Testing for the sandwich-form covariance matrix of the quasi-maximum likelihood estimator

Abstract

This study tests for the sandwich-form asymptotic covariance matrices entailed by conditionally heteroskedastic and/or autocorrelated regression errors or conditionally uncorrelated homoskedastic errors. In doing so, we enable the empirical researcher to estimate the asymptotic covariance matrix of the quasi-maximum likelihood estimator by supposing a possibly misspecified model for error distribution. Accordingly, we provide test methodologies by extending the approaches in Cho and White (in: Chang Y, Fomby T, Park JY (eds) Advances in econometrics: essays in honor of Peter CB Phillips. Emerald Group Publishing Limited, West Yorkshire, 2014) and Cho and Phillips (J Econ 202:45–56, 2018a) to detect the influence of heteroskedastic and/or autocorrelated regression errors on the asymptotic covariance matrix. In particular, we establish a sequential testing procedure to achieve our goal. We affirm the theory on our test statistics through simulation and apply the test statistics to energy price growth rate data for illustrative purposes; here, we also apply our test methodology to test the fully correct model hypothesis.

Appendix: Proofs

Proof of Theorem 1

For each $i=1$ and 2, Theorem 2 of CP implies that for $j=1$ and 2, the local alternative approximations of $\widehat{{\mathfrak {B}}}_{j,n}^{(i)}$, $\widehat{{\mathfrak {S}}}_{j,n}^{(i)}$, and $\widehat{{\mathfrak {E}}}_{j,n}^{(i)}$ are equivalent and obtained as $\frac{1}{2} \mathrm {tr}[({\mathbf {V}}_{*}^{(i)} + \sqrt{n}{\mathbf {K}}_{o,n}^{(i)})^{2}] + o_{{\mathbb {P}}}(1)$, where ${\mathbf {K}}_{o,n}^{(i)} := {\mathbf {M}}_{o,n}^{(i)} + \sum _{j=1}^{d} ( \widehat{\theta }_{j,n} - \theta _{j*} ){\mathbf {S}}_{j,*}^{(i)}$, ${\mathbf {M}}_{o,n}^{(i)} := ({\mathbf {Q}}_{*}^{(i)})^{-1}({\mathbf {Q}}_{n}^{(i)}(\varvec{\xi }_{*}) - {\mathbf {P}}_{n}^{(i)}(\varvec{\xi }_{*}) - {\mathbf {Q}}_{*n}^{(i)} + {\mathbf {P}}_{*n}^{(i)})$, and ${\mathbf {S}}_{j*}^{(i)} := ({\mathbf {P}}_{*}^{(i)})^{-1} [{\partial }/({\partial \theta _{j}}) {\mathbf {Q}}^{(i)}(\varvec{\xi }_{*}) - {\partial }/({\partial \theta _{j}}) {\mathbf {P}}_{*}^{(i)}(\varvec{\xi }_{*})]$. Furthermore, the symmetry between ${\mathbf {P}}_{*}^{(i)}$ and ${\mathbf {Q}}_{*}^{(i)}$ implies that for $j=1, 2$, the local alternative approximations of $\widetilde{{\mathfrak {B}}}_{j,n}^{(i)}$, $\widetilde{{\mathfrak {S}}}_{j,n}^{(i)}$, and $\widetilde{{\mathfrak {E}}}_{j,n}^{(i)}$ are equivalently obtained as $\frac{1}{2} \mathrm {tr}[(\widetilde{{\mathbf {V}}}_{*}^{(i)} + \sqrt{n}\widetilde{{\mathbf {K}}}_{o,n}^{(i)})^{2}] + o_{{\mathbb {P}}}(1)$, where $\widetilde{{\mathbf {V}}}_{*}^{(i)} := ({\mathbf {P}}_{*}^{(i)})^{-1} {\bar{{\mathbf {P}}}}_{*}^{(i)} - ({\mathbf {Q}}_{*}^{(i)})^{-1} {\bar{{\mathbf {Q}}}}_{*}^{(i)}$, $\widetilde{{\mathbf {K}}}_{o,n}^{(i)} := \widetilde{{\mathbf {M}}}_{o,n}^{(i)} + \sum _{j=1}^{d} ( \widehat{\theta }_{j,n} - \theta _{j*} ) \widetilde{{\mathbf {S}}}_{j,*}$, $\widetilde{{\mathbf {M}}}_{o,n}^{(i)} := ({\mathbf {P}}_{*}^{(i)})^{-1} ({\mathbf {P}}_{n}^{(i)}(\varvec{\xi }_{*}) - {\mathbf {Q}}_{n}^{(i)}(\varvec{\xi }_{*}) - {\mathbf {P}}_{*n}^{(i)} + {\mathbf {Q}}_{*n}^{(i)})$, and $\widetilde{{\mathbf {S}}}_{j*}^{(i)} := ({\mathbf {Q}}_{*}^{(i)})^{-1} [ {\partial }/({\partial \theta _{j}}) {\mathbf {P}}_{*}^{(i)}(\varvec{\xi }_{*}) - {\partial }/({\partial \theta _{j}}) {\mathbf {Q}}^{(i)}(\varvec{\xi }_{*}) ]$. That is, ${\mathbf {V}}_{*}^{(i)} = -\widetilde{{\mathbf {V}}}_{*}^{(i)}$ and ${\mathbf {K}}_{o,n}^{(i)} = -{\mathbf {K}}_{o,n}^{(i)}$, so that for $j=1$ and 2, the local alternative approximations of $\widehat{{\mathfrak {B}}}_{j,n}^{(i)}$, $\widehat{{\mathfrak {S}}}_{j,n}^{(i)}$, and $\widehat{{\mathfrak {E}}}_{j,n}^{(i)}$ are equivalent to the local alternative approximations of $\widetilde{{\mathfrak {B}}}_{j,n}^{(i)}$, $\widetilde{{\mathfrak {S}}}_{j,n}^{(i)}$, and $\widetilde{{\mathfrak {E}}}_{j,n}^{(i)}$. Therefore, it now follows that

$$\begin{aligned}&\widehat{{\mathfrak {M}}}_{n}^{(i)}\\&\quad := \max _{j=1,2}\left[ \widehat{{\mathfrak {B}}}_{j,n}^{(i)}, \widehat{{\mathfrak {S}}}_{j,n}^{(i)}, \widehat{{\mathfrak {E}}}_{j,n}^{(i)}, \widetilde{{\mathfrak {B}}}_{j,n}^{(i)}, \widetilde{{\mathfrak {S}}}_{j,n}^{(i)}, \widetilde{{\mathfrak {E}}}_{j,n}^{(i)} \right] = \frac{1}{2} \mathrm {tr}[({\mathbf {V}}_{*}^{(i)} + \sqrt{n}{\mathbf {K}}_{o,n}^{(i)})^{2}] + o_{{\mathbb {P}}}(1). \end{aligned}$$

In addition, Corollary 1 of CP implies that $\frac{1}{2} \mathrm {tr}[(\sqrt{n} {\mathbf {K}}_{o,n}^{(i)})^{2}] \Rightarrow \varvec{{\mathcal {Z}}}^{(i)\prime }\varvec{{\varOmega }}_{*}^{(i)} \varvec{{\mathcal {Z}}}^{(i)}$, suggesting that $\frac{1}{2} \mathrm {tr}[({\mathbf {V}}_{*}^{(i)} + \sqrt{n}{\mathbf {K}}_{o,n}^{(i)})^{2}] \Rightarrow (\varvec{{\mathcal {Z}}}^{(i)\prime } + {\mathbf {V}}_{*}^{(i)\prime }\varvec{{\varOmega }}_{*}^{(i)-1/2}) \varvec{{\varOmega }}_{*}^{(i)} (\varvec{{\mathcal {Z}}}^{(i)} + \varvec{{\varOmega }}_{*}^{(i)-1/2} {\mathbf {V}}_{*}^{(i)})$. Note that ${\ddot{{\mathbf {V}}}}_{*}^{(i)} := \varvec{{\varOmega }}_{*}^{(i)-1/2} {\mathbf {V}}_{*}^{(i)}$, so that $\widehat{{\mathfrak {M}}}_{n}^{(i)} \Rightarrow (\varvec{{\mathcal {Z}}}^{(i)} + {\ddot{{\mathbf {V}}}}_{*}^{(i)})'\varvec{{\varOmega }}_{*}^{(i)}(\varvec{{\mathcal {Z}}}^{(i)} + {\ddot{{\mathbf {V}}}}_{*}^{(i)})$, as desired. $\square$

Proof of Corollary 1

For each $i=1$ and 2, given the null hypothesis ${\mathcal {H}}_{0}^{(i)}$, ${\bar{{\mathbf {P}}}}_{*}^{(i)} = {\bar{{\mathbf {Q}}}}_{*}^{(i)} = {\mathbf {0}}$. Therefore, Theorem 1 implies that $\widehat{{\mathfrak {M}}}_{n}^{(i)} \Rightarrow \varvec{{\mathcal {Z}}}^{(i)\prime } \varvec{{\varOmega }}_{*}^{(i)} \varvec{{\mathcal {Z}}}^{(i)}$. $\square$

Proof of Theorem 2

For $i = 1$ and 2, we first note that $\widehat{{\mathfrak {M}}}_{n}^{(i)} = \max [ {\dot{{\mathfrak {M}}}}_{n}^{(i)}, {\ddot{{\mathfrak {M}}}}_{n}^{(i)}]$, where ${\dot{{\mathfrak {M}}}}_{n}^{(i)} := \max _{j=1,2}[\widehat{{\mathfrak {B}}}_{j,n}^{(i)}, \widehat{{\mathfrak {S}}}_{j,n}^{(i)},$ $\widehat{{\mathfrak {E}}}_{j,n}^{(i)}]$ and ${\ddot{{\mathfrak {M}}}}_{n}^{(i)} := \max _{j=1,2}[\widetilde{{\mathfrak {B}}}_{j,n}^{(i)}, \widetilde{{\mathfrak {S}}}_{j,n}^{(i)}, \widetilde{{\mathfrak {E}}}_{j,n}^{(i)}]$. Here, the leading term of ${\dot{{\mathfrak {M}}}}_{n}^{(i)}$ is determined by ${\dot{\mu }}_{*}^{(i)} := \max _{j=1, 2}[{\dot{{\mathfrak {B}}}}_{j,*}^{(i)}, {\dot{{\mathfrak {S}}}}_{j,*}^{(i)}, {\dot{{\mathfrak {E}}}}_{j,*}^{(i)}]$, where ${\dot{{\mathfrak {B}}}}_{2*}^{(i)} := ({\dot{\delta }}_{*}^{(i)})^{2} + 2{\dot{\zeta }}_{*}^{(i)}$, ${\dot{{\mathfrak {S}}}}_{1*}^{(i)}:= ({\dot{\delta }}_{*}^{(i)})^{2} + 2{\dot{\gamma }}_{*}^{(i)}$, and ${\dot{{\mathfrak {S}}}}_{2*}^{(i)}:= ({\dot{\eta }}_{*}^{(i)})^{2} + 2{\dot{\zeta }}_{*}^{(i)}$. Here, $({\dot{\delta }}_{*}^{(i)})^{2}$ is dominated by $({\dot{\tau }}_{*}^{(i)})^{2}$ or $({\dot{\eta }}_{*}^{(i)})^{2}$ from the fact that ${\dot{\delta }}_{*}^{(i)} \in [{\dot{\eta }}_{*}^{(i)}, {\dot{\tau }}_{*}^{(i)}]$ and ${\dot{\eta }}_{*}^{(i)} \ge -1$. Therefore, it now follows that ${\dot{\mu }}_{*}^{(i)} = \max [{\dot{{\mathfrak {B}}}}_{1,*}^{(i)}, {\dot{{\mathfrak {S}}}}_{2,*}^{(i)}, {\dot{{\mathfrak {E}}}}_{1,*}^{(i)}, {\dot{{\mathfrak {E}}}}_{2,*}^{(i)} ]$.

Likewise, if we let ${\ddot{\mu }}_{*}^{(i)}$ be the leading term of ${\ddot{{\mathfrak {M}}}}_{n}^{(i)}$, ${\ddot{\mu }}_{*}^{(i)} = \max [{\ddot{{\mathfrak {B}}}}_{1,*}^{(i)}, {\ddot{{\mathfrak {S}}}}_{2,*}^{(i)}, {\ddot{{\mathfrak {E}}}}_{1,*}^{(i)}, {\ddot{{\mathfrak {E}}}}_{2,*}^{(i)}]$, where ${\ddot{{\mathfrak {S}}}}_{2,*}^{(i)}:= ({\ddot{\eta }}_{*}^{(i)})^{2} + 2{\ddot{\gamma }}_{*}^{(i)}$. Therefore, it now follows that $\mu _{*}^{(i)} = \max [{\dot{{\mathfrak {B}}}}_{1,*}^{(i)}, {\dot{{\mathfrak {S}}}}_{2,*}^{(i)}, {\dot{{\mathfrak {E}}}}_{1,*}^{(i)}, {\dot{{\mathfrak {E}}}}_{2,*}^{(i)}, {\ddot{{\mathfrak {B}}}}_{1,*}^{(i)}, {\ddot{{\mathfrak {S}}}}_{2,*}^{(i)},$ ${\ddot{{\mathfrak {E}}}}_{1,*}^{(i)}, {\ddot{{\mathfrak {E}}}}_{2,*}^{(i)}]$, and further ${\dot{{\mathfrak {S}}}}_{2,*}^{(i)} \le \max [{\dot{{\mathfrak {E}}}}_{1,*}^{(i)}, {\ddot{{\mathfrak {B}}}}_{1,*}^{(i)}]$ and ${\ddot{{\mathfrak {S}}}}_{2,*}^{(i)} \le \max [{\ddot{{\mathfrak {E}}}}_{1,*}^{(i)}, {\dot{{\mathfrak {B}}}}_{1,*}^{(i)}]$. Therefore, we can further simplify $\mu _{*}^{(i)}$: $\mu _{*}^{(i)} = \max [{\dot{{\mathfrak {B}}}}_{1,*}^{(i)}, {\dot{{\mathfrak {E}}}}_{1,*}^{(i)}, {\dot{{\mathfrak {E}}}}_{2,*}^{(i)},$ ${\ddot{{\mathfrak {B}}}}_{1,*}^{(i)}, {\ddot{{\mathfrak {E}}}}_{1,*}^{(i)}, {\ddot{{\mathfrak {E}}}}_{2,*}^{(i)}]$. This completes the proof. $\square$