On divergence tests for composite hypotheses under composite likelihood

Abstract

It is well-known that in some situations it is not easy to compute the likelihood function as the datasets might be large or the model is too complex. In that contexts composite likelihood, derived by multiplying the likelihoods of subjects of the variables, may be useful. The extension of the classical likelihood ratio test statistics to the framework of composite likelihoods is used as a procedure to solve the problem of testing in the context of composite likelihood. In this paper we introduce and study a new family of test statistics for composite likelihood: Composite $\phi$ -divergence test statistics for solving the problem of testing a simple null hypothesis or a composite null hypothesis. To do that we introduce and study the asymptotic distribution of the restricted maximum composite likelihood estimate.

Appendix: Proofs

1.1 Proof of Theorem 1

Under the standard regularity assumptions of asymptotic statistics (cf. Serfling 1980, p. 144 and Pardo 2006, p. 58), we have

$$\begin{aligned} \frac{\partial D_{\phi }(\varvec{\theta },\varvec{\theta } _{0})}{\partial \varvec{\theta } }=\int _{ \mathbb {R} ^{m}}\frac{\partial \mathcal {CL}(\varvec{\theta },\varvec{y} )}{\partial \varvec{\theta }}\phi ^{\prime }\left( \frac{ \mathcal {CL}(\varvec{\theta },\varvec{y})}{ \mathcal {CL}(\varvec{\theta }_{0},\varvec{y})} \right) d\varvec{y}, \end{aligned}$$

therefore

$$\begin{aligned} \left. \frac{\partial D_{\phi }(\varvec{\theta },\varvec{ \theta }_{0})}{\partial \varvec{\theta }}\right| _{\varvec{\theta }= \varvec{\theta }_{0}}=\phi ^{\prime }\left( 1\right) \int _{ \mathbb {R} ^{m}}\left. \frac{\partial \mathcal {CL}(\varvec{\theta }, \varvec{y})}{\partial \varvec{\theta }}\right| _{ \varvec{\theta }=\varvec{\theta }_{0}}d\varvec{y}=\varvec{0} _{p}. \end{aligned}$$

On the other hand,

$$\begin{aligned} \frac{\partial ^{2}D_{\phi }(\varvec{\theta },\varvec{ \theta }_{0})}{\partial \varvec{\theta }\partial \varvec{\theta }^{T} }&= \int _{ \mathbb {R} ^{m}}\frac{\partial ^{2}\mathcal {CL}(\varvec{\theta }, \varvec{y})}{\partial \varvec{\theta }\partial \varvec{ \theta }^{T}}\phi ^{\prime }\left( \frac{\mathcal {CL}(\varvec{\theta },\varvec{y})}{\mathcal {CL}(\varvec{\theta }_{0},\varvec{y})}\right) d\varvec{y} \\&\quad + \int _{ \mathbb {R} ^{m}}\frac{\partial \mathcal {CL}(\varvec{\theta },\varvec{y} )}{\partial \varvec{\theta }}\frac{\partial \mathcal {CL}( \varvec{\theta },\varvec{y})}{\partial \varvec{ \theta }^{T}}\frac{1}{\mathcal {CL}(\varvec{\theta }_{0}, \varvec{y})}\phi ^{\prime \prime }\left( \frac{\mathcal {CL}( \varvec{\theta },\varvec{y})}{\mathcal {CL}( \varvec{\theta }_{0},\varvec{y})}\right) d \varvec{y} \end{aligned}$$

and

$$\begin{aligned} \left. \frac{\partial ^{2}D_{\phi }(\varvec{\theta }, \varvec{\theta }_{0})}{\partial \varvec{\theta }\partial \varvec{ \theta }^{T}}\right| _{\varvec{\theta }=\varvec{\theta } _{0}}=\phi ^{\prime \prime }\left( 1\right) \int _{ \mathbb {R} ^{m}}\left. \frac{\partial c\ell (\varvec{\theta },\varvec{y })}{\partial \varvec{\theta }}\frac{\partial c\ell (\varvec{ \theta },\varvec{y})}{\partial \varvec{\theta }^{T} }\right| _{\varvec{\theta }=\varvec{\theta }_{0}}\mathcal {CL}( \varvec{\theta }_{0},\varvec{y})d\varvec{y} =\phi ^{\prime \prime }\left( 1\right) \varvec{J}(\varvec{ \theta }_{0}). \end{aligned}$$

Then, from

$$\begin{aligned} D_{\phi }(\widehat{\varvec{\theta }}_{c},\varvec{\theta } _{0})=\frac{\phi ^{\prime \prime }\left( 1\right) }{2}(\widehat{\varvec{ \theta }}_{c}-\varvec{\theta }_{0})^{T}\varvec{J}(\varvec{\theta }_{0}\mathbf {\mathbf {)(}}\widehat{\varvec{\theta }} _{c}-\varvec{\theta }_{0})+o(n^{-1/2}) \end{aligned}$$

the desired result is obtained. The value of k comes from

$$\begin{aligned} k=\mathrm {rank}\left( \varvec{G}_{*}^{-1}(\varvec{\theta }_{0} )\varvec{J}^{T}(\varvec{\theta }_{0}\mathbf { )}\varvec{G}_{*}^{-1}(\varvec{\theta }_{0}) \right) = \mathrm {rank}(\varvec{J}(\varvec{\theta }_{0})). \end{aligned}$$

1.2 Proof of Theorem 2

A first order Taylor expansion gives

$$\begin{aligned} D_{\phi }(\widehat{\varvec{\theta }}_{c},\varvec{\theta } _{0})=D_{\phi }(\varvec{\theta }^{*},\varvec{\theta } _{0})+\varvec{q}^{T}(\widehat{\varvec{\theta }}_{c}-\varvec{ \theta }^{*})+o(\Vert \widehat{\varvec{\theta }}_{c}- \varvec{\theta }^{*}\Vert ). \end{aligned}$$

But

$$\begin{aligned} \sqrt{n}(\widehat{\varvec{\theta }}_{c}-\varvec{\theta })\underset{ n\rightarrow \infty }{\overset{\mathcal {L}}{\longrightarrow }}\mathcal {N} \left( \varvec{0},\varvec{G}_{*}^{-1}(\varvec{\theta } )\right) \end{aligned}$$

and $\sqrt{n}o(\left\| \widehat{\varvec{\theta }}_{c}-\varvec{ \theta }^{*}\right\| )=o_{p}(1).$ Now the result follows.

1.3 Proof of Theorem 3

Following Sen and Singer (1993, pp. 242–243), let $\varvec{\theta }_{n}= \varvec{\theta }+n^{-1/2}\varvec{v}$, where $\left\| \varvec{v }\right\| <K^{*}$, $0<K^{*}<\infty$. Consider now the following Taylor expansion of the partial derivative of the composite log-density,

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum \limits _{i=1}^{n}\left. \frac{\partial }{\partial \varvec{\theta }}c\ell (\varvec{\theta },\varvec{y}_{i} )\right| _{\varvec{\theta }=\varvec{\theta }_{n}}=\frac{ 1}{\sqrt{n}}\sum \limits _{i=1}^{n}\frac{\partial }{\partial \varvec{ \theta }}c\ell (\varvec{\theta },\varvec{y}_{i}) + \frac{1}{n}\sum \limits _{i=1}^{n}\left. \frac{\partial ^{2}}{\partial \varvec{\theta }\mathbf {\partial }\varvec{\theta }^{T}}c\ell ( \varvec{\theta },\varvec{y}_{i})\right| _{ \varvec{\theta }=\varvec{\theta }_{n}^{*}}\sqrt{n}\left( \varvec{\theta }_{n}-\varvec{\theta }\right) , \end{aligned}$$

(30)

where $\varvec{\theta }_{n}^{*}$ belongs to the line segment joining $\varvec{\theta }$ and $\varvec{\theta }_{n}$. Then, observing that (cf. Theorem 2.3.6 of Sen and Singer 1993, p. 61)

$$\begin{aligned} \frac{1}{n}\sum \limits _{i=1}^{n}\frac{\partial ^{2}}{\partial \varvec{ \theta }\mathbf {\partial }\varvec{\theta }^{T}}c\ell (\varvec{\theta },\varvec{y}_{i})\overset{P}{\underset{n\rightarrow \infty }{\longrightarrow }}E_{\varvec{\theta }}\left[ \frac{\partial ^{2} }{\partial \varvec{\theta }\mathbf {\partial }\varvec{\theta }^{T}} c\ell (\varvec{\theta },\varvec{Y})\right] =E_{ \varvec{\theta }}\left[ \frac{\partial }{\partial \varvec{\theta }} \varvec{u}^{T}(\varvec{\theta },\varvec{Y}) \right] {=}-\varvec{H}(\varvec{\theta }), \end{aligned}$$

Eq. (30) leads

$$\begin{aligned} \left. \frac{1}{\sqrt{n}}\sum \limits _{i=1}^{n}\frac{\partial }{\partial \varvec{\theta }}c\ell (\varvec{\theta },\varvec{y}_{i} )\right| _{\varvec{\theta }=\varvec{\theta }_{n}}=\frac{ 1}{\sqrt{n}}\sum \limits _{i=1}^{n}\frac{\partial }{\partial \varvec{ \theta }}c\ell (\varvec{\theta },\varvec{y}_{i}\mathbf {)-} \varvec{H}(\varvec{\theta })\sqrt{n}\left( \varvec{\theta }_{n}- \varvec{\theta }\right) +o_{P}(1). \end{aligned}$$

(31)

Since $\varvec{G}(\varvec{\theta })=\frac{\partial \varvec{g} ^{T}(\varvec{\theta })}{\partial \varvec{\theta }}$ is continuous in $\varvec{\theta }$, it is true that,

$$\begin{aligned} \varvec{g}(\varvec{\theta }_{n})=\varvec{G}^{T}(\varvec{ \theta })\sqrt{n}\left( \varvec{\theta }_{n}-\varvec{\theta }\right) +o_{P}(1). \end{aligned}$$

(32)

Since, the restricted maximum composite likelihood estimator $\widetilde{ \varvec{\theta }}_{rc}$ should satisfy the likelihood equations

$$\begin{aligned} \sum \limits _{i=1}^{n}\frac{\partial }{\partial \varvec{\theta }}c\ell ( \varvec{\theta },\varvec{y}_{i})+\varvec{G}( \varvec{\theta })\mathbf {\lambda }= & {} \varvec{0}_{p}, \\ \varvec{g}(\varvec{\theta })= & {} \varvec{0}_{r}, \end{aligned}$$

and in view of (31) and (32) it holds that

$$\begin{aligned}&\displaystyle \frac{1}{\sqrt{n}}\sum \limits _{i=1}^{n}\frac{\partial }{\partial \varvec{\theta }}c\ell (\varvec{\theta },\varvec{y}_{i})-\varvec{H}(\varvec{\theta })\sqrt{n}\left( \widetilde{ \varvec{\theta }}_{rc}-\varvec{\theta }\right) +\varvec{G}( \varvec{\theta })\frac{1}{\sqrt{n}}\overline{\mathbf {\lambda }} _{n}+o_{P}(1) =\varvec{0}_{p},&\\&\displaystyle \varvec{G}^{T}(\varvec{\theta })\sqrt{n}(\widetilde{\varvec{ \theta }}_{rc}-\varvec{\theta })+o_{P}(1) =\varvec{0}_{p}.&\end{aligned}$$

In matrix notation it may be re-expressed as

$$\begin{aligned} \left( \begin{array}{cc} \varvec{H}(\varvec{\theta }) &{}\quad -\varvec{G}(\varvec{\theta }) \\ -\varvec{G}^{T}(\varvec{\theta }) &{}\quad \varvec{0}_{r\times r} \end{array} \right) \left( \begin{array}{c} \sqrt{n}(\widetilde{\varvec{\theta }}_{rc}-\varvec{\theta })\\ n^{-1/2}\overline{\mathbf {\lambda }}_{n} \end{array} \right) =\left( \begin{array}{c} \frac{1}{\sqrt{n}}\sum \limits _{i=1}^{n}\frac{\partial }{\partial \varvec{\theta }}c\ell (\varvec{\theta },\varvec{y}_{i} ) \\ \varvec{0}_{r} \end{array} \right) +o_{P}(1). \end{aligned}$$

Then

$$\begin{aligned} \left( \begin{array}{c} \sqrt{n}(\widetilde{\varvec{\theta }}_{rc}-\varvec{\theta })\\ n^{-1/2}\overline{\mathbf {\lambda }}_{n} \end{array} \right) =\left( \begin{array}{cc} \varvec{P}(\varvec{\theta }) &{}\quad \varvec{Q}(\varvec{\theta }) \\ \varvec{Q}^{T}(\varvec{\theta }) &{}\quad \varvec{R}(\varvec{\theta }) \end{array} \right) \left( \begin{array}{c} \frac{1}{\sqrt{n}}\sum \limits _{i=1}^{n}\frac{\partial }{\partial \varvec{\theta }}c\ell (\varvec{\theta },\varvec{y}_{i} ) \\ \varvec{0}_{r} \end{array} \right) +o_{P}(1), \end{aligned}$$

(33)

where

$$\begin{aligned} \left( \begin{array}{cc} \varvec{P}(\varvec{\theta }) &{}\quad \varvec{Q}(\varvec{\theta }) \\ \varvec{Q}^{T}(\varvec{\theta }) &{} \quad \varvec{R}( \varvec{\theta }) \end{array} \right) =\left( \begin{array}{cc} \varvec{H}(\varvec{\theta }) &{}\quad -\varvec{G}(\varvec{\theta }) \\ -\varvec{G}^{T}(\varvec{\theta }) &{}\quad \varvec{0}_{r\times r} \end{array} \right) ^{-1}. \end{aligned}$$

This last equation implies (cf. Sen and Singer 1993, p. 243, Eq. (5.6.24)),

$$\begin{aligned} \varvec{P}(\varvec{\theta })= & {} \varvec{H}^{-1}(\varvec{\theta } )\left( \varvec{I}_{p}-\varvec{G}(\varvec{\theta })\left( \varvec{G}^{T}(\varvec{\theta })\varvec{H} ^{-1}(\varvec{\theta })\varvec{G}(\varvec{\theta } )\right) ^{-1}\varvec{G}^{T}(\varvec{\theta })\varvec{H} ^{-1}(\varvec{\theta })\right) , \\ \varvec{Q}(\varvec{\theta })= & {} -\varvec{H}^{-1}(\varvec{\theta })\varvec{G}(\varvec{\theta })\left( \varvec{G}^{T}( \varvec{\theta })\varvec{H}^{-1}(\varvec{\theta }) \varvec{G}(\varvec{\theta })\right) ^{-1},\\ \varvec{R}(\varvec{\theta })= & {} -\left( \varvec{G}^{T}( \varvec{\theta })\varvec{H}^{-1}(\varvec{\theta }) \varvec{G}(\varvec{\theta })\right) ^{-1}. \end{aligned}$$

Based on the central limit theorem (Theorem 3.3.1 of Sen and Singer 1993, p. 107) and the Cramér–Wald theorem (Theorem 3.2.4 of Sen and Singer 1993, p. 106) it is obtained

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum \limits _{i=1}^{n}\frac{\partial }{\partial \varvec{\theta }}c\ell (\varvec{\theta },\varvec{y}_{i} )\overset{\mathcal {L}}{\underset{n\rightarrow \infty }{ \longrightarrow }}\mathcal {N}\left( \varvec{0}_{p},\textit{Var}_{\varvec{ \theta }}[\varvec{u}(\varvec{\theta },\varvec{Y}\mathbf { )}]\right) . \end{aligned}$$

with $\textit{Var}_{\varvec{\theta }}[\varvec{u}(\varvec{\theta }, \varvec{Y})]=\varvec{J}(\varvec{\theta } )$. Then, it follows from (33) that

$$\begin{aligned} \left( \begin{array}{c} \sqrt{n}(\widetilde{\varvec{\theta }}_{rc}-\varvec{\theta })\\ n^{-1/2}\overline{\mathbf {\lambda }}_{n} \end{array} \right) \overset{\mathcal {L}}{\underset{n\rightarrow \infty }{ \longrightarrow }}\mathcal {N}\left( \varvec{0},\varvec{\Sigma } \right) , \end{aligned}$$

with

$$\begin{aligned} \varvec{\Sigma }=\left( \begin{array}{cc} \varvec{P}(\varvec{\theta }) &{}\quad \varvec{Q}(\varvec{\theta }) \\ \varvec{Q}^{T}(\varvec{\theta }) &{}\quad \varvec{R}( \varvec{\theta }) \end{array} \right) \left( \begin{array}{cc} \varvec{J}(\varvec{\theta }) &{}\quad \varvec{0}_{p\times r} \\ \varvec{0}_{r\times p} &{}\quad \varvec{0}_{r\times r} \end{array} \right) \left( \begin{array}{cc} \varvec{P}^{T}(\varvec{\theta }) &{}\quad \varvec{Q}(\varvec{\theta }) \\ \varvec{Q}^{T}(\varvec{\theta })&{}\quad \varvec{R}^{T}( \varvec{\theta }) \end{array} \right) , \end{aligned}$$

or

$$\begin{aligned} \varvec{\Sigma }=\left( \begin{array}{cc} \varvec{P}(\varvec{\theta })\varvec{J}(\varvec{\theta }) \varvec{P}^{T}(\varvec{\theta }) &{}\quad \varvec{P}(\varvec{\theta })\varvec{J}(\varvec{\theta })\varvec{Q}(\varvec{\theta }) \\ \varvec{Q}^{T}(\varvec{\theta })\varvec{J}(\varvec{\theta }) \varvec{P}^{T}(\varvec{\theta }) &{}\quad \varvec{Q}^{T}( \varvec{\theta })\varvec{J}(\varvec{\theta })\varvec{Q}( \varvec{\theta }) \end{array} \right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \sqrt{n}(\widetilde{\varvec{\theta }}_{rc}-\varvec{\theta })\overset{ \mathcal {L}}{\underset{n\rightarrow \infty }{\longrightarrow }}\mathcal {N} \left( \varvec{0}_{p},\varvec{P}(\varvec{\theta })\varvec{J}( \varvec{\theta })\varvec{P}^{T}(\varvec{\theta })\right) , \end{aligned}$$

with

$$\begin{aligned} \varvec{P}(\varvec{\theta })&=\varvec{H}^{-1}(\varvec{ \theta })\left( \varvec{I}_{p}-\varvec{G}(\varvec{\theta } )\left( \varvec{G}^{T}(\varvec{\theta }) \varvec{H}^{-1}(\varvec{\theta })\varvec{G}(\varvec{ \theta })\right) ^{-1}\varvec{G}^{T}(\varvec{\theta }) \varvec{H}^{-1}(\varvec{\theta })\right) \\&=\varvec{H}^{-1}(\varvec{\theta })-\varvec{H}^{-1}(\varvec{ \theta })\varvec{G}(\varvec{\theta })\left( \varvec{G}^{T} (\varvec{\theta })\varvec{H}^{-1}(\varvec{ \theta })\varvec{G}(\varvec{\theta })\right) ^{-1} \varvec{G}^{T}(\varvec{\theta })\varvec{H}^{-1}(\varvec{ \theta }) \\&=\varvec{H}^{-1}(\varvec{\theta })+\varvec{Q}(\varvec{ \theta })\varvec{G}^{T}(\varvec{\theta }) \varvec{H}^{-1}(\varvec{\theta }), \end{aligned}$$

and the proof of the lemma is now completed.

1.4 Proof of Lemma 4

Based on Eq. (33), above,

$$\begin{aligned} \sqrt{n}(\widetilde{\varvec{\theta }}_{rc}-\varvec{\theta })= \varvec{P}(\varvec{\theta })\frac{1}{\sqrt{n}}\sum \limits _{i=1}^{n} \frac{\partial }{\partial \varvec{\theta }}c\ell (\varvec{\theta } ,\varvec{y}_{i})+o_{P}(1). \end{aligned}$$

(34)

The Taylor series expansion (30) gives that

$$\begin{aligned} \varvec{0}= & {} \left. \frac{1}{\sqrt{n}}\sum \limits _{i=1}^{n}\frac{\partial }{\partial \varvec{\theta }}c\ell (\varvec{\theta }, \varvec{y}_{i})\right| _{\varvec{\theta }=\widehat{ \varvec{\theta }}_{c}}=\frac{1}{\sqrt{n}}\sum \limits _{i=1}^{n}\frac{ \partial }{\partial \varvec{\theta }}c\ell (\varvec{\theta }, \varvec{y}_{i}) + \frac{1}{n}\sum \limits _{i=1}^{n}\left. \frac{ \partial ^{2}}{\partial \varvec{\theta }\mathbf {\partial }\varvec{ \theta }^{T}}c\ell (\varvec{\theta },\varvec{y}_{i}\mathbf {) }\right| _{\varvec{\theta }=\varvec{\theta }_{n}^{*}}\sqrt{n}(\widehat{\varvec{\theta }}_{c}-\varvec{\theta }), \end{aligned}$$

or

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum \limits _{i=1}^{n}\frac{\partial }{\partial \varvec{\theta }}c\ell (\varvec{\theta },\varvec{y}_{i} )=\mathbf {-}\frac{1}{n}\sum \limits _{i=1}^{n}\left. \frac{\partial ^{2}}{\partial \varvec{\theta }\mathbf {\partial }\varvec{\theta }^{T} }c\ell (\varvec{\theta },\varvec{y}_{i}) \right| _{\varvec{\theta }=\varvec{\theta }_{n}^{*}}\sqrt{n}( \widehat{\varvec{\theta }}_{c}-\varvec{\theta }), \end{aligned}$$

where $\varvec{\theta }_{n}^{*}$ belongs to the line segment joining $\varvec{\theta }$ and $\widehat{\varvec{\theta }}_{c}$. Taking into account Theorem 2.3.6 of Sen and Singer (1993, p. 61),

$$\begin{aligned} \frac{1}{n}\sum \limits _{i=1}^{n}\left. \frac{\partial ^{2}}{\partial \varvec{\theta }\mathbf {\partial }\varvec{\theta }^{T}}c\ell ( \varvec{\theta },\varvec{y}_{i})\right| _{ \varvec{\theta }=\varvec{\theta }_{n}^{*}}\overset{P}{\underset{ n\rightarrow \infty }{\longrightarrow }}-\varvec{H}(\varvec{\theta } ), \end{aligned}$$

and the above two equations lead

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum \limits _{i=1}^{n}\frac{\partial }{\partial \varvec{\theta }}c\ell (\varvec{\theta },\varvec{y}_{i} )=\varvec{H}(\varvec{\theta })\sqrt{n}(\widehat{\varvec{ \theta }}_{c}-\varvec{\theta })+o_{P}(1). \end{aligned}$$

(35)

Equations (34), (35) and the fact that $\varvec{P}(\varvec{\theta })=\varvec{H}^{-1}(\varvec{\theta })+ \varvec{Q}(\varvec{\theta })\varvec{G}^{T}( \varvec{\theta })\varvec{H}^{-1}(\varvec{\theta })$ give that

$$\begin{aligned} \sqrt{n}(\widetilde{\varvec{\theta }}_{rc}-\varvec{\theta })&= \varvec{P}(\varvec{\theta })\frac{1}{\sqrt{n}}\sum \limits _{i=1}^{n} \frac{\partial }{\partial \varvec{\theta }}c\ell (\varvec{\theta } ,\varvec{y}_{i})+o_{P}(1) \\&=\varvec{P}(\varvec{\theta })\varvec{H}(\varvec{\theta }) \sqrt{n}(\widehat{\varvec{\theta }}_{c}-\varvec{\theta })+o_{P}(1) \\&=\left( \varvec{H}^{-1}(\varvec{\theta })+\varvec{Q}( \varvec{\theta })\varvec{G}^{T}(\varvec{\theta })\varvec{H}^{-1}(\varvec{\theta })\right) \varvec{H}( \varvec{\theta })\sqrt{n}(\widehat{\varvec{\theta }}_{c}-\varvec{ \theta })+o_{P}(1), \end{aligned}$$

which completes the proof of the lemma.

1.5 Proof of Theorem 5

A second order Taylor expansion of $D_{\phi }(\widehat{\varvec{\theta }} _{rc},\widetilde{\varvec{\theta }}_{rc})$, considered as a function of $\widehat{\varvec{\theta }}_{c}$, around $\widetilde{\varvec{\theta }} _{rc}$, gives

$$\begin{aligned} D_{\phi }(\widehat{\varvec{\theta }}_{c},\widetilde{\varvec{\theta }} _{rc})&=D_{\phi }(\widehat{\varvec{\theta }}_{rc},\widetilde{\varvec{ \theta }}_{rc})+\left. \frac{\partial }{\partial \varvec{\theta }}D_{\phi }(\varvec{\theta },\widetilde{\varvec{\theta }} _{rc})\right| _{\varvec{\theta }=\widetilde{\varvec{\theta }}_{rc}}(\widehat{\varvec{\theta }}_{c}-\widetilde{\varvec{\theta }} _{rc}) \\&\quad +\frac{1}{2}(\widehat{\varvec{\theta }}_{c}-\widetilde{\varvec{ \theta }}_{rc})^{T}\left. \frac{\partial ^{2}}{\partial \varvec{\theta } \mathbf {\partial } \varvec{\theta }^{T}}D_{\phi }(\varvec{\theta }, \widetilde{\varvec{\theta }}_{rc})\right| _{\theta = \widetilde{\varvec{\theta }}_{rc}}\!\!\!(\widehat{\varvec{\theta }}_{c}- \widetilde{\varvec{\theta }}_{rc}){+}o(\Vert \widehat{\varvec{\theta }} _{c}-\widetilde{\varvec{\theta }}_{rc}\Vert ^{2}). \end{aligned}$$

Based on Pardo (2006, pp. 411–412), we obtain $D_{\phi }(\widetilde{\varvec{\theta }}_{rc},\widetilde{\varvec{\theta }}_{rc})=0$, $\left. \frac{\partial }{\partial \varvec{\theta }}D_{\phi }( \varvec{\theta },\widetilde{\varvec{\theta }} _{rc})\right| _{\varvec{\theta }=\widetilde{\varvec{\theta }} _{rc}}=\varvec{0}_{p}$ and $\left. \frac{\partial ^{2}}{\partial \varvec{\theta }\mathbf {\partial }\varvec{\theta }^{T}}D_{\phi }( \varvec{\theta },\widetilde{\varvec{\theta }} _{rc})\right| _{\varvec{\theta }=\widetilde{\varvec{\theta }} _{rc}}=\phi ^{\prime \prime }(1)\varvec{J}(\widetilde{\varvec{\theta }}_{rc})$. Then, the above equation leads

$$\begin{aligned} \frac{2n}{\phi ^{\prime \prime }(1)}D_{\phi }(\widehat{\varvec{\theta }} _{c},\widetilde{\varvec{\theta }}_{rc})=\sqrt{n}(\widehat{\varvec{ \theta }}_{c}-\widetilde{\varvec{\theta }}_{rc})^{T}\varvec{J}( \widetilde{\varvec{\theta }}_{rc})\sqrt{n}(\widehat{\varvec{\theta }} _{c}-\widetilde{\varvec{\theta }}_{rc})+no(\Vert \widehat{\varvec{ \theta }}_{c}-\widetilde{\varvec{\theta }}_{rc}\Vert ^{2}), \end{aligned}$$

or

$$\begin{aligned} T_{\phi ,n}(\widehat{\varvec{\theta }}_{c},\widetilde{ \varvec{\theta }}_{rc})=\sqrt{n}(\widehat{\varvec{\theta }}_{c}- \widetilde{\varvec{\theta }}_{rc})^{T}\varvec{J}(\widetilde{ \varvec{\theta }}_{rc})\sqrt{n}(\widehat{\varvec{\theta }}_{c}- \widetilde{\varvec{\theta }}_{rc})+no(\Vert \widehat{\varvec{\theta } }_{c}-\widetilde{\varvec{\theta }}_{rc}\Vert ^{2}). \end{aligned}$$

(36)

On the other hand (cf., Pardo 2006, p. 63),

$$\begin{aligned} no(||\widehat{\varvec{\theta }}_{c}-\widetilde{\varvec{\theta }} _{rc}||^{2})\le no(\Vert \widehat{\varvec{\theta }}_{c}-\varvec{ \theta }\Vert ^{2})+no(||\widetilde{\varvec{\theta }}_{rc}-\varvec{ \theta }||^{2}), \end{aligned}$$

and $no(\Vert \widehat{\varvec{\theta }}_{c}-\varvec{\theta }\Vert ^{2})=o_{P}(1)$, $no(||\widetilde{\varvec{\theta }}_{rc}-\varvec{ \theta }||^{2})=o_{P}(1)$. Therefore, $o(||\widehat{\varvec{\theta }} _{c}-\widetilde{\varvec{\theta }}_{rc}||^{2})=o_{P}(1)$. To apply the Slutsky’s theorem, it remains to obtain the asymptotic distribution of the quantity

$$\begin{aligned} \sqrt{n}(\widehat{\varvec{\theta }}_{c}-\widetilde{\varvec{\theta }} _{rc})^{T}\varvec{J}(\widetilde{\varvec{\theta }}_{rc})\sqrt{n}( \widehat{\varvec{\theta }}_{c}-\widetilde{\varvec{\theta }}_{rc}). \end{aligned}$$

From Lemma 4 it is immediately obtained that

$$\begin{aligned} \sqrt{n}(\widehat{\varvec{\theta }}_{c}-\widetilde{\varvec{\theta }} _{rc})=\varvec{Q}(\varvec{\theta })\varvec{G}^{T} (\varvec{\theta })\sqrt{n}(\widehat{\varvec{\theta }}_{c}-\varvec{\theta })+o_{P}(1). \end{aligned}$$

On the other hand, we know that

$$\begin{aligned} \sqrt{n}(\widehat{\varvec{\theta }}_{c}-\varvec{\theta })\overset{ \mathcal {L}}{\underset{n\rightarrow \infty }{\longrightarrow }}\mathcal {N} \left( \varvec{0}_{p},\varvec{G}_{*}^{-1}(\varvec{\theta } )\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \sqrt{n}(\widehat{\varvec{\theta }}_{c}-\widetilde{\varvec{\theta }} _{rc})\overset{\mathcal {L}}{\underset{n\rightarrow \infty }{\longrightarrow } }\mathcal {N}\left( \varvec{0}_{p},\varvec{Q}(\varvec{ \theta }\mathbf {)}\varvec{G}^{T}(\varvec{\theta } )\varvec{G}_{*}^{-1}(\varvec{\theta }) \varvec{G}(\varvec{\theta })\varvec{Q}^{T}( \varvec{\theta })\right) , \end{aligned}$$

and taking into account (36) and Corollary 2.1 of Dic and Gunst (1985), $T_{\phi ,n}({\widehat{\varvec{\theta }}_{c}},{\widetilde{ \varvec{\theta }}_{rc}})$ converge in law to the random variable $\sum _{i=1}^{k}\beta _{i}Z_{i}^{2}$, where $\beta _{i}$, $i=1,\ldots ,k$, are the eigenvalues of the matrix $\varvec{J}(\varvec{\theta })\varvec{Q} (\varvec{\theta }\mathbf {)}\varvec{G}^{T}( \varvec{\theta })\varvec{G}_{*}^{-1}(\varvec{\theta })\varvec{G}(\varvec{\theta })\varvec{ Q}^{T}(\varvec{\theta })$ and

$$\begin{aligned} k=\mathrm {rank}\left( \varvec{Q}(\varvec{\theta }\mathbf { )}\varvec{G}^{T}(\varvec{\theta }) \varvec{G}_{*}^{-1}(\varvec{\theta })\varvec{G} (\varvec{\theta })\varvec{Q}^{T}(\varvec{ \theta })\varvec{J}(\varvec{\theta })\varvec{Q}( \varvec{\theta }\mathbf {)}\varvec{G}^{T}( \varvec{\theta })\varvec{G}_{*}^{-1}(\varvec{\theta })\varvec{G}(\varvec{\theta })\varvec{ Q}^{T}(\varvec{\theta })\right) . \end{aligned}$$

1.6 Proof of Theorem 6

The result follows in a straightforward manner by considering a first order Taylor expansion of $D_{\phi }(\widehat{\varvec{\theta }}_{c},\widetilde{ \varvec{\theta }}_{rc})$, which yields

$$\begin{aligned} D_{\phi }(\widehat{\varvec{\theta }}_{c},\widetilde{\varvec{\theta }} _{rc})=D_{\phi }(\varvec{\theta },\varvec{\theta }^{*})+ \varvec{t}^{T}(\widehat{\varvec{\theta }}_{c}-\varvec{\theta })+ \varvec{s}^{T}(\widetilde{\varvec{\theta }}_{rc}-\varvec{\theta } ^{*})+o(\Vert \widehat{\varvec{\theta }}_{c}-\varvec{\theta }\Vert +\Vert \widetilde{\varvec{\theta }}_{rc}-\varvec{ \theta }^{*}\Vert ). \end{aligned}$$

1.7 Calculation of the integral $I_{a}$ in (28)

The integral $I_{a}$ is given by

$$\begin{aligned} I_{a}=\int \limits _{ \mathbb {R} ^{q}}\mathcal {CL}(\widehat{\mu },\widehat{\sigma },\widehat{\rho }, \varvec{y})^{a}\mathcal {CL}(\widehat{\mu },\sigma _{0},\rho _{0}, \varvec{y})^{1-a}d\varvec{y}, \quad a\ne 0,1, \end{aligned}$$

where

$$\begin{aligned} \mathcal {CL}(\mu ,\sigma ,\rho ,\varvec{y})= & {} L_{\rho ,\sigma }(1)\exp \left\{ -\frac{1}{2\sigma ^{2}}\left[ \sum \limits _{r=2}^{q}(y_{r-1}-\mu )^{2}+\sum \limits _{r=2}^{q}(y_{r}-\mu )^{2}\right. \right. \\&\left. \left. -2\rho \sum \limits _{r=2}^{q}(y_{r-1}-\mu )(y_{r}-\mu )\right] \right\} , \end{aligned}$$

and

$$\begin{aligned} L_{\rho ,\sigma }(s)=\frac{(1-\rho ^{2})^{s(q-1)/2}}{(2\pi )^{s(q-1)/2}\sigma ^{2(q-1)s}}. \end{aligned}$$

(37)

Then,

$$\begin{aligned} I_{a}= & {} L_{\widehat{\rho },\widehat{\sigma }}(a)L_{\rho _{0},\sigma _{0}}(1-a) \nonumber \\&\times \int \limits _{ \mathbb {R} ^{q}}\exp \left\{ -\frac{a}{2\widehat{\sigma }^{2}}\left[ \sum \limits _{r=2}^{q}(y_{r-1}-\widehat{\mu })^{2}+\!\!\sum \limits _{r=2}^{q}(y_{r}- \widehat{\mu })^{2}-2\widehat{\rho }\sum \limits _{r=2}^{q}(y_{r-1}-\widehat{ \mu })(y_{r}-\widehat{\mu })\right] \right\} \nonumber \\&\times \exp \left\{ -\frac{(1-a)}{2\sigma _{0}^{2}}\left[ \sum \limits _{r=2}^{q}(y_{r-1}-\widehat{\mu })^{2}+\!\!\sum \limits _{r=2}^{q}(y_{r}-\widehat{\mu })^{2}-2\rho _{0}\sum \limits _{r=2}^{q}(y_{r-1}-\widehat{\mu })(y_{r}-\widehat{\mu }) \right] \right\} \nonumber \\&\quad dy_{1}\ldots dy_{q}. \end{aligned}$$

(38)

But, if

$$\begin{aligned} E_{1}= & {} \exp \left\{ -\frac{a}{2\widehat{\sigma }^{2}}\left[ \sum \limits _{r=2}^{q}(y_{r-1}-\widehat{\mu })^{2}+\sum \limits _{r=2}^{q}(y_{r}-\widehat{\mu })^{2}-2\widehat{\rho } \sum \limits _{r=2}^{q}(y_{r-1}-\widehat{\mu })(y_{r}-\widehat{\mu })\right] \right\} , \\ E_{2}= & {} \exp \left\{ -\frac{(1-a)}{2\sigma _{0}^{2}}\left[ \sum \limits _{r=2}^{q}(y_{r-1}-\widehat{\mu })^{2}+\sum \limits _{r=2}^{q}(y_{r}- \widehat{\mu })^{2}-2\rho _{0}\sum \limits _{r=2}^{q}(y_{r-1}-\widehat{\mu } )(y_{r}-\widehat{\mu })\right] \right\} , \end{aligned}$$

then

$$\begin{aligned} E_{1}\times E_{2}= & {} \exp \left\{ -\frac{1}{2}\left( \frac{a}{\widehat{ \sigma }^{2}}-\frac{1-a}{\sigma _{0}^{2}}\right) \left[ \sum \limits _{r=2}^{q}(y_{r-1}-\widehat{\mu })^{2}+\sum \limits _{r=2}^{q}(y_{r}- \widehat{\mu })^{2}\right] \right. \nonumber \\&\left. +\left( \frac{a\widehat{\rho }}{\widehat{\sigma } ^{2}}+\frac{(1-a)\rho _{0}}{\sigma _{0}^{2}}\right) \sum \limits _{r=2}^{q}(y_{r-1}-\widehat{\mu })(y_{r}-\widehat{\mu })\right\} \nonumber \\= & {} \exp \left\{ -\frac{1}{2}\left( \frac{a}{\widehat{\sigma }^{2}}-\frac{1-a }{\sigma _{0}^{2}}\right) \left[ \sum \limits _{r=2}^{q}(y_{r-1}-\widehat{\mu })^{2}+\sum \limits _{r=2}^{q}(y_{r}-\widehat{\mu })^{2}\right. \right. \nonumber \\&\left. \left. -2\frac{\frac{a\widehat{\rho }}{\widehat{ \sigma }^{2}}+\frac{(1-a)\rho _{0}}{\sigma _{0}^{2}}}{\frac{a}{\widehat{ \sigma }^{2}}-\frac{1-a}{\sigma _{0}^{2}}}\sum \limits _{r=2}^{q}(y_{r-1}- \widehat{\mu })(y_{r}-\widehat{\mu })\right] \right\} \nonumber \\= & {} \exp \left\{ -\frac{1}{2\sigma _{*}^{2}}\left[ \sum \limits _{r=2}^{q}(y_{r-1}-\widehat{\mu })^{2}+\sum \limits _{r=2}^{q}(y_{r}- \widehat{\mu })^{2}\right. \right. \nonumber \\&\left. \left. -2\rho ^{*}\sum \limits _{r=2}^{q}(y_{r-1}-\widehat{ \mu })(y_{r}-\widehat{\mu })\right] \right\} , \end{aligned}$$

(39)

with

$$\begin{aligned} \frac{1}{\sigma _{*}^{2}}=\frac{a}{\widehat{\sigma }^{2}}-\frac{1-a}{ \sigma _{0}^{2}}=\frac{a\sigma _{0}^{2}-(1-a)\widehat{\sigma }^{2}}{\widehat{ \sigma }^{2}\sigma _{0}^{2}}, \end{aligned}$$

or

$$\begin{aligned} \sigma _{*}^{2}=\frac{\widehat{\sigma }^{2}\sigma _{0}^{2}}{a\sigma _{0}^{2}-(1-a)\widehat{\sigma }^{2}}, \end{aligned}$$

(40)

and

$$\begin{aligned} \rho ^{*}=\frac{\frac{a\widehat{\rho }}{\widehat{\sigma }^{2}}+\frac{ (1-a)\rho _{0}}{\sigma _{0}^{2}}}{\frac{a}{\widehat{\sigma }^{2}}-\frac{1-a}{ \sigma _{0}^{2}}}=\frac{\frac{a\widehat{\rho }\sigma _{0}^{2}+(1-a)\rho _{0} \widehat{\sigma }^{2}}{\widehat{\sigma }^{2}\sigma _{0}^{2}}}{\frac{a\sigma _{0}^{2}-(1-a)\widehat{\sigma }^{2}}{\widehat{\sigma }^{2}\sigma _{0}^{2}}}= \frac{a\widehat{\rho }\sigma _{0}^{2}+(1-a)\rho _{0}\widehat{\sigma }^{2}}{ a\sigma _{0}^{2}-(1-a)\widehat{\sigma }^{2}}. \end{aligned}$$

(41)

Based on (38)–(41),

$$\begin{aligned} I_{a}= & {} \frac{L_{\widehat{\rho },\widehat{\sigma }}(a)L_{\rho _{0},\sigma _{0}}(1-a)}{L_{\rho ^{*},\sigma ^{*}}(1)} \\&\times \int \limits _{ \mathbb {R} ^{q}}L_{\rho ^{*},\sigma ^{*}}(1)\exp \left\{ -\frac{1}{2\sigma _{*}^{2}}\left[ \sum \limits _{r=2}^{q}(y_{r-1}-\widehat{\mu } )^{2}+\sum \limits _{r=2}^{q}(y_{r}-\widehat{\mu })^{2}\right. \right. \nonumber \\&\left. \left. -2 \rho ^{*}\sum \limits _{r=2}^{q}(y_{r-1}-\widehat{\mu })(y_{r}-\widehat{\mu })\right] \right\} d\varvec{y}, \end{aligned}$$

and taking into account that the last integral is equal to one,

$$\begin{aligned} I_{a}=\frac{L_{\widehat{\rho },\widehat{\sigma }}(a)L_{\rho _{0},\sigma _{0}}(1-a)}{L_{\rho ^{*},\sigma ^{*}}(1)}, \end{aligned}$$

(42)

with $L_{\rho ,\sigma }(s),$ defined by (37). After some algebraic manipulations, (42) leads to the explicit expression of the integral $I_{a}$, given by (28).