HAL Id: hal-00004619
https://hal.archives-ouvertes.fr/hal-00004619
Submitted on 1 Apr 2005
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
L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires
BSDE with quadratic growth and unbounded terminal value
Philippe Briand, Ying Hu
To cite this version:
Philippe Briand, Ying Hu. BSDE with quadratic growth and unbounded terminal value. Probability Theory and Related Fields, Springer Verlag, 2006, 136 (4), pp.604-618. �10.1007/s00440-006-0497-0�.
�hal-00004619�
ccsd-00004619, version 1 - 1 Apr 2005
BSDE with quadratic growth and unbounded terminal value
Philippe Briand Ying Hu
IRMAR, Universit´ e Rennes 1, 35 042 RENNES Cedex, FRANCE philippe.briand@univ-rennes1.fr ying.hu@univ-rennes1.fr
Abstract. In this paper, we study the existence of solution to BSDE with quadratic growth and unbounded terminal value. We apply a localization procedure together with a priori bounds. As a byproduct, we apply the same method to extend a result on BSDEs with integrable terminal condition.
1. Introduction.
In this paper we are concerned with real valued backward stochastic differential equations – BSDEs for short in the remaining –
Y t = ξ + Z T
t
f (s, Y s , Z s ) ds − Z T
t
Z s · dB s , 0 ≤ t ≤ T
where (B t ) t≥0 is a standard brownian motion. Such equations have been extensively studied since the first paper of E. Pardoux and S. Peng [PP90]. The full list of contributions is too long to give and we will only quote results in our framework.
Our setting is mainly the following : the generator, namely the function f , is of quadratic growth in the variable z and the terminal condition, the random variable ξ, will not be bounded. BSDEs with quadratic growth have been first studied by Magdalena Kobylanski in her PhD (see [Kob97, Kob00]) and then by Jean-Pierre Lepeltier and Jaime San Martin in [LSM98]. We should point out that BSDEs with quadratic growth in the variable z have found applications in control and finance, see, e.g., Bismut [Bis78], El Karoui, Rouge [EKR00], Hu, Imkeller, Muller [HIM05], . . .
All the results on BSDEs with quadratic growth require that the terminal condition ξ is a bounded random variable. The boundedness of the terminal condition appears, from the point of view of the applications, to be restrictive and, moreover, from a theoretical point of view, is not necessary to obtain a solution. Indeed, let us consider the following well known equation :
Y t = ξ + 1 2
Z T t
|Z s | 2 ds − Z T
t
Z s dB s , 0 ≤ t ≤ T ;
the change of variables P t = e Y
t, Q t = e Y
tZ t , leads to the equation P t = e ξ −
Z T t
Q s dB s which has a solution as soon as e ξ is integrable.
On this simple example we see that the existence of exponential moments of the terminal condition is sufficient to construct a solution to our BSDE. Our paper will be focused on the theoretical study of these BSDEs but with unbounded terminal value with only exponential moments.
To fill the gap between boundedness and existence of exponential moments, we will use an approach based upon a localization procedure together with a priori bounds. Let us quickly explain how it works on a simple example. Let f : R × R d −→ R be a continuous function and ξ be a nonnegative terminal condition such that
|f(y, z)| ≤ 1
2 |z| 2 , E h
e ξ i
< ∞, and let us try to construct a solution to the BSDE
Y t = ξ + Z T
t
f (Y s , Z s ) ds − Z T
t
Z s · dB s , 0 ≤ t ≤ T.
As mentioned before, BSDEs with quadratic growth in the variable z can be solved when the terminal solution is bounded. That is why we introduce (Y n , Z n ) as the minimal solution to the BSDE
Y t n = ξ ∧ n + Z T
t
f (Y s n , Z s n ) ds − Z T
t
Z s n · dB s , and of course we want to pass to the limit when n → ∞ in this equation.
The process Y n is known to be bounded but the estimate depends on kξ ∧ nk ∞ and thus is far from being useful when ξ is not bounded. The first step of our approach consists in finding an estimation for Y n independent of n. In this example, we can use the explicit formula mentioned before to show that
0 ≤ − ln E
e −(ξ∧n) | F t
≤ Y t n ≤ ln E
e ξ∧n | F t
≤ ln E e ξ | F t
. With these inequalities in hands, we introduce the stopping time
τ k = inf n
t ∈ [0, T ] : ln E
e ξ | F t
≥ k o
∧ T
and instead of working on the time interval [0, T ] we will restrict ourselves to [0, τ k ] by considering the BSDE
Y t∧τ n
k= Y τ n
k+ Z T ∧τ
kt∧τ
kf (Y s n , Z s n ) ds − Z T ∧τ
kt∧τ
kZ s n · dB s , 0 ≤ t ≤ T.
By construction, we have sup n sup t Y t∧τ n
k∞ ≤ k. This last property together with the
fact that the sequence (Y n ) n≥1 is nondecreasing allows us, with the help of a result of
Kobylanski, to pass to the limit when n → ∞, k being fixed and then to send k to infinity to get a solution.
The rest of the paper is organized as follows. Next section is devoted to the notations we use during this text. In Section 3, we claim our main result that we prove in Section 4.
Section 5 is devoted to some additional results on BSDEs with quadratic growth in z.
Finally, in the last section, we apply the same approach to study BSDEs with terminal value in L 1 .
2. Notations.
For the remaining of the paper, let us fix a nonnegative real number T > 0.
First of all, B = {B t } t≥0 is a standard brownian motion with values in R d defined on some complete probability space (Ω, F, P). {F t } t≥0 is the augmented natural filtration of B which satisfies the usual conditions. In this paper, we will always use this filtration. P denotes the sigma-field of predictable subsets of [0, T ] × Ω.
As mentioned in the introduction, a BSDE is an equation of the following type Y t = ξ +
Z T t
f (s, Y s , Z s ) ds − Z T
t
Z s · dB s , 0 ≤ t ≤ T. (1) f is called the generator and ξ the terminal condition.
Let us recall that a generator is a random function f : [0, T ] × Ω × R × R d −→ R which is measurable with respect to P × B(R) × B(R d ) and a terminal condition is simply a real F T –measurable random variable.
By a solution to the BSDE (1) we mean a pair (Y, Z) = {(Y t , Z t )} t∈[0,T ] of predictable processes with values in R × R d such that P–a.s., t 7−→ Y t is continuous, t 7−→ Z t belongs to L 2 (0, T ), t 7−→ f (t, Y t , Z t ) belongs to L 1 (0, T ) and P–a.s.
Y t = ξ + Z T
t
f (s, Y s , Z s ) ds − Z T
t
Z s · dB s , 0 ≤ t ≤ T.
We will use the notation BSDE(ξ, f ) to say that we consider the BSDE whose generator is f and whose terminal condition is ξ; Y f (ξ), Z f (ξ)
means a solution to the BSDE(ξ, f ).
Y f (ξ), Z f (ξ)
is said to be minimal if P -a.s., for each t ∈ [0, T ], Y t f (ξ) ≤ Y t g (ζ ) whenever P–a.s. ξ ≤ ζ and f (t, y, z) ≤ g(t, y, z) for all (t, y, z). Y f (ξ), Z f (ξ)
is said to be minimal in some space B if it belongs to this space and the previous property holds true as soon as (Y g (ζ), Z g (ζ)) ∈ B.
For any real p > 0, S p denotes the set of real-valued, adapted and c` adl` ag processes {Y t } t∈[0,T] such that
kY k S
p:= E h
sup t∈[0,T ] |Y t | p i 1∧1/p
< +∞.
If p ≥ 1, k · k S
pis a norm on S p and if p ∈ (0, 1), (X, X ′ ) 7−→
X − X ′
S
pdefines a
distance on S p . Under this metric, S p is complete.
M p denotes the set of (equivalent classes of) predictable processes {Z t } t∈[0,T ] with values in R d such that
kZ k M
p:= E Z T
0
|Z s | 2 ds p/2 1∧1/p
< +∞.
For p ≥ 1, M p ( R n ) is a Banach space endowed with this norm and for p ∈ (0, 1), M p is a complete metric space with the resulting distance.
We set S = ∪ p>1 S p , M = ∪ p>1 M p and denote by S ∞ the set of predictable bounded processes. Finally, let us recall that a continuous process {Y t } t∈[0,T] belongs to the class (D) if the family {Y τ : τ stopping time bounded by T } is uniformly integrable.
3. Quadratic BSDEs.
In this section, we consider BSDE(ξ, f ) when the generator f has a linear growth in y and a quadratic growth in z. We denote (H1) the assumption: there exist α ≥ 0, β ≥ 0 and γ > 0 such that P –a.s.
∀t ∈ [0, T ], (y, z) 7−→ f (t, y, z) is continuous,
∀(t, y, z) ∈ [0, T ] × R × R d , |f (t, y, z)| ≤ α + β |y| + γ
2 |z| 2 . (H1) Concerning the terminal condition ξ, we will assume that
E h
e γe
βT|ξ| i
< +∞. (H2)
We will use also a stronger assumption on the integrability of ξ namely
∃λ > γe βT , E h
e λ |ξ| i
< +∞. (H3)
It is clear that we can assume without loss of generality that α ≥ β/γ.
As we explained in the introduction, our method relies heavily on a priori estimate. To obtain such estimations, we will use the change of variable P t = e γY
t, Q t = γe γY
tZ t ; if (Y, Z) is a solution to the BSDE(ξ, f ), (P, Q) solves the BSDE
P t = e γξ + Z T
t
F (s, P s , Q s ) ds − Z T
t
Q s · dB s , 0 ≤ t ≤ T, with the function F defined by
F (s, p, q) = 1 p>0
γp f
s, ln p γ , q
γp
− 1 2
|q| 2 p
. (2)
In view of the growth of the generator f , we have F (s, p, q) ≤ 1 p>0 p(αγ + β| ln p|). For notational convenience, we denote by H the function
∀p ∈ R , H(p) = p (αγ + β ln p) 1 [1,+∞) (p) + γα1 (−∞,1) (p).
It is straightforward to check that, since α ≥ β/γ, H is convex and locally Lipschitz continuous and that, for any real p > 0, p (αγ + β| ln p|) ≤ H(p). Thus we deduce the inequality
∀s ∈ [0, T ], ∀p ∈ R, ∀q ∈ R d , F (s, p, q) ≤ H(p). (3) To get an upper bound for Y t , the idea is to compare more or less P t with φ t (ξ) where, for any real z, {φ t (z)} 0≤t≤T stands for the solution to the differential equation
φ t = e γz + Z T
t
H(φ s ) ds, 0 ≤ t ≤ T. (4)
Using the convexity of H, we will able to prove that P t ≤ E (φ t (ξ) | F t ) , Y t ≤ 1
γ ln E (φ t (ξ) | F t ) .
Before proving this result rigorously, let us recall that the differential equation (4) can be solved easily. Indeed, we have, for any z ≥ 0,
φ t (z) = exp γα e β(T −t) − 1 β
! exp
zγe β(T −t)
, if β > 0, and φ t (z) = e γα(T −t) e γz if β = 0.
Let us consider the case where z < 0. If e γz + T γα ≤ 1 then the solution is φ t = e γz + γα(T − t)
and otherwise there exists 0 < S < T such that e γz + γα(T − S) = 1 and φ t = [e γz + γα(T − t)] 1 (S,T ] (t) + exp γα e β(S−t) − 1
β
!
1 [0,S] (t).
It is plain to check that t 7→ φ t (z) is decreasing and that z 7→ φ t (z) is increasing and continuous.
Lemma 1. Let the assumption (H1) hold and let ξ be a bounded F T –measurable random variable.
If (Y, Z) is a solution to the BSDE(ξ, f ) in S ∞ × M 2 then
− 1
γ ln E (φ t (−ξ) | F t ) ≤ Y t ≤ 1
γ ln E (φ t (ξ) | F t ) . Proof. Let us set Φ t = E (φ t (ξ) | F t ). We have
Φ t = E
e γξ + Z T
t
H(φ s (ξ)) ds F t
= E
e γξ + Z T
t
E (H(φ s (ξ)) | F s ) ds F t
. Thus writing the bounded brownian martingale
E
e γξ + Z T
0
E (H(φ s (ξ)) | F s ) ds F t
= E
e γξ + Z T
0
E (H(φ s (ξ)) | F s ) ds
+ Z t
0
Ψ s ·dB s
(Φ, Ψ) solves the BSDE Φ t = e γξ +
Z T t
E (H(φ s (ξ)) | F s ) ds − Z T
t
Ψ s · dB s .
On the other hand, if (Y, Z ) ∈ S ∞ × M 2 is a solution of (1), setting as before P t = e γY
t, Q t = γe γY
tZ t , we have
P t = e γξ + Z T
t
F (s, P s , Q s ) ds − Z T
t
Q s · dB s , with F defined by (2).
It follows that Φ t − P t =
Z T
t
(H(Φ s ) − H(P s )) ds + Z T
t
R s ds − Z T
t
(Ψ s − Q s ) · dB s where, in view of the inequality (3) and since H is convex,
R s = E (H(φ s (ξ)) | F s ) − H ( E (φ s (ξ) | F s )) + H(P s ) − F(s, P s , Q s ) is a nonnegative process.
H is only locally Lipschitz but since Φ and P are bounded we can apply the comparison theorem to get P t ≤ Φ t and Y t ≤ γ 1 ln Φ t .
Finally, since the function −f (t, −y, −z) still satisfies the assumption (H1), we get also the inequality −Y t ≤ 1 γ ln E (φ t (−ξ) | F t ).
We are now in position to prove that under the assumptions described before the BSDE (1) has at least a solution.
Theorem 2. Let the assumptions (H1) and (H2) hold. Then the BSDE (1) has at least a solution (Y, Z) such that :
− 1
γ ln E (φ t (−ξ) | F t ) ≤ Y t ≤ 1
γ ln E (φ t (ξ) | F t ) . (5) If moreover, (H3) holds, then Z belongs to M 2 .
Proof of the last part of Theorem 2. If (Y, Z) is a solution to the BSDE (1) such that the inequalities (5) hold, then
|Y t | ≤ 1
γ ln E (φ 0 (|ξ|) | F t )
and, under the assumption (H3), we deduce that, for some p > 1, E
h
sup t∈[0,T] e pγ|Y
t| i
< +∞.
For n ≥ 1, let τ n be the following stopping time τ n = inf
t ≥ 0 :
Z t 0
e 2γ|Y
s| |Z s | 2 ds ≥ n
∧ T,
and let us consider the function from R + into itself defined by u(x) = 1
γ 2 (e γx − 1 − γx) .
x 7−→ u(|x|) is C 2 and we have from Itˆo’s formula, with the notation sgn(x) = −1 x≤0 +1 x>0 , u(|Y 0 |) = u(|Y t∧τ
n|) +
Z t∧τ
n0
u ′ (|Y s |) sgn(Y s )f (s, Y s , Z s ) − 1
2 u ′′ (|Y s |)|Z s | 2
ds
− Z t∧τ
n0
u ′ (|Y s |) sgn(Y s )Z s · dB s . It follows from (H1) since u ′ (x) ≥ 0 for x ≥ 0 that
u(|Y 0 |) ≤ u(|Y t∧τ
n|) + Z t∧τ
n0
u ′ (|Y s |) (α + β|Y s |) ds − Z t∧τ
n0
u ′ (|Y s |) sgn(Y s )Z s · dB s
− 1 2
Z t∧τ
n0
u ′′ (|Y s |) − γ u ′ (|Y s |)
|Z s | 2 ds.
Moreover, we have (u ′′ − γu ′ )(x) = 1 for x ≥ 0 and, taking expectation of the previous inequality, we get
1 2 E
Z T ∧τ
n0
|Z s | 2 ds
≤ E
"
1 γ 2 sup
t∈[0,T ]
e γ|Y
t| + 1 γ
Z T 0
e γ|Y
s| (α + β|Y s |) ds
#
Fatou’s lemma together with the fact that e γ|Y
t| ∈ S p gives the result.
4. Proof of Theorem 2.
Let us first construct a solution to the BSDE (1) in the case where ξ is nonnegative.
For each n ∈ N ∗ , we set ξ n = ξ ∧ n. Then it is known from [Kob00, Theorem 2.3] that the BSDE
Y t n = ξ n + Z T
t
f (s, Y s n , Z s n ) ds − Z T
t
Z s n · dB s , 0 ≤ t ≤ T has a minimal solution (Y n , Z n ) in S ∞ × M 2 . Lemma 1 implies the inequalities
− 1
γ ln E (φ t (−ξ n ) | F t ) ≤ Y t n ≤ 1
γ ln E (φ t (ξ n ) | F t ) . Since we consider only minimal solutions, we have,
∀t ∈ [0, T ], Y t n ≤ Y t n+1 . We define Y = sup n≥1 Y n .
Since 0 ≤ φ t (ξ n ) ≤ φ 0 (|ξ|) and 0 ≤ φ t (−ξ n ) ≤ φ 0 (|ξ|), we deduce from the dominated convergence theorem, noting that the random variable φ 0 (|ξ|) is integrable by (H2), that
− 1
γ ln E (φ t (−ξ) | F t ) ≤ Y t ≤ 1
γ ln E (φ t (ξ) | F t ) .
In particular, we have lim t→T Y t = ξ = Y T . Indeed, for each S < T , lim sup
t→T
Y t ≤ lim sup
t→T
1
γ ln E (φ t (ξ) | F t ) ≤ lim
t→T
1
γ ln E (φ S (ξ) | F t ) = 1
γ ln φ S (ξ), and lim S→T 1
γ ln φ S (ξ) = ξ. We can do the same for lim inf.
Let us introduce the following stopping time : τ k = inf
t ∈ [0, T ] : 1
γ ln E (φ 0 (|ξ|) | F t ) ≥ k
∧ T.
Then (Y k n , Z k n ) := (Y t∧τ n
k
, Z t n 1 t≤τ
k) satisfies the following BSDE Y k n = ξ n k +
Z T t
1 s≤τ
kf (s, Y k n (s), Z k n (s)) ds − Z T
t
Z k n (s) · dB s , where of course ξ k n = Y k n (T ) = Y τ n
k.
We are going to pass to the limit when n tends to +∞ for k fixed in this last equation.
The key point is that Y k n is increasing in n and remains bounded by k. At this stage, let us mention a mere generalization of Proposition 2.4 in [Kob00].
Lemma 3 ([Kob00]). Let (ξ n ) n≥1 be a sequence of F T –measurable bounded random variables and (f n ) n≥1 be a sequence of generators which are continuous with respect to (y, z).
We assume that (ξ n ) n≥1 converges P–a.s. to ξ, that (f n ) n≥1 converges locally uniformly in (y, z) to the generator f , and also that
1. sup n≥1 kξ n k ∞ < +∞ ;
2. sup n≥1 |f n (t, y, z)| satisfies the inequality in (H1).
If for each n ≥ 1, the BSDE(ξ n , f n ) has a solution in S ∞ × M 2 , such that Y f
n(ξ n )
n≥1
is nondecreasing (respectively nonincreasing), then P–a.s.
Y t f
n(ξ n )
n≥1 converges uni- formly on [0, T ] to Y t = sup n≥1 Y t f
n(ξ n ) (respectively Y t = inf n≥1 Y t f
n(ξ n )), Z f
n(ξ n )
n≥1
converges to some Z in M 2 and (Y, Z) is a solution to BSDE(ξ, f ) in S ∞ × M 2 . Proof. It follows from Lemma 1 that there exists r > 0 such that, P–a.s.
∀n ≥ 1, ∀t ∈ [0, T ],
Y t f
n(ξ n ) ≤ r.
Let us consider the continuous function ρ(x) = xr/ max(r, |x|). Since ρ(x) = x for |x| ≤ r, Y f
n(ξ n ), Z f
n(ξ n )
solves the BSDE(ξ n , g n ) where g n (t, y, z) = f n (t, ρ(y), z). Obviously, we have, for each n ≥ 1,
|g n (t, y, z)| ≤ α + β r + γ
2 |z| 2 ,
and thus we can apply the result of Kobylanski.
Setting Y k (t) = sup n Y k n (t), it follows from the previous lemma that there exists a process Z k ∈ M 2 such that lim n Z k n = Z k in M 2 and (Y k , Z k ) solves the BSDE
Y k (t) = ξ k + Z T
t
1 s≤τ
kf (s, Y k (s), Z k (s)) ds − Z T
t
Z k (s) · dB s , (6) where ξ k = sup n Y τ n
k.
But τ k ≤ τ k+1 , and thus we get, coming back to the definition of Y k , Z k and Y , Y t∧τ
k= Y k+1 (t ∧ τ k ) = Y k (t), Z k+1 (t) 1 t≤τ
k= Z k (t).
As τ k → T and the Y k ’s are continuous processes we deduce in particular that Y is continuous on [0, T ). On the other hand, as mentioned before lim t→T Y t = ξ and Y T is equal to ξ by construction. Thus Y is a continuous process on the closed interval [0, T ].
Then we define Z on (0, T ) by setting :
Z t = Z k (t), if t ∈ (0, τ k ).
From (6), (Y, Z) satisfies:
Y t∧τ
k= Y τ
k+ Z τ
kt∧τ
kf (s, Y s , Z s )ds − Z τ
kt∧τ
kZ s · dB s . (7)
Finally, we have P(
Z T 0
|Z s | 2 ds = ∞) = P Z T
0
|Z s | 2 ds = ∞, τ k = T
+ P Z T
0
|Z s | 2 ds = ∞, τ k < T
≤ P Z τ
k0
|Z k (s)| 2 ds = ∞
+ P(τ k < T ), and we deduce that, P–a.s.
Z T 0
|Z s | 2 ds < ∞.
By sending k to infinity in (7), we deduce that (Y, Z) is a solution of (1).
Let us explain quickly how to extend this construction to the general case. Let us fix n ∈ N ∗ and p ∈ N ∗ and set ξ n,p = ξ + ∧ n − ξ − ∧ p. Let us consider, (Y n,p , Z n,p ) the minimal bounded solution to the BSDE
Y t n,p = ξ n,p + Z T
t
f (s, Y s n,p , Z s n,p ) ds − Z T
t
Z s n,p · dB s , 0 ≤ t ≤ T which satisfies
− 1
γ ln E (φ t (−ξ n,p ) | F t ) ≤ Y t n,p ≤ 1
γ ln E (φ t (ξ n,p ) | F t ) . We have,
∀t ∈ [0, T ], Y t n,p+1 ≤ Y t n,p ≤ Y t n+1,p ,
and we define Y p = sup n≥1 Y n,p so that Y t p+1 ≤ Y t p and Y t = inf p≥1 Y t p . By the dominated convergence theorem, we have
− 1
γ ln E (φ t (−ξ) | F t ) ≤ Y t ≤ 1
γ ln E (φ t (ξ) | F t ) , and in particular, we have lim t→T Y t = ξ = Y T .
(Y t∧τ n,p
k, Z t n,p 1 t≤τ
k) solves the BSDE Y t∧τ n,p
k= Y τ n,p
k+
Z T t
1 s≤τ
kf (s, Y s n,p , Z s n,p ) ds − Z T
t
Z s n,p 1 s≤τ
k· dB s . But, once again Y t∧τ n,p
k
is increasing in n and decreasing in p and remains bounded by k.
Arguing as before, setting Y k (t) = inf p sup n Y t∧τ n,p
k
, there exists a process Z k such that lim p lim n Z n,p (s)1 s≤τ
k= Z k (s) and (Y k , Z k ) still solves the BSDE (6). The rest of the proof is unchanged.
5. Additional results on quadratic BSDEs.
5.1. Minimal solution. In this section, we give some complements on BSDEs with quadratic growth in z.
Proposition 4. Let (H1) holds and assume moreover that there exists an integer r ≥ 0 such that P–a.s.
f (t, y, z) ≥ −r (1 + |y| + |z|) .
Let us assume also that (H3) holds for ξ + and that, for some p > 1, ξ − ∈ L p . Then BSDE(ξ, f ) has a minimal solution in S .
Proof. For each n ≥ r, let us consider the function f n (t, y, z) = inf n
f (t, p, q) + n|p − y| + n|q − z| : (p, q) ∈ Q 1+d o .
Then f n is well defined and it is globally Lipschitz continuous with constant n. More- over (f n ) n≥r is increasing and converges pointwise to f . Dini’s theorem implies that the convergence is also uniform on compact sets. We have also, for all n ≥ r,
−r(1 + |y| + |z|) ≤ f n (t, y, z) ≤ f (t, y, z)
Let (Y n , Z n ) be the unique solution in S p × M p to BSDE(ξ, f n ). It follows from the classical comparison theorem that
Y t r ≤ Y t n ≤ Y t n+1 .
Let us prove that Y t n ≤ γ 1 ln E (φ t (ξ) | F t ). To do this let us recall that, since f n is Lipschitz,
Y t n = lim m→+∞ Y t f
n(ξ m ) where ξ m = ξ 1 |ξ|≤m . Moreover E (φ t (ξ m ) | F t ) −→ E (φ t (ξ) | F t )
a.s. since sup m≥1 |φ t (ξ m )| ≤ φ 0 (ξ + ) which is integrable. Thus we have only to prove
that Y t f
n(ξ m ) ≤ γ 1 ln E (φ t (ξ m ) | F t ). We keep the notations of the beginning of Section 4.
(Φ, Ψ) is solution to the BSDE Φ t = e γξ
m+
Z T
t
H(Φ s ) ds + Z T
t
Γ s ds − Z T
t
Ψ s · dB s ,
where Γ s = E (H(φ s (ξ)) | F s ) − H(Φ s ) is a nonnegative process since H is convex.
It follows by setting U t = 1 γ ln Φ t , V t = γΦ Ψ
tt