Cox, Ingersoll和Ross^[1-3]引入了Cox-Ingersoll-Ross(CIR)过程, 该过程可用来研究利率和期权等金融模型. 设$ B(t) $是标准布朗运动, CIR过程$ R = \{X(t), t \ge 0\} $是指以下随机微分方程(SDE)的解,

$ \begin{equation} dX(t) = (k-\theta X(t))dt+\sigma \sqrt{X(t)}dB(t), \; t\geq0. \end{equation} $

(1.1)

该方程很难得到解析解, 高等人^[4]得到了CIR过程的期望和方差并进一步考虑数值解. CIR过程最初是作为利率随时间变化的模型提出的, 但实际上金融市场存在动态特征有所谓“记忆现象”, 是无法通过CIR过程反映出来的. 因此, 需要引入分数Cox-Ingersoll-Ross(fCIR) 过程, 通过用分数布朗运动(fBm)$ B^{H} $ = $ \{B^{H}(t), t\ge 0\} $代替方程(1.1) 中的标准布朗运动(Bm)得到fCIR模型, 其中$ B^{H} $ = $ \{B^{H}(t), t \ge 0\} $是Hurst参数为$ H $的fBm, 其协方差函数为$ EB^{H}(t)B^{H}(s) $ = $ \frac{1}{2}(t^{2H}+s^{2H}-\mid t-s\mid^{2H}) $. 该fCIR过程满足以下随机微分方程(SDE)

$ \begin{equation} dX(t) = (k-\theta X(t))dt+\sigma \sqrt{X(t)}\circ dB^{H}(t), \; t\geq0, \end{equation} $

(1.2)

其中$ X(0)>0 $, 常数$ k, \; \theta\in {R} $, $ \sigma>0 $, $ B^{H} $ = $ \{B^{H}(t), t\ge 0\} $是分数布朗运动. $ \int_{0}^{t}\sqrt{X(s)}\circ dB^{H}(s) $为关于分数布朗运动的路径Stratonovich积分.

Melnikov等人^[5]讨论了布朗运动和分数布朗运动驱动的随机微分方程解的存在唯一性, 给出了特殊情况下, 即当$ k = 0 $和$ H>\frac{1}{2} $时, 方程(1.2)存在唯一解; Mishura等人6, 7]给出了fCIR过程的定义, 利用Lamperti变换证明了在第一次到达零时刻之前, 该fCIR过程是分数Ornstein-Uhlenbeck(fOU)过程的平方, Cheridito等人^[8]得到fOU过程的相关统计性质. 在$ k>0 $的情况下, Mishura等人^[7]得到当$ H>\frac{1}{2} $时, 该fCIR过程是严格正的, 永远不会到达零. 对于fCIR过程是严格正的另外一个充分条件是如果该方程满足Feller条件, 即方程的系数满足$ 2k\geq \sigma^{2} $ ^[9].

由于方程(1.2)为非线性随机微分方程, 难以直接求得它的解以及解的期望和方差, 本文通过以下的线性随机微分方程

$ \begin{equation} dY(t) = \frac{1}{{2}}(k-\theta Y(t))dt +\frac{\sigma}{{2}}dB^{H}(t)\; , \; Y(0)>0, \end{equation} $

(1.3)

通过Lamperti变换, 也就是令随机过程$ X $ = $ \{X(t), t \ge 0\} $为过程$ Y(t) $的平方, 当该过程第一次到达零的时刻之前, 证明它满足以下随机微分方程

$ \begin{equation} dX(t) = (k\sqrt{X(t)}-\theta X(t))dt+\sigma \sqrt{X(t)}\circ dB^{H}(t), \; t\geq0. \end{equation} $

(1.4)

由于方程(1.3)是线性随机微分方程, 通过计算高斯过程的期望和方差可求得一类形如方程(1.4)的非线性随机微分方程解的期望和方差. 特别的, 当$ k = 0 $时, 利用随机过程$ Y = \{Y(t), t \ge 0\} $是高斯过程, 可求得fCIR过程的期望和方差, 从而进一步回答了fCIR过程是fOU过程的平方^[6], 那么它的相关统计性质是什么的问题. 最后文章模拟了fCIR过程理论期望和方差, 发现跟用蒙特卡洛方法的仿真模拟是接近的.

文章组织如下: 第二部分证明方程(1.4)的唯一解为方程(1.3)的解的平方, 利用高斯分布的平方为非中心卡方分布, 计算第一次到达零时刻之前的随机过程$ X = \{X(t), t \ge 0\} $的期望和方差; 第三部分计算第一次到达零时刻之前fCIR过程的期望和方差并且对该过程的期望和方差进行模拟.

令$ (\Omega, F, P) $是一个概率空间, $ H\in(0, 1) $为Hurst参数, 考虑过程$ Y = \{Y(t), t\ge 0\} $满足以下随机微分方程(SDE) :

$ \begin{equation} dY(t) = \frac{1}{{2}}(k-\theta Y(t))dt +\frac{\sigma}{{2}} dB^{H}(t), \; Y(0)>0, \end{equation} $

(2.1)

其中$ \theta, \; k\in {R}, \; \sigma>0 $且$ {B^{H}(t)} $是Hurst参数为$ H\in(0, 1) $的分数布朗运动.

定义2.1^[6]   若$ \{X(t), \; t \ge 0\} $, $ \{Y(t), t\ge 0\} $为定义在$ (\Omega, \; F, \; P) $上的随机过程, 如果对任意的分割$ 0 = t_{0}<t_{1}<t_{2}<...<t_{n-1}<t_{n} = t $, 当区间长度$ \max\limits_{1\leq i\leq n}|t_i-t_{i-1}| $趋于0时,

$ \begin{equation*} \sum\limits_{k = 1}^{n}\frac{X(t_{k})+X(t_{k-1})}{2}(Y(t_{k})-Y(t_{k-1})) \end{equation*} $

的极限存在, 则该极限定义为$ \int_{0}^{t}{X（(s)）}\circ d{Y(s)} $的路径Stratonovich积分.

定义2.2    令$ \tau: = \inf{(s>0:Y(s) = 0)} $为过程$ \{Y(t), t \ge 0\} $第一次到达零的时刻.

定义2.3    $ \{Y(t), t \ge 0\} $为满足方程(2.1)的随机过程, $ \tau $为过程$ \{Y(t), t \ge 0\} $第一次到达零的时刻. 定义随机过程$ \{X(t), t \ge 0\} $为

$ \begin{equation} X(t)(\omega) = {Y(t)}^{2}{{\rm I}_{\{ t < \tau(\omega)\}}}. \end{equation} $

(2.2)

定理2.4   令$ \tau: = \inf{(s>0:Y(s) = 0)} $. 对于$ 0\leq t\leq\tau $, 定义2.3中的随机过程$ \{X(t), t \ge 0\} $满足以下随机微分方程:

$ \begin{equation} dX(t) = (k\sqrt{X(t)}-\theta X(t))dt+\sigma \sqrt{X(t)}\circ dB^{H}(t), \end{equation} $

(2.3)

其中$ X(0) = Y(0)^{2}\ge 0, \int_{0}^{t}\sqrt{X(s)}\circ dB^{H}(s) $为关于分数布朗运动的路径Stratonovich积分.

证  对任意$ t<\tau(\omega) $, 根据方程(2.1)和方程(2.2),

$ \begin{equation} X(t) = {Y(t)}^{2} = (\sqrt{X(0)}+\frac{1}{{2}}\int_{0}^{t}(k-\; \theta Y(s))ds +\frac{\sigma}{{2}} B^{H}(t))^{2}. \end{equation} $

(2.4)

考虑区间$ [0, t] $的任意分割: $ 0 = t_{0}<t_{1}<t_{2}<...<t_{n-1}<t_{n} = t, $利用方程(2.4), 有:

$ \begin{align*} X(t) = &\sum\limits_{i = 1}^{n}(X(t_{i})-X(t_{i-1}))+X(0) \\ = & \sum\limits_{i = 1}^{n}([\sqrt{X(0)}+\frac{1}{{2}}\int_{0}^{t_{i}}(k-\; \theta Y(s))ds +\frac{\sigma}{{2}} B^{H}(t_{i})]^{2}\\&-[\sqrt{X(0)}+\frac{1}{{2}}\int_{0}^{t_{i-1}}(k-\; \theta Y(s))ds +\frac{\sigma}{{2}} B^{H}(t_{i-1})]^{2})+X(0). \end{align*} $

展开得下列式子

$ \begin{align*} X(t) = & X(0)+ \sum\limits_{i = 1}^{n}[2\sqrt{X(0)}+\frac{1}{{2}}(\int_{0}^{t_{i}}(k-\; \theta Y(s))ds \\&+\int_{0}^{t_{i-1}}(k-\; \theta Y(s))ds) +\frac{\sigma}{{2}}( B^{H}(t_{i})+ B^{H}(t_{i-1}))] \\& \times[\frac{1}{{2}}\int_{t_{i-1}}^{t_{i}}(k-\; \theta Y(s))ds+\frac{\sigma}{{2}}( B^{H}(t_{i})- B^{H}(t_{i-1}))]. \end{align*} $

进一步计算得

$ \begin{align*} X(t) = &X(0)+ \sum\limits_{i = 1}^{n}\sqrt{X(0)}\int_{t_{i-1}}^{t_{i}}(k-\; \theta Y(s))ds \\&+\frac{1}{{4}}\sum\limits_{i = 1}^{n}(\int_{0}^{t_{i}}(k-\; \theta Y(s))ds +\int_{0}^{t_{i-1}}(k-\; \theta Y(s))ds)\times\int_{t_{i-1}}^{t_{i}}(k-\; \theta Y(s))ds \\&+\frac{\sigma}{{4}}\sum\limits_{i = 1}^{n}( B^{H}(t_{i})+ B^{H}(t_{i-1})) \int_{t_{i-1}}^{t_{i}}(k-\; \theta Y(s))ds \\&+\sigma\sqrt{X(0)}\sum\limits_{i = 1}^{n}( B^{H}(t_{i})- B^{H}(t_{i-1}))+\frac{\sigma^{2}}{{4}}\sum\limits_{i = 1}^{n}( B^{H}(t_{i})- B^{H}(t_{i-1}))( B^{H}(t_{i})+ B^{H}(t_{i-1})) \\&+\frac{\sigma}{{4}}\sum\limits_{i = 1}^{n}(\int_{0}^{t_{i}}(k-\; \theta Y(s))ds +\int_{0}^{t_{i-1}}(k-\; \theta Y(s))ds)( B^{H}(t_{i})- B^{H}(t_{i-1})). \end{align*} $

令$ \bigtriangleup t = \max\limits_{1\leq i\leq n}|t_i-t_{i-1}|\rightarrow $0, 前三项极限和

$ \begin{align*} &\sum\limits_{i = 1}^{n}\sqrt{X(0)}\int_{t_{i-1}}^{t_{i}}(k-\; \theta Y(s))ds \\&+\frac{1}{{4}}\sum\limits_{i = 1}^{n}(\int_{0}^{t_{i}}(k+\; \theta Y(s))ds +\int_{0}^{t_{i-1}}(k-\; \theta Y(s))ds)\times\int_{t_{i-1}}^{t_{i}}(k-\; \theta Y(s))ds \\&+\frac{\sigma}{{4}}\sum\limits_{i = 1}^{n}( B^{H}(t_{i})+ B^{H}(t_{i-1})) \int_{t_{i-1}}^{t_{i}}(k-\; \theta Y(s))ds \\\rightarrow &\int_{0}^{t}(k-\; \theta Y(s))(\sqrt{X(0)}+\frac{1}{{2}}\int_{0}^{s}(k-\; \theta Y(u))du+\frac{\sigma}{{2}} B^{H}(s))ds \\ = &\int_{0}^{t}(kY(s)-\; \theta Y^{2}(s))ds = \int_{0}^{t}(k\sqrt{X(s)}-\; \theta X(s))ds, \end{align*} $

后三项极限和为

$ \begin{align*} &\sigma\sqrt{X(0)}\sum\limits_{i = 1}^{n}( B^{H}(t_{i})- B^{H}(t_{i-1}))+\frac{\sigma^{2}}{{4}}\sum\limits_{i = 1}^{n}( B^{H}(t_{i})- B^{H}(t_{i-1}))( B^{H}(t_{i})+ B^{H}(t_{i-1})) \\&+\frac{\sigma}{{4}}\sum\limits_{i = 1}^{n}(\int_{0}^{t_{i}}(k-\; \theta Y(s))ds +\int_{0}^{t_{i-1}}(k-\; \theta Y(s))ds)( B^{H}(t_{i})- B^{H}(t_{i-1})) \\\rightarrow &\sigma\int_{0}^{t}(\sqrt{X(0)}+\frac{1}{2}\int_{0}^{s}(k-\; \theta Y(u))du+\frac{\sigma}{2}B^{H}(s))\circ dB^{H}(s) \\ = &\sigma\int_{0}^{t}Y(s)\circ dB^{H}(s) = \sigma\int_{0}^{t}\sqrt{X(s)}\circ d B^{H}(s)\; . \end{align*} $

因此定义2.3中的过程$ X = \{X(t), t \ge 0\} $满足以下形式的随机微分方程

$ \begin{equation*} X(t) = X(0)+\int_{0}^{t}(k\sqrt{X(s)}-\; \theta X(s))ds+\sigma\int_{0}^{t}\sqrt{X(s)}\circ d B^{H}(s), \end{equation*} $

其中$ \int_{0}^{t}{X((s))}\circ dB^{H}(s) $为路径Stratonovich积分. 证毕.

过程$ X = \{X(t), t \ge 0\} $是过程$ Y = \{Y(t), t \ge 0\} $在第一次到达零时刻之前的平方, 而满足SDE (2.1) 的随机过程$ Y = \{Y(t), t \ge 0\} $为可求解的线性随机微分方程, 因此可通过较为简单的随机过程$ Y = \{Y(t), t \ge 0\} $的期望和方差计算求得随机过程$ X = \{X(t), t \ge 0\} $的期望和方差.

考虑随机过程$ Y = \{Y(t), t \ge 0\} $所满足的SDE(2.1), 可求出解析解为

$ \begin{align*} Y(t)& = Y(0)e^{-\frac{1}{2}\theta t}+\frac{1}{2}ke^{-\frac{1}{2}\theta t}\int_{0}^{t} e^{\frac{1}{2}\theta s}ds+\frac{\sigma}{2}e^{-\frac{1}{2}\theta t}\int_{0}^{t}e^{\frac{1}{2}\theta s}dB^{H}(s) \\& = Y(0)e^{-\frac{1}{2}\theta t}+\frac{k}{\theta} (1-e^{-\frac{1}{2}\theta t})+\frac{\sigma}{2}e^{-\frac{1}{2}\theta t}\int_{0}^{t}e^{\frac{1}{2}\theta s}dB^{H}(s). \end{align*} $

期望为

$ \begin{align} E[Y(t)]& = E[ Y(0)e^{-\frac{1}{2}\theta t}]+E[\frac{k}{\theta} (1-e^{-\frac{1}{2}\theta t})]+E [\frac{\sigma}{2}e^{-\frac{1}{2}\theta t}\int_{0}^{t}e^{\frac{1}{2}\theta s}dB^{H}(s)]\notag \\& = Y(0)e^{-\frac{1}{2}\theta t}+\frac{k}{\theta} (1-e^{-\frac{1}{2}\theta t}). \end{align} $

(2.5)

方差为

$ \begin{align} &Var[Y(t)]\notag\\ = &E[Y^{2}(t)]-E[Y(t)]^{2}\notag \\ = &E[Y^{2}(0)e^{-\theta t}+2Y(0)e^{-\frac{1}{2}\theta t}\frac{k}{\theta} (1-e^{-\frac{1}{2}\theta t})+Y(0)e^{-\theta t}\sigma\int_{0}^{t}e^{\frac{1}{2}\theta s}dB^{H}(s)+\frac{k^{2}}{\theta^{2}} (1-e^{-\frac{1}{2}\theta t})^{2}\notag \\&+\frac{k}{\theta} (1-e^{-\frac{1}{2}\theta t})\sigma e^{-\frac{1}{2}\theta t}\int_{0}^{t}e^{\frac{1}{2}\theta s}dB^{H}(s)+\frac{\sigma^{2}}{4}e^{-\theta t}(\int_{0}^{t}e^{\frac{1}{2}\theta s}dB^{H}(s))^{2}]-E[Y(t)]^{2}\notag \\ = &Y^{2}(0)e^{-\theta t}+2Y(0)e^{-\frac{1}{2}\theta t}\frac{k}{\theta} (1-e^{-\frac{1}{2}\theta t})+\frac{k^{2}}{\theta^{2}} (1-e^{-\frac{1}{2}\theta t})^{2} +\frac{\sigma^{2}}{4}E[e^{-\theta t}(\int_{0}^{t}e^{\frac{1}{2}\theta s}dB^{H}(s))^{2}]\notag \\&-Y^{2}(0)e^{-\theta t}-2Y(0)e^{-\frac{1}{2}\theta t}\frac{k}{\theta} (1-e^{-\frac{1}{2}\theta t})-\frac{k^{2}}{\theta^{2}} (1-e^{-\frac{1}{2}\theta t})^{2}\notag \\ = &\frac{\sigma^{2}}{4}E[e^{-\theta t}(\int_{0}^{t}e^{\frac{1}{2}\theta s}dB^{H}(s))^{2}]. \end{align} $

(2.6)

下面计算(2.6)中的期望,

$ \begin{align} &E[e^{-\theta t}(\int_{0}^{t}e^{\frac{1}{2}\theta s}dB^{H}(s))^{2}]\\ = &E[e^{-\theta t}(e^{\frac{1}{2}\theta t}B^{H}(t)-\frac{1}{2}\theta\int_{0}^{t}e^{\frac{1}{2}\theta s}B^{H}(s)ds)^{2}]\notag \\ = &E[e^{-\theta t}(e^{\theta t}(B^{H}(t))^{2}-\theta e^{\frac{1}{2}\theta t}\int_{0}^{t}e^{\frac{1}{2}\theta s}B^{H}(t)B^{H}(s)ds+\frac{1}{4}\theta^{2}(\int_{0}^{t}e^{\frac{1}{2}\theta s}B^{H}(s)ds)^{2})]\notag \\ = &E[(B^{H}(t))^{2}]-\theta\int_{0}^{t}e^{-\frac{1}{2}\theta (t-s)}E[B^{H}(t)B^{H}(s)]ds\notag +\frac{1}{4}\theta^{2} e^{-\theta t}\int_{0}^{t}\int_{0}^{t}e^{\frac{1}{2}\theta s+\frac{1}{2}\theta u}E[B^{H}(s)B^{H}(u)]dsdu\notag \\ = &t^{2H}-\frac{1}{2}\theta \int_{0}^{t}e^{-\frac{1}{2}\theta (t-s)}(t^{2H}+s^{2H}-(t-s)^{2H})]ds\notag \\&+\frac{1}{8}\theta^{2} e^{-\theta t}\int_{0}^{t}\int_{0}^{t}e^{\frac{1}{2}\theta s+\frac{1}{2}\theta u}(s^{2H}+u^{2H}-|s-u|^{2H})dsdu\notag \\ = &t^{2H}-t^{2H}(1-e^{-\frac{1}{2}\theta t})-\frac{1}{2}\theta e^{-\frac{1}{2}\theta t} \int_{0}^{t}e^{\frac{1}{2}\theta s}s^{2H}ds+\frac{1}{2}\theta e^{-\frac{1}{2}\theta t} \int_{0}^{t}e^{\frac{1}{2}\theta(t-u)}u^{2H}du\notag \\&+\frac{1}{4}\theta^{2}e^{-\theta t}\int_{0}^{t}e^{\frac{1}{2}\theta s}s^{2H}ds\int_{0}^{t}e^{\frac{1}{2}\theta u}du-\frac{1}{8}\theta^{2}e^{-\theta t}\int_{0}^{t}\int_{0}^{t}e^{\frac{1}{2}\theta s+\frac{1}{2}\theta u}|s-u|^{2H}dsdu\notag \\ = &t^{2H}e^{-\frac{1}{2}\theta t}-\frac{1}{2}\theta e^{-\frac{1}{2}\theta t} \int_{0}^{t}e^{\frac{1}{2}\theta s}s^{2H}ds+\frac{1}{2}\theta e^{-\frac{1}{2}\theta t} \int_{0}^{t}e^{\frac{1}{2}\theta(t-u)}u^{2H}du\notag \\&+\frac{1}{2}\theta e^{-\theta t}(e^{\frac{1}{2}\theta t}-1)\int_{0}^{t}e^{\frac{1}{2}\theta s}s^{2H}ds-\frac{1}{8}\theta^{2}e^{-\theta t}\int_{0}^{t}\int_{0}^{t}e^{\frac{1}{2}\theta s+\frac{1}{2}\theta u}|s-u|^{2H}dsdu\notag \end{align} $

(2.7)

$ \begin{align*} = &t^{2H}e^{-\frac{1}{2}\theta t}+\frac{1}{2}\theta e^{-\frac{1}{2}\theta t} \int_{0}^{t}e^{\frac{1}{2}\theta(t-u)}u^{2H}du\notag \\&-\frac{1}{2}\theta e^{-\theta t}\int_{0}^{t}e^{\frac{1}{2}\theta s}s^{2H}ds-\frac{1}{8}\theta^{2}e^{-\theta t}\int_{0}^{t}\int_{0}^{t}e^{\frac{1}{2}\theta s+\frac{1}{2}\theta u}|s-u|^{2H}dsdu. \end{align*} $

计算上面的最后一项得

$ \begin{align*} &\frac{1}{8}\theta^{2} e^{-\theta t}\int_{0}^{t}\int_{0}^{t}e^{\frac{1}{2}\theta s+\frac{1}{2}\theta u}|s-u|^{2H}dsdu \\ = &\frac{1}{8}\theta^{2} e^{-\theta t}(\int_{0}^{t}\int_{0}^{s}e^{\frac{1}{2}\theta s+\frac{1}{2}\theta u}(s-u)^{2H}duds+\int_{0}^{t}\int_{s}^{t}e^{\frac{1}{2}\theta s+\frac{1}{2}\theta u}(u-s)^{2H}duds) \\ = &\frac{1}{4}\theta^{2} e^{-\theta t}(\int_{0}^{t}\int_{0}^{s}e^{\frac{1}{2}\theta s+\frac{1}{2}\theta u}(s-u)^{2H}duds = \frac{1}{4}\theta^{2} e^{-\theta t}\int_{0}^{t}\int_{0}^{s}e^{\theta s-\frac{1}{2}\theta v}v^{2H}dvds \\ = &\frac{1}{4}\theta\int_{0}^{t}e^{-\frac{1}{2}\theta v}v^{2H}dv-\frac{1}{4}\theta e^{-\theta t} \int_{0}^{t}e^{\frac{1}{2}\theta v}v^{2H}dv. \end{align*} $

将结果带入(2.7)有

$ \begin{align*} &E[e^{-\theta t}(\int_{0}^{t}e^{\frac{1}{2}\theta s}dB^{H}(s))^{2}] \\ = &t^{2H}e^{-\frac{1}{2}\theta t}+\frac{1}{2}\theta e^{-\frac{1}{2}\theta t} \int_{0}^{t}e^{\frac{1}{2}\theta(t-s)}s^{2H}ds \\&-\frac{1}{2}\theta e^{-\theta t}\int_{0}^{t}e^{\frac{1}{2}\theta s}s^{2H}ds-\frac{1}{4}\theta\int_{0}^{t}e^{-\frac{1}{2}\theta s}s^{2H}ds+\frac{1}{4}\theta e^{-\theta t} \int_{0}^{t}e^{\frac{1}{2}\theta s}s^{2H}ds \\ = &t^{2H}e^{-\frac{1}{2}\theta t}+\frac{1}{4}\theta\int_{0}^{t}e^{-\frac{1}{2}\theta s}s^{2H}ds-\frac{1}{4}\theta e^{-\theta t} \int_{0}^{t}e^{\frac{1}{2}\theta s}s^{2H}ds \\ = &t^{2H}e^{-\frac{1}{2}\theta t}-\frac{1}{2}\int_{0}^{t}s^{2H}d(e^{-\frac{1}{2}\theta s}+e^{-\theta t+\frac{1}{2}\theta s}) = H\int_{0}^{t}s^{2H-1}(e^{-\frac{1}{2}\theta s}+e^{-\theta t+\frac{1}{2}\theta s})ds. \end{align*} $

再带入(2.6)有

$ \begin{equation} Var[Y(t)] = \frac{\sigma^{2}}{4}E[e^{-\theta t}(\int_{0}^{t}e^{\frac{1}{2}\theta s}dB^{H}(s))^{2}]\notag = \frac{\sigma^{2}}{4}H\int_{0}^{t}s^{2H-1}(e^{-\frac{1}{2}\theta s}+e^{-\theta t+\frac{1}{2}\theta s})ds. \end{equation} $

(2.8)

$ \{Y(t), t \ge 0\} $服从高斯分布$ N( Y(0)e^{-\frac{1}{2}\theta t}+\frac{k}{\theta} (1-e^{-\frac{1}{2}\theta t}), \frac{\sigma^{2}}{4}H\int_{0}^{t}s^{2H-1}(e^{-\frac{1}{2}\theta s}+e^{-\theta t+\frac{1}{2}\theta s})ds) $, 则在到达零时刻之前过程$ \{X(t), t \ge 0\} $服从自由度为1的非中心卡方分布$ \chi^{2}(1, Y^{2}(0)e^{-\theta t}+\frac{k^{2}}{\theta^{2}}(1-e^{-\frac{1}{2}\theta t})^{2}+2Y(0)\frac{k}{\theta}e^{-\frac{1}{2}\theta t}(1-e^{-\frac{1}{2}\theta t}) $).

由(2.5)和(2.8)有

$ \begin{align*} &E[X(t)]\\ = &Y^{2}(0)e^{-\theta t}+\frac{k^{2}}{\theta^{2}}(1-e^{-\frac{1}{2}\theta t})^{2}+2Y(0)\frac{k}{\theta} e^{-\frac{1}{2}\theta t}(1-e^{-\frac{1}{2}\theta t})+\frac{\sigma^{2}}{4}H\int_{0}^{t}s^{2H-1}(e^{-\frac{1}{2}\theta s}+e^{-\theta t+\frac{1}{2}\theta s})ds, \\ \end{align*} $

$ \begin{align*} &Var[X(t)]\\ = &\sigma^{2} H(Y^{2}(0)e^{-\theta t}+\frac{k^{2}}{\theta^{2}}(1-e^{-\frac{1}{2}\theta t})^{2}+2Y(0)\frac{k}{\theta} e^{-\frac{1}{2}\theta t}(1-e^{-\frac{1}{2}\theta t}))\int_{0}^{t}s^{2H-1}(e^{-\frac{1}{2}\theta s}+e^{-\theta t+\frac{1}{2}\theta s})ds\\ &+\frac{\sigma^{4}}{8}H^{2}(\int_{0}^{t}s^{2H-1}(e^{-\frac{1}{2}\theta s}+e^{-\theta t+\frac{1}{2}\theta s})ds)^{2}. \end{align*} $

当k = 0时, 考虑直到第一次到达零时刻为止, 分数Ornstein-Uhlenbeck(fOU)过程$ \{Y(t), t\ge 0\} $满足以下随机微分方程:

$ \begin{equation} dY(t) = -\frac{1}{{2}}\; \theta Y(t)dt +\frac{\sigma}{{2}} dB^{H}(t), \; Y(0)>0, \end{equation} $

(3.1)

其中$ \theta\in {R} $, $ \sigma>0 $且$ {B^{H}(t)} $是Hurst参数$ H $的分数布朗运动. 令$ \tau $为过程$ \{Y(t), t \ge 0\} $到达零的第一时刻. 对所有$ t\geq 0, \; \omega \in {\Omega} $: $ X(t)(\omega) = {Y^{2}(t)}{{\rm I}_{\{ \left| t \right| < \tau(\omega)\}. }} $

由定理2.4, 过程$ X(t) $满足以下随机微分方程: $ dX(t) = -\theta X(t)dt+\sigma \sqrt{X(t)}\circ dB^{H}(t), $其中$ X(0) = Y(0)^{2}\ge 0 $, 积分$ \int_{0}^{T}{\sqrt{ X ((s)})}\circ dB^{H}(s) $为关于分数布朗运动的路径Stratonovich积分. 考虑关于fCIR过程的第一次到达零时刻的问题对应于fOU过程的第一次到达零时刻问题. 因为fOU过程是高斯过程, Mishura等人^[6]通过对高斯分布的估计, 证明了在$ \theta<0 $时, 有限时间内第一次到达零时刻的概率等于1; 在$ \theta>0 $时, 第一次到达零时刻的概率是正的, 但是小于1, 且给出了这个概率的上界.

由fOU过程和fCIR过程在到达零之前的关系, 下面通过fOU过程的期望和方差来求fCIR过程的期望和方差.考虑fOU过程所满足的SDE(3.1), 可求出解析解为$ Y(t) = Y(0)e^{-\frac{1}{2}\theta t}+\frac{\sigma}{2} e^{-\frac{1}{2}\theta t} \int_{0}^{t} e^{\frac{1}{2}\theta s}dB^{H}(s). $期望

$ \begin{equation} \tag{$3.2$} E[Y(t)] = E[ Y(0)e^{-\frac{1}{2}\theta t}]+E[\frac{\sigma}{2} \int_{0}^{t} e^{-\frac{1}{2}\theta(t-s)}dB^{H}(t)]\notag = Y(0)e^{-\frac{1}{2}\theta t}. \end{equation} $

(3.2)

由Kukush等人^[10]的引理5.1, 方差

$ \begin{equation} Var[Y(t)] = \frac{\sigma^{2}}{4}H\int_{0}^{t}s^{2H-1}( e^{-\frac{1}{2}\theta s}+e^{-\frac{1}{2}\theta(2t-s)})ds. \end{equation} $

(3.3)

$ \{Y(t), t \ge 0\} $为fOU过程, 服从高斯分布$ Y(t)\sim (Y(0)e^{-\frac{1}{2}\theta t}, \frac{\sigma^{2}}{4}H\int_{0}^{t}s^{2H-1}( e^{-\frac{1}{2}\theta s}+e^{-\frac{1}{2}\theta(2t-s)})ds) $, 则到达零时刻之前fCIR过程服从自由度为1的非中心卡方分布$ \chi^{2}(1, Y(0)^{2}e^{-\theta t}) $. 通过(3.2)和(3.3)计算可得

$ \begin{align*} &E[X(t)]\\ = &Y^{2}(0)e^{-\theta t}+\frac{\sigma^{2}}{4}H\int_{0}^{t}s^{2H-1}( e^{-\frac{1}{2}\theta s}+e^{-\frac{1}{2}\theta(2t-s)})ds, \\ &Var[X(t)]\\ = &\sigma^{2}HY^{2}(0)e^{-\theta t}\int_{0}^{t}s^{2H-1}( e^{-\frac{1}{2}\theta s}+e^{-\frac{1}{2}\theta(2t-s)})ds+\frac{\sigma^{4}}{8}H^{2}(\int_{0}^{t}s^{2H-1}( e^{-\frac{1}{2}\theta s}+e^{-\frac{1}{2}\theta(2t-s)})ds)^{2}. \end{align*} $

可知这个结果跟第二部分的计算结果取$ k = 0 $时是一致的.

现在用Euler-Maruyama(EM)方法^[12]来模拟过程$ Y^{2}(t) $和fCIR过程$ X(t) $的路径, 在时间[0, T]中$ \delta t = \frac{T}{N} $, 用函数fbm1d^[12]模拟分数布朗运动$ B^{H}(t) $, 其增量为$ \Delta B^{H}_j = B^{H}_{j+1}-B^{H}_{j}, $

过程$ Y(t) $和过程$ X(t) $的步长分别为

$ \begin{align*} \Delta Y_j& = Y_{j+1}-Y_{j} = -\frac{1}{{2}}aY_{j}\Delta t +\frac{\sigma}{2}\Delta B^{H}, \; \; \; Y(0)>0, \\ \Delta X_j& = X_{j+1}-X_{j} = -aX_{j}\Delta t+\sigma \sqrt{X_{j}}\circ\Delta B^{H}, \; \; \; t\geq 0, \end{align*} $

其中$ j = 1, 2, ..., L $.

现在模拟过程$ Y^{2}(t) $和过程$ X(t) $的期望和方差, 令$ a = 1, \sigma = 0.3, X_{0} = 1, H = 0.7 $, 已知当$ k = 0 $时, 随机过程$ X = \{X(t), t\ge 0\} $的期望和方差分别为

$ \begin{align*} &E[X(t)]\\ = &Y^{2}(0)e^{-\theta t}+\frac{\sigma^{2}}{4}H\int_{0}^{t}s^{2H-1}( e^{-\frac{1}{2}\theta s}+e^{-\frac{1}{2}\theta(2t-s)})ds, \\ &Var[X(t)]\\ = &\sigma^{2}HY^{2}(0)e^{-\theta t}\int_{0}^{t}s^{2H-1}( e^{-\frac{1}{2}\theta s}+e^{-\frac{1}{2}\theta(2t-s)})ds+\frac{\sigma^{4}}{8}H^{2}(\int_{0}^{t}s^{2H-1}( e^{-\frac{1}{2}\theta s}+e^{-\frac{1}{2}\theta(2t-s)})ds)^{2}. \end{align*} $

用蓝色实线代表模拟过程$ X(t) $的期望, 用绿色实线代表模拟过程$ Y^{2}(t) $的期望, 用红色实线代表期望$ E[X(t)] $, 用青色实线代表过程$ X(t) $的方差, 用粉色实线代表方差$ Var[X(t)] $, 如图 1和图 2所示.

由图像和表格可知, 模拟的结果与理论上的期望和方差具有较大拟合度.