BSDEs driven by two mutually independent fractional Brownian motions with stochastic Lipschitz coefficients

Backward stochastic differential equations (BSDEs in short) were first introduced by Pardoux and Peng [7]. They proved the celebrated existence and uniqueness result under Lipschitz assumption. This pioneer work was extensively used in many fields like stochastic interpretation of solutions of PDEs and financial mathematics.

Few years later, several authors investigated BSDEs with respect to fractional Brownian motion $\begin{matrix} (B_{t}^{H})_{t \geq 0} \end{matrix}$ $\begin{array}{} \big(B^H_t\big)_{t\geq 0} \end{array}$ with Hurst parameter H. Since B^H is not a semimartingale when H ≠ $\begin{matrix} \frac{1}{2} \end{matrix}$ $\begin{array}{} \frac{1}{2} \end{array}$, we cannot use the beautiful classical theory of stochastic calculus to define the fractional stochastic integral. It is a significant and challenging problem to extend the results in the classical stochastic calculus to this fractional Brownian motion. Essentially, two different types of integrals with respect to a fractional Brownian motion have been defined and studied. The first one is the pathwise Riemann-Stieltjes integral (see Young [10]). This integral has a proprieties of Stratonovich integral, which leads to difficulties in applications. The second one, introduced in Decreusefond and Ustunel [3] is the divergence operator (or Skorohod integral), defined as the adjoint of the derivative operator in the framework of the Malliavin calculus. Since this stochastic integral satisfies the zero mean property and it can be expressed as the limit of Riemann sums defined using Wick products, it was later developed by many authors.

Recently, new classes of BSDEs driven by both standard and fractional Brownian motions were introduced by Fei et al [4]. They established the existence and uniqueness of solutions.

In this paper, our aim is to generalize the result established in [2] to the following equation called fractional BSDE under stochastic conditions on the generator:

$\begin{matrix} Y_{t} = ξ + \int_{t}^{T} f (s, η_{s}, Y_{s}, Z_{1, s}, Z_{2, s} d s - \int_{t}^{T} Z_{1, s} d B_{1, t}^{H_{1}} - \int_{t}^{T} Z_{2, s} d B_{2, t}^{H_{2}}, t \in [0, T], \end{matrix}$ $$\begin{array}{} \displaystyle Y_t = \xi + \int_t^Tf(s, \eta_s,Y_s, Z_{1,s},Z_{2,s}ds - \int_t^TZ_{1,s}dB_{1,t}^{H_1} - \int_t^T Z_{2,s}dB_{2,t}^{H_2},\quad t\in [0, T], \end{array}$$

where $\begin{matrix} {(B_{1, t}^{H_{1}})}_{t \geq 0} and {(B_{2, t}^{H_{2}})}_{t \geq 0} \end{matrix}$ $\begin{array}{} \left(B_{1,t}^{H_1}\right)_{t\geq 0}\text{ and }\left(B_{2,t}^{H_2}\right)_{t\geq 0} \end{array}$ are two mutually independent fractional Brownian motions. The novelty in these types of stochastic equations lies in the fact of coupling two mutually independent fractional Brownian motions. In this work, the authors established some properties of solutions of a fractional BSDE with Lipschitz coefficients. By the help of the fixed point principle, we establish existence and uniqueness of solutions.

The paper is organized as follows: In Section 2, we introduce some preliminaries, before studying the solvability of our equation under Lipschitz conditions on the generator in Section 3. Using this result, we prove existence and uniqueness of the solution with a coefficient satisfying rather weaker conditions.

Fractional Stochastic calculus

Let us assume given two mutually independent fractional Brownian motions $\begin{matrix} B^{H} \in \{B_{1}^{H_{1}}, B_{2}^{H_{2}}\} \end{matrix}$ $\begin{array}{} B^{H}\in\left\{B_1^{H_1}, B_2^{H_2}\right\} \end{array}$ with Hurst parameter H ≥ $\begin{matrix} \frac{1}{2} \end{matrix}$ $\begin{array}{} \frac{1}{2} \end{array}$ is given.

Let Ώ be a non-empty set, ℱ a σ−algebra of sets Ώ, P a probability measure defined on ℱ and {ℱ_t, t ∈ [0, T]} a σ−algebra generated by both fractional Brownian motions.

The triplet (Ώ, ℱ, P) defines a probability space and E the mathematical expectation with respect to the probability measure P.

The fractional Brownian motion B^H is a zero mean Gaussian process with the covariance function

$\begin{matrix} E [B_{t}^{H} B_{s}^{H}] = \frac{1}{2} (t^{2 H} + s^{2 H} - | t - s |^{2 H}), t, s \geq 0. \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}[B_t^{H}B_s^{H}]=\frac{1}{2}\left(t^{{2H}} + s^{{2H}} -|t-s|^{{2H}}\right), \quad t,s\geq0. \end{array}$$

Denote ϕ(t, s) = H(2H − 1)∣t − s∣^2H−2, (t, s) ∈ R².

Let ξ and η be measurable functions on [0, T]. Define

$\begin{matrix} 〈 ξ, η 〉_{t} = \int_{0}^{t} \int_{0}^{t} ϕ (u, v) ξ (u) η (v) d u d v and {∥ξ∥}_{t}^{2} = 〈 ξ, ξ 〉_{t} . \end{matrix}$ $$\begin{array}{} \displaystyle \langle \xi,\eta\rangle_t =\int_0^t\int_0^t\phi(u,v)\xi(u)\eta(v)dudv \quad\text{and}\quad \left\|\xi\right\|_t^2 = \langle \xi ,\xi \rangle_t. \end{array}$$

Note that, for any t ∈ [0, T], 〈ξ, η〉_t is a Hilbert scalar product. Let 𝓗 be the completion of the set of continuous functions under this Hilbert norm ∥⋅∥_t and (ξ_n)_n be a sequence in 𝓗 such that 〈ξ_i,ξ_j〉_T = δ_ij.

Let $\begin{matrix} P_{T}^{H} \end{matrix}$ $\begin{array}{} \mathscr{P}_T^{H} \end{array}$ be the set of all polynomials of fractional Brownian motion $\begin{matrix} (B_{t}^{H})_{t \geq 0} \end{matrix} .$ $\begin{array}{} \big(B^H_t\big)_{t\geq 0}. \end{array}$ Namely, $\begin{matrix} P_{T}^{H} \end{matrix}$ $\begin{array}{} \mathscr{P}_T^{H} \end{array}$ contains all elements of the form

$\begin{matrix} F (ω) = f (\int_{0}^{T} ξ_{1} (t) d B_{t}^{H}, \int_{0}^{T} ξ_{2} (t) d B_{t}^{H}, \dots, \int_{0}^{T} ξ_{n} (t) d B_{t}^{H}) \end{matrix}$ $$\begin{array}{} \displaystyle F(\omega)=f\left(\int_0^T\xi_1(t)dB_t^{H}, \int_0^T\xi_2(t)dB_{t}^{H},\dots, \int_0^T\xi_n(t)dB_{t}^{H} \right) \end{array}$$

where f is a polynomial function of n variables.

The Malliavin derivative $\begin{matrix} D_{t}^{H} \end{matrix}$ $\begin{array}{} D_t^{H} \end{array}$ of F is given by

$\begin{matrix} D_{s}^{H} F = \sum_{j = 1}^{n} \frac{\partial f}{\partial x_{j}} (\int_{0}^{T} ξ_{1} (t) d B_{t}^{H}, \int_{0}^{T} ξ_{2} (t) d B_{t}^{H}, \dots, \int_{0}^{T} ξ_{n} (t) d B_{t}^{H}) ξ_{j} (s), 0 \leq s \leq T . \end{matrix}$ $$\begin{array}{} \displaystyle D_s^{H}F=\sum_{j=1}^n\frac{\partial f}{\partial x_j}\left(\int_0^T\xi_1(t)dB_t^{H}, \int_0^T\xi_2(t)dB_t^{H},\dots, \int_0^T\xi_n(t)dB_t^{H} \right)\xi_j(s), \quad 0\leq s\leq T. \end{array}$$

Now we introduce the Malliavin ϕ-derivative $\begin{matrix} D_{t}^{H} \end{matrix}$ $\begin{array}{} \mathbb{D}_t^{H} \end{array}$ of F by

$\begin{matrix} D_{t}^{H} F = \int_{0}^{T} ϕ (t, s) D_{s}^{H} F d s . \end{matrix}$ $$\begin{array}{} \displaystyle \mathbb{D}_t^{H}F=\int_0^T\phi(t,s)D_s^{H}Fds. \end{array}$$

We have the following (see[[5], Proposition 6.25]):

Theorem 2.1

Let F : (Ω, ℱ, P) ⟶ 𝓗 be a stochastic processes such that

$\begin{matrix} E ({∥F∥}_{T}^{2} + \int_{0}^{T} \int_{0}^{T} | D_{s}^{H} F_{t} |^{2} d s d t) < + \infty . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left(\left\|F\right\|_T^2 +\int_0^T\int_0^T |\mathbb{D}_s^{H} F_t|^2 ds dt\right) \lt +\infty. \end{array}$$

Then, the Itô-Skorohod-type stochastic integral denoted by $\begin{matrix} \int_{0}^{T} F_{s} d B_{s}^{H} \end{matrix}$ $\begin{array}{} \int_0^TF_sdB_s^{H} \end{array}$ exists in L²(Ω, ℱ, P) and satisfies

$\begin{matrix} E (\int_{0}^{T} F_{s} d B_{s}^{H}) = 0 a n d E {(\int_{0}^{T} F_{s} d B_{s}^{H})}^{2} = E (∥ F ∥_{T}^{2} + \int_{0}^{T} \int_{0}^{T} D_{s}^{H} F_{t} D_{t}^{H} F_{s} d s d t) . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left( \int_0^TF_sdB_s^{H}\right)=0 \quad{and}\quad {\bf E} \left( \int_0^TF_sdB_s^{H}\right)^2= {\bf E}\left(\Vert F\Vert_T^2 + \int_0^T\int_0^T\mathbb{D}_s^{H} F_t\mathbb{D}_t^{H} F_sdsdt \right). \end{array}$$

Let us recall the fractional Itô formula (see[[4], Theorem 3.1]).

Theorem 2.2

Let σ₁, σ₂ ∈ 𝓗 be deterministic continuous functions. Denote

$\begin{matrix} X_{t} = X_{0} + \int_{0}^{t} α (s) d s + \int_{0}^{t} σ_{1} (s) d B_{1, s}^{H_{1}} + \int_{0}^{t} σ_{2} (s) d B_{2, s}^{H_{2}}, \end{matrix}$ $$\begin{array}{} \displaystyle X_t= X_0 + \int_0^t\alpha (s)ds + \int_0^t\sigma_1(s)dB_{1,s}^{H_1} + \int_0^t\sigma_2(s)dB_{2,s}^{H_2}, \end{array}$$

where X₀ is a constant and α(t) is a deterministic function with $\begin{matrix} \int_{0}^{t} | α (s) | d s < + \infty . \end{matrix}$ $\begin{array}{} \int_0^t|\alpha(s)|ds \lt +\infty. \end{array}$

Let F(t, x) be continuously differentiable with respect to t and twice continuously differentiable with respect to x. Then

$\begin{matrix} F (t, X_{t}) = F (0, X_{0}) + \int_{0}^{t} \frac{\partial F}{\partial s} (s, X_{s}) d s + \int_{0}^{t} \frac{\partial F}{\partial x} (s, X_{s}) d X_{s} \\ + \frac{1}{2} \int_{0}^{t} \frac{\partial^{2} F}{\partial x^{2}} (s, X_{s}) [\frac{d}{d s} {∥σ_{1}∥}_{s}^{2} + \frac{d}{d s} {∥σ_{2}∥}_{s}^{2}] d s, 0 \leq t \leq T . \end{matrix}$ $$\begin{array}{} \displaystyle F(t,X_t)= F(0,X_0) + \int_0^t\frac{\partial F}{\partial s}(s,X_s)ds + \int_0^t\frac{\partial F}{\partial x}(s,X_s)dX_s \\\displaystyle\qquad\quad\,\,\,\,+\frac{1}{2}\int_0^t\frac{\partial^2 F}{\partial x^2}(s,X_s)\left[\frac{d}{ds}\left\|\sigma_1\right\|_s^2 + \frac{d}{ds}\left\|\sigma_2\right\|_s^2\right]ds,\quad 0\leq t\leq T. \end{array}$$

Let us finish this section by giving a fractional Itô chain rule (see[[4], Theorem 3.2]).

Theorem 2.3

Assume that for j = 1, 2, the processes μ_j, α_j and ϑ_j, satisfy

$\begin{matrix} E [\int_{0}^{T} μ_{j}^{2} (s) d s + \int_{0}^{T} α_{j}^{2} (s) d s + \int_{0}^{T} ϑ_{j}^{2} (s) d s] < + \infty . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left[\int_0^T\mu_j^2(s)ds + \int_0^T\alpha_j^2(s)ds + \int_0^T\vartheta_j^2(s)ds\right] \lt +\infty. \end{array}$$

Suppose that $\begin{matrix} D_{t}^{H_{1}} α_{j} (s) a n d D_{t}^{H_{2}} ϑ_{j} (s) \end{matrix}$ $\begin{array}{} \mathbb{D}_t^{H_1}\alpha_j(s){\,\, and\,\, }\mathbb{D}_t^{H_2}\vartheta_j(s) \end{array}$ are continuously differentiable with respect to (s, t) ∈ [0, T]² for almost all ω ∈ Ω. Let X_t and Y_t be two processes satisfying

$\begin{matrix} X_{t} = X_{0} + \int_{0}^{t} μ_{1} (s) d s + \int_{0}^{t} α_{1} (s) d B_{1, s}^{H_{1}} + \int_{0}^{t} ϑ_{1} (s) d B_{2, s}^{H_{2}}, 0 \leq t \leq T, \\ Y_{t} = Y_{0} + \int_{0}^{t} μ_{2} (s) d s + \int_{0}^{t} α_{2} (s) d B_{1, s}^{H_{1}} + \int_{0}^{t} ϑ_{2} (s) d B_{2, s}^{H_{2}}, 0 \leq t \leq T . \end{matrix}$ $$\begin{array}{} \displaystyle X_t = X_0 + \int_0^t\mu_1(s)ds + \int_0^t\alpha_1(s)dB_{1,s}^{H_1} + \int_0^t\vartheta_1(s)dB_{2,s}^{H_2}, \quad\quad 0\leq t\leq T,\\\displaystyle\, Y_t = Y_0 + \int_0^t\mu_2(s)ds + \int_0^t\alpha_2(s)dB_{1,s}^{H_1} + \int_0^t\vartheta_2(s)dB_{2,s}^{H_2}, \quad\quad 0\leq t\leq T. \end{array}$$

If the following conditions hold:

$\begin{matrix} E [\int_{0}^{T} | D_{t}^{H_{1}} α_{i} (s) |^{2} d s d t] < + \infty a n d E [\int_{0}^{T} | D_{t}^{H_{2}} ϑ_{i} (s) |^{2} d s d t] < + \infty \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left[\int_0^T|\mathbb{D}_t^{H_1}\alpha_i(s)|^2dsdt\right] \lt +\infty\quad{and}\quad {\bf E}\left[\int_0^T|\mathbb{D}_t^{H_2}\vartheta_i(s)|^2dsdt\right] \lt +\infty \end{array}$$

then

$\begin{matrix} X_{t} Y_{t} = X_{0} Y_{0} + \int_{0}^{t} X_{s} d Y_{s} + \int_{0}^{t} Y_{s} d X_{s} \\ + \int_{0}^{t} [α_{1} (s) D_{s}^{H_{1}} Y_{s} + α_{2} (s) D_{s}^{H_{1}} X_{s} + ϑ_{1} (s) D_{s}^{H_{2}} Y_{s} + ϑ_{2} (s) D_{s}^{H_{2}} X_{s}] d s, \end{matrix}$ $$\begin{array}{} \displaystyle X_tY_t = X_0Y_0 +\int_0^tX_sdY_s + \int_0^tY_sdX_s \\\displaystyle\qquad\,\,+\int_0^t\left[\alpha_1(s)\mathbb{D}_s^{H_1}Y_s + \alpha_2(s)\mathbb{D}_s^{H_1}X_s + \vartheta_1(s)\mathbb{D}_s^{H_2}Y_s + \vartheta_2(s)\mathbb{D}_s^{H_2}X_s \right]ds, \end{array}$$

which may be written formally as

$\begin{matrix} d (X_{t} Y_{t}) = X_{t} d Y_{t} + Y_{t} d X_{t} + [α_{1} (t) D_{t}^{H_{1}} Y_{t} + α_{2} (t) D_{t}^{H_{1}} X_{t} + ϑ_{1} (t) D_{t}^{H_{2}} Y_{t} + ϑ_{2} (t) D_{t}^{H_{2}} X_{t}] d t . \end{matrix}$ $$\begin{array}{} \displaystyle d\left(X_tY_t\right) = X_tdY_t + Y_tdX_t +\left[\alpha_1(t)\mathbb{D}_t^{H_1}Y_t + \alpha_2(t)\mathbb{D}_t^{H_1}X_t + \vartheta_1(t)\mathbb{D}_t^{H_2}Y_t + \vartheta_2(t)\mathbb{D}_t^{H_2}X_t \right]dt. \end{array}$$

Fractional BSDEs

3.1

Definitions and notations

Let T > 0 be fixed throughout this paper. Let $\begin{matrix} {\{B_{1, t}^{H_{1}}\}}_{t \in [0, T]} and {\{B_{2, t}^{H_{2}}\}}_{t \in [0, T]} \end{matrix}$ $\begin{array}{} \left\{B_{1,t}^{H_1}\right\}_{t\in[0,T]}\text{ and }\left\{B_{2,t}^{H_2}\right\}_{t\in[0,T]} \end{array}$ be two mutually independent fractional Brownian motions processes, with respectively H₁ ≥ $\begin{matrix} \frac{1}{2} \end{matrix}$ $\begin{array}{} \frac{1}{2} \end{array}$ and H₂ ≥ $\begin{matrix} \frac{1}{2} \end{matrix}$ $\begin{array}{} \frac{1}{2} \end{array}$, defined on a probability space (Ώ,ℱ,P). Let N denote the class of P-null sets of ℱ.

we define

$\begin{matrix} F_{s} = F_{0, s}^{B_{1}^{H_{1}}} \lor F_{s, T}^{B_{2}^{H_{2}}}, s \in [0, T], \end{matrix}$ $$\begin{array}{} \displaystyle {\mathscr F}_{s}={\mathscr F}_{0,s}^{B_{1}^{H_1}} \vee {\mathscr F}_{s,T}^{B_{2}^{H_2}},\quad s\in[0,T], \end{array}$$

where for any process {ψ_t}_t≥0, $\begin{matrix} F_{s, t}^{ψ} \end{matrix}$ $\begin{array}{} {\mathscr F}^\psi_{s,t} \end{array}$ = σ{ψ_r − ψ_s, s ≤ r ≤ t} ∨ 𝒩.

For every ℱ-adapted random process α = (α(t))_t≥0 with positive values, we define an increasing process (A(t))_t≥0 by setting A(t) = $\begin{matrix} \int_{0}^{t} α^{2} (s) d s . \end{matrix}$ $\begin{array}{} \int_0^t\alpha^2(s)ds. \end{array}$

For a fixed β > 0, we will use the following sets:

$\begin{matrix} C_{pol}^{1, 2} ([0, T] \times R) \end{matrix}$ $\begin{array}{} \mathscr{C}_{\mbox{pol}}^{1,2}\left([0, T]\times {\bf R}\right) \end{array}$ is the space of all 𝒞^1,2-functions over [0, T] × R, which together with their derivative is of polynomial growth.

ℒ²(β, ℱ_t, R) = {ξ : Ώ → R ∣ ξ is ℱ_t − measurable, E [e^βA(T)∣ξ∣²] < +∞},

𝒱_[0,T] = {Y = ψ(⋅, η) : ψ ∈ 𝒞^1,2([0, T] × R), $\begin{matrix} \frac{\partial ψ}{\partial t} \end{matrix}$ $\begin{array}{} \frac{\partial \psi}{\partial t} \end{array}$ is bounded, t ∈ [0, T]},

$\begin{matrix} {\tilde{V}}_{[0, T]}^{β} and {\tilde{V}}_{[0, T]}^{a, β} \end{matrix}$ $\begin{array}{} \widetilde{\cal V}_{[0,T]}^{\beta} \text{ and }\widetilde{\cal V}_{[0,T]}^{a,\beta} \end{array}$ are the completion of 𝒱_[0,T] under the following norm

$\begin{matrix} ∥ Y ∥_{α, β} = {(E \int_{0}^{T} α^{2} (t) e^{β A (t)} | Y_{t} |^{2} d t)}^{1 / 2}, ∥ Z ∥_{β} = {(E \int_{0}^{T} e^{β A (t)} | Z_{t} |^{2} d t)}^{1 / 2} \end{matrix}$ $$\begin{array}{} \displaystyle \Vert Y\Vert_{\alpha,\beta} =\left({\bf E}\int_0^{T}\alpha^2(t) e^{\beta A(t)}|Y_t|^2dt\right)^{1/2}, \quad \Vert Z\Vert_\beta =\left({\bf E}\int_0^{T} e^{\beta A(t)}|Z_t|^2dt\right)^{1/2} \end{array}$$

$\begin{matrix} B^{2} ([0, T], R) = {\tilde{V}}_{[0, T]}^{α, β} \times {\tilde{V}}_{[0, T]}^{β} \times {\tilde{V}}_{[0, T]}^{β} \end{matrix}$ $\begin{array}{} {\mathscr B}^2{([0,T],{\bf R})}=\widetilde{\cal V}_{[0,T]}^{\alpha,\beta}\times\widetilde{\cal V}_{[0,T]}^{\beta}\times\widetilde{\cal V}_{[0,T]}^{\beta} \end{array}$ is a Banach space with the norm

$\begin{matrix} ∥ (Y, Z_{1}, Z_{2}) ∥_{B}^{2} = ∥ Y ∥_{α, β}^{2} + ∥ Z_{1} ∥_{β}^{2} + ∥ Z_{2} ∥_{β}^{2} . \end{matrix}$ $$\begin{array}{} \displaystyle \Vert (Y,Z_1,Z_2)\Vert^2_{\mathscr B} =\Vert Y\Vert^2_{\alpha,\beta} + \Vert Z_1\Vert^2_\beta +\Vert Z_2\Vert^2_\beta. \end{array}$$

Let us consider

$\begin{matrix} η_{t} = η_{0} + \int_{0}^{t} b (s) d s + \int_{0}^{t} σ_{1} (s) d B_{1, s}^{H_{1}} + \int_{0}^{t} σ_{2} (s) d B_{2, s}^{H_{2}}, 0 \leq t \leq T \end{matrix}$ $$\begin{array}{} \displaystyle \eta_t= \eta_0 + \int_0^tb(s)ds + \int_0^t\sigma_1(s)dB_{1,s}^{H_1} +\int_0^t\sigma_2(s)dB_{2,s}^{H_2}, \quad 0\le t\le T \end{array}$$

where the coefficients η₀, b, σ₁ and σ₂ satisfy:

η₀ is a given constant and b : [0, T] → R is a deterministic continuous function,

σ₁, σ₂:[0, T] → R are deterministic differentiable continuous functions, and σ₁(t) ≠ 0, and σ₂(t) ≠ 0 such that

$\begin{matrix} | σ |_{t}^{2} = {∥σ_{1}∥}_{t}^{2} + {∥σ_{2}∥}_{t}^{2}, 0 \leq t \leq T, \end{matrix}$ $$\begin{array}{} \displaystyle |\sigma|_t^2=\left\|\sigma_1\right\|_t^2 + \left\|\sigma_2\right\|_t^2, \quad 0\le t\le T, \end{array}$$(3.1)

$\begin{matrix} where {∥σ_{i}∥}_{t}^{2} = H_{i} (2 H_{i} - 1) \int_{0}^{t} \int_{0}^{t} | u - v |^{2 H_{i} - 2} σ_{i} (u) σ_{i} (v) d u d v, i = 1, 2. \end{matrix}$ $$\begin{array}{} \displaystyle \text{where} \quad\quad\quad\left\|\sigma_i\right\|_{t}^2=H_i(2H_i-1)\int_0^t\int_0^t |u-v|^{{2H_i}-2}\sigma_i(u)\sigma_i(v)dudv,\quad i=1,2. \end{array}$$

The next Remark will be useful in the sequel.

Remark 3.1

There exists a constant C₀ ∈ (0, 1) such that $\begin{matrix} inf_{0 \leq t \leq T} \frac{{\hat{σ}}_{i} (t)}{σ_{i} (t)} \geq C_{0} \end{matrix}$ $\begin{array}{} \inf_{0\le t\le T} \frac{\hat \sigma_i(t)}{\sigma_i(t)}\geq C_0 \end{array}$

where $\begin{matrix} {\hat{σ}}_{i} (t) = \int_{0}^{t} ϕ (t, v) σ_{i} (v) d v i = 1, 2. \end{matrix}$ $\begin{array}{} \hat{\sigma}_i(t)=\int_0^t\phi(t,v)\sigma_i(v)dv \quad i=1,2. \end{array}$

We are interested in the following one-dimensional fractional BSDE:

$\begin{matrix} Y_{t} = ξ + \int_{t}^{T} f (s, η_{s}, Y_{s}, Z_{1, s}, Z_{2, s}) d s - \int_{t}^{T} Z_{1, s} d B_{1, s}^{H_{1}} - \int_{t}^{T} Z_{2, s} d B_{2, s}^{H_{2}}, t \in [0, T] . \end{matrix}$ $$\begin{array}{} \displaystyle Y_t = \xi + \int_t^Tf(s, \eta_s,Y_s, Z_{1,s},Z_{2,s})ds - \int_t^TZ_{1,s}dB_{1,s}^{H_1} - \int_t^TZ_{2,s}dB_{2,s}^{H_2},\quad t\in [0, T]. \end{array}$$(3.2)

Definition 3.2

A triplet of processes (Y, Z₁, Z₂) is called a solution to fractional BSDE (3.2), if (Y, Z₁, Z₂) ∈ ℬ²([0, T], R) and satisfies eq.(3.2).

The next proposition will be useful in the sequel.

Proposition 3.3

Let (Y, Z₁, Z₂) be a solution of the fractional BSDE (3.2). Then for almost t ∈ [0, T], we have

$\begin{matrix} D_{t}^{H_{1}} Y_{t} = \frac{{\hat{σ}}_{1} (t)}{σ_{1} (t)} Z_{1, t} a n d D_{t}^{H_{2}} Y_{t} = \frac{{\hat{σ}}_{2} (t)}{σ_{2} (t)} Z_{2, t} . \end{matrix}$ $$\begin{array}{} \displaystyle \mathbb{D}^{H_1}_tY_t=\frac{\hat \sigma_1(t)}{\sigma_1(t)}Z_{1,t}\quad{and} \quad \mathbb{D}^{H_2}_tY_t=\frac{\hat \sigma_2(t)}{\sigma_2(t)}Z_{2,t}. \end{array}$$

3.2

The case of stochastic Lipschitz coefficient

3.2.1

Aassumptions

In the following, we assume that f satisfies assumptions (H1):

(H1.1)

There exist three non-negative processes {μ(t)}_0≤t≤T, {ν(t)}_0≤t≤T and {ϑ(t)}_0≤t≤T such that:

for any t ∈ [0, T], μ(t), ν(t), ϑ(t) are ℱ_t-measurable,

(H1.1)

for any t ∈ [0, T], $\begin{matrix} x, y, y^{'}, z_{1}, z_{1}^{'}, z_{2}, z_{2}^{'} \in R, \end{matrix}$ $\begin{array}{} x,y,y',z_1,z'_1,z_2,z'_2\in{\bf R}, \end{array}$ we have

$\begin{matrix} |f (t, x, y, z_{1}, z_{2}) - f (t, x, y^{'}, z_{1}^{'}, z_{2}^{'})| \leq μ (t) | y - y^{'} | + ν (t) | z_{1} - z_{1}^{'} | + ϑ (t) | z_{2} - z_{2}^{'} | . \end{matrix}$ $$\begin{array}{} \displaystyle \left|f\left(t, x, y,z_1,z_2 \right) - f\left(t, x, y',z'_1,z'_2\right)\right|\leq \mu(t)|y\!-\!y'| + \nu(t)|z_1-z_1'| + \vartheta(t)|z_2-z_2'|. \end{array}$$

(H1.2)

for any t ∈ [0, T], α²(t) = μ(t)+ν²(t)+ ϑ²(t) > 0.

3.2.2

Existence and uniqueness of the solution

The main result of this section is the following theorem:

Theorem 3.4

Let the assumptions (H1) be satisfied. Then the fractional BSDE (3.2) has a unique solution (Y, Z₁, Z₂) in the space ℬ²([0, T],R).

Proof

Let us consider the mapping Γ: ℬ²([0, T],R) → ℬ²([0, T],R) driven by (U, V₁, V₂) ⟼ Γ(U, V₁, V₂) = (Y, Z₁, Z₂).

We will show that the mapping Γ is a contraction, where (Y, Z₁, Z₂) is a solution of the following fractional BSDE:

$\begin{matrix} Y_{t} = \int_{t}^{T} f (s, η_{s}, U_{s}, V_{1, s}, V_{2, s}) d s - \int_{t}^{T} Z_{1, s} d B_{1, s}^{H_{1}} - \int_{t}^{T} Z_{2, s} d B_{2, s}^{H_{2}}, t \in [0, T] . \end{matrix}$ $$\begin{array}{} \displaystyle Y_t = \int_t^Tf(s, \eta_s,U_s, V_{1,s},V_{2,s})ds - \int_t^T Z_{1,s}dB_{1,s}^{H_1} - \int_t^T Z_{2,s}dB_{2,s}^{H_2},\quad t\in [0, T]. \end{array}$$(3.3)

Let us define for a process δ ∈ {Y, Z₁, Z₂, U, V₁, V₂}, δ = δ − δ′ where $\begin{matrix} δ^{'} \in {\tilde{V}}_{[0, T]}^{β} \end{matrix}$ $\begin{array}{} \delta'\in\widetilde{\mathscr V}_{[0,T]}^{\beta} \end{array}$ and the function

$\begin{matrix} Δ f (t) = f (t, η_{t}, U_{t}, V_{1, t}, V_{2, t}) - f (t, η_{t}, U_{t}^{'}, V_{1, t}^{'}, V_{2, t}^{'}) . \end{matrix}$ $$\begin{array}{} \displaystyle \Delta f(t)\!=\!f(t,\!\eta_t, U_t, V_{1,t}, V_{2,t}) -f(t,\!\eta_t, U_t', V_{1,t}', V_{2,t}'). \end{array}$$

Then, the triplet (Y, Z₁, Z₂) solves the fractional BSDE

$\begin{matrix} {\bar{Y}}_{t} = \int_{t}^{T} Δ f (s) d s - \int_{t}^{T} {\bar{Z}}_{1, s} d B_{1, s}^{H_{1}} - \int_{t}^{T} {\bar{Z}}_{2, s} d B_{2, s}^{H_{2}}, t \in [0, T] . \end{matrix}$ $$\begin{array}{} \displaystyle \overline Y_t= \int_t^T\Delta f(s)ds - \int_t^T\overline Z_{1,s}dB_{1,s}^{H_1} - \int_t^T\overline Z_{2,s}dB_{2,s}^{H_2},\quad t\in [0, T]. \end{array}$$(3.4)

By the fractional Itô chain rule, we have

$\begin{matrix} | {\bar{Y}}_{t} |^{2} = 2 \int_{t}^{T} {\bar{Y}}_{s} Δ f (s) d s - 2 \int_{t}^{T} {\bar{Z}}_{1, s} D_{s}^{H_{1}} {\bar{Y}}_{s} d s - 2 \int_{t}^{T} {\bar{Z}}_{2, s} D_{s}^{H_{2}} {\bar{Y}}_{s} d s \\ - 2 \int_{t}^{T} {\bar{Y}}_{s} {\bar{Z}}_{1, s} d B_{1, s}^{H_{1}} - 2 \int_{t}^{T} {\bar{Y}}_{s} {\bar{Z}}_{2, s} d B_{2, s}^{H_{2}} . \end{matrix}$ $$\begin{array}{} \displaystyle |\overline Y_t|^2 = 2 \int_t^T \overline Y_s\Delta f(s)ds - 2 \int_t^T\overline Z_{1,s}\mathbb{D}^{H_1}_s\overline Y_sds - 2 \int_t^T\overline Z_{2,s}\mathbb{D}^{H_2}_s\overline Y_sds \\\displaystyle \qquad\,\,\,-2 \int_t^T\overline Y_s\overline Z_{1,s}dB_{1,s}^{H_1} - 2 \int_t^T\overline Y_s\overline Z_{2,s}d B_{2,s}^{H_2}. \end{array}$$

Applying the Itô formula to e^βA(t)∣Y_t∣², we obtain that

$\begin{matrix} e^{β A (t)} | {\bar{Y}}_{t} |^{2} = 2 \int_{t}^{T} e^{β A (s)} {\bar{Y}}_{s} Δ f (s) d s - 2 \int_{t}^{T} e^{β A (s)} {\bar{Z}}_{1, s} D_{s}^{H_{1}} {\bar{Y}}_{s} d s - 2 \int_{t}^{T} e^{β A (s)} {\bar{Z}}_{2, s} D_{s}^{H_{2}} {\bar{Y}}_{s} d s \\ - 2 \int_{t}^{T} e^{β A (s)} {\bar{Y}}_{s} {\bar{Z}}_{1, s} d B_{1, s}^{H_{1}} - 2 \int_{t}^{T} e^{β A (s)} {\bar{Y}}_{s} {\bar{Z}}_{2, s} d B_{2, s}^{H_{2}} - β \int_{t}^{T} e^{β A (s)} α^{2} (s) | {\bar{Y}}_{s} |^{2} d s . \end{matrix}$ $$\begin{array}{} \displaystyle e^{\beta A(t)}|\overline Y_t|^2 = 2\int_t^Te^{\beta A(s)}\overline Y_s\Delta f(s)ds - 2\int_t^Te^{\beta A(s)}\overline Z_{1,s}\mathbb{D}^{H_1}_s\overline Y_sds - 2\int_t^Te^{\beta A(s)}\overline Z_{2,s}\mathbb{D}^{H_2}_s\overline Y_sds\\\displaystyle\qquad\qquad\,\,\,\,\,-2\int_t^Te^{\beta A(s)}\overline Y_s\overline Z_{1,s}dB_{1,s}^{H_1} - 2\int_t^Te^{\beta A(s)}\overline Y_s\overline Z_{2,s}dB_{2,s}^{H_2} - \beta\int_t^T e^{\beta A(s)}\alpha^2(s)|\overline Y_s|^2ds. \end{array}$$

It is known that, by Proposition 3.3, $\begin{matrix} D_{s}^{H_{1}} {\bar{Y}}_{s} = \frac{{\hat{σ}}_{1} (s)}{σ_{1} (s)} {\bar{Z}}_{1, s} and D_{s}^{H_{2}} {\bar{Y}}_{s} = \frac{{\hat{σ}}_{2} (s)}{σ_{2} (s)} {\bar{Z}}_{2, s} . \end{matrix}$ $\begin{array}{} \mathbb{D}^{H_1}_s\overline Y_s=\frac{\hat{\sigma}_1(s)}{\sigma_1(s)}\overline Z_{1,s}\text{ and }\mathbb{D}^{H_2}_s\overline Y_s=\frac{\hat{\sigma}_2(s)}{\sigma_2(s)}\overline Z_{2,s}. \end{array}$

Then, we have

$\begin{matrix} E [e^{β A (t)} | {\bar{Y}}_{t} |^{2}] + β E \int_{t}^{T} e^{β A (s)} α^{2} (s) | {\bar{Y}}_{s} |^{2} d s + 2 E \int_{t}^{T} e^{β A (s)} [\frac{{\hat{σ}}_{1} (s)}{σ_{1} (s)} | {\bar{Z}}_{1, s} |^{2} + \frac{{\hat{σ}}_{2} (s)}{σ_{2} (s)} | {\bar{Z}}_{2, s} |^{2}] d s \\ = 2 E \int_{t}^{T} e^{β A (s)} {\bar{Y}}_{s} Δ f (s) d s . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left[e^{\beta A(t)}|\overline Y_t|^2\right] + \beta{\bf E}\int_t^T e^{\beta A(s)}\alpha^2(s)|\overline Y_s|^2ds + 2{\bf E}\int_t^Te^{\beta A(s)}\left[\frac{\hat{\sigma}_1(s)}{\sigma_1(s)}|\overline Z_{1,s}|^2 + \frac{\hat{\sigma}_2(s)}{\sigma_2(s)}|\overline Z_{2,s}|^2\right]ds \\\displaystyle \qquad\qquad\qquad\,\,\,= 2{\bf E}\int_t^Te^{\beta A(s)}\overline Y_s\Delta f(s)ds. \end{array}$$(3.5)

Using standard estimates and assumption (H1.1), we obtain that

$\begin{matrix} 2 E \int_{t}^{T} e^{β A (s)} {\bar{Y}}_{s} Δ f (s) d s \leq 2 E \int_{t}^{T} e^{β A (s)} | {\bar{Y}}_{s} | [μ (s) | {\bar{U}}_{s} | + ν (s) | {\bar{V}}_{1, s} | + ϑ (s) | {\bar{V}}_{2, s} |] d s \\ \leq \frac{1}{C_{0}} E \int_{t}^{T} e^{β A (s)} [μ (s) + ν^{2} (s) + ϑ^{2} (s)] | {\bar{Y}}_{s} |^{2} d s \\ + C_{0} E \int_{t}^{T} e^{β A (s)} μ (s) | {\bar{U}}_{s} |^{2} d s + C_{0} E \int_{t}^{T} e^{β A (s)} [| {\bar{V}}_{1, s} |^{2} + | {\bar{V}}_{2, s} |^{2}] d s, \end{matrix}$ $$\begin{array}{} \displaystyle 2{\bf E}\int_t^T\!e^{\beta A(s)}\overline Y_s\Delta f(s)ds \leq 2{\bf E}\int_t^T\!e^{\beta A(s)}|\overline Y_s|\bigg[\mu(s)|\overline U_s|+ \nu(s)|\overline V_{1,s}| + \vartheta(s)|\overline V_{2,s}| \bigg]ds\\\displaystyle\qquad\qquad\qquad\qquad\qquad\,\,\,\leq \frac{1}{C_0}{\bf E}\int_t^T\!e^{\beta A(s)}\bigg[\mu(s)+ \nu^2(s)+\vartheta^2(s)\bigg]|\overline Y_s|^2ds\\\displaystyle\qquad\qquad\qquad\qquad\qquad\,\,\,+ C_0{\bf E}\int_t^T\!e^{\beta A(s)}\mu(s)|\overline U_s|^2ds + C_0{\bf E}\int_t^T\!e^{\beta A(s)}\bigg[|\overline V_{1,s}|^2 + |\overline V_{2,s}|^2\bigg]ds, \end{array}$$

and using in addition asumption (H1.2),

$\begin{matrix} 2 E \int_{t}^{T} e^{β A (s)} {\bar{Y}}_{s} Δ f (s) d s \leq C_{0} E \int_{t}^{T} e^{β A (s)} [α^{2} (s) | {\bar{U}}_{s} |^{2} + | {\bar{V}}_{1, s} |^{2} + | {\bar{V}}_{2, s} |^{2}] d s \\ + \frac{1}{C_{0}} E \int_{t}^{T} e^{β A (s)} α^{2} (s) | {\bar{Y}}_{s} |^{2} d s \end{matrix}$ $$\begin{array}{} \displaystyle 2{\bf E}\int_t^T\!e^{\beta A(s)}\overline Y_s\Delta f(s)ds \leq C_0{\bf E}\int_t^T\!e^{\beta A(s)}\bigg[\alpha^2(s)|\overline U_s|^2 +|\overline V_{1,s}|^2 + |\overline V_{2,s}|^2\bigg]ds\\\displaystyle \qquad\qquad\qquad\qquad\qquad\,\,\,\,+\frac{1}{C_0}{\bf E}\int_t^T\!e^{\beta A(s)}\alpha^2(s)|\overline Y_s|^2ds \end{array}$$(3.6)

Using the abovementioned inequality, from (3.5) we deduce that

$\begin{matrix} E [e^{β A (t)} | {\bar{Y}}_{t} |^{2}] + β E \int_{t}^{T} e^{β A (s)} α^{2} (s) | {\bar{Y}}_{s} |^{2} d s + 2 E \int_{t}^{T} e^{β A (s)} [\frac{{\hat{σ}}_{1} (s)}{σ_{1} (s)} | {\bar{Z}}_{1, s} |^{2} + \frac{{\hat{σ}}_{2} (s)}{σ_{2} (s)} | {\bar{Z}}_{2, s} |^{2}] d s \\ \leq C_{0} E \int_{t}^{T} e^{β A (s)} [α^{2} (s) | {\bar{U}}_{s} |^{2} + | {\bar{V}}_{1, s} |^{2} + | {\bar{V}}_{2, s} |^{2}] d s \\ + \frac{1}{C_{0}} E \int_{t}^{T} e^{β A (s)} α^{2} (s) | {\bar{Y}}_{s} |^{2} d s \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left[e^{\beta A(t)}|\overline Y_t|^2\right] + \beta{\bf E}\int_t^T e^{\beta A(s)}\alpha^2(s)|\overline Y_s|^2ds + 2{\bf E}\int_t^Te^{\beta A(s)}\bigg[\frac{\hat{\sigma}_1(s)}{\sigma_1(s)}|\overline Z_{1,s}|^2 + \frac{\hat{\sigma}_2(s)}{\sigma_2(s)}|\overline Z_{2,s}|^2\bigg]ds \\\displaystyle\qquad\qquad\qquad\,\, \leq C_0{\bf E}\int_t^T\!e^{\beta A(s)}\bigg[\alpha^2(s)|\overline U_s|^2 +|\overline V_{1,s}|^2 + |\overline V_{2,s}|^2\bigg]ds\\\displaystyle\qquad\qquad\qquad\,\, +\frac{1}{C_0}{\bf E}\int_t^T\!e^{\beta A(s)}\alpha^2(s)|\overline Y_s|^2ds \end{array}$$(3.7)

By Remark 3.1, we obtain

$\begin{matrix} E [e^{β A (t)} | {\bar{Y}}_{t} |^{2}] + β E \int_{t}^{T} e^{β A (s)} α^{2} (s) | {\bar{Y}}_{s} |^{2} d s + 2 C_{0} E \int_{t}^{T} e^{β A (s)} [| {\bar{Z}}_{1, s} |^{2} + | {\bar{Z}}_{2, s} |^{2}] d s \\ \leq C_{0} E \int_{t}^{T} e^{β A (s)} [α^{2} (s) | {\bar{U}}_{s} |^{2} + | {\bar{V}}_{1, s} |^{2} + | {\bar{V}}_{2, s} |^{2}] d s \\ + \frac{1}{C_{0}} E \int_{t}^{T} e^{β A (s)} α^{2} (s) | {\bar{Y}}_{s} |^{2} d s \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left[e^{\beta A(t)}|\overline Y_t|^2\right] + \beta{\bf E}\int_t^T e^{\beta A(s)}\alpha^2(s)|\overline Y_s|^2ds + 2C_0{\bf E}\int_t^Te^{\beta A(s)}\bigg[|\overline Z_{1,s}|^2 + |\overline Z_{2,s}|^2\bigg]ds \\\displaystyle\qquad\qquad\qquad\,\,\leq C_0{\bf E}\int_t^T\!e^{\beta A(s)}\bigg[\alpha^2(s)|\overline U_s|^2 +|\overline V_{1,s}|^2 + |\overline V_{2,s}|^2\bigg]ds\\\displaystyle\qquad\qquad\qquad\,\,+\frac{1}{C_0}{\bf E}\int_t^T\!e^{\beta A(s)}\alpha^2(s)|\overline Y_s|^2ds \end{array}$$(3.8)

Taking β such that $\begin{matrix} β = 2 C_{0} + \frac{1}{C_{0}}, \end{matrix}$ $\begin{array}{} \beta =2C_0 +\frac{1}{C_0}, \end{array}$ we get

$\begin{matrix} E \int_{0}^{T} e^{β A (s)} [α^{2} (s) | {\bar{Y}}_{s} |^{2} + | {\bar{Z}}_{1, s} |^{2} + | {\bar{Z}}_{2, s} |^{2}] d s \leq \frac{1}{2} E \int_{0}^{T} e^{β A (s)} [α^{2} (s) | {\bar{U}}_{s} |^{2} + | {\bar{V}}_{1, s} |^{2} + | {\bar{V}}_{2, s} |^{2}] d s . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\int_0^{T}\!e^{\beta A(s)}\bigg[\alpha^2(s)|\overline Y_s|^2+|\overline Z_{1,s}|^2+|\overline Z_{2,s}|^2\bigg]ds \leq \frac{1}{2}{\bf E}\int_0^{T}\!e^{\beta A(s)}\bigg[\alpha^2(s)|\overline U_s|^2+ |\overline V_{1,s}|^2+|\overline V_{2,s}|^2\bigg]ds. \end{array}$$(3.9)

Thus, the mapping (U, V₁, V₂) ⟼ Γ(U, V₁, V₂) = (Y, Z₁, Z₂) determined by the fractional BSDE (3.2) is a strict contraction on ℬ²([0, T],R). Using the fixed point principle, we deduce the solution to the fractional BSDE (3.2) that exists uniquely. This completes the proof.□

3.3

The case of weak stochastic Lipschitz coefficient

3.3.1

Aassumptions

In the following, we assume that f satisfies assumptions (H2):

(H2.1)

There exist three non-negative processes {μ(t)}_0≤t≤T, {ν(t)}_0≤t≤T and {ϑ(t)}_0≤t≤T such that:

for any t ∈ [0, T], μ(t), ν(t), ϑ(t) are ℱ_t-measurable,

for any t ∈ [0, T], $\begin{matrix} x, y, y^{'}, z_{1}, z_{1}^{'}, z_{2}, z_{2}^{'} \end{matrix}$ $\begin{array}{} \displaystyle x,y,y',z_1,z'_1,z_2,z'_2 \end{array}$ ∈ R, we have

$\begin{matrix} | f (t, x, y, z_{1}, z_{2}) - f (t, x, y^{'}, z_{1}^{'}, z_{2}^{'}) | & \leq μ^{\frac{1}{2}} (t) ρ^{\frac{1}{2}} (t, | y - y^{'} |^{2}) \\ + ν (t) | z_{1} - z_{1}^{'} | + ϑ (t) | z_{2} - z_{2}^{'} |, \end{matrix}$ $$\begin{array}{} \displaystyle \bigg|f\left(t, x, y,z_1,z_2 \right) - f\left(t, x, y',z'_1,z'_2\right)\bigg|\!\!\!\!&\displaystyle\leq \mu^{\frac{1}{2}}(t)\rho^{\frac{1}{2}}\left(t,|y\!-\!y^\prime|^2\right)\\ &\displaystyle+ \nu(t)|z_1-z_1^\prime| + \vartheta(t)|z_2-z_2^\prime|, \end{array}$$

where ρ(t, ν) : [0, T] × R₊ → R₊ satisfies:

For fixed t ∈ [0, T], ρ(t, ⋅) is a continuous, concave and nondecreasing such that

$\begin{matrix} ρ (t, 0) = 0, and \forall α > 0 α ρ (t, ν) = ρ (t, α ν) . \end{matrix}$ $$\begin{array}{} \displaystyle \rho(t, 0) = 0,\quad\quad\text{and}\quad\forall\alpha \gt 0\quad \alpha\rho(t, \nu)=\rho(t, \alpha\nu). \end{array}$$

The ordinary differential equation (ODE)

$\begin{matrix} ν^{'} (t) = - ρ (t, ν (t)), v (T) = 0, \end{matrix}$ $$\begin{array}{} \displaystyle \nu^\prime(t) = -\rho(t, \nu(t)), \quad v(T) =0, \end{array}$$(3.10)

has a unique solution ν(t) = 0, 0 ≤ t ≤ T.

There exist two continuous and non-negative functions a and b such that

$\begin{matrix} ρ (t, ν) \leq a (t) + b (t) ν and \int_{0}^{T} [a (t) + b (t)] d t < \infty . \end{matrix}$ $$\begin{array}{} \displaystyle \rho(t, \nu)\le a(t)+ b(t)\nu \quad \text{and} \quad \int_0^{T} [a(t) + b(t)] dt \lt \infty. \end{array}$$

(H2.2)

for any t ∈[0, T], α²(t) = μ(t) + ν²(t) + ϑ²(t) > 0.

(H2.3)

The integrability condition holds:

$\begin{matrix} E \int_{0}^{T} e^{β A (t)} \frac{{|f (t, η_{t}, 0, 0)|}^{2}}{α^{2} (t)} d t < + \infty, t \in [0, T] . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\int_0^{T}\!e^{\beta A(t)}\frac{\left|f(t,\eta_t,0,0)\right|^2}{\alpha^2(t)}dt \lt +\infty,\quad t\in[0,T]. \end{array}$$

3.3.2

Existence and uniqueness of the solution

For n ≥ 1, we can construct the Picard approximate sequence of eq.(3.2) as follows

$\begin{matrix} \{\begin{matrix} Y_{t}^{0} = 0, Z_{t, 1}^{0} = 0, Z_{t, 2}^{0} = 0 t \in [0, T] \\ Y_{t}^{n} = ξ + \int_{t}^{T} f (s, η_{s}, Y_{s}^{n - 1}, Z_{1, s}^{n}, Z_{2, s}^{n}) d s - \int_{t}^{T} Z_{1, s}^{n} d B_{1, s}^{H_{1}} - \int_{t}^{T} Z_{2, s}^{n} d B_{2, s}^{H_{2}}, t \in [0, T] . \end{matrix} \end{matrix}$ $$\begin{array}{} \displaystyle \left\{ \begin{array}{} Y_t^0=0,\quad Z_{t,1}^0=0 ,\quad Z_{t,2}^0=0 \quad t\in [0,T]\\ Y_t^n = \xi + \!\int_t^Tf(s, \eta_s, Y_s^{n-1}, Z_{1,s}^n, Z_{2,s}^n)ds - \int_t^TZ_{1,s}^ndB_{1,s}^{H_1} - \!\int_t^TZ_{2,s}^ndB_{2,s}^{H_2}, \quad t\in[0,T]. \\ \end{array} \right. \end{array}$$(3.11)

Thanks to Theorem 3.4, this sequence is well defined.

Lemma 3.5

Assume that assumptions (H2) are true. Then for all n, m ≥ 1 and t ∈ [0, T], we have

$\begin{matrix} E [e^{β A (t)} | Y_{t}^{n + m} - Y_{t}^{n} |^{2}] \leq \int_{t}^{T} ρ (s, E [e^{β A (s)} | Y_{s}^{n + m - 1} - Y_{s}^{n - 1} |^{2}]) d s . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left[e^{\beta A(t)}|Y_t^{n+m} - Y_t^{n}|^2\right] \le \int_t^{T} \rho\left(s, {\bf E}\left[e^{\beta A(s)}|Y_s^{n+m-1} - Y_s^{n-1}|^2\right]\right)ds. \end{array}$$

Proof

Let us define for a process δ ∈ {Y, Z₁, Z₂}, n, m ≥ 1, δ^n,m = δ^n+m − δⁿ and the function Δ f^(n,m) (s) = $\begin{matrix} f (s, η_{s}, Y_{s}^{n + m - 1}, Z_{1, s}^{n + m}, Z_{2, s}^{n + m}) - f (s, η_{s}, Y_{s}^{n - 1}, Z_{1, s}^{n}, Z_{2, s}^{n}) \end{matrix}$ $\begin{array}{} \displaystyle f(s, \eta_s, Y_s^{n+m-1}, Z_{1,s}^{n+m}, Z_{2,s}^{n+m}) - f(s, \eta_s, Y_s^{n-1}, Z_{1,s}^n, Z_{2,s}^n) \end{array}$.

Then, it is obvious that $\begin{matrix} ({\bar{Y}}^{n, m}, {\bar{Z}}_{1}^{n, m}, {\bar{Z}}_{2}^{n, m}) \end{matrix}$ $\begin{array}{} \displaystyle (\overline Y^{n,m}, \overline Z_1^{n,m}, \overline Z_2^{n,m}) \end{array}$ solves the fractional BSDE

$\begin{matrix} {\bar{Y}}_{t}^{n, m} = \int_{t}^{T} Δ f^{(n, m)} (s) d s - \int_{t}^{T} {\bar{Z}}_{1, s}^{n, m} d B_{1, s}^{H_{1}} - \int_{t}^{T} {\bar{Z}}_{2, s}^{n, m} d B_{2, s}^{H_{2}}, t \in [0, T] . \end{matrix}$ $$\begin{array}{} \displaystyle \overline Y_t^{n,m}=\int_t^T\Delta f^{(n, m)}(s)ds -\int_t^T\overline Z_{1,s}^{n,m}dB_{1,s}^{H_1} - \int_t^T\overline Z_{2,s}^{n,m}dB_{2,s}^{H_2}, \quad t\in[0,T]. \end{array}$$(3.12)

By the fractional Itô chain rule, we have

$\begin{matrix} | {\bar{Y}}_{t}^{n, m} |^{2} & = 2 \int_{t}^{T} {\bar{Y}}_{s}^{n, m} Δ f^{(n, m)} (s) d s - 2 \int_{t}^{T} {\bar{Z}}_{1, s}^{n, m} D_{s}^{H_{1}} {\bar{Y}}_{s}^{n, m} d s - 2 \int_{t}^{T} {\bar{Z}}_{2, s}^{n, m} D_{s}^{H_{2}} {\bar{Y}}_{s}^{n, m} d s \\ - 2 \int_{t}^{T} {\bar{Y}}_{s}^{n, m} {\bar{Z}}_{1, s}^{n, m} d B_{1, s}^{H_{1}} - 2 \int_{t}^{T} {\bar{Y}}_{s}^{n, m} {\bar{Z}}_{2, s}^{n, m} d B_{2, s}^{H_{2}} . \end{matrix}$ $$\begin{array}{} \displaystyle |\overline Y_t^{n,m}|^2\!\!\!\!\! &\displaystyle= 2\int_t^T \overline Y_s^{n,m}\Delta f^{(n, m)}(s)ds - 2\int_t^T\overline Z_{1,s}^{n,m}\mathbb{D}^{H_1}_s\overline Y_s^{n,m}ds - 2\int_t^T\overline Z_{2,s}^{n, m}\mathbb{D}^{H_2}_s\overline Y_s^{n,m}ds \\ &\displaystyle-2\int_t^T\overline Y_s^{n,m}\overline Z_{1,s}^{n, m}dB_{1,s}^{H_1} - 2\int_t^T\overline Y_s^{n,m}\overline Z_{2,s}^{n,m}d B_{2,s}^{H_2}. \end{array}$$

Applying Itô formula to $\begin{matrix} e^{β A (t)} | {\bar{Y}}_{t}^{n, m} |^{2} \end{matrix}$ $\begin{array}{} \displaystyle e^{\beta A(t)}|\overline Y_t^{n,m}|^2 \end{array}$, we obtain that

$\begin{matrix} e^{β A (t)} | {\bar{Y}}_{t}^{n, m} |^{2} & = 2 \int_{t}^{T} e^{β A (s)} {\bar{Y}}_{s}^{n, m} Δ f^{(n, m)} (s) d s - 2 \int_{t}^{T} e^{β A (s)} {\bar{Z}}_{1, s}^{n, m} D_{s}^{H_{1}} {\bar{Y}}_{s}^{n, m} d s \\ - 2 \int_{t}^{T} e^{β A (s)} {\bar{Z}}_{2, s}^{n, m} D_{s}^{H_{2}} {\bar{Y}}_{s}^{n, m} d s - 2 \int_{t}^{T} e^{β A (s)} {\bar{Y}}_{s}^{n, m} {\bar{Z}}_{1, s}^{n, m} d B_{1, s}^{H_{1}} \\ - 2 \int_{t}^{T} e^{β A (s)} {\bar{Y}}_{s}^{n, m} {\bar{Z}}_{2, s}^{n, m} d B_{2, s}^{H_{2}} - β \int_{t}^{T} e^{β A (s)} α^{2} (s) | {\bar{Y}}_{s}^{n, m} |^{2} d s . \end{matrix}$ $$\begin{array}{} \displaystyle e^{\beta A(t)}|\overline Y_t^{n,m}|^2\!\!\!\! &\displaystyle= 2\int_t^Te^{\beta A(s)}\overline Y_s^{n,m}\Delta f^{(n, m)}(s)ds - 2\int_t^Te^{\beta A(s)}\overline Z_{1,s}^{n,m}\mathbb{D}^{H_1}_s\overline Y_s^{n,m}ds \\ &\displaystyle- 2\int_t^Te^{\beta A(s)}\overline Z_{2,s}^{n, m}\mathbb{D}^{H_2}_s\overline Y_s^{n,m}ds -2\int_t^Te^{\beta A(s)}\overline Y_s^{n,m}\overline Z_{1,s}^{n, m}dB_{1,s}^{H_1}\\ &\displaystyle- 2\int_t^Te^{\beta A(s)}\overline Y_s^{n,m}\overline Z_{2,s}^{n,m}dB_{2,s}^{H_2} - \beta\int_t^T e^{\beta A(s)}\alpha^2(s)|\overline Y_s^{n,m}|^2ds. \end{array}$$

By Proposition 3.3, we have

$\begin{matrix} E [e^{β A (t)} | {\bar{Y}}_{t}^{n, m} |^{2}] & + E \int_{t}^{T} e^{β A (s)} [β α^{2} (s) | {\bar{Y}}_{s}^{n, m} |^{2} + 2 \frac{{\hat{σ}}_{1} (s)}{σ_{1} (s)} | {\bar{Z}}_{1, s}^{n, m} |^{2} + 2 \frac{{\hat{σ}}_{2} (s)}{σ_{2} (s)} | {\bar{Z}}_{2, s}^{n, m} |^{2}] d s \\ = 2 E \int_{t}^{T} e^{β A (s)} {\bar{Y}}_{s}^{n, m} Δ f^{(n, m)} (s) d s . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left[e^{\beta A(t)}|\overline Y_t^{n,m}|^2\right] &\!\!\!\!\displaystyle+ {\bf E}\int_t^T e^{\beta A(s)}\bigg[\beta\alpha^2(s)|\overline Y_s^{n,m}|^2 + 2\frac{\hat{\sigma}_1(s)}{\sigma_1(s)}|\overline Z_{1,s}^{n,m}|^2+ 2\frac{\hat{\sigma}_2(s)}{\sigma_2(s)}|\overline Z_{2,s}^{n, m}|^2\bigg]ds \\ &\displaystyle= 2{\bf E}\int_t^Te^{\beta A(s)}\overline Y_s^{n,m}\Delta f^{(n, m)}(s)ds. \end{array}$$(3.13)

Applying standard estimates and asumption (H2.1), we obtain that

$\begin{matrix} \begin{aligned} 2 E \int_{t}^{T} e^{β A (s)} {\bar{Y}}_{s}^{n, m} & Δ f^{(n, m)} (s) d s \leq 2 E \int_{t}^{T} e^{β A (s)} | {\bar{Y}}_{s}^{n, m} | μ^{\frac{1}{2}} (s) ρ^{\frac{1}{2}} (s, | Y_{s}^{n + m - 1} - Y_{s}^{n - 1} |^{2}) d s \\ + 2 E \int_{t}^{T} e^{β A (s)} | {\bar{Y}}_{s}^{n, m} | [ν (s) | {\bar{Z}}_{1, s}^{n, m} | + ϑ (s) | {\bar{Z}}_{2, s}^{n, m} |] d s \\ \leq \frac{1}{C_{0}} E \int_{t}^{T} e^{β A (s)} [μ (s) + ν^{2} (s) + ϑ^{2} (s)] | {\bar{Y}}_{s}^{n, m} | d s \\ + C_{0} E \int_{t}^{T} e^{β A (s)} [ρ (s, | Y_{s}^{n + m - 1} - Y_{s}^{n - 1} |^{2}) + | {\bar{Z}}_{1, s}^{n, m} |^{2} + | {\bar{Z}}_{2, s}^{n, m} |^{2}] d s \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle 2{\bf E}\int_t^Te^{\beta A(s)}\overline Y_s^{n,m}&\Delta f^{(n, m)}(s)ds \leq 2{\bf E}\int_t^Te^{\beta A(s)}|\overline Y_s^{n,m}| \mu^{\frac{1}{2}}(s)\rho^{\frac{1}{2}}(s,|Y_s^{n+m-1}-Y_s^{n-1}|^2)ds\\ &+ 2{\bf E}\int_t^Te^{\beta A(s)}|\overline Y_s^{n,m}| \bigg[\nu(s)|\overline Z_{1,s}^{n,m}|+\vartheta(s)|\overline Z_{2,s}^{n, m}|\bigg]ds\\ &\leq \frac{1}{C_0}{\bf E}\int_t^Te^{\beta A(s)}\left[\mu(s)+ \nu^2(s)+\vartheta^2(s)\right]|\overline Y_s^{n,m}|ds\\ &+ C_0{\bf E}\int_t^Te^{\beta A(s)}\bigg[\rho(s,|Y_s^{n+m-1}-Y_s^{n-1}|^2) + |\overline Z_{1,s}^{n,m}|^2+|\overline Z_{2,s}^{n, m}|^2\bigg]ds \end{split} \end{array}$$

Therefore, we can write

$\begin{matrix} \begin{aligned} 2 E \int_{t}^{T} e^{β A (s)} {\bar{Y}}_{s}^{n, m} & Δ f^{(n, m)} (s) d s \leq \frac{1}{C_{0}} E \int_{t}^{T} e^{β A (s)} α^{2} (s) | {\bar{Y}}_{s}^{n, m} | d s + C_{0} E \int_{t}^{T} e^{β A (s)} | {\bar{Z}}_{1, s}^{n, m} |^{2} d s \\ + C_{0} E \int_{t}^{T} e^{β A (s)} | {\bar{Z}}_{2, s}^{n, m} |^{2} d s + C_{0} E \int_{t}^{T} e^{β A (s)} ρ (s, | Y_{s}^{n + m - 1} - Y_{s}^{n - 1} |^{2}) d s . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle 2{\bf E}\int_t^Te^{\beta A(s)}\overline Y_s^{n,m}&\Delta f^{(n, m)}(s)ds\leq \frac{1}{C_0}{\bf E}\int_t^Te^{\beta A(s)}\alpha^2(s)|\overline Y_s^{n,m}|ds +C_0{\bf E}\int_t^Te^{\beta A(s)}|\overline Z_{1,s}^{n,m}|^2ds \\ &+ C_0{\bf E}\int_t^Te^{\beta A(s)}|\overline Z_{2,s}^{n, m}|^2ds+ C_0{\bf E}\int_t^Te^{\beta A(s)}\rho(s,|Y_s^{n+m-1}-Y_s^{n-1}|^2)ds. \end{split} \end{array}$$

Using the abovementioned inequality and Remark 3.1, from (3.13), we deduce that

$\begin{matrix} \begin{aligned} E [e^{β A (t)} | {\bar{Y}}_{t}^{n, m} |^{2}] & + β E \int_{t}^{T} e^{β A (s)} α^{2} (s) | {\bar{Y}}_{s}^{n, m} |^{2} d s + 2 C_{0} E \int_{t}^{T} e^{β A (s)} [| {\bar{Z}}_{1, s}^{n, m} |^{2} + | {\bar{Z}}_{2, s}^{n, m} |^{2}] d s \\ \leq \frac{1}{C_{0}} E \int_{t}^{T} e^{β A (s)} α^{2} (s) | {\bar{Y}}_{s}^{n, m} | d s + C_{0} E \int_{t}^{T} e^{β A (s)} [| {\bar{Z}}_{1, s}^{n, m} |^{2} + | {\bar{Z}}_{2, s}^{n, m} |^{2}] d s \\ + C_{0} E \int_{t}^{T} e^{β A (s)} ρ (s, | Y_{s}^{n + m - 1} - Y_{s}^{n - 1} |^{2}) d s . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}\left[e^{\beta A(t)}|\overline Y_t^{n,m}|^2\right] &+ \beta{\bf E}\int_t^T e^{\beta A(s)}\alpha^2(s)|\overline Y_s^{n,m}|^2ds + 2C_0{\bf E}\int_t^T e^{\beta A(s)}\bigg[|\overline Z_{1,s}^{n,m}|^2+ |\overline Z_{2,s}^{n, m}|^2\bigg]ds \\ &\leq \frac{1}{C_0}{\bf E}\int_t^Te^{\beta A(s)}\alpha^2(s)|\overline Y_s^{n,m}|ds + C_0{\bf E}\int_t^Te^{\beta A(s)}\bigg[|\overline Z_{1,s}^{n,m}|^2+|\overline Z_{2,s}^{n, m}|^2\bigg]ds\\ &+ C_0{\bf E}\int_t^Te^{\beta A(s)}\rho(s,|Y_s^{n+m-1}-Y_s^{n-1}|^2)ds. \end{split} \end{array}$$

Choosing β such that β > $\begin{matrix} \frac{1}{C_{0}} \end{matrix}$ $\begin{array}{} \displaystyle \frac{1}{C_0} \end{array}$, we have

$\begin{matrix} E [e^{β A (t)} | Y_{t}^{n + m} - Y_{t}^{n} |^{2}] \leq C_{0} \int_{t}^{T} ρ (s, E [e^{β A (s)} | Y_{s}^{n + m - 1} - Y_{s}^{n - 1} |^{2}]) d s . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left[e^{\beta A(t)}|Y_t^{n+m} - Y_t^{n}|^2\right] \le C_0\int_t^{T} \rho\left(s, {\bf E}\left[e^{\beta A(s)}|Y_s^{n+m-1} - Y_s^{n-1}|^2\right]\right)ds. \end{array}$$

Finally, by Remark 3.1 we obtain

□

Lemma 3.6

Let the assumption (H2) be satisfied. Then there exists a constant M ≥ 0 and T₁ ∈ [0, T] such that

$\begin{matrix} \forall n \geq 1, E [e^{β A (t)} | Y_{t}^{n} |^{2}] \leq M, t \in [T_{1}, T] . \end{matrix}$ $$\begin{array}{} \displaystyle \forall n\ge1,\quad {\bf E} \left[e^{\beta A(t)}|Y_t^{n}|^2\right]\le M, \qquad t\in[ T_1 ,T]. \end{array}$$

Proof

Using the same method as in the proof of Lemma 3.5, we obtain that

$\begin{matrix} \begin{aligned} E [e^{β A (t)} | Y_{t}^{n} |^{2}] & + β E \int_{t}^{T} e^{β A (s)} | Y_{s}^{n} |^{2} d s + 2 C_{0} E \int_{t}^{T} e^{β A (s)} [| Z_{1, s}^{n} |^{2} + | Z_{2, s}^{n} |^{2}] d s \\ \leq E [e^{β A (T)} | ξ |^{2}] + 2 E \int_{t}^{T} e^{β A (s)} Y_{s}^{n} f (s, η_{s}, Y_{s}^{n - 1}, Z_{1, s}^{n}, Z_{2, s}^{n}) d s . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}\left[e^{\beta A(t)}|Y_t^n|^2\right] &+ \beta{\bf E}\int_t^Te^{\beta A(s)}|Y_s^n|^2ds +2 C_0{\bf E}\int_t^Te^{\beta A(s)}\bigg[ | Z_{1,s}^{n}|^2+|Z_{2,s}^{n}|^2\bigg]ds\nonumber\\ &\le{\bf E}\left[e^{\beta A(T)}|\xi|^2\right] + 2{\bf E}\int_t^Te^{\beta A(s)}Y_s^nf(s,\eta_s,Y_s^{n-1}, Z_{1,s}^n, Z_{2,s}^n)ds. \end{split} \end{array}$$(3.14)

Applying standard estimates and asumption (H2.1), we obtain that

$\begin{matrix} \begin{aligned} 2 E \int_{t}^{T} e^{β A (s)} Y_{s}^{n} f (s, η_{s}, Y_{s}^{n - 1}, Z_{1, s}^{n}, Z_{2, s}^{n}) d s & \leq C_{0} E \int_{t}^{T} e^{β A (s)} [ρ (s, | Y_{s}^{n - 1} |^{2}) + | {\bar{Z}}_{1, s}^{n, m} |^{2} + | Z_{2, s}^{n} |^{2}] d s \\ + E \int_{t}^{T} e^{β A (s)} \frac{| f (s, η_{s}, 0, 0, 0) |^{2}}{α^{2} (s)} d s & + (1 + \frac{1}{C_{0}}) E \int_{t}^{T} e^{β A (s)} α^{2} (s) | Y_{s}^{n} | d s \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle 2{\bf E}\int_t^Te^{\beta A(s)}Y_s^nf(s,\eta_s,Y_s^{n-1}, Z_{1,s}^n, Z_{2,s}^n)ds &\le C_0{\bf E}\int_t^Te^{\beta A(s)}\bigg[\rho(s,|Y_s^{n-1}|^2) + |\overline Z_{1,s}^{n,m}|^2+|Z_{2,s}^{n}|^2\bigg]ds\\ + {\bf E}\int_t^Te^{\beta A(s)}\frac{|f(s,\eta_s,0,0,0)|^2}{\alpha^2(s)}ds &+ \left(1+\frac{1}{C_0}\right){\bf E}\int_t^Te^{\beta A(s)}\alpha^2(s)|Y_s^{n}|ds \end{split} \end{array}$$

Using the abovementioned inequality, from (3.14) we deduce that

$\begin{matrix} \begin{aligned} E [e^{β A (t)} | Y_{t}^{n} |^{2}] & + [β - (1 + \frac{1}{C_{0}})] E \int_{t}^{T} e^{β A (s)} | Y_{s}^{n} |^{2} d s + C_{0} E \int_{t}^{T} e^{β A (s)} [| Z_{1, s}^{n} |^{2} + | Z_{2, s}^{n} |^{2}] d s \\ \leq Λ_{t} + C_{0} \int_{t}^{T} ρ (s, E [e^{β A (s)} | Y_{s}^{n - 1} |^{2}]) d s, \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}\left[e^{\beta A(t)}|Y_t^n|^2\right] &+ [\beta -(1+\frac{1}{C_0})]{\bf E}\int_t^Te^{\beta A(s)}|Y_s^n|^2ds + C_0{\bf E}\int_t^Te^{\beta A(s)}\bigg[ | Z_{1,s}^{n}|^2+|Z_{2,s}^{n}|^2\bigg]ds\\ &\le \Lambda_t + C_0\int_t^T\rho(s,{\bf E} \left[e^{\beta A(s)}|Y_s^{n-1}|^2\right])ds, \end{split} \end{array}$$(3.15)

where $\begin{matrix} Λ_{t} = E (e^{β A (T)} | ξ |^{2} + \int_{t}^{T} e^{β A (s)} \frac{| f (s, η_{s}, 0, 0, 0) |^{2}}{α^{2} (s)} d s) \end{matrix}$ $\begin{array}{} \displaystyle \Lambda_t={\bf E}\left(e^{\beta A(T)}|\xi|^2 + \int_t^Te^{\beta A(s)}\frac{|f(s,\eta_s,0,0,0)|^2}{\alpha^2(s)}ds \right) \end{array}$.

Choosing β such that β − $\begin{matrix} \frac{1}{C_{0}} \end{matrix}$ $\begin{array}{} \displaystyle \frac{1}{C_0} \end{array}$ > 1 we have

$\begin{matrix} E [e^{β A (t)} | Y_{t}^{n} |^{2}] \leq Λ_{t} + \int_{t}^{T} ρ (s, E [e^{β A (s)} | Y_{s}^{n - 1} |^{2}]) d s, t \in [0, T] . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left[e^{\beta A(t)}|Y_t^n|^2\right]\le \Lambda_t + \int_t^T\rho(s,{\bf E} \left[e^{\beta A(s)}|Y_s^{n-1}|^2\right])ds,\quad t\in[0,T]. \end{array}$$(3.16)

$\begin{matrix} Finally let M = 2 Λ_{0} + 2 \int_{0}^{T} a (s) d s \geq 0. \end{matrix}$ $$\begin{array}{} \displaystyle \text{Finally let}\quad\quad\quad\quad M=2\Lambda_0 + 2\int_0^Ta(s)ds\geq 0. \end{array}$$(3.17)

Arguing as in [[9], Lemma 2], we choose T₁ such that

$\begin{matrix} Λ_{0} + \int_{t}^{T} ρ (s, M) d s \leq M, t \in [T_{1}, T] . \end{matrix}$ $$\begin{array}{} \displaystyle \Lambda_0 + \int_t^T\rho(s,M)ds\leq M, \quad\quad t\in[T_1,T]. \end{array}$$(3.18)

By inequality (3.16) and (3.18), for t ∈ [T₁, T], we have

$\begin{matrix} E [e^{β A (t)} {|Y_{t}^{1}|}^{2}] \leq Λ_{t} + \int_{t}^{T} ρ (s, 0) d s \leq Λ_{0} \leq M, \\ E [e^{β A (t)} {|Y_{t}^{2}|}^{2}] \leq Λ_{t} + \int_{t}^{T} ρ (s, E [e^{β A (s)} | Y_{s}^{1} |^{2}]) d s \leq Λ_{0} + \int_{t}^{T} ρ (s, M) d s \leq M, \\ E [e^{β A (t)} {|Y_{t}^{3}|}^{2}] \leq Λ_{t} + \int_{t}^{T} ρ (s, E [e^{β A (s)} | Y_{s}^{2} |^{2}]) d s \leq Λ_{0} + \int_{t}^{T} ρ (s, M) d s \leq M . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left[e^{\beta A(t)}\left|Y_t^1\right|^2\right] \leq \Lambda_t + \int_t^T\rho\left(s, 0 \right)ds \leq \Lambda_0 \leq M,\\ \displaystyle {\bf E}\left[e^{\beta A(t)}\left|Y_t^2\right|^2\right] \leq \Lambda_t + \int_t^T\rho\left(s,{\bf E} [e^{\beta A(s)}|Y_s^1|^2]\right)ds\leq \Lambda_0 + \int_t^T\rho\left(s, M\right)ds \leq M,\\\displaystyle {\bf E}\left[e^{\beta A(t)}\left|Y_t^3\right|^2\right] \leq \Lambda_t + \int_t^T\rho\left(s,{\bf E} [e^{\beta A(s)}|Y_s^2|^2]\right)ds \leq\Lambda_0 + \int_t^T\rho\left(s, M \right)ds \leq M. \end{array}$$

Hence, by induction, one can prove that for all n ≥ 1,

$\begin{matrix} E [e^{β A (t)} | Y_{t}^{n} |^{2}] \leq M, T_{1} \leq t \leq T . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E} \left[e^{\beta A(t)}|Y_t^{n}|^2\right]\le M, \quad T_1\leq t\leq T. \end{array}$$

□

The main result of this section is the following theorem:

Theorem 3.7

Let the assumptions (H2) be satisfied. Then, the fractional BSDE (3.2) has a unique solution (Y, Z₁, Z₂) in the space ℬ²([0, T], R).

Proof

We split the proof into two parts.

(i) Existence

Using the constant M given by (3.17), we consider the sequence (φ_n)_n≥1 given by

$\begin{matrix} φ_{0} (t) = \int_{t}^{T} ρ (s, M) d s, φ_{n + 1} (t) = \int_{t}^{T} ρ (s, φ_{n} (s)) d s, n \geq 0, t \in [T_{1}, T] . \end{matrix}$ $$\begin{array}{} \displaystyle \varphi_0(t) =\int_t^{T}\rho\left(s, M\right)ds, \quad\quad \varphi_{n+1}(t) = \int_t^{T}\rho\left(s, \varphi_{n}(s)\right)ds,\quad n\geq 0,\quad t\in[T_1,T]. \end{array}$$

Then for all t ∈ [T₁, T], from the proof of Lemma 3.6, one can deduce that

$\begin{matrix} \begin{aligned} φ_{0} (t) & = \int_{t}^{T} ρ (s, M) d s \leq M, \\ φ_{1} (t) & = \int_{t}^{T} ρ (s, φ_{0} (s)) d s \leq \int_{t}^{T} ρ (s, M) d s = φ_{0} (t) \leq M, \\ φ_{2} (t) & = \int_{t}^{T} ρ (s, φ_{1} (s)) d s \leq \int_{t}^{T} ρ (s, φ_{0} (s)) d s = φ_{1} (t) \leq M . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle \varphi_0(t) &= \int_t^T\rho\left(s, M\right)ds\leq M,\\ \varphi_1(t) &= \int_t^T\rho\left(s, \varphi_0(s) \right)ds\leq \int_t^T\rho\left(s, M\right)ds= \varphi_0(t)\leq M,\\ \varphi_2(t) &= \int_t^T\rho\left(s, \varphi_1(s) \right)ds\leq \int_t^T\rho\left(s, \varphi_0(s)\right)ds= \varphi_1(t)\leq M. \end{split} \end{array}$$

By induction, one can prove that for all n ≥ 1, φ_n(t) satisfies

$\begin{matrix} 0 \leq φ_{n + 1} (t) \leq φ_{n} (t) \leq \cdot \cdot \cdot \leq φ_{1} (t) \leq φ_{0} (t) \leq M . \end{matrix}$ $$\begin{array}{} \displaystyle 0\leq\varphi_{n+1}(t)\leq\varphi_n(t)\leq\cdot\cdot\cdot\leq\varphi_1(t)\leq\varphi_0(t)\leq M. \end{array}$$

Then {φ_n(t), t ∈ [T₁, T]}_n≥1 is uniformly bounded. On the other hand, for all n ≥ 1 and t₁, t₂ ∈ [T₁, T], we obtain

$\begin{matrix} |φ_{n} (t_{1}) - φ_{n} (t_{2})| = |\int_{t_{1}}^{t_{2}} ρ (s, φ_{n - 1} (s)) d s| \leq |\int_{t_{1}}^{t_{2}} ρ (s, M) d s| . \end{matrix}$ $$\begin{array}{} \displaystyle \left|\varphi_{n}(t_1) - \varphi_{n}(t_2)\right|=\left|\int_{t_1}^{t_2}\rho\left(s, \varphi_{n-1}(s)\right)ds \right|\leq\left|\int_{t_1}^{t_2}\rho\left(s, M\right)ds \right|. \end{array}$$

Since, for fixed v, ∈ $\begin{matrix} \int_{0}^{T} ρ (s, v) d s < + \infty \end{matrix}$ $\begin{array}{} \int_0^T\rho\left(s, v\right)ds \lt + \infty \end{array}$. So

$\begin{matrix} sup_{n} |φ_{n} (t_{1}) - φ_{n} (t_{2})| \to 0 a s |t_{1} - t_{2}| \to 0, \end{matrix}$ $$\begin{array}{} \displaystyle \sup_n \left|\varphi_{n}(t_1) - \varphi_{n}(t_2)\right|\rightarrow 0 \quad as\quad \left|t_1 - t_2\right|\rightarrow 0, \end{array}$$

which means that {φ_n(t), t ∈ [T₁, T]}_n≥1 is an equicontinuous family of function. Therefore, by the Arzelá-Ascoli theorem, we can define by φ(t) the limit function of (φ_n(t))_n≥1.

By (3.10), one knows that φ(t) = 0, t ∈ [T₁, T].

Now for all t ∈[T₁, T], n, m ≥ 1, in view of Lemmas 3.5 and 3.6, we have

$\begin{matrix} \begin{aligned} E [e^{β A (t)} {|Y_{t}^{n}|}^{2}] \leq M, \\ E [e^{β A (t)} {|Y_{t}^{1 + m} - Y_{t}^{1}|}^{2}] \leq \int_{t}^{T} ρ (s, E [e^{β A (s)} {|Y_{s}^{m}|}^{2}]) d s \leq \int_{t}^{T} ρ (s, M) d s = φ_{0} (t) \leq M, \\ E [e^{β A (t)} {|Y_{t}^{2 + m} - Y_{t}^{2}|}^{2}] \leq \int_{t}^{T} ρ (s, E [e^{β A (s)} {|Y_{s}^{1 + m} - Y_{s}^{1}|}^{2}]) d s \leq φ_{1} (t) \leq M, \\ E [e^{β A (t)} {|Y_{t}^{3 + m} - Y_{t}^{3}|}^{2}] \leq \int_{t}^{T} ρ (s, E [e^{β A (s)} {|Y_{s}^{2 + m} - Y_{s}^{2}|}^{2}]) d s \leq φ_{2} (t) \leq M . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle &{\bf E}\left[ e^{\beta A(t)}\left|Y_t^n\right|^2\right] \leq M,\\ &{\bf E}\left[e^{\beta A(t)}\left|Y_t^{1+m} - Y_t^1 \right|^2\right] \leq \int_t^T\rho\left(s,{\bf E} [e^{\beta A(s)}\left|Y_s^m\right|^2]\right)ds \leq\int_t^T\rho\left(s, M\right)ds = \varphi_0(t) \leq M,\\ &{\bf E}\left[e^{\beta A(t)}\left|Y_t^{2+m} - Y_t^2 \right|^2\right] \leq\int_t^T\rho\left(s,{\bf E}\left[e^{\beta A(s)}\left|Y_s^{1+m} - Y_s^1 \right|^2\right] \right)ds \leq \varphi_1(t) \leq M,\\ &{\bf E}\left[ e^{\beta A(t)}\left|Y_t^{3+m} - Y_t^3 \right|^2\right] \leq \int_t^T\rho\left(s, {\bf E}\left[e^{\beta A(s)}\left|Y_s^{2+m} - Y_s^2 \right|^2\right] \right)ds \leq \varphi_2(t) \leq M. \end{split} \end{array}$$

By induction, we can derive that

$\begin{matrix} m \geq 1, E [e^{β A (t)} {|Y_{t}^{n + m} - Y_{t}^{n}|}^{2}] \leq φ_{n - 1} (t), t \in [T_{1}, T] . \end{matrix}$ $$\begin{array}{} \displaystyle m\geq 1,\quad{\bf E}\left[e^{\beta A(t)}\left|Y_t^{n+m} - Y_t^n \right|^2\right] \leq \varphi_{n-1}(t), \quad t\in[T_1,T]. \end{array}$$

Therefore we have

$\begin{matrix} sup_{t \in [T_{1}, T]} E [e^{β A (t)} {|Y_{t}^{n + m} - Y_{t}^{n}|}^{2}] \leq sup_{t \in [T_{1}, T]} φ_{n - 1} (t) = φ_{n - 1} (T_{1}) \to 0 n \to \infty . \end{matrix}$ $$\begin{array}{} \displaystyle \sup_{t\in[T_1,T]}{\bf E}\left[e^{\beta A(t)}\left|Y_t^{n+m} - Y_t^n \right|^2\right]\leq\sup_{t\in[T_1,T]} \varphi_{n-1}(t)=\varphi_{n-1}(T_1) \rightarrow 0 \quad n\rightarrow \infty. \end{array}$$

Exploiting the argument developed in [[1], Theorem 3.9] we prove that the sequence $\begin{matrix} (Y^{n}, Z_{1}^{n}, Z_{2}^{n}) \end{matrix}$ $\begin{array}{} \displaystyle \left(Y^n,Z_1^n,Z_2^n\right) \end{array}$ is a Cauchy sequence in ℬ²([T₁, T], R). Letting n → + ∞ in eq.(3.11), we obtain

$\begin{matrix} Y_{t} = ξ + \int_{t}^{T} f (s, η_{s}, Y_{s}, Z_{1, s}, Z_{2, s}) d s - \int_{t}^{T} Z_{1, s} d B_{1, s}^{H_{1}} - \int_{t}^{T} Z_{2, s} d B_{2, s}^{H_{2}}, T_{1} \leq t \leq T . \end{matrix}$ $$\begin{array}{} \displaystyle Y_t = \xi + \!\int_t^Tf(s,\eta_s, Y_s, Z_{1,s},Z_{2,s})ds - \int_t^TZ_{1,s}dB_{1,s}^{H_1} - \!\int_t^TZ_{2,s}dB_{2,s}^{H_2}, \quad T_1\leq t\leq T. \end{array}$$

In other words, we have shown the existence of the solution (Y, Z₁, Z₂) to fractional BSDE (3.2) on [T₁, T]. Finally, by iteration, one can deduce the existence on [T − λ(T − T₁), T], for each λ, and therefore the existence on the whole [0, T].

(ii) Uniqueness

Let $\begin{matrix} {(Y_{t}^{i}, Z_{1, t}^{i}, Z_{2, t}^{i})}_{0 \leq t \leq T} \end{matrix}$ $\begin{array}{} \displaystyle \left(Y_t^i, Z_{1,t}^i, Z_{2,t}^i\right)_{0\leq t\leq T} \end{array}$, i = 1, 2, be two solutions of fractional BSDE (3.2).

Using the same method as in the proof of Lemma (3.5), we have

$\begin{matrix} \begin{aligned} E [e^{β A (t)} | Y_{t}^{1} - Y_{t}^{2} |^{2}] & + C_{0} E \int_{t}^{T} e^{β A (s)} | Z_{1, s}^{1} - Z_{1, s}^{2} |^{2} d s + C_{0} E \int_{t}^{T} e^{β A (s)} | Z_{2, s}^{1} - Z_{2, s}^{2} |^{2} d s \\ \leq \int_{t}^{T} ρ (s, E [e^{β A (s)} | Y_{s}^{1} - Y_{s}^{2} |^{2}]) d s, t \in [0, T] . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}\left[e^{\beta A(t)}|Y_t^1 - Y_t^2|^2\right] &+ C_0{\bf E}\int_t^Te^{\beta A(s)}| Z_{1,s}^1 - Z_{1,s}^2|^2ds + C_0{\bf E}\int_t^Te^{\beta A(s)}|Z_{2,s}^1 - Z_{2,s}^2|^2ds \\ &\leq \int_t^T\rho(s,{\bf E} \left[e^{\beta A(s)}|Y_s^1 - Y_s^2|^2\right])ds,\quad t\in[0,T]. \end{split} \end{array}$$(3.19)

Therefore

$\begin{matrix} \begin{aligned} E [e^{β A (t)} | Y_{t}^{1} - Y_{t}^{2} |^{2}] \leq \int_{t}^{T} ρ (s, E [e^{β A (s)} | Y_{s}^{1} - Y_{s}^{2} |^{2}]) d s, t \in [0, T] \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}\left[e^{\beta A(t)}|Y_t^{1} - Y_t^{2}|^2\right] \le \int_t^{T} \rho\left(s, {\bf E}\left[e^{\beta A(s)}|Y_s^{1} - Y_s^{2}|^2\right]\right)ds,\quad t\in[0,T] \end{split} \end{array}$$(3.20)

From the comparison theorem of ODE, we know that $\begin{matrix} E [e^{β t} | Y_{t}^{1} - Y_{t}^{2} |^{2}] \leq r (t) \end{matrix}$ $\begin{array}{} \displaystyle {\bf E}\left[e^{\beta t}|Y_t^1 - Y_t^2|^2\right]\leq r(t) \end{array}$, where r(t) is the maximum of solution of (3.10) on [0, T]. As a consequence, we have $\begin{matrix} Y_{t}^{1} = Y_{t}^{2} \end{matrix}$ $\begin{array}{} \displaystyle Y_t^1=Y_t^2 \end{array}$ for t ∈[0, T]. From (3.19), we deduce $\begin{matrix} (Z_{1, t}^{1}, Z_{2, t}^{1}) = (Z_{1, t}^{2}, Z_{2, t}^{2}) \end{matrix}$ $\begin{array}{} \displaystyle \left(Z_{1,t}^1, Z_{2,t}^1\right)= \left(Z_{1,t}^2, Z_{2,t}^2\right) \end{array}$ for t ∈[0, T]. This completes the proof. □

eISSN:: 2444-8656
Language:: English

Publication timeframe:: Volume Open
Journal Subjects:: Life Sciences, other, Mathematics, Applied Mathematics, General Mathematics, Physics

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BSDEs driven by two mutually independent fractional Brownian motions with stochastic Lipschitz coefficients

Published Online: Jun 25, 2019

Page range: 139 - 150

Received: Mar 09, 2019

Accepted: Apr 25, 2019

DOI: https://doi.org/10.2478/AMNS.2019.1.00014

Keywordsbackward stochastic differential equation, stochastic Lipschitz coefficients, Malliavin derivative and fractional Itô’s formula

© 2019 Sadibou Aidara and Yaya Sagna, published by Sciendo

This work is licensed under the Creative Commons Attribution 4.0 International License.

Keywords
backward stochastic differential equation, stochastic Lipschitz coefficients, Malliavin derivative and fractional Itô’s formula