Testing equality between several populations covariance operators

Abstract

In many situations, when dealing with several populations, equality of the covariance operators is assumed. An important issue is to study whether this assumption holds before making other inferences. In this paper, we develop a test for comparing covariance operators of several functional data samples. The proposed test is based on the Hilbert–Schmidt norm of the difference between estimated covariance operators. In particular, when dealing with two populations, the test statistic is just the squared norm of the difference between the two covariance operators estimators. The asymptotic behaviour of the test statistic under both the null hypothesis and local alternatives is obtained. The computation of the quantiles of the null asymptotic distribution is not feasible in practice. To overcome this problem, a bootstrap procedure is considered. The performance of the test statistic for small sample sizes is illustrated through a Monte Carlo study and on a real data set.

Appendix

Proof of Theorem 1. Denote as ${\mathcal {F}}^{k}={\mathcal {F}}\times \dots \times {\mathcal {F}}$ the $k-$th dimensional product space of identical copies of ${\mathcal {F}}$ and consider the process $\mathbf V_{k,n}=\left( \sqrt{n} (\widetilde{\varvec{\varGamma }}_1-\varvec{\varGamma }_1), \dots , \right. \left. \sqrt{n} (\widetilde{\varvec{\varGamma }}_k-\varvec{\varGamma }_k)\right) ^{{t}}$. Using that $\sqrt{n}_i\left( {\widetilde{\varvec{\varGamma }}_i}-\varvec{\varGamma }_i\right) \buildrel {D}\over \longrightarrow \mathbf U_i$, the independence of the estimated operators and the fact that $n_i/n\rightarrow \tau _i\in (0,1)$, we get that $\mathbf V_{k,n} \buildrel {D}\over \longrightarrow \mathbf V=(\mathbf V_1,\ldots ,\mathbf V_k)^{{t}}$, where $\mathbf V_i=\tau _i^{-1/2}\mathbf U_i$ are independent random processes of ${\mathcal {F}}$ with covariance operators ${\tau _i}^{-1}\varvec{\varUpsilon }_i$. Hence, $\mathbf V_{k,n}$ converges in distribution to a zero mean Gaussian random element $\mathbf V=(\mathbf V_1,\ldots ,\mathbf V_k)^{{t}}\in {\mathcal {F}}^{k}$ with covariance operator $\widetilde{\varvec{\varUpsilon }}=\text{ diag }\left( {\tau _1}^{-1}\varvec{\varUpsilon }_1,\dots , {\tau _k}^{-1}\varvec{\varUpsilon }_k\right)$.

Let $A:{\mathcal {F}}^{k}\rightarrow {\mathcal {F}}^{k-1}$ be the linear operator defined as $A(V_1,\ldots ,V_k)=(V_2-V_1,\ldots ,V_k-V_1)$ and denote as $A^*:{\mathcal {F}}^{k-1}\rightarrow {\mathcal {F}}^{k}$ its adjoint operator. The continuous map theorem guarantees that $A\mathbf V_{k,n} \buildrel {D}\over \longrightarrow \mathbf W$, where $\mathbf W=(W_1,\dots , W_{k-1})^{{t}}= A \mathbf V$ is a zero mean Gaussian random element of ${\mathcal {F}}^{k-1}$ with covariance operator $\varvec{\varUpsilon }_{{w}}=A\widetilde{\varvec{\varUpsilon }}\, A^*$. Moreover, we also obtain that $n \sum _{j=2}^k\Vert (\widetilde{\varvec{\varGamma }}_j-\varvec{\varGamma }_j)-(\widetilde{\varvec{\varGamma }}_1-\varvec{\varGamma }_1)\Vert _{\mathcal {F}}^2 \buildrel {D}\over \longrightarrow \sum _{j=1}^{k-1} \Vert W_j\Vert _{\mathcal {F}}^2=\Vert \mathbf W\Vert _{{\mathcal {F}}^{k-1}}^2$. Let $\upsilon _\ell \in {\mathcal {F}}^{k-1}$ be the orthonormal eigenfunctions of $\varvec{\varUpsilon }_{{w}}$ related to the eigenvalues $\theta _\ell$ ordered in decreasing order. Since $\mathbf W$ is a zero mean Gaussian random element of ${\mathcal {F}}^{k-1}$ with covariance operator $\varvec{\varUpsilon }_{{w}}$, $\mathbf W$ can be written as $\sum _{\ell \,\ge 1} \theta _\ell ^{1/2} Z_\ell \,\upsilon _\ell$ where $Z_\ell$ are i.i.d. random variables such that $Z_\ell \sim N(0,1)$. Hence, $\Vert \mathbf W\Vert _{{\mathcal {F}}^{k-1}}^2=\sum _{\ell \,\ge 1} \theta _\ell Z_\ell ^2$, which leads to the desired result.

It only remains to show (7). Straightforward calculations allow to show that the adjoint operator $A^*:{\mathcal {F}}^{k-1}\rightarrow {\mathcal {F}}^k$ is given by $A^*(w_1,\dots ,w_{k-1})\;=\;(-\sum _{i=1}^{k-1}w_i,w_1,\dots ,w_{k-1})$. Hence, as $\mathbf U_1,\ldots ,\mathbf U_k$ are independent, we obtain that

$$\begin{aligned} \varvec{\varUpsilon }_{{w}}(w_1,\dots ,w_{k-1})&=(A\widetilde{\varvec{\varUpsilon }}\, A^*)(w_1,\dots ,w_{k-1})\\&= \left( \frac{1}{\tau _2}\varvec{\varUpsilon }_2(w_1)+\frac{1}{\tau _1}\varvec{\varUpsilon }_1 \left( \sum _{i=1}^{k-1}w_i\right) ,\dots , \frac{1}{\tau _{k}}\varvec{\varUpsilon }_{k} (w_{k-1}) \right. \\&\left. +\frac{1}{\tau _1}\varvec{\varUpsilon }_1 \left( \sum _{i=1}^{k-1}w_i\right) \right) , \end{aligned}$$

concluding the proof.$\square$

Proof of Corollary 1. Consider the process $\mathbf U_{i,n_i}=\sqrt{n}_i (\widehat{\varvec{\varGamma }}_i-\varvec{\varGamma }_i)$. The independence of the samples and among populations together with the results stated in Dauxois et al. (1982), allow to show that $\mathbf U_{i,n_i}$ are independent and converge in distribution to independent zero mean Gaussian random elements $\mathbf U_i$ of ${\mathcal {F}}$ with covariance operator $\varvec{\varUpsilon }_i$ defined in (1). The result follows now from Theorem 1.$\square$

Proof of Theorem 2. Using that $n_i/n\rightarrow \tau _i$, we get immediately that $\sqrt{n} \left( \widetilde{\varvec{\varGamma }}_i-\varvec{\varGamma }_1\right) \buildrel {D}\over \longrightarrow \varvec{\varDelta }_i+ (1/\sqrt{\tau _i})\mathbf U_i$ where $\mathbf U_i$ is a zero mean Gaussian random element with covariance operator $\varvec{\varUpsilon }_i$ and for $i=1$, $\varvec{\varDelta }_1=\mathbf O$ is the null operator. The fact that the estimators are independent implies that $\mathbf U_i$ can be chosen to be independent so, as in the proof of Theorem 1, we have that $\mathbf V_{k,n}=\left( \sqrt{n} (\widetilde{\varvec{\varGamma }}_1-\varvec{\varGamma }_1), \dots , \sqrt{n} (\widetilde{\varvec{\varGamma }}_k-\varvec{\varGamma }_1)\right) ^{{t}} \buildrel {D}\over \longrightarrow \mathbf V=(\mathbf V_1,\ldots ,\mathbf V_k)^{{t}}$, where $\mathbf V_i= \varvec{\varDelta }_i+ (1/\sqrt{\tau _i})\mathbf U_i$ are independent random processes of ${\mathcal {F}}$ with mean $\varvec{\varDelta }_i$ and covariance operators ${\tau _i}^{-1}\varvec{\varUpsilon }_i$. Hence, $\mathbf V_{k,n}$ converges in distribution to a Gaussian random element $\mathbf V=(\mathbf V_1,\ldots ,\mathbf V_k)^{{t}}\in {\mathcal {F}}^{k}$ with mean $\varvec{\varDelta }=(\varvec{\varDelta }_1,\dots , \varvec{\varDelta }_k)^{{t}}$ and covariance operator $\widetilde{\varvec{\varUpsilon }}=\text{ diag }\left( {\tau _1}^{-1}\varvec{\varUpsilon }_1,\dots , {\tau _k}^{-1}\varvec{\varUpsilon }_k\right)$.

As in the proof of Theorem 1, define $A:{\mathcal {F}}^{k}\rightarrow {\mathcal {F}}^{k-1}$ as the linear operator $A(V_1,\ldots ,V_k)=(V_2-V_1,\ldots ,V_k-V_1)$. Then, $A\mathbf V_{k,n} \buildrel {D}\over \longrightarrow \mathbf W$, where $\mathbf W=(W_1,\dots , W_{k-1})^{{t}}= A \mathbf V$ is a Gaussian random element of ${\mathcal {F}}^{k-1}$ with mean $A\varvec{\varDelta }$ and covariance operator $A\widetilde{\varvec{\varUpsilon }}\, A^*$. Using that $\varvec{\varDelta }_1$ is the null operator, we obtain that $A\varvec{\varDelta }=(\varvec{\varDelta }_2,\dots , \varvec{\varDelta }_k)=\varvec{\varDelta }^{(k-1)}$. Moreover, from the proof of Theorem 1 we get that $A\widetilde{\varvec{\varUpsilon }}\, A^*=\varvec{\varUpsilon }_{{w}}$. Let $\upsilon _\ell \in {\mathcal {F}}^{k-1}$ be the orthonormal eigenfunctions of $\varvec{\varUpsilon }_{{w}}$ related to the eigenvalues $\theta _\ell$ ordered in decreasing order. Since $\mathbf W- \varvec{\varDelta }^{(k-1)}$ is a zero mean Gaussian random element of ${\mathcal {F}}^{k-1}$ with covariance operator $\varvec{\varUpsilon }_{{w}}$, $\mathbf W- \varvec{\varDelta }^{(k-1)}$ can be written as $\sum _{\ell \,\ge 1} \theta _\ell ^{1/2} Z_\ell \,\upsilon _\ell$ where $Z_\ell$ are i.i.d. random variables such that $Z_\ell \sim N(0,1)$. On the other hand, we have the expansion $\varvec{\varDelta }^{(k-1)}=\sum _{\ell \,\ge 1} \eta _\ell \,\upsilon _\ell$, so that $\mathbf W= \sum _{\ell \,\ge 1} \left( \eta _\ell +\theta _\ell ^{1/2} Z_\ell \right) \upsilon _\ell$ and $\Vert \mathbf W\Vert _{{\mathcal {F}}^{k-1}}^2=\sum _{\ell \,\ge 1} \left( \eta _\ell +\theta _\ell ^{1/2} Z_\ell \right) ^2=\sum _{\ell \ge 1} \theta _\ell \left( \eta _\ell \,\theta _\ell ^{-1/2}+ Z_\ell \right) ^2$, which concludes the proof since $T_{k,n}=n \sum _{j=2}^k\Vert (\widehat{\varvec{\varGamma }}_j-\varvec{\varGamma }_1)-(\widehat{\varvec{\varGamma }}_1-\varvec{\varGamma }_1)\Vert _{\mathcal {F}}^2 \buildrel {D}\over \longrightarrow \sum _{j=1}^{k-1} \Vert W_j\Vert _{\mathcal {F}}^2=\Vert \mathbf W\Vert _{{\mathcal {F}}^{k-1}}^2$.$\square$

Proof of Proposition 1. The results in Dauxois et al. (1982) entail that $\sqrt{n_1} \left( \widehat{\varvec{\varGamma }}_{1}-\varvec{\varGamma }_1\right) \buildrel {D}\over \longrightarrow \mathbf U_1$, where $\mathbf U_1$ a zero mean Gaussian random element with covariance operator $\varvec{\varUpsilon }_1$ so, we only have to prove the result for $i\ge 2$. We have the following decomposition for $\sqrt{n_i}(\widehat{\varvec{\varGamma }}_i-\varvec{\varGamma }_1)$,

$$\begin{aligned} \sqrt{n_i}(\widehat{\varvec{\varGamma }}_i-\varvec{\varGamma }_1)\!\!= & {} \sqrt{n_i}\left( \frac{1}{n_i}\sum _{j=1}^{n_i}(X_{i,j}- \overline{X}_{i})\otimes (X_{i,j}-\overline{X}_{i})-\varvec{\varGamma }_1\right) \\= & {} \sqrt{n_i}(\widetilde{\varvec{\varGamma }}_i-\varvec{\varGamma }_1)+{n^{-1/4}}{\sqrt{n_i}}\;\widehat{\varvec{\varGamma }}_{i,WR}+{n^{-1/4}}{\sqrt{n_i}}\;\widehat{\varvec{\varGamma }}_{i,RW}+{n^{-1/2}}\sqrt{n_i}\;\widehat{\varvec{\varDelta }}_i \end{aligned}$$

where

$$\begin{aligned}&\widetilde{\varvec{\varGamma }}_i = \frac{1}{n_i}\sum _{j=1}^{n_i}(W_{i,j}-\overline{W}_{i})\otimes (W_{i,j}-\overline{W}_{i}),\quad \widehat{\varvec{\varDelta }}_i= \frac{1}{n_i}\sum _{j=1}^{n_i}(R_{i,j}-\overline{R}_{i})\otimes (R_{i,j}-\overline{R}_{i})\,,\\&\widehat{\varvec{\varGamma }}_{i,WR} = \frac{1}{n_i}\sum _{j=1}^{n_i}(W_{i,j}-\overline{W}_{i})\otimes (R_{i,j}-\overline{R}_{i}) \quad \text{ and }\\&\widehat{\varvec{\varGamma }}_{i,RW} = \frac{1}{n_i}\sum _{j=1}^{n_i}(R_{i,j}-\overline{R}_{i})\otimes (W_{i,j}-\overline{W}_{i})\,. \end{aligned}$$

Using that $W_{i,j}\sim W_i$ and that the covariance operator of $W_i$ is $\varvec{\varGamma }_1$, from the results in Dauxois et al. (1982) we get that $\sqrt{n_i} \left( \widetilde{\varvec{\varGamma }}_{i}-\varvec{\varGamma }_1\right) \buildrel {D}\over \longrightarrow \mathbf U_i$, where $\mathbf U_i$ a zero mean Gaussian random element with covariance operator $\varvec{\varUpsilon }_i$ given in (11).

It is worth noticing that $\widehat{\varvec{\varGamma }}_{i,WR}$ and $\widehat{\varvec{\varGamma }}_{i,RW}$ are estimators of the cross covariance operators $\varvec{\varGamma }_{i,WR}=\mathbb {E}\left\{ (W_{i}-\mathbb {E}{W}_{i})\otimes \right.$ $\left. (R_{i}- \mathbb {E}{R}_{i})\right\}$ and $\varvec{\varGamma }_{i,RW}=\mathbb {E}\left\{ (R_{i}-\mathbb {E}{R}_{i})\otimes \right. \left. (W_{i}-\mathbb {E}{W}_{i})\right\}$, respectively. The independence between $W_{i}$ and $R_{i}$ entails that $\varvec{\varGamma }_{i,WR}$ is the null operator, which implies that $\sqrt{n_i}\;\widehat{\varvec{\varGamma }}_{i,WR}$ is bounded in probability, so that $n^{-1/4}\sqrt{n_i}\;\widehat{\varvec{\varGamma }}_{i,WR} \buildrel {p}\over \longrightarrow 0$. Similarly, we obtain that $n^{-1/4}\sqrt{n_i}\;\widehat{\varvec{\varGamma }}_{i,RW} \buildrel {p}\over \longrightarrow 0$.

Finally, using the law of large numbers we have that $\widehat{\varvec{\varDelta }}_i$, the empirical covariance operator of $R_{i}$, converges in probability to $\varvec{\varDelta }_i$, so ${n^{-1/2}}\sqrt{n_i}\;\widehat{\varvec{\varDelta }}_i \buildrel {p}\over \longrightarrow \tau _i^{1/2}\varvec{\varDelta }_i$, concluding the proof of a). $\square$

Proof of Proposition 2. As in the proof of Proposition 1, we only have to prove the result for $i\ge 2$. Using the Karhunen–Loéve representation, we can write

$$\begin{aligned} X_{1,j}= & {} \mu _1+\;\sum _{\ell =1}^\infty \lambda _\ell ^{\frac{1}{2}}\,\,f_{1\ell j}\,\,\phi _\ell ,\;1\le j\le n_1\\ X_{i,j}= & {} \mu _i+\;\sum _{\ell =1}^\infty \lambda _\ell ^{\frac{1}{2}}\,\left( 1+\frac{\Delta _{i,\ell }}{\sqrt{n}}\right) ^{\frac{1}{2}}\,f_{i\ell j}\,\phi _\ell \,, \quad 1\le j\le n_i\,,\quad 2\le i\le k, \end{aligned}$$

where $f_{i\ell j}\sim f_{i\ell }$ in (12). For $1\le j\le n_i$, let $Z_{i,j}=\mu _i+\;\sum _{\ell =1}^\infty \lambda _\ell ^{\frac{1}{2}}\,f_{i\ell j}\,\phi _\ell =\mu _i+Z_{0,i,j}$. Denote as

$$\begin{aligned} V_{i,j}= X_{i,j}-Z_{i,j}=\sum _{\ell =1}^\infty \lambda _\ell ^{\frac{1}{2}}\left[ \left( 1+\frac{\Delta _{i,\ell }}{\sqrt{n}}\right) ^{\frac{1}{2}}-1\right] \,f_{i\ell j}\,\phi _\ell \;. \end{aligned}$$

Define the following operators that will be used in the sequel $\widetilde{\varvec{\varGamma }}_i= (1/{n_i}) \sum _{j=1}^{n_i}$ $\left( X_{i,j}-\mu _i\right) \otimes \left( X_{i,j}-\mu _i\right)$ , $\widehat{\varvec{\varGamma }}_{Z_0}= (1/{n_i}) \sum _{j=1}^{n_i} Z_{0,i,j}\otimes Z_{0,i,j}$, $\widehat{\varvec{\varGamma }}_V = (1/{n_i}) \sum _{j=1}^{n_i}$ $V_{i,j}\otimes V_{i,j}$ and finally, $\widetilde{\mathbf A}= (1/{n_i}) \sum _{j=1}^{n_i} (Z_{0,i,j}\otimes V_{i,j}+V_{i,j}\otimes Z_{0,i,j})$, where we avoid the index i for the sake of simplicity. Using that $X_{i,j}-\mu _i=Z_{0, i, j} +V_{i, j}$, we obtain the following expansion $\widetilde{\varvec{\varGamma }}_i=\widehat{\varvec{\varGamma }}_{Z_0}+ \widehat{\varvec{\varGamma }}_V+\widetilde{\mathbf A}$.

The proof will be carried out in several steps, by showing that

$$\begin{aligned} \sqrt{n_i}(\widehat{\varvec{\varGamma }}_i- \widetilde{\varvec{\varGamma }}_i)= & {} o_{\mathbb {P}}(1)\\ \sqrt{n_i} \,\widehat{\varvec{\varGamma }}_V= & {} o_{\mathbb {P}}(1)\\ \sqrt{n_i}\, \widetilde{\mathbf A}&\buildrel {p}\over \longrightarrow&\tau _i^{\frac{1}{2}} \varvec{\varDelta }_i \\ \sqrt{n_i}\,(\widehat{\varvec{\varGamma }}_{Z_0}-\varvec{\varGamma }_1)&\buildrel {D}\over \longrightarrow&\mathbf U_i, \end{aligned}$$

(15)

where $\mathbf U_i$ is a zero mean Gaussian random element with covariance operator $\varvec{\varUpsilon }_i$. Using that, for all $2\le i \le k$, the covariance operator of $Z_{0,i,j}$ is $\varvec{\varGamma }_1$, (18) follows from Dauxois et al. (1982).

We will derive (15). Noticing that $\widehat{\varvec{\varGamma }}_i- \widetilde{\varvec{\varGamma }}_i=-\left( \overline{X}_{i}-\mu _i\right) \otimes \left( \overline{X}_{i}-\mu _i\right)$, it will be enough to prove that $\sqrt{n_i} \left( \overline{X}_{i}-\mu _i\right) =\sqrt{n_i} \left( \overline{Z}_{0,i} +\overline{V}_i \right) =O_\mathbb {P}(1)$, with $\overline{Z}_{0,i}=(1/{n_i}) \sum _{j=1}^{n_i} Z_{0,i,j}$ and $\overline{V}_i=(1/{n_i}) \sum _{j=1}^{n_i} V_{i,j}$.

By the central limit theorem in Hilbert spaces, we get that $\sqrt{n_i}\,\overline{Z}_{0,i}$ converges in distribution, which entails that the process is tight, i.e., $\sqrt{n_i}\,\overline{Z}_{0,i}=O_{\mathbb {P}}(1)$.

We have that

$$\begin{aligned} \left( 1+\frac{\Delta _{i,\ell }}{\sqrt{n}}\right) ^{\frac{1}{2}}-1= \frac{1}{\sqrt{n}}\, \frac{\Delta _{i,\ell }}{\left( 1+\frac{\Delta _{i,\ell }}{\sqrt{n}}\right) ^{\frac{1}{2}}+1}= a_{i,\ell ,n}\frac{\Delta _{i,\ell }}{\sqrt{n}} \end{aligned}$$

(16)

where $0\le a_{i,\ell ,n}\le 1$.

To derive that $\sqrt{n_i}\,\overline{V}_i=O_\mathbb {P}(1)$, we will further show that $\sqrt{n_i}\,\overline{V}_i=o_\mathbb {P}(1)$. To do so, notice that $\mathbb {E}\,\Vert \overline{V}_i\Vert ^2 =({1}/{n_i})\sum _{\ell =1}^\infty \lambda _\ell \left[ \left( 1+({\Delta _{i,\ell }}/{\sqrt{n}})\right) ^{\frac{1}{2}}-1\right] ^2$. Using (19), we get that $\mathbb {E}(\Vert \sqrt{n_i} \,\overline{V}_i\Vert ^2)\le ({1}/{n}) \sum _{\ell =1}^\infty \lambda _\ell \Delta _{i,\ell }^2$, concluding the proof of (15).

To obtain (16), notice that (19) entails that $V_{i,j}\otimes V_{i,j} =(1/{n})\sum _{\ell , s} \lambda _\ell ^{\frac{1}{2}}\lambda _s^{\frac{1}{2}} a_{i,\ell ,n} a_{i,s,n}$ $\Delta _{i,s} \Delta _{i,\ell } f_{i \ell j} f_{i s j}\phi _\ell \otimes \phi _s$,

so if we denote as $U_{\ell s}=({1}/{n_i})\sum _{j=1}^{n_i}f_{i \ell j}\, \,f_{i sj}$, we get that

$$\begin{aligned} \widehat{\varvec{\varGamma }}_V= \frac{1}{n} \sum _{\ell , s} \lambda _\ell ^{\frac{1}{2}}\lambda _s^{\frac{1}{2}}a_{i,\ell ,n} \, a_{i,s,n} \Delta _{i,s} \Delta _{i,\ell } \, U_{\ell s}\,\;\phi _\ell \otimes \phi _s\,. \end{aligned}$$

Recall that $f_{i \ell j}\sim f_{i \ell }$ and $\mathbb {E}(f_{i \ell }f_{i s})=\delta _{\ell s}$, where $\delta _{\ell s}=1$ if $\ell =s$ and 0 otherwise. Hence, we have that $\mathbb {E}(U_{\ell s})= \delta _{\ell s}$ which implies that

$$\begin{aligned} \mathbb {E}(U^2_{\ell s})= & {} {Var}(U_{\ell s})+ \mathbb {E}^2 (U_{\ell s})=\frac{1}{n_i}{Var}(f_{i\ell }f_{is}) + \delta _{\ell s}\nonumber \\\le & {} \frac{1}{n_i}\mathbb {E}(f^2_{i\ell }f^2_{is}) + \delta _{\ell s} \le \frac{1}{n_i}\sigma _{4,i,\ell } \sigma _{4,i,s} + \delta _{\ell s}, \end{aligned}$$

(17)

where the last bound follows from the Cauchy–Schwartz inequality and the fact that $\sigma ^2_{4,i,s}=\mathbb {E}(f^4_{is})$. Hence, using (20) and the fact that $0\le a_{i,\ell ,n}\le 1$, we obtain the bound

$$\begin{aligned} \mathbb {E}(n_i\Vert \widehat{\varvec{\varGamma }}_V\Vert _{\mathcal {F}}^2)\le & {} \frac{n_i}{n^2}\sum _{\ell , s} \lambda _\ell \lambda _s \Delta _{i,s}^2 \Delta _{i,\ell }^2\, \mathbb {E}(U^2_{\ell s})\\\le & {} \frac{n_i}{n^2}\sum _{\ell , s} \lambda _\ell \lambda _s \,\Delta _{i,\ell }^2 \Delta _{i,s}^2\left( \frac{1}{n_i}\sigma _{4,i,\ell }\sigma _{4,i,s}+ \delta _{\ell s}\right) \\= & {} \frac{1}{n^2}\left( \sum _{\ell } \lambda _\ell \,{\Delta _{i,\ell }^2} {\sigma _{4,i,\ell }}\right) ^2+\frac{1}{n} \sum _{\ell } \lambda _\ell ^2\,\Delta _{i,\ell }^4\,. \end{aligned}$$

Therefore, from the fact that $\sum _{\ell } \lambda _\ell ^2\,\Delta _{i,\ell }^4\le \left( \sum _{\ell } \lambda _\ell \,\Delta _{i,\ell }^2\right) ^2<\infty$ we get that $\mathbb {E}({n_i}\Vert \widehat{\varvec{\varGamma }}_V\Vert _{\mathcal {F}}^2)\rightarrow 0$, concluding the proof of (16).

Finally, to derive (17) we perform the decomposition

$$\begin{aligned} \widetilde{\mathbf A}=\frac{1}{\sqrt{n}} \sum _{\ell , s} \lambda _\ell ^{\frac{1}{2}}\lambda _s^{\frac{1}{2}}a_{i,s,n} \Delta _{i,s} \, U_{\ell s}\,\left( \phi _\ell \;\otimes \phi _s+\phi _s\;\otimes \phi _\ell \right) =\widetilde{\mathbf A}_1+\widetilde{\mathbf A}_2 \end{aligned}$$

where $U_{\ell s}=({1}/{n_i})\sum _{j=1}^{n_i}f_{i \ell j}\, \,f_{isj}$, as before. We will show that $\sqrt{n_i}\left( \widetilde{\mathbf A}_j-\mathbb {E}(\widetilde{\mathbf A}_j)\right)$ $\buildrel {p}\over \longrightarrow 0$, for $j=1,2$ which entails that $\sqrt{n_i}\left( \widetilde{\mathbf A}-\mathbb {E}(\widetilde{\mathbf A})\right) \buildrel {p}\over \longrightarrow 0$. We will only prove that $\sqrt{n_i}\left( \widetilde{\mathbf A}_1-\mathbb {E}(\widetilde{\mathbf A}_1)\right) \buildrel {p}\over \longrightarrow 0$, since the other one follows similarly. From (20), we get that ${Var}(U_{\ell s})\le ({1}/{n_i})\sigma _{4,i,\ell } \sigma _{4,i,s}$ which together with the fact that $0\le a_{i,\ell ,n}\le 1$ leads to

$$\begin{aligned} \mathbb {E}(n_i\Vert \widetilde{\mathbf A}_1-\mathbb {E}(\widetilde{\mathbf A}_1)\Vert _{\mathcal {F}}^2)= & {} \frac{n_i}{n} \sum _{\ell , s} \lambda _\ell \lambda _s a_{i,s,n}^2 \Delta _{i,s}^2 {Var}(U_{\ell s})\\\le & {} \frac{1}{n} \sum _{\ell , s} \lambda _\ell \lambda _s \Delta _{i,s}^2 \sigma _{4,i,\ell } \sigma _{4,i,s} \\= & {} \frac{1}{\sqrt{n}}\left( \sum _{\ell }\lambda _\ell \sigma _{4,i,\ell }\right) \left( \sum _{\ell }\lambda _\ell \sigma _{4,i,\ell }\Delta _{i,\ell }^2\right) \end{aligned}$$

so that $\sqrt{n_i}\left( \widetilde{\mathbf A}_1-\mathbb {E}(\widetilde{\mathbf A}_1)\right) \buildrel {p}\over \longrightarrow 0$, as desired. Besides, using that $\mathbb {E}(U_{\ell s})=\delta _{\ell s}$, we get that

$$\begin{aligned} \mathbb {E}(\sqrt{n_i}\, \widetilde{\mathbf A})= & {} \frac{2\, \sqrt{n_i}}{\sqrt{n}} \sum _{\ell } \lambda _\ell \, a_{i,\ell ,n} \Delta _{i,\ell } \; \phi _\ell \otimes \phi _\ell \rightarrow \tau _i^{\frac{1}{2}} \sum _{\ell =1}^\infty \lambda _\ell {\Delta _{i,\ell }}\phi _\ell \otimes \phi _\ell = \tau _i^{\frac{1}{2}} \varvec{\varDelta }_i \end{aligned}$$

where we have used that $a_{i,\ell ,n} \rightarrow 1/2$, as $n\rightarrow \infty$ and $\sum _{\ell =1}^\infty \lambda _\ell |{\Delta _{i,\ell }}|<\infty$. This concludes the proof of (17). The proof of Proposition 2a) follows now combining (15) to (18). $\square$

Proof of Theorem 3. Recall that $\widetilde{X}_n=(X_{1,1},\ldots ,X_{1,n_1},\dots , X_{k,1},\ldots ,X_{k,n_k})$. Let $\widetilde{Z}_n=(Z_1,\ldots ,Z_{q_n})$ and $\widetilde{Z}=\{Z_\ell \}_{\ell \,\ge 1}$ with $Z_i\sim N(0,1)$ independent. Define $\widehat{{\mathcal {U}}}_n(\widetilde{X}_n,\widetilde{Z}_n)=\sum _{\ell =1}^{q_n}\widehat{\theta }_\ell Z_\ell ^2$, ${{\mathcal {U}}}_n(\widetilde{Z}_n)=\sum _{\ell =1}^{q_n} \theta _\ell Z_\ell ^2$ and ${{\mathcal {U}}}(\widetilde{Z})=\sum _{\ell =1}^{\infty } \theta _\ell Z_\ell ^2$. It is worth noticing that $\widehat{{\mathcal {U}}}_n$ has the same distribution as ${\mathcal {U}}^*_n$.

First notice that, for any $\ell$, $|\widehat{\theta }_\ell -\theta _\ell | \le \Vert \widehat{\varvec{\varUpsilon }}_{{w}}-\varvec{\varUpsilon }_{{w}}\Vert _{{\mathcal {G}}^{k-1}}$ (see, for instance, Kato (1966)), which implies that

$$\begin{aligned} \sum _{\ell =1}^{q_n}\vert \widehat{\theta }_\ell -\theta _\ell \vert \le \frac{q_n}{\sqrt{n}}\,\sqrt{n} \Vert \widehat{\varvec{\varUpsilon }}_{{w}}-\varvec{\varUpsilon }_{{w}}\Vert _{{\mathcal {G}}^{k-1}}\;. \end{aligned}$$

(18)

On the other hand, we have

$$\begin{aligned} \mathbb {E}\left[ |\widehat{{\mathcal {U}}}_n -{{\mathcal {U}}}|\vert \widetilde{X}_n\right] =\mathbb {E}\left[ |\widehat{{\mathcal {U}}}_n -{{\mathcal {U}}}_n+{{\mathcal {U}}}_n-{{\mathcal {U}}}|\;\vert \tilde{X}_n\right] \le \sum _{\ell =1}^{q_n}\vert \widehat{\theta }_\ell -\theta _\ell \vert +\sum _{\ell >q_n}\theta _\ell \end{aligned}$$

which together with (21), the fact that $\sqrt{n} \Vert \widehat{\varvec{\varUpsilon }}_{{w}}-\varvec{\varUpsilon }_{{w}}\Vert =O_{\mathbb {P}}(1)$, $q_n/\sqrt{n}\rightarrow 0$ and $\sum _{\ell \,\ge 1} \theta _\ell <\infty$ implies that

$$\begin{aligned} \mathbb {E}\left[ |\widehat{{\mathcal {U}}}_n -{{\mathcal {U}}}|\;\vert \widetilde{X}_n \right] \buildrel {p}\over \longrightarrow 0\,. \end{aligned}$$

(19)

We also have the following inequalities

$$\begin{aligned} \mathbb {P}(\widehat{{\mathcal {U}}}_n\le t\vert \widetilde{X}_n)= & {} \mathbb {P}(\widehat{{\mathcal {U}}}_n\le t \cap |\widehat{{\mathcal {U}}}_n-{{\mathcal {U}}}|<\epsilon \;\vert \widetilde{X}_n)+ \mathbb {P}(\widehat{{\mathcal {U}}}_n\le t \cap |\widehat{{\mathcal {U}}}_n-{{\mathcal {U}}}|>\epsilon \;\vert \widetilde{X}_n)\\\le & {} \mathbb {P}( {{\mathcal {U}}}\le t +\epsilon )+ \mathbb {P}(|\widehat{{\mathcal {U}}}_n-{{\mathcal {U}}}|>\epsilon \;\vert \widetilde{X}_n)\\\le & {} F_{{\mathcal {U}}}( t +\epsilon )+ \frac{1}{\epsilon }\mathbb {E}(|\widehat{{\mathcal {U}}}_n-{{\mathcal {U}}}|\;\vert \widetilde{X}_n)\le F_{{\mathcal {U}}}(t)+\Delta _\epsilon (t)+\frac{1}{\epsilon }\mathbb {E}(|\widehat{{\mathcal {U}}}_n-{{\mathcal {U}}}|\;\vert \widetilde{X}_n), \end{aligned}$$

where $\Delta _\epsilon (t)=\sup _{\vert \delta \vert \le \epsilon }\vert F_{{\mathcal {U}}}( t+\delta ) -F_{{\mathcal {U}}}(t)\vert$. Besides,

$$\begin{aligned} \mathbb {P}(\widehat{{\mathcal {U}}}_n\le t\;\vert \widetilde{X}_n)= & {} \mathbb {P}(\widehat{{\mathcal {U}}}_n\le t \cap |\widehat{{\mathcal {U}}}_n-{{\mathcal {U}}}|<\epsilon \;\vert \widetilde{X}_n)+ \mathbb {P}(\widehat{{\mathcal {U}}}_n\le t \cap |\widehat{{\mathcal {U}}}_n-{{\mathcal {U}}}|>\epsilon \;\vert \widetilde{X}_n)\\\ge & {} \mathbb {P}( {{\mathcal {U}}}\le t-\epsilon \cap |\widehat{{\mathcal {U}}}_n-{{\mathcal {U}}}|<\epsilon \;\vert \widetilde{X}_n)\\\ge & {} F_{{\mathcal {U}}}( t -\epsilon )- \frac{1}{\epsilon } \mathbb {E}(|\widehat{{\mathcal {U}}}_n-{{\mathcal {U}}}|\vert \widetilde{X}_n)\ge F_{{\mathcal {U}}}(t) -\Delta _\epsilon (t)-\frac{1}{\epsilon }\mathbb {E}(|\widehat{{\mathcal {U}}}_n-{{\mathcal {U}}}|\vert \widetilde{X}_n)\;. \end{aligned}$$

Therefore,

$$\begin{aligned} |\mathbb {P}(\widehat{{\mathcal {U}}}_n\le t\;\vert \widetilde{X}_n)-F_{{\mathcal {U}}}(t)|\le \Delta _\epsilon (t)+\frac{1}{\epsilon }\mathbb {E}(|\widehat{{\mathcal {U}}}_n-{{\mathcal {U}}}|\;\vert \widetilde{X}_n)\;. \end{aligned}$$

As we mentioned in Remark 1, $F_{{\mathcal {U}}}$ is a continuous distribution function on $\mathbb {R}$ and so uniformly continuous; hence, $\lim _{\epsilon \rightarrow 0}\,\sup _{t\in \mathbb {R}} \;\Delta _\epsilon (t)=0$, which together with (22) implies that $\rho _{{k}}( F_{{{\mathcal {U}}}^*_n\vert \widetilde{X}_n}, F_{{\mathcal {U}}})=\sup _{t}|\mathbb {P}(\widehat{{\mathcal {U}}}_n\le t\;\vert \tilde{X}_n)-F_{{\mathcal {U}}}(t)| \buildrel {p}\over \longrightarrow 0$. $\square$