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arXiv 2307.04510q-fin.MF
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用于摆动期权定价的线性回归与神经网络近似分析

An analysis of linear regression and neural networks approximation for the pricing of swing options

Christian Yeo

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33 pages
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AI总结：

该研究分析摆动期权定价中线性回归与神经网络的近似方法，证明两类方法下近似解随基函数数量、蒙特卡洛样本量增大的收敛性，并给出线性回归的收敛速率。

摘要：

线性回归最初被引入用于美式期权定价，此后其应用范围已扩展至摆动期权定价领域。摆动期权的价格可被视为反向动态规划原理的解，该原理涉及一个被称为延续价值的条件期望。使用线性回归对延续价值进行近似包含两层近似。首先，延续价值被替换为在由有限个$m$平方可积函数张成的子空间上的正交投影，从而得到摆动价值函数的第一阶近似$V^m$。本文证明，当回归函数选择得当时，随着$m \to + \infty$，$V^m$收敛于摆动期权的实际价格$V$。当用神经网络替代经典回归函数时，也证明了类似的结果。针对这两种方法（线性回归和神经网络），我们分析了第二层近似，即使用蒙特卡洛模拟实际计算摆动期权价格并得到近似值$V^{m, N}$（其中$N$表示蒙特卡洛样本量）。特别地，我们证明了对于两种方法，以及在线性回归中采用希尔伯特基假设的情况下，当$N \to + \infty$时，$V^{m, N} \to V^m$。此外，在线性回归情形下，证明了收敛速度为$\mathcal{O}\big(\frac{1}{\sqrt{N}} \big)$阶。

英文摘要：

Linear regression, firstly introduced for the pricing of American-style options, has since been expanded to include swing options pricing. Swing options price may be viewed as the solution to a Backward Dynamic Programming Principle, which involves a conditional expectation known as the continuation value. The approximation of the continuation value using linear regression involves two levels of approximation. First, the continuation value is replaced by an orthogonal projection over a subspace spanned by a finite set of $m$ squared-integrable functions yielding a first approximation $V^m$ of the swing value function. In this paper, we prove that, with well-chosen regression functions, $V^m$ converges to the swing actual price $V$ as $m \to + \infty$. A similar result is proved when classic regression functions are replaced by neural networks. For both methods (linear regression and neural networks), we analyze the second level of approximation involving practical computation of the swing price using Monte Carlo simulations and yielding an approximation $V^{m, N}$ (where $N$ denotes the Monte Carlo sample size). Especially, we prove that $V^{m, N} \to V^m$ as $N \to + \infty$ for both methods and using a Hilbert basis assumption in the linear regression. Besides, a convergence rate of order $\mathcal{O}\big(\frac{1}{\sqrt{N}} \big)$ is proved in the linear regression case.

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