Likelihood-based inference for dynamic panel data models

Abstract

This paper considers maximum likelihood (ML)-based inferences for dynamic panel data models. We focus on the analysis of the panel data with a large number (N) of cross-sectional units and a small number (T) of repeated time series observations for each cross-sectional unit. We examine several different ML estimators and their asymptotic and finite-sample properties. Our major finding is that when data follow unit-root processes without or with drifts, the ML estimators have singular information matrices. This is a case of Sargan (Econometrica 51:1605–1634, 1983) in which the first-order condition for identification fails, but parameters are identified. The ML estimators are consistent, but they have non-standard asymptotic distributions, and their convergence rates are lower than N^1/2. In addition, the sizes of usual Wald statistics based on the estimators are distorted even asymptotically, and they reject the unit-root hypothesis too often. However, following Rotnitzky et al. (Bernoulli 6:243–284, 2000) we show that likelihood ratio (LR) tests for unit root follow mixtures of chi-square distributions. Our Monte Carlo experiments show that the LR tests with the p-values from the mixed distributions are much better sized than the Wald tests, although they tend to slightly over-reject the unit-root hypothesis in small samples. It is also shown that the LR tests for unit roots have good finite-sample power properties.

Appendix

Proof of Equation (12): Define $P_{T} = T^{ - 1} 1_{T} 1^{\prime}_{T}$,$Q_{T} = I_{T} - P_{T}$; and

$$D_{T - 1}^{{\prime }} = \left( {\begin{array}{*{20}c} { - 1} & 1 & 0 & {...} & 0 & 0 \\ 0 & { - 1} & 1 & {...} & 0 & 0 \\ : & : & : & {} & : & : \\ 0 & 0 & 0 & {...} & 1 & 0 \\ 0 & 0 & 0 & {...} & { - 1} & 1 \\ \end{array} } \right)_{(T - 1) \times T} .$$

Observe that $D_{T - 1}^{{\prime }} (y_{i1} ,...,y_{iT} )^{{\prime }} = \Delta y_{i}$ and $D_{T - 1}^{{\prime }} 1_{T}$ = $0_{(T - 1) \times 1}$. By Rao (1973, p. 77), we can have $Q_{T} = D_{T - 1} B_{T - 1}^{ - 1} D_{T - 1}^{{\prime }}$. We can also show

$$\overline{{\Delta_{0} y_{i} }} \equiv T^{ - 1} \Sigma_{t = 1}^{T} (y_{it} - y_{i0} ) = \Delta y_{i1} + k_{T - 1}^{{\prime }} \Delta y_{i} .$$

Similarly,$\overline{{\Delta_{0} y_{i, - 1} }} = k_{T - 1}^{{\prime }} \Delta y_{i, - 1}$. Using these results, we obtain

$$\begin{aligned} &(\Delta_{0} y_{i} - \Delta_{0} y_{i, - 1} \delta - p_{i} 1_{T} )^{{\prime }} (\Delta_{0} y_{i} - \Delta_{0} y_{i, - 1} \delta - p_{i} 1_{T} ) \hfill \\ &\quad = (\Delta_{0} y_{i} - \Delta_{0} y_{i, - 1} \delta - p_{i} 1_{T} )^{{\prime }} Q_{T} (\Delta_{0} y_{i} - \Delta_{0} y_{i, - 1} \delta - p_{i} 1_{T} ) \hfill \\ & \qquad + (\Delta_{0} y_{i} - \Delta_{0} y_{i, - 1} \delta - p_{i} 1_{T} )^{{\prime }} P_{T} (\Delta_{0} y_{i} - \Delta_{0} y_{i, - 1} \delta - p_{i} 1_{T} ) \hfill \\ & \quad= (y_{i} - y_{i, - 1} \delta )^{{\prime }} Q_{T} (y_{i} - y_{i, - 1} \delta ) + T(\overline{{\Delta_{0} y_{i} }} - \overline{{\Delta_{0} y_{i, - 1} }} \delta - p_{i} )^{2} \hfill \\ &\quad = (\Delta y_{i} - \Delta y_{i, - 1} \delta )^{{\prime }} B_{T - 1}^{ - 1} (\Delta y_{i} - \Delta y_{i, - 1} \delta ) + T(\Delta y_{i1} + k_{T - 1}^{{\prime }} (\Delta y_{i} - \Delta y_{i, - 1} \delta ) - p_{i} )^{2} . \hfill \\ \end{aligned}$$

Proof of Proposition 1

A straight algebra shows

$$\begin{aligned} & f_{FE,i} (\delta ,\nu ,p_{i} )f(p_{i} |\omega_{HPT} ,\nu ) \hfill \\ & \quad = \frac{1}{{(2\pi )^{T/2} v^{T/2} }}\exp \left[ { - \frac{1}{2}\left( {\Delta y_{i} - \Delta y_{i, - 1} \delta } \right)^{{\prime }} B_{T - 1}^{ - 1} \left( {\Delta y_{i} - \Delta y_{i, - 1} \delta } \right) - \frac{T}{2\nu }(p_{i} - g_{i} )^{2} } \right] \hfill \\ & \qquad \times \frac{1}{{(2\pi )^{1/2} \nu^{1/2} (\omega_{HPT} )^{1/2} }}\exp \left( { - \frac{1}{{2\nu \omega_{HPT} }}p_{i}^{2} } \right) \hfill \\ & = \frac{1}{{(2\pi )^{T/2} v^{T/2} }}\exp \left[ { - \frac{1}{2}\left( {\Delta y_{i} - \Delta y_{i, - 1} \delta } \right)^{{\prime }} B_{T - 1}^{ - 1} \left( {\Delta y_{i} - \Delta y_{i, - 1} \delta } \right)} \right] \hfill \\ & \qquad \times \frac{1}{{(2\pi )^{1/2} v^{1/2} (\omega_{HPT} )^{1/2} }}\exp \left[ { - \frac{T}{2\nu }p_{i}^{2} - \frac{1}{{2\nu \omega_{HPT} }}p_{i}^{2} + \frac{T}{\nu }p_{i} g_{i} - \frac{T}{2\nu }g_{i}^{2} } \right]. \hfill \\ \end{aligned}$$

where $g_{i} = \Delta y_{i1} + k_{T - 1}^{{\prime }} (\Delta y_{i} - \Delta y_{i, - 1} \delta)$. We can also show:

$$\begin{aligned} & \frac{1}{{(2\pi )^{1/2} v^{1/2} (\omega_{HPT} )^{1/2} }}\exp \left[ { - \frac{T}{2\nu }p_{i}^{2} - \frac{1}{{2\nu \omega_{HPT} }}p_{i}^{2} + \frac{T}{\nu }p_{i} g_{i} - \frac{T}{2\nu }g_{i}^{2} } \right] \hfill \\ &\quad = (\xi_{HPT,T} )^{ - 1/2} \frac{{(\xi_{HPT,T} )^{1/2} }}{{(2\pi )^{1/2} v^{1/2} (\omega_{HPT} )^{1/2} }}\exp \left[ { - \frac{{\xi_{HPT,T} }}{{2\nu \omega_{HPT} }}\left( {p_{i} - \frac{{T\omega_{HPT} }}{{\xi_{HPT,T} }}g_{i} } \right)^{2} } \right] \hfill \\ & \qquad \times \exp \left[ { - \frac{T}{{2\nu \xi_{HPT,T} }}g_{i}^{2} } \right]. \hfill \\ \end{aligned}$$

Thus,

$$\begin{aligned} & \int_{ - \infty }^{\infty } {f_{FE,i} (\delta ,\nu ,p_{i} )f_{i} (p_{i} |\omega_{HPT} ,\nu )dp_{i} } \hfill \\ &\quad = \frac{1}{{(2\pi )^{T/2} \nu^{T/2} (\xi_{HPT,T} )^{1/2} }}\exp \left[ { - \frac{1}{2\nu }(\Delta y_{i} - \Delta y_{i, - 1} \delta )^{\prime}B_{T - 1}^{ - 1} (\Delta y_{i} - \Delta y_{i,t - 1} \delta ) - \frac{T}{{2\nu \xi_{HPT,T} }}g_{i}^{2} } \right] \hfill \\ & \qquad \times \int_{ - \infty }^{\infty } {\frac{{(\xi_{HPT,T} )^{1/2} }}{{(2\pi )^{1/2} \nu^{1/2} (\omega_{HPT} )^{1/2} }}\exp \left[ { - \frac{{\xi_{HPT,T} }}{{2\nu \omega_{HPT} }}\left( {p_{i} - \frac{{T\omega_{HPT} }}{{\xi_{HPT,T} }}g_{i}^{2} } \right)^{2} } \right]} dr_{i} \hfill \\ &\quad = \frac{1}{{(2\pi )^{T/2} \nu^{T/2} (\xi_{HPT,T} )^{1/2} }}\exp \left[ { - \frac{1}{2\nu }(\Delta y_{i} - \Delta y_{i, - 1} \delta )^{\prime}B_{T - 1}^{ - 1} (\Delta y_{i} - \Delta y_{i,t - 1} \delta ) - \frac{T}{{2\nu \xi_{HPT,T} }}g_{i}^{2} } \right] \hfill \\ &\quad = f_{HPT,i} (r_{i} |\theta_{HPT} ). \hfill \\ \end{aligned}$$

The following lemmas are useful to prove the propositions in Sect. 3. The first two lemmas provide alternative forms of $\Omega_{T} (\omega )$ and $[\Omega_{T} (\omega )]^{ - 1}$.

Lemma A.1

Let $\Omega_{s} (\omega )$ be the s × s (s = 2, …, T) matrix of the form of $\Omega_{T} (\omega )$ in Sect. 2. Let:

$$L_{s}^{{\prime }} = \left( {\begin{array}{*{20}c} 1 & 0 & 0 & {...} & 0 & 0 \\ { - 1} & 1 & 0 & {...} & 0 & 0 \\ 0 & { - 1} & 1 & {...} & 0 & 0 \\ : & : & : & {} & : & : \\ 0 & 0 & 0 & {...} & 1 & 0 \\ 0 & 0 & 0 & {...} & { - 1} & 1 \\ \end{array} } \right)_{s \times s} ;\,c_{s} = \left( {\begin{array}{*{20}c} 1 \\ 0 \\ 0 \\ 0 \\ : \\ 0 \\ \end{array} } \right)_{s \times s} .$$

Then,$\Omega_{s} (\omega ) = L_{s}^{{\prime }} L_{s} + \omega c_{s} c_{s}^{{\prime }}$.

Lemma A.2

$[\Omega_{T} (\omega )]^{ - 1} = (L_{T}^{{\prime }} L_{T} )^{ - 1} - (T\omega /(T\omega + 1))\overline{k}_{T} \overline{k}_{T}^{{\prime }}$, where $\overline{k}_{T}^{{\prime }} = (1,k_{T - 1}^{{\prime }} )^{{\prime }}$.

Lemma A.3

$B_{T - 1}^{ - 1} = (\tilde{D}^{\prime}_{T - 1} \tilde{D}_{T - 1} )^{ - 1} - Tm_{T - 1} m^{\prime}_{T - 1}$, where $m_{T - 1} = T^{ - 1} (1,2,\,\,...\,\,,\,\,T - 1)^{\prime}$ and $\tilde{D}_{T - 1}$ is the square matrix of the first $(T - 1)$ columns of $D^{\prime}_{T - 1}$. In addition, ${\text{trace}}\left( {B_{T - 1}^{ - 1} } \right) = (T - 1)(T + 1)/6$; and ${\text{trace}}\left( {k_{T - 1} k^{\prime}_{T - 1} } \right) = (T - 1)(2T - 1)/(6T)$.

Lemma A.4

The first-order and second-order derivatives of $\ell_{RE,i} (\theta )$ are given:

$$\begin{aligned} \ell_{RE,i,\delta } & = \frac{1}{\nu }\Delta y_{i, - 1}^{{\prime }} B_{T - 1}^{ - 1} (\Delta y_{i} - \Delta y_{i, - 1} \delta ) + \frac{T}{{\nu \xi_{T} }}\Delta y_{i, - 1}^{{\prime }} k_{T - 1} d_{i} (\psi ,\delta ); \hfill \\ \ell_{RE,i,\omega } &= - \frac{T}{{2\xi_{T} }} + \frac{{T^{2} }}{{2\nu \xi_{T}^{2} }}(d_{i} (\psi ,\delta ))^{2} ; \hfill \\ \ell_{RE,i,\nu } &= - \frac{T}{2\nu } + \frac{1}{{2\nu^{2} }}\left( {\Delta y_{i} - \Delta y_{i, - 1} \delta } \right)^{\prime } B_{T - 1}^{ - 1} \left( {\Delta y_{i} - \Delta y_{i, - 1} \delta } \right) + \frac{T}{{2\nu^{2} \xi_{T} }}\left( {d_{i} (\psi ,\delta )} \right)^{2} ; \hfill \\ \ell_{RE,i,\psi } &= \frac{T}{{\nu \xi_{T} }}y_{i0} d_{i} (\psi ,\delta ); \hfill \\ \ell_{RE,i,\delta \delta } &= - \frac{1}{\nu }\Delta y_{i, - 1}^{{\prime }} B_{T - 1}^{ - 1} \Delta y_{i, - 1} - \frac{T}{{\nu \xi_{T} }}\Delta y_{i, - 1}^{{\prime }} k_{T - 1} k_{T - 1}^{{\prime }} \Delta y_{i, - 1} ; \hfill \\ \ell_{RE,i,\delta \nu } &= - \frac{1}{{\nu^{2} }}\Delta y_{i, - 1}^{{\prime }} B_{T - 1}^{ - 1} (\Delta y_{i} - \Delta y_{i, - 1} \delta ) - \frac{T}{{\nu^{2} \xi_{T} }}\Delta y_{i, - 1}^{{\prime }} k_{T - 1} d_{i} (\psi ,\delta ); \hfill \\ \ell_{RE,i,\delta \omega } &= - \frac{{T^{2} }}{{\nu \xi_{T}^{2} }}\Delta y_{i, - 1}^{{\prime }} k_{T - 1} d_{i} (\psi ,\delta );\,\ell_{RE,i,\delta \psi } = - \frac{T}{{\nu \xi_{T} }}\Delta y_{i, - 1}^{{\prime }} k_{T - 1} y_{i0} ; \hfill \\ \ell_{RE,i,\nu \nu } &= \frac{T}{{2\nu^{2} }} - \frac{1}{{\nu^{3} }}(\Delta y_{i} - \Delta y_{i, - 1} \delta )^{{\prime }} B_{T - 1}^{ - 1} (\Delta y_{i} - \Delta y_{i, - 1} \delta ) - \frac{T}{{\nu^{3} \xi_{T} }}\left( {d_{i} (\psi ,\delta )} \right)^{2} ; \hfill \\ \ell_{RE,i,\nu \omega } & = - \frac{{T^{2} }}{{2\nu^{2} \xi_{T}^{2} }}\left( {d_{i} (\psi ,\delta )} \right)^{2} ;\,\ell_{RE,i,\nu \psi } = - \frac{T}{{\nu^{2} \xi_{T} }}d_{i} (\psi ,\delta )y_{i0} ; \hfill \\ \ell_{RE,i,\omega \omega } & = \frac{{T^{2} }}{{2\xi_{T}^{2} }} - \frac{{T^{3} }}{{\nu \xi_{T}^{3} }}\left( {d_{i} (\psi ,\delta )} \right)^{2} ;\,\ell_{RE,i,\lambda \psi } = - \frac{{T^{2} }}{{\nu \xi_{T}^{2} }}d_{i} (\psi ,\delta )y_{i0} ;\,\ell_{RE,i,\psi \psi } = - \frac{T}{{\nu \xi_{T} }}(y_{i0} )^{2} , \hfill \\ \end{aligned}$$

where $d_{i} (\psi ,\delta ) = \Delta y_{i1} - \psi y_{i0} + k^{\prime}_{T - 1} (\Delta y_{i} - \delta \Delta y_{i, - 1})$.

Lemma A.5

Under the RE assumption,

$$\begin{aligned} {\text{E}}\left( {\Delta y_{i, - 1}^{{\prime }} B_{T - 1}^{ - 1} \Delta y_{i, - 1} } \right) & = \psi^{2} \sigma_{{y_{0} }}^{2} c_{T - 1}^{{\prime }} H_{T - 1} (\delta_{o} )B_{T - 1}^{ - 1} H_{T - 1} (\delta_{o} )^{{\prime }} c_{T - 1} \hfill \\ &\qquad + \nu \times {\text{trace}}\left( {B_{T - 1}^{ - 1} H_{T - 1} (\delta_{o} )^{{\prime }} \Omega_{T - 1,o} H_{T - 1} (\delta_{o} )} \right); \hfill \\ \end{aligned}$$

(28)

$$\begin{aligned} {\text{E}}\left( {\Delta y_{i, - 1}^{{\prime }} k_{T - 1} k_{T - 1}^{{\prime }} \Delta y_{i, - 1} } \right) & = \psi^{2} \sigma_{{y_{0} }}^{2} c_{T - 1}^{{\prime }} H_{T - 1} (\delta_{o} )k_{T - 1} k_{T - 1}^{{\prime }} H_{T - 1} (\delta_{o} )^{{\prime }} c_{T - 1} \hfill \\ & \qquad + \nu_{o} \times {\text{trace}}\left( {k_{T - 1}^{{\prime }} H_{T - 1} (\delta_{o} )^{{\prime }} \Omega_{T - 1,o} H_{T - 1} (\delta_{o} )k_{T - 1} } \right); \hfill \\ \end{aligned}$$

(29)

$${\text{E}}\left( {\Delta y_{i, - 1}^{{\prime }} B_{T - 1}^{ - 1} \Delta \varepsilon_{i} } \right) = - \nu_{o} b(\delta_{o} )^{{\prime }} ;$$

(30)

$${\text{E}}\left( {\Delta y_{i, - 1}^{{\prime }} k_{T - 1} \left( {\Delta y_{i1} - \psi y_{i0} + k_{T - 1}^{{\prime }} (\Delta y_{i} - \Delta y_{i, - 1} \delta_{o} )} \right)} \right) = \frac{{\nu_{o} \xi_{T,o} }}{T}b(\delta_{o} )^{{\prime }} ;$$

(31)

$${\text{E}}\left[ {(\Delta y_{i} - \Delta y_{i, - 1} \delta_{o} )^{{\prime }} B_{T - 1}^{ - 1} (\Delta y_{i} - \Delta y_{i, - 1} \delta_{o} )} \right] = (T - 1)\nu_{o} ;$$

(32)

$${\text{E}}[(\Delta y_{i1} - \psi_{o} y_{i0} + k_{T - 1}^{{\prime }} (\Delta y_{i} - \Delta y_{i, - 1} \delta_{o} ))^{2} ] = \frac{{\xi_{T,o} \nu_{o} }}{T},$$

(33)

where $\xi_{T,0} = T\omega_{o} + 1$,$\Omega_{T - 1,o} = \Omega_{T - 1} (\omega_{o})$,$b(\delta )$ is defined in (13),$b(\delta )^{\prime} = db/d\delta$, and

$$H_{T - 1} (\delta )^{\prime} = \left( {\begin{array}{*{20}c} 1 & 0 & 0 & {...} & 0 \\ \delta & 1 & 0 & {...} & 0 \\ : & : & : & {} & : \\ {\delta^{T - 3} } & {\delta^{T - 4} } & {\delta^{T - 5} } & {...} & 0 \\ {\delta^{T - 2} } & {\delta^{T - 3} } & {\delta^{T - 4} } & {...} & 1 \\ \end{array} } \right)_{(T - 1) \times (T - 1)} .$$

Proof

Define $h_{i,T - 1} = (u_{i} ,\Delta \varepsilon_{i2} ,...,\Delta \varepsilon_{i,T - 1})$, so that ${\text{Var}}(h_{i,T} ) = \nu_{o} \Omega_{T - 1,o}$, and

$$\delta y_{i, - 1} = H_{T - 1} (\delta_{o} )^{{\prime }} c_{T - 1} \psi_{o} y_{i0} + H_{T - 1} (\delta_{o} )^{{\prime }} h_{i,T - 1} .$$

Then,

$$\begin{aligned} & {\text{E}}\left( {\Delta y_{i, - 1}^{{\prime }} B_{T - 1}^{ - 1} \Delta y_{i, - 1} } \right) \hfill \\ & \quad = \psi^{2} \sigma_{{y_{0} }}^{2} c_{T - 1}^{{\prime }} H_{T - 1} (\delta_{o} )B_{T - 1}^{ - 1} H_{T - 1} (\delta_{o} )^{{\prime }} c_{T - 1}\\ &\qquad + {\text{E}}\left( {h_{i,T - 1}^{{\prime }} H_{T - 1} (\delta_{o} )B_{T - 1}^{ - 1} H_{T - 1} (\delta_{o} )^{{\prime }} h_{i,T - 1} } \right) \hfill \\ &\quad = \psi^{2} \sigma_{{y_{0} }}^{2} c_{T - 1}^{{\prime }} H_{T - 1} (\delta_{o} )B_{T - 1}^{ - 1} H_{T - 1} (\delta_{o} )^{{\prime }} c_{T - 1}\\ &\qquad + {\text{trace}}\left( {{\text{E}}\left( {H_{T - 1} (\delta_{o} )B_{T - 1}^{ - 1} H_{T - 1} (\delta_{o} )^{{\prime }} h_{i,T - 1} h_{i,T - 1}^{{\prime }} } \right)} \right) \hfill \\ & \quad = \psi^{2} \sigma_{{y_{0} }}^{2} c_{T - 1}^{{\prime }} H_{T - 1} (\delta_{o} )B_{T - 1}^{ - 1} H_{T - 1} (\delta_{o} )^{{\prime }} c_{T - 1}\\ &\qquad+ \nu_{o} \times {\text{trace}}\left( {B_{T - 1}^{ - 1} H_{T - 1} (\delta_{o} )^{{\prime }} \Omega_{T - 1,o} H_{T - 1} (\delta_{o} )} \right). \hfill \\ \end{aligned}$$

(29)–(33) can be obtained by the similar method.

Lemma A.6

Under the RE assumption,

$${\text{E}}\left( { - \ell_{RE,i,\theta \theta } (\theta_{o} )} \right) = \left( {\begin{array}{*{20}c} {B + A} & 0 & {\frac{{Tb(\delta_{o} )^{{\prime }} }}{{\xi_{T,o} }}} & {\frac{{T\psi_{o} \sigma_{{y_{0} }}^{2} }}{{\nu_{o} \xi_{T,o} }}b(\delta_{o} )^{{\prime }} } \\ 0 & {\frac{T}{{2\nu_{o}^{2} }}} & {\frac{T}{{2\nu \xi_{T,o} }}} & 0 \\ {\frac{{Tb(\delta_{o} )^{{\prime }} }}{{\xi_{T,o} }}} & {\frac{T}{{2\nu_{o} \xi_{T,o} }}} & {\frac{{T^{2} }}{{2\xi_{T,o}^{2} }}} & 0 \\ {\frac{{T\psi \sigma_{{y_{0} }}^{2} }}{{\nu_{o} \xi_{T,o} }}b(\delta_{o} )^{{\prime }} } & 0 & 0 & {\frac{{T\sigma_{{y_{0} }}^{2} }}{{\nu_{o} \xi_{T,o} }}} \\ \end{array} } \right),$$

(34)

where

$$\begin{aligned} A &= {\text{trace}}\left( {B_{T - 1}^{ - 1} H_{T - 1} (\delta_{o} )^{{\prime }} \Omega_{T - 1,0} H_{T - 1} (\delta_{o} )} \right)\\ &\quad + \frac{T}{{\xi_{T,o} }}{\text{trace}}\left( {k_{T - 1}^{{\prime }} H_{T - 1} (\delta_{o} )^{{^{{\prime }} }} \Omega_{T - 1,o} H_{T - 1} (\delta_{o} )k_{T - 1} } \right); \hfill \\ B &= \frac{1}{{v_{o} }}\psi^{2} \sigma_{{y_{0} }}^{2} c_{T - 1}^{{\prime }} H_{T - 1} (\delta_{o} )B_{T - 1}^{ - 1} H_{T - 1} (\delta_{o} )^{{\prime }} c_{T - 1}\\ &\quad + \frac{T}{{\xi_{T,0} v_{0} }}\psi^{2} \sigma_{0}^{2} c_{T - 1}^{{\prime }} H_{T - 1} (\delta_{o} )^{{\prime }} k_{T - 1} k_{T - 1}^{{\prime }} H_{T - 1} (\delta_{o} )^{{\prime }} c_{T - 1} . \hfill \\ \end{aligned}$$

Proof of Proposition 2

When $\theta_{o} = \theta_{*} = (1,\nu_{*} ,0,0)^{\prime}$, Lemma A.1 and the definition of $H_{T - 1} (\delta)$ in Lemma 5 imply that $H_{T - 1} (1)^{\prime}\Omega_{T - 1} (0)H_{T - 1} (1) = I_{T - 1}$. This is so because $\Omega_{T - 1} (0)$ = $L^{\prime}_{T - 1} L_{T - 1}$ and $H_{T - 1} (1) = (L^{\prime}_{T - 1} )^{ - 1}$. In addition, in (34), B = 0 because $\psi_{o} = 0$. Using Lemma A.3 and the fact that $\xi_{T,o} = 1$ when $\theta_{o} = \theta_{*}$, we can show A = $T(T - 1)/2$ = $Tb(1)^{\prime}$. Substituting these results,$\psi_{o} = 0$, and $\xi_{T,o} = 1$, into (34), we have:

$${\text{E}}\left( { - \ell_{RE,i,\theta \theta } (\theta_{*} )} \right) = \left( {\begin{array}{*{20}c} {Tb(1)^{{\prime }} } & 0 & {Tb(1)^{{\prime }} } & 0 \\ 0 & {\frac{T}{{2\nu_{o}^{2} }}} & {\frac{T}{{2\nu_{o} }}} & 0 \\ {Tb(1)^{{\prime }} } & {\frac{T}{{2\nu_{o} }}} & {\frac{{T^{2} }}{2}} & 0 \\ 0 & 0 & 0 & {\frac{{T\sigma_{{y_{0} }}^{2} }}{{\nu_{o} }}} \\ \end{array} } \right),$$

(35)

which has a zero determinant. The likelihood theory indicates ${\text{E}}\left( {H_{RE,i} (\theta_{*} ) + B_{RE,i} (\theta_{*} )} \right) = 0$, at $\theta_{o} = \theta_{*}$. Thus,${\text{E}}\left( {B_{RE,i} (\theta_{*} )} \right)$ must be also singular.

Proof of Proposition 3

We first derive ${\text{E}}(\ell_{RE,i} (\theta ))$ under $H_{0}^{UND}$. We can show that

$$\begin{aligned} {\text{E}}\left( {\left( {d_{i} (\psi ,\delta )} \right)^{2} } \right) & = \nu_{*} \frac{(T + 1)(2T + 1)}{{6T}} - 2\nu_{*} \frac{(T - 1)(T + 1)}{{3T}}\delta + \sigma_{{y_{0} }}^{2} \psi^{2}\\ & \quad + v_{*} \frac{(T - 1)(2T - 1)}{{6T}}\delta^{2} ; \hfill \\ {\text{E}}\left( {\Delta \varepsilon_{i} (\delta )^{{\prime }} B_{T - 1}^{ - 1} \Delta \varepsilon_{i} (\delta )} \right) &= \frac{{\nu_{*} }}{6}\left( {(T - 1)(T + 1) - 2(T - 1)(T - 2)\delta + (T - 1)(T + 1)\delta^{2} } \right), \hfill \\ \end{aligned}$$

where $d_{i} (\psi ,\delta ) = \Delta y_{i1} - \psi y_{io} + k^{\prime}_{T - 1} (\Delta y_{i} - \Delta y_{i, - 1} \delta )$ and $\Delta \varepsilon_{i} (\delta ) = \Delta y_{i} - \Delta y_{i, - 1} \delta$. Thus, we have

$$\begin{aligned} {\text{E}}\left( {\ell_{RE,i} (\delta ,\nu ,\omega ,\psi )} \right) & = - \frac{T}{2}\ln (\pi ) - \frac{T}{2}\ln (\nu ) - \frac{1}{2}\ln \xi_{T} \hfill \\ & \quad - \frac{1}{12}\frac{{\nu_{*} }}{\nu }\left\{ {(T - 1)(T + 1) - 2(T - 1)(T - 2)\delta + (T - 1)(T + 1)\delta^{2} } \right\} \hfill \\ & \quad - \frac{1}{12}\frac{{\nu_{*} }}{{\nu \xi_{T} }}\left\{ \begin{array}{l} (T + 1)(2T + 1) - 4(T - 1)(T + 1)\delta \hfill \\ + \sigma_{{y_{0} }}^{2} \psi^{2} + (T - 1)(2T - 1)\delta^{2} \hfill \\ \end{array} \right\}, \hfill \\ \end{aligned}$$

where $\xi_{T} = T\omega + 1$. We now concentrate out $\psi$,$\omega$, and $\nu$ from ${\text{E}}(\ell_{RE,i} (\delta ,\nu ,\omega ,\psi ))$. Since $\psi = 0$ maximizes ${\text{E}}(\ell_{RE,i} (\delta ,\nu ,\omega ,\psi ))$, we can have the concentrated value of ${\text{E}}(\ell_{RE,i} (\delta ,\nu ,\omega ,\psi ))$:

$$\begin{aligned} {\text{E}}(\ell_{RE,i}^{c} \left( {\delta ,\nu ,\omega } \right)) & = - \frac{T}{2}\ln (\pi ) - \frac{T}{2}\ln (\nu ) - \frac{1}{2}\ln \xi_{T} \hfill \\ & \quad - \frac{1}{12}\frac{{\nu_{*} }}{\nu }\left( {(T - 1)(T + 1) - 2(T - 1)(T - 2)\delta + (T - 1)(T + 1)\delta^{2} } \right) \hfill \\ & \quad - \frac{1}{12}\frac{{\nu_{*} }}{{\nu \xi_{T} }}E\left( {(T + 1)(2T + 1) - 4(T - 1)(T + 1)\delta + (T - 1)(2T - 1)\delta^{2} } \right). \hfill \\ \end{aligned}$$

Now, solve the first-order condition with respect to $\xi_{T}$ (instead of $\omega$):

$\frac{{\partial {\text{E}}(\ell_{RE,i}^{c} \left( {\delta ,\nu ,\omega } \right))}}{{\partial \xi_{T} }} = - \frac{1}{2\xi } + \frac{1}{12}\frac{{\nu_{*} }}{{\nu \xi_{T}^{2} }}\left( \begin{gathered} (T + 1)(2T + 1) - 4(T - 1)(T + 1)\delta \hfill \\ + (T - 1)(2T - 1)\delta^{2} \hfill \\ \end{gathered} \right) = 0$.

Then we obtain:

$$\xi_{T} = \frac{1}{6}\frac{{\nu_{*} }}{\nu }\left( {(T + 1)(2T + 1) - 4(T - 1)(T + 1)\delta + (T - 1)(2T - 1)\delta^{2} } \right).$$

Thus,

$$\begin{aligned} {\text{E}}(\ell_{RE,i}^{c} \left( {\delta ,\nu } \right)) & = - \frac{T}{2}\ln (\pi ) - \frac{T}{2}\ln (\nu ) - \frac{1}{2}\ln \left( {\frac{{\nu_{*} }}{\nu }} \right) - \frac{1}{2}\ln \left( \frac{1}{6} \right) \hfill \\ & \quad - \frac{1}{2}\ln \left( {\frac{(T + 1)(2T + 1)}{6} - 2\frac{(T - 1)(T + 1)}{3}\delta + \frac{(T - 1)(2T - 1)}{6}\delta^{2} } \right) \hfill \\ & \quad - \frac{1}{12}\frac{{\nu_{*} }}{\nu }\left( {(T - 1)(T + 1) - 2(T - 1)(T - 2)\delta + (T - 1)(T + 1)\delta^{2} } \right). \hfill \\ \end{aligned}$$

Now, solving

$$\begin{aligned} &\frac{{\partial {\text{E}}\left( {\ell_{RE,i}^{c} \left( {\delta ,\nu } \right)} \right)}}{\partial \nu }\\ &\quad = - \frac{T - 1}{{2\nu }} + \frac{1}{12}\frac{{\nu_{*} }}{{\nu^{2} }}\left( {(T - 1)(T + 1) - 2(T - 1)(T - 2)\delta + (T - 1)(T + 1)\delta^{2} } \right)\\ &\quad= 0,\end{aligned}$$

we have:

$$\nu = \nu_{*} \frac{1}{T - 1}\left( {\frac{(T - 1)(T + 1)}{6} - 2\frac{(T - 1)(T - 2)}{6}\delta + \frac{(T - 1)(T + 1)}{6}\delta^{2} } \right).$$

Substituting this solution to ${\text{E}}(\ell_{RE,i}^{c} (\delta ,\nu ))$ yields

$$\begin{aligned} {\text{E}}(\ell_{RE,i}^{c} \left( \delta \right)) &= - \frac{T}{2}\ln (\pi ) - \frac{T}{2}\ln (\nu_{*} ) - \frac{T - 1}{2}\ln \left( {\frac{1}{T - 1}} \right) - \frac{1}{2}\ln \left( \frac{1}{6} \right) \hfill \\ & \quad - \frac{T - 1}{2}\ln \left( {\frac{(T - 1)(T + 1)}{6} - 2\frac{(T - 1)(T - 2)}{6}\delta + \frac{(T - 1)(T + 1)}{6}\delta^{2} } \right) \hfill \\ & \quad - \frac{1}{2}\ln \left( {\frac{(T + 1)(2T + 1)}{6} - 2\frac{(T - 1)(T + 1)}{3}\delta + \frac{(T - 1)(2T - 1)}{6}\delta^{2} } \right). \hfill \\ \end{aligned}$$

Then, a little algebra shows:

$$\frac{{\partial {\text{E}}(\ell_{RE,i}^{c} \left( \delta \right))}}{\partial \delta } = - \frac{{\left( {\delta - 1} \right)^{3} T\left( {T + 1} \right)\left( {2T - 1} \right)}}{{\left[ {(2T - 1)\left\{ {\delta - \frac{2(T + 1)}{{2T - 1}}} \right\}^{2} + \frac{3}{2T - 1}} \right]\left[ {(T + 1)\left\{ {\delta - \frac{T - 2}{{T + 1}}} \right\}^{2} + \frac{3(T - 1)}{{T + 2}}} \right]}}.$$

Since the denominator of $\partial {\text{E}}(\ell_{RE,i}^{c} (\delta ))/\partial \delta$ is positive for any T ≥ 2, it is positive for $\delta < 1$ and negative for $\delta = 1$. The derivative equals zero only at $\delta = 1$. This indicates that $\delta = 1$ is the global maximum point.

Proof of Proposition 4

It can be shown that when $\theta_{o} = \theta_{*}$,

$$\begin{aligned} {\text{E}}\left( {\ell_{Lan,i} (\delta ,\nu )} \right) & = b(\delta ) - \frac{T - 1}{2}\ln (\nu ) \hfill \\ &\quad - \frac{{\nu_{*} }}{2\nu }\left( {\frac{(T - 1)(T + 1)}{6} - \frac{(T - 1)(T - 2)}{3}\delta + \frac{(T - 1)(T + 1)}{6}\delta^{2} } \right). \hfill \\ \end{aligned}$$

Without loss of generality, we set $\nu_{*} = 1$. Then, the first-order maximization condition with respect to $\nu$ yields

$$\nu = \frac{1}{T - 1}\left( {\frac{(T - 1)(T + 1)}{6} - \frac{(T - 1)(T - 2)}{3}\delta + \frac{(T - 1)(T + 1)}{6}\delta^{2} } \right).$$

Substituting this into the expected value of $E(\ell_{Lan,i} (\delta ,\nu ))$, we can get

$$\begin{aligned} {\text{E}}\left( {\ell_{Lan,i}^{c} (\delta )} \right) &= b(\delta ) - \frac{T - 1}{2}\ln \left( {\frac{1}{T - 1}} \right) - \frac{T - 1}{2} \hfill \\ &\quad - \frac{T - 1}{2}\ln \left( {\frac{(T - 1)(T + 1)}{6} - \frac{(T - 1)(T - 2)}{3}\delta + \frac{(T - 1)(T + 1)}{6}\delta^{2} } \right). \hfill \\ \end{aligned}$$

If we differentiate the concentrated likelihood function, we yield.

$\begin{aligned} \frac{{\partial {\text{E}}\left( {\ell_{Lan,i}^{c} (\delta )} \right)}}{\partial \delta } = & \, b(\delta )^{{\prime }} - \frac{{\left( {T - 1} \right)\left( {\left( {T + 1} \right)\delta - T\left( {T - 2} \right)} \right)}}{{\left( {T + 1} \right)\delta^{2} - 2T\left( {T - 2} \right)\delta + \left( {T + 1} \right)}}; \\ \frac{{\partial^{2} {\text{E}}\left( {\ell_{Lan,i}^{c} (\delta )} \right)}}{{\partial \delta^{2} }} = &\, b(\delta )^{{{\prime \prime }}} - \frac{{2\left( {T - 1} \right)\left( {T\left( {\delta - 1} \right) + \delta + 2} \right)^{2} }}{{\left( {\left( {T\left( {\delta - 2} \right) + 4 + \delta } \right)\delta + \left( {T + 1} \right)} \right)^{2} }} \\ & \quad + \frac{{\left( {T - 1} \right)\left( {T + 1} \right)}}{{\left( {\left( {T\left( {\delta - 2} \right) + 4 + \delta } \right)\delta + \left( {T + 1} \right)} \right)}}. \\ \frac{{\partial^{3} {\text{E}}\left( {\ell_{Lan,i}^{c} (\delta )} \right)}}{{\partial \delta^{3} }} = &\, b(\delta )^{{{\prime \prime \prime }}} + \frac{{8\left( {T - 1} \right)\left( {T\left( {\delta - 1} \right) + \delta + 2} \right)^{3} }}{{\left( {\left( {T\left( {\delta - 2} \right) + 4 + \delta } \right)\delta + \left( {T + 1} \right)} \right)^{3} }} \\ & \quad - \frac{{6\left( {T - 1} \right)\left( {T + 1} \right)\left( {T\left( {\delta - 1} \right) + \delta + 2} \right)}}{{\left( {\left( {T + 1} \right)\delta^{2} - 2\left( {T - 2} \right)\delta + \left( {T + 1} \right)} \right)^{2} }}. \\ \end{aligned}$

It can be shown that when $\delta = 1$,

$b(1)^{{\prime }} = \frac{T - 1}{2};b(1)^{{{\prime \prime }}} = \frac{(T - 1)(T - 2)}{6};b(1)^{{{\prime \prime \prime }}} = \frac{(T - 1)(T - 2)(T - 3)}{{12}}$.

Using these results, we can easily show that at $\delta = 1$,

$$\frac{{\partial {\text{E}}\left( {\ell_{Lan,i}^{c} (1)} \right)}}{\partial \delta } = \frac{{\partial^{2} {\text{E}}\left( {\ell_{Lan,i}^{c} (1)} \right)}}{{\partial \delta^{2} }} = 0,$$

but,

$\frac{{\partial^{3} E\left( {\ell_{Lan,i}^{c} (1)} \right)}}{{\partial \delta^{3} }} = \frac{{\left( {T - 1} \right)\left( {T - 2} \right)\left( {T - 3} \right)}}{12} - \frac{{\left( {T - 1} \right)\left( {T + 1} \right)}}{2} \ne 0$.

Thus, $\delta = 1$ is an inflexion point of $\text{E}\left( {\ell_{Lan,i}^{c} (\delta )} \right)$.

Proof of Proposition 5

At $\theta = \theta_{*}$, using the alternative representation of $[\Omega_{T} (\omega )]^{ - 1}$ given in Lemma A.2, we can have

$$\begin{aligned} \ell_{RE,i,\nu } (\theta_{*} ) & = - \frac{T}{2\nu } + \frac{1}{{2\nu^{2} }}\left( {\begin{array}{*{20}c} {\Delta y_{i1} } \\ {\Delta y_{i} - \Delta y_{i, - 1} } \\ \end{array} } \right)^{{\prime }} (L_{T}^{{\prime }} L_{T} )^{ - 1} \left( {\begin{array}{*{20}c} {\Delta y_{i1} } \\ {\Delta y_{i} - \Delta y_{i, - 1} } \\ \end{array} } \right) \hfill \\ & = - \frac{T}{2\nu } + \frac{1}{{2\nu^{2} }}r_{i}^{{\prime }} L_{T} (L_{T}^{{\prime }} L_{T} )^{ - 1} L_{T}^{{\prime }} r_{i} = - \frac{T}{2\nu } + \frac{1}{{2\nu^{2} }}\Sigma_{i} r_{i}^{{\prime }} r_{i} , \hfill \\ \end{aligned}$$

where $r_{i} = (\Delta y_{i1} ,...,\Delta y_{iT} )^{{\prime }}$ and $\nu = \nu_{*}$. Also, we have:

$$\begin{aligned} \ell_{RE,i,\omega } (\theta_{*} ) & = - \frac{T}{2} + \frac{{T^{2} }}{2\nu }\left( {\Delta y_{i1} + k_{T - 1}^{{\prime }} (\Delta y_{i} - \Delta y_{i, - 1} )} \right)^{2} = - \frac{T}{2} + \frac{{T^{2} }}{2\nu }\left( {\overline{k}_{T}^{{\prime }} L_{T}^{{\prime }} r_{i} } \right)^{2} \hfill \\ & = - \frac{T}{2} + \frac{{T^{2} }}{2\nu }r_{i}^{{\prime }} L_{T} \overline{k}_{T} \overline{k}_{T}^{{\prime }} L_{T}^{{\prime }} r_{i} = - \frac{T}{2} + \frac{1}{2\nu }r_{i}^{{\prime }} 1_{T} 1_{T}^{{\prime }} r_{i} . \hfill \\ \end{aligned}$$

Thus,

$$\ell_{RE,i,\omega } (\theta_{*} ) - v\ell_{RE,i,\nu } (\theta_{*} ) = - \frac{1}{2\nu }r_{i}^{{\prime }} (I_{T} - 1_{T} 1_{T}^{{\prime }} )r_{i} = \frac{1}{v}\Sigma_{s = 2}^{T} \Sigma_{t = 1}^{s - 1} \Delta y_{it} \Delta y_{is} .$$

Now,

$$\begin{aligned}\ell_{RE,i,\delta } (\theta_{*} ) &= \frac{1}{\nu } \left( {\begin{array}{*{20}c} 0 \\ {\Delta y_{i, - 1} } \\ \end{array} } \right)^{{\prime }} (L_{T}^{{\prime }} L_{T} )^{ - 1} \left( {\begin{array}{*{20}c} {\Delta y_{i1} } \\ {\Delta y_{i} - \Delta y_{i, - 1} } \\ \end{array} } \right)^{{\prime }}\\ & = \frac{1}{\nu }\left( {\begin{array}{*{20}c} 0 \\ {\Delta y_{i1} } \\ : \\ {\Sigma_{t = 1}^{T - 1} \Delta y_{it} } \\ \end{array} } \right)^{{\prime }} \left( {\begin{array}{*{20}c} {\Delta y_{i1} } \\ {\Delta y_{i2} } \\ : \\ {\Delta y_{iT} } \\ \end{array} } \right) = \frac{1}{\nu }\Sigma_{s = 2}^{T} \Sigma_{t = 1}^{s - 1} \Delta y_{it} \Delta y_{is} .\end{aligned}$$

Thus,$\ell_{RE,i,\delta } (\theta_{*} ) - \ell_{RE,i,\omega } (\theta_{*} ) + \nu_{*} \ell_{RE,i,\nu } (\theta_{*} ) = 0.$

Proof of Proposition 6

We first check whether (i) whether $\tilde{\ell }_{RE,i,\delta \delta } (\theta_{*} )$ equals zero or is linearly related to $\tilde{\ell }_{RE,i,\delta } (\theta_{*} )$ and $s_{i,2}$. Note that

$$\tilde{\ell }_{{RE,i,\nu_{r} }} (\theta_{*} ) = \ell_{RE,i,\nu } (\theta_{*} );\tilde{\ell }_{{RE,i,\omega_{r} }} (\theta_{*} ) = \ell_{RE,i,\omega } (\theta_{*} );\tilde{\ell }_{RE,i,\psi } (\theta_{*} ) = \ell_{RE,i,\psi } (\theta_{*} ).$$

Since $\tilde{\ell }_{RE,i} (\delta ,\nu_{r} ,\omega_{r} ,\psi ) = \ell_{RE,i} (\delta ,\nu (\nu_{r} ,\delta ),\omega (\omega_{r} ,\delta ),\psi)$, we have:

$$\begin{aligned} \widetilde{\ell }_{RE,i,\delta } & = \ell_{RE,i,\delta } - \ell_{RE,i,\omega } + \nu_{r} \ell_{RE,i,\nu } ; \hfill \\ &\quad\widetilde{\ell }_{RE,i,\delta \delta } = \ell_{RE,i,\delta \delta } - \ell_{RE,i,\delta \omega } + \nu_{r} \ell_{RE,i,\delta \nu } - \ell_{RE,i,\omega \delta } + \ell_{RE,i,\omega \omega } - \nu_{r} \ell_{RE,i,\omega \nu } \hfill \\ & \quad + \nu_{r} \left( {\ell_{RE,i,\nu \delta } - \ell_{RE,i,\nu \omega } + \nu_{r} \ell_{RE,i,\nu \nu } } \right) \hfill \\ & = \left( {\ell_{RE,i,\delta \delta } - 2\ell_{RE,i,\delta \omega } + \ell_{RE,i,\omega \omega } } \right) + 2\nu_{r} \left( {\ell_{RE,i,\delta \nu } - \ell_{RE,i,\omega \nu } } \right) + \nu_{r}^{2} \ell_{RE,i,\nu \nu } . \hfill \\ \end{aligned}$$

Then, with a little algebra and Lemma A.3, we can show that at $\theta_{r} = \theta_{*}$,

$$\begin{aligned} \tilde{\ell }_{RE,i,\delta \delta }^{{}} (\theta_{*} ) & = \left( {\ell_{RE,i,\delta \delta } (\theta_{*} ) - 2\ell_{RE,i,\delta \omega } (\theta_{*} ) + \ell_{RE,i,\omega \omega } (\theta_{*} )} \right) \hfill \\ & \quad+ 2\nu_{*} \left( {\ell_{RE,i,\delta \nu } (\theta_{*} ) - \ell_{RE,\omega \nu } (\theta_{*} )} \right) + \nu_{*}^{2} \ell_{RE,i,\nu \nu } (\theta_{*} ) \hfill \\ & = \left( \begin{array}{l} - \frac{1}{{\nu_{*} }}\Delta y_{i, - 1}^{{\prime }} B_{T - 1}^{ - 1} \Delta y_{i, - 1} - \frac{T}{{\nu_{*} }}\Delta y_{i, - 1}^{{\prime }} k_{T - 1}^{{\prime }} k_{T - 1} \Delta y_{i, - 1} \hfill \\ + \frac{{2T^{2} }}{{\nu_{*} }}\Delta y_{i, - 1}^{{\prime }} k_{T - 1} (\Delta y_{i1} + k_{T - 1}^{{\prime }} (\Delta y_{i} - \Delta y_{i, - 1} )) - \frac{{T^{3} }}{{\nu_{*} }}\left( {\Delta y_{i1} + k_{T - 1}^{{\prime }} (\Delta y_{i} - \Delta y_{i, - 1} )} \right)^{2} \hfill \\ \end{array} \right) \hfill \\ & \quad + \frac{T(T + 1)}{2} \hfill \\ & = \left( { - \frac{1}{{\nu_{*} }}\Delta y_{i, - 1}^{{\prime }} \left( {\tilde{D}_{T}^{\prime } \tilde{D}_{T} } \right)^{ - 1} \Delta y_{i, - 1} - \frac{2T}{{\nu_{*} }}\Delta y_{i, - 1}^{{\prime }} m_{T - 1} (\Delta y_{iT} ) - \frac{T}{{\nu_{*} }}(\Delta y_{iT} )^{2} } \right) + \frac{T(T + 1)}{2}, \hfill \\ \end{aligned}$$

which is neither zero, nor a linear combination of $\tilde{\ell }_{RE,i,\delta } (\theta_{*})$ and $s_{i,2}$.

We now check (ii) whether $\tilde{\ell }_{RE,i,\delta \delta \delta } (\theta_{*} )$ is a linear function of $s_{i}$. We can show

$$\begin{aligned} \widetilde{\ell }_{RE,i,\delta \delta \delta } &= \left( {\ell_{RE,i,\delta \delta \delta } - 3\ell_{RE,,\delta \delta \omega } + 3\ell_{RE,i,\omega \omega \omega } - \ell_{RE,i,\omega \omega \omega } } \right) \hfill \\ & \quad+ 3\nu_{r} \left( {\ell_{RE,.i,\delta \delta \nu } - 2\ell_{RE,i,\delta \omega \omega \nu } + \ell_{RE,i,\omega \omega \nu } } \right)\\ &\quad + 3\nu_{r}^{2} \left( {\ell_{RE,i,\delta \nu \nu } - \ell_{RE,i,\omega \nu \nu } } \right) + \nu_{r}^{3} \left( {\ell_{RE,i,\nu \nu \nu } } \right). \hfill \\ \end{aligned}$$

But, at $\theta = \theta_{*}$,

$$\begin{aligned} \ell_{RE,i,\delta \delta \delta }^{{}} (\theta_{*} ) &= 0;\,\ell_{RE,i,\delta \delta \omega }^{{}} = \frac{{T^{2} }}{{\nu_{*} }}\Delta y_{i, - 1}^{{\prime }} k_{T - 1} k_{T - 1}^{{\prime }} \Delta y_{i, - 1} ; \hfill \\ \ell_{RE,i,\omega \omega \delta } (\theta_{*} ) &= \frac{{2T^{3} }}{{\nu_{*} }}\Delta y_{i, - 1}^{{\prime }} k_{T - 1} \left( {\Delta y_{i1} + k_{T}^{{\prime }} (\Delta y_{i} - \Delta y_{i, - 1} )} \right); \hfill \\ \ell_{RE,i,\omega \omega \omega } (\theta_{*} ) & = - T^{3} + \frac{{3T^{4} }}{{\nu_{*} }}(\Delta y_{i1} + k_{T - 1}^{{\prime }} (\Delta y_{i} - \Delta y_{i, - 1} ))^{2} ; \hfill \\ \ell_{RE,i,\delta \delta \nu } (\theta_{*} ) &= - \frac{1}{{\nu_{*} }}\ell_{RE,i,\delta \delta } (\theta_{*} );\,\ell_{RE,i,\delta \omega \nu } (\theta_{*} ) = - \frac{1}{{\nu_{*} }}\ell_{RE,i,\delta \omega } (\theta_{*} ); \hfill \\ \ell_{RE,i,\omega \omega \nu } (\theta_{*} ) &= - \frac{1}{{\nu_{*} }}\ell_{RE,i,\omega \omega } (\theta_{*} ) + \frac{{T^{2} }}{{2\nu_{*} }};\,\ell_{RE,i,\delta \nu \nu } (\theta_{*} ) = - \frac{2}{{\nu_{*} }}\ell_{RE,i,\delta \nu } (\theta_{*} ); \hfill \\ \ell_{RE,i,\nu \nu \omega } (\theta_{*} ) &= - \frac{2}{{\nu_{*} }}\ell_{RE,i,\nu \omega } (\theta_{*} );\,\ell_{RE,i,\nu \nu \nu } (\theta_{*} ) = - \frac{3}{{\nu_{*} }}\ell_{RE,i,\nu \nu } (\theta_{*} ) + \frac{T}{{2\nu_{*}^{3} }}. \hfill \\ \end{aligned}$$

Using these results, we can show:

$$\begin{aligned} \tilde{\ell }_{RE,i,\delta \delta \delta }^{{}} (\theta_{*} ) & = \frac{{2T^{3} + 3T^{2} + T}}{2} - \frac{{3T^{2} }}{{\nu_{*} }}\Delta y_{i, - 1}^{{\prime }} m_{T - 1} m_{T - 1}^{^{\prime}} \Delta y_{i, - 1} \hfill \\ & \quad - \frac{{6T^{2} }}{{\nu_{*} }}(m_{T - 1}^{{\prime }} \Delta y_{i, - 1} )\Delta y_{iT} - \frac{{3T^{2} }}{{\nu_{*} }}(\Delta y_{iT} )^{2} - 3\tilde{\ell }_{RE,i,\delta \delta } (\theta_{*} ), \hfill \\ \end{aligned}$$

which is neither zero, nor a linear combination of $s_{i}$. For example, when T = 2:

$$\begin{aligned} \tilde{\ell }_{RE,i,\delta \delta \delta } (\theta_{*} ) & = 15 - \frac{3}{{\nu_{*} }}\left( {\Delta y_{i1} } \right)^{2} - \frac{12}{{\nu_{*} }}\Delta y_{i1} \Delta y_{i2} - \frac{12}{{\nu_{*} }}(\Delta y_{i2} )^{2} - 3\tilde{\ell }_{RE,i,\delta \delta } (\theta_{*} ); \hfill \\ \tilde{\ell }_{RE,i,\delta \delta } (\theta_{*} ) &= \left( { - \frac{1}{{\nu_{*} }}\left( {\Delta y_{i1} } \right)^{2} - \frac{2}{{\nu_{*} }}\Delta y_{i1} \Delta y_{i2} - \frac{2}{{\nu_{*} }}(\Delta y_{i2} )^{2} } \right) + 3; \hfill \\ \tilde{\ell }_{{RE,i,\nu_{r} }} (\theta_{*} ) &= \ell_{RE,i,\nu } (\theta_{*} ) = - \frac{1}{{\nu_{*} }} + \frac{1}{{2\nu_{*}^{2} }}\left( {(\Delta y_{i1} )^{2} + (\Delta y_{i2} )^{2} } \right) \hfill \\ \tilde{\ell }_{{RE,i,\omega_{r} }} (\theta_{*} ) &= \ell_{RE,i,\omega } (\theta_{*} ) = - 1 + \frac{1}{{2\nu_{*} }}(\Delta y_{i1} + \Delta y_{i2} )^{2} ; \hfill \\ \tilde{\ell }_{{RE,i,\psi_{r} }} (\theta_{*} ) &= \ell_{RE,i,\psi } (\theta_{*} ) = \frac{1}{{\nu_{*} }}y_{i0} \left( {\Delta y_{i1} + \Delta y_{i2} } \right). \hfill \\ \end{aligned}$$

All these derivatives are linearly independent.

We now consider the asymptotic distribution of the ML estimator of $\theta_{r}$. Let $L_{N} (\theta_{r})$ = $\Sigma_{i = 1}^{N} \tilde{\ell }_{RE,i} (\theta_{r} ),$ where $\theta_{r} = (\delta ,\varphi^{\prime}_{r} )^{\prime}.$ For convenience, redefine $\varphi_{r}$ as $\varphi_{r} = (\omega_{r} ,\psi ,\nu_{r} )^{\prime}$ instead of $(\nu_{r} ,\omega_{r} ,\psi )^{\prime}$. Correspondingly, we also reorder $s_{i,2}$ and $Z_{2}$. Partition $Z_{2}$ into $Z_{2} = (Z_{21} ,Z_{22} )^{\prime}$, where $Z_{21}$ is a 2 × 1 vector and $Z_{22}$ is a scalar. We also partition $\Upsilon$ and $\Upsilon^{ - 1}$, accordingly:

$$\begin{aligned} \Upsilon &= \left( {\begin{array}{*{20}c} {\Upsilon_{11}^{{}} } & {\Upsilon_{21}^{\prime } } \\ {\Upsilon_{21} } & {\Upsilon_{22} } \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {\Upsilon_{11} } & {\Upsilon_{21,1}^{\prime } } & {\Upsilon_{22,1}^{\prime } } \\ {\Upsilon_{21,1} } & {\Upsilon_{21,21} } & {\Upsilon_{22,21}^{\prime } } \\ {\Upsilon_{22,1} } & {\Upsilon_{22,21} } & {\Upsilon_{22,22} } \\ \end{array} } \right);\,\Upsilon^{ - 1} = \left( {\begin{array}{*{20}c} {\Upsilon^{11} } & {(\Upsilon^{21} )^{\prime } } \\ {\Upsilon^{21} } & {\Upsilon^{22} } \\ \end{array} } \right)\\ &= \left( {\begin{array}{*{20}c} {\Upsilon^{11} } & {\Upsilon^{21,1 \prime } } & {\Upsilon^{22,1\prime } } \\ {\Upsilon^{21,1} } & {\Upsilon^{21,21} } & {(\Upsilon^{22,21} )^{\prime } } \\ {\Upsilon_{22,1} } & {\Upsilon_{22,21} } & {\Upsilon^{22,22} } \\ \end{array} } \right).\end{aligned}$$

Given the conditions (i) – (ii) are satisfied, Theorem 3 of RCBR implies that in the bounded neighborhood of $\theta_{*}$, the log-likelihood function can be approximated by:

$$\begin{aligned} L_{N} (\theta_{r} ) &= L_{N} (\theta_{*} ) + N^{1/2} \left( {\begin{array}{*{20}c} {(\delta - 1)^{2} } \\ {\varphi_{r} - \varphi^{*} } \\ \end{array} } \right)^{\prime } \Upsilon \left( {\begin{array}{*{20}c} {Z_{1} } \\ {Z_{2} } \\ \end{array} } \right) - \frac{1}{2}N\left( {\begin{array}{*{20}c} {(\delta - 1)^{2} } \\ {\varphi_{r} - \varphi^{*} } \\ \end{array} } \right)^{\prime } \Upsilon \left( {\begin{array}{*{20}c} {(\delta - 1)^{2} } \\ {\varphi_{r} - \varphi^{*} } \\ \end{array} } \right) + o_{p} (1) \hfill \\ & = L_{N} (\theta_{*} ) + N^{1/2} \left( {\begin{array}{*{20}c} {(\delta - 1)^{2} } \\ {\varphi_{r} - \varphi^{*} } \\ \end{array} } \right)^{\prime } (\Upsilon^{ - 1} )^{ - 1} \left( {\begin{array}{*{20}c} {Z_{1} } \\ {Z_{2} } \\ \end{array} } \right)\\ &\quad - \frac{1}{2}N\left( {\begin{array}{*{20}c} {(\delta - 1)^{2} } \\ {\varphi_{r} - \varphi^{*} } \\ \end{array} } \right)^{\prime } (\Upsilon^{ - 1} )^{ - 1} \left( {\begin{array}{*{20}c} {(\delta - 1)^{2} } \\ {\varphi_{r} - \varphi^{*} } \\ \end{array} } \right) + o_{p} (1) \hfill \\ \end{aligned}$$

(36)

Suppose that $Z_{1} > 0$. Then, it is straightforward to show that the ML estimator of $\theta_{r}$ equals

$$\left( {\begin{array}{*{20}c} {N^{1/2} (\hat{\delta } - 1)^{2} } \\ {N^{1/2} (\hat{\varphi }_{r} - \varphi^{*} )} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {Z_{1} } \\ {Z_{2} } \\ \end{array} } \right) + o_{p} (1).$$

That is,

$$\left( {\begin{array}{*{20}c} {N^{1/4} (\hat{\delta } - 1)} \\ {N^{1/2} (\hat{\varphi }_{r} - \varphi^{*} )} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {( - 1)^{B} \sqrt {Z_{1} } } \\ {Z_{2} } \\ \end{array} } \right) + o_{p} (1).$$

(37)

This is so because the solution for $N^{1/4} (\hat{\delta } - 1)$ has two solutions and both solutions have equal chances. Now, suppose $Z_{1} < 0$. Then, we do not have an interior solution for $N^{1/2} (\hat{\delta } - 1)^{2} > 0$. The corner solution for $N^{1/2} (\hat{\delta } - 1)^{2}$ equals zero. For this case, we have:

$$\left( {\begin{array}{*{20}c} {N^{1/4} (\hat{\delta } - 1)} \\ {N^{1/2} (\hat{\varphi }_{r} - \varphi^{*} )} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} 0 \\ {Z_{2} - \Upsilon^{21} (\Upsilon^{11} )^{ - 1} Z_{1} } \\ \end{array} } \right) + o_{p} (1).$$

(38)

Thus, we obtain (17).

We now derive the asymptotic distributions of LR test statistics. Substituting (37) and (38) into (36), we have:

$$2[L_{N} (\hat{\theta }_{r} ) - L_{N} (\theta_{*} )] = 1(Z_{1} > 0) \times Z^{\prime}(\Upsilon^{ - 1} )^{ - 1} Z + 1(Z_{2} < 0) \times Z_{2}^{{*{\prime }}} \Upsilon_{22} Z_{2}^{*} ,$$

(39)

where 1(·) is the index function,$\Upsilon_{22} = [\Upsilon^{22} - \Upsilon^{21} (\Upsilon^{11} )^{ - 1} \Upsilon^{12} ]^{ - 1}$ and $Z_{2}^{*} = Z_{2} - \Upsilon^{21} (\Upsilon^{11} )^{ - 1} Z_{1}$.

A tedious algebra shows:

$$\begin{aligned} Z(\Upsilon^{ - 1} )^{ - 1} Z &= {\text{Z}}_{{1}}^{\prime } (\Upsilon^{11} )^{ - 1} Z_{1} + Z_{2}^{*\prime } \Upsilon_{22} Z_{2}^{*} \hfill \\ & = {\text{Z}}_{{1}}^{\prime } (\Phi_{11} )^{ - 1} Z_{1} + Z_{21}^{**\prime } (\Phi_{21,21} )^{ - 1} Z_{21}^{**} + Z_{22}^{*\prime}(\Phi_{22,22} )^{ - 1} Z_{22}^{**} , \hfill \\ \end{aligned}$$

(40)

where $\Phi_{11} = \Upsilon^{11}$,$\Phi_{21,21} = \Upsilon^{21,21} - \Upsilon^{21,1} (\Upsilon^{11} )^{ - 1} \Upsilon^{21,1\prime }$,$Z_{21}^{**} = Z_{21} - \Upsilon^{21,1} (\Upsilon^{11} )^{ - 1} Z_{1}$,

$$\begin{aligned} \Phi_{22,22} & = \Upsilon^{22,22} - \left( {\begin{array}{*{20}c} {\Upsilon^{22,1} } & {\Upsilon^{22,21} } \\ \end{array} } \right) \left( {\begin{array}{*{20}c} {\Upsilon^{11} } & {\Upsilon^{21,1\prime } } \\ {\Upsilon^{21,1} } & {\Upsilon^{21,21} } \\ \end{array} } \right)^{ - 1} \left( {\begin{array}{*{20}c} {\Upsilon^{{22,1{\prime }}} } \\ {\Upsilon^{{22,21{\prime }}} } \\ \end{array} } \right); \hfill \\ Z_{22}^{**} &= Z_{22} - \left( {\begin{array}{*{20}c} {\Upsilon^{22,1} } & {\Upsilon^{22,21} } \\ \end{array} } \right)\left( {\begin{array}{*{20}c} {\Upsilon^{11} } & {\Upsilon^{21,1\prime } } \\ {\Upsilon^{21,1} } & {\Upsilon^{21,21} } \\ \end{array} } \right)^{ - 1} \left( {\begin{array}{*{20}c} {Z_{1} } \\ {Z_{21} } \\ \end{array} } \right). \hfill \\ \end{aligned}$$

Notice that $Z_{1}$,$Z_{21}^{**}$ and $Z_{22}^{**}$ are uncorrelated, and

$$Z_{1} \sim N(0,\Phi_{11} );\,Z_{21}^{**} \sim N(0,\Upsilon_{21,21} );\,Z_{22}^{**} \sim N(0,\Upsilon_{22,22} ).$$

(41)

Using (40), we can rewrite (39) as:

$$\begin{aligned} 2[L_{N} (\hat{\theta }_{r} ) - L_{N} (\theta_{*} )] & = 1(Z_{1} > 0) \times (Z_{1}^{\prime } \Phi_{11}^{ - 1} Z_{1} + Z_{21}^{**\prime } \Phi_{21,21}^{ - 1} Z_{21}^{**} + Z_{22}^{**\prime } \Phi_{22,22}^{ - 1} Z_{22}^{**} ) \hfill \\ &\quad + 1(Z_{1} < 0) \times (Z_{21}^{**\prime } \Phi_{21,21}^{ - 1} Z_{21}^{**} + Z_{22}^{**\prime } \Phi_{22,22}^{ - 1} Z_{22}^{**} ). \hfill \\ \end{aligned}$$

(42)

Let $\tilde{\theta }_{r} = (1,1,0,\tilde{\nu })^{\prime}$ be the restricted ML estimator of $\theta_{r}$ which maximizes (36) with the restrictions,$\delta = 1$ and $\omega = \psi = 0$. It can be shown that:

$$N^{1/2} (\tilde{\nu } - \nu_{*} ) = Z_{22}^{**} + o_{p} (1).$$

Substituting this solution into (36), we can have

$$\begin{aligned} 2[L_{N} (\tilde{\theta }_{r} ) - L_{N} (\theta_{*} )] & = 2N^{1/2} \left( {\begin{array}{*{20}c} 0 \\ {0_{2 \times 1} } \\ {\tilde{\nu } - \nu_{*} } \\ \end{array} } \right)^{\prime } \Upsilon \left( {\begin{array}{*{20}c} {Z_{1} } \\ {Z_{21} } \\ {Z_{22} } \\ \end{array} } \right)\\ &\quad - N\left( {\begin{array}{*{20}c} 0 \\ {0_{2 \times 1} } \\ {\tilde{\nu } - \nu *} \\ \end{array} } \right)^{\prime } \Upsilon \left( {\begin{array}{*{20}c} 0 \\ {0_{2 \times 1} } \\ {\tilde{\nu } - \nu *} \\ \end{array} } \right) + o_{p} (1) \hfill \\ & = Z_{22}^{**\prime } \Phi_{22,22}^{ - 1} Z_{22}^{**} + o_{p} (1) \hfill \\ \end{aligned}$$

(43)

Using (41)-(43), we can show that when $\theta_{r,o} = \theta_{*}$, the LR test statistic for testing the joint hypotheses of $\delta_{o} = 1$,$\omega_{r,o} = 0$, and $\psi_{o} = 0$ follows a mixed Chi-square distribution:

$$\begin{aligned} & 2[L_{N} (\hat{\theta }_{r} ) - L_{N} (\tilde{\theta }_{r} )] = 1(Z_{1} > 0) \times (Z_{1}^{\prime } (\Phi_{11} )^{ - 1} Z_{1} + Z_{21}^{**\prime } (\Phi_{21,21} )^{ - 1} Z_{21}^{**} ) \hfill \\ & \quad + 1(Z_{1} < 0) \times (Z_{21}^{**\prime } (\Phi_{21,21} )^{ - 1} Z_{21}^{**} ) + o_{p} (1) \hfill \\ & \to_{d} B \times \chi^{2} (3) + (1 - B) \times \chi^{2} (2). \hfill \\ \end{aligned}$$

Finally, consider the LR statistic for testing the hypothesis of $\delta_{o} = 1$. Let ${\mathop{\theta }\limits^{\frown}}_{r} = (1,{{\mathop{\varphi }\limits^{\frown}}^{\prime}}_{r} )^{\prime}$ be restricted ML estimator that maximize (36) with the restriction $\delta_{o} = 1$. We can show that

$$N^{1/2} ({\mathop{\varphi }\limits^{\frown}}_{r} - \varphi_{*} ) = Z_{2}^{*} + o_{p} (1).$$

Substituting this solution into (3) yields:

$$\begin{aligned} 2[L_{N} (\tilde{\theta }_{r} ) - L_{N} (\theta_{*} )] &= 2\sqrt N \left( {\begin{array}{*{20}c} 0 \\ {{\mathop{\varphi }\limits^{\frown}}_{r} - \varphi_{*} } \\ \end{array} } \right)^{\prime } \Upsilon \left( {\begin{array}{*{20}c} {Z_{1} } \\ {Z_{2} } \\ \end{array} } \right)\\ &\quad- N\left( {\begin{array}{*{20}c} 0 \\ {{\mathop{\varphi }\limits^{\frown}}_{r} - \varphi_{*} } \\ \end{array} } \right)^{\prime } \Upsilon \left( {\begin{array}{*{20}c} 0 \\ {{\mathop{\varphi }\limits^{\frown}}_{r} - \varphi_{*} } \\ \end{array} } \right) + o_{p} (1) \hfill \\ & = Z_{2}^{*\prime } \Upsilon_{22} Z_{2}^{*} + o_{p} (1) = Z_{21}^{**\prime } \Phi_{21,21}^{ - 1} Z_{21}^{**} + Z_{22}^{**\prime } \Phi_{22,22}^{ - 1} Z_{22}^{**} + o_{p} (1) \hfill \\ \end{aligned}$$

(44)

Then, using (41), (42) and (44), we have:

$$2[L_{N} (\hat{\theta }_{r} ) - L_{N} ({\mathop{\theta }\limits^{\frown}}_{r} )] = 1(Z_{1} > 0) \times Z_{1}^{\prime } \Phi_{11}^{ - 1} Z_{1} + o_{p} (1) \to_{d} B\chi^{2} (1).$$

This completes our proof.

Proof of Proposition 7

It is easy to see that $(\ell_{RE,i,\delta }^{D} (\phi_{*}^{D} ) - a_{*} \ell_{RE,i,\beta }^{D} (\phi_{*}^{D} ))$,$\ell_{RE,i,\nu }^{D} (\phi_{*}^{D})$, and $\ell_{RE,i,\omega }^{D} (\phi_{*}^{D} )$ are the same as $\ell_{RE,i,\delta } (\theta_{*})$,$\ell_{RE,i,\nu } (\theta_{*})$, and $\ell_{RE,i,\omega }^{{}} (\theta_{*} )$ with the $\Delta y_{it}$ replaced by $\Delta y_{it} - a$. Thus, we can obtain the result by the same method used for Propositions 5.

Proof of Proposition 8

Define

$$L_{T}^{HD\prime } = \left( {\begin{array}{*{20}c} 1 & 0 & 0 & 0 & {...} & 0 \\ { - 1} & 1 & 0 & 0 & {...} & 0 \\ 1 & { - 2} & 1 & 0 & {...} & 0 \\ 0 & 1 & { - 2} & 1 & {...} & 0 \\ : & : & : & : & {} & : \\ 0 & 0 & 0 & 0 & {...} & 1 \\ \end{array} } \right)_{T \times T} ;\,J_{T} = \left( {\begin{array}{*{20}c} {I_{2} } \\ {0_{(T - 2) \times 2} } \\ \end{array} } \right);\,G_{T} = \left( {\begin{array}{*{20}c} 1 & 0 \\ 1 & 1 \\ 1 & 2 \\ : & : \\ 1 & {T - 1} \\ \end{array} } \right),$$

We can easily show that $G = (L_{T}^{HD\prime } )^{ - 1} J_{T}$ and $\Omega_{T}^{HD} (W) = L_{T}^{HD\prime } L_{T}^{HD} + J_{T} WJ^{\prime}_{T}$. Let

$$\left( {\begin{array}{*{20}c} {g_{11} } & {g_{12} } \\ {g_{12} } & {g_{22} } \\ \end{array} } \right) \equiv G^{\prime}_{T} G_{T} = \left( {\begin{array}{*{20}c} T & {\frac{T(T - 1)}{2}} \\ {\frac{T(T - 1)}{2}} & {\frac{T(T - 1)(2T - 1)}{6}} \\ \end{array} } \right).$$

Then, we have:

$$\begin{aligned} (\Omega_{T}^{HD} )^{ - 1} & = \left[ {L_{T}^{HD\prime } L_{T}^{HD} + J_{T} WJ^{\prime}_{T} } \right]^{ - 1} \hfill \\ & = \left( {L_{T}^{HD\prime } L_{T}^{HD} } \right)^{ - 1} - (L_{T}^{HD\prime } L_{T}^{HD} )^{ - 1} J_{T} W[J^{\prime}_{T} (L_{T}^{HD\prime } L_{T}^{HD} )^{ - 1} J_{T} W + I_{2} ]^{ - 1} J^{\prime}_{T} (L_{T}^{HD\prime } L_{T}^{HD} )^{ - 1} \hfill \\ & = (L_{T}^{HD} )^{ - 1} (L_{T}^{HD\prime } )^{ - 1} - (L_{T}^{HD} )^{ - 1} G_{T} W[G^{\prime}_{T} G_{T} W + I_{2} ]^{ - 1} G^{\prime}_{T} (L_{T}^{HD\prime } )^{ - 1} . \hfill \\ \end{aligned}$$

Thus, the log-likelihood function can be written:

$$\begin{aligned} \ell_{RE,i}^{HD} & = - \frac{1}{2}\ln (2\pi ) - \frac{T}{2}\ln (\nu ) - \frac{1}{2}\ln [\det (L_{T}^{HD\prime } L_{T}^{HD} + J_{T} WJ^{\prime}_{T} )] \hfill \\ & \quad - \frac{1}{2\nu }\left( {\begin{array}{*{20}c} {\Delta y_{i1} - a_{1} - \psi_{1} y_{i0} } \\ {\Delta y_{i2} - a_{2} - \psi_{2} y_{i0} - \Delta y_{i1} \delta } \\ {\Delta^{2} y_{i} - \Delta^{2} y_{i, - 1} \delta } \\ \end{array} } \right)^{\prime } (L_{T}^{HD\prime } L_{T}^{HD}\\ &\quad + J_{T} WJ^{\prime}_{T} )^{ - 1} \left( {\begin{array}{*{20}c} {\Delta y_{i1} - a_{1} - \psi_{1} y_{i0} } \\ {\Delta y_{i2} - a_{2} - \psi_{2} y_{i0} - \Delta y_{i1} \delta } \\ {\Delta^{2} y_{i} - \Delta^{2} y_{i, - 1} \delta } \\ \end{array} } \right) \hfill \\ \end{aligned}$$

Define $d_{it} = \Delta y_{it} - a_{1,*} - \psi_{1,*} y_{i0}$. By a tedious but straightforward algebra, we can show at $\phi^{HD} = \phi_{*}^{HD}$:

$$\begin{gathered} \ell_{RE,i,\delta }^{HD} = \frac{1}{\nu }\Sigma_{t = 1}^{T - 1} \Sigma_{s = 1}^{t} \Delta y_{is} d_{i,t + 1} - \frac{{\omega_{11}^{{}} }}{{\nu (g_{11} \omega_{11}^{{}} + 1)}}(\Sigma_{t = 1}^{T - 1} (T - t)\Delta y_{it} )\Sigma_{t = 1}^{T} d_{it} ; \hfill \\ \ell_{{RE,i,a_{2} }}^{HD} = \frac{1}{\nu }\Sigma_{t = 2}^{T} (t - 1)d_{it} - \frac{{w_{11} }}{{\nu (g_{11} \omega_{11}^{{}} + 1)}}\frac{T(T - 1)}{2}\Sigma_{t = 1}^{T} d_{it} ; \hfill \\ \ell_{{RE,i,\psi_{2} }}^{HD} = \frac{1}{\nu }\Sigma_{t = 2}^{T} (t - 1)d_{it} y_{i0} - \frac{{w_{11}^{{}} }}{{\nu (g_{11} \omega_{11}^{{}} + 1)}}\frac{T(T - 1)}{2}\Sigma_{t = 1}^{T} d_{it} y_{i0} , \hfill \\ \end{gathered}$$

where $\nu = \nu_{*}$ and $\omega_{11}^{{}} = \omega_{11,*}^{{}}$. Thus, we have:

$$\begin{aligned}&\ell_{RE,i,\delta }^{HD} - a_{1} \ell_{{RE,i,a_{2} }}^{HD} - \psi_{1} \ell_{{RE,i,\psi_{2} }}^{HD}\\ &\quad = \frac{1}{\nu }\Sigma_{t = 1}^{T - 1} \Sigma_{s = 1}^{t} d_{i,s} d_{i,t + 1} - \frac{{\omega_{11}^{{}} }}{{\nu (g_{11} \omega_{11}^{{}} + 1)}}(\Sigma_{t = 1}^{T - 1} (T - t)d_{it} )(\Sigma_{t = 1}^{T} d_{it} ). \end{aligned}$$

We also can show:

$$\begin{aligned} \ell_{RE,i,\nu }^{HD} & = - \frac{T}{2\nu } + \frac{1}{{2\nu^{2} }}\Sigma_{t = 1}^{T} d_{it}^{2} - \frac{{\omega_{11}^{{}} }}{{2\nu^{2} (g_{11} \omega_{11}^{{}} + 1)}}(\Sigma_{t = 1}^{T} d_{it} )^{2} ; \hfill \\ \ell_{{RE,i,\omega_{11}}}^{HD} &= - \frac{1}{2}\frac{{g_{11} }}{{g_{11} \omega_{11}^{{}} + 1}} - \frac{{g_{11} \omega_{11}^{{}} }}{{2\nu (g_{11} \omega_{11}^{{}} + 1)^{2} }}(\Sigma_{t = 1}^{T} d_{it} )^{2}\\ &\qquad + \frac{1}{{2\nu (g_{11} \omega_{11}^{{}} + 1)}}(\Sigma_{t = 1}^{T} d_{it} )^{2} ; \hfill \\ \ell_{{RE,i,\omega_{12}}}^{HD} &= - \frac{{g_{12} }}{{(g_{11} \omega_{11}^{{}} + 1)}} - \frac{{g_{12} \omega_{11}^{{}} }}{{\nu (g_{11} \omega_{11}^{{}} + 1)^{2} }}(\Sigma_{t = 1}^{T} d_{it} )^{2}\\& \qquad+ \frac{1}{{\nu (g_{11} \omega_{11}^{{}} + 1)}}(\Sigma_{t = 1}^{T} d_{it} )(\Sigma_{t = 1}^{T - 1} td_{i,t + 1} ). \hfill \\ \end{aligned}$$

But,

$$\begin{aligned} & \ell_{RE,i,\delta }^{HD} + \nu \ell_{RE,i,\nu }^{HD} - (g_{11} \omega_{11} + 1)\ell_{{RE,i,\omega_{11}}}^{HD} \hfill \\ & \quad = \frac{1}{2\nu }\left[ \begin{array}{l} \Sigma_{t = 1}^{T} d_{it}^{2} + 2\Sigma_{t = 1}^{T} \Sigma_{s = 1}^{t - 1} d_{is} d_{it} \hfill \\ - (\Sigma_{t = 1}^{T} d_{it} )^{2} \hfill \\ \end{array} \right] - \frac{{\omega_{11} \left( {\Sigma_{t = 1}^{T - 1} (T - t)d_{it} } \right)\left( {\Sigma_{t = 1}^{T} d_{it} } \right)}}{{\nu (g_{11} \omega_{11} + 1)}}\\ &\qquad + \frac{{(g_{11} - 1)\omega_{11}^{{}} \left( {\Sigma_{t = 1}^{T} d_{it} } \right)^{2} }}{{2\nu (g_{11} \omega_{11}^{{}} + 1)}} \hfill \\ & \quad = - \frac{{\omega_{11} \left( {\Sigma_{t = 1}^{T - 1} (T - t)d_{it} } \right)\left( {\Sigma_{t = 1}^{T} d_{it} } \right)}}{{\nu (g_{11} \omega_{11}^{{}} + 1)}} + \frac{{(g_{11} - 1)\omega_{11}^{{}} \left( {\Sigma_{t = 1}^{T} d_{it} } \right)^{2} }}{{2\nu (g_{11} \omega_{11}^{{}} + 1)}} \hfill \\ &\quad = - \frac{{\omega_{11} \left( {(T - 1)\Sigma_{t = 1}^{T} d_{it} - \Sigma_{t = 1}^{T} (t - 1)d_{it} } \right)\left( {\Sigma_{t = 1}^{T} d_{it} } \right)}}{{\nu (g_{11} \omega_{11} + 1)}} + \frac{{(g_{11} - 1)\omega_{11}^{{}} \left( {\Sigma_{t = 1}^{T} d_{it} } \right)^{2} }}{{2\nu (g_{11} \omega_{11} + 1)}} \hfill \\ &\quad = - \frac{{(T - 1)\omega_{11} \left( {\Sigma_{t = 1}^{T} d_{it} } \right)^{2} }}{{\nu (g_{11} \omega_{11} + 1)}} + \frac{{\omega_{11} \left( {\Sigma_{t = 1}^{T} (t - 1)d_{it} } \right)\left( {\Sigma_{t = 1}^{T} d_{it} } \right)}}{{\nu (g_{11} \omega_{11} + 1)}}\\ &\qquad + \frac{{(g_{11} - 1)\omega_{11} \left( {\Sigma_{t = 1}^{T} d_{it} } \right)^{2} }}{{2\nu (g_{11} \omega_{11} + 1)}} \hfill \\ & \quad = - \frac{{(T - 1)\omega_{11} }}{{2\nu (T\omega_{11}^{{}} + 1)}}\left( {\Sigma_{t = 1}^{T} d_{it} } \right)^{2} + \frac{{\omega_{11}^{{}} }}{{\nu (T\omega_{11}^{{}} + 1)}}\left( {\Sigma_{t = 1}^{T} (t - 1)d_{it} } \right)\left( {\Sigma_{t = 1}^{T} d_{it} } \right), \hfill \\ \end{aligned}$$

where the last equality results from g₁₁ = T. Now observe that:

$$\begin{aligned}(T - 1)\ell_{{RE,i,\omega_{11}}}^{HD} - \ell_{{RE,i,\omega_{12}}}^{HD} &= \frac{(T - 1)}{{2\nu (T\omega_{11}^{{}} + 1)}}\left( {\Sigma_{t = 1}^{T} d_{it} } \right)^{2}\\ &\quad - \frac{1}{{\nu (T\omega_{11}^{{}} + 1)}}\left( {\Sigma_{t = 1}^{T} d_{it} } \right)\left( {\Sigma_{t = 1}^{T} (t - 1)d_{it} } \right). \end{aligned}$$

Thus, we have:

$$\begin{aligned} & \ell_{RE,i,\delta }^{HD} + \nu \ell_{RE,i,\nu }^{HD} - (T\omega_{11} + 1)\ell_{{RE,i,\omega_{11}}}^{HD} + \omega_{11} [(T - 1)\ell_{{RE,i,\omega_{11}}}^{HD} - \ell_{{RE,i,\omega_{12}}}^{HD} ] \hfill \\ & \quad = \ell_{RE,i,\delta }^{HD} + \nu \ell_{RE,i,\nu }^{HD} - (\omega_{11} + 1)\ell_{{RE,i,\omega_{11}}}^{HD} - \omega_{11} \ell_{{RE,i,\omega_{12}}}^{HD} = 0. \hfill \\ \end{aligned}$$

An alternative representation of the log-likelihood function (26) is given:

$$\begin{aligned} \ell_{RE,i}^{HD} & = - \frac{1}{2}\ln (2\pi ) - \frac{1}{2}\ln [\det (B_{T - 2}^{HD} )] - \frac{T}{2}\ln (\nu ) - \frac{1}{2}\ln [\det (\Xi )] \hfill \\ & \qquad- \frac{1}{2\nu }(\Delta^{2} y_{i} - \Delta^{2} y_{i, - 1} \delta )^{\prime}(B_{T - 2}^{HD} )^{ - 1} (\Delta^{2} y_{i} - \Delta^{2} y_{i, - 1} \delta ) \hfill \\ &\qquad + \frac{1}{2\nu }\left( {\left( {\begin{array}{*{20}c} {\Delta y_{i1} - a_{1} - \psi_{1} y_{i0} } \\ {\Delta y_{i2} - a_{2} - \psi_{2} y_{i0} - \delta \Delta y_{i1} } \\ \end{array} } \right) + K_{T - 2}^{HD\prime } (\Delta^{2} y_{i} - \Delta^{2} y_{i, - 1} \delta )} \right)^{\prime } \hfill \\ & \qquad \times \Xi^{ - 1} \left( {\left( {\begin{array}{*{20}c} {\Delta y_{i1} - a_{1} - \psi_{1} y_{i0} } \\ {\Delta y_{i2} - a_{2} - \psi_{2} y_{i0} - \delta \Delta y_{i1} } \\ \end{array} } \right) + K_{T - 2}^{HD\prime } (\Delta^{2} y_{i} - \Delta^{2} y_{i, - 1} \delta )} \right), \hfill \\ \end{aligned}$$

where

$$\begin{gathered} C_{T - 2}^{HD} = \left( {\begin{array}{*{20}c} { - 1} & 3 \\ 0 & { - 1} \\ 0 & 0 \\ : & : \\ 0 & 0 \\ \end{array} } \right)_{{\left( {T - 2} \right)x2}} ;\,B_{T - 2}^{HD} = \left( {\begin{array}{*{20}c} 6 & { - 4} & 1 & 0 & {...} & 0 \\ { - 4} & 6 & { - 4} & 1 & {...} & 0 \\ 1 & { - 4} & 6 & { - 4} & {...} & 0 \\ : & : & : & : & {} & : \\ 0 & 0 & 0 & 0 & {...} & { - 4} \\ 0 & 0 & 0 & {} & {...} & 6 \\ \end{array} } \right)_{{\left( {T - 2} \right)x\left( {T - 2} \right)}} ;\\ K_{T - 2}^{HD} = (B_{T - 2}^{HD} )^{ - 1} (C_{T - 2}^{HD} ); \hfill \\ \Xi \equiv \left( {\begin{array}{*{20}c} {\xi_{11,T} } & {\xi_{12,T} } \\ {\xi_{12,T} } & {\xi_{22,T} } \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {\omega_{11} + 1} & {\omega_{12} - 1} \\ {\omega_{12} - 1} & {\omega_{22} + 2} \\ \end{array} } \right) - (C_{T - 2}^{HD} )^{\prime}(B_{T - 2}^{HD} )^{ - 1} (C_{T - 2}^{HD} ). \hfill \\ \end{gathered}$$

Our unreported simulations show that it is much easier to minimize the above log-likelihood function with respect to $(\delta ,\nu ,\xi_{T,11} ,\xi_{T,12} ,\xi_{T,22} ,\psi_{1} ,\psi_{2} ,a_{1} ,a_{2} )^{\prime}$ than to minimize the equivalent log-likelihood function (26) with respect to $\phi^{HD} = (\delta ,\nu ,\omega_{11} ,\omega_{12} ,\omega_{13} ,\psi_{1} ,\psi_{2} ,a_{1} ,a_{2} )^{\prime}$.