Three-Layer Artificial Neural Network for Pricing Multi-Asset European Option

Abstract

This paper studies an artificial neural network (ANN) for multi-asset European options. Firstly, a simple three-layer ANN-3 is established with undetermined weights and bias. Secondly, the time–space discrete PDE of the multi-asset option is given and the corresponding discrete data are fed into the ANN-3. Then, using least squares error as the objective function, the weights and bias of ANN-3 are trained well. Numerical examples are carried out to confirm the stability, accuracy and efficiency. Experiments show the ANN’s relative error is about $0.8\%$. This method can be extended into multi-layer ANN-$q(q>3)$ and extended into American options.

1. Introduction

In the last 30 years, there have been several methods for option pricing. For example, the radial basis function (RBF) method for options with a few assets [1,2,3,4,5,6,7,8,9,10,11]; analytical and semi-analytical methods for simple options [12,13]; finite difference method (FDM) for lower-dimensional option PDEs [14,15,16,17,18]; Monte Carlo simulation for complex options [19,20,21]; and Laplace transform or Mellin transform scheme [16,22,23,24,25] for classical options. Wu et al. discuss option pricing by the willow tree (WT) method for generalized hyperbolic Leύy processes [26]. However, for RBF and FD methods, only lower-dimensional ($d\le 2$) problems can be solved easily, and for the analytical or semi-analytical method, only some simple or classical options can be determined. In addition, although the Monte Carlo (MC) algorithm and the WT method are very simple and they have a wide application background, the MC method has low computational efficiency (consumes a lot of CPU time) and the WT method is of low convergence order, which limits their application.

In recent years, some of the literature starts to consider artificial neural networks (ANN) to solve option problems. Anderson and Ulrych discuss an accelerated American option pricing with deep neural networks [27]. Caverhill et al. give a neural network approach for option pricing and hedging [28]. Gan and Liu discuss option pricing based on the residual neural network [29]. Glau et al. discuss neural network expression rates and applications of the deep parametric PDE method in counter-party credit risk [30]. He and Guan study the parameter estimation method of the option pricing model based on the convolutional neural network in high-frequency financial trading [31]. Kapllani and Teng consider deep learning algorithms for solving high-dimensional nonlinear backward stochastic differential equations [32]. Lee and Son propose predicting arbitrage-free American option prices using an artificial neural network with pseudo inputs [33]. Mary and Salchenberger give a neural network model for estimating option prices [34]. Shvimer et al. discuss pricing options with a new hybrid neural network model [35]. Teng et al. discuss the combination of the neural network and classical volatility prediction model [36]. Tung et al. use a self-organizing neural-fuzzy semantic network for financial volatility trading [37]. Umeorah, Mashele and Agbaeze give barrier option pricing with neural networks [38]. Wang proposes a neural network forecasting model for the stock index option price [39]. However, none of the above literature is a neural network algorithm that is based on partial differential equations.

Unlike applying ANN to simple fitting problems, option pricing does not have natural samples, and we only know that the option satisfies a specific partial differential equation (PDE). To apply the ANN to option pricing, a key procedure is to discretize PDE and therefore make it a minimum optimization problem. To the best of our knowledge, this is the first time that ANN has been considered for option pricing from a multi-dimensional PDE.

In this paper, we propose an ANN to price the multi-dimensional option. The PDEs governed by the multi-asset option are discretized first, then a three-layer artificial neural network (ANN-3) is established. By minimizing the target error, all parameters (weights and bias) in ANN are trained well, then the neural network is applied. ANN-3 is easy to implement and has a powerful ability for multi-asset option pricing. Additionally, we propose some skills to accelerate the ANN training process.

The rest of this paper is formed as follows. In Section 2, the PDE governed by multi-asset option valuing is stated. In Section 3, a computational frame of the artificial neural network is proposed. We find out the computational formulas of derivatives for ANN weights and bias, and then we give the update formulas for the weights and bias. Section 4 gives some numerical examples to confirm the efficiency, accuracy and convergence of ANN. In Section 5, we give some conclusions and the future work. In Appendix A, some appendices are given, including algorithms and computational results.

2. PDE of Multi-Asset Option Values

Let $\mathit{S}={(S_1\left(t\right),S_2\left(t\right),\dots ,S_d\left(t\right))}^{\prime }$, with ${(·)}^{\prime }$ denoting the transposition of $(·)$, be d-asset prices at time t. Let $x_I\left(t\right)=log{S}_I\left(t\right),I=1,2,\dots ,d$ be modeled by Brownian motion together with a drift term under no-arbitrage assumption, i.e., $x_I\left(t\right)$ are controlled by stochastic differential equations (SDEs),

$$\begin{array}{c}\hfill \mathrm{d}{x}_I\left(t\right)=(r-q_I)\mathrm{d}t+{\sigma }_I\mathrm{d}{W}_t^{\left(I\right)},I=1,2,\dots ,d.\end{array}$$

(1)

In SDEs (1), r represents the risk-free interest, $q_I$ represents the dividend yield, and ${\sigma }_I$ is the volatility of asset I. $W_t^{\left(I\right)},I=1,2,\dots ,d$, are d standard Brownian motions, and it is assumed that

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\mathbb{E}\left(\mathrm{d}{W}_t^{\left(I\right)}\right)=0,Var\left(\mathrm{d}{W}_t^{\left(I\right)}\right)=\mathrm{d}t,I=1,2,\dots ,d,\hfill \\ Cov\left(\mathrm{d}{W}_t^{\left(I\right)},\mathrm{d}{W}_t^{\left(J\right)}\right)={\rho }_{IJ}\mathrm{d}t,I,J=1,2,\dots ,d,I\ne J,\hfill \end{array}\right.\end{array}$$

(2)

where ${\rho }_{IJ}$ is the coefficients of association between $\mathrm{d}{W}_t^{\left(I\right)}$ and $\mathrm{d}{W}_t^{\left(J\right)}$.

Let $u(\tau ,\mathit{x})$ be the option with $\tau =T-t$ and d log underlying $\mathit{x}={(x_1\left(t\right),x_2\left(t\right),\dots ,x_d\left(t\right))}^{\prime }\in \Omega ={(-\infty ,+\infty )}^d$. Using the risk-free hedging theorem (see [40]), one can obtain the multi-dimensional PDE, named Black–Scholes equation, with the d-asset option value as

$$\begin{array}{c}\hfill {\displaystyle \frac{\partial u}{\partial \tau }}(\tau ,\mathit{x})=\mathcal{A}u(\tau ,\mathit{x}),\tau \in (0,T),\mathit{x}\in \Omega ,\end{array}$$

(3)

where the linear differential operator $\mathcal{A}$ is defined as

$$\begin{array}{c}\hfill \mathcal{A}:=\sum _{I,J=1}^d{a}_{IJ}{\displaystyle \frac{{\partial }^2}{\partial x_I\partial x_J}}+\sum _{I=1}^d{b}_I{\displaystyle \frac{\partial }{\partial x_I}}-r,\end{array}$$

(4)

with coefficients $a_{IJ}:={\displaystyle \frac{1}{2}}{\rho }_{IJ}{\sigma }_I{\sigma }_J$ and $b_I:=r-q_I-{\displaystyle \frac{1}{2}}{a}_{II}$.

For the European option, $u(\tau ,\mathit{x})$ has initial values (also known as terminal conditions, or payoff function, for time $\tau =0$)

$$\begin{array}{c}\hfill u(0,\mathit{x})=\Pi \left(\mathit{x}\right).\end{array}$$

(5)

System (3)–(5) can be written as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle \frac{\partial u}{\partial \tau }}(\tau ,\mathit{x})-\mathcal{A}u(\tau ,\mathit{x})=0,\tau \in (0,T],\mathit{x}\in \Omega ,\hfill \\ u(0,\mathit{x})=\Pi \left(\mathit{x}\right),\mathit{x}\in \Omega .\hfill \end{array}\right.\end{array}$$

(6)

For put options and geometric mean payoff function $\Pi \left(\mathit{x}\right)$, the initial values of u are taken as

$$\begin{array}{c}\hfill u(0,\mathit{x})=\Pi \left(\mathit{x}\right)=max\left(0,K-e^{{\sum }_{I=1}^d{\alpha }_I{x}_I}\right),0<{\alpha }_I<1,\sum _{I=1}^d{\alpha }_I=1,\forall \mathit{x}\in \Omega ,\end{array}$$

(7)

with strike price K. Ina general case, system (6) has no analytical solution, and a certain numerical method is needed to find the option value $u(\tau ,\mathit{x})$ in domain $(0,T]\times \Omega$.

With geometric mean payoff function $\Pi \left(\mathit{x}\right)$ defined by (7), the put option governed by system (6) has the analytical solution (see [40]),

$$\begin{array}{c}\hfill u(\tau ,\mathit{x})=K{e}^{-r\tau }\Phi (-{\widehat{d}}_2)-e^{{\sum }_{I=1}^d{\alpha }_I{x}_I}\Phi (-{\widehat{d}}_1),\end{array}$$

(8)

where $\Phi (·)$ is the cumulative normal distribution function and

$$\begin{array}{ccc}\hfill & & {\widehat{d}}_1={\displaystyle \frac{1}{\widehat{\sigma }\sqrt{\tau }}}\left[\sum _{I=1}^d{\alpha }_I{x}_I-lnK+(r-\widehat{q}+{\widehat{\sigma }}^2/2)\tau \right],{\widehat{d}}_2={\widehat{d}}_1-\widehat{\sigma }\sqrt{\tau },\hfill \end{array}$$

(9)

with

$$\begin{array}{c}\hfill {\widehat{\sigma }}^2=\sum _{I,J=1}^d{a}_{IJ}{\alpha }_I{\alpha }_J,\widehat{q}=\sum _{I=1}^d{\alpha }_I(q_I+a_{II}/2)-{\widehat{\sigma }}^2/2.\end{array}$$

(10)

However, for the put option with arithmetic mean payoff function,

$$\begin{array}{c}\hfill \Pi \left(\mathit{x}\right)=max\left(0,K-\sum _{I=1}^d{\alpha }_I{e}^{x_I}\right),0<{\alpha }_I<1,\sum _I^d{\alpha }_I=1,\forall \mathit{x}\in \Omega ,\end{array}$$

(11)

the option $u(\tau ,\mathit{x})$ has no analytical solution. So, the numerical scheme is not a feasible method to price the multi-asset option governed by (6). The application of artificial neural networks to option pricing is discussed in the next section.

3. Computational Frame of ANN

An artificial neural network (ANN) is defined as in Figure 1. In this figure, the ANN-3 includes an input layer, one hidden layer and an output layer. For example, the neural units in the three layers are $n_1=3,n_2=4,n_3=1$, and we see in Section 4 that this structure is powerful for some examples.

The evolution of the 3-layer artificial neural network is represented as matrix expressions,

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\mathit{O}}^{\left(p\right)}={\mathit{\omega }}^{(p-1)}{\mathit{U}}^{(p-1)}+{\mathit{\theta }}^{\left(p\right)},p=2,3,\hfill \\ {\mathit{U}}^{\left(p\right)}=f_p\left({\mathit{O}}^{\left(p\right)}\right),p=2,3,\hfill \end{array}\right.\end{array}$$

(12)

where ${\mathit{O}}^{\left(p\right)}$ is the input data in the layer p, ${\mathit{U}}^{\left(p\right)}$ is the output in the layer p, and $f_p$ is the transform function(or activation function). The algebraic expression is given by

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle O_j^{\left(p\right)}=\sum _{i=1}^{n_{p-1}}{\omega }_{ji}^{(p-1)}{U}_i^{(p-1)}+{\theta }_j^{\left(p\right)},p=2,3;j=1,2,\dots ,n_p,}\hfill \\ U_j^{\left(p\right)}=f_p\left(O_j^{\left(p\right)}\right),p=2,3;j=1,2\dots ,n_p,\hfill \end{array}\right.\end{array}$$

(13)

where $U^{\left(1\right)}$ is a $(d+1)$ input data column vector ${(x_1,x_2,\dots ,x_d,\tau )}^{\prime }$ m and $f_p(·)(p=2,3)$ are the transform functions, i.e.,

$$\begin{array}{c}\hfill f_2\left(x\right)={\displaystyle \frac{1}{1+e^{-x}}}\left(\mathrm{sigmoid}\mathrm{function}\right),f_3\left(x\right)=x\left(\mathrm{linear}\mathrm{function}\right).\end{array}$$

(14)

We see the derivatives of transform functions $f_p(·)$ are

$$\begin{array}{c}\hfill {\displaystyle \frac{\partial f_2\left(x\right)}{\partial x}}=f_2\left(x\right)(1-f_2\left(x\right)),{\displaystyle \frac{\partial f_3\left(x\right)}{\partial x}}=1.\end{array}$$

(15)

The weights and bias in ANN are defined as

$$\begin{array}{c}\hfill {\mathit{\omega }}^{\left(2\right)}=\left[\begin{array}{cccc}{\omega }_{11}^{\left(2\right)}& {\omega }_{12}^{\left(2\right)}& \dots & {\omega }_{1n_2}^{\left(2\right)}\end{array}\right],{\mathit{\omega }}^{\left(1\right)}=\left[\begin{array}{cccc}{\omega }_{11}^{\left(1\right)}& {\omega }_{12}^{\left(1\right)}& \dots & {\omega }_{1n_1}^{\left(1\right)}\\ {\omega }_{21}^{\left(1\right)}& {\omega }_{22}^{\left(1\right)}& \dots & {\omega }_{2n_1}^{\left(1\right)}\\ ⋮& ⋮& ⋮& ⋮\\ {\omega }_{n_{2}1}^{\left(1\right)}& {\omega }_{n_{2}2}^{\left(1\right)}& \dots & {\omega }_{n_2{n}_1}^{\left(1\right)}\end{array}\right],{\mathit{\theta }}^{\left(\mathit{p}\right)}=\left[\begin{array}{c}{\theta }_1^{\left(p\right)}\\ {\theta }_2^{\left(p\right)}\\ ⋮\\ {\theta }_{n_p}^{\left(p\right)}\end{array}\right]\end{array}$$

(16)

for $p=2,3$. We call (12) or (13) an ANN-3 artificial neural network. Using the ANN-3 network has the following merits:  

(1)

It is relatively simple to compute the partial derivatives, $\Delta {\mathit{\omega }}^{\left(i\right)}$ and $\Delta {\mathit{\theta }}^{\left(j\right)}$, of ${\mathit{\omega }}^{\left(i\right)}$ for $i=1,2$ and ${\mathit{\theta }}^{\left(j\right)}$ for $j=2,3$ in ANN-3. Then we can easily update the parameters ${\mathit{\omega }}^{\left(i\right)}={\mathit{\omega }}^{\left(i\right)}-\eta \Delta {\mathit{\omega }}^{\left(i\right)}$ and ${\mathit{\theta }}^{\left(j\right)}={\mathit{\theta }}^{\left(j\right)}-\eta \Delta {\mathit{\theta }}^{\left(j\right)}$ with learning rate $\eta$.

(2)

For deep learning network ANN-$q,q>3$, it becomes very complicated to compute the partial derivatives of ${\mathit{\omega }}^{\left(i\right)}$ and ${\mathit{\theta }}^{\left(j\right)}$. So, we use ANN-q for $q=3$.

(3)

For option pricing, we do not need deep learning networks. ANN-$q(q>3)$ and the shallow artificial neural networks (ANN-3) are enough to solve the problem.

The output of the network is denoted by

$$\begin{array}{c}\hfill U^{\left(3\right)}:=U_1^{\left(3\right)}\left(\mathit{X}\right),{\mathit{U}}^{\left(2\right)}=\left\{U_1^{\left(2\right)}\left(\mathit{X}\right),U_2^{\left(2\right)}\left(\mathit{X}\right),\dots ,U_{n_2}^{\left(2\right)}\left(\mathit{X}\right)\right\},U^{\left(1\right)}=\mathit{X},\end{array}$$

(17)

with input data $\mathit{X}={\left[x_1,x_2,\dots ,x_d,\tau \right]}^{\prime }$. We denote by $U_1^{\left(3\right)}\left(\mathit{X}\right):=net\left({\mathit{\omega }}^{\left(1\right)},{\mathit{\omega }}^{\left(2\right)},{\mathit{\theta }}^{\left(2\right)},{\mathit{\theta }}^{\left(3\right)},\mathit{X}\right)$ the network output with input data $\mathit{X}={\left(x_1,x_2,\cdots ,x_d,\tau \right)}^{\prime }$ under parameters ${\mathit{\omega }}^{\left(1\right)},{\mathit{\omega }}^{\left(2\right)},{\mathit{\theta }}^{\left(2\right)}$ and ${\mathit{\theta }}^{\left(3\right)}$. Furthermore, we denote by $U^{(p,k)}=U^{\left(p\right)}\left({\mathit{X}}^{\left(k\right)}\right),p=1,2,3$ with N space and time input data

$$\begin{array}{c}\hfill {\mathit{X}}^{\left(k\right)}={\left[x_1^{\left(k\right)},x_2^{\left(k\right)},\dots ,x_d^{\left(k\right)},{\tau }^{\left(k\right)}\right]}^T,k=1,2,\dots ,N.\end{array}$$

(18)

The purpose of our ANN structure is to determine the optimal weights ${\mathit{\omega }}^{\left(1\right)},{\mathit{\omega }}^{\left(2\right)}$ and bias ${\mathit{\theta }}^{\left(2\right)},{\mathit{\theta }}^{\left(3\right)}$. For this purpose, we use an iteration algorithm, i.e., we modify the values of the parameters

$$\left({\mathit{\omega }}^{\left(1\right)},{\mathit{\omega }}^{\left(2\right)},{\mathit{\theta }}^{\left(2\right)},{\mathit{\theta }}^{\left(3\right)}\right)$$

until some objective function $MSE$ is less than the pre-specified error $\epsilon$.

For the convenience, we list some of the symbols in this paper in

Table 1. Some symbols for ANN.

Name	Meaning
$N^{\prime }$	The number of data that are the initial conditions
N	The number of input data for the training of ANN
$N-N^{\prime }$	The input number of data for discrete PDE
$net({\mathit{\omega }}^{\left(1\right)},{\mathit{\omega }}^{\left(2\right)},{\mathit{\theta }}^{\left(2\right)},{\mathit{\theta }}^{\left(3\right)})$	ANN with weights $({\mathit{\omega }}^{\left(1\right)},{\mathit{\omega }}^{\left(2\right)})$ and bias $({\mathit{\theta }}^{\left(2\right)},{\mathit{\theta }}^{\left(3\right)})$
$U^3\left(\mathit{X}\right)=net\left({\mathit{\omega }}^{\left(1\right)},{\mathit{\omega }}^{\left(2\right)},{\mathit{\theta }}^{\left(2\right)},{\mathit{\theta }}^{\left(3\right)},\mathit{X}\right)$	ANN output with input data $\mathit{X}={(x_1,x_2,\dots ,x_d,\tau )}^{\prime }$
$\eta$	Learning rate of ANN
${\mathit{O}}_k^{\left(p\right)}$	Input data at the p layer ($p=2,q$)
${\mathit{U}}_k^{\left(p\right)}$	Output data at the p layer ($p=1,2,q$)
$ER{R}^{\left(payoff\right)}$	Error for initial values
$ER{R}^{\left(PDE\right)}$	Error for discrete PDE
$\Delta {\mathit{\theta }}^{(payoff,·)}$, $\Delta {\mathit{\theta }}^{(PDE,·)}$	Partial derivatives for $\mathit{\theta }$
$\Delta {\mathit{\omega }}^{(payoff,·)}$, $\Delta {\mathit{\omega }}^{(PDE,·)}$	Partial derivatives for $\mathit{\omega }$
D	Partial derivatives for $\mathit{x}$ with $D=D_{\tau },D=D_I$ and $D=D_{IJ}$

.

In the next paragraphs, we give the definition of objective function for the network, the mean square error (MSE),

$$\begin{array}{c}\hfill MSE=MS{E}^{\left(payoff\right)}+MS{E}^{\left(PDE\right)},\end{array}$$

(19)

with $MS{E}^{\left(payoff\right)}$ being the error at points ${\mathit{X}}^{\left(k\right)}=\left(x_1^{\left(k\right)},\dots ,x_d^{\left(k\right)},{\tau }_1\right)$ with $k=1,2,\dots ,N^{\prime }$ and ${\tau }_1=0$, and $MS{E}^{\left(PDE\right)}$ being the error of the PDE (6) with ${\mathit{X}}^{\left(k\right)}=\left(x_1^{\left(k\right)},\dots ,x_d^{\left(k\right)},{\tau }^{\left(k\right)}\right)$ and ${\tau }^{\left(k\right)}$ being one of the values

$$\begin{array}{c}\hfill {\tau }^{\left(k\right)}=\Delta \tau ,\dots ,{\tau }_j,\dots ,{\tau }_M=T.\end{array}$$

(20)

Firstly, we consider the $MS{E}^{\left(payoff\right)}$ at ${\tau }_1=0$.

For the initial conditions, we have $U_1^{\left(3\right)}\left({\mathit{X}}^{\left(k\right)}\right)=\Pi \left({\mathit{x}}^{\left(k\right)}\right):=y^{\left(k\right)},k=1,2,\dots ,N^{\prime }$. Then the $MS{E}^{\left(payoff\right)}$ of ANN at ${\tau }_1=0$ is defined by

$$\begin{array}{c}\hfill MS{E}^{\left(payoff\right)}={\displaystyle \frac{1}{2N^{\prime }}}\sum _{k=1}^{N^{\prime }}{\left(U_1^{(3,k)}-y^{\left(k\right)}\right)}^2.\end{array}$$

(21)

The partial derivatives of $MS{E}^{\left(payoff\right)}$ with respect to ${\mathit{\omega }}^{\left(2\right)}=\left[{\omega }_{11}^{\left(2\right)},\dots ,{\omega }_{1\ell }^{\left(12\right)},\dots ,{\omega }_{1n_2}^{\left(2\right)}\right]$ are

$$\begin{array}{ccc}\hfill & & {\displaystyle \frac{\partial MS{E}^{\left(payoff\right)}}{\partial {\omega }_{1\ell }^{\left(2\right)}}}={\displaystyle \frac{\partial MS{E}^{\left(payoff\right)}}{\partial U_1^{(3,k)}}}{\displaystyle \frac{\partial U_1^{(3,k)}}{\partial O_1^{(3,k)}}}{\displaystyle \frac{\partial O_1^{(3,k)}}{\partial {\omega }_{1\ell }^{\left(2\right)}}}={\displaystyle \frac{1}{N^{\prime }}}\sum _{k=1}^{N^{\prime }}\left(U_1^{(3,k)}-y^{\left(k\right)}\right)U_{\ell }^{(2,k)},\hfill \end{array}$$

(22)

for $\ell =1,2,\dots ,n_2$, and the partial derivative with respect to ${\theta }_1^{\left(3\right)}$ is

$$\begin{array}{ccc}& & {\displaystyle \frac{\partial MS{E}^{\left(payoff\right)}}{\partial {\theta }_1^{\left(3\right)}}}={\displaystyle \frac{\partial MS{E}^{\left(payoff\right)}}{\partial U_1^{(3,k)}}}{\displaystyle \frac{\partial U_1^{(3,k)}}{\partial O_1^{(3,k)}}}{\displaystyle \frac{\partial O_1^{(3,k)}}{\partial {\theta }_1^{\left(3\right)}}}={\displaystyle \frac{1}{N^{\prime }}}\sum _{k=1}^{N^{\prime }}\left(U_1^{(3,k)}-y^{\left(k\right)}\right).\hfill \end{array}$$

(23)

The partial derivatives of $MS{E}^{\left(payoff\right)}$ with respect to ${\omega }_{j\ell }^{\left(1\right)}$ are

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial MS{E}^{\left(payoff\right)}}{\partial {\omega }_{j\ell }^{\left(1\right)}}}& =& {\displaystyle \frac{\partial MS{E}^{\left(payoff\right)}}{\partial U_1^{(3,k)}}}{\displaystyle \frac{\partial U_1^{(3,k)}}{\partial O_1^{(3,k)}}}{\displaystyle \frac{\partial O_1^{(3,k)}}{\partial U_j^{(2,k)}}}{\displaystyle \frac{\partial U_j^{(2,k)}}{\partial O_j^{(2,k)}}}{\displaystyle \frac{\partial O_j^{(2,k)}}{\partial {\omega }_{j\ell }^{\left(1\right)}}}\hfill \\ & =& {\displaystyle \frac{1}{N^{\prime }}}\sum _{k=1}^{N^{\prime }}\left(U_1^{(3,k)}-y^{\left(k\right)}\right){\omega }_{1j}^{\left(2\right)}{U}_j^{(2,k)}\left(1-U_j^{(2,k)}\right)U_{\ell }^{(1,k)},\hfill \end{array}$$

(24)

for $j=1,2,\dots ,n_2$ and $\ell =1,2,\dots ,n_1$. The partial derivatives of $MS{E}^{\left(payoff\right)}$ with respect to ${\theta }_j^{\left(2\right)}$ are

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial MS{E}^{\left(payoff\right)}}{\partial {\theta }_j^{\left(2\right)}}}& =& {\displaystyle \frac{\partial MS{E}^{\left(payoff\right)}}{\partial U_1^{(3,k)}}}{\displaystyle \frac{\partial U_1^{(3,k)}}{\partial O_1^{(3,k)}}}{\displaystyle \frac{\partial O_1^{(3,k)}}{\partial U_j^{(2,k)}}}{\displaystyle \frac{\partial U_j^{(2,k)}}{\partial O_j^{(2,k)}}}{\displaystyle \frac{\partial O_j^{(2,k)}}{\partial {\theta }_j^{\left(2\right)}}}\hfill \\ & =& {\displaystyle \frac{1}{N^{\prime }}}\sum _{k=1}^{N^{\prime }}\left(U_1^{(3,k)}-y^{\left(k\right)}\right){\omega }_{1j}^{\left(2\right)}{U}_j^{(2,k)}\left(1-U_j^{(2,k)}\right),\hfill \end{array}$$

(25)

for $j=1,2,\dots ,n_2$. Let

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\Delta {\mathit{\omega }}^{(payoff,2)}={\left[{\displaystyle \frac{\partial MS{E}^{\left(payoff\right)}}{\partial {\omega }_{1j}^{\left(2\right)}}}\right]}_{1\times n_2},\hfill \\ \Delta {\mathit{\omega }}^{(payoff,1)}={\left[{\displaystyle \frac{\partial MS{E}^{\left(payoff\right)}}{\partial {\omega }_{j\ell }^{\left(1\right)}}}\right]}_{n_2\times n_1}\hfill \\ {\displaystyle \Delta {\mathit{\theta }}^{(payoff,3)}=\frac{1}{N^{\prime }}\sum _{k=1}^N\left(U^{(3,k)}-y^{\left(k\right)}\right),}\hfill \\ \Delta {\mathit{\theta }}^{(payoff,2)}={\left[{\displaystyle \frac{\partial MS{E}^{\left(payoff\right)}}{\partial {\theta }_j^{\left(2\right)}}}\right]}_{n_2\times 1},\hfill \end{array}\right.\end{array}$$

(26)

with the definition of partial derivatives (22)–(25).

Secondly, we discuss the $MS{E}^{\left(PDE\right)}$ at discrete points ${\mathit{X}}^{\left(k\right)},k=N^{\prime }+1,2,\dots ,N$.

The discrete error $MS{E}^{\left(PDE\right)}$ of PDE at ${\mathit{X}}^{\left(k\right)}$ for $k=N^{\prime }+1,\dots ,N$ is defined by

$$\begin{array}{ccc}\hfill MS{E}^{\left(PDE\right)}& :=& {\displaystyle \frac{1}{2(N-N^{\prime })}}\sum _{k=N^{\prime }+1}^{N}ER{R}_k^2,\hfill \end{array}$$

(27)

where

$$\begin{array}{ccc}\hfill ER{R}_k& :=& {\left[{\displaystyle \frac{\partial U(\tau ,\mathit{x})}{\partial \tau }}-\mathcal{A}U(\tau ,\mathit{x})\right]}_{\mathit{X}={\mathit{X}}^{\left(k\right)}}\hfill \\ & \approx & D_{\tau }{U}^{(3,k)}-\sum _{I=1}^d\sum _{J=1}^d{a}_{IJ}{D}_{IJ}{U}^{(3,k)}-\sum _{I=1}^d{b}_I{D}_I{U}^{(3,k)}+r{U}^{(3,k)},\hfill \end{array}$$

(28)

for $k=N^{\prime }+1,\dots ,N$. Here, differential operators $D_{\tau }:={\displaystyle \frac{\partial }{\partial \tau }},D_I:={\displaystyle \frac{\partial }{\partial x_I}}$ and $D_{IJ}:={\displaystyle \frac{{\partial }^2}{\partial x_I\partial x_J}}$. The partial derivatives are defined as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle D_{IJ}{U}_1^{(3,k)}:=\frac{{\partial }^2}{\partial x_I\partial x_J}{U}_1^{(3,k)}=\frac{\partial }{\partial x_J}\left(\frac{\partial }{\partial x_I}{U}_1^{(3,k)}\right),}\hfill \\ {\displaystyle D_I{U}_1^{(3,k)}:=\frac{\partial }{\partial x_I}{U}_1^{(3,k)}=\frac{\partial U_1^{(3,k)}}{\partial O_1^{(3,k)}}\sum _{j=1}^{n_2}\frac{\partial O_1^{(3,k)}}{\partial U_j^{(2,k)}}\frac{\partial U_j^{(2,k)}}{\partial O_j^{(2,k)}}\frac{\partial O_j^{(2,k)}}{\partial U_I^{(1,k)}},}\hfill \\ {\displaystyle D_{\tau }{U}_1^{(3,k)}:=\frac{\partial }{\partial \tau }{U}_1^{(3,k)}=\frac{\partial U_1^{(3,k)}}{\partial O_1^{(3,k)}}\sum _{j=1}^{n_2}\frac{\partial O_1^{(3,k)}}{\partial U_j^{(2,k)}}\frac{\partial U_j^{(2,k)}}{\partial O_j^{(2,k)}}\frac{\partial O_j^{(2,k)}}{\partial U_{d+1}^{(1,k)}},}\hfill \end{array}\right.\end{array}$$

(29)

for $k=N^{\prime }+1,2,\cdots ,N$, which induces

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle D_{IJ}{U}_1^{(3,k)}=\sum _{j=1}^{n_2}{\omega }_{1j}^{\left(2\right)}{\omega }_{jI}^{\left(1\right)}{\omega }_{jJ}^{\left(1\right)}{U}_j^{(2,k)}\left(1-U_j^{(2,k)}\right)\left(1-2U_j^{(2,k)}\right),}\hfill \\ {\displaystyle D_I{U}_1^{(3,k)}=\sum _{j=1}^{n_2}{\omega }_{1j}^{\left(2\right)}{\omega }_{jI}^{\left(1\right)}{U}_j^{(2,k)}\left(1-U_j^{(2,k)}\right),}\hfill \\ {\displaystyle D_{\tau }{U}_1^{(3,k)}=\sum _{j=1}^{n_2}{\omega }_{1j}^{\left(2\right)}{\omega }_{j,d+1}^{\left(1\right)}{U}_j^{(2,k)}\left(1-U_j^{(2,k)}\right),}\hfill \end{array}\right.\end{array}$$

(30)

for $I,J=1,2,\dots ,d$.

Define the partial derivatives

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial MS{E}^{\left(PDE\right)}}{\partial {\omega }_{j\ell }^{\left(m\right)}}}& =& {\displaystyle \frac{1}{2(N-N^{\prime })}}\sum _{k=N^{\prime }+1}^N{\displaystyle \frac{\partial ER{R}_k^2}{\partial {\omega }_{j\ell }^{\left(m\right)}}},\hfill \end{array}$$

(31)

for $m=2,1$, $k=N^{\prime }+1,\dots ,N,j=1,2,\dots ,n_{m+1}$ and $\ell =1,2,\dots ,n_m$. Define the partial derivatives

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\partial MS{E}^{\left(PDE\right)}}{\partial {\theta }_j^{\left(m\right)}}}& =& {\displaystyle \frac{1}{2(N-N^{\prime })}}\sum _{k=N^{\prime }+1}^N{\displaystyle \frac{\partial ER{R}_k^2}{\partial {\theta }_j^{\left(m\right)}}},\hfill \end{array}$$

(32)

for $m=3,2$, $j=1,2,\dots ,n_m$. The partial derivatives ${\displaystyle \frac{1}{2}}{\displaystyle \frac{\partial ER{R}_k^2}{\partial {\omega }_{j\ell }^{\left(m\right)}}}$ are found as

$$\begin{array}{ccc}\hfill & & {\displaystyle \frac{1}{2}}{\displaystyle \frac{\partial ER{R}_k^2}{\partial {\omega }_{j\ell }^{\left(m\right)}}}=ER{R}_k{\displaystyle \frac{\partial ER{R}_k}{\partial {\omega }_{j\ell }^{\left(m\right)}}}\hfill \\ & =& ER{R}_k\left\{{\displaystyle \frac{\partial \left(D_{\tau }{U}_1^{(3,k)}\right)}{\partial {\omega }_{j\ell }^{\left(m\right)}}}-\sum _{I=1}^d\sum _{J=1}^d{a}_{IJ}{\displaystyle \frac{\partial \left(D_{IJ}{U}_1^{(3,k)}\right)}{\partial {\omega }_{j\ell }^{\left(m\right)}}}-\sum _{I=1}^d{b}_I{\displaystyle \frac{\partial \left(D_I{U}_1^{(3,k)}\right)}{\partial {\omega }_{j\ell }^{\left(m\right)}}}+r{\displaystyle \frac{\partial U_1^{(3,k)}}{\partial {\omega }_{j\ell }^{\left(m\right)}}}\right\},\hfill \end{array}$$

(33)

for $m=1,2$, for $j=1,2,\dots ,n_{m+1}$ and $\ell =1,2,\dots ,n_m$. The partial derivatives ${\displaystyle \frac{1}{2}}{\displaystyle \frac{\partial ER{R}_k^2}{\partial {\theta }_j^{\left(m\right)}}}$ are found as

$$\begin{array}{ccc}\hfill & & {\displaystyle \frac{1}{2}}{\displaystyle \frac{\partial ER{R}_k^2}{\partial {\theta }_j^{\left(m\right)}}}=ER{R}_k{\displaystyle \frac{\partial ER{R}_k}{\partial {\theta }_j^{\left(m\right)}}}\hfill \\ & =& ER{R}_k\left\{{\displaystyle \frac{\partial D_{\tau }{U}_1^{(3,k)}}{\partial {\theta }_j^{\left(m\right)}}}-\sum _{I=1}^d\sum _{J=1}^d{a}_{IJ}{\displaystyle \frac{\partial \left(D_{IJ}{U}_1^{(3,k)}\right)}{\partial {\theta }_j^{\left(m\right)}}}-\sum _{I=1}^d{b}_I{\displaystyle \frac{\partial \left(D_I{U}_1^{(3,k)}\right)}{\partial {\theta }_j^{\left(m\right)}}}+r{\displaystyle \frac{\partial U_1^{(3,k)}}{\partial {\theta }_j^{\left(m\right)}}}\right\},\hfill \end{array}$$

(34)

for $j=1,2,\dots ,n_m$ and $m=2,3$.

In (33), the partial derivatives ${\displaystyle \frac{\partial {\left(D{U}_1\right)}^{(3,k)}}{\partial {\omega }_{j\ell }^{\left(m\right)}}}$ with $D=D_{\tau },D=D_{IJ}$ or $D=D_I$ are defined as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle \frac{\partial {\left(D{U}_1\right)}^{(3,k)}}{\partial {\omega }_{1j}^{\left(2\right)}}={\left(D{U}_1\right)}_j^{(2,k)},}\hfill \\ {\displaystyle \frac{\partial {\left(D{U}_1\right)}^{(3,k)}}{\partial {\omega }_{j\ell }^{\left(1\right)}}={\omega }_{1j}^{\left(2\right)}{\left(D{U}_j\right)}^{(2,k)}\left(1-{\left(D{U}_j\right)}^{(2,k)}\right){\left(D{U}_{\ell }\right)}^{(1,k)},}\hfill \end{array}\right.\end{array}$$

(35)

for $j=1,2,\cdots ,n_2$ and $\ell =1,2,\cdots ,n_1$.

In (34), the partial derivatives ${\displaystyle \frac{\partial {\left(D{U}_1\right)}^{(3,k)}}{\partial {\theta }_j^{\left(m\right)}}}$ with $D=D_{\tau },D=D_{IJ}$ or $D=D_I$ are defined as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle \frac{\partial {\left(D{U}_1\right)}^{(3,k)}}{\partial {\theta }_1^{\left(3\right)}}=1,}\hfill \\ {\displaystyle \frac{\partial {\left(D{U}_1\right)}^{(3,k)}}{\partial {\theta }_j^{\left(2\right)}}={\omega }_{1j}^{\left(2\right)}{\left(D{U}_j\right)}^{(2,k)}\left(1-{\left(D{U}_j\right)}^{(2,k)}\right),}\hfill \end{array}\right.\end{array}$$

(36)

for $j=1,2,\cdots ,n_2$.

Define the increments of ${\mathit{\omega }}^{(PDE,i)}$ and ${\mathit{\theta }}^{(PDE,i)}$ as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle \Delta {\mathit{\omega }}^{(PDE,2)}=\frac{1}{2(N-N^{\prime })}\sum _{k=N^{\prime }+1}^N{\left[\frac{\partial ER{R}_k^{\left(2\right)}}{\partial {\omega }_{1j}^{\left(2\right)}}\right]}_{1\times n_2},}\hfill \\ {\displaystyle \Delta {\mathit{\omega }}^{(PDE,1)}=\frac{1}{2(N-N^{\prime })}\sum _{k=N^{\prime }+1}^N{\left[\begin{array}{c}{\displaystyle \frac{\partial ER{R}_k^{\left(2\right)}}{\partial {\omega }_{ij}^{\left(1\right)}}}\end{array}\right]}_{n_2\times n_1},}\hfill \\ \Delta {\mathit{\theta }}^{(PDE,3)}=1,\hfill \\ {\displaystyle \Delta {\mathit{\theta }}^{(PDE,2)}=\frac{1}{2(N-N^{\prime })}\sum _{k=N^{\prime }+1}^N{\left[\frac{\partial ER{R}_k^{\left(2\right)}}{\partial {\theta }_j^{\left(2\right)}}\right]}_{n_2\times 1},}\hfill \end{array}\right.\end{array}$$

(37)

with definitions (31) and (32).

Then, we update the network’s weights ${\mathit{\omega }}^{\left(2\right)},{\mathit{\omega }}^{\left(1\right)},{\mathit{\theta }}^{\left(3\right)}$ and ${\mathit{\theta }}^{\left(2\right)}$ as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\mathit{\omega }}^{\left(2\right)}={\mathit{\omega }}^{\left(2\right)}-\eta \left(\Delta {\mathit{\omega }}^{(payoff,2)}+\Delta {\mathit{\omega }}^{(PDE,2)}\right),\hfill \\ {\mathit{\omega }}^{\left(1\right)}={\mathit{\omega }}^{\left(1\right)}-\eta \left(\Delta {\mathit{\omega }}^{(payoff,1)}+\Delta {\mathit{\omega }}^{(PDE,1)}\right),\hfill \\ {\mathit{\theta }}^{\left(3\right)}={\mathit{\theta }}^{\left(3\right)}-\eta (\Delta {\mathit{\theta }}^{(payoff,3)}+\Delta {\mathit{\theta }}^{(PDE,3)}),\hfill \\ {\mathit{\theta }}^{\left(2\right)}={\mathit{\theta }}^{\left(2\right)}-\eta (\Delta {\mathit{\theta }}^{(payoff,2)}+\Delta {\mathit{\theta }}^{(PDE,2)}),\hfill \end{array}\right.\end{array}$$

(38)

where $\eta =constant$ is the learning rate and $\Delta {\mathit{\omega }}^{(*,j)}$ and $\Delta {\mathit{\theta }}^{(*,j)}$ are defined by (26) and (37).

Finally, define the total error as in (19) and

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle MS{E}^{\left(payoff\right)}=\frac{1}{2N^{\prime }}\sum _{k=1}^{N^{\prime }}{\left(U_1^{(3,k)}-y^{\left(k\right)}\right)}^2,(\mathrm{from}\mathrm{the}\mathrm{initial}\mathrm{constrains}),}\hfill \\ {\displaystyle MS{E}^{\left(PDE\right)}=\frac{1}{2(N-N^{\prime })}\sum _{k=N^{\prime }+1}^N{\left(ER{R}_k\right)}^2,(\mathrm{from}\mathrm{the}\mathrm{discrete}\mathrm{PDE}),}\hfill \end{array}\right.\end{array}$$

(39)

with $ER{R}_k$ being defined by (28). To minimize the objective function $MSE$, we have

$$\begin{array}{c}\hfill {\left\{{\mathit{\omega }}^{\left(i\right)},{\mathit{\theta }}^{\left(j\right)}\right\}}_{i=1,2;j=2,3}=argminMSE=argmin\left[MS{E}^{\left(payoff\right)}+MS{E}^{\left(PDE\right)}\right].\end{array}$$

(40)

and we obtain the trained network $net\left({\mathit{\omega }}^{\left(1\right)},{\mathit{\omega }}^{\left(2\right)},{\mathit{\theta }}^{\left(2\right)},{\mathit{\theta }}^{\left(3\right)},{\left\{U^{(1,k)}\right\}}_{k=1}^N\right)$ with input samples ${\left\{U^{(1,k)}\right\}}_{k=1}^N$ and optimal parameters $({\mathit{\omega }}^{\left(1\right)},{\mathit{\omega }}^{\left(2\right)},{\mathit{\theta }}^{\left(2\right)},{\mathit{\theta }}^{\left(3\right)})$ by iterating algorithm (38). Finally, we can compute

$$\begin{array}{c}\hfill U_1^{\left(3\right)}\left(\mathit{X}\right)=net\left({\mathit{\omega }}^{\left(1\right)},{\mathit{\omega }}^{\left(2\right)},{\mathit{\theta }}^{\left(2\right)},{\mathit{\theta }}^{\left(3\right)},\mathit{X}\right),\end{array}$$

(41)

for any input data $\mathit{X}={(x_1,x_2,\cdots ,x_d,\tau )}^{\prime }$.

The Algorithm 1 in Appendix A.1 gives the detailed procedure of the artificial neural network for option pricing.

4. Numerical Examples

4.1. Parameter Setting and Selection of Learning Rate

In computational experiments, we set parameters for the multi-asset option as follows. The dimension of PDE is set as $d=2,3,4$. The coefficients are set as ${\rho }_{ij}=0.8(i\ne j)$ and ${\rho }_{ii}=1$ for $i,j=1,2,\dots ,d$. The dividend yields are set as $q_i=0.002$, and risk-free interest is set as $r=0.05$. The volatilities are set as ${\sigma }_i=0.1$ for $i=1,2,\dots ,d$. Strike prices are set as $K=8,10$. Asset ratios are set as ${\alpha }_i=1/d$ for $i=1,2,\dots ,d$. The maturity date is taken as $T=0.4$.

The discrete stock prices are taken as $x_i=log\left(0.02\right)+{\displaystyle \frac{i}{n}}\left[log\left(70\right)-log\left(0.02\right)\right]$ with $i=1,2,\dots ,n$ and $n=5$. The time to maturity $\tau$ is discretized as ${\tau }_j=j\Delta \tau$ with $j=0,1,\dots ,M-1$, $M=T/\Delta \tau$ and $\Delta \tau =0.1$. The input data for network training are expressed by $\mathit{X}={[x_1^{\left(k\right)},x_2^{\left(k\right)},\dots ,x_d^{\left(k\right)},{\tau }^{\left(k\right)}]}^{\prime }$ with each $x_i^{\left(k\right)}$ being one of $x_i$ and each ${\tau }^{\left(k\right)}$ being one of ${\tau }_j$. So, the total number of training data is $n^{d}M$. The input data $\mathit{X}={[x_1^{\left(k\right)},x_2^{\left(k\right)},\dots ,x_d^{\left(k\right)},0]}^{\prime },k=1,2,\dots ,N^{\prime }$ are taken as the initial data with network output being the payoff functions $\Pi \left(\mathit{X}\right)$ (see (7) and (11)). The reminder data $\mathit{X}={[x_1^{\left(k\right)},x_2^{\left(k\right)},\dots ,x_d^{\left(k\right)},{\tau }^{\left(k\right)}]}^{\prime },k=N^{\prime }+1,\dots ,N$ with ${\tau }^{\left(k\right)}\ne 0$ are taken as the input data corresponding to the discrete PDE.

The network layers are taken as $n_1=d+1,n_2=5$ and $n_3=1$, and the training number is taken as $L=1000$ in the ANN structure.

Throughout the training process, we use the following techniques:

(1)

The learning rate $\eta$ is set as $\eta =1$ initially. When the objective function is not decreasing, we set $\eta =0.95\eta$. Throughout the training process, we use a factor of 0.95 to reduce the learning rate when the error does not decrease, repeatedly.

(2)

To speed the training process, at each iteration, we only use partial training data to update the weights ${\mathit{\omega }}^{\left(i\right)}$ and the bias ${\mathit{\theta }}^{\left(i\right)}$. Simply, at the $i^{\mathrm{th}}$ iteration, we use the data sequence labeled by $mod(i,N^{\prime }):N^{\prime }:N$ as the input data of ANN.

(3)

The simulation result is $U^3\left(\mathit{X}\right)=net\left({\mathit{\omega }}^{\left(1\right)},{\mathit{\omega }}^{\left(2\right)},{\mathit{\theta }}^{\left(1\right)},{\mathit{\theta }}^{\left(2\right)},\mathit{X}\right)$ for any input data $\mathit{X}$ (see expression (41)) contained within the envelope of $\{{\mathit{X}}^{\left(k\right)},k=1,2,\cdots ,N\}$, with trained ${\mathit{\omega }}^{\left(i\right)}$ and ${\mathit{\theta }}^{\left(j\right)}$.

  The Monte Carlo algorithm for arithmetic mean payoff function is described as in Algorithm A2 in Appendix A.1. For geometric mean payoff function, we modify the simulated option $P^{\left(k\right)}=max(0,K-{\sum }_{i=1}^d{e}^{{\alpha }_i{x}_{M-1,i}^{\left(k\right)}})$ at Step 3.

4.2. Numerical Results with Geometric Mean Payoff Function

With geometric mean payoff function (7), the multi-asset put option has analytical solution (8). We use some numerical examples to illustrate the errors between ANN numerical solutions and analytical options.

Table A1. Computational results of ANN for geometric mean payoff function with $d=2$. $[{\mathit{x}}^{\prime },\tau ]$ is the input data. ANN and Anal. are the ANN solution and analytical solution, respectively. ERR and RE are the absolute error and relative error between ANN solution and analytical solution, respectively. MSE is the error of training.

Input $[{\mathit{x}}^{\prime },\mathit{\tau }]$	ANN	Anal.	ERR	RE
	$K=8,MSE=1.8\times {10}^{-3}$
$(-3.91,-3.91,+0.40)$	$+7.8216$	$+7.7888$	$+0.0328$	$+0.0042$
$(-3.91,-1.87,+0.40)$	$+7.7861$	$+7.9078$	$-0.1217$	$-0.0156$
$(-3.91,+0.17,+0.40)$	$+7.6878$	$+7.7360$	$-0.0482$	$-0.0063$
$(-3.91,+2.21,+0.40)$	$+7.4150$	$+7.4818$	$-0.0668$	$-0.0090$
$(-1.87,-3.91,+0.40)$	$+7.7861$	$+7.7835$	$+0.0027$	$+0.0003$
$(+0.17,+0.17,+0.40)$	$+6.6584$	$+6.6113$	$+0.0471$	$+0.0071$
$(+2.21,+0.17,+0.40)$	$+4.5601$	$+4.5837$	$-0.0236$	$-0.0052$
	$K=10,MSE=4.3\times {10}^{-4}$
$(-3.91,-1.87,+0.40)$	$+9.7465$	$+9.7205$	$+0.0260$	$+0.0027$
$(-3.91,+0.17,+0.40)$	$+9.6482$	$+9.6928$	$-0.0447$	$-0.0046$
$(-3.91,+2.21,+0.40)$	$+9.3754$	$+9.3301$	$+0.0452$	$+0.0048$
$(-1.87,-3.91,+0.40)$	$+9.7465$	$+9.7531$	$-0.0065$	$-0.0007$
$(-1.87,+0.17,+0.40)$	$+9.3754$	$+9.4092$	$-0.0339$	$-0.0036$
$(-1.87,+2.21,+0.40)$	$+8.6188$	$+8.6425$	$-0.0237$	$-0.0028$
$(+0.17,-3.91,+0.40)$	$+9.6482$	$+9.6565$	$-0.0084$	$-0.0009$
$(+0.17,-1.87,+0.40)$	$+9.3754$	$+9.3850$	$-0.0097$	$-0.0010$
$(+0.17,+0.17,+0.40)$	$+8.6188$	$+8.6722$	$-0.0535$	$-0.0062$
$(+0.17,+2.21,+0.40)$	$+6.5205$	$+6.5223$	$-0.0018$	$-0.0003$
$(+2.21,-3.91,+0.40)$	$+9.3754$	$+9.3442$	$+0.0311$	$+0.0033$
$(+2.21,-1.87,+0.40)$	$+8.6188$	$+8.6436$	$-0.0248$	$-0.0029$
$(+2.21,+0.17,+0.40)$	$+6.5205$	$+6.5357$	$-0.0152$	$-0.0023$

,

Table A2. Computational results of ANN for geometric mean payoff function with $d=3.$.

Input $[{\mathit{x}}^{\mathit{T}},\mathit{\tau }]$	ANN	Anal.	ERR	RE
	$K=8,MSE=8.1\times {10}^{-4}$
$(-3.91,-3.91,-3.91,+0.40)$	$+7.8216$	$+7.8601$	$-0.0385$	$-0.0049$
$(-3.91,+0.17,+0.17,+0.40)$	$+7.5379$	$+7.4850$	$+0.0529$	$+0.0070$
$(-3.91,+0.17,+2.21,+0.40)$	$+7.2422$	$+7.2064$	$+0.0358$	$+0.0049$
$(-1.87,-3.91,-1.87,+0.40)$	$+7.7637$	$+7.7131$	$+0.0505$	$+0.0065$
$(-1.87,-3.91,+0.17,+0.40)$	$+7.6878$	$+7.6655$	$+0.0222$	$+0.0029$
$(-1.87,-1.87,-3.91,+0.40)$	$+7.7637$	$+7.7690$	$-0.0053$	$-0.0007$
$(-1.87,-1.87,-1.87,+0.40)$	$+7.6878$	$+7.7568$	$-0.0691$	$-0.0090$
$(-1.87,-1.87,+2.21,+0.40)$	$+7.2422$	$+7.2045$	$+0.0377$	$+0.0052$
$(-1.87,+0.17,-1.87,+0.40)$	$+7.5379$	$+7.5624$	$-0.0245$	$-0.0032$
$(-1.87,+0.17,+0.17,+0.40)$	$+7.2422$	$+7.1991$	$+0.0431$	$+0.0060$
$(-1.87,+2.21,-1.87,+0.40)$	$+7.2422$	$+7.2154$	$+0.0268$	$+0.0037$
$(+0.17,-3.91,-1.87,+0.40)$	$+7.6878$	$+7.6450$	$+0.0427$	$+0.0056$
$(+0.17,-1.87,-3.91,+0.40)$	$+7.6878$	$+7.6628$	$+0.0250$	$+0.0032$
$(+0.17,-1.87,+0.17,+0.40)$	$+7.2422$	$+7.2991$	$-0.0569$	$-0.0079$
$(+0.17,-1.87,+2.21,+0.40)$	$+6.6584$	$+6.6864$	$-0.0280$	$-0.0042$
$(+0.17,+2.21,-1.87,+0.40)$	$+6.6584$	$+6.6696$	$-0.0112$	$-0.0017$
$(+2.21,-3.91,-1.87,+0.40)$	$+7.5379$	$+7.5412$	$-0.0033$	$-0.0004$
$(+2.21,-3.91,+2.21,+0.40)$	$+6.6584$	$+6.7058$	$-0.0474$	$-0.0071$
$(+2.21,-1.87,-3.91,+0.40)$	$+7.5379$	$+7.6035$	$-0.0655$	$-0.0087$
$(+2.21,-1.87,+2.21,+0.40)$	$+5.5060$	$+5.4715$	$+0.0345$	$+0.0063$
$(-3.91,-3.91,-1.87,+0.40)$	$+9.7625$	$+9.8370$	$-0.0745$	$-0.0076$
$(-3.91,-3.91,+0.17,+0.40)$	$+9.7241$	$+9.7084$	$+0.0157$	$+0.0016$
$(-3.91,-3.91,+2.21,+0.40)$	$+9.6482$	$+9.5665$	$+0.0817$	$+0.0085$
$(-3.91,-1.87,-3.91,+0.40)$	$+9.7625$	$+9.8223$	$-0.0598$	$-0.0061$
$(-3.91,-1.87,-1.87,+0.40)$	$+9.7241$	$+9.6967$	$+0.0274$	$+0.0028$
$(-3.91,+0.17,-3.91,+0.40)$	$+9.7241$	$+9.6769$	$+0.0471$	$+0.0048$
$(-3.91,+0.17,+2.21,+0.40)$	$+9.2026$	$+9.1913$	$+0.0113$	$+0.0012$
$(-3.91,+2.21,+0.17,+0.40)$	$+9.2026$	$+9.1926$	$+0.0099$	$+0.0011$
$(-1.87,-3.91,-3.91,+0.40)$	$+9.7625$	$+9.7628$	$-0.0003$	$-0.0000$
$(-1.87,-3.91,+0.17,+0.40)$	$+9.6482$	$+9.7135$	$-0.0654$	$-0.0068$
$(-1.87,-3.91,+2.21,+0.40)$	$+9.4983$	$+9.5020$	$-0.0037$	$-0.0004$
$(-1.87,-1.87,-1.87,+0.40)$	$+9.6482$	$+9.6656$	$-0.0175$	$-0.0018$
$(-1.87,-1.87,+0.17,+0.40)$	$+9.4983$	$+9.5573$	$-0.0589$	$-0.0062$
$(-1.87,-1.87,+2.21,+0.40)$	$+9.2026$	$+9.2010$	$+0.0016$	$+0.0002$
$(-1.87,+0.17,-1.87,+0.40)$	$+9.4983$	$+9.4822$	$+0.0161$	$+0.0017$
$(-1.87,+0.17,+2.21,+0.40)$	$+8.6188$	$+8.6905$	$-0.0717$	$-0.0083$
$(-1.87,+2.21,-1.87,+0.40)$	$+9.2026$	$+9.2577$	$-0.0552$	$-0.0060$
$(-1.87,+2.21,+2.21,+0.40)$	$+7.4664$	$+7.5078$	$-0.0415$	$-0.0056$
$(+0.17,-3.91,-3.91,+0.40)$	$+9.7241$	$+9.7144$	$+0.0097$	$+0.0010$
$(+0.17,-3.91,-1.87,+0.40)$	$+9.6482$	$+9.6062$	$+0.0419$	$+0.0043$
$(+0.17,-3.91,+0.17,+0.40)$	$+9.4983$	$+9.5576$	$-0.0593$	$-0.0062$
$(+0.17,-1.87,-3.91,+0.40)$	$+9.6482$	$+9.5883$	$+0.0599$	$+0.0062$
$(+0.17,-1.87,-1.87,+0.40)$	$+9.4983$	$+9.4604$	$+0.0379$	$+0.0040$
$(+0.17,-1.87,+0.17,+0.40)$	$+9.2026$	$+9.2699$	$-0.0674$	$-0.0073$
$(+0.17,-1.87,+2.21,+0.40)$	$+8.6188$	$+8.5845$	$+0.0343$	$+0.0040$
$(+0.17,+0.17,-3.91,+0.40)$	$+9.4983$	$+9.4349$	$+0.0635$	$+0.0067$
$(+0.17,+0.17,-1.87,+0.40)$	$+9.2026$	$+9.2763$	$-0.0738$	$-0.0080$
$(+0.17,+2.21,-3.91,+0.40)$	$+9.2026$	$+9.1713$	$+0.0313$	$+0.0034$
$(+2.21,-3.91,-3.91,+0.40)$	$+9.6482$	$+9.7166$	$-0.0684$	$-0.0071$
$(+2.21,-3.91,-1.87,+0.40)$	$+9.4983$	$+9.5741$	$-0.0758$	$-0.0080$
$(+2.21,-3.91,+2.21,+0.40)$	$+8.6188$	$+8.5569$	$+0.0619$	$+0.0072$
$(+2.21,-1.87,-3.91,+0.40)$	$+9.4983$	$+9.5719$	$-0.0735$	$-0.0077$
$(+2.21,+0.17,-3.91,+0.40)$	$+9.2026$	$+9.2670$	$-0.0644$	$-0.0070$

and

Table A3. Computational results of ANN for geometric mean payoff function with $d=4$. $[{\mathit{x}}^{\prime },\tau ]$ are the input data. ANN and Anal. are the ANN solution and analytical solution, respectively. ERR and RE are the absolute error and relative error between ANN solution and analytical solution, respectively. MSE is the error of training.

Input $[{\mathit{x}}^{\mathit{T}},\mathit{\tau }]$	ANN	Anal.	ERR	RE
	$K=8,MSE=1.2\times {10}^{-3}$
$(-3.91,-3.91,+2.21,-1.87,+0.40)$	$+7.6878$	$+7.6748$	$+0.0129$	$+0.0017$
$(-3.91,-3.91,+2.21,+0.17,+0.40)$	$+7.5854$	$+7.6169$	$-0.0314$	$-0.0041$
$(-3.91,-3.91,+2.21,+2.21,+0.40)$	$+7.4150$	$+7.4759$	$-0.0609$	$-0.0082$
$(-3.91,-1.87,-3.91,-3.91,+0.40)$	$+7.8083$	$+7.7742$	$+0.0341$	$+0.0044$
$(-3.91,-1.87,-3.91,-1.87,+0.40)$	$+7.7861$	$+7.7877$	$-0.0016$	$-0.0002$
$(-3.91,-1.87,-1.87,-1.87,+0.40)$	$+7.7492$	$+7.7526$	$-0.0034$	$-0.0004$
$(-3.91,-1.87,-1.87,+0.17,+0.40)$	$+7.6878$	$+7.7240$	$-0.0363$	$-0.0047$
$(-3.91,-1.87,-1.87,+2.21,+0.40)$	$+7.5854$	$+7.5666$	$+0.0188$	$+0.0025$
$(-3.91,-1.87,+0.17,-3.91,+0.40)$	$+7.7492$	$+7.7100$	$+0.0393$	$+0.0051$
$(-3.91,-1.87,+0.17,-1.87,+0.40)$	$+7.6878$	$+7.6942$	$-0.0064$	$-0.0008$
$(-3.91,-1.87,+0.17,+0.17,+0.40)$	$+7.5854$	$+7.6411$	$-0.0557$	$-0.0073$
$(-3.91,-1.87,+0.17,+2.21,+0.40)$	$+7.4150$	$+7.4632$	$-0.0482$	$-0.0065$
$(-3.91,-1.87,+2.21,-3.91,+0.40)$	$+7.6878$	$+7.6420$	$+0.0458$	$+0.0060$
$(-3.91,-1.87,+2.21,-1.87,+0.40)$	$+7.5854$	$+7.5844$	$+0.0010$	$+0.0001$
$(-3.91,-1.87,+2.21,+0.17,+0.40)$	$+7.4150$	$+7.4604$	$-0.0455$	$-0.0061$
$(-1.87,-1.87,-3.91,+0.17,+0.40)$	$+7.6878$	$+7.6981$	$-0.0104$	$-0.0014$
$(-1.87,-1.87,-3.91,+2.21,+0.40)$	$+7.5854$	$+7.4987$	$+0.0867$	$+0.0114$
$(-1.87,-1.87,-1.87,-3.91,+0.40)$	$+7.7492$	$+7.7232$	$+0.0261$	$+0.0034$
$(-1.87,-1.87,-1.87,-1.87,+0.40)$	$+7.6878$	$+7.7151$	$-0.0273$	$-0.0036$
$(-1.87,-1.87,-1.87,+0.17,+0.40)$	$+7.5854$	$+7.6486$	$-0.0632$	$-0.0083$
$(-1.87,-1.87,-1.87,+2.21,+0.40)$	$+7.4150$	$+7.4078$	$+0.0071$	$+0.0010$
$(-1.87,-1.87,+0.17,-3.91,+0.40)$	$+7.6878$	$+7.6679$	$+0.0199$	$+0.0026$
$(-1.87,-1.87,+0.17,-1.87,+0.40)$	$+7.5854$	$+7.6244$	$-0.0390$	$-0.0051$
$(-1.87,-1.87,+0.17,+0.17,+0.40)$	$+7.4150$	$+7.5068$	$-0.0918$	$-0.0124$
$(-1.87,-1.87,+0.17,+2.21,+0.40)$	$+7.1311$	$+7.1738$	$-0.0427$	$-0.0060$
$(-1.87,-1.87,+2.21,-3.91,+0.40)$	$+7.5854$	$+7.5571$	$+0.0283$	$+0.0037$
$(-1.87,-1.87,+2.21,-1.87,+0.40)$	$+7.4150$	$+7.4324$	$-0.0174$	$-0.0024$
$(-1.87,-1.87,+2.21,+0.17,+0.40)$	$+7.1311$	$+7.1799$	$-0.0488$	$-0.0068$
$(-1.87,-1.87,+2.21,+2.21,+0.40)$	$+6.6584$	$+6.6481$	$+0.0103$	$+0.0015$
$(-1.87,+0.17,-3.91,-3.91,+0.40)$	$+7.7492$	$+7.7233$	$+0.0259$	$+0.0033$
$(-1.87,+0.17,-3.91,-1.87,+0.40)$	$+7.6878$	$+7.6863$	$+0.0015$	$+0.0002$
$(+0.17,+0.17,-3.91,+0.17,+0.40)$	$+7.4150$	$+7.3730$	$+0.0420$	$+0.0057$
$(+0.17,+0.17,-3.91,+2.21,+0.40)$	$+7.1311$	$+7.1468$	$-0.0157$	$-0.0022$
$(+0.17,+0.17,-1.87,-3.91,+0.40)$	$+7.5854$	$+7.6143$	$-0.0289$	$-0.0038$
$(+0.17,+0.17,-1.87,-1.87,+0.40)$	$+7.4150$	$+7.5071$	$-0.0922$	$-0.0124$
$(+0.17,+0.17,-1.87,+0.17,+0.40)$	$+7.1311$	$+7.1774$	$-0.0463$	$-0.0065$
$(+0.17,+0.17,-1.87,+2.21,+0.40)$	$+6.6584$	$+6.6715$	$-0.0132$	$-0.0020$
$(+0.17,+0.17,+0.17,-3.91,+0.40)$	$+7.4150$	$+7.4622$	$-0.0472$	$-0.0064$
$(+0.17,+0.17,+0.17,-1.87,+0.40)$	$+7.1311$	$+7.2425$	$-0.1114$	$-0.0156$
$(+0.17,+0.17,+0.17,+0.17,+0.40)$	$+6.6584$	$+6.6995$	$-0.0412$	$-0.0062$
$(+0.17,+0.17,+2.21,-3.91,+0.40)$	$+7.1311$	$+7.1357$	$-0.0046$	$-0.0006$
$(+0.17,+2.21,+2.21,-1.87,+0.40)$	$+5.8711$	$+5.7729$	$+0.0982$	$+0.0167$
$(+2.21,-3.91,+0.17,-1.87,+0.40)$	$+7.4150$	$+7.4679$	$-0.0530$	$-0.0071$
$(+2.21,-3.91,+0.17,+0.17,+0.40)$	$+7.1311$	$+7.2295$	$-0.0984$	$-0.0138$
$(+2.21,-3.91,+0.17,+2.21,+0.40)$	$+6.6584$	$+6.6742$	$-0.0158$	$-0.0024$
$(+2.21,-3.91,+2.21,-3.91,+0.40)$	$+7.4150$	$+7.3577$	$+0.0572$	$+0.0077$
$(+2.21,-3.91,+2.21,-1.87,+0.40)$	$+7.1311$	$+7.0954$	$+0.0357$	$+0.0050$
$(+2.21,-3.91,+2.21,+0.17,+0.40)$	$+6.6584$	$+6.6215$	$+0.0369$	$+0.0055$
$(+2.21,-3.91,+2.21,+2.21,+0.40)$	$+5.8711$	$+5.7745$	$+0.0966$	$+0.0164$
$(+2.21,-1.87,-3.91,-3.91,+0.40)$	$+7.6878$	$+7.6959$	$-0.0082$	$-0.0011$
	$K=10,MSE=2.2\times {10}^{-3}$
$(-3.91,-1.87,-1.87,+0.17,+0.40)$	$+9.6482$	$+9.6399$	$+0.0082$	$+0.0009$
$(-3.91,-1.87,-1.87,+2.21,+0.40)$	$+9.5458$	$+9.6132$	$-0.0674$	$-0.0071$
$(-3.91,-1.87,+0.17,-3.91,+0.40)$	$+9.7096$	$+9.6602$	$+0.0494$	$+0.0051$
$(-3.91,+2.21,+0.17,+2.21,+0.40)$	$+8.6188$	$+8.5567$	$+0.0620$	$+0.0072$
$(-3.91,+2.21,+2.21,-3.91,+0.40)$	$+9.3754$	$+9.3482$	$+0.0271$	$+0.0029$
$(-3.91,+2.21,+2.21,-1.87,+0.40)$	$+9.0915$	$+8.9321$	$+0.1594$	$+0.0175$
$(-3.91,+2.21,+2.21,+0.17,+0.40)$	$+8.6188$	$+8.6074$	$+0.0114$	$+0.0013$
$(-1.87,+0.17,-3.91,-1.87,+0.40)$	$+9.6482$	$+9.6558$	$-0.0076$	$-0.0008$
$(-1.87,+0.17,-3.91,+0.17,+0.40)$	$+9.5458$	$+9.5892$	$-0.0434$	$-0.0045$
$(-1.87,+0.17,-3.91,+2.21,+0.40)$	$+9.3754$	$+9.4482$	$-0.0729$	$-0.0078$
$(-1.87,+0.17,-1.87,-3.91,+0.40)$	$+9.6482$	$+9.6601$	$-0.0119$	$-0.0012$
$(-1.87,+0.17,-1.87,-1.87,+0.40)$	$+9.5458$	$+9.6054$	$-0.0596$	$-0.0062$
$(-1.87,+0.17,-1.87,+0.17,+0.40)$	$+9.3754$	$+9.4869$	$-0.1116$	$-0.0119$
$(-1.87,+0.17,-1.87,+2.21,+0.40)$	$+9.0915$	$+9.2449$	$-0.1534$	$-0.0169$
$(+0.17,-1.87,+0.17,-3.91,+0.40)$	$+9.5458$	$+9.6244$	$-0.0786$	$-0.0082$
$(+0.17,-1.87,+0.17,+2.21,+0.40)$	$+8.6188$	$+8.5543$	$+0.0645$	$+0.0075$
$(+0.17,+0.17,-3.91,-3.91,+0.40)$	$+9.6482$	$+9.6409$	$+0.0073$	$+0.0008$
$(+0.17,+0.17,-3.91,-1.87,+0.40)$	$+9.5458$	$+9.5239$	$+0.0219$	$+0.0023$
$(+0.17,+0.17,-3.91,+0.17,+0.40)$	$+9.3754$	$+9.3272$	$+0.0481$	$+0.0051$
$(+0.17,+0.17,-3.91,+2.21,+0.40)$	$+9.0915$	$+9.1138$	$-0.0224$	$-0.0025$
$(+2.21,-3.91,+2.21,-1.87,+0.40)$	$+9.0915$	$+9.0316$	$+0.0599$	$+0.0066$
$(+2.21,-3.91,+2.21,+0.17,+0.40)$	$+8.6188$	$+8.7206$	$-0.1018$	$-0.0118$
$(+2.21,-3.91,+2.21,+2.21,+0.40)$	$+7.8315$	$+7.9283$	$-0.0968$	$-0.0124$
$(+2.21,-1.87,-3.91,-3.91,+0.40)$	$+9.6482$	$+9.6509$	$-0.0028$	$-0.0003$
$(+2.21,-1.87,-3.91,-1.87,+0.40)$	$+9.5458$	$+9.5469$	$-0.0011$	$-0.0001$
$(+2.21,-1.87,-3.91,+0.17,+0.40)$	$+9.3754$	$+9.3579$	$+0.0175$	$+0.0019$
$(+2.21,-1.87,-1.87,-3.91,+0.40)$	$+9.5458$	$+9.5765$	$-0.0307$	$-0.0032$
$(+2.21,-1.87,-1.87,-1.87,+0.40)$	$+9.3754$	$+9.4073$	$-0.0320$	$-0.0034$
$(+2.21,-1.87,+0.17,-3.91,+0.40)$	$+9.3754$	$+9.4532$	$-0.0778$	$-0.0083$
$(+2.21,-1.87,+0.17,-1.87,+0.40)$	$+9.0915$	$+9.0265$	$+0.0649$	$+0.0071$
$(+2.21,-1.87,+0.17,+0.17,+0.40)$	$+8.6188$	$+8.5463$	$+0.0725$	$+0.0084$

list the ANN solutions (labeled with ‘ANN’) and the analytical solutions (labeled with ‘Anal.’) with $d=2,3,4$, time to expiry date $\tau =0.4$ and different strike prices of $K=8,10$. In Table A1, Table A2 and Table A3, the fourth column and the fifth column are the absolute errors (labeled with ‘ERR’) and the relative errors (labeled with ‘RE’), respectively. Figure A1 plots the analytical solutions and ANN solutions at different asset points with parameters: $T=0.4$, $K=8,10$ and $d=2$.

From Table A1, Table A2 and Table A3, we see the absolute errors are about $0.05$ and the relative errors are about $0.08\%$ between ANN solutions and analytical solutions, which illustrate that our ANN scheme is efficient and accurate. Moreover, the whole results of calculation are quite stable, there is no unstable result.

The Figure A2 records the error evolution path with respect to training number L. We see from this figure, the errors are decreasing quickly as the iteration number L increases. As iteration number L increases, the Figure A3 records the learning rate path $\eta$. The learning rate $\eta$ decreases with the increase in L from 1 to 0.1.

Table A1, Table A2 and Table A3 and Figure A1, Figure A2 and Figure A3 show that our ANN-3 is efficient, accurate and computationally stable for option pricing for geometric mean payoff function.

Lastly, according to the results of numerical examples, we give a conjecture about the errors between ANN solutions and analytical solutions of multi-asset options, which is the following conjecture.

Conjecture 1.

Assume $U_1^{\left(3\right)}\left(\mathit{X}\right)=net({\mathit{\omega }}^{\left(1\right)},{\mathit{\omega }}^{\left(2\right)},{\mathit{\theta }}^{\left(2\right)},{\mathit{\theta }}^{\left(3\right)},\mathit{X})$ is the network solution and $u\left(\mathit{X}\right)$ is the exact solution of multi-asset option. Then, the error between $U_1^{\left(3\right)}\left(\mathit{X}\right)$ and $u\left(\mathit{X}\right)$ can be estimated by

$$\begin{array}{c}\hfill |U_1^{\left(3\right)}\left(\mathit{X}\right)-u\left(\mathit{X}\right)|\le O\left({\displaystyle \frac{C}{n_2}}\right),\end{array}$$

(42)

for any asset prices and any remainder time $\mathit{X}=\left[{\mathit{x}}^{\prime },\tau \right]$, where $n_2$ is the number of neurons in ANN hidden layer, and C is a certain constant.

Figure A4 plots the error vs. the number $n_2$ of ANN, from which we see the ANN error is about $O\left({\displaystyle \frac{C}{n_2}}\right)$ with C being a positive constant, which confirms Conjecture 1.

4.3. Numerical Results with Arithmetic Mean Payoff Function

In this subsection, we list some results of the put options with arithmetic mean payoff function $\Pi \left(\mathit{x}\right)=max\left({\displaystyle 0,\sum _{i=1}^d{\alpha }_i{e}^{x_i}}\right)$. This type of option has no analytical solution. We give the Monte Carlo simulation (the procedure can be seen in Algorithm A2) as the compared results.

Table A4. Computational results of ANN for arithmetic mean payoff function with $d=2$. $[{\mathit{x}}^{\prime },\tau ]$ are the input data. ANN and Anal. are the ANN solution and analytical solution, respectively. ERR and RE are the absolute error and relative error between ANN solution and analytical solution, respectively. MSE is the error of training.

Input $[{\mathit{x}}^{\mathit{T}},\mathit{\tau }]$	ANN	Anal.	ERR	RE
	$K=8,MSE=3.14\times {10}^{-4}$
$(-3.91,-1.87,+0.40)$	$+7.7546$	$+7.8076$	$-0.0530$	$-0.0068$
$(-1.87,-3.91,+0.40)$	$+7.7546$	$+7.7609$	$-0.0063$	$-0.0008$
$(-1.87,+0.17,+0.40)$	$+7.1722$	$+7.1619$	$+0.0103$	$+0.0014$
$(+0.17,-1.87,+0.40)$	$+7.1725$	$+7.2073$	$-0.0348$	$-0.0049$
$(+2.21,+0.17,+0.40)$	$+2.6955$	$+2.6926$	$+0.0029$	$+0.0011$
	$K=10,MSE=2.85\times {10}^{-4}$
$(-3.91,+0.17,+0.40)$	$+9.1994$	$+9.1818$	$+0.0176$	$+0.0019$
$(+0.17,-3.91,+0.40)$	$+9.1997$	$+9.1792$	$+0.0205$	$+0.0022$
$(+0.17,-1.87,+0.40)$	$+9.1325$	$+9.1688$	$-0.0363$	$-0.0040$
$(+0.17,+0.17,+0.40)$	$+8.6171$	$+8.5883$	$+0.0288$	$+0.0033$
$(+2.21,-3.91,+0.40)$	$+5.2365$	$+5.2662$	$-0.0297$	$-0.0057$

,

Table A5. Computational results of ANN for arithmetic mean payoff function with $d=3$. $[{\mathit{x}}^{\prime },\tau ]$ are the input data. ANN and Anal. are the ANN solution and analytical solution, respectively. ERR and RE are the absolute error and relative error between ANN solution and analytical solution, respectively. MSE is the error of training.

Input $[{\mathit{x}}^{\mathit{T}},\mathit{\tau }]$	ANN	MC	ERR	RE
	$K=8,MSE=4.8\times {10}^{-4}$
$(-3.91,-3.91,-1.87,+0.40)$	$+7.7769$	$+7.8250$	$-0.0481$	$-0.0062$
$(-3.91,-3.91,+0.17,+0.40)$	$+7.4334$	$+7.4477$	$-0.0143$	$-0.0019$
$(-3.91,-1.87,-3.91,+0.40)$	$+7.7769$	$+7.7968$	$-0.0199$	$-0.0026$
$(-3.91,+0.17,-3.91,+0.40)$	$+7.4333$	$+7.3982$	$+0.0351$	$+0.0047$
$(-3.91,+0.17,+0.17,+0.40)$	$+7.0453$	$+7.0257$	$+0.0196$	$+0.0028$
$(-3.91,+0.17,+2.21,+0.40)$	$+4.4034$	$+4.4039$	$-0.0005$	$-0.0001$
$(-1.87,-3.91,-1.87,+0.40)$	$+7.7323$	$+7.7277$	$+0.0045$	$+0.0006$
$(-1.87,-3.91,+0.17,+0.40)$	$+7.3887$	$+7.3561$	$+0.0326$	$+0.0044$
$(-1.87,-1.87,-3.91,+0.40)$	$+7.7323$	$+7.7070$	$+0.0253$	$+0.0033$
$(-1.87,-1.87,+0.17,+0.40)$	$+7.3441$	$+7.3741$	$-0.0300$	$-0.0041$
$(-1.87,+0.17,-3.91,+0.40)$	$+7.3886$	$+7.3410$	$+0.0476$	$+0.0064$
$(-1.87,+2.21,+0.17,+0.40)$	$+4.3591$	$+4.3728$	$-0.0137$	$-0.0031$
$(+0.17,-3.91,-1.87,+0.40)$	$+7.3887$	$+7.3364$	$+0.0522$	$+0.0071$
$(+0.17,-3.91,+2.21,+0.40)$	$+4.4034$	$+4.3813$	$+0.0220$	$+0.0050$
$(+0.17,-1.87,-3.91,+0.40)$	$+7.3887$	$+7.3382$	$+0.0505$	$+0.0068$
$(+0.17,-1.87,-1.87,+0.40)$	$+7.3443$	$+7.3378$	$+0.0065$	$+0.0009$
	$K=10,MSE=1.1\times {10}^{-3}$
$(-3.91,-1.87,+2.21,+0.40)$	$+6.7066$	$+6.6975$	$+0.0092$	$+0.0014$
$(-3.91,+0.17,-3.91,+0.40)$	$+9.3938$	$+9.3207$	$+0.0731$	$+0.0078$
$(-3.91,+2.21,+0.17,+0.40)$	$+6.3626$	$+6.4076$	$-0.0450$	$-0.0071$
$(-3.91,+2.21,+2.21,+0.40)$	$+3.7206$	$+3.6841$	$+0.0365$	$+0.0098$
$(-1.87,+0.17,+0.17,+0.40)$	$+8.9609$	$+9.0426$	$-0.0816$	$-0.0091$
$(+0.17,-3.91,-1.87,+0.40)$	$+9.3489$	$+9.3431$	$+0.0058$	$+0.0006$
$(+0.17,-3.91,+0.17,+0.40)$	$+9.0053$	$+8.9258$	$+0.0795$	$+0.0088$
$(+0.17,-3.91,+2.21,+0.40)$	$+6.3624$	$+6.4253$	$-0.0629$	$-0.0099$
$(+0.17,-1.87,-3.91,+0.40)$	$+9.3492$	$+9.3104$	$+0.0387$	$+0.0041$
$(+0.17,-1.87,+0.17,+0.40)$	$+8.9609$	$+8.9799$	$-0.0190$	$-0.0021$
	$K=10,MSE=1.1\times {10}^{-3}$
$(+0.17,+0.17,-1.87,+0.40)$	$+8.9606$	$+8.9319$	$+0.0287$	$+0.0032$
$(+0.17,+0.17,+2.21,+0.40)$	$+5.9758$	$+5.9343$	$+0.0415$	$+0.0069$
$(+0.17,+2.21,-3.91,+0.40)$	$+6.3625$	$+6.4112$	$-0.0487$	$-0.0077$
$(+0.17,+2.21,+0.17,+0.40)$	$+5.9747$	$+6.0245$	$-0.0498$	$-0.0083$
$(+2.21,-3.91,+0.17,+0.40)$	$+6.3634$	$+6.3904$	$-0.0270$	$-0.0043$
$(+2.21,-1.87,+0.17,+0.40)$	$+6.3175$	$+6.3166$	$+0.0009$	$+0.0001$
$(+2.21,+0.17,-3.91,+0.40)$	$+6.3621$	$+6.3359$	$+0.0262$	$+0.0041$
$(+2.21,+0.17,-1.87,+0.40)$	$+6.3182$	$+6.2769$	$+0.0413$	$+0.0065$

and

Table A6. Computational results of ANN for arithmetic mean payoff function with $d=4$.

Input $[{\mathit{x}}^{\mathit{T}},\mathit{\tau }]$	ANN	MC	ERR	RE
	$K=8,MSE=5.84\times {10}^{-4}$
$(-3.91,-3.91,-3.91,-3.91,+0.40)$	$+7.8216$	$+7.8081$	$+0.0135$	$+0.0017$
$(-3.91,-3.91,-3.91,+0.17,+0.40)$	$+7.5304$	$+7.5918$	$-0.0614$	$-0.0082$
$(-3.91,-3.91,-1.87,-1.87,+0.40)$	$+7.7546$	$+7.7980$	$-0.0434$	$-0.0056$
$(-3.91,-3.91,-1.87,+0.17,+0.40)$	$+7.4970$	$+7.5068$	$-0.0098$	$-0.0013$
$(-3.91,-3.91,+0.17,+2.21,+0.40)$	$+5.2569$	$+5.2893$	$-0.0324$	$-0.0062$
$(-3.91,-3.91,+2.21,-3.91,+0.40)$	$+5.5484$	$+5.5281$	$+0.0204$	$+0.0037$
$(-3.91,-3.91,+2.21,-1.87,+0.40)$	$+5.5154$	$+5.5128$	$+0.0026$	$+0.0005$
$(-3.91,-3.91,+2.21,+2.21,+0.40)$	$+3.2779$	$+3.2645$	$+0.0134$	$+0.0041$
$(-3.91,-1.87,-3.91,-3.91,+0.40)$	$+7.7881$	$+7.8140$	$-0.0259$	$-0.0033$
$(-1.87,-3.91,-3.91,-3.91,+0.40)$	$+7.7881$	$+7.7979$	$-0.0098$	$-0.0013$
$(-1.87,-3.91,-1.87,-3.91,+0.40)$	$+7.7546$	$+7.7171$	$+0.0375$	$+0.0048$
$(-1.87,-3.91,-1.87,+0.17,+0.40)$	$+7.4633$	$+7.5152$	$-0.0519$	$-0.0070$
$(-1.87,-3.91,+0.17,-3.91,+0.40)$	$+7.4970$	$+7.4455$	$+0.0515$	$+0.0069$
$(-1.87,-3.91,+0.17,-1.87,+0.40)$	$+7.4634$	$+7.4627$	$+0.0007$	$+0.0001$
$(-1.87,-3.91,+0.17,+0.17,+0.40)$	$+7.2058$	$+7.1455$	$+0.0603$	$+0.0084$
$(-1.87,-3.91,+2.21,+0.17,+0.40)$	$+5.2245$	$+5.2417$	$-0.0172$	$-0.0033$
$(-1.87,-1.87,-3.91,-3.91,+0.40)$	$+7.7546$	$+7.8123$	$-0.0577$	$-0.0074$
$(-1.87,-1.87,-1.87,-3.91,+0.40)$	$+7.7211$	$+7.7554$	$-0.0343$	$-0.0044$
$(-1.87,-1.87,+0.17,-3.91,+0.40)$	$+7.4635$	$+7.5239$	$-0.0604$	$-0.0081$
$(-1.87,-1.87,+0.17,+0.17,+0.40)$	$+7.1722$	$+7.2123$	$-0.0401$	$-0.0056$
$(+0.17,-1.87,-1.87,+0.17,+0.40)$	$+7.1721$	$+7.2436$	$-0.0716$	$-0.0100$
$(+0.17,-1.87,+0.17,+0.17,+0.40)$	$+6.9147$	$+6.9364$	$-0.0217$	$-0.0031$
$(+0.17,-1.87,+2.21,+0.17,+0.40)$	$+4.9332$	$+4.9557$	$-0.0224$	$-0.0045$
$(+0.17,+0.17,-3.91,-3.91,+0.40)$	$+7.2391$	$+7.2214$	$+0.0176$	$+0.0024$
$(+0.17,+0.17,-3.91,+0.17,+0.40)$	$+6.9485$	$+6.9426$	$+0.0058$	$+0.0008$
$(+0.17,+0.17,-3.91,+2.21,+0.40)$	$+4.9672$	$+4.9248$	$+0.0424$	$+0.0085$
$(+0.17,+0.17,-1.87,-3.91,+0.40)$	$+7.2057$	$+7.2566$	$-0.0509$	$-0.0071$
$(+0.17,+0.17,-1.87,+0.17,+0.40)$	$+6.9147$	$+6.9308$	$-0.0161$	$-0.0023$
$(+0.17,+0.17,-1.87,+2.21,+0.40)$	$+4.9327$	$+4.9307$	$+0.0019$	$+0.0004$
$(+0.17,+0.17,+0.17,-3.91,+0.40)$	$+6.9483$	$+6.9758$	$-0.0276$	$-0.0040$
$(+0.17,+0.17,+0.17,-1.87,+0.40)$	$+6.9144$	$+6.9505$	$-0.0361$	$-0.0052$
$(+0.17,+0.17,+0.17,+0.17,+0.40)$	$+6.6572$	$+6.6243$	$+0.0329$	$+0.0049$
$(+0.17,+0.17,+0.17,+2.21,+0.40)$	$+4.6761$	$+4.7129$	$-0.0368$	$-0.0079$
$(+0.17,+0.17,+2.21,-3.91,+0.40)$	$+4.9654$	$+4.9871$	$-0.0217$	$-0.0044$
$(+0.17,+0.17,+2.21,-1.87,+0.40)$	$+4.9319$	$+4.9594$	$-0.0275$	$-0.0056$
$(+2.21,+0.17,-3.91,+0.17,+0.40)$	$+4.9657$	$+4.9267$	$+0.0390$	$+0.0079$
$(+2.21,+0.17,+0.17,-3.91,+0.40)$	$+4.9660$	$+4.9243$	$+0.0418$	$+0.0084$
$(+2.21,+0.17,+0.17,-1.87,+0.40)$	$+4.9321$	$+4.9334$	$-0.0012$	$-0.0003$
$(+2.21,+0.17,+0.17,+0.17,+0.40)$	$+4.6752$	$+4.6722$	$+0.0030$	$+0.0006$
$(+2.21,+0.17,+2.21,-1.87,+0.40)$	$+2.9518$	$+2.9571$	$-0.0053$	$-0.0018$
$(+2.21,+0.17,+2.21,+0.17,+0.40)$	$+2.6944$	$+2.6819$	$+0.0126$	$+0.0047$
$(+2.21,+2.21,-3.91,+0.17,+0.40)$	$+2.9846$	$+2.9906$	$-0.0060$	$-0.0020$
	$K=10,MSE=8.3\times {10}^{-4}$
$(-3.91,+0.17,-1.87,-1.87,+0.40)$	$+9.4237$	$+9.4729$	$-0.0492$	$-0.0052$
$(-3.91,+0.17,-1.87,+0.17,+0.40)$	$+9.1661$	$+9.0905$	$+0.0756$	$+0.0082$
$(-3.91,+0.17,+0.17,-1.87,+0.40)$	$+9.1663$	$+9.1235$	$+0.0428$	$+0.0047$
$(-3.91,+0.17,+0.17,+2.21,+0.40)$	$+6.9271$	$+6.9686$	$-0.0415$	$-0.0060$
$(-3.91,+0.17,+2.21,+0.17,+0.40)$	$+6.9274$	$+6.9813$	$-0.0539$	$-0.0078$
$(-3.91,+2.21,-3.91,+0.17,+0.40)$	$+7.2174$	$+7.2644$	$-0.0470$	$-0.0065$
$(-3.91,+2.21,+0.17,+0.17,+0.40)$	$+6.9261$	$+6.8745$	$+0.0516$	$+0.0074$
$(-1.87,-3.91,-3.91,-3.91,+0.40)$	$+9.7485$	$+9.6798$	$+0.0687$	$+0.0070$
$(-1.87,+0.17,-3.91,-1.87,+0.40)$	$+9.4238$	$+9.4567$	$-0.0329$	$-0.0035$
$(-1.87,+0.17,-3.91,+2.21,+0.40)$	$+7.1845$	$+7.2401$	$-0.0556$	$-0.0077$
$(-1.87,+0.17,-1.87,-3.91,+0.40)$	$+9.4237$	$+9.4359$	$-0.0121$	$-0.0013$
$(-1.87,+0.17,-1.87,+0.17,+0.40)$	$+9.1326$	$+9.1322$	$+0.0004$	$+0.0000$
$(-1.87,+0.17,+0.17,-1.87,+0.40)$	$+9.1326$	$+9.1468$	$-0.0142$	$-0.0016$
$(-1.87,+0.17,+2.21,-3.91,+0.40)$	$+7.1845$	$+7.1557$	$+0.0287$	$+0.0040$
$(-1.87,+2.21,+0.17,+0.17,+0.40)$	$+6.8937$	$+6.8708$	$+0.0229$	$+0.0033$
$(+0.17,-3.91,-3.91,-3.91,+0.40)$	$+9.4908$	$+9.4429$	$+0.0479$	$+0.0050$
$(+0.17,-3.91,-3.91,-1.87,+0.40)$	$+9.4573$	$+9.5198$	$-0.0625$	$-0.0066$
$(+0.17,-3.91,-1.87,-3.91,+0.40)$	$+9.4574$	$+9.4519$	$+0.0055$	$+0.0006$
$(+0.17,-3.91,-1.87,+0.17,+0.40)$	$+9.1662$	$+9.1604$	$+0.0058$	$+0.0006$
$(+0.17,-3.91,+0.17,-3.91,+0.40)$	$+9.1998$	$+9.1222$	$+0.0776$	$+0.0084$
$(+0.17,-3.91,+2.21,-3.91,+0.40)$	$+7.2180$	$+7.2115$	$+0.0065$	$+0.0009$
$(+2.21,-1.87,-3.91,+0.17,+0.40)$	$+7.1853$	$+7.2001$	$-0.0149$	$-0.0021$
$(+2.21,-1.87,+0.17,-3.91,+0.40)$	$+7.1846$	$+7.1990$	$-0.0144$	$-0.0020$
$(+2.21,-1.87,+0.17,+0.17,+0.40)$	$+6.8925$	$+6.9404$	$-0.0479$	$-0.0069$
$(+2.21,-1.87,+2.21,-3.91,+0.40)$	$+5.2034$	$+5.2511$	$-0.0477$	$-0.0092$
$(+2.21,+0.17,-3.91,+0.17,+0.40)$	$+6.9269$	$+6.9182$	$+0.0087$	$+0.0013$
$(+2.21,+0.17,+0.17,-3.91,+0.40)$	$+6.9265$	$+6.9487$	$-0.0222$	$-0.0032$
$(+2.21,+0.17,+0.17,+0.17,+0.40)$	$+6.6357$	$+6.5998$	$+0.0359$	$+0.0054$
$(+2.21,+2.21,+0.17,+0.17,+0.40)$	$+4.6533$	$+4.6242$	$+0.0291$	$+0.0063$
$(+2.21,+2.21,+0.17,+2.21,+0.40)$	$+2.6712$	$+2.6801$	$-0.0089$	$-0.0033$

list the compared results of ANN solutions (labeled with ‘ANN’) and Monte Carlo simulation (labeled with ‘MC’). From these tables, we see the absolute errors (labeled with ‘ERR’) are less than $2\times {10}^{-2}$ and the relative errors (labeled with ‘RE’) are less than $0.8\%$, which illustrate that our ANN algorithm is effective and accurate. Figure A5 plots the compared solutions obtained by the ANN method and the Monte Carlo simulation with $d=2,T=0.4$ and $K=8$.

Figure A6 records the errors with respect to iteration number L. From this figure, we see the errors are decreasing quickly as the iteration number L increases. Figure A7 records the learning rate path as the iteration number L increases. From this figure, we see that as the iteration number progresses, the learning rate decreases from 1 to 0.01.

Table A4, Table A5 and Table A6 and Figure A5, Figure A6 and Figure A7 illustrate that our ANN-3 is efficient, accurate and computationally stable for option pricing for arithmetic mean payoff function.

5. Conclusions

This paper gives an artificial neural network (ANN) with three layers to price the European multi-asset options with the input data $\mathit{X}$ from discrete PDE. By setting the numbers $(n_1,n_2,n_3)$ of the ANN structure, the multi-asset options are simulated correctly. The key technology is obtaining the discrete formula of PDE and then computing the partial derivatives of the objective function with respect to weights ${\omega }_{ij}^{\left(k\right)}$ and bias ${\theta }_j^{\left(k\right)}$. By setting the learning parameter $\eta$, the ANN is trained well and the optimal parameters ${\omega }_{ij}^{\left(k\right)}$ and bias ${\theta }_j^{\left(k\right)}$ are then obtained. Lastly, the option prices are simulated by inputting the asset prices and remainder time into the trained ANN.

Numerical examples are carried for two options, geometric mean payoff functions and arithmetic mean payoff functions. These examples show that the three-layer ANN has powerful capability for multi-asset option pricing. The error and relative errors of ANN options are very small. In these experiments, we use a variable learning rate $\eta$, which ensures that our objective functions continue to decline. In addition, we use partial input data to train the network, thereby reducing the amount of computation in ANN.

In future, we will establish a deep learning ANN with a multi-layer (i.e., $p>3$) structure for option pricing. The challenge is how to compute the partial derivatives with respect to parameters ${\omega }_{ij}^{\left(k\right)}$ and ${\theta }_j^{\left(k\right)}$. Moreover, we will consider the ANN to price multi-asset complex options, such as American options, Asian options and barrier options, which is a bit complicated.