Proofs of the main results
1.1 Proof of Theorem 1
From (8), we have
$$\begin{aligned} d_{KL}(\widehat{\varvec{p}},\varvec{p}(\varvec{\theta }))&=\sum _{i=1}^{I}\sum _{j=1}^{2}\frac{n_{ij}}{K}\log \left( \dfrac{\frac{n_{ij} }{K_{i}}}{F_{j}(IT_{i};\,\varvec{x}_i,\varvec{\theta })}\right) \\&=\sum _{i=1}^{I}\sum _{j=1}^{2}\frac{n_{ij}}{K}\log \left( \frac{n_{i}}{K_{i} }\right) -\sum _{i=1}^{I}\sum _{j=1}^{2}\frac{n_{ij}}{K}\log \left( F_{j}(IT_{i};\,\varvec{x}_i,\varvec{\theta })\right) \\&=c-\frac{1}{K}\sum _{i=1}^{I}\left\{ n_{i}\log \left( F(IT_{i};\,\varvec{x}_i,\varvec{\theta })\right) +(K_{i}-n_{i})\log \left( R(IT_{i};\,\varvec{x}_i,\varvec{\theta })\right) \right\} \\&=c-\frac{1}{K}\log \left( \prod _{i=1}^{I}F(IT_{i};\,\varvec{x}_i,\varvec{\theta })^{n_{i}}R^{K_{i}-n_{i}}(IT_{i};\,\varvec{x}_i,\varvec{\theta })\right) \\&=c-\frac{1}{K}\log \mathcal {L}(\varvec{\theta } ) , \end{aligned}$$
where \(c=\sum _{i=1}^{I}\sum _{j=1}^{2}\frac{n_{ij}}{K}\log \left( \frac{n_{i} }{K_{i}}\right) \) does not depend on the vector of parameters \(\varvec{a} \).
1.2 Proof of Theorem 6
Since
$$\begin{aligned} \frac{\partial }{\partial \varvec{\theta }}\sum _{i=1}^{I}\frac{K_{i}}{K}d_{\beta }^{*}(\widehat{\varvec{p}}_{i},\varvec{\pi }_{i}(\varvec{\theta }))&=\sum _{i=1}^{I}\frac{K_{i}}{K}\frac{\partial }{\partial \varvec{\theta } }d_{\beta }^{*}(\widehat{\varvec{p}}_{i},\varvec{\pi }_{i} (\varvec{\theta }))\\ \end{aligned}$$
and
$$\begin{aligned}&\frac{\partial }{\partial \varvec{\theta }}d_{\beta }^{*} (\widehat{\varvec{p}}_{i},\varvec{\pi }_{i}(\varvec{\theta }))\\&\quad =\left( \frac{\partial }{\partial \varvec{\theta }}\pi _{i1}^{\beta +1} (\varvec{\theta })+\frac{\partial }{\partial \varvec{\theta }}\pi _{i2} ^{\beta +1}(\varvec{\theta })\right) -\frac{\beta +1}{\beta }\left( \widehat{p}_{i1}\frac{\partial }{\partial \varvec{\theta }}\pi _{i1}^{\beta }(\varvec{\theta })+\widehat{p}_{i2}\frac{\partial }{\partial \varvec{\theta } }\pi _{i2}^{\beta }(\varvec{\theta })\right) \\&\quad =\left( \beta +1\right) \left( \pi _{i1}^{\beta }(\varvec{\theta } )-\pi _{i2}^{\beta }(\varvec{\theta })-\widehat{p}_{i1}\pi _{i1}^{\beta -1}(\varvec{\theta })+\widehat{p}_{i2}\pi _{i2}^{\beta -1}(\varvec{\theta })\right) \frac{\partial }{\partial \varvec{\theta }}\pi _{i1}(\varvec{\theta })\\&\quad =\left( \beta +1\right) \left( \left( \pi _{i1}(\varvec{\theta })-\widehat{p}_{i1}\right) \pi _{i1}^{\beta -1}(\varvec{\theta })-\left( \pi _{i2}(\varvec{\theta })-\widehat{p}_{i2}\right) \pi _{i2}^{\beta -1}(\varvec{\theta })\right) \frac{\partial }{\partial \varvec{\theta }} \pi _{i1}(\varvec{\theta })\\&\quad =\left( \beta +1\right) \left( \left( \pi _{i1}(\varvec{\theta })-\widehat{p}_{i1}\right) \pi _{i1}^{\beta -1}(\varvec{\theta })+\left( \pi _{i1}(\varvec{\theta })-\widehat{p}_{i1}\right) \pi _{i2}^{\beta -1}(\varvec{\theta })\right) \frac{\partial }{\partial \varvec{\theta }} \pi _{i1}(\varvec{\theta })\\&\quad =\left( \beta +1\right) \left( \pi _{i1}(\varvec{\theta } )-\widehat{p}_{i1}\right) \left( \pi _{i1}^{\beta -1}(\varvec{\theta } )+\pi _{i2}^{\beta -1}(\varvec{\theta })\right) \frac{\partial }{\partial \varvec{\theta }}\pi _{i1}(\varvec{\theta })\\&\quad =\left( \beta +1\right) \left( \pi _{i1}(\varvec{\theta } )-\widehat{p}_{i1}\right) \left( \pi _{i1}^{\beta -1}(\varvec{\theta } )+\pi _{i2}^{\beta -1}(\varvec{\theta })\right) f\left( IT_{i};\,\varvec{x}_i,\varvec{\theta }\right) \frac{\partial f\left( IT_{i};\,\varvec{x}_i,\varvec{\theta }\right) }{\partial \varvec{\theta }}, \end{aligned}$$
we obtain
$$\begin{aligned} \frac{\partial }{\partial \varvec{\theta }}\sum _{i=1}^{I}\frac{K_{i}}{K}d_{\beta }^{*}(\widehat{\varvec{p}}_{i},\varvec{\pi }_{i}(\varvec{\theta }))& = \frac{\beta +1}{K}\sum _{i=1}^{I}\left( K_{i}\pi _{i1}(\varvec{\theta })-n_{i}\right) \left( \pi _{i1}^{\beta -1}(\varvec{\theta })+\pi _{i2}^{\beta -1}(\varvec{\theta })\right) \\&\quad f\left( IT_{i};\,\varvec{x}_i,\varvec{\theta }\right) \frac{\partial f\left( IT_{i};\,\varvec{x}_i,\varvec{\theta }\right) }{\partial \varvec{\theta }}. \end{aligned}$$
In a similar way, we can get the derivative with respect to \(\varvec{b}\), and then, the required results follow.
1.3 Proof of Theorem 7
Let us denote
$$\begin{aligned} \varvec{u}_{ij}(\varvec{\theta })&=\left( \frac{\partial \log \pi _{ij}(\varvec{\theta })}{\partial \varvec{a}},\frac{\partial \log \pi _{ij}(\varvec{\theta })}{\partial \varvec{b}}\right) ^{T}= \left( \frac{1}{\pi _{ij}(\varvec{\theta })}\frac{\partial \pi _{ij}(\varvec{\theta })}{\partial \varvec{a}},\frac{1}{\pi _{ij}(\varvec{\theta })}\frac{\partial \pi _{ij}(\varvec{\theta } )}{\partial \varvec{b}}\right) ^{T}\\&=\left( \frac{(-1)^{j+1}}{\pi _{ij}(\varvec{\theta })}l_{i}\varvec{x} _{i},\frac{(-1)^{j+1}}{\pi _{ij}(\varvec{\theta })}s_{i}\varvec{x} _{i}\right) ^{T}, \end{aligned}$$
with
$$\begin{aligned} l_{i}& = \,\alpha _{i} \left\{ - \Psi \left( \alpha _{i}\right) \pi _{i1}(\varvec{\theta }) +\log \left( \frac{IT_i}{\lambda _{i}}\right) \pi _{i1}(\varvec{\theta })\right. \nonumber \\&\quad \left. -\,\frac{\left( \frac{IT_i}{\lambda _{i}}\right) ^{\alpha _i}}{\alpha _i^2 \Gamma (\alpha _i)} {_2}F_2\left( \alpha _i,\alpha _i;1+\alpha _i,1+\alpha _i;-\frac{IT_i}{\lambda _{i}}\right) \right\} \end{aligned}$$
(27)
and
$$\begin{aligned} s_{i}=-f\left( IT_{i};\,\varvec{x}_i,\varvec{\theta }\right) IT_{i};\, \end{aligned}$$
(28)
see Balakrishnan and Ling (2014) for more details.
Upon using Theorem 3.1 of Ghosh and Basu (2013), we have
$$\begin{aligned} \sqrt{K}\left( \widehat{\varvec{\theta }}_{\beta }-\varvec{\theta } _{0}\right) \overset{\mathcal {L}}{\underset{K\mathcal {\rightarrow } \infty }{\longrightarrow }}\mathcal {N}\left( \varvec{0}_{2J},\varvec{J} _{\beta }^{-1}(\varvec{\theta }_{0})\varvec{K}_{\beta } (\varvec{\theta }_{0})\varvec{J}_{\beta }^{-1}(\varvec{\theta } _{0})\right) , \end{aligned}$$
where
$$\begin{aligned} \varvec{J}_{\beta }(\varvec{\theta })&=\sum _{i=1}^{I} \sum _{j=1}^{2}\frac{K_i}{K}\varvec{u}_{ij}(\varvec{\theta })\varvec{u}_{ij} ^{T}(\varvec{\theta })\pi _{ij}^{\beta +1}(\varvec{\theta }),\\ \varvec{K}_{\beta }(\varvec{\theta })&=\left( \sum _{i=1}^{I}\sum _{j=1}^{2}\frac{K_i}{K}\varvec{u}_{ij}(\varvec{\theta })\varvec{u} _{ij}^{T}(\varvec{\theta })\pi _{ij}^{2\beta +1}(\varvec{\theta } )-\sum _{i=1}^{I}\frac{K_i}{K}\varvec{\xi }_{i,\beta }(\varvec{\theta })\varvec{\xi }_{i,\beta }^{T}(\varvec{\theta })\right) , \end{aligned}$$
with
$$\begin{aligned} \varvec{\xi }_{i,\beta }(\varvec{\theta })&=\sum _{j=1}^{2}\varvec{u} _{ij}(\varvec{\theta })\pi _{ij}^{\beta +1}(\varvec{\theta })\\&=\left( l_{i}\varvec{x}_{i},s_{i}\varvec{x}_{i}\right) ^{T} \sum _{j=1}^{2}(-1)^{j+1}\pi _{ij}^{\beta }(\varvec{\theta }). \end{aligned}$$
Now, for \(\varvec{u}_{ij}(\varvec{\theta })\varvec{u} _{ij}^{T}(\varvec{\theta })\), we have
$$\begin{aligned} \varvec{u}_{ij}(\varvec{\theta })\varvec{u}_{ij}^{T}(\varvec{\theta })=\frac{1}{\pi _{ij}^{2}(\varvec{\theta })}\left( \begin{array} [c]{cc} l_{i}^{2}\varvec{x}_{i}^{T}\varvec{x}_{i} &\quad l_{i}s_{i}\varvec{x} _{i}^{T}\varvec{x}_{i}\\ l_{i}s_{i}\varvec{x}_{i}^{T}\varvec{x}_{i} &\quad s_{i}^{2}\varvec{x} _{i}^{T}\varvec{x}_{i} \end{array} \right) =\frac{1}{\pi _{ij}^{2}(\varvec{\theta })}\varvec{M}_{i}, \end{aligned}$$
with
$$\begin{aligned} \varvec{M}_{i}=\left( \begin{array} [c]{cc} l_{i}^{2}\varvec{x}_{i}^{T}\varvec{x}_{i} &\quad l_{i}s_{i}\varvec{x} _{i}^{T}\varvec{x}_{i}\\ l_{i}s_{i}\varvec{x}_{i}^{T}\varvec{x}_{i} &\quad s_{i}^{2}\varvec{x} _{i}^{T}\varvec{x}_{i} \end{array} \right) . \end{aligned}$$
(29)
It then follows that
$$\begin{aligned} \varvec{J}_{\beta }(\varvec{\theta })&=\sum _{i=1} ^{I}\frac{K_i}{K}\varvec{M}_{i}\sum _{j=1}^{2}\pi _{ij}^{\beta -1}(\varvec{\theta })\\&=\sum _{i=1}^{I}\frac{K_i}{K}\varvec{M}_{i}\left( \pi _{i1}^{\beta -1}(\varvec{\theta })+\pi _{i2}^{\beta -1}(\varvec{\theta })\right) . \end{aligned}$$
In a similar manner,
$$\begin{aligned} \varvec{\xi }_{i,\beta }(\varvec{\theta })\varvec{\xi }_{i,\beta } ^{T}(\varvec{\theta })=\varvec{M}_{i}\left( \sum _{j=1}^{2} (-1)^{j+1}\pi _{ij}^{\beta }(\varvec{\theta })\right) ^{2} \end{aligned}$$
and
$$\begin{aligned} \varvec{K}_{\beta }(\varvec{\theta })=\sum _{i=1} ^{I}\frac{K_i}{K}\varvec{M}_{i}\left( \sum _{j=1}^{2}\pi _{ij}^{2\beta -1} (\varvec{\theta })-\left( \sum _{j=1}^{2}(-1)^{j+1}\pi _{ij}^{\beta }(\varvec{\theta })\right) ^{2}\right) . \end{aligned}$$
Since
$$\begin{aligned} \sum _{j=1}^{2}\pi _{ij}^{2\beta -1}(\varvec{\theta })-\left( \sum _{j=1} ^{2}(-1)^{j+1}\pi _{ij}^{\beta }(\varvec{\theta })\right) ^{2}=\pi _{i1}(\varvec{\theta })\pi _{i2}(\varvec{\theta })\left( \pi _{i1} ^{\beta -1}(\varvec{\theta })+\pi _{i2}^{\beta -1}(\varvec{\theta })\right) ^{2}, \end{aligned}$$
we have
$$\begin{aligned} \varvec{K}_{\beta }(\varvec{\theta })=\sum _{i=1} ^{I}\frac{K_i}{K}\varvec{M}_{i}\pi _{i1}(\varvec{\theta })\pi _{i2}(\varvec{\theta })\left( \pi _{i1}^{\beta -1}(\varvec{\theta })+\pi _{i2}^{\beta -1}(\varvec{\theta })\right) ^{2}. \end{aligned}$$
1.4 Proof of Theorem 9
Let \(\varvec{\theta }_{0}\in \Theta \) be the true value of parameter \( \varvec{\theta }\). It is clear that
$$\begin{aligned} \varvec{m}\left( \widehat{\varvec{\theta }}_{\beta }\right)&= {}\, \varvec{ m}\left( \varvec{\theta }_{0}\right) +\varvec{M}^{T}\left( \widehat{ \varvec{\theta }}_{\beta }\right) \left( \widehat{\varvec{\theta }}_{\beta }- \varvec{\theta }_{0}\right) +o_{p}\left( \left\| \widehat{\varvec{\theta }} _{\beta }-\varvec{\theta }_{0}\right\| \right) \\& = \,\varvec{M}^{T}\left( \widehat{\varvec{\theta }}_{\beta }\right) \left( \widehat{\varvec{\theta }}_{\beta }-\varvec{\theta }_{0}\right) +o_{p}\left( K^{-1/2}\right) . \end{aligned}$$
But, under \(H_{0},\)
$$\begin{aligned} \sqrt{K}\left( \widehat{\varvec{\theta }}_{\beta }-\varvec{\theta }_{0}\right) \underset{K\rightarrow \infty }{\overset{\mathcal {L}}{\longrightarrow }} \mathcal {N}\left( \varvec{0}_{J+1},\varvec{\Sigma }_{\beta }\left( \widehat{\varvec{\theta }}_{\beta }\right) \right) . \end{aligned}$$
Therefore, under \(H_{0},\)
$$\begin{aligned} \sqrt{K}\varvec{m}\left( \widehat{\varvec{\theta }}_{\beta }\right) \underset{K\rightarrow \infty }{\overset{\mathcal {L}}{\longrightarrow }} \mathcal {N}\left( \varvec{0}_{r},\varvec{M}^{T}\left( \varvec{\theta } _{0}\right) \varvec{\Sigma }_{\beta }\left( \widehat{\varvec{\theta }} _{\beta }\right) \varvec{M}\left( \varvec{\theta }_{0}\right) \right) \end{aligned}$$
and taking into account that \(rank(\varvec{M}\left( \varvec{\theta } _{0}\right) )=r\), we get
$$\begin{aligned} K\varvec{m}\left( \widehat{\varvec{\theta }}_{\beta }\right) ^{T}\left( \varvec{M}^{T}\left( \varvec{\theta }_{0}\right) \varvec{\Sigma } _{\beta }\left( \varvec{\theta }_{0}\right) \varvec{M}\left( \varvec{\theta } _{0}\right) \right) ^{-1}\varvec{m}\left( \widehat{\varvec{\theta }} _{\beta }\right) \underset{K\rightarrow \infty }{\overset{\mathcal {L}}{ \longrightarrow }}\chi _{r}^{2}. \end{aligned}$$
But, \(\left( \varvec{M}\left( \widehat{\varvec{\theta }}_{\beta }\right) ^{T}\varvec{\Sigma }_{\beta }\left( \widehat{\varvec{\theta }}_{\beta }\right) \varvec{M}\left( \widehat{\varvec{\theta }}_{\beta }\right) \right) ^{-1}\) is a consistent estimator of \(\left( \varvec{M}\left( \varvec{\theta }_{0}\right) ^{T}\varvec{\Sigma }_{\beta }\left( \varvec{\theta }_{0}\right) \varvec{M}\left( \varvec{\theta }_{0}\right) \right) ^{-1}\). Therefore,
$$\begin{aligned} W_{K}\left( \widehat{\varvec{\theta }}_{\beta }\right) \underset{K\rightarrow \infty }{\overset{\mathcal {L}}{\longrightarrow }}\chi _{r}^{2}. \end{aligned}$$
1.5 Proof of Theorem 10
Under the assumption that
$$\begin{aligned} \widehat{\varvec{\theta }}_{\beta }\underset{K\rightarrow \infty }{\overset{P}{ \longrightarrow }}\varvec{\theta }^{*}, \end{aligned}$$
the asymptotic distribution of \(\ell _{\beta }\left( \widehat{\varvec{\theta }} _{1},\widehat{\varvec{\theta }}_{2}\right) \) coincides with the asymptotic distribution of \(\ell _{\beta }\left( \widehat{\varvec{\theta }}_{1}, \varvec{\theta }^{*}\right) .\) A first-order Taylor expansion of \(\ell _{\beta }\left( \widehat{\varvec{\theta }}_{\beta },\varvec{\theta }\right) \) at \(\widehat{\varvec{\theta }}_{\beta }\), around \(\varvec{\theta }^{*}\), gives
$$\begin{aligned} \left( \ell _{\beta }\left( \widehat{\varvec{\theta }}_{\beta },\varvec{\theta } ^{*}\right) -\ell _{\beta }\left( \varvec{\theta }^{*},\varvec{\theta } ^{*}\right) \right) =\left. \frac{\partial \ell _{\beta }\left( \varvec{\theta },\varvec{\theta }_{*}\right) }{\partial \varvec{\theta }^{T}} \right| _{\varvec{\theta }=\varvec{\theta }_{*}}\left( \widehat{ \varvec{\theta }}_{\beta }-\varvec{\theta }^{*}\right) +o_{p}(K^{-1/2}). \end{aligned}$$
Now, the result follows since
$$\begin{aligned} \sqrt{K}\left( \widehat{\varvec{\theta }}_{\beta }-\varvec{\theta }^{*}\right) \underset{K\rightarrow \infty }{\overset{\mathcal {L}}{ \longrightarrow }}\mathcal {N}\left( \varvec{0}_{J+1},\varvec{\Sigma } _{\beta }\left( \varvec{\theta }^{*}\right) \right) . \end{aligned}$$
1.6 Proof of Theorem 12
A Taylor series expansion of \(\varvec{m}(\widehat{\varvec{\theta }}_{\beta })\) around \(\varvec{\theta }_{n}\) yields
$$\begin{aligned} \varvec{m}(\widehat{\varvec{\theta }}_{\beta })=\varvec{m}(\varvec{ \theta }_{n})+\varvec{M}^{T}(\varvec{\theta }_{n})(\widehat{\varvec{\theta }} _{\beta }-\varvec{\theta }_{n})+o\left( \left\| \widehat{\varvec{\theta }} _{\beta }-\varvec{\theta }_{n}\right\| \right) . \end{aligned}$$
From (23), we have
$$\begin{aligned} \varvec{m}(\widehat{\varvec{\theta }}_{\beta })=K^{-1/2}\varvec{M}^{T}( \varvec{\theta }_{0})\varvec{d}+\varvec{M}^{T}(\varvec{\theta }_{n})( \widehat{\varvec{\theta }}_{\beta }-\varvec{\theta }_{n})+o\left( \left\| \widehat{\varvec{\theta }}_{\beta }-\varvec{\theta }_{n}\right\| \right) +o\left( \left\| \varvec{\theta }_{n}-\varvec{\theta }_{0}\right\| \right) . \end{aligned}$$
As
$$\begin{aligned} \sqrt{K}(\widehat{\varvec{\theta }}_{\beta }-\varvec{\theta }_{n})\underset{ n\rightarrow \infty }{\overset{\mathcal {L}}{\longrightarrow }}\mathcal {N}( \varvec{0}_{J+1},\varvec{\Sigma }_{\beta }(\varvec{\theta }_{0})) \end{aligned}$$
and \(\sqrt{K}\left( o\left( \left\| \widehat{\varvec{\theta }}_{\beta }- \varvec{\theta }_{n}\right\| \right) +o\left( \left\| \varvec{\theta }_{n}- \varvec{\theta }_{0}\right\| \right) \right) =o_{p}\left( 1\right) \), we have
$$\begin{aligned} \sqrt{n}\varvec{m}(\widehat{\varvec{\theta }}_{\beta })\underset{ n\rightarrow \infty }{\overset{\mathcal {L}}{\longrightarrow }}\mathcal {N}( \varvec{M}^{T}(\varvec{\theta }_{0})\,\varvec{d},\varvec{M}^{T}( \varvec{\theta }_{0})\varvec{\Sigma }_{\beta }(\varvec{\theta }_{0}) \varvec{M}(\varvec{\theta }_{0})). \end{aligned}$$
We can observe from the relationship \(\varvec{d}^{*}=\varvec{M}( \varvec{\theta }_{0})^{T}\varvec{d}\), if \(\varvec{m}(\varvec{\theta } _{n})=n^{-1/2}\varvec{d}^{*}\), that
$$\begin{aligned} \sqrt{n}\varvec{m}(\widehat{\varvec{\theta }}_{\beta })\underset{ n\rightarrow \infty }{\overset{\mathcal {L}}{\longrightarrow }}\mathcal {N}( \varvec{d}^{*},\varvec{M}^{T}(\varvec{\theta }_{0})\varvec{ \Sigma }_{\beta }(\varvec{\theta }_{0})\varvec{M}(\varvec{\theta }_{0})). \end{aligned}$$
So, the quadratic form here is
$$\begin{aligned} W_{K}\left( \widehat{\varvec{\theta }}_{\beta }\right) =\varvec{Z}^{T} \varvec{Z} \end{aligned}$$
with
$$\begin{aligned} \varvec{Z}=\sqrt{n}\varvec{m}(\widehat{\varvec{\theta }}_{\beta })\left( \varvec{M}^{T}(\varvec{\theta }_{0})\varvec{\Sigma }_{\beta }( \varvec{\theta }_{0})\varvec{M}(\varvec{\theta }_{0})\right) ^{-1/2} \end{aligned}$$
and
$$\begin{aligned} \varvec{Z}\underset{n\rightarrow \infty }{\overset{\mathcal {L}}{ \longrightarrow }}\mathcal {N}\left( \left( \varvec{M}^{T}(\varvec{\theta } _{0})\varvec{\Sigma }_{\beta }(\varvec{\theta }_{0})\varvec{M}( \varvec{\theta }_{0})\right) ^{-1/2}\varvec{M}(\varvec{\theta }_{0})^{T} \varvec{d},\varvec{I}_{r}\right) , \end{aligned}$$
where \(\varvec{I}_{r}\) is the identity matrix or order r. Hence, the required the result follows immediately, and the non-centrality parameter is
$$\begin{aligned} \varvec{d}^{T}\varvec{M}(\varvec{\theta }_{0})\left( \varvec{M} ^{T}(\varvec{\theta }_{0})\varvec{\Sigma }_{\beta }(\varvec{\theta }_{0}) \varvec{M}(\varvec{\theta }_{0})\right) ^{-1}\varvec{M}(\varvec{\theta } _{0})^{T}\varvec{d}=\varvec{d}^{*T}\left( \varvec{M}^{T}( \varvec{\theta }_{0})\varvec{\Sigma }_{\beta }(\varvec{\theta }_{0}) \varvec{M}(\varvec{\theta }_{0})\right) ^{-1}\varvec{d}^{*}. \end{aligned}$$
1.7 Proof of Theorem 13
The influence function of \(W_K(\widehat{\varvec{\theta }}_{\beta })\), with respect to the \(j_0\)th observation of the \(i_0\)th group of observations, is defined as
$$\begin{aligned}\mathcal {IF}\left( i_{0},j_{0},x,T_{\beta },\varvec{F}_{\underline{K},\varvec{\theta }}\right) =\left. \frac{\partial W_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) }{\partial \varepsilon }\right| _{\varepsilon =0^{+}}\end{aligned}$$
where
$$\begin{aligned}&\frac{\partial W_{K}\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) }{\partial \varepsilon }\\&\quad =2\varvec{m}\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \left( \varvec{M}^T\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \varvec{\Sigma }\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \varvec{M}\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \right) ^{-1}\\&\qquad \times \varvec{M}^T\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \frac{\partial \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) }{\partial \varepsilon }\\&\qquad +\varvec{m}^T\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \frac{\partial }{\partial \varepsilon }\left( \varvec{M}^T\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \varvec{\Sigma }\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \varvec{M}\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \right) ^{-1} \\&\qquad \times \varvec{m}\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \end{aligned}$$
and
$$\begin{aligned}&\left. \frac{\partial W_{K}\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) }{\partial \varepsilon }\right| _{\varepsilon =0^{+}}\\&\quad =2\varvec{m}\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x}\right) \right) \left( \varvec{M}^T\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x}\right) \right) \varvec{\Sigma }\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x}\right) \right) \varvec{M}\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x}\right) \right) \right) ^{-1}\\&\qquad \times \varvec{M}^T\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \mathcal {IF}\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \\&\qquad +\varvec{m}^T\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x}\right) \right) \left. \frac{\partial }{\partial \varepsilon }\left( \varvec{M}^T\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \varvec{\Sigma }\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x}\right) \right) \varvec{M}\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x}\right) \right) \right) ^{-1}\right| _{\varepsilon =0^{+}} \\&\qquad \times \varvec{m}\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x}\right) \right) \end{aligned}$$
For \(\varvec{\theta }=\varvec{\theta }_0\), \(\varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta }_0,x}\right) =\varvec{\theta }_0\) and \(\varvec{m}^T(\varvec{\theta }_0)=0\). Therefore,
$$\begin{aligned} \mathcal {IF}\left( i_{0},j_{0},x,T_{\beta },\varvec{F}_{\underline{K},\varvec{\theta }_0}\right) =0, \end{aligned}$$
in a similar way, we have
$$\begin{aligned} \mathcal {IF}\left( i_{0},j_{0},\underline{x},T_{\beta },\varvec{F}_{\underline{K},\varvec{\theta }_0}\right) =0 \end{aligned}$$
In order to get the second-order influence function, with respect to the jth observation of the ith group of observation, we must get
$$\begin{aligned} \mathcal {IF}_2\left( i_{0},j_{0},x,T_{\beta },\varvec{F}_{\underline{K},\varvec{\theta }_0}\right) =\left. \frac{\partial ^2}{\partial \varepsilon } W_K(\varvec{F}_{\underline{K}},\varvec{\theta }_0,x,\varepsilon )\right| _{\varepsilon =0^{+}} \end{aligned}$$
We can express
$$\begin{aligned} \frac{\partial ^2 W_K(\varvec{F}_{\underline{K}},\varvec{\theta },x,\varepsilon )}{\partial \varepsilon }&=\varvec{l}(K,\varvec{\theta },x,\varepsilon ) +2\frac{\partial \varvec{U}_\beta (\varvec{F}_{\underline{K}},\varvec{\theta },x,\varepsilon )}{\partial \varepsilon }\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \\&\quad \times \left( \varvec{M}^T\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \varvec{\Sigma }\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \right. \\&\qquad \left. \varvec{M}\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \right) ^{-1}\varvec{M}^T\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \\&\quad \times \frac{\partial \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) }{\partial \varepsilon } \end{aligned}$$
With \(\varvec{l}(K,\varvec{\theta },x,\varepsilon )\), we denote all the terms which contain the expression \(\varvec{m}\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta },x,\varepsilon }\right) \right) \), because for \(\varvec{\theta }=\varvec{\theta }_0\) and \(\varepsilon =0\), we have \(\varvec{m}\left( \varvec{U}_{\beta }\left( \varvec{F}_{\underline{K},\varvec{\theta }_0,x}\right) \right) =0\). Therefore, we have
$$\begin{aligned} \mathcal {IF}_2\left( i_{0},j_{0},x,T_{\beta },\varvec{F}_{\underline{K},\varvec{\theta }_0}\right)&=\left. \frac{\partial ^2}{\partial \varepsilon } W_K(\varvec{F}_{\underline{K}},\varvec{\theta }_0,x,\varepsilon )\right| _{\varepsilon =0^{+}}\\&=2\mathcal {IF}^T\left( i_{0},j_{0},x,T_{\beta },\varvec{F}_{\underline{K},\varvec{\theta }_0}\right) \varvec{M}(\varvec{\theta }_0)\left( \varvec{M}^T(\varvec{\theta }_0)\varvec{\Sigma }(\varvec{\theta }_0)\varvec{M}(\varvec{\theta }_0) \right) ^{-1}\\&\quad \times \varvec{M}^T(\varvec{\theta }_0)\mathcal {IF}\left( i_{0},j_{0},x,T_{\beta },\varvec{F}_{\underline{K},\varvec{\theta }_0}\right) . \end{aligned}$$
In a similar way, we obtain the expression of \(\mathcal {IF}_2\left( \underline{x},T_{\beta },\varvec{F}_{\underline{K},\varvec{\theta }_0}\right) \).