k θk+1(θk,lk, lk+1),θk+1,lk+1)ηk+1(lk, dlk+1)dθk+1
fort∈[θk, T]is a submartingale for anyνk∈ AkFand a true martingale forπ∗,kif it is an optimal
strategy for (2.19). Again, since we work on a quadratic minimization approach, the value process
Vtk(x,θk,lk, νk)should admit the quadratic form decomposition given by
Vtk(x,θk,lk, νk) =vtk,θk,lk,(X k,x t (θk,lk)−Ytk,θk,lk) 2 +ξk,θk,lk t ,∀k= 0, . . . , n−1
We search also a triplevk,θk,lk, Yk,θk,lk, ξk,θk,lkfor allk= 0, . . . , n−1, in the form
(Ek) dvk,θk,lk t vk,θk,lk t =−gk,θk,lk,(1) t (v k,θk,lk t , β k,θk,lk t )dt+β k,θk,lk t dWt dYk,θk,lk t =−g k,θk,lk,(2) t (Y k,θk,lk t , Z k,θk,lk t )dt+Z k,θk,lk t dWt dξk,θk,lk t =−g k,θk,lk,(3) t (ξ k,θk,lk t , R k,θk,lk t )dt+R k,θk,lk t dWt (2.20)
Then, by using the above submartingale and martingale property of the dynamic programming principle and sinceVTk(x,θk,lk, νk) =
XTk,x(θk,lk)−HTk(θk,lk)
2
αkT(θk,lk) by (2.19), we
see from Itô calculus (see again Proposition 3.5 of Goutte and Ngoupeyou [3] for more details) that the triple vk,θk,lk, Yk,θk,lk, ξk,θk,lk satisfies (2.20) for allt ∈ [θ
k, T]with terminal
con-ditionsvk,θk,lk
T = αkT(θk,lk),YTk,θk,lk = HTk(θk,lk)andξTk,θk,lk = 0. And the corresponding
coefficients of the BSDEs are given by the following equations:
gk,θk,lk,(1) t = Z E (1 +vJ,k,θk,lk t )ηk+1(lk, dlk+1) − µkt +σtkβk,θk,lk t + R E(1 +v J,k,θk,lk t )γtk(θk,lk, lk+1)ηk+1(lk, dlk+1) 2 (σtk)2+R E(1 +v J,k,θk,lk t )(γtk(θk,lk, lk+1))2ηk+1(lk, dlk+1) , gk,θk,lk,(2) t = β k,θk,lk t Z k,θk,lk t + Z E UJ,k,θk,lk t (1 +v J,k,θk,lk t )ηk+1(lk, dlk+1) + −R EU J,k,θk,lk t (1 +v J,k,θk,lk t )γtk(θk,lk, lk+1)ηk+1(lk, dlk+1)−σktZ k,θk,lk t (σk t)2+ R E(1 +v J,k,θk,lk t )(γkt(θk,lk, lk+1))2ηk+1(lk, dlk+1) × µkt +σtkβk,θk,lk t + Z E (1 +vJ,k,θk,lk t )γkt(θk,lk, lk+1)ηk+1(lk, dlk+1)
and gk,θk,lk,(3) t = v k,θk,lk t [ Z E (UJ,k,θk,lk t ) 2 (1 +vJ,k,θk,lk t )ηk+1(lk, dlk+1) + (Ztk,θk,lk)2 − −R E(1 +v J,k,θk,lk t )U J,k,θk,lk t γtk(θk,lk, lk+1)ηk+1(lk, dlk+1)−σtkZ k,θk,lk t 2 (σkt)2+R E(1 +v J,k,θk,lk t )(γtk(θk,lk, lk+1))2ηk+1(lk, dlk+1) ] where 1 +vJ,k,θk,lk = v k+1,θk+1,lk+1 vk,θk,lk t and UJ,k,θk,lk =Yk+1,θk+1,lk+1−Yk,θk,lk.
The optimal strategyπk,∗(such thatVtk(x,θk,lk, πk,∗)is a true martingale) is given by
πtk,∗(θk,lk) = 1 (σkt)2+R E(1 +v J,k,θk,lk t )γtk(θk,lk, lk+1)2ηk+1(lk, dlk+1) h σktZk,θk,lk t −K k,θk,lk t (µkt +σtkβ k,θk,lk t ) + R E(X k,x t (θk,lk)v k+1,θk+1,lk+1 t −Yk+1,θk+1,lk+1v k+1,θk+1,lk+1 t )γtk(θk,lk, lk+1)ηk+1(lk, dlk+1) vk,θk,lk t # (2.21) withKk,θk,lk t :=X k,x
t (θk,lk)−Yk,θk,lk. Again, we obtain a linear form of the optimal strategy
with respect to X. We will refer in the sequel to this problem as the(Ek)problem,k∈ {0, . . . , n−
1}.
Remark 2.4. For all(Ek)problems,k ∈ {0,1, . . . n}, we work in the time interval given for all t∈[θk, T]. Hence for the particular case where we take the value function fort=θk, we obtain
thatVtk=θ
k(x,θk,lk, ν
k) :=Vk
θk(x,θk,lk, ν
k) =Vk(x,θ
k,lk), where we recall thatxis the value
ofXkinθk, soXθkk =x.
Hence,(Ek)and(En)define thus a recursive system of families of BSDEs, indexed by(θ,l)∈
∆n(T)×En, and the rest of this paper is devoted first to prove the existence of a solution of these
system of BSDEs, and then to its uniqueness via verification theorem relating the solution to the value function 2.19 and 2.16.
2.1 Existence of a solution of the recursive system of BSDEs
The generators of our recursive system of BSDEs (2.17) and (2.20) are not trivial since the co-efficientsgk,θk,lk, k ∈ {0, . . . , n} are not standards. Hence, we give a Theorem to insure that
recursive BSDEs solutions exist and stay in their own solution’s space for allk ∈ {0,1, . . . , n}. Let consider the family{Q(θ),(θ,l)∈[0, T]×En}of probability measures such that the Radon Nikodym density ofQ(θ,l)with respect toP onFT is given by
ZTQ(θ,l) := dQ(θ,l) dP |FT = exp " Z T θ µn s(θ,l) σn s(θ,l) dWs− 1 2 Z T θ µn s(θ,l) σn s(θ,l) 2 ds # . (2.22) Theorem 2.1. For allk∈ {0,1, . . . , n}andt∈[θk, T], we have that
1. There exists a couple vk,θk,lk
t , β k,θk,lk
t
∈ S∞ ×BMO of the first BSDE of (2.20) (if k6=n) and(2.17)(ifk=n) and there exists constantsδ1kandδk2 such that
0< δ1k≤vk,θk,lk
Moreover, for the casek=n, we have an explicit solution which is vtn,θ,l= E ZTQ(θ,l) ZtQ(θ,l) !2 1 αT(θ,l) |Ft −1 (2.23)
2. There exists a coupleYk,θk,lk
t , Z k,θk,lk
t
∈ S∞×BMOsolution of the second BSDE of
(2.20)(ifk6=n) and(2.17)(ifk=n). Moreover, for the casek =n, we have an explicit solution which is Ytn,θ,l =E " ZTQ(θ,l) ZtQ(θ,l)H n T(θ,l) Ft # =EQ(θ,l)hHTn(θ,l) Ft i . (2.24)
3. There exists a coupleξk,θk,lk
t , R k,θk,lk
t
∈ S∞×BMOsolution of the third BSDE of (2.20)
(ifk6=n) and(2.17)(ifk=n). Moreover, for the casek=n, we have an explicit solution which isξn,t θ,l = 0since the market is complete (i.e. we are after the last default).
Proof. For each BSDE, we will proceed in a backward recursive proof.
First BSDE: (En) problem: If k = n(i.e. we are after the last default), the market is com-plete. Following (2.22) and from Itô calculus, we get that
" (ZtQ(θ,l))2 vtn,θ,l # t∈[θn,T] is a P-martingale. Using its terminal conditionvTn,θ,l =αT(θ,l)we finally obtain, for all
t∈[θn, T], that vtn,θ,l= E ZTQ(θ,l) ZtQ(θ,l) !2 1 αT(θ,l) |Ft −1 .
Moreover, under integrability condition 1.7, the martingale µσnn((θθ,,ll)).W is BMO. This
implies that the family {Q(θ,l),(θ,l)∈∆n(T)×En} of measures of probability, such that the Radon Nikodym density ofQ(θ,l)with respect toP is given by (2.22), satisfies the reverse Holder inequality R2(P). Hence there exists a positive constant
c4 such that for all stopping time τ ≤ T we have E[
ZTQ(θ,l)2|F
τ]
ZτQ(θ,l)2
≤ c4. This result implies in particular that for all t ∈ [0, T], Z
Q t (θ,l)2
E[ZTQ(θ,l)2|Ft] ≥
1
c4 > 0. We conclude
by Assumption 1.2 there exists a constant δ1nsuch thatvn,θ,l ≥ δ1n. Moreover using Jensen’s inequality and Assumption 1.2, there exists a positive constant δn2 such that for allt∈[0, T]:vtn,θ,l ≤δn2.
(Ek) problems: Now, assume that the solution exists for˜k:=k+1withk∈ {0,1, . . . , n−
1}(our recursive hypothesis), we have to show that it is still true fork˜−1 :=k. We prove that the problem is equivalent to a problem of BSDE with quadratic growth and bounded terminal condition, therefore using Kobylanski’s results in [8], we will get the result. Hence, the proof is divided in two parts. Firstly, we will give results for a mod-ified quadratic BSDE. Secondly, we will use comparison theorem of quadratic BSDE to show that the first component solution of the modified BSDE is non negative and we
will conclude the proof. Let define, so, the modified BSDE fork∈ {0,1, . . . , n−1} given by: dvk,θk,lk t =−g k,θk,lk,(1) t (v k,θk,lk t , β k,θk,lk t )dt+β k,θk,lk t dWt (2.25)
with generator given by
gk,θk,lk,(1) t = Z E vk+1,θk+1,lk+1 t ηk+1(lk, dlk+1) − µk t|v k,θk,lk t |+σtkβ k,θk,lk t + R Ev k+1,θk+1,lk+1 t γtk(θk,lk, lk+1)ηk+1(lk, dlk+1) 2 (σtk)2|vk,θk,lk t |+ R Ev k+1,θk+1,lk+1 t (γtk(θk,lk, lk+1))2ηk+1(lk, dlk+1) .
Using our recursive hypothesis that there exists constantsδk1+1andδ2k+1such that
0< δ1k+1≤vk+1,θk+1,lk+1
t ≤δk2+1.
and Assumption 1.2, we have that there exists a constantC >0such that:
|gk,θk,lk,(1) t | ≤ C 1 +|vk,θk,lk t |+|β k,θk,lk t | 2 . (2.26)
Therefore this coefficient follows a quadratic growth (with respect to βk,θk,lk) and
linear growth (with respect tovk,θk,lk), using Kobylanski Theorem [8], there exists a
pair(vk,θk,lk, βk,θk,lk)∈ S∞×BMOsolution of this modified BSDE. Let now find a
suitable lower bound of the coefficientgk,θk,lk,(1). Let first define:
ekt = Z E vk+1,θk+1,lk+1 t γtk(θk,lk, lk+1)ηk+1(lk, dlk+1) , lkt = 2 µkt σtk + σktekt dkt ! (2.27) dkt = Z E vk+1,θk+1,lk+1 t (γtk(θk,lk, lk+1))2ηk+1(lk, dlk+1) and ckt = 2µktekt dkt + µkt σtk !2 (2.28) Using (2.26), we find−gk,θk,lk,(1)=K0 t +Kt1+Kt2+Kt3where Kt0=− Z E vk+1,θk+1,lk+1 t ηk+1(lk, dlk+1) Kt1= µk,θk,lk t v k,θk,lk t +ekt 2 (σk,θk,lk t )2|v k,θk,lk t |+dkt ≤ µ k,θk,lk t σk,θk,lk t !2 |vk,θk,lk t |+ 2µk,θk,lk t |v k,θk,lk t |ekt dkt + (ekt)2 dkt Kt2= (σ k,θk,lk t β k,θk,lk t ) 2 (σk,θk,lk t )2|v k,θk,lk t |+dkt ≤ |β k,θk,lk t | 2 |vk,θk,lk t | and Kt3= 2σ k,θk,lk t β k,θk,lk t (µ k,θk,lk t |v k,θk,lk t |+ekt) (σk,θk,lk t )2|v k,θk,lk t |+dkt ≤2µ k,θk,lk t σk,θk,lk t βk,θk,lk t +2 σk,θk,lk t β k,θk,lk t ekt dk t .
Since the processesµk, σk, γk, vk+1,θk+1,lk+1 are bounded from Assumption 1.2 and
our recursive hypothesis at stepk+ 1, we conclude that the processes lk andckare bounded too. Using the expressions ofK0,K1,K2andK3, we obtain:
−gk,θk,lk,(1) ≤ |β k,θk,lk t | 2 |vk,θk,lk t | +ckt|vk,θk,lk t |+ltkβ k,θk,lk t + (ekt)2 dkt − Z E vk+1,θk+1,lk+1 t ηk+1(lk, dlk+1)
Using Cauchy’s inequality on the expression ofekt, we find: (ekt)2 = Z E vk+1,θk+1,lk+1 t γtk(θk,lk, lk+1)ηk+1(lk, dlk+1) 2 ≤ Z E vk+1,θk+1,lk+1 t (γtk(θk,lk, lk+1)) 2 ηk+1(lk, dlk+1) Z E vk+1,θk+1,lk+1 t ηk+1(lk, dlk+1) then we get: (ekt)2 dk t − Z E vk+1,θk+1,lk+1 t ηk+1(lk, dlk+1) = R Ev k+1,θk+1,lk+1 t γtk(θk,lk, lk+1)ηk+1(lk, dlk+1) 2 R Ev k+1,θk+1,lk+1 t (γtk(θk,lk, lk+1)) 2 ηk+1(lk, dlk+1) − Z E vk+1,θk+1,lk+1 t ηk+1(lk, dlk+1)≤0.
Hence we obtain a suitable lower boundf¯tkof the generatorgk,t (1)
gk,θk,lk,(1) ≥f¯k t :=− ckt|v k,θk,lk t |+lktβ k,θk,lk t + |βk,θk,lk t | 2 |vk,θk,lk t | .
Hence if we consider now the following BSDE:
dY¯t= cktY¯t+lktZ¯tk+ |Z¯tk|2 ¯ Yt ! dt+ ¯ZtkdWt, Y¯T =αkT(θk,lk)∈(0,1).
then from Proposition 5.1 of [10], there exists a pair( ¯Y ,Z¯) ∈ S∞×BMOsolution of the BSDE:
dY¯t=−f¯tkdt+ ¯ZtdWt, Y¯T =αkT(θk,lk).
withY¯ ≥δ1kand the coefficientf¯ksatisfies a quadratic growth (with respect toZ¯) and linear growth (with respect toY¯). Sincegk,θk,lk,(1) ≥f¯k, applying finally comparison
theorem of Kobylanski [8], then the first component’s solution of the modified BSDE (2.25) gives
vk,θk,lk
t ≥Y¯t≥δk1 >0.
Therefore the modified BSDE is equivalent to the first BSDE of the (Ek) problem (2.20), then we get the proof of the existence of the solution of this first BSDE. Moreover, to obtain the upper boundδ2kofvk,θk,lk
t , we take the terminal condition of
the corresponding BSDE:vk,θk,lk
T = αTk(θk,lk) := δ2k. These prove that there exist well constantsδ1kandδ2ksuch that
0< δ1k≤vk,θk,lk
t ≤δ2k.
Second BSDE: (En) problem: Following the resolution of the existence of the first BSDE for
k = nand (2.22), we obtain an explicit solution of the second BSDE which is given by Ytn,θ,l =E " ZTQ(θ,l) ZtQ(θ,l)H n T(θ,l) Ft # . (2.29)
Since for all(θ,l) ∈∆n(T)×En,Hn(θ,l) ∈L∞by assumption on the contingent
claim, then from (2.29), we findYtn,θ,l ∈ S∞. Moreover, we have a representation Theorem Ytn,θ,l =HT(θ,l)− Z T t Zsn,θ,ldWsQ(θ,l), t∈[θn, T] (2.30) whereWsQ(θ,l) =Ws−µ n,θ,l s σsn,θ,l
is aQ(θ,l)Brownian motion. For any stopping times
θn≤τ ≤T and from (2.30), there exists a constantd >0such that
EQ(θ,l) " Z T τ Zsn,θ,l2ds|Fτ # ≤EQ(θ,l)Hn,θ,l T 2 |Fτ ≤d.
ThenZn,θ,l.WQ(θ,l)is aBMO-martingale under the probability measureQ(θ,l), so
Zn,θ,l.W is aBMOmartingale under the probability measureP from Kazamaki [7] Theorem 3.3. Therefore we concludeZn,θ,l ∈BMO.
(Ek) problems: Now, assume that the solution exists for˜k:=k+1withk∈ {0,1, . . . , n−
1} (our recursive hypothesis), we have to show that it is still true fork˜−1 := k. We would like now to prove that Yk,θk,lk
t , Z k,θk,lk
t
∈ S∞ ×BMO for all k ∈
{0,1, . . . , n}. We can actually prove the existence of the solution of the second BSDE, since the solution of the first one exists. Given the solution of the first BSDE, the coefficient of the second one is linear. Therefore, we can characterize explicitly the solution.
Step 1: Preliminary results.
Given the explicit formula of the coefficientgk,θk,lk,(2)in (2.20), we get
gk,θk,lk,(2) t =a k,θk,lk t Z k,θk,lk t +κ k,θk,lk t Y k,θk,lk t + Λ k,θk,lk t . with ak,θk,lk t =β k,θk,lk t −σ k,θk,lk t µk,θk,lk t +σ k,θk,lk t β k,θk,lk t + R Eγtk(θk,lk, lk+1)(1 +vtJ,θk,lk)ηk+1(lk, dlk+1) (σk,θk,lk t ) 2 +R E(1 +v J,θk,lk t )(γtk(θk,lk, lk+1))2ηk+1(lk, dlk+1) , κk,θk,lk t = − Z E (1 +vJ,θk,lk t )ηk+1(lk, dlk+1) + Z E (1 +vJ,θk,lk t )γtk(θk,lk, lk+1)ηk+1(lk, dlk+1) × µk,θk,lk t +σ k,θk,lk t β k,θk,lk t + R E(1 +v J,θk,lk t )γtk(θk,lk, lk+1)ηk+1(lk, dlk+1) (σk,θk,lk t ) 2 +R E(1 +v J,θk,lk t )(γtk(θk,lk, lk+1)) 2 ηk+1(lk, dlk+1) and Λk,θk,lk t = Z E (1 +vJ,θk,lk t )Y k+1,θk+1,lk+1 t ηk+1(lk, dlk+1) − Z E (1 +vJ,θk,lk t )Y k+1,θk+1,lk+1 t γtk(θk,lk, lk+1)ηk+1(lk, dlk+1) × µk,θk,lk t +σ k,θk,lk t β k,θk,lk t + R E(1 +v J,θk,lk t )γtk(θk,lk, lk+1)ηk+1(lk, dlk+1) (σk,θk,lk t ) 2 +R E(1 +v J,θk,lk t )(γtk(θk,lk, lk+1)) 2 ηk+1(lk, dlk+1) .
Under Assumption 1.2 and the integrability condition 1.7, coefficientsσk,θk,lk,µk,θk,lk
andγkare bounded. Moreover from the solution of the first BSDE and the boundness of the processes vk+1,θk+1,lk+1 and Yk+1,θk+1,lk+1
t (recursive hypothesis), we have
that the processes vJ,k,θk,lk are bounded for all l
k ∈ Ek andβk,θk,lk.W is aBMO
martingale.
Therefore we deduce that the martingalesΛk,θk,lk.W,ak,θk,lk.W andκk,θk,lk.W are BMO under the probability measure P. Let define the probability measure Q ∼ P
with Radon Nikodym density on FT defined by ZTQ = E(ak,θk,lk.W)
T. Since the
martingaleak,θk,lk.W isBMO, the processZQ
t =E
h
ZTQ|Ftiis uniformly integrable and from Theorem 3.3 of Kazamaki [7], the martingaleκk,θk,lk.W is stillBMOunder
the probability measureQ. Therefore, there exists a non negative constantcsuch that EQ
hRT
t |κk,sθk,lk|
2
ds|Fti≤c, for allθk≤t≤T andk∈ {0,1, . . . , n}.
Step 2: Integrability of the adjoint processΓ: Let define for allk∈ {0,1, . . . , n}
e Γt:= exp Z t 0 κk,θk,lk s ds .
We prove thatΓe ∈Lp(Q)for anyp >1andδ >0: e ΓT e Γt p = exp p Z T t κk,θk,lk s ds ! ≤exp Z T t δ(κk,θk,lk s )2+ p2 4δ ! ds ! ≤ exp p 2 4δT ! exp δ Z T t (κk,θk,lk s )2ds ! .
Since there exists a non-negative constantcsuch that EQ " Z T t |κk,θk,lk s | 2 ds|Ft # ≤c
we deduce form Proposition 3.1 in Appendix that there exists0 ≤δ ≤ 1
c2 such that EQ h exp(RT t δ|(κsk,θk,lk)2|ds)|Ft i ≤ 1
1−δc2. Therefore we conclude there exists a non
negative constantC1 such that EQ " e ΓT e Γt p |Ft # ≤C1. (2.31)
Step 3: The solution of the BSDE. Let define now fork∈ {0,1, . . . , n−1}
Yk,θk,lk t =EQ " 1 e Γt e ΓTHTk(θk,lk) + Z T t e ΓsΛk,θk,lk s ds ! |Ft # , θk≤t≤T. (2.32) SinceΓ =ZQΓe, using Bayes formula equation (2.32) is equivalent to
Yk,θk,lk t =E " 1 Γt ΓTH k T(θk,lk) + Z T t ΓsΛsds ! |Ft # , t≤T. (2.33)
Moreover sinceΛk,θk,lkis bounded andHk
T(θk,lk)∈L∞, there exists a non negative
constantCsuch that
|Yk,θk,lk t | ≤CEQ e ΓT e Γt + Z T t e Γs e Γt 2 + (Λk,θk,lk s )2 ds Ft
Since the processΛk,θk,lk.WQis aBMOmartingale under the probability measureQ
and using (2.31),there exists a constantC >¯ 0such that:
|Yk,θk,lk
t | ≤C,¯ t≤T.
Let considerYk,θk,lk defined by (2.32), then the process
e ΓtYk,θk,lk t + Z t 0 Λk,θk,lk s Γesds=EQ " e ΓTHTk(θk,lk) + Z T 0 e ΓsΛk,θk,lk s ds|Ft #
is a squared integrableQ-martingale sinceHkis bounded by assumption,Λk,θk,lk.W
is BMO andΓe satisfies (2.31). Therefore from representation theorem, there exists
a process Z¯ ∈ H2 such thatd(
e ΓtYk,θk,lk t + Rt 0ΓesΛsk,θk,lkds) = ¯ZtdWtQ. Setting Zk,θk,lk = Z¯ e
Γ,using integration by part formula we find:
dYk,θk,lk t =−(Λ k,θk,lk t +Z k,θk,lk t a k,θk,lk t +κ k,θk,lk t Y k,θk,lk t )dt+Z k,θk,lk t dWt, YTk,θk,lk =H k T(θk,lk).
Applying Itô’s formula, we find
d(Yk,θk,lk t ) 2 = 2Yk,θk,lk t [−(Λ k,θk,lk t +κ k,θk,lk t Y k,θk,lk t )dt+Z k,θk,lk t .dW Q t ]+(Z k,θk,lk t ) 2 dt,
therefore, for any stopping timeσ, we find:
EQ " Z T σ (Zk,θk,lk t ) 2 dt Fσ # ≤EQ " (HTk(θk,lk)) 2 + 2 Z T σ 2Yk,θk,lk s (Λk,sθk,lk+κsk,θk,lkYsk,θk,lk)ds Fσ # .
SinceHk, Yk,θk,lk are bounded,Λk,θk,lk.WQandκk,θk,lk.WQareBMOmartingales
under the probability measure Q, we conclude Zk,θk,lk.WQ is a BMO martingale
measure underQthenZk,θk,lk.W is aBMOmartingale under the probability measure P from Kazamaki [7] Theorem 3.3. Therefore we conclude (Yk,θk,lk, Zk,θk,lk) ∈
S∞×BMOis a solution of the second BSDE.
Third BSDE: (En) problem: Sinceg3t,θ,l≡0, we have directlyξtn,θ,l ≡0.
(Ek) problems: Now, assume that the solution exists for˜k:=k+1withk∈ {0,1, . . . , n−
1}(our recursive hypothesis), we have to show that it is still true fork˜−1 := k. It lets to prove thatξk,θk,lk
t , R k,θk,lk
t
∈ S∞×BMO. Since, for allk∈ {0,1, . . . , n}, all the terms appearing in the coefficientgk,θk,lk,(3)
t are bounded andZk,θk,lk ∈BMO
by previous step, we conclude using representation Theorem that(ξk,θk,lk, Rk,θk,lk)∈
2.2 BSDEs characterization by verification theorem
Now, we show that the triplevk,θk,lk, Yk,θk,lk, ξk,θk,lk, appearing in the quadratic
decomposi-tion form, soludecomposi-tion to the recursive system indexed BSDEs provides actually the soludecomposi-tion to the global optimal investment problem in terms of the value functionsVk,k∈ {0,1, . . . , n}in (2.16) and (2.19). As a byproduct, we will obtain the existence of the optimal strategyπk,∗.
Theorem 2.2. The value functionsVk ,k = 0, . . . , ndefined in(2.16)and(2.19)are given, for allt∈[θk, T], by Vtk(x,θk,lk, νk) =vtk,θk,lk(X k,x t (θk,lk)−Ytk,θk,lk) 2 +ξk,θk,lk t (2.34)
for allx ∈ R, (θk,lk) ∈ ∆k×Ek, νk ∈ AkF where (vk,θk,lk, Yk,θk,lk, ξk,θk,lk) is the unique
solution of the recursive triple BSDEs systems given for allk={0,1, . . . , n}in 2.17 and 2.20. In particular, the solution of the Mean Variance Hedging problem is given by
V0H(x) = inf π∈AGE h (HT −XTx,π)2 i =v00(x−Y00)2+ξ00, x∈R. (2.35)
where the triple v0, Y00, ξ00
is solution of the recursive system of BSDEs:(En): (2.17)and(Ek):
(2.20),k∈ {0,1, . . . , n−1}.
Moreover, there exists an optimal strategyπ∗:= π0,∗, π1,∗, . . . , πn,∗
given by: πtk,∗(θk,lk) = 1 (σkt)2+R E(1 +v J,k,θk,lk t )γtk(θk,lk, lk+1)2ηk+1(lk, dlk+1) h σktZk,θk,lk t −K k,θk,lk t (µkt +σtkβ k,θk,lk t ) + R E(X k,x t (θk,lk)v k+1,θk+1,lk+1 t −Yk+1,θk+1,lk+1v k+1,θk+1,lk+1 t )γtk(θk,lk, lk+1)ηk+1(lk, dlk+1) vk,θk,lk t # (2.36) withKk,θk,lk t :=X k,x
t (θk,lk)−Yk,θk,lk. And for the after last default problem:
πtn,∗(θ,l) = 1 (σn
t)2
h
σtnZtn,θ,l−Xtn,x(θ,l)−Ytn,θ,l µnt +σtnβtn,θ,li (2.37)
Remark 2.5. Following(2.35), we can give some financial comments of our quadratic decompo-sition form:
– The processv0doesn’t depend on the payoffH. Moreover, we have that v00 =V00(1) := inf
π∈AGE
h
XT1,πi2.
Thereforev0is related to the minimal variance of a pure investment on the assetSwith an
initial wealthx= 1.
– The processY0is the quadratic approximation price of the option H.
– The processξ0represents the incompleteness of this market. Since if the market is complete
(as in the(En)problem) then this process vanishes.
Proof. Step1: We begin by proving for allk={0,1, . . . , n},t∈[θk, T]andνk ∈ AkF, that
vk,θk,lk t (X k,x t (θk,lk)−Ytk,θk,lk) 2 +ξk,θk,lk t ≤Vtk(x,θk,lk, νk) (2.38)
