在快速发展的金融市场, 金融衍生品的定价问题一直以来都是人们关注的焦点, 其中亚式期权作为股票期权衍生的一种新式期权, 其特殊性在于它是通过相关证券在合同期间某段时间内的平均价格来决定回报, 这就既减少了合同期内价格浮动对期权定价的影响, 同时也能较为准确地反映资产价格的浮动趋势. 在经典的Black-Scholes期权定价模型中, 亚式期权的定价问题等价于布朗运动指数泛函的分布问题, 可以表示为以下数学形式, 若$ \tilde{B}(t) $为标准布朗运动

$ \begin{equation} \begin{aligned} S(t)& = S(0)\exp(\sigma\tilde{B}(t)+(\mu-\frac{\sigma^{2}}{2})t), t\geq0, \sigma\in\mathbb{R}, \mu\in\mathbb{R}, \\ A(t)& = \frac{1}{t}\int_{0}^{t}S(u)\mathrm{du}, t>0. \end{aligned} \end{equation} $

(1.1)

其中$ S(t) $为定价过程, $ A(t) $为平均定价过程.

近些年来, 众多学者对布朗运动指数泛函的相关问题进行了广泛研究. Marc.Yor[1], Dufresne.Daniel[2-3]研究了指数泛函的分布问题并将其应用于风险理论与年金问题中. Tamás.Szabados[4-6]利用对称随机游动构造出指数泛函的离散化形式, 进一步研究其矩问题.

然而Black-Scholes定价公式本身的一些假设与现实存在差距. 比如金融资产价格服从对数正态分布, 且波动率为常数. 但实际上, 波动率可能是不确定的, 不一定是常数. $ G- $布朗运动的参数在一个区间内变化, 更加符合复杂多变的金融市场. 在不确定问题、风险测度和金融中的超套期保值问题的驱动下, 彭实戈[7-8]院士提出了次线性空间的概念, 这是概率空间的一个推广. $ G- $正态分布与$ G- $布朗运动在次线性期望理论中起着正态分布与布朗运动在线性期望中相同重要的作用.

本文将布朗运动指数泛函推广至相应$ G- $布朗运动的情况. 第二节中我们将首先介绍$ G- $布朗运动的基本定义与相关性质. 第三节中我们将借助拉普拉斯变换, 讨论$ G- $布朗运动指数泛函$ A_{t} = \int_{0}^{t}\exp(\lambda(B_{s}+\mu s))\mathrm {ds} $的$ n $阶矩的上下界, 其中$ B_{t} $服从$ G- $正态分布$ N(0, [\underline{\sigma}^{2}t, \overline{\sigma}^{2}t])(0<\underline{\sigma}<\overline{\sigma}) $. 第四节中我们在Tam$ \acute{a} $s.Szabados[4]的基础上, 借助对称随机游动的“扭曲与伸缩”, 构造$ G- $布朗运动的近似序列，进而推导出$ Y = \int_{0}^{\infty}\exp(B_{t}+\mu t)\mathrm {dt} $的离散化形式$ Y_{m, n} $, 当$ n\rightarrow \infty $时, $ Y_{m, n}\rightarrow Y_{m}, \ Y_{m} = 2^{-2m}(1+\xi_{m1}+\xi_{m1}\xi_{m2}+\cdots+\xi_{m1}\xi_{m2}\cdots) $. 当$ m\rightarrow \infty $时, $ Y_{m}\rightarrow Y $, 进而计算出$ Y $的p阶矩的上下界.

定义2.1  令$ \Omega $为给定非空集合, $ \mathbb{H} $为定义在$ \Omega $上的由全体实值函数组成的线性空间, 且满足$ 1\in\mathbb{H} $; 若$ X\in\mathbb{H} $, 则$ \vert X\vert\in\mathbb{H} $; 若$ \varphi\in C_{b, Lip}(\mathbb{R}^{d}), \ X_{i}\in\mathbb{H}, \ i = 1, 2, \cdots, d, $则有$ \varphi(X_{1}, X_{2}, \cdots, X_{d})\in\mathbb{H} $. 其中$ C_{b, Lip(\mathbb{R}^{d})} $表示$ \mathbb{R}^{d} $上的全体有界Lipschitz函数的集合. 如果定义在$ \mathbb{H} $上的函数$ \hat{\mathbb{\mathbb{E}}} $满足对任意$ X, Y\in\mathbb{H} $, 有

(1) 单调性: 若$ X\geq Y $, 那么$ \hat{\mathbb{\mathbb{E}}}[X]\geq\hat{\mathbb{\mathbb{E}}}[Y] $,

(2) 保常性: 对任意$ c\in\mathbb{R} $, 有$ \hat{\mathbb{\mathbb{E}}}[c] = c $,

(3) 次可加性: $ \hat{\mathbb{\mathbb{E}}}[X+Y]\leq\hat{\mathbb{\mathbb{E}}}[X]+\hat{\mathbb{\mathbb{E}}}[Y] $,

(4) 正齐次性: 若$ \lambda\geq0 $, 则$ \hat{\mathbb{\mathbb{E}}}[\lambda X] = \lambda\hat{\mathbb{\mathbb{E}}}[X] $,

则称函数$ \hat{\mathbb{\mathbb{E}}} $为次线性期望. $ (\Omega, \mathbb{H}, \hat{\mathbb{\mathbb{E}}}) $为次线性期望空间, 其中$ \mathbb{H} $被看作是$ \Omega $上的随机变量空间.

定义2.2  在次线性期望空间$ (\Omega, \mathbb{H}, \hat{\mathbb{\mathbb{E}}}) $中, 随机过程$ \{B_{t}, t\geq0\}\in\mathbb{H} $叫做$ G- $布朗运动, 如果满足以下条件

(1) $ B_{0} = 0 $, (2) 对每个$ t, s\geq0 $, 增量$ B_{t+s}-B_{t} $服从$ G- $正态分布$ N(0, [\underline{\sigma}^{2}s, \overline{\sigma}^{2}s]) $分布且其与$ (B_{t_1}, B_{t_2}, \cdots, B_{t_n}) $独立, 其中$ n\in\mathbb{N}, \ 0\leq t_{1}\leq t_{2}\leq\cdots\leq t_{n}\leq t. $

$ G- $布朗运动满足以下性质

(1) 对$ \xi\in\mathbb{L}^{2}(\Omega_t) $, 有$ \hat{\mathbb{\mathbb{E}}}[\xi(B_{T}-B_{t})] = 0, \ 0\leq t\leq T $,

(2) 对$ \mathscr{B}(\Omega_{t}) $可测实值有界函数$ \xi $, 满足$ \hat{\mathbb{\mathbb{E}}}[\xi^{2}(B_{T}-B_{t})^{2}]\leq\overline{\sigma}^{2}(T-t)\hat{\mathbb{\mathbb{E}}}[\xi^{2}], \ 0\leq t\leq T $,

(3) 对$ t\geq0 $, 有$ \hat{\mathbb{\mathbb{E}}}[B_{t}] = \hat{\mathbb{\mathbb{E}}}[-B_{t}] = 0 $,

(4) 当$ t\rightarrow \infty $时, $ B_{t} $拟必然H$ \ddot{o} $lder连续, 其中$ \delta<\frac{1}{2} $.

定理3.1   若$ \mu\geq0, \ n\in\mathbb{N} $, 且$ \alpha>\varphi(\overline{\sigma}(\mu+n)) $, 则有

$ \begin{eqnarray} \begin{split} n!\frac{1}{\prod\limits_{j = 0}^{n}(\alpha-\varphi(\underline{\sigma}(\mu+j)))} \leq&\int_{0}^{\infty}\exp(-\alpha t)\hat{\mathbb{E}}[(\int_{0}^{t}\exp(B_{s})\mathrm {ds})^{n}\exp(\mu B_{t})]\\ \leq& n!\frac{1}{\prod\limits_{j = 0}^{n}(\alpha-\varphi(\overline{\sigma}(\mu+j)))}. \end{split} \end{eqnarray} $

(3.1)

证   首先对任意$ x\in\mathbb{R} $, 定义$ \varphi(x) = \frac{x^{2}}{2}. $

若$ B_{t} $为$ G- $布朗运动, 服从$ G- $正态分布$ N(0, [\underline{\sigma}^{2}t, \overline{\sigma}^{2}t]) $, 对任意$ \lambda\in\mathbb{R} $

$ \begin{eqnarray} \exp(t\varphi(\underline{\sigma}\lambda))\leq\hat{\mathbb{E}}[\exp(\lambda B_{t})]\leq \exp(t\varphi(\overline{\sigma}\lambda)), \end{eqnarray} $

(3.2)

若$ \tilde{B_{t}} $为标准布朗运动, 则有$ \mathbb{E}[\exp(\lambda\tilde{B_{t}})] = \exp(t\varphi(\lambda)). $

接下来定义

$ \begin{eqnarray} \begin{split} \phi_{n, t}(\mu)& = \hat{\mathbb{E}}[(\int_{0}^{t}\exp(B_{s})\mathrm {ds})^{n}\exp(\mu B_{t})] \\ & = n!\hat{\mathbb{E}}[\int_{0}^{t}\int_{0}^{s_1}\int_{0}^{s_2}\cdots\int_{0}^{s_{n-1}}\exp(B_{s_{1}}+B_{s_{2}}+\cdots+B_{s_{n}}+\mu B_{t})\mathrm {ds}_{1}\mathrm {ds}_{2}\cdots \mathrm {ds}_{n}]. \end{split} \end{eqnarray} $

(3.3)

由公式(3.2) 可得

$ \begin{eqnarray} \begin{split} &\hat{\mathbb{E}}[\exp(B_{s_1}+\cdots+B_{s_n}+\mu B_{t})]\\ = &\hat{\mathbb{E}}[\exp(\mu(B_t-B_{s_1})+(\mu+1)(B_{s_1}-B_{s_2})+\cdots+(\mu+n)B_{s_n})] \\ \leq& \exp(\varphi(\overline{\sigma}\mu)(t-s_1)+\varphi(\overline{\sigma}(\mu+1))(s_1-s_2)+\cdots+\varphi(\overline{\sigma}(\mu+n))s_n). \end{split} \end{eqnarray} $

(3.4)

将公式(3.4) 代入公式(3.3) 得

$ \begin{eqnarray} \begin{split} &\int_{0}^{\infty}\exp(-\alpha t)\phi_{n, t}(\mu)\mathrm {dt} \\ \leq& n!\int_{0}^{\infty}\exp(-\alpha t)\int_{0}^{s_1}\cdots\int_{0}^{s_{n-1}}\exp(\varphi(\overline{\sigma}\mu)(t-s_1)+\varphi(\overline{\sigma}(\mu+1))(s_1-s_2)+\\ &\cdots+\varphi(\overline{\sigma}(\mu+n))s_n) \mathrm {ds}_{1}\cdots \mathrm {ds}_{n}\mathrm {dt} \\ = &n!\int_{0}^{\infty}\exp(-(\alpha-\varphi(\overline{\sigma}(\mu+n)))s_{n})\mathrm {ds}_{n}\\ &\int_{s_{n}}^{\infty}\exp(-(\alpha-\varphi(\overline{\sigma}(\mu+n-1)))(s_{n}-s_{n-1}))\mathrm {ds}_{n-1} \\ &\cdots\int _{s_{1}}^{\infty}\exp(-(\alpha-\varphi(\overline{\sigma}(\mu)))(t-s_{1}))\mathrm {dt} \\ = &n!\frac{1}{\prod\limits_{j = 0}^{n}(\alpha-\varphi(\overline{\sigma}(\mu+j)))}. \end{split} \end{eqnarray} $

(3.5)

由公式(3.5) 可知, 当$ \alpha>\varphi(\overline{\sigma}(\mu+n)) = \frac{\overline{\sigma}^{2}(\mu+n)^{2}}{2} $时, 公式(3.1) 右边得证.

同理当$ \alpha>\varphi(\underline{\sigma}(\mu+n)) = \frac{\underline{\sigma}^{2}(\mu+n)^{2}}{2} $时, 公式(3.1) 左边得证.

定理3.2   若$ \mu\geq0, n\in\mathbb{N} $, 且$ t\geq0 $, 则有

$ \begin{eqnarray} \begin{split} \mathbb{E}[p_{n}^{(\mu)}(\exp(\underline{\sigma}\tilde{B_{t}}))\exp(\underline{\sigma}\mu\tilde{B_{t}})]\leq& \hat{\mathbb{E}}[(\int_{0}^{t}\exp(B_{s})\mathrm {ds})^{n}\exp(\mu B_{t})]\\ \leq&\mathbb{E}[P_{n}^{(\mu)}(\exp(\overline{\sigma}\tilde{B_{t}}))\exp(\overline{\sigma}\mu\tilde{B_{t}})]. \end{split} \end{eqnarray} $

(3.6)

其中

$ \begin{eqnarray} \begin{split} P_{n}^{\mu}(z) = n!\sum\limits_{j = 0}^{n}C_{j}^{(\mu)}z^{j}, \ \ C_{j}^{(\mu)} = \prod\limits_{\substack{k\neq j\\0\leq k\leq n}}(\varphi(\overline{\sigma}(\mu+j))-\varphi(\overline{\sigma}(\mu+k)))^{-1}, \\ p_{n}^{(\mu)}(z) = n!\sum\limits_{j = 0}^{n}c_{j}^{(\mu)}z^{j}, \ \ c_{j}^{(\mu)} = \prod\limits_{\substack{k\neq j\\0\leq k\leq n}}(\varphi(\underline{\sigma}(\mu+j))-\varphi(\underline{\sigma}(\mu+k)))^{-1}. \end{split} \end{eqnarray} $

(3.7)

证

$ \begin{eqnarray} \frac{1}{\prod\limits_{j = 0}^{n}(\alpha-\varphi(\overline{\sigma}(\mu+j)))} = \sum\limits_{j = 0}^{n}C_{j}^{(\mu)}\frac{1}{(\alpha-\varphi(\overline{\sigma}(\mu+j)))}. \end{eqnarray} $

(3.8)

将公式(3.8) 代入公式(3.5) 得到

$ \begin{eqnarray} \int_{0}^{\infty}\exp(-\alpha t)\phi_{n, t}(\mu)\mathrm {dt}\leq n!\sum\limits_{j = 0}^{n}C_{j}^{(\mu)}\int_{0}^{\infty}\exp(-\alpha t)\exp(\varphi(\overline{\sigma}(\mu+j))t)\mathrm {dt}. \end{eqnarray} $

(3.9)

从而可以推出

$ \begin{eqnarray} \begin{split} \phi_{n, t}(\mu)&\leq n!\sum\limits_{j = 0}^{n}C_{j}^{(\mu)}\exp(\varphi(\overline{\sigma}(\mu+j))t) = n!\sum\limits_{j = 0}^{n}C_{j}^{(\mu)}\mathbb{E}[\exp(\overline{\sigma}j\tilde{B_{t}})\exp(\overline{\sigma}\mu\tilde{B_{t}})]\\ & = \mathbb{E}[P_{n}^{(\mu)}(\exp(\overline{\sigma}\tilde{B_{t}}))\exp(\overline{\sigma}\mu\tilde{B_{t}})]. \end{split} \end{eqnarray} $

(3.10)

同理可得

$ \begin{eqnarray} \mathbb{E}[p_{n}^{(\mu)}(\exp(\underline{\sigma}\tilde{B_{t}}))\exp(\underline{\sigma}\mu\tilde{B_{t}})]\leq\phi_{n, t}(\mu). \end{eqnarray} $

(3.11)

定理3.3   若$ \lambda\in\mathbb{R}, \ n\in\mathbb{N} $, 且$ t\geq0 $, 则有

$ \begin{eqnarray*} \frac{\mathbb{E}[P_{n}(\exp(\underline{\sigma}\lambda \tilde{B_{t}}))]}{\underline{\sigma}^{2n}}\leq\lambda^{2n}\hat{\mathbb{E}}[(\int_{0}^{t}\exp(\lambda B_{s})\mathrm {ds})^{n}]\leq \frac{\mathbb{E}[P_{n}(\exp(\overline{\sigma}\lambda \tilde{B_{t}}))]}{\overline{\sigma}^{2n}}, \end{eqnarray*} $

其中$ P_{n}(z) = 2^{n}(-1)^{n}\{\frac{1}{n!}+2\sum\limits_{j = 1}^{n}\frac{n!(-z)^{j}}{(n-j)!(n+j)!}\} $.

证  对公式(3.6), 取$ \mu = 0 $, 且将$ B_{s} $替换为$ \lambda B_{s} $, 则有

$ \begin{eqnarray} \hat{\mathbb{E}}[(\int_{0}^{t}\exp(\lambda B_{s})\mathrm {ds})^{n}]\leq \mathbb{E}[P_{n}^{(0)}(\exp(\overline{\sigma}\lambda\tilde{B_{t}}))], \end{eqnarray} $

(3.12)

其中$ P_{n}^{(0)}(z) = n!\sum\limits_{j = 0}^{n}C_{j}^{(0)}z^{j}, \ C_{j}^{(0)} = \prod\limits_{\substack{k\neq j\\0\leq k\leq n}}(\varphi(\overline{\sigma}\lambda j)-\varphi(\overline{\sigma}\lambda k))^{-1} $.

当$ j = 0 $时,

$ \begin{eqnarray*} C_{0}^{(0)} = \frac{2^{n}(-1)^{n}}{\overline{\sigma}^{2}\lambda^{2}(n!)^{2}}. \end{eqnarray*} $

当$ 1\leq j\leq n $时,

$ \begin{eqnarray*} C_{j}^{(0)} = \prod\limits_{\substack{k\neq j\\0\leq k\leq n}}(\frac{\overline{\sigma}^{2}\lambda^{2} j^{2}-\overline{\sigma}^{2}\lambda^{2} k^{2}}{2})^{-1} = \prod\limits_{\substack{k\neq j\\0\leq k\leq n}}\frac{2^{n}(-1)^{n}}{\overline{\sigma}^{2}\lambda^{2}(k-j)(k+j)} = \frac{2^{n}(-1)^{n-j}2}{\overline{\sigma}^{2n}\lambda^{2n}(n-j)!(n+j)!}. \end{eqnarray*} $

进而可得

$ \begin{eqnarray} \mathbb{E}[P_{n}^{(0)}(\exp(\overline{\sigma}\lambda\tilde{B_{t}}))] = \frac{2^{n}(-1)^{n}}{\overline{\sigma}^{2n}\lambda^{2n}}\{\frac{1}{n!}+2\sum\limits_{j = 1}^{n}\frac{n!(-\mathbb{E}( \overline{\sigma}\lambda\tilde{B_{t}}))^{j}}{(n-j)!(n+j)!}\}. \end{eqnarray} $

(3.13)

记$ P_{n}(z) = 2^{n}(-1)^{n}\{\frac{1}{n!}+2\sum\limits_{j = 1}^{n}\frac{n!(-z)^{j}}{(n-j)!(n+j)!}\} $, 代入公式(3.12) 可得

$ \begin{eqnarray} \lambda^{2n}\hat{\mathbb{E}}[(\int_{0}^{t}\exp(\lambda B_{s})\mathrm {ds})^{n}]\leq \frac{\mathbb{E}[P_{n}(\exp(\lambda \tilde{B_{t}}))]}{\overline{\sigma}^{2n}}. \end{eqnarray} $

(3.14)

同理可得

$ \begin{eqnarray} \frac{\mathbb{E}[P_{n}(\exp(\underline{\sigma}\lambda\tilde{B_{t}}))]}{\underline{\sigma}^{2n}}\leq \lambda^{2n}\hat{\mathbb{E}}[(\int_{0}^{t}\exp(\lambda B_{s})\mathrm {ds})^{n}]. \end{eqnarray} $

(3.15)

定理3.4   若$ \lambda\in\mathbb{R}, \mu\in\mathbb{R}, n\in\mathbb{N} $, 且$ t\geq0 $, 则有

$ \begin{eqnarray} \begin{split} \mathbb{E}[p_{n}^{(\nu, \mu)}(\exp(\underline{\sigma}\tilde{B_t}+\mu t))\exp(\nu(\underline{\sigma}{\tilde{B_{t}}+\mu t}))]&\leq \hat{\mathbb{E}}[(\int_{0}^{t}\exp(B_{s}+\mu s)\mathrm {ds})^{n}\exp(\nu (B_{t}+\mu t))] \\&\leq \mathbb{E}[P_{n}^{(\nu, \mu)}(\exp(\overline{\sigma}\tilde{B_t}+\mu t))\exp(\nu(\overline{\sigma}{\tilde{B_{t}}+\mu t}))]. \end{split} \end{eqnarray} $

(3.16)

证   记

$ \begin{eqnarray} \phi_{n, t}(\nu, \mu)& = \hat{\mathbb{E}}[(\int_{0}^{t}\exp(B_{s}+\mu s)\mathrm {ds})^{n}\exp(\nu(B_{t}+\mu t))]. \end{eqnarray} $

(3.17)

结合定理3.1可推出

$ \begin{eqnarray} \begin{split} \int_{0}^{\infty}\exp(-\alpha t)\phi_{n, t}(\nu)\mathrm {dt} \leq& n!\frac{1}{\prod\limits_{j = 0}^{n}(\alpha-\varphi(\overline{\sigma}(\nu+j))-\mu(\nu+j))}\\ = &n!\sum\limits_{j = 1}^{n}C_{j}^{(\nu, \mu)}\int_{0}^{\infty}\exp(-\alpha t)\exp(\varphi(\overline{\sigma}(\nu+j))t+\mu(\nu+j)t)\mathrm {dt}, \end{split} \end{eqnarray} $

(3.18)

其中

$ \begin{eqnarray} \begin{split} \frac{1}{\prod\limits_{j = 0}^{n}(\alpha-\varphi(\overline{\sigma}(\nu+j))-\mu(\nu+j))}& = \sum\limits_{j = 0}^{n}C_{j}^{(\nu, \mu)}\frac{1}{\alpha-\varphi(\overline{\sigma}(\nu+j))-\mu(\nu+j)}, \\ C_{j}^{(\nu, \mu)}& = \prod\limits_{\substack{k\neq j\\0\leq k\leq n}}(\varphi(\overline{\sigma}(\nu+\frac{\mu}{\overline{\sigma}^{2}}+j))-\varphi(\overline{\sigma}(\nu+\frac{\mu}{\overline{\sigma}^{2}}+k)))^{-1}. \end{split} \end{eqnarray} $

(3.19)

进而将公式(3.19) 代入公式(3.18) 可推出

$ \begin{eqnarray} \begin{split} \phi_{n, t}(\nu, \mu)&\leq n!\sum\limits_{j = 0}^{n}C_{j}^{(\nu, \mu)}\exp(\varphi(\overline{\sigma}(\nu+j))t+\mu(\nu+j)t)) \\& = n!\sum\limits_{j = 0}^{n}C_{j}^{(\nu, \mu)}\mathbb{E}[\exp(j(\overline{\sigma}\tilde{B_{t}} +\mu t))\exp(\nu(\overline{\sigma}\tilde{B_{t}}+\mu t))] \\ & = \mathbb{E}[P_{n}^{(\nu, \mu)}(\exp(\overline{\sigma}\tilde{B_t}+\mu t))\exp(\nu(\overline{\sigma}{\tilde{B_{t}}+\mu t}))], \\ P_{n}^{(\nu, \mu)}(z)& = n!\sum\limits_{j = 0}^{n}C_{j}^{(\nu, \mu)}z^{j}. \end{split} \end{eqnarray} $

(3.20)

同理可得

$ \begin{eqnarray} \begin{split} &\mathbb{E}[p_{n}^{(\nu, \mu)}(\exp(\underline{\sigma}\tilde{B_t}+\mu t))\exp(\nu(\underline{\sigma}{\tilde{B_{t}}+\mu t}))]\leq\phi_{n, t}(\nu, \mu), \\ &p_{n}^{(\nu, \mu)}(z) = n!\sum\limits_{j = 0}^{n}c_{j}^{(\nu, \mu)}z^{j}, \\ &c_{j}^{(\nu, \mu)} = \prod\limits_{\substack{k\neq j\\0\leq k\leq n}}(\varphi(\underline{\sigma}(\nu+\frac{\mu}{\underline{\sigma}^{2}}+j))-\varphi(\underline{\sigma}(\nu+\frac{\mu}{\underline{\sigma}^{2}}+k)))^{-1}. \end{split} \end{eqnarray} $

(3.21)

定理3.5   若$ \lambda\in\mathbb{R}, \ n\in\mathbb{N} $, 且$ t\geq0 $, 则有

$ \begin{eqnarray} \begin{split} \mathbb{E}[p_{n}^{*}(\exp(\lambda(\underline{\sigma}\tilde{B_{t}}+\mu t)))] \leq \hat{\mathbb{E}}[(\int_{0}^{t}\exp(\lambda(B_{s}+\mu s))\mathrm {ds})^{n}] \leq \mathbb{E}[P_{n}^{*}(\exp(\lambda(\overline{\sigma}\tilde{B_{t}}+\mu t)))]. \end{split} \end{eqnarray} $

(3.22)

证   对定理3.4, 取$ \nu = 0 $, 将$ B_{s}+\mu s $替换为$ \lambda(B_{s}+\mu s) $, 则有

$ \begin{eqnarray} \begin{split} &\hat{\mathbb{E}}[(\int_{0}^{t}\exp(\lambda(B_{s}+\mu s))\mathrm {ds})^{n}]\\\leq& \{\frac{n!2^{n}(-1)^{n}}{(\overline{\sigma}^{2}\lambda^{2})^{n}\prod\limits_{i = 1}^{n}(\frac{2\mu}{\overline{\sigma}^{2}}+i)n!} +\frac{n!2^{n}(-1)^{n-j}}{(\overline{\sigma}^{2} \lambda^{2})^{n}}\sum\limits_{j = 1}^{n}\frac{(\frac{2\mu}{\overline{\sigma}^{2}}+2j)\prod\limits_{i = 1}^{j}(\frac{2\mu}{\overline{\sigma}^{2}+i-1})}{\prod\limits_{i = 1}^{j}(\frac{2\mu}{\overline{\sigma}^{2}}+n+i)!(j-1)!(n-j)!} \}\\ &\exp(\frac{\lambda^{2}\overline{\sigma}^{2}j^{2}t}{2})\exp(\lambda\mu jt). \end{split} \end{eqnarray} $

(3.23)

记

$ \begin{eqnarray} \begin{split} P_{n}^{*}(z)& = \{\frac{n!2^{n}(-1)^{n}}{(\overline{\sigma}^{2}\lambda^{2})^{n}\prod\limits_{i = 1}^{n}(\frac{2\mu}{\overline{\sigma}^{2}}+i)n!}+\frac{n!2^{n}(-1)^{n-j}}{(\overline{\sigma}^{2} \lambda^{2})^{n}}\sum\limits_{j = 1}^{n}\frac{(\frac{2\mu}{\overline{\sigma}^{2}}+2j)\prod\limits_{i = 1}^{j}(\frac{2\mu}{\overline{\sigma}^{2}+i-1})}{\prod\limits_{i = 1}^{j}(\frac{2\mu}{\overline{\sigma}^{2}}+n+i)!(j-1)!(n-j)!} \}z^{j}, \\ p_{n}^{*}(z)& = \{\frac{n!2^{n}(-1)^{n}}{(\underline{\sigma}^{2}\lambda^{2})^{n}\prod\limits_{i = 1}^{n}(\frac{2\mu}{\underline{\sigma}^{2}}+i)n!}+\frac{n!2^{n}(-1)^{n-j}}{(\underline{\sigma}^{2} \lambda^{2})^{n}}\sum\limits_{j = 1}^{n}\frac{(\frac{2\mu}{\underline{\sigma}^{2}}+2j)\prod\limits_{i = 1}^{j}(\frac{2\mu}{\underline{\sigma}^{2}+i-1})}{\prod\limits_{i = 1}^{j}(\frac{2\mu}{\underline{\sigma}^{2}}+n+i)!(j-1)!(n-j)!} \}z^{j}. \end{split} \end{eqnarray} $

(3.24)

进而可得

$ \begin{eqnarray} \begin{split} \mathbb{E}[p_{n}^{*}(\exp(\lambda(\underline{\sigma}\tilde{B_{t}}+\mu t)))] \leq \hat{\mathbb{E}}[(\int_{0}^{t}\exp(\lambda(B_{s}+\mu s))\mathrm {ds})^{n}] \leq \mathbb{E}[P_{n}^{*}(\exp(\lambda(\overline{\sigma}\tilde{B_{t}}+\mu t)))]. \end{split} \end{eqnarray} $

(3.25)

参考Tamás.Szabados[5]提出的方法, 我们可以借助“扭曲与收缩”对称随机游动来定义标准布朗运动的一种近似序列, 即将单位一的步长压缩至$ 1/2^{m}, m = 1, 2, \cdots $, 对应完成该步长的时间被压缩至$ 1/2^{2m}, m = 1, 2, \cdots $.接下来我们将简述这一方法.

定义序列$ \{\tilde{S}_{m}(k), k\geq 0\} $满足以下条件

$ \begin{eqnarray} \begin{split} \tilde{S}_{m}(0)& = 0, \tilde{S}_{m}(n) = \tilde{X}_{m}(1)+\tilde{X}_{m}(2)+\cdots+\tilde{X}_{m}(n), n\geq 1, \\ \tilde{S}_{m}(t)& = \tilde{S}_{m}(n)+(t-n)\tilde{X}_{m}(n+1), n\leq t< n+1. \end{split} \end{eqnarray} $

(4.1)

其中$ \tilde{X}_{m}(k), m\geq 0, k\geq 1 $是独立同分布随机游动且满足

$ \begin{eqnarray*} \mathbb{E}(\tilde{X}_{m}(k)) = 0, \mathbb{D}(\tilde{X}_{m}(k)) = 1, \mathbb{P}(\tilde{X}_{m}(k) = \pm 1) = \frac{1}{2}. \end{eqnarray*} $

接下来可以定义标准布朗运动$ \tilde{B}_{t} $的m阶近似序列

$ \begin{eqnarray} \tilde{B}_{m}(t) = 2^{-m}\tilde{S}_{m}(t2^{2m}), t\geq 0, \end{eqnarray} $

(4.2)

即$ \tilde{B}_{m}(t) $为标准布朗运动$ \tilde{B}_{t} $在时刻$ j/2^{m}, j = 1, 2\cdots $的一个“分割模型”.

引理4.1    定义序列$ \tilde{Y}_{m} = 2^{-2m}(\sum\limits_{k = 0}^{\infty}\exp(2^{-m}\tilde{S}_{m}(k)+\mu k2^{-2m})) $, 当$ m\rightarrow \infty $时, $ \tilde{Y}_{m} $几乎处处收敛于$ \tilde{Y} = \int_{0}^{\infty}\exp(\tilde{B}_{t}+\mu t)\mathrm {dt} $.

参考[7]注释3.4知存在序列$ X_{m}(k) $服从$ G- $正态分布$ N(0, [\underline{\sigma}^{2}, \overline{\sigma}^{2}]) $, $ \frac{1}{\sqrt{n}}\sum\limits_{i}^{n}X_{m}(i) $依分布收敛于$ X $, $ X $服从$ G- $正态分布$ N(0, [\underline{\sigma}^{2}, \overline{\sigma}^{2}]) $. 从而可推出序列$ \{S_{m}(k), k\geq 0\} $满足以下条件

$ \begin{eqnarray} \begin{split} S_{m}(0)& = 0, S_{m}(n) = X_{m}(1)+X_{m}(2)+\cdots+X_{m}(n), n\geq 1, \\ S_{m}(t)& = S_{m}(n)+(t-n)X_{m}(n+1), n\leq t< n+1. \end{split} \end{eqnarray} $

(4.3)

$ \{B_{m}(t) = 2^{-m}S_{m}(t2^{2m}), t\geq 0\} $为$ G- $布朗运动$ B_{t} $的m阶近似序列.

推论4.2   定义序列$ Y_{m} = 2^{-2m}(\sum\limits_{k = 0}^{\infty}\exp(2^{-m}S_{m}(k)+\mu k2^{-2m})) $, 当$ m\rightarrow \infty $时, $ Y_{m} $几乎处处收敛于$ Y = \int_{0}^{\infty}\exp(B_{t}+\mu t)\mathrm {dt} $, 进而我们完成了$ G- $布朗运动指数泛函离散化形式的构造.

定理4.3   对任意正整数$ p $, 当$ \mu<-\frac{\overline{\sigma}^{2}}{2}p $时, $ \hat{\mathbb{E}}(Y_{m}^p) = \frac{1}{1-\hat{\mathbb{E}}(\xi_{m}^{p})}\sum\limits_{k = 0}^{p-1}\binom{p}{k}2^{-2m(p-k)}\hat{\mathbb{E}}(\xi_{m}^{k})\mathbb{E}(Y_{m}^{k}) $.

证  对任意$ m\geq 0, \ n\geq1 $, 定义序列$ Y_{m, n} $满足$ Y_{m, n} = 2^{-2m}\sum\limits_{k = 0}^{n}\exp(2^{-m}S_{m}(k)+\mu k2^{-2m}) = 2^{-2m}(1+\xi_{m1}+\xi_{m1}\xi_{m2}+\cdots+\xi_{m1}\xi_{m2}\cdots\xi_{mn}), $其中$ Y_{m, 0} = 2^{-2m}, \xi_{mj} = \exp(2^{-m}X_{m}(j)+\mu 2^{-2m}) $. 当$ n\rightarrow \infty $时, $ Y_{m, n}\rightarrow Y_{m}, \ \ Y_{m} = 2^{-2m}(1+\xi_{m1}+\xi_{m1}\xi_{m2}+\cdots+\xi_{m1}\xi_{m2}\cdots), $且$ Y_{m} $满足分布自相似性

$ \begin{eqnarray} Y_{m}\overset{d}{ = }2^{-2m}+\xi_{m}Y_{m}, \end{eqnarray} $

(4.4)

$ \xi_{m}\overset{d}{ = }\xi_{m1} $, 且$ \xi_{m} $与$ Y_{m} $相互独立, 其中$ \overset{d}{ = } $表示同分布. 对任意$ m\geq 0, \ k $为整数

$ \begin{eqnarray} \begin{split} \hat{\mathbb{E}}(\xi_{m}^{k})& = \hat{\mathbb{E}}(\exp(2^{-m}kX_{m}(1)+\mu k2^{-2m})) \\ &\leq \exp(\frac{\overline{\sigma}^{2}k^{2}2^{-2m}}{2})\exp(\mu k2^{-2m}) = \exp(k2^{-2m}(\frac{\overline{\sigma}^{2}}{2}k+\mu)). \end{split} \end{eqnarray} $

(4.5)

当$ \frac{\overline{\sigma}^{2}}{2}k+\mu<0 $, 即$ \mu<-\frac{\overline{\sigma}^{2}}{2}k $时, $ \hat{\mathbb{E}}(\xi_{m}^{k})<1 $. 当$ m\rightarrow \infty $时, 由$ \exp(x) = 1+x+o(x) $可得

$ \begin{eqnarray} \hat{\mathbb{E}}(\xi_{m}^{k})<1+k2^{-2m}(\frac{\overline{\sigma}^{2}}{2}k+\mu)+o(2^{-2m}). \end{eqnarray} $

(4.6)

当$ \mathbb{E}(\xi_{m}^{p})<1, \ p\geq1 $时, 对公式(4.4) 两边取$ p $阶矩可得

$ \begin{eqnarray} \begin{split} \hat{\mathbb{E}}(Y_{m}^p)& = \sum\limits_{k = 0}^{p-1}\binom{p}{k}2^{-2m(p-k)}\hat{\mathbb{E}}(\xi_{m}^{k})\hat{\mathbb{E}}(Y_{m}^{k})+\hat{\mathbb{E}}(\xi_{m}^{p})\hat{\mathbb{E}}(Y_{m}^{p}), \\ \hat{\mathbb{E}}(Y_{m}^p)& = \frac{1}{1-\hat{\mathbb{E}}(\xi_{m}^{p})}\sum\limits_{k = 0}^{p-1}\binom{p}{k}2^{-2m(p-k)}\hat{\mathbb{E}}(\xi_{m}^{k})\hat{\mathbb{E}}(Y_{m}^{k}). \end{split} \end{eqnarray} $

(4.7)

定理4.4   对任意正整数$ p $, 当$ \mu<-\frac{\overline{\sigma}^{2}}{2}p $时, $ \frac{1}{\prod\limits_{k = 1}^{p}-(\frac{\underline{\sigma}^{2}}{2}k+\mu)}<\lim\limits_{m\rightarrow \infty}\hat{\mathbb{E}}(Y_{m}^{p})<\frac{1}{\prod\limits_{k = 1}^{p}-(\frac{\overline{\sigma}^{2}}{2}k+\mu)}<\infty. $

证   由公式(4.6) 知, 当$ m\rightarrow \infty $时

$ \begin{eqnarray} \hat{\mathbb{E}}(\xi_{m}^{p})\leq \exp(p2^{-2m}(\frac{\overline{\sigma}^{2}}{2}p+\mu))\leq 1+p2^{-2m}(\frac{\overline{\sigma}^{2}}{2}p+\mu)+o(2^{-2m}). \end{eqnarray} $

(4.8)

进而可得

$ \begin{eqnarray} &&\frac{1}{1-\hat{\mathbb{E}}(\xi_{m}^{p})}<\frac{1}{-p2^{-2m}(\frac{\overline{\sigma}^{2}}{2}p+\mu)}, \end{eqnarray} $

(4.9)

$ \begin{eqnarray} &&\lim\limits_{m\rightarrow \infty}\hat{\mathbb{E}}(\xi_{m}^{k}) = 1, \ 0\leq k\leq p-2, \\ &&\lim\limits_{m\rightarrow \infty}\frac{\binom{p}{k}2^{-2m(p-k)}}{p2^{-2m}} = 0. \end{eqnarray} $

(4.10)

将(4.8)–(4.10) 式代入公式(4.7) 可得

$ \begin{eqnarray} \lim\limits_{m\rightarrow \infty}\hat{\mathbb{E}}(Y_{m}^{p})<\lim\limits_{m\rightarrow \infty}\frac{1}{-(\frac{\overline{\sigma}^{2}}{2}p+\mu)}\hat{\mathbb{E}}(Y_{m}^{p-1}). \end{eqnarray} $

(4.11)

进而通过迭代可得

$ \lim\limits_{m\rightarrow \infty}\hat{\mathbb{E}}(Y_{m}^{p})<\frac{1}{\prod\limits_{k = 1}^{p}-(\frac{\overline{\sigma}^{2}}{2}k+\mu)}<\infty. $

同理当$ m\rightarrow \infty $时, $ \hat{\mathbb{E}}(\xi_{m}^{p})>1+p2^{-2m}(\frac{\underline{\sigma}^{2}}{2}p+\mu)+o(2^{2m}) $, 可得

$ \frac{1}{\prod\limits_{k = 1}^{p}-(\frac{\underline{\sigma}^{2}}{2}k+\mu)}<\lim\limits_{m\rightarrow \infty}\hat{\mathbb{E}}(Y_{m}^{p}). $