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Stability of solutions of BSDEs with random terminal time
Sandrine Toldo
To cite this version:
Sandrine Toldo. Stability of solutions of BSDEs with random terminal time. ESAIM: Probability and
Statistics, EDP Sciences, 2006, 10, pp.141-163. �hal-00004726v2�
ccsd-00004726, version 2 - 15 Sep 2005
Stability of solutions of BSDEs with random terminal time
Sandrine TOLDO
∗IRMAR, Universit´ e Rennes 1, Campus de Beaulieu, 35042 Rennes Cedex, France
Abstract
In this paper, we study the stability of the solutions of Backward Stochastic Differential Equations (BSDE for short) with an almost surely finite random terminal time. More precisely, we are going to show that if (
Wn) is a sequence of scaled random walks or a sequence of martingales that converges to a Brownian motion
Wand if (
τn) is a sequence of stopping times that converges to a stopping time
τ, then the solution of the BSDE driven by
Wnwith random terminal time
τnconverges to the solution of the BSDE driven by
Wwith random terminal time
τ.
Keywords : Backward Stochastic Differential Equations (BSDE), Stability of BSDEs, Weak convergence of filtrations, Stopping times
Introduction
We want to make an approximation of the solutions of a backward stochastic differential equa- tion (BSDE for short) with an almost surely finite random terminal time τ like
Y
t∧τ= ξ + Z
τt∧τ
f (s, Y
s, Z
s)ds − Z
τt∧τ
Z
sdW
s, t > 0.
The robustness of numerical methods to approximate solutions of BSDEs had already been studied in the case of a deterministic terminal time. Antonelli and Kohatsu-Higa in [1] and also Coquet, Mackeviˇcius and M´emin in [6] and [7] proposed approximation schemes that use discretization of filtrations and convergence of filtrations. Briand, Delyon and M´emin in [3]
and Ma, Protter, San Mart´ın and Torres in [13] have approximated the Brownian motion by a scaled random walk. Then, in [4], Briand, Delyon and M´emin have studied another case : they approach the Brownian motion by a sequence of martingales.
We are interested in BSDEs with random terminal time because there have strong links with Partial Differential Equations as it is explained by Peng in [14]. As for BSDEs with de- terministic terminal time, we study the robustness of numerical methods to approximate the solutions of those BSDEs. In this paper, we shall approximate the Brownian motion either by a scaled random walk, either by a sequence of martingales. In this study, we need moreover to approximate the random almost surely finite terminal time τ by a sequence of stopping times (τ
n)
n.
In Section 1, we approximate the Brownian motion by a scaled random walk. First, we shall state the problem and study the properties of existence and uniqueness of the solutions of the BSDEs. Then, we will deal with the convergence of the solutions. To end this part, we give an example of the convergence result for hitting times.
In Section 2, we approximate the Brownian motion by a sequence of martingales. We shall see some generalizations of the results of Section 1 : existence and uniqueness of the solutions of the BSDEs under study, convergence of the solutions. Moreover, we will illustrate these results by the case of discretizations of a Brownian motion and hitting times.
For technical reasons, we need some results about convergence of stopped filtrations. So, in Appendix A, we will deal with stopped filtrations and stopped processes. We are going to
∗Email address : sandrine.toldo@math.univ-rennes1.fr
establish a link between the convergence of a sequence of stopped processes and the convergence of the associated stopped filtrations.
In what follows, we are given a probability space (Ω, A , P ). Unless otherwise specified, every σ-field will be supposed to be included in A , every process will be indexed by R
+and taking values in R , every filtration will be indexed by R
+. D = D ( R
+) denotes the space of c` adl` ag functions from R
+to R . We endow D with the Skorokhod topology. If X is a process and τ a stopping time, we denote by X
τthe corresponding stopped process, i.e. for every t, X
tτ= X
t∧τ.
Definition 1 Let ( F
t)
t>0be a filtration and τ a F -stopping time. We define the σ-field F
τby F
τ= { A ∈ F
∞: A ∩ { τ 6 s } ∈ F
s, ∀ s } where F
∞= _
t
F
tand the stopped filtration F
τby
F
tτ= F
τ∧t= { A ∈ F
t: A ∩ { τ 6 s } ∈ F
s, ∀ s 6 t } , for every t.
For technical background about Skorokhod topology, the reader may refer to Billingsley [2] or Jacod and Shiryaev [11].
1 Stability of BSDEs when the Brownian motion is ap- proximated by a scaled random walk
1.1 Statement of the problem
Let f : R
2→ R be a Lipschitz function : there exists K ∈ R
+such that, for every (y, z), (y
′, z
′) ∈ R
2, we have :
| f (y, z) − f (y
′, z
′) | 6 K[ | y − y
′| + | z − z
′| ].
For clarity’s sake, we consider a time-independent generator f but the results of Section 1 re- main true if f is time-dependent as it is explained in Remark 4.
We also suppose that f is bounded and that f fills the following property of monotonicity w.r.t. y : there exists µ > 0 such that
∀ (y, z), (y
′, z) ∈ R
+× R
2, (y − y
′)(f (y, z) − f (y
′, z)) 6 − µ(y − y
′)
2.
Let W be a Brownian motion and F its natural filtration. Let τ be a F -stopping time almost surely finite.
We consider the following stochastic differential equation : Y
t∧τ= ξ +
Z
τt∧τ
f (Y
s, Z
s)ds − Z
τt∧τ
Z
sdW
s, t > 0, (1) where ξ is a bounded F
τ-mesurable random variable.
Definition 2 We call a solution of the BSDE (1) a pair (Y, Z ) of progressively measurable processes verifying the equation (1) such that Y
t= ξ and Z
t= 0 on the set { t > τ } and :
E
sup
t∈R+
e
−2µt| Y
t|
2< + ∞ and ∀ t, E Z
t∧τ0
| Z
t|
2dt
< + ∞ .
According to Theorem 2.1 of Royer in [15], the BSDE (1) has a unique pair solution (Y, Z) in the set of processes such that Y is continuous and uniformly bounded.
In this section, we approximate equation (1) on the following way. We consider the sequence of scaled random walks (W
n)
n>1defined by :
W
tn= 1
√ n
[nt]
X
k=1
ε
nk, t > 0
where (ε
nk)
k∈N∗is a sequence of i.i.d. symmetric Bernoulli variables. Let F
nbe the natural filtrations of W
n, n > 1. We have F
tn= σ(ε
nk, k 6 [nt]). Let (τ
n)
nbe a sequence of bounded ( F
n)-stopping times. For each n, we can find T
n∈ N such that τ
n6 T
n.
Then, for each n, we consider the following equation : y
nkn∧τn
= y
nk+1n ∧τn
+
n11
{τn>kn}f y
nkn∧τn
, z
nk+1 n ∧τn− z
nk+1 n ∧τn√1
n
ε
nk+1, k = 0, . . . , (n − 1)T
ny
nτn= ξ
n(2) where (ξ
n) is a sequence of ( F
τnn)-measurable integrable random variables.
By a solution of equation (2), we mean a discrete process { y
nk n, z
nk+1n
}
k>0that satisfies (2), such that y
nkn
= ξ
nand z
nkn
= 0 on the set { τ
n<
nk} and such that { y
nkn∧τn
, z
nk+1n ∧τn
}
k>0is F
n,τn-adapted.
Proposition 3 Equation (2) has a unique solution (y
n, z
n).
Proof
We are going to build this solution on a converse iterative way.
For k = nT
n, let us put y
nkn∧τn
= ξ
nand z
nk+1n ∧τn
= 0.
Let us suppose that, for a given k, we have built y
nk+1n ∧τn
, z
nk+2 n ∧τn. Using equation (2), we are going to determinate
y
nkn∧τn
, z
nk+1 n ∧τn. Let us begin by giving an expression of z
nk+1n ∧τn
. Multiplying equation (2) by √
nε
nk+11
{τn>nk}and taking the conditional expectation with re- spect to F
k/nn,τn, we have :
z
nk+1n ∧τn
1
{τn>kn}= E h z
nk+1n ∧τn
1
{τn>kn}F
n,τk nn
i
= √
n E h y
nk+1n ∧τn
ε
nk+11
{τn>kn}F
n,τk nn
i + 1
√ n E h
1
{τn>kn}f y
nkn∧τn
, z
nk+1 n ∧τnε
nk+1F
n,τk nn
i
− √ n E h
y
nkn∧τn
ε
nk+11
{τn>kn}F
n,τk nn
i
= √ n E h y
nk+1n ∧τn
ε
nk+1F
n,τk nn
i 1
{τn>nk}because y
nkn∧τn
and z
nk+1n ∧τn
must be F
n,τk nn
-measurable, τ
nis a F
n-stopping time and ε
nk+1is independent from F
n,τk nn
and is centered.
Moreover, by definition of a solution, for every k, z
nk+1n ∧τn
= z
nk+1n ∧τn
1
{τn>nk}. So, we put : z
nk+1n ∧τn
= √ n E h
y
nk+1n ∧τn
ε
nk+1F
nkn
i 1
{τn>nk}.
We point out that z
nk+1n ∧τn
is F
n,τk nn
-measurable.
Now, we have to determinate y
nk n∧τn. On the set { τ
n<
kn} , y
nkn∧τn
= y
nk+1n ∧τn
= y
nτn= ξ
n. On { τ
n>
kn
} , y
nkn∧τn
= y
nk+1n ∧τn
+
1nf y
nkn∧τn
, z
nk+1 n ∧τn− z
nk+1 n ∧τn√1n
ε
nk+1and we can write it y
nkn∧τn
= ϕ y
nkn∧τn
with ϕ(y) = y
nk+1n ∧τn
+
1nf y, z
nk+1n ∧τn
− z
nk+1 n ∧τn√1
n
ε
nk+1. As f is K-Lipschitz in y, we have, for every y, y
′:
| ϕ(y) − ϕ(y
′) | 6 K
n | y − y
′| .
So, for n large enough (notice that this range does not depend on k),
Kn< 1 and ϕ is a contraction. Then, the equation y
nkn∧τn
= ϕ y
nkn∧τn
has a unique solution for n large enough, according to a fix point Theorem. By construction, y
nkn∧τn
is F
n,τk+1nn
-measurable. But, using the predictable representation property, we have y
nk+1n ∧τn
− z
nk+1 n ∧τn√1
n
ε
nk+1= E [y
nk+1n ∧τn
|F
n,τk nn
].
So y
nkn∧τn
is independant from ε
nk+1and y
nkn∧τn
is F
n,τk nn
-measurable.
Hence, the equation (2) has a unique solution.
Now, we define continuous time processes Y
nand Z
nby Y
tn= y
n[nt]n ∧τn
, Z
tn= z
n⌊nt⌋n ∧τn
, for every t ∈ R
+where ⌊ x ⌋ = (x − 1)
+if x is an integer, [x] otherwise. The processes Y
nand Z
nare constant on the intervals [k/n, (k + 1)/n[ and ]k/n, (k + 1)/n] respectively and satisfy the following equation :
Y
tn= ξ
n+ Z
τnt∧τn
f (Y
(sn∧τn)−, Z
sn∧τn)dA
ns− Z
τnt∧τn
Z
sn∧τndW
sn, (3) where A
ns=
[ns]n.
Remark 4 If f is time-dependent such that for every (y, z), we suppose that { f (t, y, z) }
t>0is progressively measurable, bounded, K-Lipschitz in y and z and verify the condition of mono- tonicity w.r.t. y given before. Under these assumptions, all the results of Section 1 remain true and the proofs are the same. In that case, Equation (2) becomes :
y
nkn∧τn
= y
nk+1n ∧τn
+ 1
n 1
{τn>nk}f k
n ∧ τ
n, y
nkn∧τn
, z
nk+1 n ∧τn− z
nk+1 n ∧τn√ 1 n ε
nk+1, k = 0, . . . , (n − 1)T
ny
τnn= ξ
n.
1.2 Convergence of the solutions
The aim of this section is to prove the following result of convergence of the solutions : Theorem 5 Let (Y, Z) be the solution of the BSDE (1) and (Y
n, Z
n) be the processes constant on the intervals [k/n, (k+1)/n[ and ]k/n, (k+1)/n] respectively solving equation (3). We suppose that there exists δ > 0 such that sup
nE [ | ξ
n|
1+δ]
1+δ1< + ∞ and sup
nE [ | τ
n|
1+δ]
1+δ1< + ∞ and that we have the convergences ξ
n− →
Pξ, τ
n− →
Pτ and W
n− →
PW . Then
∀ L ∈ N , sup
t∈[0,L]
| Y
tn∧τn− Y
t∧τ| + Z
τ∧τn0
| Z
tn∧τn− Z
t∧τ|
2dt − →
P0
which we shall denote by (Y
n, Z
n) → (Y, Z) and also
∀ L ∈ N , sup
t∈[0,L]
Z
t∧τn 0Z
sndW
sn− Z
t∧τ0
Z
sdW
s− →
P0.
According to this Theorem, we have the convergence in probability of the solutions under the rather strong assumption that the scaled random walks converge in probability to the Brownian motion and the random terminal times also converge. Actually, with Donsker’s Theorem, we have the convergence in law of the scaled random walks to the Brownian motion. If we only have this convergence in law, we obtain the following corollary :
Corollary 6 Let (Y, Z) be the solution of the BSDE (1) and (Y
n, Z
n) be the processes constant on the intervals [k/n, (k + 1)/n[ and ]k/n, (k + 1)/n] respectively solution of the equation (3).
We suppose that ∀ n, ∀ k, ε
nk= ε
k, ξ = g(W ) and ξ
n= g(W
n) with g bounded continuous. We assume that there exists δ > 0 such that sup
nE [ | τ
n|
1+δ]
1+δ1< + ∞ . We also suppose that we have the convergence (W
n, τ
n) −→
L(W, τ). Then
Y
.n∧τn, R
.∧τn0
Z
sndW
snn
converges in law to Y
.∧τ, R
.∧τ0
Z
sdW
sfor the Skorokhod topology.
Proof
According to the Skorokhod representation Theorem, in a space ( ˜ Ω, A ˜ , P ˜ ), we can find ( ˜ W
n, τ ˜
n) with the same law as (W
n, τ
n) and ( ˜ W , τ ˜ ) with the same law as (W, τ) such that ( ˜ W
n, τ ˜
n) −−→
a.s.( ˜ W , τ). We denote by ˜ ˜ F the natural filtration of ˜ W and by ˜ F
nthe natural filtrations of the W ˜
n.
Let us show that ˜ τ is an ˜ F -stopping time. Let us fix t > 0 and note that { τ 6 t } ∈ F
tsince τ is a F -stopping time. Then, for every ε > 0, we can find h : R
k→ R measurable and s
1, . . . , s
kin [0, t] such that
Z
| 1
{τ6t}− h(W
s1, . . . , W
sk) | d P < ε.
(W, τ ) ∼ ( ˜ W , τ) so ˜ 1
{τ6t}− h(W
s1, . . . , W
sk) ∼ 1
{˜τ6t}− h( ˜ W
s1, . . . , W ˜
sk). Then, Z
| 1
{τ6t˜ }− h( ˜ W
s1, . . . , W ˜
sk) | d ˜ P = Z
| 1
{τ6t}− h(W
s1, . . . , W
sk) | d P . So, for every ε > 0, we can find h : R
k→ R measurable and s
1, . . . , s
kin [0, t] such that
Z
| 1
{˜τ6t}− h( ˜ W
s1, . . . , W ˜
sk) | d P ˜ < ε.
Hence, { ˜ τ 6 t } ∈ F ˜
t. Then ˜ τ is an ˜ F -stopping time. On the same way, the ˜ τ
n’s are ˜ F
n-stopping times.
Let (Y
′n, Z
′n) be the solution of Y
t′n= g( ˜ W
n) +
Z
˜τnt∧τ˜n
f (Y
(s′n∧˜τn)−, Z
s′n∧τ˜n)dA
ns− Z
τ˜nt∧τ˜n
Z
s′n∧τ˜nd W ˜
snand (Y
′, Z
′) be the solution of
Y
t′∧˜τ= g( ˜ W ) + Z
τ˜t∧τ˜
f (Y
s′, Z
s′)ds − Z
˜τt∧˜τ
Z
s′d W ˜
s, t > 0.
All the assumptions of Theorem 5 are filled. So we have the convergence in probability of Y
.′∧nτ˜n, R
.∧τ˜n0
Z
s′nd W ˜
snn
to
Y
.′∧τ˜, R
.∧˜τ 0Z
s′d W ˜
s. Then, denoting by (Y
.n∧τn, Z
.n∧τn) the solution
of (1) and by (Y
.∧τ, Z
.∧τ) the solution of (3), since (Y
.n∧τn, Z
.n∧τn, W
n, τ
n) has the same law as
(Y
.′∧n˜τn, Z
.′∧nτ˜n, W ˜
n, τ ˜
n) and (Y
.∧τ, Z
.∧τ, W, τ) has the same law as (Y
.′∧˜τ, Z
.′∧τ˜, W , ˜ τ), we have the ˜
convergence of
Y
.n∧τn, R
.∧τn0
Z
sndW
snn
to
Y
.∧τ, R
.∧τ 0Z
sdW
sin distribution for the Skorokhod
topology.
We point out that in this corollary, it is necessary to have the joint convergence of ((W
n, τ
n))
nto (W, τ). In section 1.3, we shall see an example where the stopping times are hitting times.
To prove the first convergence of Theorem 5, we are going to use the Picard approximations (Y
p, Z
p) and ((Y
n,p, Z
n,p))
ndefined on the following way :
Y
tp+1∧τ= ξ + Z
τt∧τ
f (Y
sp, Z
sp)ds − Z
τt∧τ
Z
sp+1dW
s, ∀ t > 0, (4) Y
tn,p+1= ξ
n+
Z
τnt∧τn
f (Y
sn,p−, Z
sn,p)dA
ns− Z
τnt∧τn
Z
sn,p+1dW
sn, (5) with Y
0= Z
0= Y
n,0= Z
n,0= 0 and A
nt=
[nt]n.
We have existence and uniqueness of adapted pairs (Y
p, Z
p) and ((Y
n,p, Z
n,p))
nby induc- tion on p, using respectively Theorem 2.1 of Royer in [15] and Proposition 3.
We write :
Y
n− Y = (Y
n− Y
n,p) + (Y
n,p− Y
p) + (Y
p− Y ), Z
n− Z = (Z
n− Z
n,p) + (Z
n,p− Z
p) + (Z
p− Z).
We are going to prove successively the convergence of the Picard approximations to the solutions, i.e. for every n, (Y
n,p, Z
n,p) → (Y
n, Z
n) and (Y
p, Z
p) → (Y, Z ), and then check that the first convergence of the theorem is true for the Picard approximations, i.e. for every p, (Y
n,p, Z
n,p) → (Y
p, Z
p).
Most of the proof uses the same arguments as in the proof of Theorem 2.1 of Briand, Delyon and M´emin in [3] but there are some technical difficulties due to the stopping times.
The arguments of Lemma 4.1 in [3] are still true when we replace a deterministic terminal time by a bounded random terminal time. So we have :
∀ L, sup
n
E
"
sup
t∈[0,L]
| Y
τnn∧t− Y
τn,pn∧t|
2+ Z
+∞0
| Z
tn− Z
tn,p|
2dt
#
−−−−−→
p→+∞
0. (6) With a truncation argument (using the fact that τ is almost surely finite), we deduce the convergence of (Y
p, Z
p) to (Y, Z), ie
∀ L, E
"
sup
t∈[0,L]
| Y
τp∧t− Y
τ∧t|
2+ Z
+∞0
| Z
tp− Z
t|
2dt
#
−−−−−→
p→+∞
0. (7)
Now, let us show that for every p, (Y
n,p, Z
n,p) → (Y
p, Z
p) as n → + ∞ that is
∀ L ∈ N , sup
t∈[0,L]
| Y
tn,p∧τn− Y
tp∧τ| + Z
τn0
| Z
tn,p∧τn− Z
tp∧τ|
2dt − →
P0 as n → ∞ . (8) We argue by induction on p.
- The property is true for p = 0 because Y
0= Z
0= Y
n,0= Z
n,0= 0.
- We suppose that, for given p, (8) holds. Let us prove that (8) is still true for p + 1. The proof will be given through three steps.
In a first step, we introduce the sequence (M
n)
nof processes defined by M
tn= Y
tn,p+1∧τn+
Z
t∧τn 0f (Y
sn,p, Z
sn,p)dA
nsand the process M defined by
M
t= Y
tp+1∧τ+ Z
t∧τ0
f (Y
sp, Z
sp)ds.
In Lemma 7, we prove that (M
.n∧τn)
nis a sequence of ( F
n,τn)-martingales. Then, with Lemmas 9 and 10, we show that we have the following convergence :
∀ L, sup
t∈[0,L]
| M
tn∧τn− M
t∧τ| − →
P0 as n → ∞ . The aim of the second step is to prove Lemma 12 :
Z
τn0
| Z
tn,p+1∧τn− Z
tp+1∧τ|
2dt − →
P0 as n → ∞ .
At last, in a third step (Lemma 13), we deal with the convergence of Y
n,p+1to Y
p+1:
∀ L, sup
t∈[0,L]
| Y
tn,p+1∧τn− Y
tp+1∧τ| − →
P0 as n → ∞ . Step 1 : Let (M
n)
nbe the processes defined by M
tn= Y
tn,p+1∧τn+ R
t∧τn0
f (Y
sn,p, Z
sn,p)dA
ns. Lemma 7 For each n, M
nis a F
n,τn-martingale.
Proof
Let us show that M
tn= M
0n+ R
t∧τn0
Z
sn,p+1∧τndW
sn. Using (5), M
0n+
Z
t∧τn 0Z
sn,p+1∧τndW
sn= Y
0n,p+1+ Z
t∧τn0
Z
sn,p+1∧τndW
sn= ξ
n+ Z
τn0
f (Y
sn,p−, Z
sn,p)dA
ns− Z
τn0
Z
sn,p+1dW
sn+ Z
t∧τn0
Z
sn,p+1∧τndW
sn= Y
tn,p+1∧τn+ Z
t∧τn0
f (Y
sn,p−, Z
sn,p)dA
ns= M
tn. Then, for every n, as the process (M
0n+ R
t0
Z
sn,p+1∧τndW
sn)
t>0is a F
n-martingale, the stopped process (M
0n+ R
t∧τn0
Z
sn,p+1∧τndW
sn)
t>0= (M
tn)
t>0is a F
n,τn-martingale.
So, we have M
tn= E [M
τnn|F
tn,τn] with M
τnn= Y
τn,p+1n+ R
τn0
f (Y
sn,p−, Z
sn,p)dA
ns, for every n, for every t.
Before proving the convergence of (M
τnn)
n, let us see a lemma that links stopped integrals and integrals of stopped processes :
Lemma 8 For every t, we have the following relations : R
t∧τ0
f (Y
sp, Z
sp)ds = R
t0
f (Y
sp∧τ, Z
sp∧τ)ds + (τ ∧ t − t)f (Y
τp, Z
τp), R
t∧τn0
f (Y
sn,p, Z
sn,p)ds = R
t0
f (Y
sn,p∧τn, Z
sn,p∧τn)ds + (τ
n∧ t − t)f (Y
τn,pn, Z
τn,pn).
Proof
Let us prove the first relation.
On { t 6 τ } ,
Z
t0
f (Y
sp∧τ, Z
sp∧τ)ds = Z
t0
f (Y
sp, Z
sp)ds = Z
t∧τ0
f (Y
sp, Z
sp)ds
and on { t > τ } , we have : Z
t0
f (Y
sp∧τ, Z
sp∧τ)ds = Z
t∧τ0
f (Y
sp∧τ, Z
sp∧τ)ds + Z
tt∧τ
f (Y
sp∧τ, Z
sp∧τ)ds
= Z
t∧τ0
f (Y
sp, Z
sp)ds + Z
tt∧τ
f (Y
τp, Z
τp)ds
= Z
t∧τ0
f (Y
sp, Z
sp)ds + (t − t ∧ τ)f (Y
τp, Z
τp).
The second equality is proved by the same arguments.
Lemma 9 (M
τnn)
nconverges in L
1to Y
τp+1+ R
τ0
f (Y
sp, Z
sp)ds.
Proof
It suffices to show that (M
τnn)
nconverges to Y
τp+1+ R
τ0
f (Y
sp, Z
sp)ds in probability. Indeed, for every n, E [ | M
τnn|
1+δ]
1+δ16 E [ | ξ
n|
1+δ]
1+δ1+ k f k
∞E [ | τ
n|
1+δ]
1+δ1. So, when we take the sup in n, sup
nE [ | M
τnn|
1+δ]
1+δ1< ∞ according to the assumptions on (ξ
n)
n, (τ
n)
nand f . So, the sequence (M
τnn) is uniformly integrable and then the convergence in probability implies the convergence in L
1.
Let us show that (M
τnn)
nconverges in probability to Y
τp+1+ R
τ0
f (Y
sp, Z
sp)ds.
M
τnn− Y
τp+1+ Z
τ0
f (Y
sp, Z
sp)ds
6 | Y
τn,p+1n− Y
τp+1| +
Z
τn0
f (Y
sn,p−, Z
sn,p)dA
ns− Z
τ0
f (Y
sp, Z
sp)ds
But,
| Y
τn,p+1n− Y
τp+1| = | ξ
n− ξ | − →
P0. (9) We are going to conclude by showing that
Z
τn0
f (Y
sn,p−, Z
sn,p)dA
ns− Z
τ0
f (Y
sp, Z
sp)ds
− →
P0.
For each n, for each ω, there exists a unique k
nsuch that τ
nis in the interval [k
n/n, (k
n+ 1)/n[. As the processes Y
n,pand Z
n,pare constant on the intervals of the form [k/n, (k + 1)/n[
and ]k/n, (k + 1)/n] respectively by construction, we have : Z
τn0
f (Y
sn,p−, Z
sn,p)dA
ns= Z
kn/n0
f (Y
sn,p, Z
sn,p)ds.
So,
Z
τn0
f (Y
sn,p−, Z
sn,p)dA
ns− Z
τ0
f (Y
sp, Z
sp)ds
=
Z
kn/n0
f (Y
sn,p, Z
sn,p)ds − Z
τ0
f (Y
sp, Z
sp)ds
6
Z
(kn/n)∧τ 0(f (Y
sn,p, Z
sn,p) − f (Y
sp, Z
sp))ds
+
Z
(kn/n)∨τ (kn/n)∧τ(f (Y
sn,p, Z
sn,p)1
{τ6kn/n}− f (Y
sp, Z
sp)1
{τ >kn/n})ds
.
Taking t = (k
n/n) ∧ τ in the two equalities of Lemma 8, we have : Z
(kn/n)∧τ0
f (Y
sp, Z
sp)ds =
Z
(kn/n)∧τ 0f (Y
sp∧τ, Z
sp∧τ)ds, Z
(kn/n)∧τ0
f (Y
sn,p, Z
sn,p)ds =
Z
(kn/n)∧τ 0f (Y
sn,p∧τn, Z
sn,p∧τn)ds.
So,
Z
(kn/n)∧τ 0(f (Y
sn,p, Z
sn,p) − f (Y
sp, Z
sp))ds
=
Z
(kn/n)∧τ 0(f (Y
sn,p∧τn, Z
sn,p∧τn) − f (Y
sp∧τ, Z
sp∧τ))ds
6
Z
(kn/n)∧τ 0K[ | Y
sn,p∧τn− Y
sp∧τ| + | Z
sn,p∧τn− Z
sp∧τ| ]ds because f is K-Lipschitz
6 K
Z
τn∧τ0
| Y
sn,p∧τn− Y
sp∧τ| ds + K Z
τ∧τn0
| Z
sn,p∧τn− Z
sp∧τ| ds.
Using the induction assumption (8), the fact that τ < + ∞ a.s. and Cauchy-Schwarz inequality, we prove that
Z
τ∧τn0
| Z
sn,p∧τn− Z
sp∧τ|
2ds − →
P0. (10) On the other hand, let us fix ε > 0 and η > 0. Since τ < + ∞ a.s., we can find T such that P [τ > T ] 6 ε. Then,
P
"
Z
τn∧τ0
| Y
sn,p∧τn− Y
sp∧τ| ds > η
#
= P
"
Z
τn∧τ0
| Y
sn,p∧τn− Y
sp∧τ| ds 1
{τ∧τn<T}> η
# + P
"
Z
τn∧τ0
| Y
sn,p∧τn− Y
sp∧τ| ds 1
{τ∧τn>T}> η
#
6 P
"
T sup
s∈[0,T]
| Y
sn,p∧τn− Y
sp∧τ| > η
#
+ P [τ ∧ τ
n> T ] 6 2ε by choice of T and using (8).
So,
Z
τn∧τ0
| Y
sn,p∧τn− Y
sp∧τ| ds − →
P0. (11) Finally, using the convergences (10) and (11), we have
Z
(kn/n)∧τ 0(f (Y
sn,p, Z
sn,p) − f (Y
sp, Z
sp))ds
− →
P0. (12)
On the other hand, | τ
n− k
n/n | 6 1/n and τ
n− →
Pτ, so we have (k
n/n) − →
Pτ. Then,
Z
(kn/n)∨τ (kn/n)∧τ(f (Y
sn,p, Z
sn,p)1
{τ6kn/n}− f (Y
sp, Z
sp)1
{τ >kn/n})ds
6 | k
n/n − τ | k f k
∞(13)
−
P→ 0.
According to the convergences (12) and (13), we have
Z
τn0
f (Y
sn,p−, Z
sn,p)dA
ns− Z
τ0
f (Y
sp, Z
sp)ds
− →
P0. (14)
Hence, according to the convergences (9) and (14), M
τnn− →
PY
τp+1+
Z
τ0
f (Y
sp, Z
sp)ds
. (15)
At last, Lemma 9 is proved.
Let M be the process defined by M
t= Y
tp+1∧τ+ R
t∧τ0
f (Y
sp, Z
sp)ds.
Lemma 10 We have the convergence
∀ L, sup
t∈[0,L]
| M
tn− M
t| − →
P0. (16) Proof
Using the same computation as for M
tnin Lemma 7, we find : M
t= M
0+
Z
t∧τ 0Z
sp+1dW
s= Y
tp+1∧τ+ Z
t∧τ0
f (Y
sp, Z
sp)ds.
So the process M is a F
τ-martingale. Then, M
t∧τ= E [M
τ|F
tτ], for every t, where M
τ= Y
τp+1+ R
τ0
f (Y
sp, Z
sp)ds. According to (15), M
τnnL1
−−→ M
τ.
If we prove that we have the convergence of the sequence of filtrations ( F
n,τn)
nto F
τ, according to Remark 1.2 in Coquet, M´emin and S lomi´ nski [8], we have M
.n− →
PM
.for the Skorokhod topology.
(W
n)
nis a sequence of processes with independant increments that converges in probability to W . So, according to Proposition 2 in [8], F
n w−→ F . Moreover, W is continuous and τ
n− →
Pτ so according to Corollary 29, F
n,τn−→ F
w τ.
Finally, we have M
.n− →
PM
.for the Skorokhod topology. Moreover, F is a Brownian filtra- tion, so, every F -martingale is continuous. In particular, E [M
τ|F
.] is continuous. The stopped process is also continuous, i.e. M
.= E [M
τ|F
.τ] is a continuous process. So, the previous con- vergence is uniform in t on every compact set and Lemma 10 is proved.
Step 2 : In this step, we shall prove the convergence of (Z
n,p+1) to Z
p+1as n → ∞ . First, let us see a lemma of convergence of quadratic variations :
Lemma 11 (W
n,τn, M
n,τn, [M
n,τn, M
n,τn], [M
n,τn, W
n,τn]) converges to (W
τ, M
τ, [M
τ, M
τ], [M
τ, W
τ]) in probability.
Proof
It suffices to follow the lines of Briand, Delyon and M´emin in the proof of Theorem 3.1 in [3].
Lemma 12 R
τn0
| Z
tn,p+1∧τn− Z
tp+1∧τ|
2ds − →
P0.
Proof
To prove this lemma, we prove that we have these two convergences : Z
τn0
| Z
tn,p+1∧τn|
2dt − Z
τn0
| Z
tp+1∧τ|
2dt − →
P0, (17)
∀ g ∈ L
2, Z
τn0
g(s)(Z
sn,p+1∧τn− Z
sp+1∧τ)ds − →
P0. (18) Then, we conclude with the following lines :
Z
τn0
| Z
tn,p+1∧τn− Z
tp+1∧τ|
2dt
= Z
τn0
| Z
tn,p+1∧τn|
2dt + Z
τn0
| Z
tp+1∧τ|
2dt − 2 Z
τn0
Z
tn,p+1∧τnZ
tp+1∧τdt
=
Z
τn0
| Z
tn,p+1∧τn|
2dt − Z
τn0
| Z
tp+1∧τ|
2dt
! + 2
Z
τn0
| Z
tp+1∧τ|
2dt − Z
τn0
Z
tn,p+1∧τnZ
tp+1∧τdt
!
− →
P0 according to (17) and (18) with g = Z
p+1. First, let us show that R
τn0
| Z
tn,p+1∧τn|
2dt − R
τn0
| Z
tp+1∧τ|
2dt − →
P0.
Z
τn0
| Z
tn,p+1∧τn|
2dt − Z
τn0
| Z
tp+1∧τ|
2dt
6
Z
τn0
| Z
tn,p+1∧τn|
2dt − Z
τn0
| Z
tn,p+1∧τn|
2dA
nt+
Z
τn0
| Z
tn,p+1∧τn|
2dA
nt− Z
τn0
| Z
tp+1∧τ|
2dt .
According to Lemma 11,
∀ L, sup
t∈[0,L]
[M
n,τn, M
n,τn]
t− [M
τ, M
τ]
t− →
P0.
But,
M
tn,τn= M
0n+ Z
t∧τn0
Z
sn,p+1dW
sn= M
0n+ 1
√ n
[n(t∧τn)]
X
k=1
Z
k/nn,p+1ε
nk.
So,
[M
n,τn, M
n,τn]
t= 1 n
[n(t∧τn)]
X
k=1
| Z
k/nn,p+1|
2= Z
t∧τn0
| Z
sn,p+1|
2dA
ns. On the other hand, M
τis a continuous martingale, so
[M
τ, M
τ]
t=< M
τ, M
τ>
t= Z
t∧τ0
| Z
sp+1|
2ds.
Then,
∀ L, sup
s∈[0,L]
Z
s∧τn0
| Z
tn,p+1|
2dA
nt− Z
s∧τ0
| Z
tp+1|
2dt
− →
P0.
In particular, using the fact that sup
nτ
nis almost surely finite because (τ
n)
nconverges in probability to the almost surely finite stopping time τ, we have
Z
τn0
| Z
tn,p+1|
2dA
nt− Z
τ∧τn0
| Z
tp+1|
2dt
− →
P0.
As τ
n− →
Pτ, by dominated convergence, we have :
Z
τn0
| Z
tp+1|
2dt − Z
τn∧τ0
| Z
tp+1|
2dt
− →
P0 as n → + ∞ .
Moreover, R
τn0
| Z
tn,p+1|
2dA
nt= R
τn0
| Z
tn,p+1|
2dt. So, the convergence (17) is proved.
It remains to prove convergence (18). Let us fix g ∈ L
2( R
+) and show that R
τn0
g(s)(Z
sn,p+1∧τn− Z
sp+1∧τ)ds − →
P0.
According to Lemma 11, we have the convergence
∀ L, sup
t∈[0,L]
[M
n,τn, W
n,τn]
t− [M
τ, W
τ]
t− →
P0.
But,
[M
n,τn, W
n,τn]
t= 1 n
[n(t∧τn)]
X
k=1
Z
k/nn,p+1= Z
t∧τn0
Z
sn,p+1dA
ns.
On the other hand, M
τand W
τare continuous martingales, so [M
τ, W
τ]
t=< M
τ, W
τ>
t=
Z
t∧τ 0Z
sp+1ds.
Then,
∀ L, sup
t∈[0,L]
Z
t∧τn 0Z
sn,p+1dA
ns− Z
t∧τ0
Z
sp+1ds
− →
P0. (19) We write :
Z
τ∧τn 0g(s)(Z
sn,p+1∧τn− Z
sp+1∧τ)ds
6
Z
τ∧τn 0g(s)Z
sn,p+1∧τnds − Z
τ∧τn0
g(s)Z
sn,p+1∧τndA
ns+
Z
τ∧τn 0g(s)Z
sn,p+1∧τndA
ns− Z
τ∧τn0
g(s)Z
sp+1∧τds
As the function g is measurable and τ is almost surely finite, using a density argument and the convergence (19), we have :
Z
τn0
g(s)Z
sn,p+1∧τndA
ns− Z
τ∧τn0
g(s)Z
sp+1∧τnds
→ 0. (20)
By dominated convergence, we deduce that
Z
τ∧τn 0g(s)Z
sn,p+1∧τnds − Z
τ∧τn0
g(s)Z
sn,p+1∧τndA
ns− →
P0.
Next, using the density of the set of continuous function with compact support in L
2, we show
that
Z
τ∧τn 0g(s)Z
sn,p+1∧τndA
ns− Z
τ∧τn0
g(s)Z
sp+1∧τds
− →
P0.
The convergence (18) is now proved.
Lemma 12 is proved.
Step 3 : Let us prove the convergence of Y
n,p+1to Y
p+1when n goes to ∞ .
Lemma 13 For every L, sup
t∈[0,L]
| Y
tn,p+1∧τn− Y
tp+1∧τ| − →
P0.
Proof
We fix L. We write :
sup
t∈[0,L]
| Y
tn,p+1∧τn− Y
tp+1∧τ|
6 sup
t∈[0,L]
| M
tn∧τn− M
t∧τ| + sup
t∈[0,L]
Z
t∧τn 0f (Y
sn,p−, Z
sn,p)dA
ns− Z
t∧τ0
f (Y
sp, Z
sp)ds
Moreover,
sup
t∈[0,L]
Z
t∧τn 0f (Y
sn,p−, Z
sn,p)dA
ns− Z
t∧τ0
f (Y
sp, Z
sp)ds
6 sup
t∈[0,L]
Z
t∧τn 0f (Y
sn,p−, Z
sn,p)dA
ns− Z
t∧τn0
f (Y
sn,p, Z
sn,p)ds
+ sup
t∈[0,L]
Z
t∧τn 0f (Y
sn,p, Z
sn,p)ds − Z
t∧τn0
f (Y
sp, Z
sp)ds
1
τn6τ+ sup
t∈[0,L]
Z
t∧τ 0f (Y
sn,p, Z
sn,p)ds − Z
t∧τ0
f (Y
sp, Z
sp)ds
1
τn>τ+ sup
t∈[0,L]
Z
t∧τn 0f (Y
sp, Z
sp)ds − Z
t∧τ0
f (Y
sp, Z
sp)ds
1
τn6τ+ sup
t∈[0,L]
Z
t∧τn 0f (Y
sn,p, Z
sn,p)ds − Z
t∧τ0
f (Y
sn,p, Z
sn,p)ds
1
τn>τBut, for every t, for every ω, we can find k
nsuch that t ∧ τ
n∈ [k
n/n, (k
n+ 1)/n[. So, we have
Z
t∧τn 0f (Y
sn,p−, Z
sn,p)dA
ns− Z
t∧τn0
f (Y
sn,p, Z
sn,p)ds
=
Z
t∧τn kn/nf (Y
sn,p, Z
sn,p)ds 6 | t ∧ τ
n− k
n/n | k f k
∞6 k f k
∞/n.
Then,
sup
t∈[0,L]
Z
t∧τn 0f (Y
sn,p−, Z
sn,p)dA
ns− Z
t∧τn0
