Detecting the direction of a signal on high-dimensional spheres: non-null and Le Cam optimality results

Abstract

We consider one of the most important problems in directional statistics, namely the problem of testing the null hypothesis that the spike direction ${\pmb \theta }$ of a Fisher–von Mises–Langevin distribution on the p-dimensional unit hypersphere is equal to a given direction ${\pmb \theta }_0$. After a reduction through invariance arguments, we derive local asymptotic normality (LAN) results in a general high-dimensional framework where the dimension $p_n$ goes to infinity at an arbitrary rate with the sample size n, and where the concentration $\kappa _n$ behaves in a completely free way with n, which offers a spectrum of problems ranging from arbitrarily easy to arbitrarily challenging ones. We identify various asymptotic regimes, depending on the convergence/divergence properties of $(\kappa _n)$, that yield different contiguity rates and different limiting experiments. In each regime, we derive Le Cam optimal tests under specified $\kappa _n$ and we compute, from the Le Cam third lemma, asymptotic powers of the classical Watson test under contiguous alternatives. We further establish LAN results with respect to both spike direction and concentration, which allows us to discuss optimality also under unspecified $\kappa _n$. To investigate the non-null behavior of the Watson test outside the parametric framework above, we derive its local asymptotic powers through martingale CLTs in the broader, semiparametric, model of rotationally symmetric distributions. A Monte Carlo study shows that the finite-sample behaviors of the various tests remarkably agree with our asymptotic results.

Appendices

Technical proofs for Sect. 2

The proof of Theorem 1 requires the following preliminary results.

Lemma 1

Let $(p_n)$ be a sequence of integers diverging to infinity and $(\kappa _n)$ be an arbitrary sequence in $(0,\infty )$. Let $L_n:=\sum _{i=1}^n V_{ni}^2/(n f_{n2})$, where we used the notation $V_{ni}=(1-(\mathbf {X}_{ni}'{\pmb \theta }_{n0})^2)^{1/2}$ and $f_{n2}=\mathrm{E}[V_{n1}^2]$. Then, $\mathrm{E} \big [ ( L_n - 1 )^2 \big ] = o(p_n^{-1})$ as $n\rightarrow \infty$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$.

Proof of Lemma 1

Since

$$\begin{aligned} \mathrm{E} \Bigg [ \bigg ( \frac{\sum _{i=1}^n V_{ni}^2}{n f_{n2}} - 1 \bigg )^2 \Bigg ]= & {} \frac{1}{f_{n2}^2} \, \mathrm{E} \Bigg [ \Bigg ( \frac{1}{n} \sum _{i=1}^{n} V_{ni}^2 - \mathrm{E}[V_{n1}^2] \Bigg )^2 \Bigg ] \nonumber \\= & {} \frac{1}{f_{n2}^2} \, \mathrm{Var}\Bigg [\frac{1}{n} \sum _{i=1}^{n} V_{ni}^2\Bigg ] = \frac{\mathrm{Var}[V_{n1}^2]}{nf_{n2}^2} = \frac{f_{n4}-f_{n2}^2}{nf_{n2}^2} \end{aligned}$$

(recall that $f_{n4}:=\mathrm{E}[V_{n1}^4]$), it is sufficient to prove that

$$\begin{aligned} \frac{f_{n4}-f_{n2}^2}{f_{n2}^2} = O(p_n^{-1}) . \end{aligned}$$

(A.1)

Now, the expression for $f_{n4}/f_{n2}^2$ in page 82 of [25] yields

$$\begin{aligned} \bigg | \frac{f_{n4}-f_{n2}^2}{f_{n2}^2} \bigg |= & {} \bigg | \frac{(p_n+1)\mathcal {I}_{\frac{p_n}{2}+1}(\kappa _n)\mathcal {I}_{\frac{p_n}{2}-1}(\kappa _n)}{(p_n-1)(\mathcal {I}_{\frac{p_n}{2}}(\kappa _n))^2} - 1 \bigg |\\= & {} \bigg |\frac{(p_n+1)(\mathcal {I}_{\frac{p_n}{2}+1}(\kappa _n)\mathcal {I}_{\frac{p_n}{2}-1}(\kappa _n)-(\mathcal {I}_{\frac{p_n}{2}}(\kappa _n))^2)}{(p_n-1)(\mathcal {I}_{\frac{p_n}{2}}(\kappa _n))^2} + \frac{2}{p_n-1} \bigg |\\\le & {} \frac{3|\mathcal {I}_{\frac{p_n}{2}+1}(\kappa _n)\mathcal {I}_{\frac{p_n}{2}-1}(\kappa _n)-(\mathcal {I}_{\frac{p_n}{2}}(\kappa _n))^2|}{(\mathcal {I}_{\frac{p_n}{2}}(\kappa _n))^2} + \frac{2}{p_n-1} \cdot \end{aligned}$$

Since $|\mathcal {I}_{\frac{p_n}{2}+1}(\kappa _n)\mathcal {I}_{\frac{p_n}{2}-1}(\kappa _n)-(\mathcal {I}_{\frac{p_n}{2}}(\kappa _n))^2|\le (\mathcal {I}_{\frac{p_n}{2}}(\kappa _n))^2/(\frac{p_n}{2}+1)$ (see (3.1)–(3.2) in [21]), the result follows. $\square$

Lemma 2

Let $(p_n)$ be a sequence of integers that diverges to infinity and $(\kappa _n)$ be an arbitrary sequence in $(0,\infty )$. Let $({\pmb \theta }_{n0})$ be a sequence such that ${\pmb \theta }_{n0}$ belongs to $\mathcal {S}^{p_n-1}$ for any n. Consider the random variables ${\widetilde{W}}_n$ and $Z_n$ introduced in Theorem 1. Then, $({\widetilde{W}}_n,Z_n)'$ is asymptotically standard bivariate normal under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$.

Proof of Lemma 2

Throughout the proof, expectations and variances are under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$ and stochastic convergences are as $n\rightarrow \infty$ under the same sequence of hypotheses, whereas $U_{ni}$, $V_{ni}$ and $\mathbf {S}_{ni}$ refer to the tangent-normal decomposition of $\mathbf {X}_{ni}$ with respect to ${\pmb \theta }_{n0}$. Letting then

$$\begin{aligned} W^*_n = \frac{\sqrt{2(p_n-1)}}{nf_{n2}} \sum _{1\le i < j\le n} V_{ni} V_{nj} \mathbf {S}_{ni}^{\prime }\mathbf {S}_{nj} , \end{aligned}$$

assume that $(W^*_n,Z_n)'$ is asymptotically standard bivariate normal. Then,

$$\begin{aligned} {\widetilde{W}}_n - W^*_n= & {} \Bigg [ \frac{\sqrt{2(p_n-1)}}{\sum _{i=1}^n V_{ni}^2} - \frac{\sqrt{2(p_n-1)}}{nf_{n2}} \Bigg ] \sum _{1\le i< j\le n} V_{ni} V_{nj} \mathbf {S}_{ni}^{\prime }\mathbf {S}_{nj}\nonumber \\= & {} \Bigg [ 1 - \frac{\sum _{i=1}^n V_{ni}^2}{nf_{n2}} \Bigg ] \times \frac{nf_{n2}}{\sum _{i=1}^n V_{ni}^2} \times \Bigg ( \frac{\sqrt{2(p_n-1)}}{nf_{n2}} \sum _{1\le i < j\le n} V_{ni} V_{nj} \mathbf {S}_{ni}^{\prime }\mathbf {S}_{nj} \Bigg ) \nonumber \\= & {} \frac{1-L_n}{L_n} \, W^*_n , \end{aligned}$$

(A.2)

where $L_n$ was introduced in Lemma 1. This lemma implies that $L_n-1$, hence also $(1-L_n)/L_n$, is $o_{\mathrm{P}}(1)$. If $(W^*_n,Z_n)'$ is indeed asymptotically standard bivariate normal, then we conclude that ${\widetilde{W}}_n-W^*_n$ is $o_{\mathrm{P}}(1)$, so that $({\widetilde{W}}_n,Z_n)'$ itself is asymptotically standard bivariate normal.

It is therefore sufficient to show that $(W^*_n,Z_n)'$ is asymptotically standard bivariate normal. We will do this by fixing $\gamma$ and $\eta$ such that $\gamma ^2+\eta ^2=1$ and by using a classical martingale Central Limit Theorem to show that $D_n:=\gamma W^*_n + \eta Z_n$ is asymptotically standard normal. To do so, let $\mathcal{F}_{n\ell }$ be the $\sigma$-algebra generated by $\mathbf {X}_{n1}, \ldots , \mathbf {X}_{n \ell }$ and denote by $\mathrm{E}_{n\ell }[.]$ the conditional expectation with respect to $\mathcal{F}_{n\ell }$. Define $D_{n\ell } := \mathrm{E}_{n\ell } [D_n]-\mathrm{E}_{n,\ell -1} [D_n]$ for $\ell =1,\ldots ,n$ and $D_{n\ell }=0$ for $\ell >n$. It is then easy to check that $D_{n\ell }=\gamma W^*_{n\ell } + \eta Z_{n\ell }$, with

$$\begin{aligned} W^*_{n\ell } := \frac{\sqrt{2(p_n-1)}}{n f_{n2}} \, \sum _{i=1}^{\ell -1} V_{ni} V_{n\ell } \mathbf {S}_{ni}^{\prime }\mathbf {S}_{n\ell } \quad \text { and } \quad Z_{n\ell } := \frac{U_{n\ell }-e_{n1}}{\sqrt{n{\tilde{e}}_{n2}}} \end{aligned}$$

for $\ell =1,\ldots ,n$ and $W^*_{n\ell }=0=Z_{n\ell }$ for $\ell >n$ ($W^*_{n1}$ is also to be understood as zero). To conclude from the martingale Central Limit Theorem in Theorem 35.12 from [6] that $D_n=\sum _{\ell =1}^\infty D_{n\ell }$ is indeed asymptotically standard normal, we need to show that (a) $\sum _{\ell =1}^n \sigma ^2_{n\ell }\rightarrow 1$ in probability, with $\sigma ^2_{n\ell }:=\mathrm{E}_{n,\ell -1} [D_{n\ell }^2]$, and that (b) $\sum _{\ell =1}^n \mathrm{E}[D_{n\ell }^2 \, {{\mathbb {I}}}[| D_{n\ell }| > \varepsilon ]]\rightarrow 0$ for any $\varepsilon >0$. Clearly, for $\ell =1,\ldots ,n$,

$$\begin{aligned} \sigma ^2_{n\ell }= & {} \gamma ^2 \mathrm{E}_{n,\ell -1} [(W^*_{n\ell })^2] + \eta ^2 \mathrm{E}_{n,\ell -1} [Z_{n\ell }^2] + 2\gamma \eta \mathrm{E}_{n,\ell -1} [W^*_{n\ell } Z_{n\ell }] \nonumber \\= & {} \gamma ^2 \mathrm{E}_{n,\ell -1} [(W^*_{n\ell })^2] + \frac{\eta ^2}{n} , \end{aligned}$$

(A.3)

so that (a) follows from Lemma A.1 in [25]. We may thus focus on (b). Since

$$\begin{aligned} \mathrm{E}_{n,\ell -1}[(W^*_{n\ell })^2] = 2 (n^2 f_{n2})^{-1} \sum _{i,j=1}^{\ell -1} V_{ni} V_{nj} \mathbf {S}_{ni}^{\prime }\mathbf {S}_{nj} , \end{aligned}$$

we obtain $\mathrm{Var}[D_{n\ell }] =\mathrm{E}[\sigma ^2_{n\ell }] =\gamma ^2 \mathrm{E}[(W^*_{n\ell })^2] + (\eta ^2/n) =2\gamma ^2 (\ell -1)/n^2 + (\eta ^2/n) \le 2/n$, which yields that there exists a constant C such that, for any $\varepsilon >0$,

$$\begin{aligned} \sum _{\ell =1}^n \mathrm{E}[D_{n\ell }^2 \; {{\mathbb {I}}}[| D_{n\ell }|> \varepsilon ]]\le & {} \sum _{\ell =1}^n \sqrt{\mathrm{E}[D_{n\ell }^4]} \, \sqrt{\mathrm{P}[|D_{n\ell }| > \varepsilon ]}\\\le & {} \frac{1}{\varepsilon } \sum _{\ell =1}^n \sqrt{\mathrm{E}[D_{n\ell }^4]} \, \sqrt{\mathrm{Var}[D_{n\ell }]} \le \frac{\sqrt{2}}{\sqrt{n}\varepsilon } \sum _{\ell =1}^n \sqrt{\mathrm{E}[D_{n\ell }^4]}\\\le & {} \frac{C}{\sqrt{n}\varepsilon } \sum _{\ell =1}^n \sqrt{\mathrm{E}[(W^*_{n\ell })^4]} + \frac{C}{\sqrt{n}\varepsilon } \sum _{\ell =1}^n \sqrt{\mathrm{E}[Z_{n\ell }^4]} . \end{aligned}$$

From (A.9) in [25], we then obtain

$$\begin{aligned} \sum _{\ell =1}^n \mathrm{E}[D_{n\ell }^2 \; {{\mathbb {I}}}[| D_{n\ell }| > \varepsilon ]]\le & {} \frac{C}{\sqrt{n}\varepsilon } \sum _{\ell =1}^n \sqrt{\frac{12}{n^4} \bigg ( \ell \, \frac{f_{n4}^2}{f_{n2}^4} + \ell ^2 \, \frac{f_{n4}}{f_{n2}^2} \bigg ) } + \frac{C\sqrt{n}}{\varepsilon } \sqrt{\frac{{\tilde{e}}_{n4}}{n^2{\tilde{e}}_{n2}^2}}\\\le & {} \frac{\sqrt{12}C}{\varepsilon } \sqrt{ \Big (\frac{f_{n4}}{nf_{n2}^2}\Big )^2 + \frac{f_{n4}}{nf_{n2}^2} } + \frac{C}{\varepsilon } \sqrt{\frac{{\tilde{e}}_{n4}}{n{\tilde{e}}_{n2}^2}} \cdot \end{aligned}$$

The result therefore follows from the fact that both $f_{n4}/f_{n2}^2$ and ${\tilde{e}}_{n4}/{\tilde{e}}_{n2}^2$ are upper-bounded by a universal constant; see Theorem S.2.1 in [13]. $\square$

Lemma 3

Let $(\nu _n)$ be a sequence in $(0,\infty )$ that diverges to $\infty$, $(a_n)$, $(b_n)$ be sequences in $(0,\infty )$ such that $\liminf a_n>0$, $b_n/\nu _n\rightarrow \xi \in [0,\infty )$ and $b_n^6=o(a_n^4\nu _n^5)$. Let $T_n$ be a sequence of random variables that is $O_{\mathrm{P}}(1)$. Then, writing,

$$\begin{aligned} H_{\nu }(x) := \frac{\int _{-1}^1 (1-t^2)^{\nu -\frac{1}{2}} \exp (x t)\, dt}{\int _{-1}^1 (1-t^2)^{\nu -\frac{1}{2}}\, dt} = \frac{\varGamma (\nu +1)\mathcal {I}_{\nu }(x)}{(x/2)^{\nu }} , \end{aligned}$$

we have that

$$\begin{aligned} a_n^2 \log H_{\nu _n}\Big (\frac{b_n T_n}{a_n}\Big ) = \frac{b_n^2 T_n^2}{4\nu _n} - \frac{b_n^4 T_n^4}{32\nu _n^3a_n^2} + \frac{\xi ^2 T_n^2}{4} + o_{\mathrm{P}}(1) \end{aligned}$$

as $n\rightarrow \infty$.

Proof of Lemma 3

The proof is based on the bounds

$$\begin{aligned} S_{\nu +\frac{1}{2},\nu +\frac{3}{2}}(x) \le \log H_\nu (x) \le S_{\nu ,\nu +2}(x) \end{aligned}$$

for any $x>0$, with $S_{\alpha ,\beta }(x) := \sqrt{x^2+\beta ^2}-\beta -\alpha \log ((\alpha +\sqrt{x^2+\beta ^2})/(\alpha +\beta ))$; see (5) in [20]. Consider

$$\begin{aligned} G_\nu (x) := \log H_\nu (x) - \frac{x^2}{4\nu } + \frac{x^4}{32\nu ^3} +\frac{x^2}{4\nu ^2} , \end{aligned}$$

along with its resulting lower and upper bounds

$$\begin{aligned} G^{\mathrm{low}}_\nu (x) := S_{\nu +\frac{1}{2},\nu +\frac{3}{2}}(x) - \frac{x^2}{4\nu } + \frac{x^4}{32\nu ^3} +\frac{x^2}{4\nu ^2} \end{aligned}$$

and

$$\begin{aligned} G^{\mathrm{up}}_\nu (x) := S_{\nu ,\nu +2}(x) - \frac{x^2}{4\nu } + \frac{x^4}{32\nu ^3} +\frac{x^2}{4\nu ^2} \cdot \end{aligned}$$

We prove the lemma by establishing that

$$\begin{aligned} a_n^2 G^{\mathrm{low/up}}_{\nu _n}\Big (\frac{b_n T_n}{a_n}\Big ) = o_{\mathrm{P}}(1) . \end{aligned}$$

(A.4)

To do so, we expand the log term in $G^{\mathrm{low/up}}_\nu (x)$ as $\log x=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3c^3}(x-1)^3$ with $c\in (1,x)$ (note that the argument of these log terms is larger than or equal to one), and we write $G^{\mathrm{low/up}}_\nu (x)=G^{\mathrm{low/up,1}}_\nu (x)+ G^{\mathrm{low/up,2}}_\nu (x)$, with

$$\begin{aligned} G^{\mathrm{low,1}}_\nu (x):= & {} \sqrt{x^2+\left( {\textstyle {\nu +\frac{3}{2}}}\right) ^2} - \left( {\textstyle {\nu +\frac{3}{2}}}\right) \nonumber \\&-\,\left( {\textstyle {\nu +\frac{1}{2}}}\right) \Bigg [ \Bigg ( \frac{\left( {\textstyle {\nu +\frac{1}{2}}}\right) +\sqrt{x^2+\left( {\textstyle {\nu +\frac{3}{2}}}\right) ^2}}{2(\nu +1)} -1 \Bigg )\\&-\,\frac{1}{2} \Bigg ( \frac{\left( {\textstyle {\nu +\frac{1}{2}}}\right) +\sqrt{x^2+\left( {\textstyle {\nu +\frac{3}{2}}}\right) ^2}}{2(\nu +1)} -1 \Bigg )^2 \Bigg ] - \frac{x^2}{4\nu } + \frac{x^4}{32\nu ^3} +\frac{x^2}{4\nu ^2} ,\\ G^{\mathrm{up,1}}_\nu (x):= & {} \sqrt{x^2+(\nu +2)^2} - (\nu +2) - \nu \Bigg [ \Bigg ( \frac{\nu +\sqrt{x^2+(\nu +2)^2}}{2(\nu +1)} -1 \Bigg )\\&-\,\frac{1}{2} \Bigg ( \frac{\nu +\sqrt{x^2+(\nu +2)^2}}{2(\nu +1)} -1 \Bigg )^2 \Bigg ] - \frac{x^2}{4\nu } + \frac{x^4}{32\nu ^3} +\frac{x^2}{4\nu ^2} ,\\ G^{\mathrm{low,2}}_\nu (x):= & {} - \frac{{\textstyle {\nu +\frac{1}{2}}}}{3(c^{\mathrm{low}})^3} \left( \frac{\left( {\textstyle {\nu +\frac{1}{2}}}\right) +\sqrt{x^2+\left( {\textstyle {\nu +\frac{3}{2}}}\right) ^2}}{2(\nu +1)} -1 \right) ^3 , \end{aligned}$$

and

$$\begin{aligned} G^{\mathrm{up,2}}_\nu (x) := - \frac{\nu }{3(c^{\mathrm{up}})^3} \Bigg ( \frac{\nu +\sqrt{x^2+(\nu +2)^2}}{2(\nu +1)} -1 \Bigg )^3 . \end{aligned}$$

Routine yet tedious computations allow to show that

$$\begin{aligned} G^{\mathrm{low,1}}_\nu (x)= & {} \frac{x^2}{4\nu ^2(\nu +1)} + \frac{(4\nu ^2+5\nu +2)x^4}{32\nu ^3(\nu +1)^2(\nu +2)} \nonumber \\&+\,\left( 1 - \frac{4(1+\frac{2}{\nu })(1+\frac{3}{2\nu })}{\Big ( \left( 1+\frac{3}{2\nu }\right) +\sqrt{\left( \frac{x}{\nu }\right) ^2+\left( 1+\frac{3}{2\nu }\right) ^2}\,\Big )^2} \right) \frac{x^4}{32(\nu +1)^2(\nu +2)}\nonumber \\ \end{aligned}$$

(A.5)

and

$$\begin{aligned} G^{\mathrm{up,1}}_\nu (x)= & {} \frac{x^2}{4\nu ^2(\nu +1)} + \frac{(4\nu ^2+5\nu +2)x^4}{32\nu ^3(\nu +1)^2(\nu +2)} \nonumber \\&+ \,\left( 1 - \frac{4}{\left( 1+\sqrt{\left( \frac{x}{\nu +2}\right) ^2+1}\, \right) ^2} \right) \frac{x^4}{32(\nu +1)^2(\nu +2)} \cdot \end{aligned}$$

(A.6)

Since both $c^{\mathrm{low}}$ and $c^{\mathrm{up}}$ are larger than one, we easily obtain

$$\begin{aligned} |G^{\mathrm{low,2}}_\nu (x)| \le {\textstyle { \left( \left( 1+\frac{3}{2\nu }\right) +\sqrt{\left( \frac{x}{\nu }\right) ^2+\left( 1+\frac{3}{2\nu }\right) ^2}\right) ^{-3} }} \frac{\left( \nu +\frac{1}{2}\right) x^6}{24\nu ^3(\nu +1)^3} \end{aligned}$$

(A.7)

and

$$\begin{aligned} |G^{\mathrm{up,2}}_\nu (x)| \le {\textstyle { \left( \left( 1+\frac{2}{\nu }\right) +\sqrt{\left( \frac{x}{\nu }\right) ^2+\left( 1+\frac{2}{\nu }\right) ^2}\right) ^{-3} }} \frac{x^6}{24\nu ^2(\nu +1)^3} \cdot \end{aligned}$$

(A.8)

Using the mean value theorem to control the last term in the righthand sides of (A.5)–(A.6), it directly follows from (A.5)–(A.8) that, under the assumptions of the lemma,

$$\begin{aligned} a_n^2 G^{\mathrm{low/up},1}_{\nu _n}\Big (\frac{b_n T_n}{a_n}\Big ) = o_{\mathrm{P}}(1) \quad \text {and} \quad a_n^2 G^{\mathrm{low/up},2}_{\nu _n}\Big (\frac{b_n T_n}{a_n}\Big ) = o_{\mathrm{P}}(1) , \end{aligned}$$

which proves (A.4), hence establishes the result. $\square$

Proof of Theorem 1

Throughout the proof, distributions and expectations are under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$ and stochastic convergences are as $n\rightarrow \infty$ under the same sequence of hypotheses. By using the fact that $\mathbf {O}{\pmb \theta }_{n0}={\pmb \theta }_{n0}=\mathbf {O}^{\prime }{\pmb \theta }_{n0}$ for any $\mathbf {O}\in SO_{p_n}({\pmb \theta }_{n0})$ and by decomposing ${\pmb \tau }_n$ into $({\pmb \tau }_n^{\prime }{\pmb \theta }_{n0}) {\pmb \theta }_{n0}+ \varPi _{{\pmb \theta }_{n0}} {\pmb \tau }_n$, with $\varPi _{{\pmb \theta }_{n0}}:=\mathbf {I}_{p_n}-{\pmb \theta }_{n0}{\pmb \theta }_{n0}^{\prime }$, (2.5) yields

$$\begin{aligned}&\frac{d\mathrm{P}^{(n)\mathbf {T}_n}_{1-\frac{1}{2} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2,\kappa _n}}{d{m_n}} \\&\quad = \frac{c_{p_n,\kappa _n}^n}{\omega _{p_n-1}^n} \, \int _{SO_{p_n}({\pmb \theta }_{n0})} \exp \big ( n \kappa _n {\bar{\mathbf {X}}}_{n}^{\prime }\mathbf {O}^{\prime }({\pmb \theta }_{n0}+\nu _n{\pmb \tau }_n) \big ) \,d\mathbf {O}\\&\quad = \frac{c_{p_n,\kappa _n}^n}{\omega _{p_n-1}^n} \exp \big ( n \kappa _n {\bar{\mathbf {X}}}_{n}^{\prime }{\pmb \theta }_{n0} \big ) \int _{SO_{p_n}({\pmb \theta }_{n0})} \exp \big ( n \kappa _n\nu _n {\bar{\mathbf {X}}}_{n}^{\prime }\mathbf {O}^{\prime }{\pmb \tau }_n \big ) \,d\mathbf {O}\\&\quad = \frac{c_{p_n,\kappa _n}^n}{\omega _{p_n-1}^n} \exp \big ( n \kappa _n {\bar{\mathbf {X}}}_{n}^{\prime }{\pmb \theta }_{n0} \big ) \int _{SO_{p_n}({\pmb \theta }_{n0})} \exp \big ( n \kappa _n\nu _n {\bar{\mathbf {X}}}_{n}^{\prime }[({\pmb \tau }_n^{\prime }{\pmb \theta }_{n0}) {\pmb \theta }_{n0}+ \mathbf {O}^{\prime }\varPi _{{\pmb \theta }_{n0}} {\pmb \tau }_n] \big ) \,d\mathbf {O}\\&\quad = \frac{c_{p_n,\kappa _n}^n}{\omega _{p_n-1}^n} \exp \big ( n \kappa _n (1+\nu _n ({\pmb \tau }_n^{\prime }{\pmb \theta }_{n0})) {\bar{\mathbf {X}}}_{n}^{\prime }{\pmb \theta }_{n0} \big ) \int _{SO_{p_n}({\pmb \theta }_{n0})} \exp \big ( n \kappa _n\nu _n {\bar{\mathbf {X}}}_{n}^{\prime }\mathbf {O}^{\prime }\varPi _{{\pmb \theta }_{n0}} {\pmb \tau }_n \big ) \,d\mathbf {O}. \end{aligned}$$

Now, since $\mathbf{O}^{\prime }\varPi _{{\pmb \theta }_{n0}}=\mathbf{O}^{\prime }\varPi _{{\pmb \theta }_{n0}}^2=\varPi _{{\pmb \theta }_{n0}} \mathbf{O}^{\prime }\varPi _{{\pmb \theta }_{n0}}$,

$$\begin{aligned}&\int _{SO_{p_n}({\pmb \theta }_{n0})} \exp \big ( n \kappa _n\nu _n {\bar{\mathbf {X}}}_{n}^{\prime }\mathbf {O}^{\prime }\varPi _{{\pmb \theta }_{n0}} {\pmb \tau }_n \big ) \,d\mathbf {O}\\&\quad = \int _{SO_{p_n}({\pmb \theta }_{n0})} \exp \big ( n \kappa _n\nu _n {\bar{\mathbf {X}}}_{n}^{\prime }\varPi _{{\pmb \theta }_{n0}} \mathbf {O}^{\prime }\varPi _{{\pmb \theta }_{n0}} {\pmb \tau }_n \big ) \,d\mathbf {O}\\&\quad = \int _{SO_{p_n}({\pmb \theta }_{n0})} \exp \bigg ( n \kappa _n\nu _n \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert \Vert \varPi _{{\pmb \theta }_{n0}}{\bar{\mathbf {X}}}_{n}\Vert \bigg (\frac{\varPi _{{\pmb \theta }_{n0}}{\bar{\mathbf {X}}}_{n}}{\Vert \varPi _{{\pmb \theta }_{n0}}{\bar{\mathbf {X}}}_{n}\Vert }\bigg )^{\prime }\bigg ( \mathbf {O}^{\prime }\frac{\varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n}{\Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert }\bigg ) \bigg ) \,d\mathbf {O}\\&\quad = \mathrm{E} \big [ \exp \big ( n \kappa _n\nu _n \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert \Vert \varPi _{{\pmb \theta }_{n0}}{\bar{\mathbf {X}}}_{n}\Vert \mathbf {v}_n^{\prime }\mathbf {S}\big ) \,|\, \mathbf {X}_{n1},\ldots ,\mathbf {X}_{nn} \big ] , \end{aligned}$$

where $\mathbf {S}$ is uniformly distributed over $\mathcal {S}^{p_n-1}_{{\pmb \theta }_{n0}}:=\{\mathbf {x}\in \mathcal {S}^{p_n-1}: \mathbf {x}'{\pmb \theta }_{n0}=0\}$ and where $\mathbf {v}_n\in \mathcal {S}^{p_n-1}_{{\pmb \theta }_{n0}}$ is arbitrary. Since $\mathbf {v}_n'\mathbf {S}$ has density $t\mapsto c_{p_n-1}(1-t^2)^{\frac{p_n-4}{2}} \, \mathbb {I}[t\in [-1,1]]$, with $c_{p_n-1}=1/\int _{-1}^1 (1-t^2)^{\frac{p_n-4}{2}}\,dt$, this yields

$$\begin{aligned}&\int _{SO_{p_n}({\pmb \theta }_{n0})} \exp \big ( n \kappa _n\nu _n {\bar{\mathbf {X}}}_{n}^{\prime }\mathbf {O}^{\prime }\varPi _{{\pmb \theta }_{n0}} {\pmb \tau }_n \big ) \,d\mathbf {O}\\&\quad = c_{p_n-1}\int _{-1}^1 \mathrm{exp}\big (n \kappa _n \nu _n \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert \Vert \varPi _{{\pmb \theta }_{n0}}{\bar{\mathbf {X}}}_{n}\Vert t \big )(1-t^2)^{\frac{p_n-4}{2}}\, dt\\&\quad = \, H_{\frac{p_n-3}{2}}(n \kappa _n \nu _n \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert \Vert \varPi _{{\pmb \theta }_{n0}}{\bar{\mathbf {X}}}_{n}\Vert ) . \end{aligned}$$

Summing up,

$$\begin{aligned} \frac{d\mathrm{P}^{(n)\mathbf {T}_n}_{1-\frac{1}{2} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2,\kappa _n}}{d{m_n}}= & {} \frac{c_{p_n,\kappa _n}^n}{\omega _{p_n-1}^n} \exp \big ( n \kappa _n {\bar{\mathbf {X}}}_{n}^{\prime }{\pmb \theta }_{n0} \big ) \nonumber \\&\times \, \exp \big ( n \kappa _n \nu _n ({\pmb \tau }_n^{\prime }{\pmb \theta }_{n0}) {\bar{\mathbf {X}}}_{n}^{\prime }{\pmb \theta }_{n0} \big ) H_{\frac{p_n-3}{2}}(n \kappa _n \nu _n \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert \Vert \varPi _{{\pmb \theta }_{n0}}{\bar{\mathbf {X}}}_{n}\Vert ).\nonumber \\ \end{aligned}$$

(A.9)

Now, with the quantity $L_n$ introduced in Lemma 1, we have

$$\begin{aligned} T_n:= & {} 1 + \frac{\sqrt{2}\,{\widetilde{W}}_n}{\sqrt{p_n-1}} = \frac{W_{n}}{p_n-1} = \frac{n^2}{\sum _{i=1}^n V_{ni}^2} \, {\bar{\mathbf {X}}}_{n}' (\mathbf {I}_{p_n}-{\pmb \theta }_{n0}{\pmb \theta }_{n0}') {\bar{\mathbf {X}}}_n \\= & {} \frac{n}{f_{n2} L_n} \, \Vert \varPi _{{\pmb \theta }_{n0}}{\bar{\mathbf {X}}}_{n}\Vert ^2 = \frac{n\kappa _n}{(p_n-1)e_{n1} L_n} \, \Vert \varPi _{{\pmb \theta }_{n0}}{\bar{\mathbf {X}}}_{n}\Vert ^2 , \end{aligned}$$

where we used the identity $f_{n2}=(p_n-1)e_{n1}/\kappa _n$; see (2.7). Therefore, (A.9) yields

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n/{\pmb \theta }_{n0};\kappa _n}= & {} \log \frac{d\mathrm{P}^{(n)\mathbf {T}_n}_{1-\frac{1}{2} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2,\kappa _n}}{d\mathrm{P}^{(n)\mathbf {T}_n}_{1,\kappa _n}}\\= & {} n \kappa _n \nu _n ({\pmb \tau }_n^{\prime }{\pmb \theta }_{n0}) {\bar{\mathbf {X}}}_{n}^{\prime }{\pmb \theta }_{n0} + \log H_{\frac{p_n-3}{2}}(n \kappa _n \nu _n \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert \Vert \varPi _{{\pmb \theta }_{n0}}{\bar{\mathbf {X}}}_{n}\Vert )\\= & {} n \kappa _n \nu _n ({\pmb \tau }_n^{\prime }{\pmb \theta }_{n0}) {\bar{\mathbf {X}}}_{n}^{\prime }{\pmb \theta }_{n0} \\&+\, \log H_{\frac{p_n-3}{2}}(n^{1/2} (p_n-1)^{1/2} \kappa _n^{1/2} \nu _n e_{n1}^{1/2} \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert L_n^{1/2} T_n^{1/2}) . \end{aligned}$$

Since ${\widetilde{W}}_n = \sqrt{(p_n-1)/2}\times ( T_n - 1 )$ is asymptotically standard normal (Lemma 2), we have that $T_n = 1 + o_{\mathrm{P}}(1)$. Moreover, it directly follows from Lemma 1 that $L_n = 1 + o_{\mathrm{P}}(1)$. Consequently, Lemma 3 shows that, if $\nu _n$ satisfies (2.8), then

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n/{\pmb \theta }_{n0};\kappa _n}= & {} n \kappa _n \nu _n ({\pmb \tau }_n^{\prime }{\pmb \theta }_{n0}) {\bar{\mathbf {X}}}_{n}^{\prime }{\pmb \theta }_{n0} + \frac{p_n-1}{2(p_n-3)} n \kappa _n \nu _n^2 e_{n1} \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert ^2 L_n T_n\\&-\, \frac{(p_n-1)^2}{4(p_n-3)^3} n^2 \kappa _n^2 \nu _n^4 e_{n1}^2 \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert ^4 L_n^{2}T_n^2 + o_{\mathrm{P}}(1) . \end{aligned}$$

Using (2.8), Lemma 1 and the fact that $T_n=1+o_{\mathrm{P}}(1)$ yields

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n/{\pmb \theta }_{n0};\kappa _n}= & {} n \kappa _n \nu _n ({\pmb \tau }_n^{\prime }{\pmb \theta }_{n0}) {\bar{\mathbf {X}}}_{n}^{\prime }{\pmb \theta }_{n0} + \frac{1}{2} n \kappa _n \nu _n^2 e_{n1} \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert ^2 T_n \nonumber \\&-\, \frac{n^2 \kappa _n^2 \nu _n^4 e_{n1}^2}{4p_n} \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert ^4 + o_{\mathrm{P}}(1) . \end{aligned}$$

(A.10)

Using the definitions of $Z_n$ and $T_n$, we obtain

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n/{\pmb \theta }_{n0};\kappa _n}= & {} n \kappa _n \nu _n ({\pmb \tau }_n^{\prime }{\pmb \theta }_{n0}) \Big (e_{n1}+\frac{\sqrt{{\tilde{e}}_{n2}}}{n^{1/2}} Z_n \Big )\\&+\, \frac{1}{2} n \kappa _n \nu _n^2 e_{n1} \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert ^2 \Big ( 1+ \frac{\sqrt{2}\,{\widetilde{W}}_n}{\sqrt{p_n-1}} \Big )\\&- \frac{n^2 \kappa _n^2 \nu _n^4 e_{n1}^2}{4p_n} \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert ^4 + o_{\mathrm{P}}(1)\\= & {} \sqrt{n} \kappa _n \nu _n \sqrt{{\tilde{e}}_{n2}} ({\pmb \tau }_n^{\prime }{\pmb \theta }_{n0}) Z_n + \frac{n \kappa _n \nu _n^2 e_{n1}}{\sqrt{2}(p_n-1)^{1/2}} \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert ^2 {\widetilde{W}}_n\\&+\, n \kappa _n \nu _n ({\pmb \tau }_n^{\prime }{\pmb \theta }_{n0}) e_{n1} + \frac{1}{2} n \kappa _n \nu _n^2 e_{n1} \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert ^2\\&- \frac{n^2 \kappa _n^2 \nu _n^4 e_{n1}^2}{4p_n} \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert ^4 + o_{\mathrm{P}}(1) . \end{aligned}$$

Using the identities

$$\begin{aligned} {\pmb \tau }_n^{\prime }{\pmb \theta }_{n0} = - \frac{1}{2} \nu _n \Vert {\pmb \tau }_n\Vert ^2 \end{aligned}$$

(A.11)

and

$$\begin{aligned} \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert ^2 = \Vert {\pmb \tau }_n\Vert ^2 - ({\pmb \tau }_n^{\prime }{\pmb \theta }_{n0})^2 = \Vert {\pmb \tau }_n\Vert ^2 \Big (1- \frac{1}{4} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 \Big ) \end{aligned}$$

(A.12)

provides

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n/{\pmb \theta }_{n0};\kappa _n}= & {} -\frac{1}{2} \sqrt{n} \kappa _n \nu _n^2 \sqrt{{\tilde{e}}_{n2}} \Vert {\pmb \tau }_n\Vert ^2 Z_n \\&+\, \frac{n \kappa _n \nu _n^2 e_{n1}}{\sqrt{2}(p_n-1)^{1/2}} \Vert {\pmb \tau }_n\Vert ^2 \Big (1- \frac{1}{4} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 \Big ) {\widetilde{W}}_n -\frac{1}{2} n \kappa _n \nu _n^2 e_{n1} \Vert {\pmb \tau }_n\Vert ^2\\&+\, \frac{1}{2} n \kappa _n \nu _n^2 e_{n1} \Vert {\pmb \tau }_n\Vert ^2 \Big (1- \frac{1}{4} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 \Big )\\&-\,\frac{n^2 \kappa _n^2 \nu _n^4 e_{n1}^2}{4p_n} \Vert {\pmb \tau }_n\Vert ^4 \Big (1- \frac{1}{4} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 \Big )^2 + o_{\mathrm{P}}(1) . \end{aligned}$$

The result then follows from (2.8) and from the tightness of ${\widetilde{W}}_n$ (Lemma 2). $\square$

Lemma 4

Let $(p_n)$ be a sequence of integers that diverges to infinity and $(\kappa _n)$ be an arbitrary sequence in $(0,\infty )$. Let $e_{n1}$ (resp., ${\tilde{e}}_{n2})$ be the expectation (resp., the variance) of the distribution with probability density function (2.6). Then, we have the following: (i) if $\kappa _n/p_n\rightarrow \infty$, then

$$\begin{aligned} e_{n1}=1+o(1) , \quad {\tilde{e}}_{n2}=O\Big (\frac{p_n}{\kappa _n^2}\Big ) \quad \text { and }\quad f_{n2}=\frac{p_n}{\kappa _n}+o\Big (\frac{p_n}{\kappa _n}\Big ) ; \end{aligned}$$

(ii) if $\kappa _n/p_n\rightarrow \xi >0$, then, letting $c_{\xi }:=\frac{1}{2}+\sqrt{\frac{1}{4} + \xi ^2}$,

$$\begin{aligned} e_{n1}\rightarrow \frac{\xi }{c_\xi }+o(1), \quad {\tilde{e}}_{n2}=O\Big (\frac{1}{p_n}\Big ) \quad \text { and }\quad f_{n2}=\frac{1}{c_\xi }+o(1) ; \end{aligned}$$

(iii) if $\kappa _n/p_n\rightarrow 0$, then

$$\begin{aligned} e_{n1}=\frac{\kappa _n}{p_n}+O\Big (\frac{\kappa _n^3}{p_n^3}\Big ) , \quad {\tilde{e}}_{n2}=\frac{1}{p_n}+o\Big (\frac{1}{p_n}\Big ) \quad \text { and }\quad f_{n2}=1+o(1) . \end{aligned}$$

Proof of Lemma 4

Denoting again as $\mathcal {I}_\nu (\cdot )$ the order-$\nu$ modified Bessel function of the first kind, we recall [see (2.7)] that

$$\begin{aligned} e_{n1} = \frac{\mathcal {I}_{\frac{p_n}{2}}(\kappa _n)}{\mathcal {I}_{\frac{p_n}{2}-1}(\kappa _n)} , \quad {\tilde{e}}_{n2} = 1-\frac{p_n-1}{\kappa _n}\, e_{n1}-e_{n1}^2 \quad \text { and }\quad f_{n2} = \frac{p_n-1}{\kappa _n}\, e_{n1} . \end{aligned}$$

In each case (i)–(iii), the claim for $f_{n2}$ directly follows from the result on $e_{n1}$, so that it is sufficient to prove the results for $e_{n1}$ and ${\tilde{e}}_{n2}$. To do so, we will use the bounds

$$\begin{aligned} R^{\mathrm{low}}_{\nu }(z):= & {} \frac{z}{\nu +1+\sqrt{(\nu +1)^2+z^2}} \le \frac{\mathcal {I}_{\nu +1}(z)}{\mathcal {I}_\nu (z)} \nonumber \\\le & {} \frac{z}{\nu +\sqrt{(\nu +2)^2+z^2}} =: R^{\mathrm{up}}_{\nu }(z) \end{aligned}$$

(A.13)

and

$$\begin{aligned} {\tilde{R}}^{\mathrm{low}}_{\nu }(z) := \frac{z}{\nu +\frac{1}{2}+\sqrt{(\nu +\frac{3}{2})^2+z^2}} \le \frac{\mathcal {I}_{\nu +1}(z)}{\mathcal {I}_\nu (z)} ; \end{aligned}$$

(A.14)

see (11) and (16) in [1], respectively. (i) The lower bound in (A.13) provides

$$\begin{aligned} e_{n1} \ge \frac{\kappa _n}{\frac{p_n}{2}+\sqrt{\big (\frac{p_n}{2}\big )^2 + \kappa _n^2}} = \frac{1}{\frac{p_n}{2\kappa _n}+\sqrt{\big (\frac{p_n}{2\kappa _n}\big )^2 + 1}} , \end{aligned}$$

which, since $e_{n1}\le 1$, establishes the result for $e_{n1}$. Making use of the bound in (A.14), we can write

$$\begin{aligned} {\tilde{e}}_{n2} \le 1 - \frac{p_n-1}{\kappa _n} {\tilde{R}}^{\mathrm{low}}_{\frac{p_n}{2}-1}(\kappa _n) - \big ({\tilde{R}}^{\mathrm{low}}_{\frac{p_n}{2}-1}(\kappa _n)\big )^2 . \end{aligned}$$

Lengthy yet quite straightforward computations allow to rewrite this as

$$\begin{aligned} {\tilde{e}}_{n2} \le \frac{ p_n }{\kappa _n^2 \Big (\frac{p_n-1}{2\kappa _n}+\sqrt{\big (\frac{p_n+1}{2\kappa _n}\big )^2 + 1}\,\Big )^2} \cdot \end{aligned}$$

It readily follows that $\kappa _n^2 {\tilde{e}}_{n2}/p_n$ is O(1), as was to be showed. Let us turn to the proof of (iii). The bounds in (A.13) readily yield

$$\begin{aligned} \frac{1}{\frac{1}{2}+\sqrt{\big (\frac{1}{2}\big )^2 + \big (\frac{\kappa _n}{p_n}\big )^2}} \le \frac{e_{n1}}{\kappa _n/p_n} \le \frac{1}{\frac{1}{2}-\frac{1}{p_n}+\sqrt{\big (\frac{1}{2}+\frac{1}{p_n}\big )^2 + \big (\frac{\kappa _n}{p_n}\big )^2}} , \end{aligned}$$

(A.15)

which provides

$$\begin{aligned} \frac{-\big (\frac{\kappa _n}{p_n}\big )^2}{\big ( \frac{1}{2}+\sqrt{\big (\frac{1}{2}\big )^2 + \big (\frac{\kappa _n}{p_n}\big )^2}\big )^2} \le \frac{e_{n1}}{\kappa _n/p_n} - 1 \le \frac{-\big (\frac{\kappa _n}{p_n}\big )^2}{\frac{1}{2}-\frac{1}{p_n}+\sqrt{\big (\frac{1}{2}+\frac{1}{p_n}\big )^2 + \big (\frac{\kappa _n}{p_n}\big )^2}} \cdot \end{aligned}$$

This proves the result for $e_{n1}$. Turning to ${\tilde{e}}_{n2}$, the bounds in (A.13) lead to

$$\begin{aligned}&1 - \frac{p_n-1}{\kappa _n} R^{\mathrm{up}}_{\frac{p_n}{2}-1}(\kappa _n) - \big (R^{\mathrm{up}}_{\frac{p_n}{2}-1}(\kappa _n)\big )^2 \le {\tilde{e}}_{n2} \le 1 \\&\quad - \frac{p_n-1}{\kappa _n} R^{\mathrm{low}}_{\frac{p_n}{2}-1}(\kappa _n) - \big (R^{\mathrm{low}}_{\frac{p_n}{2}-1}(\kappa _n)\big )^2 . \end{aligned}$$

As above, heavy but rather straightforward computations allow to rewrite this as

$$\begin{aligned} \frac{ \frac{3}{2} + \frac{1}{p_n} - \sqrt{\big (\frac{1}{2}+\frac{1}{p_n}\big )^2 + \big (\frac{\kappa _n}{p_n}\big )^2} }{p_n\Big (\frac{1}{2}-\frac{1}{p_n}+ \sqrt{\big (\frac{1}{2}+\frac{1}{p_n}\big )^2 + \big (\frac{\kappa _n}{p_n}\big )^2}\,\Big )^2} \le {\tilde{e}}_{n2} \le \frac{ 1 }{ p_n\Big (\frac{1}{2}+\sqrt{\big (\frac{1}{2}\big )^2 + \big (\frac{\kappa _n}{p_n}\big )^2}\,\Big )},\nonumber \\ \end{aligned}$$

(A.16)

which establishes that $p_n{\tilde{e}}_{n2}=1+o(1)$. Finally, the result in (ii) readily follows from (A.15) and from the upper bound in (A.16). $\square$

Proof of Theorem 2

Stochastic convergences throughout the proof are as $n\rightarrow \infty$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$.

Assume that (i) $\kappa _n/p_n\rightarrow \infty$, (ii) $\kappa _n/p_n\rightarrow \xi >0$, or (iii) $\kappa _n/p_n\rightarrow 0$ with $\sqrt{n} \kappa _n/p_n\rightarrow \infty$, and let $(\nu _n)$ be the corresponding sequence in the statement of the theorem. Using Lemma 4 and the identity $\kappa _n f_{n2}=(p_n-1)e_{n1}$, it is then easy to check that $\nu _n$ satisfies (2.8), is such that $\nu _n=o(1)$, and is asymptotically equivalent to ${\tilde{\nu }}_n=p_n^{3/4}/(\sqrt{n} \kappa _n \sqrt{f_{n2}})$ in the sense that ${\tilde{\nu }}_n/\nu _n\rightarrow 1$. Theorem 1 thus applies and yields

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n/{\pmb \theta }_{n0};\kappa _n}= & {} - \frac{p_n^{3/2}\sqrt{{\tilde{e}}_{n2}}}{2\sqrt{n} \kappa _n f_{n2}} \, \Vert {\pmb \tau }_n\Vert ^2 Z_n + \frac{p_n e_{n1}}{\sqrt{2}\kappa _n f_{n2}} \Vert {\pmb \tau }_n\Vert ^2 {\widetilde{W}}_n \\&-\, \frac{p_n^3 e_{n1}}{8n \kappa _n^3 f_{n2}^2} \Vert {\pmb \tau }_n\Vert ^4 - \frac{p_n^2e_{n1}^2}{4\kappa _n^2 f_{n2}^2} \Vert {\pmb \tau }_n\Vert ^4 +o_{\mathrm{P}}(1) . \end{aligned}$$

Using again the identity $\kappa _n f_{n2}=(p_n-1)e_{n1}$, we then obtain

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n/{\pmb \theta }_{n0};\kappa _n}= & {} - \frac{p_n^{3/2}\sqrt{{\tilde{e}}_{n2}}}{2\sqrt{n} (p_n-1) e_{n1}} \, \Vert {\pmb \tau }_n\Vert ^2 Z_n + \frac{1}{\sqrt{2}} \Vert {\pmb \tau }_n\Vert ^2 {\widetilde{W}}_n \nonumber \\&\quad - \,\frac{p_n^3}{8n \kappa _n (p_n-1)^2 e_{n1}} \Vert {\pmb \tau }_n\Vert ^4 - \frac{1}{4} \Vert {\pmb \tau }_n\Vert ^4 +o_{\mathrm{P}}(1) . \end{aligned}$$

(A.17)

The result in cases (i)–(iii) then follows from the fact that Lemma 4 implies that, in each case, the first and third term of the righthand side of (A.17) are $o_{\mathrm{P}}(1)$.

We turn to case (iv), for which $\sqrt{n}\kappa _n/p_n=\xi$ (so that, like for all subsequent cases, $\kappa _n=o(p_n)$). Then, the same argument as above allows to check that $\nu _n = p_n^{3/4}/(\sqrt{n} \kappa _n \sqrt{f_{n2}})$ still satisfies (2.8) and is such that $\nu _n=o(1)$, so that, jointly with Lemma 4, Theorem 1 provides

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n/{\pmb \theta }_{n0};\kappa _n}= & {} - \frac{p_n^{3/2} {\tilde{e}}_{n2}^{1/2}}{2\sqrt{n} \kappa _n f_{n2}} \Vert {\pmb \tau }_n\Vert ^2 Z_n + \frac{p_n e_{n1}}{\sqrt{2}\kappa _n f_{n2}} \Vert {\pmb \tau }_n\Vert ^2 {\widetilde{W}}_n\\&-\, \frac{p_n^{3} e_{n1}}{8n \kappa _n^3 f_{n2}^2} \Vert {\pmb \tau }_n\Vert ^4 - \frac{p_n^2 e_{n1}^2}{4\kappa _n^2 f_{n2}^2} \Vert {\pmb \tau }_n\Vert ^4 +o_{\mathrm{P}}(1)\\= & {} - \frac{1}{2\xi } \Vert {\pmb \tau }_n\Vert ^2 Z_n + \frac{1}{\sqrt{2}} \Vert {\pmb \tau }_n\Vert ^2 {\widetilde{W}}_n - \frac{1}{8\xi ^2} \Vert {\pmb \tau }_n\Vert ^4 - \frac{1}{4} \Vert {\pmb \tau }_n\Vert ^4 +o_{\mathrm{P}}(1) , \end{aligned}$$

as was to be shown. Consider now case (v), under which $\sqrt{n} \kappa _n/p_n\rightarrow 0$ with $\sqrt{n} \kappa _n/\sqrt{p_n}\rightarrow \infty$, which still ensures that $\nu _n=p_n^{1/4}/(n^{1/4} \sqrt{\kappa _n})$ is o(1) and satisfies (2.8). Theorem 1 applies and, by using Lemma 4 again, yields

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n/{\pmb \theta }_{n0};\kappa _n}= & {} - \frac{1}{2} \sqrt{p_n{\tilde{e}}_{n2}} \, \Vert {\pmb \tau }_n\Vert ^2 Z_n + \frac{\sqrt{n} e_{n1}}{\sqrt{2}} \Vert {\pmb \tau }_n\Vert ^2 {\widetilde{W}}_n \\&- \,\frac{p_n e_{n1}}{8\kappa _n} \Vert {\pmb \tau }_n\Vert ^4 - \frac{n e_{n1}^2}{4} \Vert {\pmb \tau }_n\Vert ^4 +o_{\mathrm{P}}(1)\\= & {} - \frac{1}{2} \Vert {\pmb \tau }_n\Vert ^2 Z_n - \frac{1}{8} \Vert {\pmb \tau }_n\Vert ^4 +o_{\mathrm{P}}(1) , \end{aligned}$$

which establishes the result in case (v). If $\sqrt{n}\kappa _n/\sqrt{p_n}=\xi$ [case (vi)], then Lemma 4 implies that $\nu _n=1$ satisfies (2.8). Theorem 1 then provides

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n/{\pmb \theta }_{n0};\kappa _n}= & {} - \frac{\sqrt{n} \kappa _n}{2\sqrt{p_n}} \sqrt{p_n{\tilde{e}}_{n2}}\, \Vert {\pmb \tau }_n\Vert ^2 Z_n + \frac{n \kappa _n e_{n1}}{\sqrt{2}p_n^{1/2}} \Vert {\pmb \tau }_n\Vert ^2 \Big (1- \frac{1}{4} \Vert {\pmb \tau }_n\Vert ^2\Big ) {\widetilde{W}}_n \nonumber \\&\quad -\, \frac{1}{8} n \kappa _n e_{n1} \Vert {\pmb \tau }_n\Vert ^4 - \frac{n^2 \kappa _n^2 e_{n1}^2}{4p_n} \Vert {\pmb \tau }_n\Vert ^4 \Big (1- \frac{1}{4} \Vert {\pmb \tau }_n\Vert ^2\Big )^2 +o_{\mathrm{P}}(1)\nonumber \\= & {} - \frac{\xi }{2} \Vert {\pmb \tau }_n\Vert ^2 Z_n - \frac{\xi ^2}{8} \Vert {\pmb \tau }_n\Vert ^4 +o_{\mathrm{P}}(1) , \end{aligned}$$

(A.18)

where we used Lemma 4. Finally, if $\sqrt{n}\kappa _n/\sqrt{p_n}=o(1)$ [case (vii)], then (2.8) again holds for $\nu _n=1$. Therefore, Theorem 1 shows that $\varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n/{\pmb \theta }_{n0};\kappa _n}$ satisfies the first equality of (A.18), hence is $o_{\mathrm{P}}(1)$. $\square$

Proof of Theorem 3

First note that, since $p_n=o(n^2)$, Lemma 4(iii) entails that

$$\begin{aligned} Z_n= & {} \frac{\sqrt{n}({\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0}- e_{n1})}{\sqrt{{\tilde{e}}_{2n}}} = \frac{\sqrt{n}({\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0}- \xi /\sqrt{n}+O(n^{-3/2}))}{\sqrt{\frac{1}{p_n}+o(\frac{1}{p_n})}} \nonumber \\= & {} \frac{\sqrt{p_n}(\sqrt{n}{\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0}- \xi +O(1/n))}{\sqrt{1+o(1)}} = \sqrt{np_n}{\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0}- \sqrt{p_n} \xi + o_{\mathrm{P}}(1)\qquad \qquad \end{aligned}$$

(A.19)

as $n\rightarrow \infty$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$. Write then

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n, \kappa _{n,s}/{\pmb \theta }_{n0},\kappa _n}= & {} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n, \kappa _{n,s}/{\pmb \theta }_{n0},\kappa _{n,s}} + \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_{n0}, \kappa _{n,s}/{\pmb \theta }_{n0},\kappa _n}\\= & {} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n, \kappa _{n,s}/{\pmb \theta }_{n0},\kappa _{n,s}} + \log \frac{d\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _{n,s}}}{d\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _{n}}}\\=: & {} L_{n1}+L_{n2}. \end{aligned}$$

Using (A.19), we obtain

$$\begin{aligned} L_{n2}= & {} n \big ( \log (c_{p_n,\kappa _{n,s}})-\log (c_{p_n, \kappa _n}) \big ) + n(\kappa _{n,s}-\kappa _n) {\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0} \\= & {} n \Big [ \log \Big (\frac{c_{p_n,0}}{c_{p_n, \kappa _n}} \Big ) - \log \Big (\frac{c_{p_n,0}}{c_{p_n, \kappa _{n,s}}} \Big ) \Big ] + s \sqrt{n p_n}\, {\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0} \\= & {} n \Big [ \log H_{\frac{p_n}{2}-1} (\kappa _n) - \log H_{\frac{p_n}{2}-1} (\kappa _{n,s}) \Big ] + s \sqrt{p_n} \xi + s Z_{n} + o_{\mathrm{P}}(1) \\=: & {} {\tilde{L}}_{n2} + s Z_{n} + o_{\mathrm{P}}(1) \end{aligned}$$

as $n\rightarrow \infty$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$. Since $p_n=o(n^2)$, we can apply Lemma 3 with $a_n=\sqrt{n}$ and $T_n \equiv 1$. This yields

$$\begin{aligned} {\tilde{L}}_{n2}= & {} \left( \frac{n \kappa _n^2}{4\left( {\textstyle {\frac{p_n}{2}}}-1\right) } - \frac{n \kappa _n^4}{32\left( {\textstyle {\frac{p_n}{2}}}-1\right) ^3} \right) - \left( \frac{n \kappa _{n,s}^2}{4\left( {\textstyle {\frac{p_n}{2}}}-1\right) } - \frac{n \kappa _{n,s}^4}{32\left( {\textstyle {\frac{p_n}{2}}}-1\right) ^3} \right) \\&+ s\sqrt{p_n} \xi + o(1)\\= & {} -\frac{n\left( \kappa _{n,s}^2-\kappa _n^2\right) }{2p_n-4} + \frac{n\left( \kappa _{n,s}^4-\kappa _n^4\right) }{4(p_n-2)^3} + s\sqrt{p_n} \xi + o(1)\\= & {} -\frac{p_n^2((\xi + s/\sqrt{p_n})^2-\xi ^2)}{2p_n-4}+ \frac{p_n^4((\xi + s/\sqrt{p_n})^4-\xi ^4)}{4n(p_n-2)^3}+s\sqrt{p_n} \xi +o(1)\\= & {} -\frac{s^2}{2}+o(1) \end{aligned}$$

as $n\rightarrow \infty$. Therefore,

$$\begin{aligned} L_{n2} = s Z_{n} - \frac{s^2}{2}+o_{\mathrm{P}}(1) \end{aligned}$$

(A.20)

as $n\rightarrow \infty$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$, which implies that the sequences of probability measures $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _{n,s}}$ and $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$ are mutually contiguous (this results from the Le Cam first lemma).

Now, denote as $e_{n1,s}$ and ${\tilde{e}}_{2n,s}$, respectively, the values of $e_{n1}$ and ${\tilde{e}}_{2n}$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _{n,s}}$. Then, proceeding as in (A.19) and using the fact that contiguity implies that (A.19) also holds under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _{n,s}}$, one obtains

$$\begin{aligned} Z_{n,s} := \frac{\sqrt{n}({\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0}- e_{n1,s})}{\sqrt{{\tilde{e}}_{2n,s}}} = \sqrt{np_n}{\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0}- \sqrt{p_n} \xi -s + o_{\mathrm{P}}(1) = Z_n - s + o_{\mathrm{P}}(1) \end{aligned}$$

as $n\rightarrow \infty$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _{n,s}}$. Consequently, Theorem 2(iv) implies that

$$\begin{aligned} L_{1n}= & {} \Vert {\pmb \tau }_n\Vert ^2 \bigg ( \frac{{\widetilde{W}}_n}{\sqrt{2}} - \frac{Z_{n,s}}{2\xi } \bigg ) -\frac{1}{2} \Vert {\pmb \tau }_n\Vert ^4 \bigg (\frac{1}{2} + \frac{1}{4\xi ^2}\bigg ) +o_{\mathrm{P}}(1) \nonumber \\= & {} \Vert {\pmb \tau }_n\Vert ^2 \bigg ( \frac{{\widetilde{W}}_n}{\sqrt{2}} - \frac{Z_n}{2\xi } \bigg ) + \frac{\Vert {\pmb \tau }_n\Vert ^2 s}{2 \xi } -\frac{1}{2} \Vert {\pmb \tau }_n\Vert ^4 \bigg (\frac{1}{2} + \frac{1}{4\xi ^2}\bigg ) +o_{\mathrm{P}}(1)\qquad \quad \end{aligned}$$

(A.21)

as $n\rightarrow \infty$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _{n,s}}$, hence, from contiguity, also under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$. Combining (A.26) and (A.21) establishes the local asymptotic quadraticity result in (2.16). Finally, the asymptotic normality result of ${\pmb \varDelta }_n$ trivially follows from Lemma 2. $\square$

The proof of Theorem 4 requires both following lemmas.

Lemma 5

Let $(p_n)$ be a sequence of integers that diverges to infinity. Let $(\kappa _n)$ and $(\kappa _{n*})$ be sequences in $(0,\infty )$ that are $o(p_n)$ and write $e_{n1}$ and ${\tilde{e}}_{n2}$ (resp., $e_{n1*}$ and ${\tilde{e}}_{n2*})$ for the corresponding moments based on $\kappa _n$ (resp., on $\kappa _{n*})$. Let $({\pmb \theta }_{n0})$, $(\nu _n)$ and $({\pmb \tau }_n)$ be sequences such that ${\pmb \theta }_{n0}$ and ${\pmb \theta }_n={\pmb \theta }_{n0}+\nu _n{\pmb \tau }_n$ belong to $\mathcal {S}^{p_n-1}$ for any n, with $({\pmb \tau }_n)$ bounded and $(\nu _n)$ such that

$$\begin{aligned} \nu _n^2 = O\Big (\frac{\sqrt{p_n}}{n \kappa _{n*} e_{n1*}}\Big ) . \end{aligned}$$

(A.22)

Then, with the same $Z_n$ and ${\widetilde{W}}_n$ as in Theorem 1, we have that

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n/{\pmb \theta }_{n0};\kappa _{n*}}= & {} - \frac{1}{2} \sqrt{n} \kappa _{n*} \nu _n^2 \sqrt{{\tilde{e}}_{n2}}\, \Vert {\pmb \tau }_n\Vert ^2 Z_n + \frac{n \kappa _{n*}\nu _n^2 e_{n1*}}{\sqrt{2}p_n^{1/2}} \Vert {\pmb \tau }_n\Vert ^2 \Big (1- \frac{1}{4} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2\Big ) {\widetilde{W}}_n \\&-\, \frac{1}{8} n \kappa _{n*}\nu _n^4 e_{n1*} \Vert {\pmb \tau }_n\Vert ^4 - \frac{n^2 \kappa _{n*}^2 \nu _n^4 e_{n1*}^2}{4p_n} \Vert {\pmb \tau }_n\Vert ^4 \Big (1- \frac{1}{4} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2\Big )^2\\&+\, \frac{1}{2} n \kappa _{n*} \nu _n^2 (e_{n1*}-e_{n1}) \Vert {\pmb \tau }_n\Vert ^2 +o_{\mathrm{P}}(1) , \end{aligned}$$

as $n\rightarrow \infty$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$.

Proof of Lemma 5

Since $\kappa _n$ and $\kappa _{n*}$ are both o(n), Lemma 4 ensures that $f_{n2}/f_{n2*}=1+o(1)$, where $f_{n2*}$ denotes the quantity $f_{n2}$ based on $\kappa _{n*}$. Using this, it can be showed along the exact same lines as in the proof of (A.10) in Theorem 1 that, as $n\rightarrow \infty$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$,

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n/{\pmb \theta }_{n0};\kappa _{n*}}= & {} n \kappa _{n*} \nu _n ({\pmb \tau }_n^{\prime }{\pmb \theta }_{n0}) {\bar{\mathbf {X}}}_{n}^{\prime }{\pmb \theta }_{n0} + \frac{1}{2} n \kappa _{n*} \nu _n^2 e_{n1*} \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert ^2 T_n\nonumber \\&-\, \frac{n^2 \kappa _{n*}^2 \nu _n^4 e_{n1*}^2}{4p_n} \Vert \varPi _{{\pmb \theta }_{n0}}{\pmb \tau }_n\Vert ^4 + o_{\mathrm{P}}(1) , \end{aligned}$$

where $T_n:=1 +\sqrt{2}\,{\widetilde{W}}_n/\sqrt{p_n-1}$. If one replaces $T_n$ by this expression and ${\bar{\mathbf {X}}}_{n}^{\prime }{\pmb \theta }_{n0}$ by $e_{n1}+\sqrt{{\tilde{e}}_{n2}}Z_n/\sqrt{n}$, then the result follows by using (A.11), (A.12), (A.22), and the tightness of ${\widetilde{W}}_n$. $\square$

The second lemma reinforces the variance result in Lemma 4(iii).

Lemma 6

Let $(p_n)$ be a sequence of integers that diverges to infinity and $(\kappa _n)$ be a sequence in $(0,\infty )$ that is $o(p_n)$. Denote as ${\tilde{e}}_{n2}$ the variance of the distribution with probability density function (2.6). Then, $\sqrt{p_n{\tilde{e}}_{n2}}-1=O(\kappa _n^2/p_n^2)$ as $n\rightarrow \infty$.

Proof of Lemma 6

In this proof, C denotes a generic constant that may differ from line to line. Since (A.16) yields

$$\begin{aligned} \frac{ \sqrt{ \frac{3}{2} + \frac{1}{p_n} - \sqrt{\big (\frac{1}{2}+\frac{1}{p_n}\big )^2 + \big (\frac{\kappa _n}{p_n}\big )^2} } }{\frac{1}{2}-\frac{1}{p_n}+ \sqrt{\big (\frac{1}{2}+\frac{1}{p_n}\big )^2 + \big (\frac{\kappa _n}{p_n}\big )^2}} - 1 \le \sqrt{p_n{\tilde{e}}_{n2}}-1 \le \frac{ 1 }{ \sqrt{\frac{1}{2}+\sqrt{\big (\frac{1}{2}\big )^2 + \big (\frac{\kappa _n}{p_n}\big )^2}} } - 1 \le 0 , \end{aligned}$$

we have

$$\begin{aligned}&|\sqrt{p_n{\tilde{e}}_{n2}}-1| \le 1 - \frac{ \sqrt{ \frac{3}{2} + \frac{1}{p_n} - \sqrt{\big (\frac{1}{2}+\frac{1}{p_n}\big )^2 + \big (\frac{\kappa _n}{p_n}\big )^2} } }{\frac{1}{2}-\frac{1}{p_n}+ \sqrt{\big (\frac{1}{2}+\frac{1}{p_n}\big )^2 + \big (\frac{\kappa _n}{p_n}\big )^2}} \\&\quad \le C \bigg \{ \Big ( {\textstyle {\frac{1}{2}}}-{\textstyle {\frac{1}{p_n}}}+ \sqrt{\big ({\textstyle {\frac{1}{2}}}+{\textstyle {\frac{1}{p_n}}}\big )^2 + \big ({\textstyle {\frac{\kappa _n}{p_n}}}\big )^2} \Big ) - \sqrt{ {\textstyle {\frac{3}{2}}} + {\textstyle {\frac{1}{p_n}}} - \sqrt{\big ({\textstyle {\frac{1}{2}}}+{\textstyle {\frac{1}{p_n}}}\big )^2 + \big ({\textstyle {\frac{\kappa _n}{p_n}}}\big )^2}} \, \bigg \}\\&\quad \le C \bigg \{ \Big ( {\textstyle {\frac{1}{2}}}-{\textstyle {\frac{1}{p_n}}}+ \sqrt{\big ({\textstyle {\frac{1}{2}}}+{\textstyle {\frac{1}{p_n}}}\big )^2 + \big ({\textstyle {\frac{\kappa _n}{p_n}}}\big )^2} \Big )^2 - \Big ( {\textstyle {\frac{3}{2}}} + {\textstyle {\frac{1}{p_n}}} - \sqrt{\big ({\textstyle {\frac{1}{2}}}+{\textstyle {\frac{1}{p_n}}}\big )^2 + \big ({\textstyle {\frac{\kappa _n}{p_n}}}\big )^2} \Big ) \bigg \} . \end{aligned}$$

Standard computations allow us to rewrite this upper-bound as

$$\begin{aligned} |\sqrt{p_n{\tilde{e}}_{n2}}-1|\le & {} C \bigg \{ {\textstyle {\frac{2(p_n-1)}{p_n}}} \sqrt{\big ({\textstyle {\frac{p_n+2}{2p_n}}}\big )^2 + \big ({\textstyle {\frac{\kappa _n}{p_n}}}\big )^2} - {\textstyle {\frac{(p_n-1)(p_n+2)-\kappa _n^2}{p_n^2}}} \bigg \}\\\le & {} C \bigg \{ {\textstyle {\frac{4(p_n-1)^2}{p_n^2}}} \Big ( \big ({\textstyle {\frac{p_n+2}{2p_n}}}\big )^2 + \big ({\textstyle {\frac{\kappa _n}{p_n}}}\big )^2 \Big ) - {\textstyle {\frac{((p_n-1)(p_n+2)-\kappa _n^2)^2}{p_n^4}}} \bigg \} \\= & {} C \Big ( 6 - {\textstyle {\frac{6}{p_n}}} - {\textstyle {\frac{\kappa _n^2}{p_n^2}}} \Big ) {\textstyle {\frac{\kappa _n^2}{p_n^2}}} \cdot \end{aligned}$$

which, for n large, is upper-bounded by $C\kappa _n^2/p_n^2$, as was to be proved. $\square$

Proof of Theorem 4

Since $\kappa _n=o(p_n)$, Lemma 4(iii) entails that

$$\begin{aligned} Z_n= & {} \frac{\sqrt{n}({\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0}- e_{n1})}{\sqrt{{\tilde{e}}_{2n}}} = \frac{\sqrt{n}({\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0}- \xi r_n/\sqrt{n}+O(r_n^3 n^{-3/2}))}{\sqrt{\frac{1}{p_n}+o(\frac{1}{p_n})}} \nonumber \\= & {} \frac{\sqrt{p_n}(\sqrt{n}{\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0}- \xi r_n+O(r_n^3 n^{-1}))}{\sqrt{1+o(1)}} = \sqrt{np_n}{\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0}\nonumber \\&- \,\xi \sqrt{p_n} r_n + O_{\mathrm{P}}\Big (\frac{\sqrt{p_n}r_n^3}{n}\Big ) \end{aligned}$$

(A.23)

as $n\rightarrow \infty$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$. Write then

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n, \kappa _{n,s,{\pmb \tau }_n}/{\pmb \theta }_{n0},\kappa _n}= & {} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n, \kappa _{n,s,{\pmb \tau }_n}/{\pmb \theta }_{n0},\kappa _{n,s,{\pmb \tau }_n}} + \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_{n0}, \kappa _{n,s,{\pmb \tau }_n}/{\pmb \theta }_{n0},\kappa _n}\\= & {} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n, \kappa _{n,s,{\pmb \tau }_n}/{\pmb \theta }_{n0},\kappa _{n,s,{\pmb \tau }_n}} + \log \frac{d\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _{n,s,{\pmb \tau }_n}}}{d\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _{n}}}\\=: & {} L_{n1}+L_{n2}. \end{aligned}$$

Letting $\rho _n:=(1-{\textstyle {\frac{1}{2}}}\nu _n^2\Vert {\pmb \tau }_n\Vert ^2)^{-1}$ and using (A.23), we obtain

$$\begin{aligned} L_{n2}= & {} n \big ( \log (c_{p_n,\kappa _{n,s,{\pmb \tau }_n}})-\log (c_{p_n, \kappa _n}) \big ) + n(\kappa _{n,s,{\pmb \tau }_n}-\kappa _n) {\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0} \nonumber \\= & {} n \Big [ \log \Big (\frac{c_{p_n,0}}{c_{p_n, \kappa _n}} \Big ) - \log \Big (\frac{c_{p_n,0}}{c_{p_n, \kappa _{n,s,{\pmb \tau }_n}}} \Big ) \Big ] + \rho _n (s\sqrt{np_n}\nonumber \\&+\, {\textstyle {\frac{1}{2}}}\xi \sqrt{n} p_n r_n\nu _n^2\Vert {\pmb \tau }_n\Vert ^2)\, {\bar{\mathbf {X}}}_n^{\prime }{\pmb \theta }_{n0} \nonumber \\= & {} n \Big [ \log H_{\frac{p_n}{2}-1} (\kappa _n) - \log H_{\frac{p_n}{2}-1} (\kappa _{n,s,{\pmb \tau }_n}) \Big ] \nonumber \\&\quad + \,\rho _n (s\sqrt{np_n}+{\textstyle {\frac{1}{2}}} \xi \sqrt{n} p_n r_n \nu _n^2\Vert {\pmb \tau }_n\Vert ^2)\, \bigg ( \frac{Z_{n}}{\sqrt{np_n}} + \frac{\xi r_n}{\sqrt{n}} +O_{\mathrm{P}}\Big (\frac{r_n^3}{n^{3/2}}\Big ) \bigg )\nonumber \\ \end{aligned}$$

(A.24)

as $n\rightarrow \infty$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$. Since $\sqrt{p_n} r_n^3$ and $p_n r_n^4 \nu _n^2$ are o(n), this yields

$$\begin{aligned} L_{n2}= & {} n \Big [ \log H_{\frac{p_n}{2}-1} (\kappa _n) - \log H_{\frac{p_n}{2}-1} (\kappa _{n,s,{\pmb \tau }_n}) \Big ]\\&+ \,\rho _n \big (s + {\textstyle {\frac{1}{2}}} \xi \sqrt{p_n}r_n\nu _n^2\Vert {\pmb \tau }_n\Vert ^2\big ) Z_n + \xi s \rho _n r_n\sqrt{p_n} + {\textstyle {\frac{1}{2}}} \xi ^2\rho _n p_n r_n^2\nu _n^2\Vert {\pmb \tau }_n\Vert ^2 + o_{\mathrm{P}}(1) \\=: & {} {\tilde{L}}_{n2} + \rho _n (s + {\textstyle {\frac{1}{2}}} \xi \sqrt{p_n}r_n\nu _n^2\Vert {\pmb \tau }_n\Vert ^2) Z_n + o_{\mathrm{P}}(1) \end{aligned}$$

as $n\rightarrow \infty$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$. Since $p_n=o(n^2 r_n^{-4})$, we can apply Lemma 3 with $a_n=\sqrt{n}$ and $T_n \equiv 1$, which provides

$$\begin{aligned} {\tilde{L}}_{n2}= & {} \Big ( \frac{n \kappa _n^2}{4({\textstyle {\frac{p_n}{2}}}-1)} - \frac{n \kappa _n^4}{32({\textstyle {\frac{p_n}{2}}}-1)^3} \Big ) - \Big ( \frac{n \kappa _{n,s,{\pmb \tau }_n}^2}{4({\textstyle {\frac{p_n}{2}}}-1)} - \frac{n \kappa _{n,s,{\pmb \tau }_n}^4}{32({\textstyle {\frac{p_n}{2}}}-1)^3} \Big ) \\&+ \,\xi s \rho _n r_n\sqrt{p_n} + {\textstyle {\frac{1}{2}}} \xi ^2\rho _n p_n r_n^2\nu _n^2\Vert {\pmb \tau }_n\Vert ^2 + o(1) \\= & {} -\frac{n(\kappa _{n,s,{\pmb \tau }_n}^2-\kappa _n^2)}{2p_n-4} + \frac{n(\kappa _{n,s,{\pmb \tau }_n}^4-\kappa _n^4)}{4(p_n-2)^3} + \xi s \rho _n r_n\sqrt{p_n}\\&+\, {\textstyle {\frac{1}{2}}} \xi ^2\rho _n p_n r_n^2\nu _n^2\Vert {\pmb \tau }_n\Vert ^2 + o(1) \\= & {} -\frac{p_n^2r_n^2 \{\rho _n^2 (\xi +s/(\sqrt{p_n}r_n))^2-\xi ^2\}}{2p_n-4} + S_n + \xi s \rho _n r_n\sqrt{p_n} \\&+ \,{\textstyle {\frac{1}{2}}} \xi ^2\rho _n p_n r_n^2\nu _n^2\Vert {\pmb \tau }_n\Vert ^2 +o(1) , \end{aligned}$$

where since $1-\rho _n=-\frac{1}{2}\rho _n\nu _n^2\Vert {\pmb \tau }_n\Vert ^2$,

$$\begin{aligned} S_n= & {} \frac{p_n^4r_n^4\{\rho _n^4 (\xi +s/(\sqrt{p_n}r_n))^4-\xi ^4\}}{4n(p_n-2)^3} = \frac{p_n^4r_n^4}{4n(p_n-2)^3} \Big ( (\rho _n^4-1) \xi ^4 \\&+\,{\textstyle { \sum _{\ell =1}^4 {4 \atopwithdelims ()\ell } \frac{s^\ell \xi ^{4-\ell }}{(\sqrt{p_n}r_n)^\ell } }} \Big )\\= & {} O\Big ( \frac{p_n r_n^4}{n} (\rho _n-1)(\rho _n+1)(\rho _n^2+1) \Big ) + O\Big ( \frac{\sqrt{p_n} r_n^3}{n} \Big ) = O\Big ( \frac{p_n r_n^4 \nu _n^2}{n} \Big )\\&+\, O\Big ( \frac{\sqrt{p_n} r_n^3}{n} \Big ) = o(1) . \end{aligned}$$

Thus, using the identities $1-\rho _n=-\frac{1}{2}\rho _n\nu _n^2\Vert {\pmb \tau }_n\Vert ^2$ and $\rho _n^2-1=\rho _n^2\nu _n^2\Vert {\pmb \tau }_n\Vert ^2-\frac{1}{4}\rho _n^2\nu _n^4\Vert {\pmb \tau }_n\Vert ^4$, we have

$$\begin{aligned} {\tilde{L}}_{n2}= & {} -\frac{(\rho _n^2-1)\xi ^2 p_n^2 r_n^2}{2p_n-4} -\frac{s^2 \rho _n^2 p_n}{2p_n-4} -\frac{2\xi s\rho _n^2 p_n^{3/2}r_n}{2p_n-4} + \xi s \rho _n r_n\sqrt{p_n} \nonumber \\&+\, {\textstyle {\frac{1}{2}}} \xi ^2\rho _n p_n r_n^2\nu _n^2\Vert {\pmb \tau }_n\Vert ^2 +o(1) \nonumber \\= & {} -\frac{1}{2} \xi ^2\rho _n^2 p_n r_n^2 \nu _n^2\Vert {\pmb \tau }_n\Vert ^2 +\frac{1}{8} \xi ^2\rho _n^2 p_n r_n^2 \nu _n^4\Vert {\pmb \tau }_n\Vert ^4 -\frac{1}{2} s^2 \rho _n^2\nonumber \\&-\, \xi s \rho _n^2 \sqrt{p_n}r_n + \xi s \rho _n r_n\sqrt{p_n}+ {\textstyle {\frac{1}{2}}} \xi ^2\rho _n p_n r_n^2\nu _n^2\Vert {\pmb \tau }_n\Vert ^2 +o(1)\nonumber \\= & {} \frac{1}{2} \xi ^2\rho _n(1-\rho _n) p_n r_n^2 \nu _n^2\Vert {\pmb \tau }_n\Vert ^2 +\frac{1}{8} \xi ^2\rho _n^2 p_n r_n^2 \nu _n^4\Vert {\pmb \tau }_n\Vert ^4\nonumber \\&-\,\frac{1}{2} s^2\rho _n^2 +\xi s\rho _n(1-\rho _n)\sqrt{p_n}r_n +o(1) \nonumber \\= & {} -\frac{1}{2} s^2 \rho _n^2 -\frac{1}{2}\xi s \rho _n^2 \sqrt{p_n}r_n \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 -\frac{1}{8} \xi ^2 \rho _n^2 p_n r_n^2 \nu _n^4\Vert {\pmb \tau }_n\Vert ^4 +o(1)\nonumber \\ \end{aligned}$$

(A.25)

as $n\rightarrow \infty$. Therefore, we proved that, as $n\rightarrow \infty$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _n}$,

$$\begin{aligned} L_{n2}= & {} \rho _n (s + {\textstyle {\frac{1}{2}}} \xi \sqrt{p_n}r_n\nu _n^2\Vert {\pmb \tau }_n\Vert ^2) Z_n\nonumber \\&-\,\frac{1}{2} s^2 \rho _n^2 -\frac{1}{2}\xi s \rho _n^2 \sqrt{p_n}r_n \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 -\frac{1}{8} \xi ^2 \rho _n^2 p_n r_n^2 \nu _n^4\Vert {\pmb \tau }_n\Vert ^4 + o_{\mathrm{P}}(1) .\nonumber \\ \end{aligned}$$

(A.26)

We turn to $L_{1n}$. Write $c_{n,s,{\pmb \tau }_n}:=n\nu _n^2 \kappa _{n,s,{\pmb \tau }_n} e_{n1,s,{\pmb \tau }_n}/\sqrt{p_n}$, where $e_{n1,s,{\pmb \tau }_n}$ and ${\tilde{e}}_{2n,s,{\pmb \tau }_n}$ denote the values of $e_{n1}$ and ${\tilde{e}}_{2n}$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0},\kappa _{n,s,{\pmb \tau }_n}}$. Since $\sqrt{p_n} r_n^2 \nu _n^2=O(1)$, (A.30) below ensures that $c_{n,s,{\pmb \tau }_n}$ is O(1). Therefore, Lemma 5 yields

$$\begin{aligned} L_{1n} = L^Z_{1n}+{\tilde{L}}^Z_{1n}+{\bar{L}}^Z_{1n}+L^W_{1n}+o_{\mathrm{P}}(1) , \end{aligned}$$

(A.27)

where we let

$$\begin{aligned} L^Z_{1n}:= & {} - \frac{1}{2} \sqrt{n} \kappa _{n,s,{\pmb \tau }_n} \nu _n^2 \sqrt{{\tilde{e}}_{n2}}\, \Vert {\pmb \tau }_n\Vert ^2 Z_{n} , \quad {\tilde{L}}^Z_{1n} := - \frac{1}{8} \sqrt{p_n}\nu _n^2 c_{n,s,{\pmb \tau }_n} \Vert {\pmb \tau }_n\Vert ^4,\\ {\bar{L}}^Z_{1n}:= & {} \frac{1}{2} n \kappa _{n,s,{\pmb \tau }_n} \nu _n^2 (e_{n1,s,{\pmb \tau }_n}-e_{n1}) \Vert {\pmb \tau }_n\Vert ^2 , \end{aligned}$$

and

$$\begin{aligned} L^W_{1n} := \frac{1}{\sqrt{2}} c_{n,s,{\pmb \tau }_n} \Vert {\pmb \tau }_n\Vert ^2 \Big (1- \frac{1}{4} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2\Big ) {\widetilde{W}}_n - \frac{1}{4} c_{n,s,{\pmb \tau }_n}^2 \Vert {\pmb \tau }_n\Vert ^4 \Big (1- \frac{1}{4} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2\Big )^2 . \end{aligned}$$

Lemma 6 provides

$$\begin{aligned} L^Z_{1n}= & {} - \frac{\sqrt{n} \nu _n^2}{2\sqrt{p_n}} \Vert {\pmb \tau }_n\Vert ^2 \Big ( \frac{\rho _n p_n r_n}{\sqrt{n}} \Big ( \xi + \frac{s}{\sqrt{p_n}r_n} \Big ) \Big ) \sqrt{p{\tilde{e}}_{n2}}\, Z_n\nonumber \\= & {} - \frac{1}{2} \rho _n \sqrt{p_n} r_n \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 \Big ( \xi + \frac{s}{\sqrt{p_n}r_n} \Big ) \Big ( 1+O\Big (\frac{\kappa _n^2}{p_n^2}\Big ) \Big ) Z_n \nonumber \\= & {} - \frac{1}{2} \xi \rho _n \sqrt{p_n} r_n \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 Z_n - \frac{1}{2} s \rho _n \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 Z_n + o_{\mathrm{P}}(1) , \end{aligned}$$

(A.28)

where we used the fact that $\sqrt{p_n}r_n^3 \nu _n^2$ is o(n).

Now, Lemma 4(iii) yields

$$\begin{aligned} e_{n1,s,{\pmb \tau }_n}= & {} \frac{\kappa _{n,s,{\pmb \tau }_n}}{p_n} + O\Big (\frac{\kappa ^3_{n,s,{\pmb \tau }_n}}{p_n^3}\Big ) = \frac{\rho _n r_n}{\sqrt{n}} \Big ( \xi + \frac{s}{\sqrt{p_n}r_n} \Big ) + O\Big ( \frac{r_n^3}{n^{3/2}} \Big ) \nonumber \\= & {} \frac{\xi \rho _n r_n}{\sqrt{n}} + \frac{s\rho _n}{\sqrt{np_n}} + O\Big ( \frac{r_n^3}{n^{3/2}} \Big ) , \end{aligned}$$

(A.29)

so that

$$\begin{aligned} c_{n,s,{\pmb \tau }_n}= & {} \frac{n \nu _n^2}{\sqrt{p_n}} \Big ( \frac{\rho _n p_n r_n}{\sqrt{n}} \Big ( \xi + \frac{s}{\sqrt{p_n}r_n} \Big ) \Big ) \Big ( \frac{\xi \rho _n r_n}{\sqrt{n}} + \frac{s\rho _n}{\sqrt{np_n}} + O\Big ( \frac{r_n^3}{n^{3/2}} \Big ) \Big ) \nonumber \\= & {} \rho _n^2 \sqrt{p_n} r_n^2 \nu _n^2 \Big ( \xi + \frac{s}{\sqrt{p_n}r_n} \Big ) \Big ( \xi + \frac{s}{\sqrt{p_n}r_n} + O\Big ( \frac{r_n^2}{n} \Big ) \Big ) , \end{aligned}$$

(A.30)

which in turn implies that

$$\begin{aligned} {\tilde{L}}^Z_{1n}= & {} - \frac{1}{8} \sqrt{p_n}\nu _n^2 \Big ( \xi ^2 \rho _n^2 \sqrt{p_n} r_n^2 \nu _n^2 + 2\xi s\rho _n^2 r_n \nu _n^2 + s^2 \frac{\rho _n^2 \nu _n^2}{\sqrt{p_n}} + O\Big ( \frac{\sqrt{p_n} r_n^4 \nu _n^2}{n} \Big ) \Big ) \Vert {\pmb \tau }_n\Vert ^4 \nonumber \\= & {} - \frac{1}{8} \xi ^2 \rho _n^2 p_n r_n^2 \nu _n^4 \Vert {\pmb \tau }_n\Vert ^4 - \frac{1}{4} \xi s\rho _n^2 \sqrt{p_n} r_n \nu _n^4 \Vert {\pmb \tau }_n\Vert ^4 - \frac{1}{8} s^2\rho _n^2 \nu _n^4 \Vert {\pmb \tau }_n\Vert ^4 + o(1). \nonumber \\ \end{aligned}$$

(A.31)

Using (A.29) and applying Lemma 4(iii) again, we obtain

$$\begin{aligned} e_{n1,s,{\pmb \tau }_n}-e_{n1}= & {} \bigg ( \frac{\xi \rho _n r_n}{\sqrt{n}} + \frac{s\rho _n}{\sqrt{np_n}} + O\Big ( \frac{r_n^3}{n^{3/2}} \Big ) \bigg ) - \frac{\kappa _{n}}{p_n} + O\Big (\frac{\kappa ^3_{n}}{p_n^3}\Big ) \\= & {} \frac{\xi (\rho _n-1) r_n}{\sqrt{n}} + \frac{s\rho _n}{\sqrt{np_n}} + O\Big ( \frac{r_n^3}{n^{3/2}} \Big ) = \frac{\xi \rho _n r_n\nu _n^2}{2\sqrt{n}} \Vert {\pmb \tau }_n\Vert ^2 + \frac{s\rho _n}{\sqrt{np_n}} \\&+\, O\Big ( \frac{r_n^3}{n^{3/2}} \Big ) , \end{aligned}$$

so that

$$\begin{aligned} {\bar{L}}^Z_{1n}= & {} \frac{1}{2} n \Big ( \frac{\rho _n p_n r_n}{\sqrt{n}} \Big ( \xi + \frac{s}{\sqrt{p_n}r_n} \Big ) \Big ) \nu _n^2 \bigg ( \frac{\xi \rho _n r_n\nu _n^2}{2\sqrt{n}} \Vert {\pmb \tau }_n\Vert ^2 + \frac{s\rho _n}{\sqrt{np_n}} + O\Big ( \frac{r_n^3}{n^{3/2}} \Big ) \bigg ) \Vert {\pmb \tau }_n\Vert ^2 \nonumber \\= & {} \frac{1}{2} \rho _n p_n r_n \sqrt{n} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 \xi \bigg ( \frac{\xi \rho _n r_n\nu _n^2}{2\sqrt{n}} \Vert {\pmb \tau }_n\Vert ^2 + \frac{s\rho _n}{\sqrt{np_n}} + O\Big ( \frac{r_n^3}{n^{3/2}} \Big ) \bigg ) \nonumber \\&+ \,\frac{1}{2} \rho _n p_n r_n \sqrt{n} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 \frac{s}{\sqrt{p_n}r_n} \bigg ( \frac{\xi \rho _n r_n\nu _n^2}{2\sqrt{n}} \Vert {\pmb \tau }_n\Vert ^2 + \frac{s\rho _n}{\sqrt{np_n}} + O\Big ( \frac{r_n^3}{n^{3/2}} \Big ) \bigg ) \nonumber \\= & {} \frac{1}{4}\xi ^2 \rho _n^2 p_n r_n^2 \nu _n^4 \Vert {\pmb \tau }_n\Vert ^4 + \frac{1}{2}\xi s\rho _n^2 \sqrt{p_n} r_n \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 \nonumber \\&+\, \frac{1}{4} \xi s \rho _n^2 \sqrt{p_n} r_n \nu _n^4 \Vert {\pmb \tau }_n\Vert ^4 + \frac{1}{2} s^2 \rho _n^2 \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 + o(1) , \end{aligned}$$

(A.32)

where we used the facts that $p_n r_n^4 \nu _n^2$ and $\sqrt{p_n}r_n^3 \nu _n^2$ are o(n).

Jointly with (A.26), (A.28), (A.31), and (A.32), this shows that

$$\begin{aligned} \varLambda ^{(n)\mathrm{inv}}_{{\pmb \theta }_n, \kappa _{n,s,{\pmb \tau }_n}/{\pmb \theta }_{n0},\kappa _n} - L^W_{1n}= & {} s \rho _n (1 - {\textstyle {\frac{1}{2}}} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2) Z_n\\&-\, \frac{1}{2} s^2 \rho _n^2 (1- \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2+{\textstyle {\frac{1}{4}}} \nu _n^4 \Vert {\pmb \tau }_n\Vert ^4) + o_{\mathrm{P}}(1)\\= & {} s Z_n - \frac{1}{2} s^2 + o_{\mathrm{P}}(1) . \end{aligned}$$

The result thus follows from the definition of $L^W_{1n}$ and the fact that (A.30) implies that $c_{n,s,{\pmb \tau }_n}=1+o(1)$ in case (a), $c_{n,s,{\pmb \tau }_n}=\xi ^2+o(1)$ in case (b), and $c_{n,s,{\pmb \tau }_n}=o(1)$ in case (c) (in each case, the asymptotic normality result of ${\pmb \varDelta }_n$ follows from Lemma 2). $\square$

Technical proofs for Sect. 3

The proof of Theorem 5 requires the following lemma.

Lemma 7

Let $\mathbf{M}_n:={\pmb \theta }_n{\pmb \theta }_n^{\prime }-{\pmb \theta }_{n0}{\pmb \theta }_{n0}^{\prime }$, where $({\pmb \theta }_n)$ and $({\pmb \theta }_{n0})$ belong to $\mathcal {S}^{p_n-1}$. Then, for any real numbers a, b, c, d, we have that $\mathrm{tr}\big [ \mathbf{M}_n^\ell ( a {\pmb \theta }_n{\pmb \theta }_n^{\prime }+ b (\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) ) \mathbf{M}_n^\ell ( c\, {\pmb \theta }_n{\pmb \theta }_n^{\prime }+ d (\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) ) \big ]$ is equal to $(ad+bc) (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)+ (a-b)(c-d) (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)^2$ for $\ell =1$ and to $(ac+bd) (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)^2$ for $\ell =2$.

Proof of Lemma 7

Direct computations yield

$$\begin{aligned} \mathbf{M}_n^2= & {} {\pmb \theta }_n{\pmb \theta }_n^{\prime }+ {\pmb \theta }_{n0}{\pmb \theta }_{n0}^{\prime }- ({\pmb \theta }_n^{\prime }{\pmb \theta }_{n0}) {\pmb \theta }_{n0}{\pmb \theta }_n^{\prime }- ({\pmb \theta }_n^{\prime }{\pmb \theta }_{n0}) {\pmb \theta }_n{\pmb \theta }_{n0}^{\prime }\quad \text { and } \\ \mathbf{M}_n^4= & {} (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2) \mathbf{M}_n^2 . \end{aligned}$$

This provides $\mathrm{tr}[\mathbf{M}^2_n] = 2(1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)$ and $\mathrm{tr}[\mathbf{M}^4_n] = 2(1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)^2 ,$ and allows to show that ${\pmb \theta }_n^{\prime }\mathbf{M}_n^2{\pmb \theta }_n = 1 - ({\pmb \theta }_n^{\prime }{\pmb \theta }_{n0})^2$ and ${\pmb \theta }_n^{\prime }\mathbf{M}_n^4{\pmb \theta }_n = (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)^2 .$ Since ${\pmb \theta }_n^{\prime }\mathbf{M}_n{\pmb \theta }_n = 1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2$, this yields

$$\begin{aligned}&\mathrm{tr}\Big [ \mathbf{M}_n ( a {\pmb \theta }_n{\pmb \theta }_n^{\prime }+ b (\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) ) \mathbf{M}_n ( c\, {\pmb \theta }_n{\pmb \theta }_n^{\prime }+ d (\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) ) \Big ]\\&\quad = \mathrm{tr}\Big [ \mathbf{M}_n ( b \mathbf{I}_{p_n} + (a-b) {\pmb \theta }_n{\pmb \theta }_n^{\prime }) \mathbf{M}_n ( d \mathbf{I}_{p_n} + (c-d) {\pmb \theta }_n{\pmb \theta }_n^{\prime }) \Big ]\\&\quad = bd\, \mathrm{tr}[\mathbf{M}_n^2 ] + b(c-d) {\pmb \theta }_n^{\prime }\mathbf{M}_n^2 {\pmb \theta }_n + (a-b)d {\pmb \theta }_n^{\prime }\mathbf{M}_n^2 {\pmb \theta }_n + (a-b)(c-d) ({\pmb \theta }_n^{\prime }\mathbf{M}_n {\pmb \theta }_n)^2\\&\quad = 2bd {\pmb \theta }_n^{\prime }\mathbf{M}_n {\pmb \theta }_n + \{ b(c-d) + (a-b)d \} {\pmb \theta }_n^{\prime }\mathbf{M}_n {\pmb \theta }_n + (a-b)(c-d) ({\pmb \theta }_n^{\prime }\mathbf{M}_n {\pmb \theta }_n)^2\\&\quad = (ad+bc) (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2) + (a-b)(c-d) (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)^2 \end{aligned}$$

and

$$\begin{aligned}&\mathrm{tr}\Big [ \mathbf{M}_n^2 ( a {\pmb \theta }_n{\pmb \theta }_n^{\prime }+ b (\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) ) \mathbf{M}_n ( c\, {\pmb \theta }_n{\pmb \theta }_n^{\prime }+ d (\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) ) \Big ]\\&\quad = \mathrm{tr}\Big [ \mathbf{M}_n^2 ( b \mathbf{I}_{p_n} + (a-b) {\pmb \theta }_n{\pmb \theta }_n^{\prime }) \mathbf{M}_n^2 ( d \mathbf{I}_{p_n} + (c-d) {\pmb \theta }_n{\pmb \theta }_n^{\prime }) \Big ]\\&\quad = bd\, \mathrm{tr}[\mathbf{M}_n^4 ] + b(c-d) {\pmb \theta }_n^{\prime }\mathbf{M}_n^4 {\pmb \theta }_n + (a-b)d {\pmb \theta }_n^{\prime }\mathbf{M}_n^4 {\pmb \theta }_n + (a-b)(c-d) ({\pmb \theta }_n^{\prime }\mathbf{M}_n^2 {\pmb \theta }_n)^2\\&\quad = 2bd ({\pmb \theta }_n^{\prime }\mathbf{M}_n {\pmb \theta }_n)^2 + \{ b(c-d) + (a-b)d \} ({\pmb \theta }_n^{\prime }\mathbf{M}_n {\pmb \theta }_n)^2 + (a-b)(c-d) ({\pmb \theta }_n^{\prime }\mathbf{M}_n {\pmb \theta }_n)^2\\&\quad = \{ ad+ bc + (a-b)(c-d) \} (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)^2\\&\quad = (ac+bd) (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)^2 , \end{aligned}$$

as was to be showed. $\square$

Proof of Theorem 5

All expectations and variances below are taken under $\mathrm{P}^{(n)}_{{\pmb \theta }_n,F_n}$, with ${\pmb \theta }_n={\pmb \theta }_{n0}+\nu _n {\pmb \tau }_n$, and stochastic convergences are under the corresponding sequence of hypotheses. We have

$$\begin{aligned} \mathrm{E}[\mathbf{X}_{ni}]=e_{n1} {\pmb \theta }_n \quad \text { and } \quad \mathrm{E}[\mathbf{X}_{ni}{} \mathbf{X}_{ni}^{\prime }] = e_{n2} {\pmb \theta }_n{\pmb \theta }_n^{\prime }+ \frac{f_{n2}}{p_n-1} (\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) ; \end{aligned}$$

see the proof of Lemma B.3 in [12]. Writing $\mathbf{W}_{ni}:=(\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) \mathbf {X}_{ni}$, this implies that

$$\begin{aligned} \mathrm{E}[\mathbf{W}_{ni}]=\mathbf{0} \quad \text { and } \quad \mathrm{E}[\mathbf{W}_{ni}{} \mathbf{W}_{ni}^{\prime }] = \frac{f_{n2}}{p_n-1} (\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) . \end{aligned}$$

(B.1)

Writing $\mathbf{M}_n={\pmb \theta }_n{\pmb \theta }_n^{\prime }-{\pmb \theta }_{n0}{\pmb \theta }_{n0}^{\prime }$ as in Lemma 7 and $\mathbf {Y}_{ni}:=(\mathbf{I}_{p_n}- {\pmb \theta }_{n0}{\pmb \theta }_{n0}^{\prime }) \mathbf {X}_{ni}$, we have $\mathbf{W}_{ni}=\mathbf {Y}_{ni}-\mathbf{M}_n \mathbf {X}_{ni}$. This allows to decompose $W^*_n$ as

$$\begin{aligned} W^*_n:= & {} \frac{\sqrt{2(p_n-1)}}{nf_{n2}} \sum _{1 \le i<j \le n} \mathbf {Y}_{ni}^{\prime }\mathbf {Y}_{nj} \\= & {} \frac{\sqrt{2(p_n-1)}}{nf_{n2}} \sum _{1 \le i <j \le n}( \mathbf{W}_{ni}^{\prime }\mathbf{W}_{nj}+\mathbf {X}_{ni}^{\prime }\mathbf{M}_n\mathbf{W}_{nj}+\mathbf{W}_{ni}^{\prime }\mathbf{M}_n\mathbf {X}_{nj}+ \mathbf {X}_{ni} ^{\prime }\mathbf{M}_n^2 \mathbf {X}_{nj})\\=: & {} W^*_{n0} + W^*_{na} + W^*_{nb} + W^*_{nc} . \end{aligned}$$

From (B.1), $\mathrm{E}[W^*_{na}]=\mathrm{E}[W^*_{nb}]=0$. Now,

$$\begin{aligned} \mathrm{Var}[W^*_{na}]= & {} \frac{2(p_n-1)}{n^2f_{n2}^2} \sum _{1 \le i<j \le n} \sum _{1 \le r<s<n} \mathrm{E}[ \mathbf {X}_{ni}^{\prime }\mathbf{M}_n\mathbf{W}_{nj} \mathbf {X}_{nr}^{\prime }\mathbf{M}_n\mathbf{W}_{ns}]\\= & {} \frac{2(p_n-1)}{n^2f_{n2}^2} \sum _{1 \le i<j \le n} \sum _{1 \le r<s<n} \mathrm{tr}[\mathrm{E}[\mathbf{M}_n\mathbf {X}_{nr}\mathbf {X}_{ni}^{\prime }\mathbf{M}_n\mathbf{W}_{nj} \mathbf{W}_{ns}^{\prime }]]\\= & {} \frac{2(p_n-1)}{n^2f_{n2}^2} \sum _{1 \le i<j \le n} \sum _{1 \le r<s<n} c_{n,ijrs} . \end{aligned}$$

Clearly, $c_{n,ijrs}=0$ if $s\ne j$. Lemma 7 entails that for $s=j$ and $r \ne i$, we have

$$\begin{aligned} c_{n,ijrs}= & {} \mathrm{tr}[\mathbf{M}_n\mathrm{E}[\mathbf {X}_{nr}\mathbf {X}_{ni}^{\prime }] \mathbf{M}_n\mathrm{E}[\mathbf{W}_{nj}{} \mathbf{W}_{nj}^{\prime }] ]\\= & {} \mathrm{tr}\Big [ \mathbf{M}_n (e_{n1}^2 {\pmb \theta }_n{\pmb \theta }_n^{\prime }) \mathbf{M}_n \Big (\frac{f_{n2}}{p_n-1} (\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime })\Big ) \Big ]\\= & {} \frac{e_{n1}^2f_{n2}}{p_n-1} (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2) - \frac{e_{n1}^2f_{n2}}{p_n-1} (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)^2 \end{aligned}$$

and that, for $s=j$ and $r=i$, we have

$$\begin{aligned} c_{n,ijrs}= & {} \mathrm{tr}[\mathbf{M}_n\mathrm{E}[\mathbf {X}_{ni}\mathbf {X}_{ni}^{\prime }] \mathbf{M}_n\mathrm{E}[\mathbf{W}_{nj}{} \mathbf{W}_{nj}^{\prime }] ]\\= & {} \mathrm{tr}\Big [ \mathbf{M}_n \Big ( e_{n2} {\pmb \theta }_n{\pmb \theta }_n^{\prime }+ \frac{f_{n2}}{p_n-1} (\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) \Big ) \mathbf{M}_n \Big (\frac{f_{n2}}{p_n-1} (\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime })\Big ) \Big ]\\= & {} \frac{e_{n2}f_{n2}}{p_n-1} (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2) - \frac{(p_n e_{n2}-1)f_{n2}}{(p_n-1)^2} (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)^2 . \end{aligned}$$

We conclude that

$$\begin{aligned} \mathrm{Var}[W^*_{na}]= & {} \frac{2(p_n-1)}{n^2f_{n2}^2} \Bigg [ \frac{n(n-1)(n-2)}{3} \bigg ( \frac{e_{n1}^2f_{n2}}{p_n-1} (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2) \\&-\, \frac{e_{n1}^2f_{n2}}{p_n-1} (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)^2 \bigg )\\&+ \,\frac{n(n-1)}{2} \bigg ( \frac{e_{n2}f_{n2}}{p_n-1} (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2) - \frac{(p_n e_{n2}-1)f_{n2}}{(p_n-1)^2} (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)^2 \bigg ) \Bigg ]\\= & {} \frac{n-1}{3n} \Bigg [ \frac{2(n-2)e_{n1}^2+3e_{n2}}{f_{n2}} (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2) \\&- \,\bigg ( \frac{2(n-2)e_{n1}^2}{f_{n2}} + \frac{3(p_n e_{n2}-1)}{(p_n-1)f_{n2}} \bigg ) (1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2)^2 \Bigg ] . \end{aligned}$$

Since ${\pmb \theta }_n'{\pmb \theta }_{n0} = ({\pmb \theta }_{n0}+\nu _n{\pmb \tau }_n)'{\pmb \theta }_{n0} = 1+\nu _n ({\pmb \tau }_n'{\pmb \theta }_{n0}) = {\textstyle {1-\frac{1}{2} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2}} ,$ we have $1 - ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2=O(\nu _n^2)$, which, by using the fact that $\nu _n=O(1)$, yields

$$\begin{aligned} \mathrm{Var}[W^*_{na}] = \frac{np_n e_{n1}^2+p_n e_{n2}+1}{p_n f_{n2}} \, O(\nu _n^2) . \end{aligned}$$

The same computations provide $\mathrm{Var}[W^*_{nb}]=\mathrm{Var}[W^*_{na}]$. Turning to $W^*_{nc}$, we directly obtain

$$\begin{aligned} \mathrm{E}[W^*_{nc}]= & {} \frac{\sqrt{2(p_n-1)}}{nf_{n2}} \times \frac{n(n-1)}{2} \, e_{n1}^2 {\pmb \theta }_n^{\prime }\mathbf{M}_n^2{\pmb \theta }_n\\= & {} \frac{(n-1)(p_n-1)^{1/2} e_{n1}^2}{\sqrt{2}f_{n2}} \, (1-({\pmb \theta }_n^{\prime }{\pmb \theta }_{n0})^2)\\= & {} \frac{np_n^{1/2}e_{n1}^2}{\sqrt{2}f_{n2}} \, \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 (1-{\textstyle {\frac{1}{4}}} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2)(1+o(1)) ; \end{aligned}$$

see the Proof of Lemma 7. As for the variance,

$$\begin{aligned} \mathrm{Var}[W^*_{nc}]= & {} \frac{2(p_n-1)}{n^2f_{n2}^2} \sum _{1 \le i<j \le n} \sum _{1 \le r<s\le n} \mathrm{Cov}[ \mathbf {X}_{ni} ^{\prime }\mathbf{M}_n^2 \mathbf {X}_{nj}, \mathbf {X}_{nr} ^{\prime }\mathbf{M}_n^2 \mathbf {X}_{ns}]\\= & {} \frac{2(p_n-1)}{n^2f_{n2}^2} \sum _{1 \le i<j \le n} \sum _{1 \le r<s\le n} \Big ( \mathrm{E}[ \mathbf {X}_{ni} ^{\prime }\mathbf{M}_n^2 \mathbf {X}_{nj} \mathbf {X}_{nr} ^{\prime }\mathbf{M}_n^2 \mathbf {X}_{ns}] \\&-\, (e_{n1}^2 {\pmb \theta }_n^{\prime }\mathbf{M}_n^2{\pmb \theta }_n)^2 \Big )\\= & {} \frac{2(p_n-1)}{n^2f_{n2}^2} \sum _{1 \le i<j \le n} \sum _{1 \le r<s\le n} \Big ( \mathrm{tr}[\mathrm{E}[\mathbf {X}_{ni}^{\prime }\mathbf{M}_n^2 \mathbf{X}_{nj} \mathbf{X}_{ns}^{\prime }\mathbf{M}_n^2 \mathbf {X}_{nr}]] \\&-\, e_{n1}^4 (1-({\pmb \theta }_n^{\prime }{\pmb \theta }_{n0})^2)^2 \Big )\\= & {} \frac{2(p_n-1)}{n^2f_{n2}^2} \sum _{1 \le i<j \le n} \sum _{1 \le r <s\le n} \Big ( d_{n,ijrs} - e_{n1}^4 (1-({\pmb \theta }_n^{\prime }{\pmb \theta }_{n0})^2)^2 \Big ) . \end{aligned}$$

We consider three cases. (1) If i, j, r, s contain two pairs of equal indices (equivalently, if $r=i$ and $s=j$), then

$$\begin{aligned} d_{n,ijrs}= & {} \mathrm{tr}[\mathbf{M}_n^2 \mathrm{E}[\mathbf {X}_{ni}\mathbf {X}_{ni}^{\prime }] \mathbf{M}_n^2 \mathrm{E}[\mathbf{X}_{nj} \mathbf{X}_{nj}^{\prime }]]\\= & {} \mathrm{tr}\bigg [ \mathbf{M}_n^2 \Big ( e_{n2}{\pmb \theta }_n{\pmb \theta }_n^{\prime }+ \frac{f_{n2}}{p_n-1}(\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) \Big ) \mathbf{M}_n^2 \Big ( e_{n2}{\pmb \theta }_n{\pmb \theta }_n^{\prime }+ \frac{f_{n2}}{p_n-1}(\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) \Big ) \bigg ]\\= & {} \Big ( e_{n2}^2 + \frac{f_{n2}^2}{(p_n-1)^2} \Big ) (1-({\pmb \theta }_n^{\prime }{\pmb \theta }_{n0})^2)^2 . \end{aligned}$$

(2) If i, j, r, s contain exactly one pair of equal indices, then

$$\begin{aligned} d_{n,ijrs}= & {} \mathrm{tr}[\mathbf{M}_n^2 \mathrm{E}[\mathbf {X}_{ni}\mathbf {X}_{ni}^{\prime }] \mathbf{M}_n^2 \mathrm{E}[\mathbf{X}_{nj} \mathbf{X}_{ns}^{\prime }]]\\= & {} \mathrm{tr}\bigg [ \mathbf{M}_n^2 \Big ( e_{n2}{\pmb \theta }_n{\pmb \theta }_n^{\prime }+ \frac{f_{n2}}{p_n-1}(\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) \Big ) \mathbf{M}_n^2 \big ( e_{n1}^2{\pmb \theta }_n{\pmb \theta }_n^{\prime }\big ) \bigg ]\\= & {} e_{n1}^2 e_{n2} (1-({\pmb \theta }_n^{\prime }{\pmb \theta }_{n0})^2)^2 . \end{aligned}$$

(3) If the indices i, j, r, s are pairwise different, then

$$\begin{aligned} d_{n,ijrs}= & {} \mathrm{tr}[\mathbf{M}_n^2 \mathrm{E}[\mathbf {X}_{nr}\mathbf {X}_{ni}^{\prime }] \mathbf{M}_n^2 \mathrm{E}[\mathbf{X}_{nj} \mathbf{X}_{ns}^{\prime }]]\\= & {} \mathrm{tr}\Big [ \mathbf{M}_n^2 \big ( e_{n1}^2{\pmb \theta }_n{\pmb \theta }_n^{\prime }\big ) \mathbf{M}_n^2 \big ( e_{n1}^2{\pmb \theta }_n{\pmb \theta }_n^{\prime }\big ) \Big ]\\= & {} e_{n1}^4 (1-({\pmb \theta }_n^{\prime }{\pmb \theta }_{n0})^2)^2 . \end{aligned}$$

Therefore,

$$\begin{aligned} \mathrm{Var}[W^*_{nc}]= & {} \frac{2(p_n-1)}{n^2f_{n2}^2} \bigg [ \frac{n(n-1)}{2} \Big ( e_{n2}^2 + \frac{f_{n2}^2}{(p_n-1)^2} \Big ) + n(n-1)(n-2) e_{n1}^2 e_{n2} \\&+\, \frac{n(n-1)(n-2)(n-3)}{4} \, e_{n1}^4 - \frac{n^2(n-1)^2}{4} \, e_{n1}^4 \bigg ] (1-({\pmb \theta }_n^{\prime }{\pmb \theta }_{n0})^2)^2\\= & {} \frac{(n-1)(p_n-1){\tilde{e}}_{n2}({\tilde{e}}_{n2}+2(n-1)e_{n1}^2)}{nf_{n2}^2} (1-({\pmb \theta }_n^{\prime }{\pmb \theta }_{n0})^2)^2 + o(1) . \end{aligned}$$

This finally yields

$$\begin{aligned} \mathrm{Var}[W^*_{nc}] = \frac{p_n {\tilde{e}}_{n2}^2+np_n e_{n1}^2 {\tilde{e}}_{n2}}{f_{n2}^2} \, O(\nu _n^4) . \end{aligned}$$

Summarizing, $W^*_{n} = W^*_{n0}+W^*_{na}+W^*_{nb}+W^*_{nc} ,$ where $W^*_{n0}$ is asymptotically standard normal (see Theorem 3.1 from [25]),

$$\begin{aligned} \mathrm{E}[W^*_{na}]= & {} \mathrm{E}[W^*_{nb}]=0 , \quad \mathrm{E}[W^*_{nc}] = \frac{np_n^{1/2}e_{n1}^2}{\sqrt{2}f_{n2}} \, \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2 (1-{\textstyle {\frac{1}{4}}} \nu _n^2 \Vert {\pmb \tau }_n\Vert ^2)(1+o(1)) ,\\ \mathrm{Var}[W^*_{na}]= & {} \mathrm{Var}[W^*_{nb}] = \frac{np_n e_{n1}^2+p_n e_{n2}+1}{p_n f_{n2}} \, O(\nu _n^2) \end{aligned}$$

and

$$\begin{aligned} \mathrm{Var}[W^*_{nc}] = \frac{p_n {\tilde{e}}_{n2}^2+np_n e_{n1}^2 {\tilde{e}}_{n2}}{f_{n2}^2} \, O(\nu _n^4) . \end{aligned}$$

We can now consider the several cases of the theorem. In cases (i)–(iii), the sequence $(\nu _n)$ involved, namely $\nu _n=\sqrt{f_{n2}}/(\sqrt{n}p_n^{1/4}e_{n1})$, is o(1), so that $\mathrm{E}[W^*_{nc}]=t^2/\sqrt{2}+o(1)$. In all three cases, one checks that $\mathrm{Var}[W^*_{n\ell }]=o(1)$ for $\ell =a,b,c$ (note that in cases (ii)–(iii), the fact that $e_{n2}\le e_{n1}$ implies that both $e_{n2}$ and ${\tilde{e}}_{n2}$ are o(1)), which establishes that $W^*_n {\mathop {\longrightarrow }\limits ^{\mathcal {D}}} \mathcal {N}\big (\frac{t^2}{\sqrt{2}},1\big )$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0}+\nu _n {\pmb \tau }_n,F_n}$. In case (iv), we have, with $\nu _n=1$, $\mathrm{E}[W^*_{nc}]= \frac{\xi ^2t^2}{\sqrt{2}} \big (1-\frac{t^2}{4} \big ) +o(1)$. Since $\sqrt{p_n} e_{n2}=o(1)$ by assumption, one can check that $\mathrm{Var}[W^*_{n\ell }]=o(1)$ for $\ell =a,b,c$, which yields $W^*_n {\mathop {\longrightarrow }\limits ^{\mathcal {D}}} \mathcal {N}\big ( \frac{\xi ^2t^2}{\sqrt{2}} \big (1-\frac{t^2}{4} \big ) ,1 \big )$ under $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0}+\nu _n {\pmb \tau }_n,F_n}$, as was to be showed. Finally, in case (v), still with $\nu _n=1$, we have $\mathrm{E}[W^*_{nc}]=o(1)$. One can again check that $\mathrm{Var}[W^*_{n\ell }]=o(1)$ for $\ell =a,b,c$, which yields that $W^*_n$ is asymptotically standard normal. This establishes the result. $\square$

We turn to the proof of Theorem 6, that will make use of the following lemma.

Lemma 8

Under $\mathrm{P}^{(n)}_{{\pmb \theta }_n,F_n}$,

$$\begin{aligned} \mathrm{E}[(\mathbf {X}_{n1}'{\pmb \theta }_{n0})^2] = e_{n2} ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 + \frac{f_{n2}}{p_n-1} (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 ) \end{aligned}$$

and

$$\begin{aligned} \mathrm{E}[(\mathbf {X}_{n1}'{\pmb \theta }_{n0})^4]= & {} e_{n4} ({\pmb \theta }_n'{\pmb \theta }_{n0})^4 + \frac{6(e_{n2}-e_{n4})}{p_n-1} ({\pmb \theta }_n'{\pmb \theta }_{n0})^2 (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 )\\&+\,\frac{3f_{n4}}{p_n^2-1} (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 )^2 . \end{aligned}$$

Proof of Lemma 8

All computations in this proof are performed under $\mathrm{P}^{(n)}_{{\pmb \theta }_n,F_n}$, which leads us to consider the tangent-decomposition $\mathbf {X}_{n1}=U_{n1}{\pmb \theta }_n+V_{n1}\mathbf {S}_{n1}$ of $\mathbf {X}_{n1}$ with respect to ${\pmb \theta }_n$. Since $\mathbf {X}_{n1}$ is rotationally symmetric with respect to ${\pmb \theta }_n$, $\mathbf {S}_{n1}$ is equal in distribution to ${\pmb \varGamma }_{{\pmb \theta }_n}\mathbf {U}_n$, where $\mathbf {U}_n$ is uniformly distributed over the unit sphere $\mathcal {S}^{p_n-2}$ in ${\mathbb {R}}^{p_n-1}$ and where ${\pmb \varGamma }_{{\pmb \theta }_n}$ is an arbitrary $p_n\times (p_n-1)$ matrix whose columns form an orthonormal basis of the orthogonal complement of ${\pmb \theta }_n$ in ${\mathbb {R}}^{p_n}$ (so that ${\pmb \varGamma }_{{\pmb \theta }_n}'{\pmb \varGamma }_{{\pmb \theta }_n}=\mathbf {I}_{p_n-1}$ and ${\pmb \varGamma }_{{\pmb \theta }_n}{\pmb \varGamma }_{{\pmb \theta }_n}'=\mathbf {I}_{p_n}-{\pmb \theta }_n{\pmb \theta }_n'$). In particular,

$$\begin{aligned} \mathrm{E}[\mathbf {S}_{n1}]=\mathbf {0} \quad \text {and} \quad \mathrm{E}[\mathbf {S}_{n1}\mathbf {S}_{n1}'] = \frac{1}{p_n-1} \, (\mathbf{I}_{p_n}- {\pmb \theta }_n{\pmb \theta }_n^{\prime }) . \end{aligned}$$

This readily yields

$$\begin{aligned} \mathrm{E}[(\mathbf {X}_{n1}'{\pmb \theta }_{n0})^2]= & {} \mathrm{E}[(U_{n1}{\pmb \theta }_n'{\pmb \theta }_{n0}+V_{n1} \mathbf {S}_{n1}'{\pmb \theta }_{n0})^2]\\= & {} \mathrm{E}[U_{n1}^2] ({\pmb \theta }_n'{\pmb \theta }_{n0})^2 + 2 \mathrm{E}[U_{n1} V_{n1}] \mathrm{E}[\mathbf {S}_{n1}'{\pmb \theta }_{n0}]({\pmb \theta }_n'{\pmb \theta }_{n0}) \\&+\,\mathrm{E}[V_{n1}^2] {\pmb \theta }_{n0}'\mathrm{E}[\mathbf {S}_{n1}\mathbf {S}_{n1}'] {\pmb \theta }_{n0} \\= & {} e_{n2} ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 + \frac{f_{n2}}{p_n-1} (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 ) . \end{aligned}$$

Using the identity $U_{n1}^2 V_{n1}^2=U_{n1}^2-U_{n1}^4$, we obtain similarly

$$\begin{aligned}&\mathrm{E}[(\mathbf {X}_{n1}'{\pmb \theta }_{n0})^4] \nonumber \\&\quad = \mathrm{E}[(U_{n1}{\pmb \theta }_n'{\pmb \theta }_{n0}+V_{n1} \mathbf {S}_{n1}'{\pmb \theta }_{n0})^4] \nonumber \\&\quad = e_{n4} ({\pmb \theta }_n'{\pmb \theta }_{n0})^4 + 6 \mathrm{E}[U_{n1}^2 V_{n1}^2 (\mathbf {S}'_{n1}{\pmb \theta }_{n0})^2] ({\pmb \theta }_n'{\pmb \theta }_{n0})^2 + f_{n4} \mathrm{E}[ (\mathbf {S}'_{n1}{\pmb \theta }_{n0})^4] \nonumber \\&\quad = e_{n4} ({\pmb \theta }_n'{\pmb \theta }_{n0})^4 + 6 (e_{n2}-e_{n4}) \, ( {\pmb \theta }_{n0}'\mathrm{E}[\mathbf {S}_{n1}\mathbf {S}_{n1}'] {\pmb \theta }_{n0}) ({\pmb \theta }_n'{\pmb \theta }_{n0})^2 + f_{n4} \mathrm{E}[ (\mathbf {S}'_{n1}{\pmb \theta }_{n0})^4] \nonumber \\&\quad = e_{n4} ({\pmb \theta }_n'{\pmb \theta }_{n0})^4 + \frac{6(e_{n2}-e_{n4})}{p_n-1} ({\pmb \theta }_n'{\pmb \theta }_{n0})^2 (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 ) + f_{n4} \mathrm{E}[ (\mathbf {S}'_{n1}{\pmb \theta }_{n0})^4] .\nonumber \\ \end{aligned}$$

(B.2)

Standard formulas for the Kronecker product yield

$$\begin{aligned}&\mathrm{E}[(\mathbf {S}'_{n1}{\pmb \theta }_{n0})^4]\nonumber \\&\quad = \mathrm{E}[({\pmb \theta }_{n0}^{\prime }\mathbf {S}_{n1}\mathbf {S}_{n1}^{\prime }{\pmb \theta }_{n0})^2 ]\\&\quad = ({\pmb \theta }_{n0}\otimes {\pmb \theta }_{n0})^{\prime }\mathrm{E}[\mathrm{vec}(\mathbf {S}_{n1}\mathbf {S}_{n1}^{\prime })\mathrm{vec}'(\mathbf {S}_{n1}\mathbf {S}_{n1}^{\prime })] ({\pmb \theta }_{n0}\otimes {\pmb \theta }_{n0}) \\&\quad = ({\pmb \theta }_{n0}\otimes {\pmb \theta }_{n0})^{\prime }({\pmb \varGamma }_{{\pmb \theta }_n}\otimes {\pmb \varGamma }_{{\pmb \theta }_n}) \mathrm{E}[ \mathrm{vec}(\mathbf {U}_{n}\mathbf {U}_{n}^{\prime }) \mathrm{vec}'(\mathbf {U}_{n}\mathbf {U}_{n}^{\prime }) ] ({\pmb \varGamma }_{{\pmb \theta }_n}^{\prime }\otimes {\pmb \varGamma }_{{\pmb \theta }_n}^{\prime }) ({\pmb \theta }_{n0}\otimes {\pmb \theta }_{n0})\\&\quad = \frac{1}{p_n^2-1} ({\pmb \theta }_{n0}\otimes {\pmb \theta }_{n0})^{\prime }({\pmb \varGamma }_{{\pmb \theta }_n}\otimes {\pmb \varGamma }_{{\pmb \theta }_n}) \big ( \mathbf {I}_{(p_n-1)^2} + \mathbf {K}_{p_n-1} + \mathbf {J}_{p_n-1} \big ) \\&\qquad ({\pmb \varGamma }_{{\pmb \theta }_n}^{\prime }\otimes {\pmb \varGamma }_{{\pmb \theta }_n}^{\prime }) ({\pmb \theta }_{n0}\otimes {\pmb \theta }_{n0}) , \end{aligned}$$

where $\mathbf {K}_{\ell }$ is the $\ell \times \ell$ commutation matrix and where we let $\mathbf {J}_{\ell }=(\mathrm{vec}\,\mathbf {I}_\ell )(\mathrm{vec}\,\mathbf {I}_\ell )'$; see [35], page 244. Using the fact that $\mathbf {K}_\ell (\mathbf {A}\otimes \mathbf {B})=(\mathbf {A}\otimes \mathbf {B})\mathbf {K}_{\ell '}$ for $\ell \times \ell '$ matrices $\mathbf {A}$ and $\mathbf {B}$, along with the identity $\mathbf {K}_1=1$, we obtain

$$\begin{aligned} \mathrm{E}[(\mathbf {S}'_{n1}{\pmb \theta }_{n0})^4]= & {} \frac{2}{p_n^2-1} ({\pmb \theta }_{n0}\otimes {\pmb \theta }_{n0})^{\prime }({\pmb \varGamma }_{{\pmb \theta }_n}\otimes {\pmb \varGamma }_{{\pmb \theta }_n}) ({\pmb \varGamma }_{{\pmb \theta }_n}^{\prime }\otimes {\pmb \varGamma }_{{\pmb \theta }_n}^{\prime }) ({\pmb \theta }_{n0}\otimes {\pmb \theta }_{n0})\\&+ \,\frac{1}{p_n^2-1} ({\pmb \theta }_{n0}\otimes {\pmb \theta }_{n0})^{\prime }\mathrm{vec}({\pmb \varGamma }_{{\pmb \theta }_n}{\pmb \varGamma }_{{\pmb \theta }_n}^{\prime }) \mathrm{vec}'({\pmb \varGamma }_{{\pmb \theta }_n}{\pmb \varGamma }_{{\pmb \theta }_n}^{\prime }) ({\pmb \theta }_{n0}\otimes {\pmb \theta }_{n0})\\= & {} \frac{3}{p_n^2-1} (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 )^2 . \end{aligned}$$

Plugging this in (B.2) provides the result. $\square$

Proof of Theorem 6

Fix a sequence of hypotheses $\mathrm{P}^{(n)}_{{\pmb \theta }_{n0}+\nu _n {\pmb \tau }_n,F_n}$ associated with a given regime (i) to (v) in Theorem 5. Throughout the proof, stochastic convergences, expectations and variances refer to this sequence of hypotheses. In view of the decomposition ${\widetilde{W}}_n - W^*_n = L_n^{-1}(1-L_n) W^*_n$ from (A.2), it is sufficient to show that $L_n$ converges to one in quadratic mean (note indeed that Theorem 5 indeed implies that $W^*_n$ is $O_{\mathrm{P}}(1)$). In order to do so, write

$$\begin{aligned}&\mathrm{E} \big [ (L_{n}-1)^2 \big ] \\&\quad = \frac{1}{f_{n2}^2} \, \mathrm{E} \Bigg [ \Bigg ( f_{n2} - \bigg [ \frac{1}{n} \sum _{i=1}^{n} V_{ni}^2\bigg ] \Bigg )^2 \Bigg ] = \frac{1}{f_{n2}^2} \, \mathrm{E} \Bigg [ \Bigg ( \bigg [ \frac{1}{n} \sum _{i=1}^{n} (\mathbf {X}_{ni}'{\pmb \theta }_{n0})^2\bigg ] - e_{n2} \Bigg )^2 \Bigg ]\\&\quad = \frac{1}{f_{n2}^2} \, \mathrm{E} \Bigg [ \Bigg ( \bigg [ \frac{1}{n} \sum _{i=1}^{n} (\mathbf {X}_{ni}'{\pmb \theta }_{n0})^2\bigg ] - \mathrm{E}[(\mathbf {X}_{n1}'{\pmb \theta }_{n0})^2] + \mathrm{E}[(\mathbf {X}_{n1}'{\pmb \theta }_{n0})^2] - e_{n2} \Bigg )^2 \Bigg ]\\&\quad \le \frac{2}{f_{n2}^2} \, \mathrm{Var} \Bigg [ \frac{1}{n} \sum _{i=1}^{n} (\mathbf {X}_{ni}'{\pmb \theta }_{n0})^2 \Bigg ] + \frac{2}{f_{n2}^2} \big ( \mathrm{E}[(\mathbf {X}_{n1}'{\pmb \theta }_{n0})^2] - e_{n2}\big )^2 \,\\&\quad \le \frac{2}{nf_{n2}^2} \, \Big ( \mathrm{E} \big [ (\mathbf {X}_{n1}'{\pmb \theta }_{n0})^4 \big ] - \big ( \mathrm{E} \big [ (\mathbf {X}_{n1}'{\pmb \theta }_{n0})^2 \big ] \big )^2 \Big ) + \frac{2}{f_{n2}^2} \big ( \mathrm{E}[(\mathbf {X}_{n1}'{\pmb \theta }_{n0})^2] - e_{n2}\big )^2 \,\\&\quad =: 2T_{na} + 2T_{nb} , \end{aligned}$$

say. Since $f_{n2}=1-e_{n2}$, Lemma 8 provides

$$\begin{aligned} \mathrm{E}[(\mathbf {X}_{n1}'{\pmb \theta }_{n0})^2] - e_{n2} = \bigg ( \frac{f_{n2}}{p_n-1}-e_{n2}\bigg ) (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 ) = \frac{1-p_ne_{n2}}{p_n-1} (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 ) , \end{aligned}$$

which yields

$$\begin{aligned} T_{nb} = \frac{(1-p_ne_{n2})^2}{(p_n-1)^2f_{n2}^2} (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 )^2 = \frac{1+p_n^2e_{n2}^2}{p_n^2 f_{n2}^2} \, O(\nu _n^4) . \end{aligned}$$

In each of the regimes considered in Theorem 5, we thus obtain that $T_{nb}=o(1)$, irrespective of the fact that $\sqrt{p_n}e_{n2}=o(1)$ or not. Turning to $T_{na}$, Lemma 8 yields

$$\begin{aligned}&nf_{n2}^2 T_{na} \\&\quad = e_{n4} ({\pmb \theta }_n'{\pmb \theta }_{n0})^4 + \frac{6(e_{n2}-e_{n4})}{p_n-1} ({\pmb \theta }_n'{\pmb \theta }_{n0})^2 (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 ) + \,\frac{3f_{n4}}{p_n^2-1} (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 )^2\\&\qquad - \,\bigg ( e_{n2}^2 ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^4 + \frac{2e_{n2}f_{n2}}{p_n-1} ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 ) + \frac{f_{n2}^2}{(p_n-1)^2} (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 )^2 \bigg )\\&\quad = (e_{n4}-e_{n2}^2) ({\pmb \theta }_n'{\pmb \theta }_{n0})^4 + \bigg ( \frac{6(e_{n2}-e_{n4})}{p_n-1} - \frac{2e_{n2} f_{n2}}{p_n-1} \bigg ) ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 )\\&\qquad + \,\bigg ( \frac{3f_{n4}}{p_n^2-1} - \frac{f_{n2}^2}{(p_n-1)^2} \bigg ) (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 )^2 . \end{aligned}$$

Using the facts that $e_{n4}-e^2_{n2} = \mathrm{Var}[U_{n1}^2] = \mathrm{Var}[V_{n1}^2] \le \mathrm{E}[V_{n1}^4] = f_{n4}$ and that $e_{n2}-e_{n4} = \mathrm{E}[U_{n1}^2(1-U_{n1}^2)] \le \mathrm{E}[1-U_{n1}^2] = f_{n2} ,$ we then obtain

$$\begin{aligned} T_{na}\le & {} \frac{f_{n4}}{nf_{n2}^2} \, ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 + \frac{6-2e_{n2}}{n(p_n-1)f_{n2}} ({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 )\\&+ \,\bigg ( \frac{3f_{n4}}{n(p_n^2-1)f_{n2}^2} - \frac{1}{n(p_n-1)^2} \bigg ) (1-({\pmb \theta }_{n0}^{\prime }{\pmb \theta }_n)^2 )^2\\= & {} o(1) + \frac{1}{np_nf_{n2}} \, O(\nu _n^2) + o(\nu _n^4) = o(1) + \frac{1}{np_nf_{n2}} \, O(\nu _n^2) . \end{aligned}$$

Trivially, we then have $T_{na}=o(1)$ in each of the regime considered in Theorem 5, still irrespective of the fact that $\sqrt{p_n}e_{n2}=o(1)$ or not. This establishes the result. $\square$