0 引言

Andreoue等^[1]在Inclán等^[2]的基础上研究了GARCH(1, 1) 模型的变点问题, Lee等^[3]提出构造RCUSUM统计量来研究GARCH(1, 1) 模型的变点问题, Chu^[4]及Kokoszka等^[5]研究了CARCH模型的单变点问题, Pedro^[6]检验了GARCH(1, 1) 模型波动率方程中的单个参数变化问题.在国内, 文献[7]对GARCH模型参数变化的残量进行检验, 文献[8]对GARCH(1, 1) 模型中变点的个数和位置都未知的多变点进行了检验, 文献[9]对GARCH模型的均值多变点进行了ANOVA检验.文献[10]提出了Sup-Wald, Sup-LM和Sup-LR方法.文献[11-13]在Sup-Wald, Sup-LM等统计性质研究方面取得了较大进展.文献[14]利用SupF检验统计量检验了ARCH(1) 模型的多变点问题.笔者在文献[15]基础上利用SupF检验统计量检验了ARMA(1, 1) 模型的多变点问题, 在此基础上本文将GARCH(1, 1) 模型转化为ARMA(1, 1) 模型, 然后构造SupF检验统计量来检验模型参数的多变点问题.

考虑GARCH(1, 1) 模型

$ {\eta _t} = {\sigma _t} \cdot {v_t},\sigma _t^2 = \left\{ \begin{array}{l} {\varphi _0} + {\varphi ^{\left( 1 \right)}}\sigma _{t - 1}^2 + {\theta ^{\left( 1 \right)}}\eta _{t - 1}^2,t = 1, \cdots ,T_1^ * ,\\ {\varphi _0} + {\varphi ^{\left( 2 \right)}}\sigma _{t - 1}^2 + {\theta ^{\left( 2 \right)}}\eta _{t - 1}^2,t = T_1^ * + 1, \cdots ,T_2^ * ,\\ \cdots \\ {\varphi _0} + {\varphi ^{\left( m \right)}}\sigma _{t - 1}^2 + {\theta ^{\left( m \right)}}\eta _{t - 1}^2,t = T_m^ * + 1, \cdots ,T. \end{array} \right. $

(1)

其中, {v_t}独立同分布, 且有

$ \left\{ {{v_t}} \right\} \sim N\left( {0,1} \right),{\phi _0} > 0,{\phi ^{\left( i \right)}} \ge 0,{\theta ^{\left( i \right)}} \ge 0,{\phi ^{\left( i \right)}} + {\theta ^{\left( i \right)}} < 1,\left( {i = 1,2, \cdots ,m} \right). $

考虑如下检验问题:(H₀):m=0, (H₁):m=k.

当(H₀)成立时, ${\phi ^{\left(1 \right)}} = {\phi ^{\left(2 \right)}} = \cdots = {\phi ^{\left(m \right)}} = \phi, {\theta ^{\left(1 \right)}} = {\theta ^{\left(2 \right)}} = \cdots = {\theta ^{\left(m \right)}} = \theta.$

当(H₁)成立时, 存在k个变点且k未知, ${\phi ^{\left(1 \right)}} \ne {\phi ^{\left(2 \right)}} \ne \cdots \ne {\phi ^{\left(m \right)}}, {\theta ^{\left(1 \right)}} \ne {\theta ^{\left(2 \right)}} \ne \cdots \ne {\theta ^{\left(m \right)}}.$

1 主要结果 1.1 模型转化及假设

考虑GARCH(1, 1) 模型

$ {\eta _t} = {\sigma _t} \cdot {v_t},\sigma _t^2 = {\varphi _0} + \varphi \eta _{t - 1}^2 + \theta \sigma _{t - 1}^2. $

(2)

令

$ {\varepsilon _t} = \eta _t^2 - \sigma _t^2 = \sigma _t^2\left( {\varepsilon _t^2 - 1} \right),{F_t} = \sigma \left( {{y_0},{\varepsilon _1},{\varepsilon _2}, \cdots ,{\varepsilon _t}} \right), $

则

$ \sigma _t^2 = \eta _t^2 - {\varepsilon _t}, $

(3)

将式(3) 代入式(2), 可得

$ \eta _t^2 - E\eta _t^2 = {\phi _0} - \left( {1 - \phi - \theta } \right)E\eta _t^2 + \left( {\phi + \theta } \right)\left( {\eta _{t - 1}^2 - E\eta _{t - 1}^2} \right) + {\varepsilon _t} - \theta {\varepsilon _{t - 1}}. $

由文献[16]可知

$ E\eta _t^2 = E\left( {\sigma _t^2} \right) = \frac{{{\phi _0}}}{{1 - \phi - \theta }}, $

则有

$ {y_t} = \left( {\varphi + \theta } \right){y_{t - 1}} - \theta {\varepsilon _{t - 1}} + {\varepsilon _t}. $

(4)

对于模型(2) 作如下假设:

(B₁) T_i=[Tλ_i], (i=1, …, m), 其中[·]表示取整运算, 1≤T_j＜T_i≤T, 1＜j＜i＜m, T₀=0,

$ {T_{m + 1}} = T,0 < {\lambda _1} < {\lambda _2} < \cdots < {\lambda _m} < 1,{\lambda _0} = 0,{\lambda _{m + 1}} = 1; $

(B₂)存在α>0, 使得$\mathop {\sup }\limits_{t \ge 1} E{\left({v_t^2 -1} \right)^{2\left({1 + \alpha } \right)}}＜\infty $, a.s., 则$E(\varepsilon _t^{2(1 + \alpha)}|{F_{t -1}}) = \sigma _t^{4(1 + \alpha)}E{({v_t}^2 -1)^{2{\rm{ }}(1 + \alpha)}}$；

(B₃):设E(v_t⁴)存在, 记E(v_t²-1)²=A, 其中A＜∞, 则$E(\varepsilon _t^2|{F_{t -1}}) = \sigma _t^4E{(v_t^2 -1)^2} = A\sigma _t^4$；

(B₄):设E(η_t⁸)＜∞.

1.2 SupF统计量

定义

$ {F_T}\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right) = \frac{{T - 2\left( {k + 1} \right)}}{{2k}}\frac{{SS{R_0} - SS{R_k}}}{{SS{R_k}}}. $

其中SSR₀表示没有变点时的总残差, 即

$ SS{R_0} = \sum\limits_{t = 1}^T {{{\left( {{y_t} - {{\hat y}_t}} \right)}^2}} . $

SSR_k表示模型在(T₁, T₂, …, T_m)下被分为m+1段后的总残差, 即

$ SS{R_k} = \sum\limits_{i = 1}^{m + 1} {\sum\limits_{t = T_{i - 1} + 1}^{Ti} {{{\left( {{y_t} - {{\hat y}_t}} \right)}^2}} } . $

其中${{\hat y}_t} = \left({\hat \varphi + \hat \theta } \right){y_{t -1}} -\hat \theta {\varepsilon _{t -1}}, \hat \varphi, \hat \theta $分别是$\phi $, θ用最小二乘法得到的估计量.

考虑检验统计量

$ \sup {F_T}\left( k \right) = \mathop {\sup }\limits_{\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right) \in {\mathit{\Lambda }_\varepsilon }} {F_T}\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right), $

其中

$ {\mathit{\Lambda }_\varepsilon } = \left\{ {\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right),\left| {{\lambda _i} - {\lambda _j}} \right| \ge \varepsilon ,{\lambda _1} \ge \varepsilon ,{\lambda _k} \le 1 - \varepsilon ,1 < j < i < m} \right\}. $

通过Monte Carlo方法对检验统计量进行数值模拟, 可得当原假设(H₀)成立时的检验统计量极限分布的渐近α分位数, 从而在置信水平α下, 原假设的拒绝域为(F_k(α), +∞).

定理1 在假设条件(B₁)~(B₄)成立以及原假设(H₀)下, 有

$ \sup {F_T}\left( k \right) = \xrightarrow{d}{F_k}{ = ^{\underline{\underline {{\text{def}}}} }}\mathop {\sup }\limits_{\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right) \in {\mathit{\Lambda }_\varepsilon }} F\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right). $

其中

$ \begin{gathered} F\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right){ = ^{\underline{\underline {{\text{def}}}} }}\frac{{3k}}{2}{{\chi ^2}}\left( {\frac{{\left[ {{{\left( {\phi + \theta } \right)}^2} - 1} \right]E\left( {\sigma _t^4y_{t - 1}^2} \right)}}{{\left( {1 + {\theta ^2}} \right)A{{\left( {E\sigma _t^4} \right)}^2}}}} \right) + \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {k + 2} \right)\sum\limits_{i = 1}^{k + 1} {\left[ {\frac{{\left[ {1 - {{\left( {\phi + \theta } \right)}^2}} \right]E\left( {\sigma _t^4y_{t - 1}^2} \right)}}{{\left( {1 + {\theta ^2}} \right){{\left( {E\sigma _t^4} \right)}^2}}}N\left( {0,{\lambda _i} - {\lambda _{i - 1}}} \right)N\left( {0,1 - \left( {{\lambda _i} - {\lambda _{i - 1}}} \right)} \right)} \right]} , \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {T \to \infty } \right). \hfill \\ \end{gathered} $

$ {\mathit{\Lambda }_\varepsilon } = \left\{ {\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right),\left| {{\lambda _i} - {\lambda _j}} \right| \geqslant \varepsilon ,1 < j < i < m,{\lambda _1} \geqslant \varepsilon ,{\lambda _k} \leqslant 1 - \varepsilon } \right\}. $

2 定理证明

为证明定理1, 首先给出如下引理.

引理1 在假设条件(B₁)~(B₄)以及原假设(H₀)下, ∀s∈[0, 1], 当T→∞时，

$ \frac{{y_{\left[ {Ts} \right]}^2}}{T}\xrightarrow{P}0, $

(5)

$ \frac{1}{T}\sum\limits_{t = 1}^{\left[ {Ts} \right]} {{\varepsilon _t}{y_{t - 1}}} \xrightarrow{P}0, $

(6)

$ \frac{1}{T}\sum\limits_{t = 1}^{\left[ {Ts} \right]} {{\varepsilon _{t - 1}}{y_{t - 1}}} \xrightarrow{P}0. $

(7)

证明 类似文献[14]可以证明引理1成立.

引理2 若假设(B₁)~(B₄)成立, 那么, 当T→∞且i>j时, 在原假设(H₀)下有

$ \frac{1}{T}\sum\limits_{t = {T_j} + 1}^{Ti} {y_t^2\xrightarrow{P}\frac{{1 + {\theta ^2}}}{{1 - {{\left( {\varphi + \theta } \right)}^2}}}A\left( {{\lambda _i} - {\lambda _j}} \right)E{\sigma ^4}} , $

(8)

$ \frac{1}{{\sqrt T }}\sum\limits_{t = {T_j} + 1}^{Ti} {{\varepsilon _t}{y_{t - 1}}\xrightarrow{d}N\left( {0,A\left( {{\lambda _i} - {\lambda _j}} \right)E\left( {{\sigma ^4}y_{t - 1}^2} \right)} \right)} . $

(9)

证明类似文献[14]可以证明引理2成立.

引理3 若假设(B₁)~(B₄)成立, 令

$ {X_{iT}} = \frac{1}{{\sqrt T }}\sum\limits_{t = {T_{i-1}} + 1}^{Ti} {{\varepsilon _t}{y_{t - 1}},{Z_i} = N\left( {0,A\left( {{\lambda _i} - {\lambda _{i - 1}}} \right)E\left( {{\sigma ^4}y_{t - 1}^2} \right)} \right)} , $

则当T→∞时, 在原假设(H₀)下{X_i}满足联合分布收敛, 即

$ \left( {{X_{1T}},{X_{2T}}, \cdots ,{X_{k + 1,T}}} \right)\xrightarrow{d}\left( {{Z_1},{Z_2}, \cdots ,{Z_{k + 1}}} \right). $

证明 类似文献[14]可以证明引理3成立.

定理1的证明

$ {F_T}\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right) = \frac{{T - 2\left( {k + 1} \right)}}{{2k}}\frac{{SS{R_0} - SS{R_k}}}{{SS{R_k}}}, $

令F_T^*=SSR₀-SSR_k, D^R(i, j)表示在原假设(H₀)下从T_i-1+1时刻到T_j时刻的残差, D^U(i, j)表示模型在被分为k+1段后从T_i-1+1时刻到T_j时刻的残差.从而有

$ F_T^ * = {D^R}\left( {1,k + 1} \right) - \sum\limits_{i = 1}^{k + 1} {{D^U}\left( {i,i} \right)} = \sum\limits_{i = 1}^{k + 1} {\left[ {{D^R}\left( {i,i} \right) - {D^U}\left( {i,i} \right)} \right]} . $

对于任意i, 有

$ \begin{array}{l} {F_{T,i}} = {D^R}\left( {i,i} \right) - {D^U}\left( {i,i} \right) = \\ \;\;\;\;\;\;\;\;\;\sum\limits_{t = T_{i - 1}}^{Ti} {{{\left[ {\left( {\hat \phi - \phi + \hat \theta - \theta } \right){y_{t - 1}} + \left( {\hat \theta - \theta } \right){\varepsilon _{t - 1}} + {\varepsilon _t}} \right]}^2}} - \\ \;\;\;\;\;\;\;\;\;\sum\limits_{t = {T_{i - 1}}}^{Ti} {{{\left[ {\left( {{{\hat \phi }_i} - {\phi _i} + {{\hat \theta }_i} - {\theta _i}} \right){y_{t - 1}} + \left( {{{\hat \theta }_i} - {\theta _i}} \right){\varepsilon _{t - 1}} + {\varepsilon _t}} \right]}^2}} . \end{array} $

其中$\hat \phi, \hat \theta $, 是在原假设(H₀)下的最小二乘估计, ${{\hat \phi }_i}, {{\hat \theta }_i}$是模型在被分为k+1段后的最小二乘估计.计算可得

$ \begin{array}{l} {F_{T,i}} = \frac{{\sum\limits_{t = {T_{i - 1}} + 1}^{Ti} {y_{t - 1}^2} }}{{{{\left( {\sum\limits_{t = 1}^{Ti} {y_{t - 1}^2} } \right)}^2}}}{\left( {\sum\limits_{t = {T_{i - 1}} + 1}^{Ti} {{\varepsilon _t}{y_{t - 1}}} + \sum\limits_{l \ne i} {\sum\limits_{t = {T_{i - 1}} + 1}^{Ti} {{\varepsilon _t}{y_{t - 1}}} } } \right)^2} + \\ \;\;\;\;\;\;\;\;\;\frac{{2\sum\limits_{t = {T_{i - 1}} + 1}^{Ti} {{\varepsilon _t}{y_{t - 1}}} }}{{\sum\limits_{t = 1}^{Ti} {y_{t - 1}^2} }}\left( {\sum\limits_{t = {T_{i - 1}} + 1}^{Ti} {{\varepsilon _t}{y_{t - 1}}} + \sum\limits_{l \ne i} {\sum\limits_{t = {T_{i - 1}} + 1}^{Ti} {{\varepsilon _t}{y_{t - 1}}} } } \right) - \frac{{3{{\left( {\sum\limits_{t = {T_{i - 1}} + 1}^{Ti} {{\varepsilon _t}{y_{t - 1}}} } \right)}^2}}}{{\sum\limits_{t = {T_{i - 1}} + 1}^{Ti} {y_{t - 1}^2} }}. \end{array} $

由引理3可知

$ \begin{gathered} \hfill \\ \left( {{X_{1T}},{X_{2T}}, \cdots ,{X_{k + 1,T}}} \right)\xrightarrow{P}\left( {{Z_1},{Z_2}, \cdots ,{Z_{k + 1}}} \right). \hfill \\ \end{gathered} $

其中

$ {X_{iT}} = \frac{1}{{\sqrt T }}\sum\limits_{t = {T_{i - 1}} + 1}^{Ti} {{\varepsilon _t}{y_{t - 1}}} ,{Z_i} = N\left( {0,A\left( {{\lambda _i} - {\lambda _{j}}} \right)E\left( {{\sigma ^4}y_{t - 1}^2} \right)} \right), $

且有

$ \begin{gathered} \frac{1}{T}\sum\limits_{t = {T_{i - 1}} + 1}^{Ti} {y_{t - 1}^2} \xrightarrow{P}\frac{{1 + {\theta ^2}}}{{1 - {{\left( {\phi + \theta } \right)}^2}}}A\left( {{\lambda _i} - {\lambda _{i - 1}}} \right)E\sigma _t^4,\left( {T \to \infty } \right), \hfill \\ {\left( {\frac{1}{T}\sum\limits_{t = {T_{i - 1}} + 1}^{Ti} {y_{t - 1}^2} } \right)^2}\xrightarrow{P}\frac{{{{\left( {1 + {\theta ^2}} \right)}^2}\left( {{\lambda _i} - {\lambda _{i - 1}}} \right) \cdot A \cdot {{\left( {E\sigma _t^4} \right)}^2}}}{{{{\left[ {1 - {{\left( {\varphi + \theta } \right)}^2}} \right]}^2}}},\left( {T \to \infty } \right). \hfill \\ \end{gathered} $

利用连续映射定理, 进一步得到

$ \begin{gathered} F_T^ * \xrightarrow{d}\sum\limits_{i = 1}^{k + 1} {\left[ {\frac{{3\left[ {1 - {{\left( {\phi + \theta } \right)}^2}} \right]E\left( {\sigma _t^4y_{t - 1}^2} \right)\left( {{\lambda _i} - {\lambda _{i - 1}} - 1} \right)}}{{\left( {1 + {\theta ^2}} \right)E\sigma _t^4}}{{\chi ^2}}\left( 1 \right)} \right] + } \hfill \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\sum\limits_{i = 1}^{k + 1} {\left[ {\frac{{2\left[ {1 - {{\left( {\phi + \theta } \right)}^2}} \right]E\left( {\sigma _t^4y_{t - 1}^2} \right)\left( {1 + \left( {{\lambda _i} - {\lambda _{i - 1}}} \right)} \right)}}{{\left( {1 + {\theta ^2}} \right)E\sigma _t^4}}N\left( {0,{\lambda _i} - {\lambda _{i - 1}}} \right)N\left( {0,1 - \left( {{\lambda _i} - {\lambda _{i - 1}}} \right)} \right)} \right].} \hfill \\ \end{gathered} $

又由于

$ \frac{1}{{T - 2\left( {k + 1} \right)}}SS{R_k}\xrightarrow{P}A \cdot E\sigma _t^4,\left( {T \to \infty } \right), $

从而有

$ {F_T}\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right)\xrightarrow{d}F\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right). $

其中

$ \begin{gathered} F{\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right)^{\underline{\underline {{\text{def}}}} }}\frac{{3k}}{2}{{\chi ^2}}\left( {\frac{{\left[ {{{\left( {\phi + \theta } \right)}^2} - 1} \right]E\left( {\sigma _t^4y_{t - 1}^2} \right)}}{{\left( {1 + {\theta ^2}} \right)A{{\left( {E\sigma _t^4} \right)}^2}}}} \right) + \hfill \\ \left( {k + 2} \right)\sum\limits_{i = 1}^{k + 1} {\left[ {\frac{{\left[ {1 - {{\left( {\theta + \varphi } \right)}^2}} \right]E\left( {\sigma _t^4y_{t - 1}^2} \right)}}{{\left( {1 + {\theta ^2}} \right)E\sigma _t^4}}N\left( {0,{\lambda _i} - {\lambda _{i - 1}}} \right)N\left( {0,1 - \left( {{\lambda _i} - {\lambda _{i - 1}}} \right)} \right)} \right]} ,\left( {T \to \infty } \right). \hfill \\ \;\;\;\;\;\;\;\;\;{\mathit{\Lambda }_\varepsilon }{\text{ = }}\left\{ {\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right),\left| {{\lambda _i} - {\lambda _j}} \right| \geqslant \varepsilon ,{\lambda _1} \geqslant \varepsilon ,{\lambda _k} \leqslant 1 - \varepsilon ,1 < j < i < m} \right\}. \hfill \\ \end{gathered} $

从而

$ \sup {F_T}\left( k \right)\xrightarrow{d}{F_k}^{\underline{\underline {{\text{def}}}} }\mathop {\sup }\limits_{\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right) \in {\mathit{\Lambda }_\varepsilon }} F\left( {{\lambda _1},{\lambda _2}, \cdots ,{\lambda _k}} \right). $

证毕.

3 数值模拟

利用Monte Carlo方法对检验统计量进行数值模拟.首先考虑GARCH(1, 1) 模型

$ {\eta _t} = \sqrt {{\varphi _0} + \varphi \sigma _{t - 1}^2 + \theta \eta _{t - 1}^2} \cdot {v_t}, $

其中, $\phi $₀>0, $\phi $≥0, θ≥0, $\phi $+θ＜1, v_t~N(0, 1).

假设当原假设(H₀)为真时生成y₁, …, y_{1 000}共1 000个样本, 其中

$ {\eta _t} = \sqrt {0.1 + 0.2\sigma _{t - 1}^2 + 0.3\eta _{t - 1}^2} \cdot {v_t},t = 1,2, \cdots ,1000. $

y₀=0是初始数据, v_t~N(0, 1), 可以得到检验统计量极限分布在检验水平分别为0.1, 0.05, 0.025, 0.01下对应的临界值分别为5.584 9, 8.659 3, 12.294 0, 18.954 2.

令$\phi $₀=0.2, $\phi $=0.25, θ=0.4, 利用模型

$ {\eta _t} = \sqrt {0.2 + 0.25\sigma _{t - 1}^2 + 0.4\eta _{t - 1}^2} \cdot {v_t},t = 1,2, \cdots ,1000. $

取样本容量分别为n=500, 800, 1 000, 重复试验400次, 在检验的显著性水平α=0.05下计算拒绝原假设的频率, 模拟结果见表 1.由表 1可知, 样本容量n越大, 经验水平越接近0.05, 检测水平失真越小.

在模拟统计量的经验势函数值时, 重复进行400次试验, 分别取样本容量依次为n=500, 800, 1 000, 在检验的显著性水平为α=0.05下, 用400次试验中拒绝原假设的频率作为经验势函数值, 其中考虑如下数据生成过程:

$ {\eta _t} = \left\{ \begin{gathered} \sqrt {{\varphi _0} + \varphi \sigma _{t - 1}^2 + \theta \eta _{t - 1}^2} \cdot {v_t},t = 1, \cdots ,\frac{n}{2}, \hfill \\ \sqrt {\varphi _0^ * + {\varphi ^ * } + \sigma _{t - 1}^2 + {\theta ^ * }\eta _{t - 1}^2} \cdot {v_t},t = \frac{n}{2} + 1, \cdots ,n. \hfill \\ \end{gathered} \right. $

其中, v_t~N(0, 1), t=1, 2, …, n, n为样本容量.

模拟过程中取$\phi $₀=0.2, $\phi $=0.1, θ=0.2, $\phi $₀^*, $\phi $^*, θ^*的取值如表 2.经过400次试验可以得到SupF检验统计量F_T的经验势函数值如表 2.

从表 2中可以看出, 当样本容量增加时, 经验势的函数值也在增加, 并且样本容量越大, 经验势函数值越接近于1, 这说明SupF统计量在GARCH(1, 1) 模型的变点检验中的可行性.