A Convolution Method for Numerical Solution of Backward Stochastic Differential Equations Based on the Fractional FFT

Abstract

BSDEs are applied in many areas, particularly in finance and economics. In this paper, we extended the convolution method to numerically solve FBSDEs. First, a generalized θ-scheme is applied to discretize the backwards component. Second, the convolution method is used to solve the conditional expectation. Third, the resulting convolution is dealt with numerically by the Fourier transform. Therefore, the fractional FFT algorithm is applied to compute the Fourier and inverse the transforms. Then, we prove some error estimates. Finally, a numerical example is implemented to test the efficiency and stability of the proposed method.

1. Introduction

BSDE has attracted great attention, resulting in many applications in the areas of finance and partial differential equations (see [1,2]). In this paper, we are concerned with numerically solving the BSDE of this type:

$$Y_t=\xi +{\displaystyle {\int }_t^{T}f(s,Y_s,Z_s)ds-{\displaystyle {\int }_t^T{Z}_{s}d{W}_s}}.$$

(1)

Let $(\mathsf{\Omega },F,{\left\{F_t\right\}}_{0\le t\le T},P)$ be a complete filtered probability space, where $W$ is a standard $n$-dimensional Brownian motion. The terminal condition $\xi =\phi (W_T)\in R^k$ is a square integrable $F_t$-measurable random variable and the driver $f$: $\left[0,T\right]\times R^k\times R^{k\times n}\to R^k$ is a functional.

The existence and uniqueness of the solution to general nonlinear BSDE (1) were proven by Paradoux and Peng [3]. Many works have extended this result. Antonelli [4] and Mao [5] introduced an adapted solution to BSDE with non-Lipschitz coefficients. In addition, from Paradoux and Peng [6], we have found that if the Cauchy problem to one-dimensional diffusion PDE

$$\{\begin{array}{l}\frac{\partial u}{\partial t}+\frac{1}{2}\frac{{\partial }^{2}u}{\partial t^2}+f(t,u,\frac{\partial u}{\partial t})=0, (t,x)\in \left[0,T\right)\times R,\\ u(T,x)=g(x),\end{array}$$

has a unique solution $u\in C^{1,2}$, then the solution $(Y,Z)$ for one-dimensional BSDE with a terminal condition has the representation

$$Y_t=u(t,W_t),$$

$$Z_t=\frac{\partial u}{\partial x}(t,W_t).$$

It is often difficult to obtain the analytic solution of BSDEs. Therefore, it is critical to address the numerical schemes. Zhao, Li and Zhang [7] proposed a type of $\theta$-scheme with four parameters to solve the backward stochastic differential equation. Zhang, Zhao and Ju [8] proposed a multistep scheme on time–space grids for solving BSDEs, and the conditional expectations of this scheme are approximated by using Lagrange interpolating polynomials with the values of the integrands. Wang, Luo and Zhao [9] proposed the Crank–Nicolson scheme to solve BSDE, and proposed using the Gauss–Hermite quadrature formula approximate conditional expectations in the numerical applications. Respectively, their numerical experiments showed that they are highly accurate. In addition, much research has theoretically proven that the high-order convergence rate with different generators does not depend on $Z_t$ (see [9,10,11,12]). The key to the numerical scheme is realizing how to discretize the conditional expectation. To date, there have been many methods to solve the problem. Zhang et al. [13] constructed a type of sparse-grid Gauss–Hermite quadrature rule and hierarchical sparse-grid interpolation to approximate this conditional expectation. In [14], Hyndman and Oyono Ngou introduced the convolution method to approximate this conditional expectation. In [15], Fang et al. applied the convolution method based on the fast Fourier transform to solve the conditional expectations. Many results are expressed using the FFT algorithm or the fractional FFT algorithm (see [14,15,16,17,18,19,20,21]). Ge and Li [17] introduced a method that uses the fractional FFT method to solve the backwards stochastic differential equation. As we know, Fourier transforms, FFTs and fractional FFTs are commonly used tools for pricing various types of options in exponential Levy asset price models (see [22,23,24,25,26,27,28,29,30]). Inspired by the literature, we propose a method of solving BSDE using the convolution method, generalized $\theta$-scheme and fractional FFT to obtain the numerical solution to BSDE. When $\theta$ takes a certain value, we can prove that the convergence rate is high for solving BSDE.

This paper is organized as follows. In Section 2, we introduce the generalized $\theta$-scheme for BSDE, which is used in Section 3 to develop the convolution method. In Section 4, we further elaborate the fractional Fourier method and its applications to finance. In Section 5, we perform a theoretical error analysis of the proposed scheme. In Section 6, we give a numerical experiment to demonstrate the stability and effectiveness of the proposed method. In Section 7, conclusions are provided.

2. Time Discretization of BSDE

In this section, we consider Zhao et al. [7] proposition of a new type of $\theta$-scheme with four parameters. This scheme expressed high-order accuracy when ${\theta }_i=\frac{1}{2}(i=1,2,3),{\theta }_4\in \left(-\frac{1}{2},\frac{1}{2}\right)$.

We assume that $L_T^2(R^2)$ is the set of $F_t-\mathrm{measurable}$ random variables which $X:\mathsf{\Omega }\to R^2$ are square integrable, and by $H_T^2(R)$ the set of predictable processes $\eta :\mathsf{\Omega }\times [0,T]\to R$ such that

$$\mathrm{E}[{\displaystyle {\int }_0^T{\left|{\eta }_t\right|}^{2}dt}]<\infty ,$$

where $|·|$ is the standard Euclidean norm in the Euclidean space $R$. The terminal condition $X:\mathsf{\Omega }\to R^2$ in Equation (1) is $F_t-\mathrm{measurable}$ and square integrable. In addition, both the terminal condition $g$ and the driver $f$ verify the Lipschitz Condition

$$\left|g(x)-g(\overline{x})\right|+\left|f(t,x,y)-f(\overline{t},\overline{x},\overline{y})\right|<C(\left|x-\overline{x}\right|+\left|y-\overline{y}\right|+\left|z-\overline{z}\right|).$$

for some constant $C>0,\forall x,\overline{x},y,\overline{y},z,\overline{z}\in R$, and $\forall t,\overline{t}\ge 0$.

Here, we are in the position of discrete BSDE by the generalized $\theta$-scheme. Given a partition $\Delta$: $0=t_0<t_1<\cdots <t_M=T$ with time steps $\Delta t_m=t_m-t_{m-1}$, denote $X_m=X_{t_m},Y_m=Y_{t_m},Z_m=Z_{t_m}$ and $\Delta W_m=W_{t_m}-W_{t_{m-1}}$. In the time interval $[0,T]$, we rewrite BSDE (1) to the following form:

$$Y_{t_{m-1}}=Y_{t_m}+{\displaystyle {\int }_{t_{m-1}}^{t_m}f(s,Y_s,Z_s)}ds+{\displaystyle {\int }_{t_{m-1}}^{t_m}{Z}_s}d{W}_s.$$

(2)

For $m=1,\cdots ,M$. Taking the conditional expectation ${\mathrm{E}}_{m-1}^x[\cdot ]$ on both sides of (2), we get

$$Y_{m-1}={\mathrm{E}}_{m-1}^x\left[Y_m\right]+{\displaystyle {\int }_{t_{m-1}}^{t_m}{\mathrm{E}}_{m-1}^x[f(s,Y_s,Z_s)]}ds,$$

where ${\mathrm{E}}_{m-1}^x[\cdot ]={\mathrm{E}}_{t_{m-1}}^x[\cdot ]=\mathrm{E}\left[\cdot |F_{t_{m-1}}=x\right]$, $f$ is a smooth function. To simplify the presentation, we set $f_s=f(s,Y_s,Z_s)$. Then we may use numerical integration to accurately approximate the integral in (2). We use the following $\theta$-scheme to approximate this integral:

$$Y_{m-1}={\mathrm{E}}_{m-1}^x\left[Y_m\right]+{\theta }_1\Delta t_m{f}_{m-1}+\left(1-{\theta }_1\right)\Delta t_m{\mathrm{E}}_{m-1}^x\left[f_m\right]+R_y^m,$$

(3)

$$R_y^m={\displaystyle {\int }_{t_{m-1}}^{t_m}\left\{{\mathrm{E}}_{m-1}^x[f_s]-(1-{\theta }_1){\mathrm{E}}_{m-1}^x\left[f_m\right]-{\theta }_1{f}_{m-1}\right\}}ds,$$

(4)

where the parameter ${\theta }_1\in \left[0,1\right]$. Multiplying $\Delta W_m$ and taking conditional expectations on both sides of (2), we have the following identities:

$${\mathrm{E}}_{{}_{m-1}}^x[Y_m\Delta W_m]=\Delta t_m{\mathrm{E}}_{m-1}^x[\nabla Y_m]=\Delta t_m{\mathrm{E}}_{m-1}^x[Z_m].$$

(5)

We deduce

$$\begin{array}{ll}{Z}_{m-1}& ={\theta }_4{\theta }_3^{-1}{\mathrm{E}}_{m-1}^x\left[Z_m\right]+({\theta }_3-{\theta }_4){\theta }_3^{-1}\Delta t_m^{-1}{\mathrm{E}}_{m-1}^x\left[Y_m\Delta W_m\right]\\ & +(1-{\theta }_2){\theta }_3^{-1}{\mathrm{E}}_{m-1}^x\left[f_m\Delta W_m\right]+{\theta }_3^{-1}\Delta t_m^{-1}{R}_Z^m,\end{array}$$

(6)

$$R_{Z1}^m={\displaystyle {\int }_{t_{m-1}}^{t_m}{\mathrm{E}}_{m-1}^x[f_s\Delta W_m]ds-(1-{\theta }_2)\Delta t_m{\mathrm{E}}_{m-1}^x\left[f_{t_m}\Delta W_m\right]},$$

(7)

$$R_{Z2}^m=-{\displaystyle {\int }_{t_{m-1}}^{t_m}\left\{{\mathrm{E}}_{m-1}^x[z_s]-(1-{\theta }_3){\mathrm{E}}_{m-1}^x\left[z_{t_m}\right]-{\theta }_1{z}_{t_{m-1}}\right\}}ds,$$

(8)

$$R_Z^m=R_{Z1}^m+R_{Z2}^m,$$

(9)

where ${\theta }_i(i=1,2)\in \left[0,1\right],{\theta }_3\in \left(0,1\right],{\theta }_4\in \left[-1,1\right]$ constrained by $\left|{\theta }_4\right|\le {\theta }_3$. Based on (3) and (6), we obtain a discrete solution $(Y_{m-1}^{\Delta },Z_{m-1}^{\Delta })$ to approximate the solution $(Y_{m-1},Z_{m-1})$ to BSDE:

$$Y_{m-1}^{\Delta }={\mathrm{E}}_{m-1}^x\left[Y_m^{\Delta }\right]+{\theta }_1\Delta t_{m}f(t_{m-1},Y_{m-1}^{\Delta },Z_{m-1}^{\Delta })+\left(1-{\theta }_1\right)\Delta t_m{\mathrm{E}}_{m-1}^x\left[(t_m,Y_m^{\Delta },Z_m^{\Delta })\right],$$

(10)

$$\begin{array}{ll}{Z}_{m-1}^{\Delta }& ={\theta }_4{\theta }_3^{-1}{\mathrm{E}}_{m-1}^x\left[Z_m^{\Delta }\right]+({\theta }_3-{\theta }_4){\theta }_3^{-1}\Delta t_m^{-1}{\mathrm{E}}_{m-1}^x\left[Y_m^{\Delta }\Delta W_m\right]\\ & +(1-{\theta }_2){\theta }_3^{-1}{\mathrm{E}}_{m-1}^x\left[f(t_m,Y_m^{\Delta },Z_m^{\Delta })\Delta W_m\right].\end{array}$$

(11)

3. Convolution Method

In this section, due to the convolution method introduced in [14], for the numerical solution to BSDE, we consider this method that involves expressing the conditional expectations in a generalized $\theta$-scheme time discretization of BSDE. Then, it can be used to calculate the Fourier transform of the approximate solution with the convolution theorem of the Fourier analysis. Finally, one can take the inverse Fourier transform of the results to recover expressions for the approximate solution that is recursive backwards in time. To implement the convolution method, we present the discretization of intermediate quadratures and their relationship to the discrete Fourier transform; this can be efficiently computed using the fractional FFT.

3.1. Convolution on the θ-Scheme

To differentiate the literature [14], we use a $\theta$-scheme as starting point for the convolution method. If we consider the generalized $\theta$-scheme of Equations (3) and (6), an approximate solution to BSDE at mesh time $t_i$ consists of real-valued functions $b_1(x),b_2(x),b_3(x),b_4(x),b_5(x)$ defined by the conditional expectation:

$$b_1(x)={\mathrm{E}}_{m-1}^x\left[Y_m\right]={\displaystyle {\int }_{-\infty }^{+\infty }{Y}_m(v)h\left(v-x\right)}dv,$$

(12)

$$b_2(x)={\mathrm{E}}_{m-1}^x\left[f_m\right]={\displaystyle {\int }_{-\infty }^{+\infty }{f}_m(v)h\left(v-x\right)}dv,$$

(13)

$$b_3(x)={\mathrm{E}}_{m-1}^x\left[Y_m\Delta W_m\right]={\displaystyle {\int }_{-\infty }^{+\infty }{Y}_m(v)(v-x)h\left(v-x\right)}dv,$$

(14)

$$b_4(x)={\mathrm{E}}_{m-1}^x\left[f_m\Delta W_m\right]={\displaystyle {\int }_{-\infty }^{+\infty }{f}_m(v)(v-x)h\left(v-x\right)}dv,$$

(15)

$$b_5(x)={\mathrm{E}}_{m-1}^x\left[Z_m\right]={\displaystyle {\int }_{-\infty }^{+\infty }{Z}_m(v)h\left(v-x\right)}dv,$$

(16)

where the function $h(x)$ is the density function of $W_{t_{i+1}}$ conditional on the value of $W_{t_i}$

$$h(x)=\frac{1}{\sqrt{2\pi \Delta t_m}}{e}^{-\frac{x^2}{2\Delta t_m}}.$$

Lord et al. [15] explained that the method relies heavily on the Fourier transformation. We know that the method to calculate the integrals (12)–(16) is available. Hyndman and Oyono Ngou [14] also stated that these convolutions suggest using Fourier transforms and the computation of the integrals via discrete Fourier transforms.

Let us introduce the Fourier transform. Denote

$$F\left[f(x)\right](\xi ):=\widehat{f}(\xi )={\displaystyle {\int }_{-\infty }^{+\infty }{e}^{-i\xi x}}f(x)dx.$$

so that the inverse Fourier transform recovers the function $f(x)$ from its Fourier transform $\widehat{f}(\xi )$ through the relation

$$f(x):=F^{-1}\left[\widehat{f}(\xi )\right](x)=\frac{1}{2\pi }{\displaystyle {\int }_{-\infty }^{+\infty }{e}^{i\xi x}}\widehat{f}(\xi )d\xi .$$

We define the damped function $f^{\alpha }(x),\alpha \in R$

$$f^{\alpha }(x)=e^{-\alpha x}f(x).$$

The reason we introduce a damping function is that we must guarantee the integration of the Fourier transform according to Lord et al. [15]. Many studies in finance have found that the numerical stability of the option price can be controlled using the damping coefficient $\alpha$ (see [15,21,26]).

Then, we obtain the Fourier transform of $b_1{}^{\alpha }(x)$ in Equation (12) with the dampening parameter.

$$\begin{array}{ll}{\widehat{b}}_1{}^{\alpha }(x)(\xi )& ={\displaystyle {\int }_{-\infty }^{+\infty }{e}^{-i\xi x}}{e}^{-\alpha x}{\displaystyle {\int }_{-\infty }^{+\infty }{Y}_m}(v)h(v-x)dvdx\\ & \stackrel{v-x=-z}{=}{\displaystyle {\int }_{-\infty }^{+\infty }{e}^{-i\xi v}}{Y}_m^{\alpha }(v)dv{\displaystyle {\int }_{-\infty }^{+\infty }{e}^{-\alpha z}}{e}^{-i\xi z}h(-z)dz\\ & ={\widehat{Y}}_m^{\alpha }(v)(\xi )\cdot \left[e^{-\alpha z}\widehat{h}(-z)\right](\xi )\\ & ={\widehat{Y}}_m^{\alpha }(v)(\xi )\cdot \varphi \left[(\xi -i\alpha )\right].\end{array}$$

(17)

The proof of $\left[e^{-\alpha z}\widehat{h}(-z)\right]$ refers to Hyndman and Oyono Ngou [14]. $\varphi (\xi )=e^{-\frac{1}{2}\Delta t_m{\xi }^2}$ is the characteristic function of the density $h$.

Similarly, the Fourier transform of $b_2{}^{\alpha }(x),b_5{}^{\alpha }(x)$ Equations (13) and (16) are given by:

$${\widehat{b}}_2{}^{\alpha }(x)(\xi )={\widehat{f}}_m^{\alpha }(\xi )\cdot \varphi (\xi -i\alpha ),$$

(18)

$${\widehat{b}}_5{}^{\alpha }(x)(\xi )={\widehat{z}}_m^{\alpha }(\xi )\cdot \varphi (\xi -i\alpha ).$$

(19)

According to this description, we can obtain the Fourier transform of $b_3{}^{\alpha }(x),b_4{}^{\alpha }(x)$ in Equations (14) and (15):

$$\begin{array}{ll}{\widehat{b}}_3{}^{\alpha }(x)(\xi )& ={\displaystyle {\int }_{-\infty }^{+\infty }{e}^{-i\xi x}}{e}^{-ax}{\displaystyle {\int }_{-\infty }^{+\infty }{Y}_m(v)\cdot }(v-x)\cdot h(v-x)dvdx\\ & \stackrel{v-x=-z}{=}{\displaystyle {\int }_{-\infty }^{+\infty }{e}^{-i\xi v}}{Y}_m{}^{\alpha }(v)dv{\displaystyle {\int }_{-\infty }^{+\infty }{e}^{-az}{e}^{-i\xi z}\cdot }(-z)\cdot h(-z)dz\\ & =i\Delta t_m(i\alpha -\xi )\cdot \left[{\widehat{Y}}_m{}^{\alpha }(v)\right](\xi )\cdot \varphi (\xi -i\alpha ).\end{array}$$

(20)

where the proof of ${\displaystyle {\int }_{-\infty }^{+\infty }{e}^{-az}{e}^{i\xi z}\cdot }(-z)\cdot h(-z)dz$ is the following:

$$\begin{array}{ll}{\displaystyle {\int }_{-\infty }^{+\infty }{e}^{-az}{e}^{i\xi z}\cdot }(-z)\cdot h(-z)dz& ={\displaystyle {\int }_{-\infty }^{+\infty }{e}^{-az}{e}^{-i\xi z}\cdot }(-z)\cdot \frac{1}{\sqrt{2\pi \Delta t_m}}\cdot e^{-\frac{z^2}{2\Delta t_m}}dz\\ & =\frac{\Delta t_m}{\sqrt{2\pi \Delta t_m}}{\displaystyle {\int }_{-\infty }^{+\infty }{e}^{-az}{e}^{-i\xi z}}d{e}^{-\frac{z^2}{2\Delta t_m}}\\ & =\frac{\Delta t_m}{\sqrt{2\pi \Delta t_m}}{\displaystyle {\int }_{-\infty }^{+\infty }{e}^{i(\xi -i\alpha )(-z)}}d{e}^{-\frac{z^2}{2\Delta t_m}}\\ & =-\frac{\Delta t_m}{\sqrt{2\pi \Delta t_m}}{\displaystyle {\int }_{-\infty }^{+\infty }{e}^{i(\xi -i\alpha )(-z)}}\cdot i(\xi -i\alpha )\cdot e^{-\frac{z^2}{2\Delta t_m}}dz\\ & =i\Delta t_m\cdot (i\alpha -\xi )\cdot \varphi (\xi -i\alpha ).\end{array}$$

Because $\left|e^{i(\xi -i\alpha )(-z)}\right|<\infty$ for any $\alpha \in R$, in the numerical implementation integrability is not necessary for the damping parameter.

Similarly,

$${\widehat{b}}_4^{\alpha }(x)(\xi )={\widehat{f}}_m^{\alpha }(\xi )\cdot i\Delta t_m\cdot (i\alpha -\xi )\cdot \varphi (\xi -i\alpha ).$$

(21)

From Equations (17)–(21), we use the inverse Fourier transform and adjust for the dampening parameter to obtain a new type of $b_1(x),b_2(x),b_3(x),b_4(x),b_5(x)$:

$$b_1(x)=e^{\alpha x}{F}^{-1}\left[{\widehat{Y}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)\right],$$

(22)

$$b_2(x)=e^{\alpha x}{F}^{-1}\left[{\widehat{f}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)\right],$$

(23)

$$b_3(x)=e^{\alpha x}{F}^{-1}\left[i\Delta t_m(i\alpha -\xi )\cdot \left[{\widehat{Y}}_m{}^{\alpha }(v)\right](\xi )\cdot \varphi \left(\xi -i\alpha \right)\right],$$

(24)

$$b_4(x)=e^{\alpha x}{F}^{-1}\left[i\Delta t_m\cdot (i\alpha -\xi )\cdot {\widehat{f}}_m^{\alpha }(\xi )\cdot \varphi (\xi -i\alpha )\right],$$

(25)

$$b_5(x)=e^{\alpha x}{F}^{-1}\left[{\widehat{z}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)\right].$$

(26)

Equations (22)–(26) are substituted into Equations (10) and (11), producing a new type of numerical solution:

$$\begin{array}{ll}{Y}_{m-1}^{\Delta }& =e^{\alpha x}{F}_{\kappa }^{-1}\left[{\widehat{Y}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)\right]+{\theta }_1\Delta t_{m}f(t_{m-1},Y_{m-1}^{\Delta },Z_{m-1}^{\Delta })\\ & +\left(1-{\theta }_1\right)\Delta t_m\left[e^{\alpha x}{F}_{\kappa }^{-1}\left[{\widehat{f}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)\right]\right],\end{array}$$

(27)

$$\begin{array}{ll}{Z}_{m-1}^{\Delta }& ={\theta }_4{\theta }_3^{-1}\left[e^{\alpha x}{F}_{\kappa }^{-1}\left[{\widehat{z}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)\right]\right]\\ & +({\theta }_3-{\theta }_4){\theta }_3^{-1}\left[e^{\alpha x}{F}_{\kappa }^{-1}\left[i(i\alpha -\xi )\cdot \left[{\widehat{Y}}_m{}^{\alpha }(v)\right](\xi )\cdot \varphi (\xi -i\alpha )\right]\right]\\ & +(1-{\theta }_2){\theta }_3^{-1}\Delta t_m\left[e^{\alpha x}{F}_{\kappa }^{-1}\left[i\cdot (i\alpha -\xi )\cdot {\widehat{f}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)\right]\right].\end{array}$$

(28)

3.2. Numerical Implementation

To solve BSDEs, based on the Equations (27) and (28), the key to the equations is the value of the conditional expectations. Therefore, we consider $b_1(x),b_2(x),$$b_3(x),b_4(x),b_5(x)$. For $b_1(x)$, we take the following measures, and $b_2(x),b_3(x),b_4(x),b_5(x)$ are similar.

For $b_1(x)$, we recall Equation (22):

$$b_1(x)=e^{\alpha x}{F}_{\kappa }^{-1}\left[{\widehat{Y}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)\right].$$

Now, we let

$$c_1(x)=F_{\kappa }^{-1}\left[{\widehat{Y}}_m^{\alpha }(\xi )\cdot \varphi \left(i\alpha -\xi \right)\right]=\frac{1}{2\pi }{\displaystyle {\int }_{-\infty }^{+\infty }{e}^{i\xi x}}{\widehat{Y}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)d\xi ,$$

(29)

$${\widehat{Y}}_m^{\alpha }(\xi )={\displaystyle {\int }_{-\infty }^{+\infty }{e}^{-ix\xi }}{Y}_m^{\alpha }(x)dx.$$

(30)

We now focus on Equations (29) and (30). To use the fractional FFT, we discretize the convolution in (29) and (30).

3.3. Discretising the Convolution

We approximate both integrals in (29) and (30) by a discrete sum, such that the fractional FFT algorithm can be used for their computation, which necessitates the use of uniform grids for $\xi ,x$:

$${\xi }_k={\xi }_0+k\Delta \xi ,$$

(31)

$$x_j=x_0+j\Delta x,$$

(32)

where ${\xi }_0=-\frac{L}{2},\Delta \xi =\frac{L}{N},x_0=-\frac{l}{2},\Delta x=\frac{l}{N},k=0,\cdots ,N-1,j=0,\cdots ,N-1$. In addition, the Nyquist relation $Ll=2\pi N$ is satisfied. However, in principle, we could use the fractional FFT algorithm, which does not require the Nyquist relation to be satisfied. Substituting (30) into (29), the integral is numerically computed by discretizing the Fourier space with a uniform grid of $N+1$ points ${\left\{{\xi }_k\right\}}_{k=0}^N$ on the interval $\left[-\frac{L}{2},\frac{L}{2}\right]$ of length $L$, where we have fixed $L>0$. In addition, we approximate (30) with a general Newton–Cotes rule:

$$\begin{array}{ll}{c}_1(x)& =F^{-1}\left[{\widehat{Y}}_m^{\alpha }(\xi )\cdot \varphi \left(i\alpha -\xi \right)\right]=\frac{1}{2\pi }{\displaystyle {\int }_{-\infty }^{+\infty }{e}^{i\xi x}{\widehat{Y}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)d\xi }\\ & \approx -\frac{1}{2\pi }{\displaystyle {\int }_{-L/2}^{L/2}{e}^{i\xi x}{\widehat{Y}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)d\xi }\\ & =\frac{\Delta \xi }{2\pi }{\displaystyle \sum _{k=0}^{\infty }{e}^{i{\xi }_{k}x}{\widehat{Y}}_m^{\alpha }(\xi )\cdot \varphi \left({\xi }_k-i\alpha \right)}.\end{array}$$

We let

$$c_1^{\prime }(x)=\frac{\Delta \xi }{2\pi }{\displaystyle \sum _{k=0}^{\infty }{e}^{i{\xi }_{k}x}{\widehat{Y}}_m^{\alpha }(\xi )\cdot \varphi \left({\xi }_k-i\alpha \right)}.$$

(33)

By the approximation of the Equation (33) with the left-rectangle rule yields, we have

$$c_1^{″}(x)=\frac{\Delta \xi }{2\pi }{\displaystyle \sum _{k=0}^{N-1}{e}^{i{\xi }_{k}x}\widehat{Y_m^{\alpha }({\xi }_k)}\cdot \varphi \left(i\alpha -{\xi }_k\right)}.$$

(34)

For Equation (30), this integral is also computed using a uniform grid of $N+1$ points ${\left\{x_j\right\}}_{j=0}^N$ on the restricted interval $\left[x_0,x_N\right]$. When we use the trapezoidal rule, we choose the weights $w_n$ as follows:

$$w_0=\frac{1}{2},w_N=\frac{1}{2},w_n=1, \mathrm{for} n=1,\cdots ,N-2.$$

(35)

Further, by approximating (30) with a general Newton–Cotes rule, we can define the DFT and its inverse of a sequence $x_p$, as follows:

$${\widehat{x}}_p:=D\left[x_p\right]=\frac{1}{N}{\displaystyle \sum _{j=0}^N{e}^{-ijp\frac{2\pi }{N}}}{x}_j,$$

(36)

$$x_p=D^{-1}{\left[\hat{x}\right]}_p=\frac{1}{N}{\displaystyle \sum _{j=0}^{N-1}{e}^{ijp\frac{2\pi }{N}}{\widehat{x}}_j}.$$

(37)

We have

$$c_1^{″}(x)=\frac{\Delta \xi }{2\pi }{\displaystyle \sum _{k=0}^{N-1}{e}^{i{\xi }_{k}x}{\widehat{Y}}_m^{\alpha }(\xi )\cdot \varphi \left({\xi }_k-i\alpha \right)},$$

$$\begin{array}{ll}{\widehat{Y}}_m^{\alpha }(\xi )& ={\displaystyle {\int }_{-\infty }^{+\infty }{e}^{-ix{\xi }_k}{Y}_m^{\alpha }(x)dx}\\ & \approx {\displaystyle {\int }_{x_0}^{x_N}{e}^{-ix{\xi }_k}{Y}_m^{\alpha }(x)dx}\\ & =\Delta x{\displaystyle \sum _{j=0}^{\infty }{w}_j{e}^{-i{x}_j{\xi }_k}{Y}_m^{\alpha }(x_j)}\\ & =\Delta x{\displaystyle \sum _{j=0}^{\infty }{w}_j{e}^{-i(x_0+j\Delta x)({\xi }_0+k\Delta \xi )}{Y}_m^{\alpha }(x_j)}\\ & =\Delta x{e}^{-i{x}_0{\xi }_k}{\displaystyle \sum _{j=0}^{\infty }{w}_j{e}^{-ij\Delta x{\xi }_0}{e}^{-ijk\frac{2\pi }{N}}{Y}_m^{\alpha }(x_j)}.\end{array}$$

(38)

We can insert (38) into (34), and then deduce

$$\begin{array}{ll}{c}_1^{‴}(x)& \approx \frac{\Delta \xi }{2\pi }{\displaystyle \sum _{k=0}^{N-1}{e}^{i{\xi }_k{x}_j}\cdot \varphi \left({\xi }_k-i\alpha \right)\Delta x{e}^{-i{x}_0{\xi }_k}{\displaystyle \sum _{j=0}^{N-1}{w}_j{e}^{-ij\Delta x{\xi }_0}{e}^{-ijk\frac{2\pi }{N}}{Y}_m^{\alpha }(x_j)}}\\ & \stackrel{e^{-i\Delta x{\xi }_0}=-1}{=}{\displaystyle \sum _{k=0}^{N-1}{e}^{i{\xi }_k{x}_j}\cdot \varphi \left({\xi }_k-i\alpha \right)\cdot e^{-i{x}_0{\xi }_k}D{\left[{\left\{{(-1)}^j{w}_j{Y}_m^{\alpha }(x_j)\right\}}_{j=0}^{N-1}\right]}_k}\\ & ={\displaystyle \sum _{k=0}^{N-1}{e}^{i{\xi }_k(x_j-x_0)}\cdot \varphi \left({\xi }_k-i\alpha \right)\cdot D{\left[{\left\{{(-1)}^j{w}_j{Y}_m^{\alpha }(x_j)\right\}}_{j=0}^{N-1}\right]}_k}\\ & ={(-1)}^j{\displaystyle \sum _{k=0}^{N-1}{e}^{ijk\frac{2\pi }{N}}\cdot \varphi \left({\xi }_k-i\alpha \right)\cdot D{\left[{\left\{{(-1)}^j{w}_j{Y}_m^{\alpha }(x_j)\right\}}_{j=0}^{N-1}\right]}_k}\\ & ={(-1)}^j{D}^{-1}{\left[{\left\{\varphi \left({\xi }_k-i\alpha \right)\cdot D{\left[{\left\{{(-1)}^j{w}_j{Y}_m^{\alpha }(x_j)\right\}}_{j=0}^{N-1}\right]}_i\right\}}_{i=0}^{N-1}\right]}_k.\end{array}$$

(39)

Other conditional expectations are also similar.

Remark 1. 

To solve Equation (27), the most difficult problem is the implicit value of $Y$inside the nonlinear function $f$. Ge et al. [17] solved this problem using the Picard iteration.

We also use this method:

$$\begin{array}{ll}{Y}_{m-1}^{p+1}& =e^{\alpha x}{F}^{-1}\left[{\widehat{Y}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)\right]+{\theta }_1\Delta t_{m}f(t_{m-1},Y_{m-1}^p,Z_{m-1})\\ & +\left(1-{\theta }_1\right)\Delta t_m\left[e^{\alpha x}{F}^{-1}\left[{\widehat{f}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)\right]\right].\end{array}$$

In addition, Ge et al. [17] in Theorem 2 prove that the Picard iteration converges to $Y$.

4. Fractional FFT

The fractional Fourier transform was introduced to the signal processing literature by Bailey and Swarztrauber [31]. Here, we apply the fractional Fourier transform to compute the conditional expectation. In addition, we also know that this method can be applied to the pricing of financial options. In this section, we illustrate the idea of the fractional Fourier transform and its application to European call options.

First, we review the fractional Fourier transform (see [32,33]). The fractional Fourier transform can be defined from different perspectives. At present, the most commonly used definition is given from the perspective of integral transformation and from the perspective of feature decomposition. We give a feature decomposition of FrFT

$$\begin{array}{ll}{F}^{\sigma }{\phi }_n(t)& ={\lambda }_n^{\sigma }{\phi }_n(t)\\ & =e^{-jn\sigma }{\phi }_n(t),\end{array}$$

where ${\phi }_n(t)$ represents the Hermite–Gauss function, ${\lambda }_n$ represents the eigenvalue, and $\sigma =\frac{\pi p}{2}$ represents the rotation angle of the transformed signal for FrFT. On the integral definition of the fractional Fourier transform

$$F^{\sigma }[x(t)]={\displaystyle {\int }_{-\infty }^{+\infty }{K}_{\sigma }(u,t)x(t)dt}=B_{\sigma }{\displaystyle {\int }_{-\infty }^{+\infty }\exp [j\frac{t^2+u^2}{2}\cot \sigma -\frac{jtu}{\sin \sigma }]x(t)dt,}$$

where $K_{\sigma }(u,t)$ represents a kernel, $B_{\sigma }=\sqrt{\frac{1-j\cot \sigma }{2\pi }}$, $\sigma =\frac{\pi p}{2}$ is not a multiple of $\pi$, and when $p=1,\sigma =\frac{\pi }{2}$, $F^{\frac{\pi }{2}}[x(t)]={\displaystyle {\int }_{-\infty }^{+\infty }\exp [-j2\pi ut]x(t)dt}$ becomes the traditional Fourier transform.

Remark 2. 

Whether it is the fractional Fourier transform or a fast Fourier transform, it is essentially a Fourier transform. The difference between the two is that the fractional Fourier transform takes the axis in the time-frequency plane and rotates it counterclockwise about the origin by an arbitrary Angle, and the choice Angle is represented as a fraction. In [33], the literature indicated that the fractional Fourier transform and the discrete fractional Fourier transform in detail.

In [17], Ge et al. converted the Fourier transform into a fractional Fourier transform. Hence, we also will convert the Fourier transform into the fractional Fourier transform. We should consider finding the integration

$$\begin{array}{ll}{\widehat{g}}^{\alpha }({\xi }_k)& ={\displaystyle {\int }_{-\frac{L}{2}}^{\frac{L}{2}}{e}^{-ix{\xi }_k}{g}^{\alpha }(x)dx}\\ & \approx △x{\displaystyle \sum _{j=0}^{N-1}{w}_j{e}^{-i{x}_j{\xi }_k}{g}^{\alpha }(x_j)},\end{array}$$

(40)

by Formula (31) and (32), Formula (40) is derived. We have

$$\begin{array}{ll}{\widehat{g}}^{\alpha }({\xi }_k)& \approx \Delta x{\displaystyle \sum _{j=0}^{N-1}{w}_j{e}^{-i{x}_j{\xi }_k}{g}^{\alpha }(x_j)}\\ & =\Delta x{\displaystyle \sum _{j=0}^{N-1}{w}_j{e}^{-i(-\frac{l}{2}+j\Delta x)(-\frac{L}{2}+k\Delta \xi )}{g}^{\alpha }(x_j)}.\end{array}$$

In order to make the above formula simpler, we simplify it

$$\begin{array}{ll}{\widehat{g}}^{\alpha }({\xi }_k)& =\Delta x{\displaystyle \sum _{j=0}^{N-1}{w}_j{e}^{-i(x_0+j\Delta x)({\xi }_0+k\Delta \xi )}{g}^{\alpha }(x_j)}\\ & =\Delta x{e}^{-i{x}_0{\xi }_0}{e}^{-ik{x}_0\Delta \xi }{\displaystyle \sum _{j=0}^{N-1}{w}_j{e}^{-ij\Delta x{\xi }_0}{e}^{-ijk\Delta x\Delta \xi }{g}^{\alpha }(x_j)}\\ & =\Delta x{e}^{-i{x}_0{\xi }_0}{e}^{-ik{x}_0\Delta \xi }{\displaystyle \sum _{j=0}^{N-1}(e^{-ijk\Delta x\Delta \xi })·(w_j{e}^{-ij\Delta x{\xi }_0}{g}^{\alpha }(x_j)})\\ & =\Delta x{e}^{-i{x}_0{\xi }_0}{e}^{-ik{x}_0\Delta \xi }{\displaystyle \sum _{j=0}^{N-1}{e}^{-2\pi ijk\sigma }·g*},\end{array}$$

(41)

where $g*=w_j{e}^{-ij\Delta x{\xi }_0}{g}^{\alpha }(x_j)$, $\sigma =\frac{\Delta x\Delta \xi }{2\pi }$.

Now, we define a vector [20]:

$$a=\left(e^{i{0}^2\sigma },e^{i{1}^2\sigma },\cdots ,e^{i{(N-1)}^2\sigma },0,\cdots ,0,e^{i{(N-1)}^2\sigma },\cdots ,e^{i{1}^2\sigma }\right),$$

where $a$ is a $\left(N^{*}\times 1\right)$ vector and it is a power of 2. Let

$$\tilde{g}=\left(e^{-i{0}^2\sigma }g*,e^{-i{1}^2\sigma }g*,\cdots ,e^{-i{(N-1)}^2\sigma }g*,0,\cdots ,0\right),$$

where $\tilde{g}$ is a $\left(N^{*}\times 1\right)$ vector. We rewrite the integration

$$\widehat{g}=\Delta x{e}^{-i{x}_0{\xi }_k}{e}^{-ij\Delta x{\xi }_0}{e}^{-i{k}^2\sigma }(F^{-1}(F(a)·F(\tilde{g}))).$$

(42)

Remark 3. 

In Equation (41), the Fourier transform exists in a fractional discrete Fourier transform. The sum of all points on the $\xi$-grid can be calculated efficiently by using the fractional fast Fourier transform algorithm. In addition, in [31], the fractional Fourier transform algorithm was given, and Bailey and Swarztrauber indicated that the fractional Fourier transform algorithms are faster than the conventional Fourier transform methods.

Next, we address the application of fast Fourier to option pricing [29]. Take the BS European call option as an example (see [27]). The BS European-type call option pricing formula is as follows [28]:

$$c=SN(d_1)-X{e}^{-r(T-t)}N(d_2),$$

where

$$\begin{array}{l}{d}_1=(\ln (S/N)+(r+{\sigma }^2/2)(T-t))/(\sigma \sqrt{T-t}),\\ d_2=d_1-(\sigma \sqrt{T-t}).\end{array}$$

The option pricing formula under risk neutral measure is

$$C_T(k)=e^{-rT}{\displaystyle {\int }_k^{+\infty }(e^x-e^k)q_T(x)dx},$$

where $C_T(k)$ is the option price of the European call option; $T$ is the maturity time; $S_T$ is the underlying asset price on the maturity date; $K$ is the execution price, and $q_T$ is the density function.

Introducing the damping parameter $\alpha$, let $C_T(k)=e^{\alpha x}{C}_T(k)$, and we obtain

$${\phi }_T(\alpha )={\displaystyle {\int }_{-\infty }^{+\infty }{e}^{iuk}{C}_T(k)dk}.$$

We perform a Fourier transform to this equation. The characteristic under the risk neutral measure is

$${\varphi }_T(u)={\displaystyle {\int }_{-\infty }^{+\infty }{e}^{iux}{q}_T(x)dx},$$

where the characteristic function is determined according to the specific process $S_T$. Therefore, the VG process was chosen in this paper (see [21,30]). We have

$$C_T(k)=\frac{e^{-\alpha k}}{2\pi }{\displaystyle {\int }_0^{+\infty }{e}^{iuk}\frac{e^{-rT}{\varphi }_T(u-(\alpha +1)i)}{(\alpha +iu)(\alpha +iu+1)}}du.$$

(43)

With Formula (41), using the fractional Fourier transform to discretize the formula, we can obtain

$$C_T(k)=\frac{e^{-\alpha k}}{\pi }{\displaystyle \sum _{j=0}^{N-1}{h}_j{e}^{i{u}_{j}k}\frac{e^{-rT}{\varphi }_T(u_j-(\alpha +1)i)}{(\alpha +i{u}_j)(\alpha +i{u}_j+1)}\delta {\eta }_j},$$

where $u_j=\delta {\eta }_j$, $\delta$ is the interval, ${\eta }_j=\{\begin{array}{l}1/2,j=0,or,i=N-1,\\ 1,j=1,2,\cdots ,N-2.\end{array}$

Remark 4. 

When using the fractional Fourier approach [21], note:

$$h_j=e^{ij\delta k_0}\frac{e^{-rT}{\varphi }_T(u_j-(\alpha +1)i)}{(\alpha +i{u}_j)(\alpha +i{u}_j+1)}\delta ,p=\lambda \delta /2\pi ,$$

$$\left|\mathrm{Re}\left(\frac{e^{-rT}{\varphi }_T(u_j-(\alpha +1)i)}{(\alpha +i{u}_j)(\alpha +i{u}_j+1)}\right)\right|\le \left|\frac{e^{-rT}{\varphi }_T(u_j-(\alpha +1)i)}{(\alpha +i{u}_j)(\alpha +i{u}_j+1)}\right|,$$

$$\left|\mathrm{Im}\left(\frac{e^{-rT}{\varphi }_T(u_j-(\alpha +1)i)}{(\alpha +i{u}_j)(\alpha +i{u}_j+1)}\right)\right|\le \left|\frac{e^{-rT}{\varphi }_T(u_j-(\alpha +1)i)}{(\alpha +i{u}_j)(\alpha +i{u}_j+1)}\right|.$$

5. Error Analysis

For simplicity, we only consider one-dimensional BSDEs. To obtain error estimates for the proposed method, we first introduce the estimates of $R_Y^n$ and $R_z^n$ defined in (3) and (6) in the following lemma (see [7]).

Lemma 1. 

Let $R_Y^n$and $R_z^n$be the truncation resulting from (3) and (6). Then, for a sufficiently small time step $\Delta t_m$, we have the following conclusions.

• For parameters ${\theta }_i\in \left[0,1\right](i=1,2)$and ${\theta }_3\in \left(0,1\right]$, if $\phi \in C_b^2$and $f(t,y,z)\in C_b^{1,2,2}$, then we have the estimates $\left|R_Y^n\right|\le C{\left(\Delta t_m\right)}^2,\left|R_z^n\right|\le C{\left(\Delta t_m\right)}^2$.
• In particular, for ${\theta }_i=\frac{1}{2}(i=1,2,3)$, if $\phi \in C_b^3$and $f(t,y,z)\in C_b^{2,4,4}$, we have $\left|R_Y^n\right|\le C{\left(\Delta t_m\right)}^3,\left|R_z^n\right|\le C{\left(\Delta t_m\right)}^3$.

where $C$is positive depending only on $T$and the upper bounds of the derivatives of the functions $f$and $\phi$.

The proposed method includes two primary types of error. The time discretization error $R_Y^n$ and $R_z^n$ can be controlled by $\Delta t_m$ (for details, see [7]). Thus, we primarily focus on this error term, which estimates truncation errors using the error analysis of [25]. This analysis is based on a Fourier series expansion of the damped value which is the conditional expectation. We now consider the problem of computational error, and the following situation.

We thus obtain a numerical solution $(Y_{m-1}^{\Delta },Z_{m-1}^{\Delta })$ to approximate the solution $(Y_{m-1},Z_{m-1})$ to BSDE. We assume that the conditional expectation solved numerically is denoted as ${\widehat{\mathrm{E}}}_{m-1}^x[\cdot ]$, thus we write the error types:

$$\begin{array}{ll}{e}_Y& =\left|{\mathrm{E}}_{m-1}^x\left[Y_m\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[Y_m\right]\right|+{\theta }_1\Delta t_m\left|f(t_{m-1},Y_{m-1},Z_{m-1})-f(t_{m-1},Y_{m-1}^{\Delta },Z_{m-1}^{\Delta })\right|\\ & +\left(1-{\theta }_1\right)\Delta t_m\left|{\mathrm{E}}_{m-1}^x\left[f_{t_m}\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[f_{t_{m-1}}\right]\right|+R_y^m,\end{array}$$

(44)

$$\begin{array}{ll}{e}_Z& ={\theta }_4{\theta }_3^{-1}\left|{\mathrm{E}}_{m-1}^x\left[Z_m\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[Z_m\right]\right|+({\theta }_3-{\theta }_4){\theta }_3^{-1}\Delta t_m^{-1}\left|{\mathrm{E}}_{m-1}^x\left[Y_m\Delta W_m\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[Y_m\Delta W_m\right]\right|\\ & +(1-{\theta }_2){\theta }_3^{-1}\left|{\mathrm{E}}_{m-1}^x\left[f_m\Delta W_m\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[f_m\Delta W_m\right]\right|+{\theta }_3^{-1}\Delta t_m^{-1}{R}_Z^m.\end{array}$$

(45)

Here, we let:

$$I_1=\left|{\mathrm{E}}_{m-1}^x\left[Y_m\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[Y_m\right]\right|,$$

$$I_2={\theta }_1\Delta t_m\left|f(t_{m-1},Y_{m-1},Z_{m-1})-f(t_{m-1},Y_{m-1}^{\Delta },Z_{m-1}^{\Delta })\right|,$$

$$I_3=\left(1-{\theta }_1\right)\Delta t_m\left|{\mathrm{E}}_{m-1}^x\left[f_{t_m}\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[f_{t_{m-1}}\right]\right|,$$

$$I_1^{\prime }={\theta }_4{\theta }_3^{-1}\left|{\mathrm{E}}_{m-1}^x\left[Z_m\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[Z_m\right]\right|,$$

$$I_2^{\prime }=({\theta }_3-{\theta }_4){\theta }_3^{-1}\Delta t_m^{-1}\left|{\mathrm{E}}_{m-1}^x\left[Y_m\Delta W_m\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[Y_m\Delta W_m\right]\right|,$$

$$I_3^{\prime }=(1-{\theta }_2){\theta }_3^{-1}\left|{\mathrm{E}}_{m-1}^x\left[f_m\Delta W_m\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[f_m\Delta W_m\right]\right|.$$

Therefore, the error term is defined by:

$$e_Y=I_1+I_2+I_3+R_y^m, e_Z=I^{\prime }+I_2^{\prime }+I_3^{‴}+{\theta }_3^{-1}\Delta t_m^{-1}{R}_Z^m.$$

(46)

We now consider $I_1$, recalling the numerical implementation of Section 3. The first approximation we make replaces $c_1(x)$ by its Fourier series expansion on $\left[-\frac{L}{2},\frac{L}{2}\right]$, where we have fixed $L>0$. Therefore, the error of the first part is defined as:

$$\left|c_1(x)-c_1^{\prime }(x)\right|=\left|\frac{1}{2\pi }{\displaystyle \underset{R\backslash [-L\backslash 2,L\backslash 2]}{\int }{e}^{i\xi x}{\widehat{Y}}_m^{\alpha }(\xi )\cdot \varphi \left(\xi -i\alpha \right)d\xi }\right|\sim O(\Delta \xi ).$$

(47)

Many studies have stated that the error can be controlled by the partition $\Delta \xi$ (see [7,8,9]). The error of the second part of $I_1$ is $\left|c_1^{\prime }(x)-c_1^{″}(x)\right|$, which is defined as:

$$\begin{array}{ll}\left|c_1^{\prime }(x)-c_1^{″}(x)\right|& =\frac{\Delta \xi }{2\pi }\left|{\displaystyle \sum _{k=0}^{\infty }{e}^{i{\xi }_{i}x}{Y}_m^{\alpha }({\xi }_k)\cdot \varphi \left({\xi }_k-i\alpha \right)}-{\displaystyle \sum _{k=0}^{N-1}{e}^{i{\xi }_{k}x}{\widehat{Y}}_m^{\alpha }({\xi }_k)\cdot \varphi \left({\xi }_k-i\alpha \right)}\right|\\ & =\frac{\Delta \xi }{2\pi }\left|{\displaystyle \sum _{k=N}^{\infty }{e}^{i{\xi }_{i}x}{\widehat{Y}}_m^{\alpha }({\xi }_k)\cdot \varphi \left({\xi }_k-i\alpha \right)}\right|.\end{array}$$

Lee [26] stated that the dampening value by the Fourier transform is restricted by two cases. Therefore, we use these conclusions.

Theorem 1. 

When $\left|{\widehat{Y}}_m^{\alpha }({\xi }_k)\right|\le \varphi (\xi )/{\xi }^{1+\beta }$, and $\varphi (\xi )$is decreasing in $\xi$with $\beta >0$. We have

$$\left|c_1^{\prime }(x)-c_1^{″}(x)\right|\le \frac{C\varphi \left(N\right)}{N^{{\beta }_1}}.$$

(48)

Proof: 

It can be seen from the assumptions that

$$\begin{array}{ll}\left|c_1^{\prime }(x)-c_1^{″}(x)\right|& =\frac{\Delta \xi }{2\pi }\left|{\displaystyle \sum _{k=N}^{\infty }{e}^{i{\xi }_{k}x}{\widehat{Y}}_m^{\alpha }({\xi }_k)\cdot \varphi \left({\xi }_k-i\alpha \right)}\right|\\ & \le \frac{\Delta \xi }{2\pi }{\displaystyle \sum _{k=N}^{\infty }\left|e^{i{\xi }_{k}x}{\widehat{Y}}_m^{\alpha }({\xi }_k)\cdot \varphi \left({\xi }_k-i\alpha \right)\right|}\\ & \le \frac{\Delta \xi }{2\pi }{\displaystyle \sum _{k=N}^{\infty }\left|\frac{\varphi ({\xi }_k)}{{({\xi }_k)}^{1+{\beta }_1}}\right|}\\ & \le C\varphi (N){\displaystyle {\int }_N^{\infty }\frac{d({\xi }_k)}{{({\xi }_k)}^{1+{\beta }_1}}}\\ & =\frac{C\varphi (N)}{N^{{\beta }_1}}.\end{array}$$

□

Now, we begin to analyze the last part of the error $I_1$. According to Section 3, we have $\left|c_1^{″}(x)-c_1^{‴}(x)\right|$. We deduce

$$\left|c_1^{″}(x)-c_1^{‴}(x)\right|=\frac{\Delta \xi }{2\pi }\left|{\displaystyle \sum _{k=0}^{N-1}{e}^{i{\xi }_k{x}_j}\cdot \varphi \left(i\alpha -{\xi }_k\right)}\cdot {\displaystyle \sum _{j=N}^{\infty }\Delta x\cdot w_j{e}^{-ij{\xi }_0\Delta x}{e}^{-ijk\frac{2\pi }{N}}{y}_m^{\alpha }(x_j)}\right|$$

Based on these conclusions, we obtain the last part of the error $I_1$. We consider $\left|{\displaystyle \sum _{j=N}^{\infty }\Delta x\cdot w_j{e}^{-ij{\xi }_0\Delta x}{e}^{-ijk\frac{2\pi }{N}}{y}_m^{\alpha }(x_j)}\right|$, because we choose the Newton–Cotes rule of $O\left(N^{-{\beta }_2}\right)$ by the article of Fang et al. [15]. Because $\left|{\displaystyle \sum _{j=N}^{\infty }\Delta x\cdot w_j{e}^{-ij{\xi }_0\Delta x}{e}^{-ijk\frac{2\pi }{N}}{y}_m^{\alpha }(x_j)}\right|$ is bounded, we have $\left|{\displaystyle \sum _{j=N}^{\infty }\Delta x\cdot w_j{e}^{-ij{\xi }_0\Delta x}{e}^{-ijk\frac{2\pi }{N}}{y}_m^{\alpha }(x_j)}\right|\le \frac{C}{N^{{\beta }_2}}$. Based on these conclusions, we obtain the following error estimate:

$$\begin{array}{ll}\left|c_1^{″}(x)-c_1^{‴}(x)\right|& =\frac{\Delta \xi }{2\pi }\left|{\displaystyle \sum _{i=0}^{N-1}{e}^{i{\xi }_k{x}_j}\cdot \varphi \left(i\alpha -{\xi }_k\right)}\cdot {\displaystyle \sum _{j=N}^{\infty }\Delta x\cdot w_j{e}^{-ij{\xi }_0\Delta x}{e}^{-ijk\frac{2\pi }{N}}{y}_m^{\alpha }(x_j)}\right|\\ & \le \frac{C}{N^{{\beta }_2}}{\displaystyle \sum _{i=0}^{N-1}\left|\varphi \left(i\alpha -{\xi }_k\right)\Delta \xi \right|}.\end{array}$$

The characteristic function can be assumed to be bounded: $\left|\varphi \left({\xi }_k-i\alpha \right)\right|\le C$, we have

$$\begin{array}{ll}\left|c_1^{″}(x)-c_1^{‴}(x)\right|& \le \frac{C}{N^{{\beta }_2}}{\displaystyle \sum _{i=0}^{N-1}\left|\varphi \left(i\alpha -{\xi }_i\right)\Delta \xi \right|}\\ & \le \frac{C}{N^{{\beta }_2}}{\displaystyle {\int }_0^{N-1}d\xi }\\ & \le \frac{C}{N^{{\beta }_2-1}}.\end{array}$$

(49)

According to the conclusions in (47)–(49) we obtain the error $I_1$. We define the error $I_1$ as $e_{I_1}$ as follows:

$$\begin{array}{ll}{e}_{I_1}& =\left|c_1(x)-c_1^{‴}(x)\right|\\ & \le \left|c_1(x)-c_1^{\prime }(x)\right|+\left|c_1^{\prime }(x)-c_1^{\prime }(x)\right|+\left|c_1^{″}(x)-c_1^{‴}(x)\right|\\ & =O\left(\Delta \xi \right)+O(C{N}^{-{\beta }_1})+O\left(C{N}^{-{\beta }_2+1}\right),\end{array}$$

(50)

where $f(t_{m-1},Y_{m-1}^{\Delta },Z_{m-1}^{\Delta })$ of the numerical value in the $I_2$ is given by the Picard iteration. The error of $I_2$ can be controlled by the Lipschitz condition of $f$. Thus, we have the following error estimate:

$$e_f=\left|f(t_{m-1},Y_{m-1},Z_{m-1})-f(t_{m-1},Y_{m-1}^{\Delta },Z_{m-1}^{\Delta })\right|\le L\left(\left|e_y\right|+\left|e_z\right|\right).$$

Assume that ${\theta }_1\Delta t_{m}L<1$ and $\Delta t_m$ is sufficiently small, we have

$$\begin{array}{ll}{e}_{I_2}& ={\theta }_1\Delta t_m\left|f(t_{m-1},Y_{m-1},Z_{m-1})-f(t_{m-1},Y_{m-1}^{\Delta },Z_{m-1}^{\Delta })\right|\\ & \le {\theta }_1\Delta t_{m}L\left(\left|e_y\right|+\left|e_z\right|\right)\le CL\left(\left|e_y\right|+\left|e_z\right|\right).\end{array}$$

For the $Y-\mathrm{component}$, the error of the last part is $I_3$. Now we analyze this part and note that

$$\begin{array}{ll}\left|{\mathrm{E}}_{m-1}^x\left[f_{t_m}\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[f_{t_{m-1}}\right]\right|& \le {\mathrm{E}}_{m-1}^x\left|L\left(\left|e_y\right|+\left|e_z\right|\right)\right|\\ & \le L{\mathrm{E}}_{m-1}^x\left|e_y\right|+L{\mathrm{E}}_{m-1}^x\left|e_Z\right|\\ & \le L\left(e_{I_1}+e_{I_1^{\prime }}\right).\end{array}$$

Thus, we obtain the error $I_3$:

$$\begin{array}{ll}{e}_{I_3}& =\left(1-{\theta }_1\right)\Delta t_m\left|{\mathrm{E}}_{m-1}^x\left[f_{t_m}\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[f_{t_{m-1}}\right]\right|\\ & \le \left(1-{\theta }_1\right)\Delta t_{m}L\left(e_{I_1}+e_{I_1^{\prime }}\right).\end{array}$$

For $z-\mathrm{component}$, $e_z=I_1^{\prime }+I_2^{\prime }+I_3^{\prime }+{\theta }_3^{-1}\Delta t_m^{-1}{R}_Z^m$ with

$$\begin{array}{ll}{e}_z& ={\theta }_4{\theta }_3^{-1}\left|{\mathrm{E}}_{m-1}^x\left[Z_m\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[Z_m\right]\right|+({\theta }_3-{\theta }_4){\theta }_3^{-1}\Delta t_m^{-1}\left|{\mathrm{E}}_{m-1}^x\left[Y_m\Delta W_m\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[Y_m\Delta W_m\right]\right|\\ & +(1-{\theta }_2){\theta }_3^{-1}\left|{\mathrm{E}}_{m-1}^x\left[f_m\Delta W_m\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[f_m\Delta W_m\right]\right|+{\theta }_3^{-1}\Delta t_m^{-1}{R}_Z^m.\end{array}$$

The error $I_1^{\prime }$ is similar to the error $I_1$, allowing us to deduce the following:

$$e_{I_1^{\prime }}=O\left(\Delta \xi \right)+O(C{N}^{-{\beta }_1^{\prime }})+O\left(C{N}^{-{\beta }_2^{\prime }+1}\right).$$

(51)

We consider the error of the second part $I_1^{\prime }$ of the z-component and note the following:

$${\mathrm{E}}_{m-1}^x\left[Y_m\Delta W_m\right]=\Delta t_m{\mathrm{E}}_{m-1}^x\left[Z_m\right].$$

Therefore, we deduce the following:

$$e_{I_2^{\prime }}=O\left(\Delta \xi \right)+O(C{N}^{-{\beta }_1^{\prime }})+O\left(C{N}^{-{\beta }_2^{\prime }+1}\right).$$

(52)

For the last part of the z-component, we deduce the following:

$$I_3^{\prime }=(1-{\theta }_2){\theta }_3^{-1}\left|{\mathrm{E}}_{m-1}^x\left[f_m\Delta W_m\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[f_m\Delta W_m\right]\right|.$$

By the Lipschitz condition for $f$, we can make the following conclusion:

$$\begin{array}{l}\left|{\mathrm{E}}_{m-1}^x\left[f_m\Delta W_m\right]-{\widehat{\mathrm{E}}}_{m-1}^x\left[f_m\Delta W_m\right]\right|\\ \le {\mathrm{E}}_{m-1}^x\left|f_m-{\widehat{f}}_m\right|\left|\Delta W_m\right|\\ \le \sup \left|f_m-{\widehat{f}}_m\right|\sqrt{\Delta t_m}\\ \le L\left(\left|e_Y\right|+\left|e_z\right|\right)\sqrt{\Delta t_m}.\end{array}$$

By the above conclusions and Lemma 1, we obtain

• For parameters ${\theta }_i\in \left[0,1\right](i=1,2)$ and ${\theta }_3\in \left(0,1\right]$, if $\phi \in C_b^2$ and $f(t,y,z)\in C_b^{1,2,2}$ we have the estimates:

  $$\begin{array}{ll}{e}_Y& =I_1+I_2+I_3+R_y^m\\ & =O\left(\Delta \xi \right)+O(C{N}^{-{\beta }_1})+O\left(C{N}^{-{\beta }_2+1}\right)+O\left({\left(\Delta \xi \right)}^2\right),\end{array}$$

  (53)

  $$\begin{array}{ll}{e}_z& =I_1^{\prime }+I_2^{\prime }+I_3^{\prime }+R_z^m\\ & =O\left(\Delta \xi \right)+O(C{N}^{-{\beta }_1^{\prime }})+O\left(C{N}^{-{\beta }_2^{\prime }+1}\right)+O\left({\left(\Delta \xi \right)}^2\right).\end{array}$$

  (54)
• In particular, for ${\theta }_i=\frac{1}{2}(i=1,2,3)$, if $\phi \in C_b^3$ and $f(t,y,z)\in C_b^{2,4,4}$, we have

  $$\begin{array}{ll}{e}_Y& =I_1+I_2+I_3+R_y^m\\ & =O\left(\Delta \xi \right)+O(C{N}^{-{\beta }_1})+O\left(C{N}^{-{\beta }_2+1}\right)+O\left({\left(\Delta \xi \right)}^3\right),\end{array}$$

  (55)

  $$\begin{array}{ll}{e}_z& =I_1^{\prime }+I_2^{\prime }+I_3^{\prime }+R_z^m\\ & =O\left(\Delta \xi \right)+O(C{N}^{-{\beta }_1^{\prime }})+O\left(C{N}^{-{\beta }_2^{\prime }+1}\right)+O\left({\left(\Delta \xi \right)}^3\right).\end{array}$$

  (56)

By the above conclusions, we know that the error is bounded, and it can be controlled by a constant.

6. Numerical Test

In this section, we present the numerical scheme for BSDE based on the fractional FFT. In Section 3, we provide the steps about numerical implementation to solve the approximate solution of BSDE. Because the goal of this study is to demonstrate the accuracy of the proposed method, we provide an example.

Example 1. 

Similar to [7,17], we consider BSDE

$$\{\begin{array}{l}-d{y}_t=f(t,W_t,y_t,z_t)dt-z_{t}d{W}_t,\\ y_T=u(T,W_T),\end{array}$$

(57)

where$W_t$is standard Brownian motion. The driver (or generator)$f(t,W_t,y_t,z_t)$of the BSDE(51) is

$$f(t,W_t,y_t,z_t)=y_t{z}_t-z_t+\frac{1}{32}{y}_t-\frac{1}{4}\sin (t+\frac{1}{4}{W}_t)\cos (t+\frac{1}{4}{W}_t)-\frac{3}{4}\cos (t+\frac{1}{4}{W}_t),$$

$$\left(y_T,z_T\right)=\left(\sin (T+\frac{1}{4}{W}_T),\frac{1}{4}\cos (T+\frac{1}{4}{W}_T)\right).$$

Then, the analytical solution of BSDE (51) is

$$\left(y_t,z_t\right)=\left(\sin (t+\frac{1}{4}{W}_t),\frac{1}{4}\cos (t+\frac{1}{4}{W}_t)\right).$$

Let $t=0,T=1$, and knowing the explicit solution $(y_0,z_0)$, we can obtain the numerical solution $(y^0,z^0)$ and the errors of $\left|y_0-y^0\right|,\left|z_0-z^0\right|$. In this numerical example, we first numerically solve the BSDE at the point $\left(W_0,t_0\right)=\left(0,0\right)$ by the generalized $\theta$-scheme with different parameters ${\theta }_i(i=1,2,3,4)$ to demonstrate the effectiveness. The errors and different time steps are listed

Table 1. The errors $\left|y_0-y^0\right|,\left|z_0-z^0\right|$, for example.

	Scheme A		Scheme B		Scheme C
	θ1 = 0.5, θ2 = 0.5, θ3 = 0.5, θ4 = 0		θ1 = 1, θ2 = 0.5, θ3 = 0.5, θ4 = 0		θ1 = 0.5, θ2 = 0.75, θ3 = 0.5, θ4 = 0
M	$\left|y_0-y^0\right|$	$\left|z_0-z^0\right|$	$\left|y_0-y^0\right|$	$\left|z_0-z^0\right|$	$\left|y_0-y^0\right|$	$\left|z_0-z^0\right|$
16	1.507 × 10−5	5.485 × 10−4	1.599 × 10−2	7.253 × 10−3	1.915 × 10−3	9.110 × 10−5
32	2.097 × 10−5	1.124 × 10−4	7.969 × 10−3	3.910 × 10−3	8.789 × 10−4	6.863 × 10−5
64	2.164 × 10−5	2.837 × 10−6	3.988 × 10−3	2.036 × 10−3	4.113 × 10−4	2.795 × 10−5
128	2.171 × 10−5	2.395 × 10−5	2.003 × 10−3	1.050 × 10−3	1.904 × 10−4	4.014 × 10−7
256	2.065 × 10−5	1.429 × 10−5	1.011 × 10−3	5.278 × 10−4	8.433 × 10−5	3.850 × 10−7
	scheme D		scheme E		scheme F
	θ1 = 0.5, θ2 = 0.5, θ3 = 0.25, θ4 = 0		θ1 = 0, θ2 = 0, θ3 = 0.5, θ4 = 0		θ1 = 1, θ2 = 1, θ3 = 1, θ4 = 0
	$\left|y_0-y^0\right|$	$\left|z_0-z^0\right|$	$\left|y_0-y^0\right|$	$\left|z_0-z^0\right|$	$\left|y_0-y^0\right|$	$\left|z_0-z^0\right|$
16	3.776 × 10−3	7.650 × 10−4	1.258 × 10−2	8.875 × 10−3	1.209 × 10−2	7.715 × 10−3
32	1.788 × 10−3	3.235 × 10−6	6.253 × 10−3	4.108 × 10−3	6.170 × 10−3	3.875 × 10−3
64	8.785 × 10−4	9.920 × 10−5	3.111 × 10−3	1.961 × 10−3	3.124 × 10−3	1.954 × 10−3
128	4.436 × 10−4	8.593 × 10−5	1.544 × 10−3	9.455 × 10−4	1.580 × 10−3	9.933 × 10−4
256	2.300 × 10−4	4.529 × 10−5	7.619 × 10−4	4.694 × 10−4	8.013 × 10−4	4.979 × 10−4
	scheme G		scheme H		scheme I
	θ1 = 0.5, θ2 = 0.5, θ3 = 0.5, θ4 = −0.25		θ1 = 0.5, θ2 = 0.5, θ3 = 0.5, θ4 = 0.25		θ1 = 1, θ2 = 1, θ3 = 1, θ4= 0.5
M	$\left|y_0-y^0\right|$	$\left|z_0-z^0\right|$	$\left|y_0-y^0\right|$	$\left|z_0-z^0\right|$	$\left|y_0-y^0\right|$	$\left|z_0-z^0\right|$
16	8.194 × 10−5	3.397 × 10−4	2.804 × 10−4	1.153 × 10−3	8.109 × 10−3	9.852 × 10−3
32	4.001 × 10−6	5.975 × 10−5	9.291 × 10−5	2.679 × 10−4	4.310 × 10−3	4.325 × 10−3
64	1.535 × 10−5	1.029 × 10−5	4.016 × 10−5	4.195 × 10−5	2.239 × 10−3	2.003 × 10−3
128	2.015 × 10−5	2.693 × 10−5	2.635 × 10−5	1.452 × 10−5	1.150 × 10−3	9.709 × 10−4
256	2.028 × 10−5	7.556 × 10−6	2.179 × 10−5	1.953 × 10−5	5.900 × 10−4	4.824 × 10−4

. From Table 1, the accuracy of the generalized $\theta$-scheme depends on the parameters ${\theta }_i$. As the time steps increases, the error decreases. Further, when ${\theta }_i=\frac{1}{2}(i=1,2,3),{\theta }_4\in \left(-\frac{1}{2},\frac{1}{2}\right)$, the errors are small.

In order to show the performance of the generalized $\theta$-scheme, we also solve the BSDE by the generalized $\theta$-scheme with different parameters ${\theta }_i(i=1,2,3,4)$ for the time steps $△t=2^{-8}$. In Figure 1, we graph the errors of the analytical and numerical solution and spacing fineness M. We find that with the increase in M, the closer the errors $\left|y_0-y^0\right|,\left|z_0-z^0\right|$ are to zero. The results show that the errors are convergent.

Remark 5. 

According to Table 1 and Figure 1, the results show that the accuracy of the convolution method with generalized$\theta$-scheme based on fractional Fourier transform is high. Especially when${\theta }_i=\frac{1}{2}(i=1,2,3),{\theta }_4\in \left(-\frac{1}{2},\frac{1}{2}\right)$, the convergence rate is faster than the other value θ, and the stability is better.

Remark 6. 

Compared to Table 1 of the literature [7], we can find that the accuracy of using the convolution with generalized θ-scheme based on fractional Fourier transform is better than using the generalized θ-scheme.

7. Conclusions

In this paper, we propose the convolution method of the numerical solution to BSDE based on the fractional FFT. In this method, we use a generalized-scheme time discretization of BSDE to express the conditional expectations; these can calculate the Fourier transform of the approximate solution with the convolution theorem. Finally, we can take the inverse Fourier transform to recover expressions for the approximate solution that are recursive backwards in time. Our method differs from the literature in [7,14,17], in that this paper combines the generalized θ-scheme of reference [7], the convolution method of reference [14] and the fractional Fourier transform of reference [17]; this becomes the new idea of this paper. It can be proved that the errors of the exact solution and numerical solution are well-controlled in theory. In addition, we give a numerical experiment to test the effectiveness and accuracy of the proposed method. The result shows that the error is second order convergent.