Proof of Theorem 3.2.1

h(x)−h(x⁰) ≤k7

x−x⁰

and |h(x)| ≤λ5

for allx, x⁰ ∈R^m.

The second main result of this work is the following:

Theorem 3.2.2. If (B1) - (B4) hold, then there exists a constantC_k,λ depending only on the coefficients ki and λi such that if T ≤ Ck,λ, then there exist two constants C₁ andC₂ such that FBSDE (3.2.2)has a unique solution (X, Y, Z) such that(X, Y, Z·W) ∈ S²(R^m)× S^∞(R^m

0)×BMO andkYk_S∞(R^m0) ≤C₁, andkZ·Wk_BMO≤C₂.

3.3 FBSDEs with superquadratic growth

3.3.1 Proof of Theorem 3.2.1

Step 1: We first assume thath, b, gare continuously differentiable in all variables.

We will show that the sequence(Xⁿ, Yⁿ, Zⁿ)given byX⁰ = 0,Y⁰ = 0,Z⁰ = 0 and

(X_tⁿ⁺¹ =x+Rt

0 b(X_uⁿ⁺¹, Y_uⁿ)du+Rt

0σudWu

Y_tⁿ⁺¹ =h(X_Tⁿ⁺¹) +RT

t g_u(X_uⁿ⁺¹, Y_uⁿ⁺¹, Z_uⁿ⁺¹)du−RT

t Z_uⁿ⁺¹dW_u, n≥1 is well defined and that there exists a constantC >0which does not depend onn such that|Zⁿ|< Cfor alln.

By [68] and [61] the processX¹is well defined, belongs toD^1,2(R^m)and its Malliavin’s derivative satisfies

D_tX_r¹= 0, 0≤r < t≤T, DtX_r¹=

r

Z

t

(∂xbDtX_u¹+∂ybDtY_u⁰)du+Dt





r

Z

t

σudWu



, 0≤t≤r≤T,

withDt(Rr

t σudWu) = σ1_[t,r], see [61, Theorem 2.2.1]. Hence, sincebis Lips-chitz continuous, by Gronwall’s inequality we have

DtX_r¹

≤e^{T k1}λ2.

Moreover, by the chain rule, see [61, Proposition 1.2.4] it follows thath(X_T¹) ∈ D^1,2(R^m

0) and D(h(X_T¹)) = ∂xh(X_T¹)DX_T¹. Therefore, h(X_T¹) has bounded Malliavin derivative since∂_xhis bounded. We then deduce from Theorem 3.A.2 and its proof that(Y¹, Z¹)exists, (Y¹, Z¹) ∈ D^1,2(R^m0)× D^1,2(R^m0×d),DY¹ is bounded and Z_t¹ = DtY_t¹. Now let n ∈ N, assume that (Xⁿ, Yⁿ, Zⁿ) ∈ D^1,2(R^m)×D^1,2(R^m

0)×D^1,2(R^m

0×d),DXⁿ, DYⁿare bounded andZ_tⁿ=D_tY_tⁿ. The processXⁿ⁺¹is well defined, belongs toD^1,2(R^m)and its Malliavin deriva-tive satisfies

DtX_rⁿ⁺¹ = 0, 0≤r < t≤T, D_tX_rⁿ⁺¹ =σ1_[t,r]+

r

Z

t

(∂_xbD_tX_uⁿ⁺¹+∂_ybD_tY_uⁿ)du, 0≤t≤r ≤T.

Since∂xb,∂yb andσ are bounded byk1, k2 andλ2 respectively, it follows from Gronwall’s inequality that

DtX_rⁿ⁺¹

≤e^{T k1}



λ2+k2 T

Z

0

|D_tY_uⁿ|du



. Hence,

D_tXⁿ⁺¹

_S∞ ≤e^{T k1}(λ₂+k₂TkD_tYⁿk_S∞)<∞. (3.3.1) By the chain rule,D(h(X_Tⁿ⁺¹))exists and is bounded. It then follows again from Theorem 3.A.2 and its proof that(Yⁿ⁺¹, Zⁿ⁺¹) exists andZⁿ⁺¹ is bounded. In addition,(Yⁿ⁺¹, Zⁿ⁺¹)are Malliavin differentiable and the derivatives satisfy, for j= 1, . . . , d,

D_t^jY_rⁿ⁺¹ = 0, D_t^jZ_rⁿ⁺¹ = 0, 0≤r < t≤T, D_t^jY_rⁿ⁺¹ =∂_xh(X_Tⁿ⁺¹)D^j_tX_Tⁿ⁺¹+

T

Z

r

∂_xgD^j_tX_uⁿ⁺¹+∂_ygD^j_tY_uⁿ⁺¹

+∂_zgD^j_tZ_uⁿ⁺¹ du−

T

Z

r

D^j_tZ_uⁿ⁺¹dW_u, 0≤t≤r ≤T.

By (A3)-(A5) and the boundedness ofZⁿ⁺¹andDXⁿ⁺¹, it follows from the same

procedure of the proof of Lemma 3.A.1 that fori= 1, . . . , m⁰;j= 1, . . . , d. Plugging the above estimate in (3.3.1), we obtain

0×d). Indeed, using (A1) we can estimate the norm of the dif-ferenceX_tⁿ⁺¹−X_tⁿas

Taking expectation on both sides and using Cauchy-Schwarz’ inequality, we have E

ChoosingT to be small enough so that2T²k₁² ≤ ¹₂, it follows

E

"

sup

0≤t≤T

|X_tⁿ⁺¹−X_tⁿ|²

#

≤4T²k²₂E

"

sup

0≤t≤T

|Y_tⁿ−Y_tⁿ⁻¹|²

#

. (3.3.2)

On the other hand, applying Itô’s formula toe^βt|Y_tⁿ⁺¹−Y_tⁿ|²,β ≥0, we have

e^βt|Y_tⁿ⁺¹−Y_tⁿ|²

=e^βT|h(X_Tⁿ⁺¹)−h(X_Tⁿ)|²−2

T

Z

t

e^βs(Y_sⁿ⁺¹−Y_sⁿ)(Z_sⁿ⁺¹−Z_sⁿ)dW_s

−

T

Z

t

e^βs(Z_sⁿ⁺¹−Z_sⁿ)²ds−

T

Z

t

βe^βs(Y_sⁿ⁺¹−Y_sⁿ)²ds

+ 2

T

Z

t

e^βs(Y_sⁿ⁺¹−Y_sⁿ)

gs(X_sⁿ⁺¹, Y_sⁿ⁺¹, Z_sⁿ⁺¹)−gs(X_sⁿ, Y_sⁿ, Z_sⁿ) ds.

Hence, due to the condition (A3) and the boundedness of(Zⁿ), it holds

e^βt|Y_tⁿ⁺¹−Y_tⁿ|²+

T

Z

t

e^βs(Z_sⁿ⁺¹−Z_sⁿ)²ds

≤e^βT

h(X_Tⁿ⁺¹)−h(X_Tⁿ)

2−2

T

Z

t

e^βs(Y_sⁿ⁺¹−Y_sⁿ)(Z_sⁿ⁺¹−Z_sⁿ)dW_s

−

T

Z

t

βe^βs(Y_sⁿ⁺¹−Y_sⁿ)²ds+ 2

T

Z

t

e^βsρ(Q)

Y_sⁿ⁺¹−Y_sⁿ

Z_sⁿ⁺¹−Z_sⁿ ds

+ 2

T

Z

t

e^βsk7

Y_sⁿ⁺¹−Y_sⁿ

X_sⁿ⁺¹−X_sⁿ ds+ 2

T

Z

t

e^βsk4

Y_sⁿ⁺¹−Y_sⁿ

2ds.

With some positive constantsα1,α2, it follows from (A5) and Young’s inequality

that

2 + 2k8 and taking expectation on both sides above, we have Next, taking conditional expectation with respect toF_tin (3.3.3),

e^βt|Y_tⁿ⁺¹−Y_tⁿ|²+E

Thus, by Burkholder-Davis-Gundy’s inequality, with a positive constantc1 and It now follows from (3.3.2) that

E TakingT small enough so that

8(c₁+ 1)e^βT(k²₅+T)T²k₂²≤ 1 2,

we obtain that (Xⁿ, Yⁿ, Zⁿ) is a Cauchy sequence in S²(R^m) × S²(R^m

0) × H²(R^m

0×d). By continuity of b, g and h we have the existence of a solution (X, Y, Z)inS²(R^m)× S²(R^m

0)× H²(R^m

0×d)of FBSDE (3.2.1) and it follows from the boundedness of(Zⁿ)thatZsatisfies (3.2.3). The uniqueness follows from the boundedness ofZ and by repeating the above arguments on the difference of two solutions.

It is easy to check thatbⁿsatisfies (A1) with the constantsk1, k2 and2λ1and that gⁿ andhⁿ satisfy (A3) - (A4) and (A5), respectively, with the same constants.

From the above argument, there existsC¯_k,λ,m⁰_,dindependent ofnsuch that ifT ≤ C¯k,λ,m⁰,d, FBSDE (3.2.1) with parameters (bⁿ, hⁿ, gⁿ) admits a unique solution (Xⁿ, Yⁿ, Zⁿ)∈ S²(R^m)× S²(R^m

0)× S^∞(R^m

0×d)and

|Z_t^ij,n| ≤2λ2m⁰e^k1T+m0k4T (k5+k3T) P⊗dt-a.e.

By the Lipschitz continuity conditions onbandhand the locally Lipschitz condi-tion ofg, the sequences(bⁿ)and(hⁿ)converge uniformly tobandh onR^m+m

0

andR^m, respectively, and (gⁿ)converges toguniformly onR^m+m

0 ×Λfor any compact subset Λ of R^m

0×d. Combining these uniform convergences with the boundedness of Zⁿ, similar to above, we can show that there exists a constant C˜_k,λ,m⁰_,ddepending only onk1, k2, k3, k4, k5, λ2, m⁰, dsuch that ifT ≤C˜_k,λ,m⁰_,d, (Xⁿ, Yⁿ, Zⁿ)is a Cauchy sequence inS²(R^m)× S²(R^m

0)× H²(R^m

0×d). Hence withC_k,λ,m⁰_,d = ¯C_k,λ,m⁰_,d∧C˜_k,λ,m⁰_,d, for anyT ≤C_k,λ,m⁰_,d, the FBSDE (3.2.2) admits a solution (X, Y, Z) ∈ S²(R^m)× S²(R^m

0)× S^∞(R^m

0×d) and|Z_t^ij| ≤ 2λ₂m⁰e^k1T+m0k4T(k₅+k₃T). The uniqueness follows from similar arguments.

Step 4: Now, assumeT > C_k,λ,m⁰_,dand the additional growth conditions onb, gandhgiven by (3.2.4) hold. Let˜hQ : R → Rbe a continuously differentiable function whose derivative is bounded by1and such that˜h⁰_Q(a) = 1for all−Q≤ a≤Qand

˜hQ(a) =







(Q+ 1) if a > Q+ 2 a if |a| ≤Q

−(Q+ 1) if a <−(Q+ 2).

An example of such a function is given by h˜_Q(a) =

( −Q²+ 2Qa−a(a−4)

/4 if a∈[Q, Q+ 2]

Q²+ 2Qa+a(a+ 4)

/4 if [−(Q+ 2),−Q], see [43]. By the assumptions (A3) the functiong˜: [0, T]×R^m×R^m

0×R^m

0×d→R defined by

˜

gt(x, y, z) :=gt(x, y, hQ(z)) (3.3.4) withh_Q(z) := (˜h_Q(z^ij))_ijis Lipschitz continuous in all variables. Thus, it follows from [20, Theorem 2.6] that the equation

(X˜_t=x+Rt

0b_u( ˜X_u,Y˜_u)du+Rt

0 σ_udW_u Y˜_t=h( ˜X_T) +RT

t ˜g_u( ˜X_u,Y˜_u,Z˜_u)du−RT

t Z˜_udW_u, t∈[0, T] (3.3.5) admits a unique solution( ˜X,Y ,˜ Z)˜ ∈ S²(R^m)× S^∞(R^m

0)× H^∞(R^m

0×d). More-over, there exists a bounded functionθ : [0, T]×R^m → R^m

0 which is Lipschitz continuous inxsuch thatY˜t=θ(t,X˜t)for allt∈[0, T]. PutN = [T /Ck,A,q,m⁰,d],

where[a]denotes the integer part ofa, andti := iCk,A,q,m⁰,d,i = 0, . . . , N and

That is,(X, Y, Z)satisfies Equation (3.3.6). Uniqueness follows from [20,

Theo-rem 2.6]. This concludes the proof.

3.3.2 Fully coupled systems

In order to consider the fully coupled forward-backward system, i.e., we allow the dependence in(x, y, z)ofbandσ, we assume boundedness conditions on the

Malliavin derivatives of the generator and the terminal condition. Under this as-sumption, we can obtain solvability on any time interval[0, T],T ∈(0,∞)for the Markovian case. Now, consider the following conditions

(A1’) b: Ω×[0, T]×R^m×R^m

0×R^m

0×d→R^mis a continuous and measurable function such that there existk₁, k₂, k₃, λ₁≥0such that

0is a continuous and measurable function such thatgⁱ_t(x, y, z) = g_tⁱ(x, y, zⁱ)fori = 1, . . . , m⁰ and there

(A5’) h: Ω×R^m →R^m

0 is continuous andF_T-measurable such thath(XT)∈ D^1,2(R^m

Theorem 3.3.1. If (A1’) - (A5’) hold, then there exist two constants C_k,A,q,m⁰_,d andε_k,A,q,m⁰_,ddepending only onk₁, k₂, k₃, k₄, k₅, k₇, k₈, k₉, A, q, m⁰, dsuch that solution of the FBSDE

(X_tⁿ⁺¹ =x+Rt and|Zⁿ⁺¹−Zⁿ|follows from exactly the same procedure as in the the proof of Theorem 3.2.1. Indeed, we have

|X_tⁿ⁺¹−X_tⁿ|² ≤6

Taking supremum with respect tot, then expectation to both sides and using Cauchy-Schwarz’ and Burkholder-Davis-Gundy’s inequalities, we have

E Hence the result follows directly from the arguments in the proof of Theorem

3.2.1.

0 is continuous and continu-ously differentiable inyandz, and is such thatg_tⁱ(x, y, z) =gⁱ_t(x, y, zⁱ)

i = 1, . . . , m⁰ and there existk7, k8, λ3 ≥0 as well as a nondecreasing

0 is continuously differentiable and such thath(X_T) ∈ D^1,2 for any X_T ∈ L²(F_T) and there exist constants k₉, λ₄, A_ij ≥ 0

Theorem 3.3.2. If (A1”) - (A5”) hold, then the FBSDE (Xt=x+Rt

0×d)satisfying

|Z_t| ≤Q:=

Proof. Consider the constant Ck,A,q,m⁰,d introduced in Theorem 3.3.1. If T ≤ C_k,A,q,m⁰_,d, the result follows from Theorem 3.3.1.

In the rest of the proof let us assume that T > C_k,A,q,m⁰_,d. The functiong˜ defined by (3.3.4) is Lipschitz continuous and differentiable in(y, z), and satisfies (A4”). Hence, by [20, Theorem 2.6] the FBSDE

(X˜_t=x+Rt

0b_u( ˜X_u,Y˜_u,Z˜_u)du+Rt

0σ_u( ˜X_u,Y˜_u)dW_u Y˜_t=h( ˜X_T) +RT

t ˜g_u( ˜X_u,Y˜_u,Z˜_u)du−RT

t Z˜_udW_u, t∈[0, T] (3.3.7) admits a unique solution( ˜X,Y ,˜ Z)˜ ∈ S²(R^m)× S^∞(R^m

0)× H^∞(R^m

0×d). More-over, by [28] the processesY˜ andZ˜ are Malliavin differentiable and we have for j= 1, . . . , d,

D_r^jY˜_t= 0, D_r^jZ˜_t= 0, 0≤t < r < T, D_r^jY˜t=D_r^jh( ˜XT) +

T

Z

t

∂yg˜uD^j_rY˜u+∂zg˜uD^j_rZ˜u+D_r^jg˜u( ˜Xy,Y˜u,Z˜u)du

−

T

Z

t

D^j_rZ˜udWu, 0≤r ≤t≤T.

Since ˜g is Lipschitz inz, ∂_z˜g( ˜X_y,Y˜_u,Z˜_u) is bounded. By (A4”) and (A5”), it follows from the same procedure as in the proof of Lemma 3.A.1 that

|D_r^jY˜_tⁱ| ≤





m⁰

X

i=1

Aij +

m⁰

X

i=1 T

Z

t

qij(s)e^{−m0k8(T−s)}ds



e^m0k8(T−t), P⊗dt-a.e., i = 1, . . . , m⁰;j = 1, . . . , d. Let C_k,A,q,m˜ 0,d be the constant given by Theorem 3.3.1 replacingA_ij by A˜_ij. One can easily check thatC_k,A,q,m˜ 0,d ≤ C_k,A,q,m⁰_,d sinceAij ≤ A˜ij. Considering a sequence0 = t0 ≤ t1 ≤ · · · ≤ tN = T with max1≤i≤N|t_i−ti−1| ≤C_k,A,q,m˜ ⁰,dsimilar to the last part of the proof of Theorem 3.2.1. SinceD_rY˜_ti ∈L^∞for allr∈[ti−1, t_i]we can get that fori= 1, . . . , Nthat

(X_t= ˜X_ti−1+Rt

ti−1b_u(X_u, Y_u, Z_u)du+Rt

ti−1σ_u(X_u, Y_u)dW_u Yt= ˜Yti+R_ti

t gu(Xu, Yu, Zu)du−R_ti

t ZudWu, t∈[ti−1, ti] has a unique solution (Xⁱ, Yⁱ, Zⁱ) ∈ S²(R^m) × S²(R^m

0) × S^∞(R^m

0×d) and Z_tⁱ

≤ Q for all t ∈ [ti−1, t_i]. By the uniqueness of FBSDE (3.3.7), we have (Xⁱ, Yⁱ, Zⁱ)1_[ti−1,ti] = ( ˜X,Y ,˜ Z˜)1_[ti−1,ti]. The result follows from a recursion

and pasting procedure as above.

Im Dokument Essays on Multidimensional BSDEs and FBSDEs (Seite 45-58)