Spurious regression due to neglected of non-stationary volatility

Abstract

This paper examines the effects of permanent changes in the variance of the errors on routine applications of standard t-ratio test in regression models. It is shown the asymptotic distribution of t-ratio test is not invariant to non-stationary in variance, and the phenomenon of spurious regression will occur independently of the structure assumed for these time series. The intuition behind this is that the non-stationary volatility can increase persistency in the level of regression errors, which then leads to spurious correlation. Monte Carlo experiment evidence indicates that, in contrast to the broken level/trend case, the presence of spurious relationship critically depends on the location and magnitude of changes, regardless of the sample size. Finally, some real data sets from the Shanghai stock database are reported for illustration.

Appendix

Proof of Lemma 1

Part (1) We consider two case: when $r\le \tau$, by Assumption 1 and a functional central limit theorem as in Hansen (1992a), we have

$$\begin{aligned} T^{-1/2}\sum ^{[Tr]}_{t=1}\lambda _{1,t}\varepsilon _{t} \Rightarrow \sigma _\varepsilon \int ^r_0\lambda _{0}d{\dot{B}}_\varepsilon (s), \end{aligned}$$

(7.1)

and for the other case $r>\tau$, the limit is

$$\begin{aligned} T^{-1/2}\sum ^{[Tr]}_{t=1}\lambda _{1,t}\varepsilon _{t} \Rightarrow \sigma _\varepsilon \int ^\tau _0\lambda _0d{\dot{B}}_{\varepsilon }(s) +\sigma _\varepsilon \int ^r_\tau \lambda _1d{\dot{B}}_{\varepsilon }(s). \end{aligned}$$

(7.2)

Therefore, we get

$$\begin{aligned} T^{-1/2}\sum ^{[Tr]}_{t=1}x_{1,t} \Rightarrow {\dot{\lambda }}\sigma _\varepsilon {\dot{\lambda }}^{-1}\int ^r_0[\lambda _0{\mathbf {1}}_{\{s\le \tau \}} +\lambda _1{\mathbf {1}}_{\{s>\tau \}}]d{\dot{B}}_{\varepsilon }(s)={\dot{\lambda }}\sigma _\varepsilon {\dot{B}}_{x}(r). \end{aligned}$$

Similarly, it is straightforward to show that

$$\begin{aligned} T^{-1}\sum ^{T}_{t=1}x^2_{1,t}\!=\!T^{-1}\sum ^{[T\tau ]}_{t=1}\lambda ^2_0\varepsilon ^2_{t} +T^{-1}\sum ^{T}_{t=[T\tau ]+1}\lambda ^2_1\varepsilon ^2_{t} \mathop {\rightarrow }\limits ^{p}\tau \lambda ^2_0\sigma ^2_\varepsilon +(1-\tau )\lambda ^2_1\sigma ^2_\varepsilon \!=\!{\dot{\lambda }}^2\sigma ^2_\varepsilon . \end{aligned}$$

Part (2) The convergence of $T^{-1/2}\sum \nolimits ^{[Tr]}_{t=1}y_{1,t}$ and $T^{-1}\sum \nolimits ^{T}_{t=1}y^2_{1,t}$ to ${\dot{\omega }}\sigma _\xi {\dot{B}}_{y}(r)$ and ${\dot{\omega }}^2\sigma ^2_\xi$ can be respectively proved by above exactly the same steps as in (1) case.

Part (3) We first deal with $\tau \le \varsigma$. Note that $\varepsilon _{1,t}$ and $\xi _{1,t}$ are independent, when $r\le \tau$, yielding

$$\begin{aligned} T^{-1/2}\sum ^{[Tr]}_{t=1}\lambda _{1,t}\varepsilon _{1,t}\omega _{1,t}\xi _{1,t} \Rightarrow \sigma _\varepsilon \sigma _\xi \int ^r_0\lambda _{0}\omega _0d{\dot{B}}_{\varepsilon \xi }(s), \end{aligned}$$

(7.3)

which follows the results in Stout (1974) and Davidson (1994).

Similarly, observing that for $\tau <r\le \varsigma$,

$$\begin{aligned} T^{-1/2}\sum ^{[Tr]}_{t=1}\lambda _{1,t}\varepsilon _{1,t}\omega _{1,t}\xi _{1,t} \Rightarrow \sigma _\varepsilon \sigma _\xi \int ^\tau _0\lambda _{0}\omega _0d{\dot{B}}_{\varepsilon \xi }(s) +\sigma _\varepsilon \sigma _\xi \int ^r_\tau \lambda _{1}\omega _0d{\dot{B}}_{\varepsilon \xi }(s),\nonumber \\ \end{aligned}$$

(7.4)

and for $\varsigma <r$,

$$\begin{aligned}&T^{-1/2}\sum ^{[Tr]}_{t=1}\lambda _{1,t}\varepsilon _{1,t}\omega _{1,t}\xi _{1,t} \Rightarrow \sigma _\varepsilon \sigma _\xi \left\{ \int ^\tau _0\lambda _{0}\omega _0d{\dot{B}}_{\varepsilon \xi }(s)\right. \nonumber \\&\quad \left. +\int ^\varsigma _\tau \lambda _{1}\omega _0d{\dot{B}}_{\varepsilon \xi }(s) +\int ^r_\varsigma \lambda _{1}\omega _1d{\dot{B}}_{\varepsilon \xi }(s)\right\} . \end{aligned}$$

(7.5)

Just as relations (7.3), (7.4) and (7.5) imply that

$$\begin{aligned} T^{-1/2}\sum ^{[Tr]}_{t=1}x_{1,t}y_{1,t} \Rightarrow \,&{\dot{\pi }}\sigma _\varepsilon \sigma _\xi {\dot{\pi }}^{-1}\int ^r_0[\lambda _0\omega _0{\mathbf {1}}_{\{s\le \tau \}} +\lambda _1\omega _0{\mathbf {1}}_{\{\tau<s\le \varsigma \}}\\&+\,\lambda _1\omega _1{\mathbf {1}}_{\{\varsigma <s\}}]d{\dot{B}}_{\varepsilon \xi }(s) =\,{\dot{\pi }}\sigma _\varepsilon \sigma _\xi {\dot{B}}_{\varepsilon \xi }(r). \end{aligned}$$

Hence, we conclude that

$$\begin{aligned} T^{-1}\sum ^{T}_{t=1}x^2_{1,t}y^2_{1,t}= & {} T^{-1}\left\{ \lambda ^2_0\omega ^2_0\sum ^{[T\tau ]}_{t=1}\varepsilon ^2_{t}\xi ^2_{t} +\lambda ^2_1\omega ^2_0\sum ^{[T\varsigma ]}_{t=[T\tau ]+1}\varepsilon ^2_{t}\xi ^2_{t} +\lambda ^2_1\omega ^2_1\sum ^{T}_{t=[T\varsigma ]+1}\varepsilon ^2_{t}\xi ^2_{t}\right\} \\&\mathop {\rightarrow }\limits ^{p}\sigma ^2_\varepsilon \sigma ^2_\xi \{\tau \lambda ^2_0\omega ^2_0+(\varsigma -\tau )\lambda ^2_1\omega ^2_0 +(1-\varsigma )\lambda ^2_1\omega ^2_1\}={\dot{\pi }}^2\sigma ^2_\varepsilon \sigma ^2_\xi . \end{aligned}$$

Thus following the proof of $\tau \le \varsigma$, one can verify the negligibility of $\tau >\varsigma$ case.

Part (4) For following convenience, let $\sigma ^2_{1,x}=\sum ^{T-1}_{k=-T+1}h\left( k/q_T\right) \gamma _{1,x}(k)$, where for $k\ge 0, \gamma _{1,x}(k)=T^{-1}\sum ^{T-k}_{t=1}E(x_{1,t}x_{1,t+k})$, while $\gamma _{1,x}(k)=\gamma _{1,x}(-k)$ for $k<0$, it suffices to prove that $\sigma ^2_{1,x}-{\dot{\lambda }}^2\sigma ^2_\varepsilon \rightarrow 0$ as $T\rightarrow \infty$.

Under the Assumptions 1 and 2, de Jong (2000) and Andrew (1991) guaranteed that $\sigma ^2_{\varepsilon ,T}\rightarrow \sigma ^2_{\varepsilon }$, where $\sigma ^2_{\varepsilon ,T}=\sum ^{T-1}_{k=-T+1}h\left( k/q_T\right) \gamma _{\varepsilon }(k), \gamma _{\varepsilon }(k)=E(\varepsilon _t\varepsilon _{t+|k|})$. Thus, we only just need to prove $\sigma ^2_{1,x}-{\dot{\lambda }}^2\sigma ^2_{\varepsilon ,T}\rightarrow 0$. The remaider of the proof is necessary to show that, as $T\rightarrow \infty$, (i) $\sigma ^2_{1,x}-{\tilde{\sigma }}^2_{1,x}\rightarrow 0$, where ${\tilde{\sigma }}^2_{1,x}=\sum ^{T-1}_{k=-T+1}h\left( k/q_T\right) \gamma _{\varepsilon }(k)T^{-1}\sum ^{T-|k|}_{t=1}\lambda ^2_t$; (ii) ${\tilde{\sigma }}^2_{1,x}-{\dot{\lambda }}^2\sigma ^2_{\varepsilon ,T}\rightarrow 0$.

We first verify the (i) case. Observe that

$$\begin{aligned} \left| \sigma ^2_{1,x}-{\tilde{\sigma }}^2_{1,x}\right|= & {} \left| \sum ^{T-1}_{k=-T+1}h\left( k/q_T\right) \frac{1}{T}\sum ^{T-|k|}_{t=1}(\lambda _t\lambda _{t+|k|}-\lambda ^2_t)\gamma _{\varepsilon }(k)\right| \nonumber \\ \!\le & {} \! \sum ^{T-1}_{k=-T+1}h\left( k/q_T\right) \frac{1}{T}\sum ^{T-|k|}_{t=1}\lambda _t|\lambda _{t+|k|}-\lambda _t| \cdot \sum ^{T-1}_{k=-T+1}|\gamma _{\varepsilon }(k)|=I\cdot { II}.\qquad \end{aligned}$$

(7.6)

By the theorem 3.1 of Phillips and Perron (1988) and corollary 14.3 of Davidson (1994), we obtain that for some $C>0, \sum ^\infty _{k=0}|\gamma _\varepsilon (k)|<C\sum ^\infty _{k=0}\varphi ^{1-d/2}_{\varepsilon ,k}<\infty$, thus $II<\infty$.

Let us now consider that $I=o_p(1)$. Define $\breve{\lambda }=\max \{\lambda _0,\lambda _1\}$, as $T\rightarrow \infty$, we have

$$\begin{aligned} \sum ^{T-|k|}_{t=1}\lambda _t|\lambda _{t+|k|}-\lambda _t|\le \breve{\lambda }\sum ^{T-|k|}_{t=1}|\lambda _{t+|k|}-\lambda _t|\le \breve{\lambda }\cdot |k||\lambda _1-\lambda _0|\le |k|\breve{\lambda }^2, \end{aligned}$$

and therefore,

$$\begin{aligned} I\le \frac{\breve{\lambda }^2}{T} \sum ^{T-1}_{k=-T+1}h\left( k/q_T\right) |k|=\frac{\breve{\lambda }^2q^2_T}{T} \sum ^{T-1}_{k=-T+1}h\left( k/q_T\right) \frac{|k|}{q_T}\frac{1}{q_T}\rightarrow 0, \end{aligned}$$

since $\sup _{T\ge 1}q^{-2}_T\sum ^{T-1}_{k=-T+1}|h(k/q_T)||k|<\infty$ and $q^2_T/T=o_p(1)$ under Assumption 2. Because $I\rightarrow 0$ and II is bounded, this implies $\sigma ^2_{1,x}-{\tilde{\sigma }}^2_{1,x}\rightarrow 0$.

Regarding (ii), by setting ${\tilde{\lambda }}^{2}=T^{-1}\sum ^T_{t=1}\lambda ^2_t$, after some algebra it follows that

$$\begin{aligned} \left| {\tilde{\sigma }}^2_{1,x}-{\dot{\lambda }}^{2}\sigma ^2_{\varepsilon ,T}\right| \le \left| {\tilde{\sigma }}^2_{1,x}-{\tilde{\lambda }}^{2}\sigma ^2_{\varepsilon ,T}\right| +\left| {\tilde{\lambda }}^{2}\sigma ^2_{\varepsilon ,T}-{\dot{\lambda }}^{2}\sigma ^2_{\varepsilon ,T}\right| . \end{aligned}$$

It is straightforward to see that,

$$\begin{aligned} \left| {\tilde{\sigma }}^2_{1,x}-{\tilde{\lambda }}^{2}\sigma ^2_{\varepsilon ,T}\right|= & {} \left| \sum ^{T-1}_{k=-T+1}h(k/q_T)\gamma _\varepsilon (k)\frac{-1}{T}\sum ^{T}_{t=T-|k|+1}\lambda ^2_t\right| \nonumber \\\le & {} \frac{\breve{\lambda }^2}{T}\sum ^{T-1}_{k=-T+1}|h(k/q_T)||k||\gamma _\varepsilon (k)|. \end{aligned}$$

(7.7)

We conclude by showing that the foregoing results continue to hold and obtain ${\tilde{\sigma }}^2_{1,x}-{\tilde{\lambda }}^{2}\sigma ^2_{\varepsilon ,T}\rightarrow 0$. Moreover, under the Assumptions 1, ${\tilde{\lambda }}^{2}\sigma ^2_{\varepsilon ,T}-{\dot{\lambda }}^{2}\sigma ^2_{\varepsilon ,T}\rightarrow 0$ follows because the limit of ${\tilde{\lambda }}^{2}$ is ${\dot{\lambda }}^2$. Therefore, the limit of (7.6) and (7.7) tend to zero and the proof is completed. $\square$

Proof of Lemma 2

The proof follows trivially from Lemma 1 and is omitted for brevity. $\square$

Proof of Theorem 1

By Lemmas 1 and 2, it guarantees that

$$\begin{aligned} T^{1/2}\hat{\beta }= & {} \frac{T^{-1/2}\sum \limits ^T_{t=1}x_{1,t}y_{1,t} -T^{-1/2}\left( T^{-1/2}\sum \limits ^T_{t=1}x_{1,t}\right) \left( T^{-1/2}\sum \limits ^T_{t=1}y_{1,t}\right) }{T^{-1}\sum \limits ^T_{t=1}x^2_{1,t}-T^{-1/2}\left( T^{-1/2}\sum \limits ^T_{t=1}x_{1,t}\right) \left( T^{-1/2}\sum \limits ^T_{t=1}x_{1,t}\right) }\nonumber \\\Rightarrow & {} \frac{{\dot{\pi }}\sigma _\xi }{{\dot{\lambda }}^2\sigma _\varepsilon }\cdot {\dot{B}}_{xy}(1). \end{aligned}$$

(7.8)

From the weak convergence $\sum ^T_{t=1}x_{1,t}=O_p(T^{1/2})$ and $\sum ^T_{t=1}y_{1,t}=O_p(T^{1/2})$, it implied that $\hat{\eta }_t=(y_{1,t}-T^{-1}\sum ^T_{i=1}y_{1,i})-\hat{\beta }(x_{1,t}-T^{-1}\sum ^T_{i=1}x_{1,i}) =y_{1,t}-\hat{\beta }x_{1,t}+o_p(1)$. Thus, we have

$$\begin{aligned} \hat{\gamma }_\eta (k)= & {} \frac{1}{T}\sum ^{T-|k|}_{t=1}\hat{\eta }_t\hat{\eta }_{t+|k|} =\frac{1}{T}\sum ^{T-|k|}_{t=1}(y_{1,t}-\hat{\beta }x_{1,t})(y_{1,t+|k|}-\hat{\beta }x_{1,t+|k|})\nonumber \\= & {} \frac{1}{T}\sum ^{T-|k|}_{t=1}\left( y_{1,t}y_{1,t+|k|}-\hat{\beta }y_{1,t}x_{1,t+|k|} -\hat{\beta }y_{1,t+|k|}x_{1,t}+\hat{\beta }^2x_{1,t}x_{1,t+|k|}\right) \nonumber \\= & {} \frac{1}{T}\sum ^{T-|k|}_{t=1}y_{1,t}y_{1,t+|k|}+\hat{\beta }^2\frac{1}{T}\sum ^{T-|k|}_{t=1}x_{1,t}x_{1,t+|k|}, \end{aligned}$$

(7.9)

which follows $T^{-1}\sum ^T_{t=1}x_{1,t}y_{1,t+|k|}=o_p(1)$ and $T^{-1}\sum ^T_{t=1}x_{1,t+|k|}y_{1,t}=o_p(1)$ as $x_{1,s}$ and $y_{1,t}$ are supposed to be independent for all $s,t\ge 1$.

Together with (2.7) and (7.9), yields that

$$\begin{aligned} \hat{\sigma }^2_\eta =\sum ^{T-1}_{k=-T+1}h\left( \frac{k}{q_T}\right) \frac{1}{T}\sum ^{T-|k|}_{t=1}y_{1,t}y_{1,t+|k|} +\hat{\beta }^2\sum ^{T-1}_{k=-T+1}h\left( \frac{k}{q_T}\right) \frac{1}{T}\sum ^{T-|k|}_{t=1}x_{1,t}x_{1,t+|k|}. \end{aligned}$$

Using the Lemma 1 and fact $\hat{\beta }$ is of $O_p(T^{-1/2})$, we obtain

$$\begin{aligned} \hat{\sigma }^2_\eta \mathop {\rightarrow }\limits ^{p}{\dot{\omega }}^2\sigma ^2_\xi +\hat{\beta }^2{\dot{\lambda }}^2\sigma ^2_\varepsilon ={\dot{\omega }}^2\sigma ^2_\xi +o_p(1). \end{aligned}$$

(7.10)

Combining these results with the asymptotic behavior of $\hat{\beta }$, we get

$$\begin{aligned} t_{\hat{\beta }}=\frac{T^{1/2}\hat{\beta }}{\hat{\sigma }_\eta \left( T^{-1}\sum ^T_{t=1}x^2_{1,t}\right) ^{1/2}+o_p(1)} \Rightarrow \frac{{\dot{\pi }}}{{\dot{\omega }}{\dot{\lambda }}}\cdot {\dot{B}}_{xy}(1). \end{aligned}$$

Finally, we also have

$$\begin{aligned} TR^2=T\hat{\beta }^2 \cdot \frac{T^{-1}\sum \limits ^T_{t=1}x^2_{1,t}+o_p(1)}{T^{-1}\sum \limits ^T_{t=1}y^2_{1,t}+o_p(1)} \mathop {\rightarrow }\limits ^{p}\frac{{\dot{\pi }}^2}{{\dot{\lambda }}^2{\dot{\omega }}^2}\cdot {\dot{B}}^2_{xy}(1). \end{aligned}$$

Hence, the proof of Theorem 1 is completed.

Proof of Theorem 2

The idea of the proof is similar to Theorem 1 but, instead of Lemma 1, Lemma 2 is needed. $\square$