Appendix: Proofs of main results
This section contains all the proofs of the main theorems stated in Sects. 3 and 4. Since the main idea of the proofs are similar to those in Koul and Song (2012), only differences are presented here for the sake of brevity. In particular, we will focus the discussion on the statistic \(T_n({\hat{\alpha }}_n,{\hat{\beta }}_n,{\hat{\theta }}_n)\), which will be decomposed into two parts, one part can be dealt with directly using Koul and Song (2012)’s argument, and another part involving all terms related to the kernel density estimator \({\hat{f}}_{{\bar{U}} n}\) has to be investigated separately. The discussions on the normalizing constants \({\hat{C}}_n\) and \({\hat{\Gamma }}_n\) are similar to Koul and Song (2012)’s argument, hence omitted for the sake of brevity.
The proof of Theorem 1:
Note that
$$\begin{aligned} \tilde{f}_{\xi ,b}(v;\hat{\beta }_n,\hat{\theta }_n)&=f_{\xi ,b} (v;\hat{\beta }_n,\hat{\theta }_n)+\iint K_b(v-u)f_\varepsilon \left( u+\hat{\beta }_n^T t,\hat{\theta }_n\right) ({\hat{f}}_{{\bar{U}}n}(t)\\&\quad -f_{{\bar{U}}}(t))dudt\\&=f_{\xi ,b}(v;\hat{\beta }_n,\hat{\theta }_n)+R_{bw}(v;{\hat{\beta }}_n,{\hat{\theta }}_n), \end{aligned}$$
then the statistic \(T_n({\hat{\alpha }}_n,{\hat{\beta }}_n,{\hat{\theta }}_n)\) defined in (8) can be written as the sum of the following three terms,
$$\begin{aligned} T_{n1}=&\int [\hat{f}_{\xi ,n}(v;\hat{\alpha }_n,\hat{\beta }_n)-f_{\xi ,b} (v;\hat{\beta }_n,\hat{\theta }_n)]^2d\Pi (v),\\ T_{n2}=&\int [R_{bw}(v;\hat{\beta }_n,\hat{\theta }_n)]^2d\Pi (v),\\ T_{n3}=&-2\int [\hat{f}_{\xi ,n}(v;\hat{\alpha }_n,\hat{\beta }_n)-f_{\xi ,b} (v;\hat{\beta }_n,\hat{\theta }_n)]R_{bw}(v;\hat{\beta }_n,\hat{\theta }_n)d\Pi (v). \ \end{aligned}$$
We start with showing \(nb^{1/2}T_{n2}=o_p(1)\). Adding and subtracting \(f_\varepsilon (u+\beta _0^Tt,\theta _0)\) from \(f_\varepsilon (u+\hat{\beta }_n^Tt,\hat{\theta }_n)\), \(E{\hat{f}}_{{\bar{U}}n}(t)\) from \({\hat{f}}_{{\bar{U}}n}(t)\), \(R_{bw}\) can be written as the sum of the following four terms:
$$\begin{aligned} R_{bw1}= & {} \iint K_b(v-u)\left[ f_\varepsilon \left( u+\hat{\beta }_n^Tt,\hat{\theta }_n\right) -f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) \right] \\&\times \left[ {\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)\right] dudt,\\ R_{bw2}= & {} \iint K_b(v-u)f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) \left( {\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)\right) dudt,\\ R_{bw3}= & {} \iint K_b(v-u)\left[ f_\varepsilon \left( u+\hat{\beta }_n^Tt,\hat{\theta }_n\right) -f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) \right] \\&\times \left[ E{\hat{f}}_{{\bar{U}}n}(t)-f_{{\bar{U}}}(t)\right] dudt,\\ R_{bw4}= & {} \iint K_b(v-u)f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) \left[ E{\hat{f}}_{{\bar{U}}n}(t)-f_{{\bar{U}}}(t)\right] dudt. \end{aligned}$$
It is well known that \(E {\hat{f}}_{{\bar{U}}n}(t)=f_{{\bar{U}}}(t)+w^2\mu _2(L)\text{ tr }(f_{{\bar{U}}}''(t))/2+o(w^2)\). Then from (f1), we can show that \( R_{bw4}=2^{-1}w^2\mu _2(L)\iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0)\text{ tr }(f_{{\bar{U}}}''(t))dudt+o(w^2). \) This, together with (g2), one can easily show that \(|R_{bw4}(v)|=O(w^2)\) uniformly on v, this in turn implies that \(\int R_{bw4}^2(v)d\Pi (v)=O(w^4).\) Hence, by assumption (b1),
$$\begin{aligned} nb^{\frac{1}{2}}\int R_{bw4}^2(v)d\Pi (v)=O(nb^{1/2}w^4)=o(1). \end{aligned}$$
(11)
Now consider \(R_{bw3}\). Denote
$$\begin{aligned}&f_\varepsilon \left( u+\hat{\beta }_n^Tt,\hat{\theta }_n\right) -f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) -(\hat{\beta }_n-\beta _0)^T\dot{f}_{\varepsilon ,\beta }\left( u+\beta _0^Tt,\theta _0\right) \\&\quad -(\hat{\theta }_n-\theta _0)^T\dot{f}_{\varepsilon ,\theta }\left( u+\beta _0^Tt,\theta _0\right) \end{aligned}$$
by \(\Delta f_\varepsilon (t,u;{\hat{\beta }}_n,{\hat{\theta }}_n)\). First we can write \(R_{bw3}\) as the sum of the following two terms,
$$\begin{aligned} R_{bw31}=\iint K_b(v-u) \Delta f_\varepsilon (t,u;{\hat{\beta }}_n,{\hat{\theta }}_n) [E{\hat{f}}_{{\bar{U}}n}(t)-f_{{\bar{U}}}(t)]dudt, \end{aligned}$$
and
$$\begin{aligned} R_{bw32}= & {} \iint K_b(v-u)\left[ (\hat{\beta }_n-\beta _0)^T\dot{f}_{\varepsilon ,\beta } \left( u+\beta _0^Tt,\theta _0\right) \right. \\&\left. +\,(\hat{\theta }_n-\theta _0)^T\dot{f}_{\varepsilon ,\theta } \left( u+\beta _0^Tt,\theta _0\right) \right] \cdot \\&\quad \left[ E{\hat{f}}_{{\bar{U}}n}(t)-f_{{\bar{U}}}(t)\right] dudt. \end{aligned}$$
By (f1), \(R_{bw31}\) is bounded above by
$$\begin{aligned}&\sup _{u,t}\left| \Delta f(t,u;{\hat{\beta }}_n,{\hat{\theta }}_n)\right| \cdot \left[ \frac{1}{2}w^2\mu _2(L) \iint K_b(v-u)\big |\text{ tr }\left( f_{{\bar{U}}}''(t)\right) \big |dudt+o(w^2)\right] \\&\quad =O_p(w^2/n), \end{aligned}$$
and \(R_{bw32}\) is bounded above by
$$\begin{aligned}&\Vert \hat{\beta }_n-\beta _0\Vert \cdot \iint K_b(v-u)\Big \Vert \dot{f}_{\varepsilon ,\beta }\left( u+\beta _0^Tt,\theta _0\right) \Big \Vert \cdot \Big |E{\hat{f}}_{{\bar{U}}n}(t)-f_{{\bar{U}}}(t)\Big |dudt\\&\quad +\Big \Vert \hat{\theta }_n-\theta _0\Big \Vert \cdot \iint K_b(v-u)\Big \Vert \dot{f}_{\varepsilon ,\theta }\left( u+\beta _0^Tt,\theta _0\right) \Big \Vert \cdot \Big |E{\hat{f}}_{{\bar{U}}n}(t)-f_{{\bar{U}}}(t)\Big |dudt. \end{aligned}$$
Note that
$$\begin{aligned}&\iint K_b(v-u)\Big \Vert \dot{f}_{\varepsilon ,\beta }\left( u+\beta _0^Tt,\theta _0\right) \Big \Vert \Big |E{\hat{f}}_{{\bar{U}}n}(t)-f_{{\bar{U}}}(t)\Big |dudt\\&\quad \le O(w^2)\iint K_b(v-u)\Big \Vert \dot{f}_{\varepsilon ,\beta }\left( u+\beta _0^Tt,\theta _0\right) \Big \Vert dudt+o(w^2)=O(w^2), \end{aligned}$$
and by changing variables, \(u=v+bx\), from (f2),
$$\begin{aligned}&\iint K_b(v-u)\Big \Vert \dot{f}_{\varepsilon ,\beta }\left( u+\beta _0^Tt,\theta _0\right) \Big \Vert dudt\\&\quad =\iint K(x)\Big \Vert \dot{f}_{\varepsilon ,\beta }\left( v+bx+\beta _0^Tt,\theta _0\right) \Big \Vert dxdt\\&\quad \le \iint K(x)\Big \Vert \dot{f}_{\varepsilon ,\beta }\left( v+\beta _0^Tt,\theta _0\right) \Big \Vert dxdt+b\iint | x|K(x)B\left( v+\beta _0^Tt,\theta _0\right) dxdt\\&\quad =\int \Big \Vert \dot{f}_{\varepsilon ,\beta }\left( v+\beta _0^Tt,\theta _0\right) \Big \Vert dt+b\int |x|K(x)dx\cdot \int B\left( v+\beta _0^Tt,\theta _0\right) dt. \end{aligned}$$
The \(\sqrt{n}\)-consistency of \({\hat{\beta }}_n\) and \({\hat{\theta }}_n\), and the integrability of \(\dot{f}_{\varepsilon ,\beta }(v+\beta _0^Tt,\theta _0)\) and \(B(v+\beta _0^Tt,\theta _0)\) with respect to t imply \( \int |R_{bw3}|^2d\Pi (v)= 2\left[ O_p\left( n^{-2}w^4\right) +O_p\left( n^{-1}w^4\right) \right] . \) Thus
$$\begin{aligned} nb^{1/2}\cdot \int |R_{bw3}|^2d\Pi (v)=nb^{1/2}O_p\left( \frac{w^4}{n^2} \right) +nb^{1/2}O_p\left( \frac{w^4}{n}\right) =o_p(1). \end{aligned}$$
(12)
Next, we consider \(R_{bw1}\). Adding and subtracting \((\hat{\beta }_n-\beta _0)^T\dot{f}_{\varepsilon ,\beta }(u+\beta _0^Tt,\theta _0) +(\hat{\theta }_n-\theta _0)^T\dot{f}_{\varepsilon ,\theta }(u+\beta _0^Tt,\theta _0)\) from \(f_\varepsilon (u+\hat{\beta }_n^Tt,\hat{\theta }_n)-f_\varepsilon (u+\beta _0^Tt,\theta _0)\), we can rewrite \(R_{bw1}\) as the summation of three terms
$$\begin{aligned} R_{bw11}&=\iint K_b(v-u)[f_\varepsilon \left( u+\hat{\beta }_n^Tt,\hat{\theta }_n\right) -f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) -(\hat{\beta }_n\\&\quad -\beta _0)^T\dot{f}_{\varepsilon ,\beta }\left( u+\beta _0^Tt, \theta _0\right) \\&\quad -(\hat{\theta }_n-\theta _0)^T\dot{f}_{\varepsilon ,\theta }\left( u+\beta _0^Tt,\theta _0\right) ] \left[ {\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)\right] dudt,\\ R_{bw12}&=(\hat{\beta }_n-\beta _0)^T\iint K_b(v-u)\dot{f}_{\varepsilon ,\beta }\left( u+\beta _0^Tt, \theta _0\right) \left[ {\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)\right] dudt,\\ R_{bw13}&=(\hat{\theta }_n-\theta _0)^T\iint K_b(v-u)\dot{f}_{\varepsilon ,\theta } \left( u+\beta _0^Tt,\theta _0\right) \left[ {\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)\right] dudt. \end{aligned}$$
From condition (f1),
$$\begin{aligned} |R_{bw11}|&\le \sup _{t,u}\Big |f_\varepsilon \left( u+\hat{\beta }_n^Tt,\hat{\theta }_n\right) -f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) -(\hat{\beta }_n-\beta _0)^T\dot{f}_{\varepsilon ,\beta } \left( u+\beta _0^Tt,\theta _0\right) \\&-(\hat{\theta }_n-\theta _0)^T\dot{f}_{\varepsilon ,\theta }\left( u+\beta _0^Tt,\theta _0\right) \Big | \iint K_b(v-u)\Big |{\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)\Big |dudt\\ =&\,O_p\left( n^{-1}\right) \int K_b(v-u)du\int \Big |{\hat{f}}_{{\bar{U}}n}(t) -E{\hat{f}}_{{\bar{U}}n}(t)\Big |dt=O_p(n^{-1}). \end{aligned}$$
To consider \(R_{bw12}\) and \(R_{bw13}\), we need an upper bound for \(E\int |{\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)|dt\). By the Cauchy–Schwarz inequality, we have \( E\int |{\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)|dt\le \int (E|{\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)|^2)^{\frac{1}{2}}dt.\) Note that \(E[{\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)]^2\) equals
$$\begin{aligned}&\frac{1}{n}\left\{ {\frac{f_{{\bar{U}}}(t)}{w^d}\int L^2(v)dv+\frac{1}{2w^{d-2}}\int L^2(v) v^Tf_{{\bar{U}}}''(\tilde{t}_1)v dv}\right. \\&\qquad \left. -\left[ f_{{\bar{U}}}(t)+\frac{w^2}{2}\int L(v)v^T f_{{\bar{U}}}''(\tilde{t}_2) vdv\right] ^2\right\} \\&\quad =\frac{f_{{\bar{U}}}(t)}{nw^d}\int L^2(v)dv+\frac{1}{2nw^{d-2}} \int L^2(v) v^Tf_{{\bar{U}}}''(\tilde{t}_1)vdv-\frac{1}{n}f_{{\bar{U}}}^2(t)\\&\qquad -\frac{w^4}{4n}\left( \int L(v)v^Tf_{{\bar{U}}}''(\tilde{t}_2) vdv\right) ^2-\frac{w^2 f_{{\bar{U}}}(t)}{n}\int L(v)v^Tf_{{\bar{U}}}'' (\tilde{t}_2)vdv. \end{aligned}$$
where \(\tilde{t}_1\) and \(\tilde{t}_2\) are between t and \(t+vw\). Then by condition (g2) and (g3), we have \(E\int |{\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)|dt=O((nw^d)^{-1/2})\). Hence \( \int |{\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)|dt=O_p \left( (nw^d)^{-1/2}\right) \).
For \(R_{bw12}\), we have
$$\begin{aligned}&\left\| \iint K_b(v-u)\dot{f}_{\varepsilon ,\beta }\left( u+\beta _0^T t,\theta _0\right) \left[ {\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)\right] dudt\right\| \\&\quad \le \iint K(x) \Big \Vert \dot{f}_{\varepsilon ,\beta }\left( v+bx+\beta _0^Tt,\theta _0\right) \Big \Vert \cdot \Big |{\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)\Big |dxdt\\&\quad \le \iint K(x)\left[ \Big \Vert \dot{f}_{\varepsilon ,\beta }\left( v+\beta _0^T t,\theta _0\right) \Big \Vert b|x|B(v+\beta _0^T t,\theta _0)\right] \\&\qquad \cdot |{\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)|dxdt. \end{aligned}$$
which has the order of \(O_p(1/\sqrt{nw^d})\) by condition (f2). This, together with the \(\sqrt{n}\)-consistency of \({\hat{\beta }}_n\), implies that \(R_{bw12}(v)=O_p((nw^{d/2})^{-1})\) uniformly for v. Similarly, we also have \(R_{bw13}(v)=O_p((nw^{d/2})^{-1})\) uniformly for v as well. Therefore,
$$\begin{aligned} nb^{1/2}\int (R_{bw1}(v))^2d\Pi (v)= nb^{1/2}\cdot O_p\left( \frac{1}{n^2w^d}\right) =O_p\left( \frac{b^{1/2}}{nw^d}\right) =o_p(1) \end{aligned}$$
(13)
from assumption (b2). Next we consider \(R_{bw2}\). Note that
$$\begin{aligned} R_{bw2}(v)&=\frac{1}{nw^d}\iint K_b(v-u)f_\varepsilon (u+\beta _0^T t,\theta _0)\\&\quad \times \left[ \sum _{i=1}^nL\left( \frac{t-\tilde{Z}_i}{w}\right) -\sum _{i=1}^n EL\left( \frac{t-\tilde{Z}_i}{w}\right) \right] dudt\\&=\frac{1}{n}\sum _{i=1}^n\iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0)[L_w(t-\tilde{Z}_i)-EL_w(t-\tilde{Z}_i)]dudt. \end{aligned}$$
Therefore,
$$\begin{aligned} E(R_{bw2}(v))^2=&\frac{1}{n}E\left[ \iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt, \theta _0)[L_w(t-\tilde{Z})\right. \\&\qquad \left. -EL_w(t-\tilde{Z})]dtdu\right] ^2\\ =&\frac{1}{n}\int \left[ \iint K(x)f_\varepsilon (v+bx+\beta _0^Tt,\theta _0)\right. \\&\quad \left. \times \left[ \frac{1}{w^d}L\left( \frac{t-z}{w}\right) -f_{{\bar{U}}}(t)+O(w^2)\right] dtdx\right] ^2f_{\tilde{Z}}(z)dz, \end{aligned}$$
which is the order of \(O(n^{-1})\), implying that
$$\begin{aligned} nb^{1/2}\int (R_{bw2}(v))^2d\Pi (v)=nb^{1/2}O_p\left( n^{-1}\right) =o_p(1). \end{aligned}$$
(14)
Therefore, combining (11), (12), (13) and (14), we eventually show that
$$\begin{aligned} nb^{1/2}T_{n2}=nb^{1/2}\int [R_{bw}(v;\hat{\beta }_n,\hat{\theta }_n)]^2d\Pi (v)=o_p(1) \end{aligned}$$
(15)
Next, we investigate the asymptotic behaviour of \(T_{n3}({\hat{\alpha }}_n,{\hat{\beta }}_n,{\hat{\theta }}_n)\). By the decomposition of \(R_{bw}\), we can see that
$$\begin{aligned} \int [\hat{f}_{\xi ,n}(v;\hat{\alpha }_n,\hat{\beta }_n)-f_{\xi ,b} (v;\hat{\beta }_n,\hat{\theta }_n)]R_{bw}(v;\hat{\beta }_n,\hat{\theta }_n)d\Pi (v)= \sum _{j=1}^4Q_{nj}, \end{aligned}$$
(16)
where \( Q_{nj}=\int [\hat{f}_{\xi ,n}(v;\hat{\alpha }_n,\hat{\beta }_n)-f_{\xi ,b}(v; \hat{\beta }_n,\hat{\theta }_n)]R_{bwj}(v;{\hat{\beta }}_n,\hat{\theta }_n)d\Pi (v). \) From Koul and Song (2012), we know that
$$\begin{aligned}&nb^{\frac{1}{2}}\left[ \int [\hat{f}_{\xi ,n}(v;\hat{\alpha }_n,\hat{\beta }_n) f_{\xi ,b}(v;\hat{\beta }_n,\hat{\theta }_n)]^2d\Pi (v)-\hat{C}_n\right] \Longrightarrow N(0,\Gamma ),\quad \nonumber \\&nb^{\frac{1}{2}}\hat{C}_n=O_p(b^{-1/2}), \end{aligned}$$
(17)
where \(\Gamma \) is defined in (10). By the Cauchy–Schwarz inequality, we can see that \(nb^{1/2}|Q_{n1}|\) is bounded above by
$$\begin{aligned}&\left\{ nb^{\frac{1}{2}}\int \left[ \hat{f}_{\xi ,n}\left( v;\hat{\alpha }_n,\hat{\beta }_n\right) -f_{\xi ,b}\left( v;\hat{\beta }_n,\hat{\theta }_n\right) \right] ^2d\Pi (v)\right\} ^{\frac{1}{2}}\\&\qquad \times \left\{ nb^{\frac{1}{2}} \int \left[ R_{bw1}\left( v;\hat{\beta }_n,\hat{\theta }_n\right) \right] ^2d\Pi (v)\right\} ^{\frac{1}{2}}\\&\quad = \left\{ nb^{\frac{1}{2}}\left[ \int [\hat{f}_{\xi ,n}(v;\hat{\alpha }_n, \hat{\beta }_n)-f_{\xi ,b}(v;\hat{\beta }_n,\hat{\theta }_n)]^2d\Pi (v)-\hat{C}_n\right] +nb^{\frac{1}{2}}\hat{C}_n\right\} ^{\frac{1}{2}}\\&\qquad \cdot \left\{ nb^{\frac{1}{2}}\int [R_{bw1}(v;\hat{\beta }_n,\hat{\theta }_n)]^2 d\Pi (v)\right\} ^{\frac{1}{2}}, \end{aligned}$$
this, together with (13), implies that we can conclude
$$\begin{aligned} nb^{1/2}|Q_{n1}|=O_p(b^{-1/4})\cdot O_p\left( \frac{b^{1/4}}{\sqrt{nw^d}}\right) =o_p(1). \end{aligned}$$
(18)
Similarly, from (12), we can show that
$$\begin{aligned} nb^{1/2}|Q_{n3}|=O_p(b^{-1/4})\cdot O_p\left( b^{1/4}w^2\right) =o_p(1). \end{aligned}$$
(19)
Now we shall show that \(nb^{1/2}Q_{nj}=o_p(1)\) holds for \(j=2, 4\). Recall the definitions of \({\hat{f}}_{\xi ,n}(v;{\hat{\alpha }}_n,{\hat{\beta }}_n)\), \(f_{\xi ,b}(v;{\hat{\beta }}_n,{\hat{\theta }}_n)\), we see that \(nb^{1/2}Q_{n2}\) can be written as
$$\begin{aligned} nb^{1/2}Q_{n2}=&\int \left[ \frac{1}{nb}\sum _{i=1}^nK\Big (\frac{v-\hat{\xi }_i}{b}\Big )-\int K_b(v-u)f_\xi (u;\hat{\beta }_n,\hat{\theta }_n)du\right] \\&\cdot \left[ \iint K_b(v-u)f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) \left( {\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)\right) dudt\right] d\Pi (v)\\ =&\int \bigg [\frac{1}{nb}\sum _{i=1}^n K\Big (\frac{v-Y_i+\hat{\alpha }_n+\hat{\beta }_n^T\bar{Z}_i}{b}\Big )-\frac{1}{nb}\sum _{i=1}^n K\Big (\frac{v-Y_i+\alpha _0+\beta _0^T\bar{Z}_i}{b}\Big )\\&+\frac{1}{nb}\sum _{i=1}^n K\Big (\frac{v-Y_i+\alpha _0+\beta _0^T\bar{Z}_i}{b}\Big )-\int K_b(v-u)f_\xi (u;\beta _0,\theta _0)du\\&+\int K_b(v-u)f_\xi (u;\beta _0,\theta _0)du-\int K_b(v-u)f_\xi (u;\hat{\beta }_n,\hat{\theta }_n)du\bigg ]\\&\cdot \bigg [\iint K_b(v-u) f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) \left[ {\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)\right] dtdu\bigg ]d\Pi (v)\\ =&\int \left[ \frac{1}{nb}\sum _{i=1}^n K\Big (\frac{v-Y_i+\alpha _0+\beta _0^T\bar{Z}_i}{b}\Big )-\int K_b(v-u)f_\xi (u;\beta _0,\theta _0)du\right] \\&\left[ \iint K_b(v-u)f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) \left( {\hat{f}}_{{\bar{U}}n}(t)-E{\hat{f}}_{{\bar{U}}n}(t)\right) dudt\right] d\Pi (v)+R_n, \end{aligned}$$
where the remainder term \(R_n\) converges to 0 faster than the first term. So, it is sufficient to consider the first term only. By the definition of \({\hat{f}}_{{\bar{U}}n}(t)\), we can rewrite the first term as \(S_{n}\),
$$\begin{aligned} S_n=&\frac{1}{n^2}\sum _{i=1}^n\sum _{j=1}^n\int \left[ \frac{1}{b}K\Big (\frac{v-\xi _i}{b}\Big )-E\frac{1}{b}K\Big (\frac{v-\xi }{b}\Big )\right] \\&\cdot \left[ \iint K_b(v-u)f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) \left[ \frac{1}{w^d}L\Big (\frac{t-\tilde{Z}_j}{w}\Big )-E\frac{1}{w^d}L\Big (\frac{t-\tilde{Z}}{w}\Big )\right] dudt\right] \\&\quad d\Pi (v). \end{aligned}$$
Recall the notation \(\xi =Y-\alpha _0-\beta _0^T\bar{Z}=\varepsilon -\beta _0^T\Big (U_1+U_2\Big )/2\), \(\tilde{Z}=(U_1-U_2)/2\). We have
$$\begin{aligned} ES_{n}&=\frac{1}{n}E\int \frac{1}{b}K\Big (\frac{v-\xi }{b}\Big )\left[ \iint K_b(v-u)f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) \frac{1}{w^d}L\Big (\frac{t-\tilde{Z}}{w}\Big )dtdu\right] d\Pi (v)\\&\quad -\frac{1}{n}\int E\frac{1}{b}K\Big (\frac{v-\xi }{b}\Big )\iint K_b(v-u)f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) E\frac{1}{w^d}L\Big (\frac{t-\tilde{Z}}{w}\Big )dtdud\Pi (v)\\&=\frac{1}{n}\int \Bigg [\int \!\!\!\int \!\!\!\int \frac{1}{b}K\Big (\frac{v-\varepsilon +\beta _0^T(u_1+u_2)/2}{b}\Big )\\&\left[ \iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0)\frac{1}{w^d}L\Big (\frac{t-(u_1-u_2)/2}{w}\Big )dtdu\right] \\&f(\varepsilon )f_{U}(u_1)f_{U}(u_2)\ \mathrm {d}\varepsilon \ \mathrm {d}u_1\ \mathrm {d}u_2\Bigg ]d\Pi (v)\\&\quad - \frac{1}{n}\int E\frac{1}{b}K\Big (\frac{v-\xi }{b}\Big )\iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0)E\frac{1}{w^d}L\Big (\frac{t-\tilde{Z}}{w}\Big )dtdud\Pi (v)\\&=\,O(n^{-1}). \end{aligned}$$
We also have
$$\begin{aligned} ES^2_{n}&=E\Bigg [\frac{1}{n^2}\sum _{i,j}\int \left[ \frac{1}{b}K\Big (\frac{v-\xi _i}{b}\Big )-E\frac{1}{b}K\Big (\frac{v-\xi }{b}\Big )\right] \\&\quad \iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0)\Bigg [\frac{1}{w^d}L\Big (\frac{t-\tilde{Z}_j}{w}\Big )-E\frac{1}{w^d}L\Big (\frac{t-\tilde{Z}}{w}\Big )\Bigg ]dudtd\Pi (v)\Bigg ]^2\\&=\frac{1}{n^4}\sum _{i,j}E\Bigg [\int \Big [\frac{1}{b}K\Big (\frac{v-\xi _i}{b}\Big )-E\frac{1}{b}K\Big (\frac{v-\xi }{b}\Big )\Big ]\\&\quad \iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0)\Big [\frac{1}{w^d}L\Big (\frac{t-\tilde{Z}_j}{w}\Big )-E\frac{1}{w^d}L\Big (\frac{t-\tilde{Z}}{w}\Big )\Big ]dudtd\Pi (v)\Bigg ]^2\\&\quad +\frac{n(n-1)}{n^4}E\Bigg [\int \Big [\frac{1}{b}K\Big (\frac{v-\xi _1}{b}\Big )-E\frac{1}{b}K\Big (\frac{v-\xi }{b}\Big )\Big ]\iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0)\\&\quad \Big [\frac{1}{w^d}L\Big (\frac{t-\tilde{Z}_2}{w}\Big )-E\frac{1}{w^d}L\Big (\frac{t-\tilde{Z}}{w}\Big )\Big ]dudtd\Pi (v)\int \Big [\frac{1}{b}K\Big (\frac{v-\xi _2}{b}\Big )-E\frac{1}{b}K\Big (\frac{v-\xi }{b}\Big )\Big ]\\&\quad \iint K_b(v-u)f_\varepsilon (u+\beta _0^T t,\theta _0)\Big [\frac{1}{w^d}L\Big (\frac{t-\tilde{Z}_1}{w}\Big )-E\frac{1}{w^d}L\Big (\frac{t-\tilde{Z}}{w}\Big )\Big ]dudtd\Pi (v)\Bigg ], \end{aligned}$$
which is the order of \(O(n^{-2})\). The expectation and variance arguments implies \(S_{n}=O_p(1/n)\). Hence
$$\begin{aligned} nb^{1/2}Q_{n2}=o_p(1). \end{aligned}$$
(20)
Finally, we are going to prove \(nb^{\frac{1}{2}}Q_{n4}=o_p(1)\). First note that \(nb^{\frac{1}{2}}Q_{n4}\) can be written as the sum of \(nb^{1/2}S_{nj}\), \(j=1,2,3\), where
$$\begin{aligned} nb^{\frac{1}{2}}S_{n1} =&\int \left[ \frac{1}{nb}\sum _{i=1}^n K\Big (\frac{v-Y_i+\alpha _0 +\beta _0^T\bar{Z}_i}{b}\Big )-\int K_b(v-u)f_\xi (u;\beta _0,\theta _0)du\right] \\&\left[ \iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0)(E {\hat{f}}_{{\bar{U}}n} (t)-f_{{\bar{U}}}(t))dudt\right] d\Pi (v),\\ nb^{\frac{1}{2}}S_{n2}=&\int \bigg [\frac{1}{nb}\sum _{i=1}^n K\Big (\frac{v-Y_i+\hat{\alpha }_n +\hat{\beta }_n^T\bar{Z}_i}{b}\Big )-\frac{1}{nb}\sum _{i=1}^n K\Big (\frac{v-Y_i +\alpha _0+\beta _0^T\bar{Z}_i}{b}\Big )\bigg ]\\&\cdot \bigg [\iint K_b(v-u) f_\varepsilon (u+\beta _0^Tt,\theta _0)[E {\hat{f}}_{{\bar{U}}n} (t)-f_{{\bar{U}}}(t)]dtdu\bigg ]d\Pi (v),\\ nb^{\frac{1}{2}}S_{n3}=&\int \bigg [\int K_b(v-u)f_\xi (u;\beta _0,\theta _0)du-\int K_b(v-u)f_\xi (u;\hat{\beta }_n,\hat{\theta }_n)du\bigg ]\\&\cdot \bigg [\iint K_b(v-u) f_\varepsilon (u+\beta _0^Tt,\theta _0)[E {\hat{f}}_{{\bar{U}}n} (t)-f_{{\bar{U}}}(t)]dtdu\bigg ]d\Pi (v). \end{aligned}$$
We can easily see that \(E S_{n1}=0\). From the boundedness of \(f_{{\bar{U}}}''(t)\), we further have
$$\begin{aligned} E S^2_{n1}&\le \,\frac{1}{n}E \Bigg \{\int \left| \frac{1}{b}K\Big (\frac{v-\xi _1}{b}\Big )-E \frac{1}{b}K\Big (\frac{v-\xi }{b}\Big )\right| \\&\quad \times \left[ \iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0)\frac{w^2}{2}\int L(z)|z^T f_{{\bar{U}}}''(\tilde{t})z|dzdudt\right] d\Pi (v)\Bigg \}^2\\&\le \frac{B^2w^4}{4n}E \Bigg \{\int \left| \frac{1}{b}K\Big (\frac{v-\xi _1}{b} \Big )-E \frac{1}{b}K\Big (\frac{v-\xi }{b}\Big )\right| \\&\quad \times \left[ \iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0)dudt\right] d\Pi (v)\Bigg \}^2 \end{aligned}$$
for some finite positive constant B. Note that
$$\begin{aligned}&E \Bigg \{\int \left| \frac{1}{b}K\Big (\frac{v-\xi }{b}\Big )-E \frac{1}{b}K\Big (\frac{v-\xi }{b}\Big )\right| \Bigg [\iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0)dudt\Bigg ]d\Pi (v)\Bigg \}^2 \end{aligned}$$
is the order of O(1), so, we have \(S_{n1}=O_p(w^2/\sqrt{n})\). Thus \(nb^{\frac{1}{2}} S_{n1}=O_p\left( \sqrt{n}b^{\frac{1}{2}}w^2\right) =O_p(\sqrt{nbw^4})=o_p(1).\) Using the Cauchy–Schwarz inequality and from the proof of Theorem 3.1 in Koul and Song (2012), we have
$$\begin{aligned} nb^{1/2}|S_{n2}|&= nb^{\frac{1}{2}}\left| \int \left[ \frac{1}{n} \sum _{i=1}^n(K_b(v-\hat{\xi _i})-K_b(v-\xi _i))\right] \right. \\&\cdot \left. \left[ \iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0)(E {\hat{f}}_{{\bar{U}}n}(t)-f_{{\bar{U}}}(t))dudt\right] d\Pi (v)\right| \\&\le \left\{ nb^{1/2}\int \left[ \frac{1}{n}\sum _{i=1}^n(K_b(v-\hat{\xi _i}) -K_b(v-\xi _i))\right] ^2d\Pi (v)\right\} ^{1/2}\\&\cdot \left\{ nb^{1/2}\int \left[ \iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0) (E {\hat{f}}_{{\bar{U}}n}(t)-f_{{\bar{U}}}(t))dudt\right] ^2d\Pi (v)\right\} ^{1/2}\\&\le o_p(1)O(\sqrt{nb^{1/2}w^4})=o_p(1), \end{aligned}$$
and
$$\begin{aligned} nb^{1/2}S_{n3}&= nb^{\frac{1}{2}}\Big |\int [ f_{\xi ,b}(v;\beta _0, \theta _0)-f_{\xi ,b} (v;\hat{\beta }_n, \hat{\theta }_n)]\\&\quad \cdot \left[ \iint K_b(v-u)f_\varepsilon (u+\beta _0^Tt,\theta _0)(E {\hat{f}}_{{\bar{U}}n}(t)-f_{{\bar{U}}}(t))dudt\right] d\Pi (v)\Big |\\&\le nb^{\frac{1}{2}}\int [(\hat{\theta }_n-\theta _0)^T\dot{f}_{\xi b\theta }(v;\beta _0,\theta _0)+(\hat{\beta }_n-\beta _0)^T\dot{f}_{\xi b\beta }(v;\beta _0,\theta _0)+O_p(1/n)]\\&\quad \cdot \left[ \iint K_b(v-u)f_\varepsilon (u+\beta _0t,\theta _0)\frac{w^2}{2}\int L(z)|z^Tf_{{\bar{U}}}''(\tilde{t})z|dzdudt\right] d\Pi (v)\\&\le O_p(nb^{\frac{1}{2}}w^2/\sqrt{n})=o_p(1). \end{aligned}$$
Therefore, we have
$$\begin{aligned} nb^{1/2}Q_{n4}=o_p(1). \end{aligned}$$
(21)
Combining (18), (19), (20) and (21), we have
$$\begin{aligned} nb^{1/2}T_{n3}=-\,2nb^{1/2}(Q_{n1}+Q_{n2}+Q_{n3}+Q_{n4})=o_p(1). \end{aligned}$$
(22)
Note that
$$\begin{aligned} nb^{1/2}(T_n({\hat{\alpha }}_n,{\hat{\beta }}_n,{\hat{\theta }}_n) -{\hat{C}}_n)=nb^{\frac{1}{2}}\left[ T_{n1}-\hat{C}_n\right] +nb^{1/2}T_{n2}+nb^{1/2}T_{n3}, \end{aligned}$$
combining (15), (17) and (22), we have
$$\begin{aligned} nb^{1/2}(T_n({\hat{\alpha }}_n,{\hat{\beta }}_n,{\hat{\theta }}_n) -{\hat{C}}_n)\Longrightarrow N(0,\Gamma ). \end{aligned}$$
This, together with \({\hat{\Gamma }}_n\rightarrow \Gamma \) in probability, which can be easily shown by the consistency of \({\hat{\alpha }}_n,{\hat{\beta }}_n\) and the kernel density estimator \({\hat{f}}_{\xi ,n}\), completes the proof of Theorem 1.
Proof of Theorem 2:
Define
$$\begin{aligned} \check{f}_{\xi ,b}(v;\beta )=\int K_b(v-u)\tilde{f}_{\xi ,a}(u;\beta )du, \quad \tilde{f}_{\xi ,a}(u;\beta )=\int f_{\varepsilon ,a}(u+\beta ^Tt){\hat{f}}_{{\bar{U}}n}(t)dt, \end{aligned}$$
By adding and subtracting \(\check{f}_{\xi ,b}(v;\hat{\beta }_n)\) from \(\hat{f}_{\xi ,n}(v;\hat{\alpha }_n,\hat{\beta }_n)-\tilde{f}_{\xi ,b}(v;\hat{\beta }_n, \hat{\theta }_n)\), we can rewrite \(T_n=T_{1n}-2T_{2n}+T_{3n}\), where
$$\begin{aligned} T_{1n}=&\int [\hat{f}_{\xi ,n}(v;\hat{\alpha }_n,\hat{\beta }_n)-\check{f}_{\xi ,b} (v;\hat{\beta }_n)]^2d\Pi (v),\\ T_{2n}=&\int [\hat{f}_{\xi ,n}(v;\hat{\alpha }_n,\hat{\beta }_n)-\check{f}_{\xi ,b} (v;\hat{\beta }_n)][\check{f}_{\xi ,b}(v;\hat{\beta }_n)-\tilde{f}_{\xi ,b} (v;\hat{\beta }_n,\hat{\theta }_n)]d\Pi (v),\\ T_{3n}=&\int [\check{f}_{\xi ,b}(v;\hat{\beta }_n)-\tilde{f}_{\xi ,b}(v;\hat{\beta }_n, \hat{\theta }_n)]^2d\Pi (v). \end{aligned}$$
Therefore,
$$\begin{aligned} \mathcal {T}_n=nb^{\frac{1}{2}}\hat{\Gamma }_n^{-\frac{1}{2}}(T_{1n}-\hat{C}_n)-2nb^{\frac{1}{2}} \hat{\Gamma }_n^{-\frac{1}{2}}T_{2n}+nb^{\frac{1}{2}} \hat{\Gamma }_n^{-\frac{1}{2}}T_{3n}. \end{aligned}$$
One can show that \(nb^{\frac{1}{2}}\hat{\Gamma }_n^{-\frac{1}{2}}(T_{1n}-\hat{C}_n)\Rightarrow N(0,1).\) The proof is similar to that of Theorem 1. Note that \( \hat{\Gamma }_n\rightarrow 2\int f_{\xi ,a}^2(v;\beta _a)\pi ^2(v)dv\int K_*^2(u)du={\tilde{\Gamma }}>0, \) where \(K_*(u)=\int K(t)K(u+t)dt\), and
$$\begin{aligned} T_{3n}=&\int \left[ \iint K_b(v-u)[f_{\varepsilon ,a}(u+\hat{\beta }_n^Tt)-f_\varepsilon (u+ \hat{\beta }_n^Tt,\hat{\theta }_n)] {\hat{f}}_{{\bar{U}}n}(t)dtdu\right] ^2d\Pi (v)\\ =&\int \left[ \iint K(x)[f_{\varepsilon ,a}(v+bx+\hat{\beta }_n^Tt)-f_\varepsilon (v+bx +\hat{\beta }_n^Tt,\hat{\theta }_n)] {\hat{f}}_{{\bar{U}}n}(t)dtdx\right] ^2d\Pi (v)\\ \rightarrow&\int \left[ \int f_{\varepsilon ,a}(v+\beta ^T_a t)f_{{\bar{U}}}(t)dt-\int f_\varepsilon (v+\beta ^T_a t, \theta _a)f_{{\bar{U}}}(t)dt\right] ^2d\Pi (v)\\ =&\int [f_{\xi ,a}(v;\beta _a)-f_\xi (v;\beta _a,\theta _a)]^2d\Pi (v)>0 \end{aligned}$$
we have \( nb^{1/2}\hat{\Gamma }_n^{-1/2}T_{3n}=nb^{1/2}{\tilde{\Gamma }}^{-1/2} \int [f_{\xi ,a}(v;\beta _a)-f_\xi (v;\beta _a,\theta _a)]^2d\Pi (v)+o_p(nb^{1/2})\) as \(n\rightarrow \infty \).
By the Cauchy–Schwarz inequality, and using the fact \(\hat{C}_n=O_p(1/(nb))\) from Koul and Song (2012), \(nb^{\frac{1}{2}}\hat{\Gamma }_n^{-1/2}|T_{2n}|\) is bounded above by
$$\begin{aligned}&[nb^{\frac{1}{2}}\hat{\Gamma }_n^{-1/2}T_{1n}]^{\frac{1}{2}}[nb^{\frac{1}{2}} \hat{\Gamma }_n^{-1/2}T_{3n}]^{\frac{1}{2}}= [nb^{\frac{1}{2}}\hat{\Gamma }_n^{-1/2} (T_{1n}-\hat{C}_n+\hat{C}_n)]^{\frac{1}{2}}[nb^{\frac{1}{2}}\hat{\Gamma }_n^{-1/2} T_{3n}]^{\frac{1}{2}}\\&\quad \le [nb^{\frac{1}{2}}\hat{\Gamma }_n^{-1/2}|T_{1n}-\hat{C}_n| +nb^{\frac{1}{2}}\hat{\Gamma }_n^{-1/2}\hat{C}_n]^{\frac{1}{2}}O_p(\sqrt{nb^{1/2}})\\&\quad = [O_p(1)+O_p(b^{-1/2})]^{\frac{1}{2}}O_p(\sqrt{nb^{1/2}})=o_p(nb^{1/2}) \end{aligned}$$
from \(nb\rightarrow \infty \) guaranteed by the assumption (b1). Therefore,
$$\begin{aligned} \mathcal {T}_n&=nb^{1/2}\hat{\Gamma }_n^{-1/2}(T_{1n}-\hat{C}_n)+nb^{1/2}\tilde{\Gamma }^{-1/2} \int [f_{\xi ,a}(v;\beta _a)-f_\xi (v;\beta _a,\theta _a)]^2d\Pi (v)\\&\quad +o_p(nb^{1/2}). \end{aligned}$$
Clearly, the right hand side of the above expression tends to \(\infty \) as \(n\rightarrow \infty \), implying that the proposed test is consistent. \(\square \)
Proof of Theorem 3:
Denote
$$\begin{aligned}&\tilde{f}_\xi ^\mathrm{loc}(v;\beta _0,\theta _0)=\int \left[ (1-\delta _n)f_\varepsilon \left( v+\beta _0^Tu, \theta _0\right) +\delta _n\varphi \left( v+\beta _0^Tu\right) \right] f_{{\bar{U}}}(u)du\\&\quad = \int f_\varepsilon \left( v+\beta _0^Tu,\theta _0\right) f_{{\bar{U}}}(u)du-\delta _n\int \left[ f_\varepsilon \left( v+\beta _0^Tu, \theta _0\right) -\varphi \left( v+\beta _0^Tu\right) \right] f_{{\bar{U}}}(u)du.\\&\tilde{f}_{\xi ,b}^\mathrm{loc}(v;\hat{\beta }_n,\hat{\theta }_n)=\int K_b(v-u)\tilde{f}_\xi ^\mathrm{loc}(u;\hat{\beta }_n,\hat{\theta }_n)du. \end{aligned}$$
Adding and subtracting \(\tilde{f}_{\xi ,b}^\mathrm{loc}(v;\hat{\beta }_n,\hat{\theta }_n)\) from \(\hat{f}_{\xi ,n}(v;\hat{\alpha }_n,\hat{\beta }_n)-\tilde{f}_{\xi ,b}(v; \hat{\beta }_n,\hat{\theta }_n)\), we can rewrite the test statistic as
$$\begin{aligned}&T_n({\hat{\alpha }}_n,{\hat{\beta }}_n,{\hat{\theta }}_n)=\int \left[ \hat{f}_{\xi ,n} (v;\hat{\alpha }_n,\hat{\beta }_n)-\tilde{f}_{\xi ,b}^\mathrm{loc}(v;\hat{\beta }_n, \hat{\theta }_n)+\tilde{f}_{\xi ,b}^\mathrm{loc}(v;\hat{\beta }_n,\hat{\theta }_n)\right. \\&\left. \quad -\tilde{f}_{\xi ,b}(v;\hat{\beta }_n, \hat{\theta }_n)\right] ^2d\Pi (v). \end{aligned}$$
Note that
$$\begin{aligned}&\tilde{f}_{\xi ,b}^\mathrm{loc}(v;\hat{\beta }_n,\hat{\theta }_n)-\tilde{f}_{\xi ,b} (v;\hat{\beta }_n,\hat{\theta }_n)=\int K_b(v-u)\left[ \tilde{f}_\xi ^\mathrm{loc}(u;\hat{\beta }_n, \hat{\theta }_n)-\tilde{f}_\xi (u;\hat{\beta }_n,\hat{\theta }_n)\right] du\\&\quad = -\int K_b(v-u)\cdot \delta _n\int \left[ f_\varepsilon \left( u+\hat{\beta }_n^Tt,\hat{\theta }_n\right) -\varphi \left( u+\hat{\beta }_n^Tt\right) \right] {\hat{f}}_{{\bar{U}}n}(t)dtdu\\&\quad =-\delta _n\iint K_b(v-u)\left[ f_\varepsilon \left( u+\hat{\beta }_n^Tt,\hat{\theta }_n\right) -\varphi \left( u +\hat{\beta }_n^Tt\right) \right] {\hat{f}}_{{\bar{U}}n}(t)dtdu\\&\quad =-\delta _n D_n(v;\hat{\beta }_n,\hat{\theta }_n). \end{aligned}$$
We can rewrite \(T_n\) as the sum of the following three terms
$$\begin{aligned} T_{n1}&=\int \left[ \hat{f}_{\xi ,n}(v;\hat{\alpha }_n,\hat{\beta }_n)-\tilde{f}_{\xi ,b}^\mathrm{loc}(v;\hat{\beta }_n,\hat{\theta }_n)\right] ^2d\Pi (v),\\ T_{n2}&=-\,2\delta _n\int \left[ \hat{f}_{\xi ,n}(v;\hat{\alpha }_n,\hat{\beta }_n) -\tilde{f}_{\xi ,b}^\mathrm{loc}(v;\hat{\beta }_n,\hat{\theta }_n)\right] D_n(v;\hat{\beta }_n,\hat{\theta }_n)d\Pi (v),\\ T_{n3}&=\delta ^2_n\int D_n^2(v;\hat{\beta }_n,\hat{\theta }_n)d\Pi (v). \end{aligned}$$
For the sake of simplicity, denote \( f_n(u,t)=(1-\delta _n)f_\varepsilon (u+\hat{\beta }_n^Tt,\hat{\theta }_n)+\delta _n \varphi (u+\hat{\beta }_n^Tt). \) Adding and subtracting \(f_{{\bar{U}}}(t)\) from \({\hat{f}}_{{\bar{U}} n}(t)\), \(T_{n1}\) can be further written as the sum of the following three terms,
$$\begin{aligned} T_{n11}=&\int \left[ \hat{f}_{\xi ,n}(v;\hat{\alpha }_n,\hat{\beta }_n) -\iint K_b(v-u)f_n(u,t)f_{{\bar{U}}}(t)dtdu\right] ^2d\Pi (v),\\ T_{n12}=&\int \left[ \iint K_b(v-u)f_n(u,t)\left[ {\hat{f}}_{{\bar{U}}n}(t) -f_{{\bar{U}}}(t)\right] dtdu\right] ^2d\Pi (v),\\ T_{n13}=&\int \left[ \hat{f}_{\xi ,n}(v;\hat{\alpha }_n,\hat{\beta }_n) -\iint K_b(v-u)f_n(u,t)f_{{\bar{U}}}(t)dtdu\right] \\&\cdot \left[ \iint K_b(v-u)f_n(u,t)\left[ {\hat{f}}_{{\bar{U}}n}(t)-f_{{\bar{U}}} (t)\right] dtdu\right] d\Pi (v). \end{aligned}$$
Similar to the derivations of \(R_{bw}(v;\hat{\beta }_n,\hat{\theta }_n)\), one can show that \(nb^{\frac{1}{2}}T_{n12}=o_p(1)\). Follow the proof of Theorem 1 in Koul and Song (2012), we can show that \( nb^{\frac{1}{2}}[T_{n11}-\hat{C}_n]\Rightarrow N(0,\Gamma )\) and using the Cauchy–Schwarz inequality, we also have \( nb^{\frac{1}{2}}T_{n13}=o_p(1).\) Therefore, we have \( nb^{\frac{1}{2}}[T_{n1}-\hat{C}_n]=nb^{\frac{1}{2}}[T_{n11}-\hat{C}_n]+o_p(1). \)
By the boundedness of \(f''(t)\) and \(\varphi ''(t)\), then we have
$$\begin{aligned}&nb^{\frac{1}{2}}\hat{\Gamma }_n^{-\frac{1}{2}}T_{n3}\\&\quad =nb^{\frac{1}{2}} \hat{\Gamma }_n^{-\frac{1}{2}}\delta _n^2\int D_n^2(v;\hat{\beta }_n,\hat{\theta }_n)d\Pi (v)=\hat{\Gamma }_n^{-\frac{1}{2}}\int D^2_n(v,\hat{\beta }_n,\hat{\theta }_n)d\Pi (v)\\&\quad =\hat{\Gamma }_n^{-\frac{1}{2}}\int \left[ \iint K_b(v-u)\left[ f_\varepsilon \left( u+\hat{\beta }^T_nt,\hat{\theta }_n\right) -\varphi \left( u+\hat{\beta }^T_nt\right) \right] {\hat{f}}_{{\bar{U}}n}(t)dtdu\right] ^2d\Pi (v)\\&\quad =\Gamma ^{-\frac{1}{2}}\int \left[ \int \left[ f_\varepsilon \left( v+\beta ^T_0 t,\theta _0\right) -\varphi \left( v+\beta ^T_0t\right) \right] f_{{\bar{U}}}(t)dt\right] ^2d\Pi (v)+o_p(1). \end{aligned}$$
Similarly, we can obtain
$$\begin{aligned} nb^{\frac{1}{2}}T_{n2}&=\sqrt{nb^{\frac{1}{2}}}\int \left[ \frac{1}{nb}\sum _{i=1}^n K\left( \frac{v-Y_i+\alpha _0+\beta ^T_0\bar{Z}_i}{b}\right) \right. \\&\quad \left. -\int K_b(v-u) \tilde{f}_\xi ^\mathrm{loc}(u;\beta _0,\theta _0)du\right] \\&\quad \times \left[ \iint K_b(v-u)\left[ f_\varepsilon \left( u+\beta _0^Tt,\theta _0\right) -\varphi \left( u+\beta _0^Tt\right) \right] f_{{\bar{U}}}(t)dudt\right] \\&\qquad d\Pi (v)+o_p(1)\\&= O_p(b^{\frac{1}{4}})=o_p(1). \end{aligned}$$
Summarizing the above results, we can conclude the proof of Theorem 3.\(\square \)
