Comparative studies on the adequacy check of parametric measurement error models with auxiliary variable

Abstract

The adequacy check of regression models is a fundamental approach to avoid model misspecifications. Three types of tests: the weighted integrated squared distance test, the U-statistic test and the empirical process based test, are very popular due to attractive theoretical merits such as consistency and satisfactory performances in practice. In this paper, we apply these three tests to check the adequacy of a mean parametric regression model with measurement error. By rigorously investigating the asymptotic properties of three testing methods under the null, local and global alternative hypotheses, we make detailed comparisons for the three tests. To the best of our knowledge, the results of these theoretical comparisons are novel. We conduct simulation studies and a real data analysis to compare the finite sample behaviors of the proposed methods.

Appendix

1.1 Estimating procedure

Set $\xi =\mathrm{E}({\tilde{\xi }}|V):=\xi (V).$ Let $\omega _v=\text{ diag }\{k((V_1-v)/h_n),\ldots , k((V_n-v)/h_n)\}$ for $v\in {\mathbb {R}}^q,$ where $k(\cdot )$ is a kernel function and $h_n$ is a bandwidth. By applying the local linear smoothing method, we can obtain the estimate ${\hat{\xi }}_n(v)$ of $\xi (v)$:

$$\begin{aligned} {\hat{\xi }}_n(v)^\top = (1,\mathbf{0 }_{1\times q})({\mathbb {V}}_v^\top \omega _v{\mathbb {V}}_v)^{-1}{\mathbb {V}}_v^\top \omega _v\tilde{{\mathbb {X}}}, \end{aligned}$$

(A.1)

where $\tilde{\mathbb {{X}}}= ({\tilde{\xi }}_{1}, \ldots , {\tilde{\xi }}_{n})^\top , {\mathbb {V}}_v= (V_{1v}, \ldots , V_{nv})^\top$ with $V_{iv} = \left( 1, (V_i-v)^\top /h_n\right) ^\top .$ Then we can define an estimator ${\hat{\theta }}_n,$ which solves the estimating equation

$$\begin{aligned} \sum _{i=1}^n{\dot{g}}_{\theta }\left( {\hat{\xi }}_n(V_i),\theta \right) \{Y_i-g({\hat{\xi }}_n(V_i),\theta )\}=\mathbf{0 }, \end{aligned}$$

(A.2)

where ${\dot{g}}_{\theta }(\xi ,\theta )$ is the partial derivative of $g(\xi ,\theta )$ with respect to $\theta .$

1.2 Conditions

We list some conditions needed to prove Theorems 1–6.

• (C1) With probability 1, both $\xi$ and V lie in compact sets. Assume that $\mathrm{E}(Y^2|\xi )<\infty ,$ and ${\mathrm{E}}({\tilde{\xi }}^2|V)<\infty .$

• (C2) (i) The function $g(\xi ,\theta )$ is three times continuously differentiable with finite first, second and third order partial derivatives respect to both $\xi$ and $\theta ;$ (ii) the function $\xi (V) = \mathrm{E}({\tilde{\xi }}|V)$ is twice continuously differentiable with finite partial derivatives.

• (C3) $\varSigma _0=\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0){\dot{g}}_{{\theta }}(\xi ,\theta _0)^\top ]$ is positive definite.

• (C4) The densities of $\xi$ and V are bounded away from zero and infinity.

• (C5) The functions $k(\cdot )$ and $\lambda (\cdot )$ are bounded kernel functions of order 2.

• (C6) (i) $\ln n/(nh^q_n)\rightarrow 0$ and $nh_n^{2q}\rightarrow 0;$ (ii) $nb_n^q\rightarrow \infty$ and $nb_n^{2q}\rightarrow 0;$ (iii) $h_nb_n^{-q/2}(\ln n)^{-1}$ $\rightarrow 0$ and $b_n^{2q}h_n^{-q/2}(\ln n)^{-1}$ $\rightarrow 0.$

Remark 1

Conditions (C1)–(C3) are necessary for the asymptotic normality of the estimator defined in (A.2). Condition (C4) aims for avoiding tedious proofs of the theorems. Conditions (C5)–(C6) are common assumptions for the convergence of the local smoothing method.

1.3 Some useful lemmas

Let $V_{i,j}=V_i-V_j$ for $i,j=1,\ldots ,n.$ We first list two lemmas needed for the proofs of main theorems.

Lemma 1

Under Conditions (C1)–(C5), (C6)(i) and the alternative hypothetical models ${\mathcal {H}}_{1n}$ (3.1) with $0\le \rho \le q/2,$ we have

$$\begin{aligned} n^{1/2}({\hat{\theta }}_{n}-\theta _0)= & {} n^{-1/2}\varSigma _0^{-1} \sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,\theta _0)\left\{ \varepsilon _i-\varDelta _{\theta _0}(\xi _i,{\tilde{\xi }}_i) \right\} \nonumber \\&\quad +b_n^{-\rho }\varSigma _0^{-1}\mathrm{E}\{{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )\}+o_p(1). \end{aligned}$$

(A.3)

Proof

We only prove the result with $\rho =0.$ The proof of the results with $0<\rho \le q/2$ is similar and we omit the details. Note that

$$\begin{aligned} A_n:=\sum _{i=1}^n{\dot{g}}_{\theta }({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})\{Y_i-g({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})\} = 0. \end{aligned}$$

Then $A_n$ can be divided into

$$\begin{aligned} n^{-1/2}A_n= & {} n^{-1/2}\sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,{\hat{\theta }}_{n})\{Y_i-g(\xi _i,{\hat{\theta }}_{n})\}\nonumber \\&\quad +\,n^{-1/2}\sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,{\hat{\theta }}_{n})\{g(\xi _i,{\hat{\theta }}_{n})-g({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})\}\nonumber \\&\quad +\, n^{-1/2}\sum _{i=1}^n\{{\dot{g}}_{\theta }({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})-{\dot{g}}_{\theta }(\xi _i,{\hat{\theta }}_{n})\}\{Y_i-g({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})\} \nonumber \\=: & {} A_{n1}+A_{n2}+A_{n3}. \end{aligned}$$

(A.4)

By some simple computations, we can decompose $A_{n1}$ into

$$\begin{aligned} A_{n1}= & {} n^{-1/2}\sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,\theta _0)\{Y_i-g(\xi _i,\theta _0)\} +n^{-1/2}\sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,\theta _0)\{g(\xi _i,\theta _0)-g(\xi _i,{\hat{\theta }}_{n})\}\nonumber \\&\quad +\, n^{-1/2}\sum _{i=1}^n\{{\dot{g}}_{\theta }(\xi _i,{\hat{\theta }}_{n})-{\dot{g}}_{\theta }(\xi _i,\theta _0)\} \{Y_i-g(\xi _i,\theta _0)\}\nonumber \\&\quad +\, n^{-1/2}\sum _{i=1}^n\{{\dot{g}}_{\theta }(\xi _i,{\hat{\theta }}_{n})-{\dot{g}}_{\theta }(\xi _i,\theta _0)\} \{g(\xi _i,\theta _0)-g(\xi _i,{\hat{\theta }}_{n})\}\nonumber \\=: & {} A_{n11}+A_{n12}+A_{n13}+A_{n14}. \end{aligned}$$

(A.5)

When the alternative hypothetical models $Y=g(\xi ,\theta _0)+n^{-1/2}b_n^{-\rho }S(\xi )+\varepsilon$ with $\rho =0$ hold, we have $Y_i-g(\xi _i,\theta _0)=n^{-1/2}S(\xi _i)+\varepsilon _i$ for $i=1,\ldots , n.$ By Taylor expansion and the law of large numbers, we have

$$\begin{aligned}&A_{n11}=n^{-1/2}\sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,\theta _0)\varepsilon _i+\mathrm{E}\{{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )\}+o_p(1),\\&A_{n12}=-\,n^{-1/2}\sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,\theta _0){\dot{g}}_{\theta }(\xi _i,\theta _0)^\top ({\hat{\theta }}_{n}-\theta _0) +o_p(1),\\&A_{n13}=\sqrt{n}({\hat{\theta }}_{n}-\theta _0)^\top \frac{1}{n}\sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,\theta _0){\dot{g}}_{\theta }(\xi _i,\theta _0) \{n^{-1/2}S(\xi _i)+\varepsilon _i\}+o_p(1)=o_p(1),\\&A_{n14}=\sqrt{n}({\hat{\theta }}_{n}-\theta _0)^\top \frac{1}{n}\sum _{i=1}^n\ddot{g}_{\theta ,\theta }(\xi _i,\theta _0) {\dot{g}}_{\theta }(\xi _i,\theta _0)^\top ({\hat{\theta }}_{n}-\theta _0)+o_p(1)=o_p(1). \end{aligned}$$

Thus it yields

$$\begin{aligned} A_{n1}= & {} -\,n^{-1/2}\sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,\theta _0)\varepsilon _i+\mathrm{E}\{{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )\}\nonumber \\&\quad -\,\varSigma _0 n^{-1/2} ({\hat{\theta }}_{n}-\theta _0) +o_p(1) \end{aligned}$$

(A.6)

with $\varSigma _0$ defined in Condition (C3). For $A_{n2},$ by Taylor expansion and Condition (C2), we can validate that

$$\begin{aligned} A_{n2}=-n^{-1/2}\sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i, {\hat{\theta }}_{n}){\dot{g}} _{\xi }(\xi _i,{\hat{\theta }}_{n})^\top \{{\hat{\xi }}_n(V_i)-\xi _i\}+o_p(1). \end{aligned}$$

Let ${\dot{g}}_{\theta }(\xi _i, \theta ){\dot{g}}_{\xi }(\xi _i,\theta )^\top =:\vartheta (\xi _i, \theta ).$ Following the result of Theorem 6 in Masry (1996), we have $\sup _v|{\hat{\xi }}_n(v) -\xi (v)|=O_P(\{\ln n /(nh^q_n)\}^{1/2} + h_n^{2}).$ Then by Taylor expansion and Conditions (C1)–(C2), $n^{-1/2}\sum _{i=1}^n\{\vartheta (\xi _i, \theta _0)-\vartheta (\xi _i, {\hat{\theta }}_{n})\}\{{\hat{\xi }}_n(V_i)-\xi _i\}=O_P(\{\ln n /(nh^q_n)\}^{1/2} + h_n^{2})=o_p(1).$ Thus we have

$$\begin{aligned} A_{n2}= & {} -n^{-1/2}\sum _{i=1}^n \vartheta (\xi _i, \theta _0)\{{\hat{\xi }}_n(V_i)-\xi _i\} +o_p(1). \end{aligned}$$

Note that under Conditions C4, C5 and C6(i),

$$\begin{aligned} \sup _v \left\| nh_n^q(1,\mathbf{0 }_{1\times q})({\mathbb {V}}_v^\top \omega _v{\mathbb {V}}_v)^{-1}- (1/f_v(v),\mathbf{0 }_{1\times q})\right\| _2 = o_p(1). \end{aligned}$$

By the definition of ${\hat{\xi }}_n(v),$ we have

$$\begin{aligned} A_{n2}= -\,n^{-1/2}\sum _{i=1}^n\vartheta (\xi _i, \theta _0)\left\{ n^{-1}h_n^{-q}\sum _{j=1}^n{\tilde{\xi }}_j k(V_{i,j}/h_n)/f_v(V_i)-\xi _i\right\} + o_p(1). \end{aligned}$$

By Conditions C2, C5 and C6(i), we can validate that

$$\begin{aligned} A_{n2}= & {} -n^{-1/2}\sum _{j=1}^{n}\left\{ n^{-1}h_n^{-q}\sum _{i=1}^n \vartheta (\xi _i, \theta _0) k(V_{i,j}/h_n)/f_v(V_i)\right\} {\tilde{\xi }}_j\nonumber \\&\quad + \,n^{-1/2}\sum _{i=1}^n \vartheta (\xi _i, \theta _0) \xi _i +o_p(1)\nonumber \\= & {} -\,n^{-1/2}\sum _{i=1}^{n}{\dot{g}}_{\theta }(\xi _i, \theta _0){\dot{g}}_{\xi }(\xi _i,\theta _0)^\top ({\tilde{\xi }}_i- \xi _i) +o_p(1). \end{aligned}$$

(A.7)

Lemma 3 which is listed below yields the result:

$$\begin{aligned} A_{n3} = o_p(1). \end{aligned}$$

(A.8)

Since $n^{1/2}A_{n}=0,$ from (A.4), (A.6), (A.7) and (A.8), we can obtain that

$$\begin{aligned}&n^{-1/2}\sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,\theta _0)\varepsilon _i -\varSigma _0 n^{1/2}({\hat{\theta }}_{n}-\theta )+\mathrm{E}\{{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )\}\\&\quad -\,n^{-1/2}\sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i, \theta _0){\dot{g}}_{\xi }(\xi _i,\theta _0)^\top ({\tilde{\xi }}_i- \xi _i)+o_p(1)=0. \end{aligned}$$

Thus Lemma 1 for $\rho =0$ is proved. $\square$

Lemma 2

Under Conditions (C1)–(C6)(i) and ${\mathcal {H}}_0,$ we have

$$\begin{aligned} n^{1/2}({\hat{\theta }}_{n}-\theta _0)=n^{-1/2}\varSigma _0^{-1} \sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,\theta _0)\left\{ \varepsilon _i- \varDelta _{\theta _0}(\xi _i,{\tilde{\xi }}_i)\right\} +o_p(1). \end{aligned}$$

(A.9)

Remark 2

Lemma 2 can be proved by letting $S(\xi )=0$ in the proof of Lemma 1. We omit the details.

Lemma 3

Under the same conditions of Lemma 1, we have

$$\begin{aligned} n^{-1/2}\sum _{i=1}^n\{{\dot{g}}_{\theta }({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})-{\dot{g}}_{\theta }(\xi _i,{\hat{\theta }}_{n})\}\{Y_i-g({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})\}=o_p(1). \end{aligned}$$

Proof

We notice that $n^{-1/2}\sum _{i=1}^n\{{\dot{g}}_{\theta }({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})-{\dot{g}}_{\theta }(\xi _i,{\hat{\theta }}_{n})\}\{Y_i-g({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})\}$ is just $A_3$ in (A.4). It can be decomposed into two parts:

$$\begin{aligned} A_3= & {} n^{-1/2}\sum _{i=1}^n\{{\dot{g}}_{\theta }({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})-{\dot{g}}_{\theta }(\xi _i,{\hat{\theta }}_{n})\}\{Y_i-g(\xi _i,{\hat{\theta }}_{n})\}\nonumber \\&\quad -\,n^{-1/2}\sum _{i=1}^n\{{\dot{g}}_{\theta }({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})-{\dot{g}}_{\theta }(\xi _i,{\hat{\theta }}_{n})\}\{g({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})-g(\xi _i,{\hat{\theta }}_{n})\} \nonumber \\=: & {} A_{n31}-A_{n32}. \end{aligned}$$

(A.10)

Let $\ddot{g}_{\theta ,\xi }(\xi ,\theta )$ be the twice partial derivatives related to $\theta$ and $\xi .$ For the first term $A_{n31},$ we have

$$\begin{aligned} A_{n31}= & {} n^{-1/2}\sum _{i=1}^n\ddot{g}_{\theta ,\xi }(\xi _i,{\hat{\theta }}_{n})^\top \{{\hat{\xi }}_n(V_i)-\xi _i\}\{n^{-1/2}S(\xi _i)+\varepsilon _i\}+o_p(1) \\= & {} n^{-1/2}\sum _{i=1}^n\ddot{g}_{\theta ,\xi }(\xi _i,\theta )^\top \{{\hat{\xi }}_n(V_i)-\xi _i\}\{n^{-1/2}S(\xi _i)+\varepsilon _i\} \\&\quad +\,n^{-1/2}\sum _{i=1}^n\{\ddot{g}_{\theta ,\xi }(\xi _i,{\hat{\theta }}_{n})-\ddot{g}_{\theta ,\xi }(\xi _i,\theta )\}^\top \{{\hat{\xi }}_n(V_i)-\xi _i\}\{n^{-1/2}S(\xi _i)+\varepsilon _i\}+o_p(1)\\=: & {} A^{[1]}_{n31}+A^{[2]}_{n31}+o_p(1). \end{aligned}$$

By the similar method to prove (A.7), we can prove that $n^{-1}\sum _{i=1}^n\ddot{g}_{\theta ,\xi }(\xi _i,\theta )^\top \{{\hat{\xi }}_n(V_i)-\xi _i\}S(\xi _i)=O_p(n^{-1/2})=o_p(1).$ So further by the definition of ${\hat{\xi }}_n(v)$ and some simple computations, it yields

$$\begin{aligned} A^{[1]}_{n31}= & {} n^{-1/2}\sum _{i=1}^n\varepsilon _i\ddot{g}_{\theta ,\xi }(\xi _i,\theta )^\top \{{\hat{\xi }}_n(V_i)-\xi _i\}+o_p(1) \\= & {} n^{-1/2}\sum _{i=1}^n\varepsilon _i\ddot{g}_{\theta ,\xi }(\xi _i,\theta )^\top \left\{ n^{-1}h_n^{-q}\sum _{j=1}^n({\tilde{\xi }}_j-\xi _i) k(V_{i,j}/h_n)/f_v(V_i)\right\} +o_p(1) \\= & {} n^{-1/2}\sum _{i=1}^n\varepsilon _i\ddot{g}_{\theta ,\xi }(\xi _i,\theta )^\top \left\{ n^{-1}h_n^{-q}\sum _{j=1,j\ne i}^n({\tilde{\xi }}_j -\xi _j) k(V_{i,i}/h_n)/f_v(V_i)\right\} \\&\quad +\,n^{-3/2}h_n^{-q}\sum _{i=1}^n\varepsilon _i\ddot{g}_{\theta ,\xi }(\xi _i,\theta )^\top ({\tilde{\xi }}_i -\xi _i) k(V_{i,j}/h_n)/f_v(V_i)\\&\quad +\,n^{-1/2}\sum _{i=1}^n\varepsilon _i\ddot{g}_{\theta ,\xi }(\xi _i,\theta )^\top \left\{ n^{-1}h_n^{-q}\sum _{j=1}^n(\xi _j-\xi _i) k(V_{i,j}/h_n)/f_v(V_i)\right\} +o_p(1)\\= & {} A^{[1]}_{n31,1}+A^{[1]}_{n31,2}+A^{[1]}_{n31,3}+o_p(1). \end{aligned}$$

We can prove that the second moment of each component of the vector $A^{[1]}_{n31,1}$ is $O(n^{-1}h_n^{-q}).$ So $A^{[1]}_{n31,1}=o_p(1)$ by Condition (C6). Note that $(nh_n^{q})^{-1/2}\sum _{i=1}^n$ $\varepsilon _i$ $\ddot{g}_{\theta ,\xi }(\xi _i,\theta )^\top ({\tilde{\xi }}_i -\xi _i)$ $k(V_{i,j}/h_n)/f_v(V_i)$ $=O_p(1).$ Then it yields $A^{[1]}_{n31,2}=O_p(n^{-1}h_n^{-q/2})=o_p(1)$ by the condition of the bandwidth. Recall that $\xi _j-\xi _i=\xi (V_j)-\xi (V_i).$ Via Taylor expansion of $\xi (V_j)-\xi (V_i),$ we can obtain that $A^{[1]}_{n31,2}=O_p(h_n^2)=o_p(1).$ Thus we have the result that $A^{[1]}_{n31}=o_p(1).$ For $A^{[2]}_{n31},$ we have $A^{[2]}_{n31}=({\hat{\theta }}_{n}-\theta )n^{-1/2}\sum _{i=1}^n \breve{g}_{\theta ,\xi ,\theta }(\xi _i,\theta )\}^\top \{{\hat{\xi }}_n(V_i)-\xi _i\}\{n^{-1/2}S(\xi _i)+\varepsilon _i\}$ with $\breve{g}_{\theta ,\xi ,\theta }(\xi _i,\theta )$ the three times partial derivative related to $\theta ,\xi ,\theta .$ By similar method to prove $A^{[1]}_{n31}=o_p(1),$ we can prove that $n^{-1/2}\sum _{i=1}^n \breve{g}_{\theta ,\xi ,\theta }(\xi _i,\theta )\}^\top \{{\hat{\xi }}_n(V_i)-\xi _i\}\{n^{-1/2}S(\xi _i)+\varepsilon _i\}=o_p(1).$ Further we can obtain $A^{[2]}_{n31}=o_p(1)$ and then $A_{n31}=o_p(1).$ For $A_{n32},$ by denoting $\tau (\xi ,\theta )={\dot{g}} _{\xi }(\xi _i,\theta )\ddot{g}_{\theta ,\xi }(\xi ,\theta )^\top ,$ we have

$$\begin{aligned} A_{n32}= & {} n^{-1/2}\sum _{i=1}^n\{{\dot{g}}_{\theta }({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})-{\dot{g}}_{\theta }(\xi _i,{\hat{\theta }}_{n})\}\{g({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})-g(\xi _i,{\hat{\theta }}_{n})\} \nonumber \\= & {} n^{-1/2}\sum _{i=1}^n\{{\hat{\xi }}_n(V_i)-\xi _i\}^\top {\dot{g}} _{\xi }(\xi _i,{\hat{\theta }}_{n})\ddot{g}_{\theta ,\xi }(\xi _i,{\hat{\theta }}_{n})^\top \{{\hat{\xi }}_n(V_i)-\xi _i\}+o_p(1) \nonumber \\= & {} n^{-1/2}\sum _{i=1}^n\{{\hat{\xi }}_n(V_i)-\xi _i\}^\top \tau (\xi _i,\theta )\{{\hat{\xi }}_n(V_i)-\xi _i\}\nonumber \\&\quad +\,n^{-1/2}\sum _{i=1}^n\{{\hat{\xi }}_n(V_i)-\xi _i\}^\top \{\tau (\xi _i,{\hat{\theta }}_{n})-\tau (\xi _i,\theta )\}\{{\hat{\xi }}_n(V_i)-\xi _i\} +o_p(1).\nonumber \\ \end{aligned}$$

(A.11)

By the existing result $\sup _v|{\hat{\xi }}_n(v) -\xi (v)|=O_P(\{\ln n /(nh^q_n)\}^{1/2} + h_n^{2}),$ we can prove that both terms in (A.11) are $o_p(1).$ Recalling that we have proved $A_{n31}=o_p(1),$ therefore we obtain that $A_{n3}=o_p(1)$ from (A.10). $\square$

Recall that $\varDelta _{\theta }(\xi ,{\tilde{\xi }}) =({\tilde{\xi }}-\xi )^\top {\dot{g}}_{\xi }(\xi ,\theta ).$ Then we prove a lemma needed in Proof of Theorem 5.

Lemma 4

Let ${\hat{\xi }}_{i}= {\hat{\xi }}_n(V_i)$ and $\lambda _{b_n}(V_{i,j})= \lambda ((V_i-V_j)/b_n)/b_n^q.$ Assume that Conditions (C1)–(C6) hold, then we have

$$\begin{aligned}&n^{-1}b_n^{q/2}\sum _{i\ne j}(\xi _i-{\hat{\xi }}_{i})^\top {\dot{g}}_{\xi }(\xi _i,\theta _0){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top (\xi _j-{\hat{\xi }}_{j})\lambda _{b_n}(V_{i,j})\nonumber \\&\quad = n^{-1}b_n^{q/2} \sum _{i\ne j} \varDelta _{\theta _0}(\xi _i,{\tilde{\xi }}_i) \varDelta _{\theta _0}(\xi _j,{\tilde{\xi }}_j)\lambda _{b_n}(V_{i,j}) + o_p(1). \end{aligned}$$

(A.12)

Proof

Let $U_n =n^{-1}b_n^{q/2}\sum _{i\ne j}(\xi _i-{\hat{\xi }}_{i})^\top {\dot{g}}_{\xi }(\xi _i,\theta _0){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top (\xi _j-{\hat{\xi }}_{j})\lambda _{b_n}(V_{i,j}).$ Then

$$\begin{aligned} U_n= & {} n^{-1}b_n^{q/2}\sum _{i\ne j}\xi _i^\top {\dot{g}}_{\xi }(\xi _i,\theta _0){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top \xi _j\lambda _{b_n}(V_{i,j})\nonumber \\&\quad -\,2n^{-1}b_n^{q/2}\sum _{i\ne j}\xi _i^\top {\dot{g}}_{\xi }(\xi _i,\theta _0){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top {\hat{\xi }}_j\lambda _{b_n}(V_{i,j})\nonumber \\&\quad +\,n^{-1}b_n^{q/2}\sum _{i\ne j}{\hat{\xi }}_i^\top {\dot{g}}_{\xi }(\xi _i,\theta _0){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top {\hat{\xi }}_j\lambda _{b_n}(V_{i,j}) \nonumber \\= & {} :U_{n1}-2U_{n2}+U_{n3}. \end{aligned}$$

(A.13)

We first focus on $U_{n2}.$ Let $k_{h_n}(V_{j,s})=k((V_j-V_s)/h_n)/h_n^q.$ Note that $\sup _j \Vert {\hat{\xi }}_j -n^{-1}\sum _{s=1}^n{\tilde{\xi }}_s k_{h_n}(V_{j,s})/f_v(V_j) \Vert _2 = o_p(1).$ Then

$$\begin{aligned} U_{n2}= & {} n^{-2}b_n^{q/2}\sum _{i=1}^n\sum _{j=1}^n \xi _i^\top {\dot{g}}_{\xi }(\xi _i,\theta _0){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top \sum _{s=1}^n{\tilde{\xi }}_s k_{h_n}(V_{j,s})\lambda _{b_n}(V_{i,j})/f_v(V_j)\\&\quad -\,n^{-2}b_n^{q/2}\sum _{i=1}^n \xi _i^\top {\dot{g}}_{\xi }(\xi _i,\theta _0){\dot{g}}_{\xi }(\xi _i,\theta _0)^\top \sum _{s=1}^n{\tilde{\xi }}_s k_{h_n}(V_{i,s})\lambda _{b_n}(V_{i,i})/f_v(V_i)+o_p(1)\\= & {} U_{n2}^{[1]} - U_{n2}^{[2]}. \end{aligned}$$

By law of large numbers, we can prove that $\frac{1}{n}\sum _{j=1}^n{\dot{g}}_{\xi }(\xi _j,\theta _0)^\top k_{h_n}(V_{j,s})\lambda _{b_n}(V_{i,j})/f_v(V_j)=(1+o_p(1)){\dot{g}}_{\xi }(\xi _s,\theta _0)^\top \lambda _{b_n}(V_{i,s}).$ For $U_{n2}^{[1]},$ we have

$$\begin{aligned} U_{n2}^{[1]}= & {} n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{s=1}^n \xi _i^\top {\dot{g}}_{\xi }(\xi _i,\theta _0)\left\{ \frac{1}{n}\sum _{j=1}^n{\dot{g}}_{\xi }(\xi _j,\theta _0)^\top k_{h_n}(V_{j,s})\lambda _{b_n}(V_{i,j})/f_v(V_j)\right\} {\tilde{\xi }}_s\\= & {} n^{-1}b_n^{q/2} \sum _{i=1}^n\sum _{s=1}^n \xi _i^\top {\dot{g}}_{\xi }(\xi _i,\theta _0){\dot{g}}_{\xi }(\xi _s,\theta _0)^\top {\tilde{\xi }}_s\lambda _{b_n}(V_{i,s})+o_p(1). \end{aligned}$$

Similarly,

$$\begin{aligned} U_{n2}^{[2]}= & {} n^{-1}b_n^{q/2}(1+o_p(1))\sum _{i=1}^n \{\xi _i^\top {\dot{g}}_{\xi }(\xi _i,\theta _0)\}^2\lambda _{b_n}(V_{i,i}). \end{aligned}$$

Thus

$$\begin{aligned} U_{n2} = n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{j\ne i}^n \xi _i^\top {\dot{g}}_{\xi }(\xi _i,\theta _0){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top {\tilde{\xi }}_j\lambda _{b_n}(V_{i,j})+o_p(1). \end{aligned}$$

(A.14)

Similar to prove (A.14), we can obtain that

$$\begin{aligned} U_{n3} = n^{-1}b_n^{q/2} \sum _{i=1}^n\sum _{j\ne i}^n {\tilde{\xi }}_i^\top {\dot{g}}_{\xi }(\xi _i,\theta _0){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top {\tilde{\xi }}_j\lambda _{b_n}(V_{i,j})+o_p(1).\nonumber \\ \end{aligned}$$

(A.15)

Recall that $\varDelta _{\theta }(\xi ,{\tilde{\xi }}) =({\tilde{\xi }}-\xi )^\top {\dot{g}}_{\xi }(\xi ,\theta ).$ Combining (A.13)–(A.15), we have

$$\begin{aligned} U_{n} = (1+o_p(1))\sum _{i=1}^n\sum _{j\ne i}^n \varDelta _{\theta _0}(\xi _i,{\tilde{\xi }}_i) \varDelta _{\theta _0}(\xi _j,{\tilde{\xi }}_j) \lambda _{b_n}(V_{i,j})+o_p(1). \end{aligned}$$

Since $n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{j\ne i}^n \varDelta _{\theta _0}(\xi _i,{\tilde{\xi }}_i) \varDelta _{\theta _0}(\xi _j,{\tilde{\xi }}_j) \lambda _{b_n}(V_{i,j}) = O_p(1)$ by verifying $\mathrm{E} (T_n) = 0$ and $\mathrm{E} (T_n^2) = O(1),$ the lemma is proved. $\square$

1.4 Proofs of the Theorems

Theorems 2.1–2.3 are special cases of Theorems 4–6 with $S(\xi )=0.$ As follows, we give the detailed proofs for Theorems 4–6.

Proof of Theorem 4

First define $B_n(v)= n^{-1/2}b_n^{-q/2}\sum _{i=1}^{n}\{Y_i-g({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})\} \lambda ((V_i-v)/b_n).$ Then $T_n^{\mathtt {WISD}}=\int B^2_n(v)dv.$ For $B_n(v),$ it can be decomposed into

$$\begin{aligned} B_n(v)= & {} n^{-1/2}b_n^{-q/2}\sum _{i=1}^{n}\{Y_i-g(\xi _i,\theta _0)\} \lambda ((V_i-v)/b_n)\nonumber \\&\quad +\,n^{-1/2}b_n^{-q/2}\sum _{i=1}^{n}\{g(\xi _i,\theta _0)-g({\hat{\xi }}_n(V_i),\theta _0)\} \lambda ((V_i-v)/b_n)\nonumber \\&\quad +\,n^{-1/2}b_n^{-q/2}\sum _{i=1}^{n}\{g({\hat{\xi }}_n(V_i),\theta _0)-g({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})\} \lambda ((V_i-v)/b_n)\nonumber \\=: & {} B_{n1}(v)+B_{n2}(v)+B_{n3}(v). \end{aligned}$$

(A.16)

For the alternative hypothetical models (3.1) when $\rho =q/2,$ we can prove that

$$\begin{aligned} B_{n1}(v)= & {} n^{-1/2}b_n^{-q/2}\sum _{i=1}^{n}\varepsilon _i \lambda ((V_i-v)/b_n)\nonumber \\&\quad +\,\mathrm{E}[S(\xi )|V=v]f_v(v) +o_p(1). \end{aligned}$$

(A.17)

For the second term $B_{n2}(v),$ we can obtain that

$$\begin{aligned} B_{n2}(v) = n^{-1/2}b_n^{-q/2}\sum _{i=1}^{n}{\dot{g}}_{\xi }(\xi _i,\theta _0)^\top \{\xi _i-{\hat{\xi }}_n(V_i)\} \lambda ((V_i-v)/b_n)+o_p(1). \end{aligned}$$

By the definition of ${\hat{\xi }}_n(v),$ it follows that

$$\begin{aligned} B_{n2}(v)= & {} - n^{-1/2}b_n^{-q/2}\sum _{i=1}^{n}\lambda ((V_i-v)/b_n){\dot{g}}_{\xi }(\xi _i,\theta _0)^\top \nonumber \\&\quad \times \,\left\{ n^{-1}h_n^{-q}\sum _{j=1}^n{\tilde{\xi }}_j k(V_{i,j}/h_n)/f_v(V_i)\right\} \nonumber \\&\quad +\,n^{-1/2}b_n^{-q/2}\sum _{i=1}^{n}\lambda ((V_i-v)/b_n){\dot{g}}_{\xi }(\xi _i,\theta _0)^\top \xi _i +o_p(1) \nonumber \\=: & {} -B^{[1]}_{n2}(v)+B^{[2]}_{n2}(v)+o_p(1). \end{aligned}$$

(A.18)

Let $\psi _v(V_i)=\lambda ((V_i-v)/b_n){\dot{g}}_{\xi }(\xi _i,\theta _0)^\top /f_v(V_i).$ For $B^{[1]}_{n2}(v),$ we can get

$$\begin{aligned} B^{[1]}_{n2}(v)= & {} n^{-1/2}b_n^{-q/2}\sum _{j=1}^{n}\left\{ n^{-1}h_n^{-q}\sum _{i=1}^n \psi _v(V_i) k(V_{i,j}/h_n)\right\} {\tilde{\xi }}_j \nonumber \\= & {} n^{-1/2}b_n^{-q/2}\sum _{j=1}^{n} f_v(V_j)\psi _v(V_j){\tilde{\xi }}_j +o_p(1). \end{aligned}$$

(A.19)

Then combining (A.18) and (A.19), we have

$$\begin{aligned} B_{n2}(v)= n^{-1/2}b_n^{-q/2}\sum _{i=1}^{n}(\xi _i -{\tilde{\xi }}_i)^\top {\dot{g}}_{\xi }(\xi _i,\theta _0) \lambda ((V_i-v)/b_n)+o_p(1). \end{aligned}$$

(A.20)

For $B_{n3}(v),$ via Taylor expansion and a simple decomposition, we have

$$\begin{aligned} B_{n3}(v)= & {} n^{-1/2}b_n^{-q/2}\sum _{i=1}^{n}{\dot{g}}_{\theta }({\hat{\xi }}_n(V_i),\theta _0)^\top (\theta _0-{\hat{\theta }}_{n})\lambda ((V_i-v)/b_n)+o_p(1) \nonumber \\= & {} n^{-1/2}b_n^{-q/2}\sum _{i=1}^{n}\{{\dot{g}}_{\theta }({\hat{\xi }}_n(V_i),\theta _0)-{\dot{g}}_{\theta }(\xi _i,\theta _0)\}^\top (\theta _0-{\hat{\theta }}_{n}) \lambda ((V_i-v)/b_n)\nonumber \\&\quad +\,n^{-1/2}b_n^{-q/2}\sum _{i=1}^{n}{\dot{g}}_{\theta }(\xi _i,\theta _0)^\top (\theta _0-{\hat{\theta }}_{n}) \lambda ((V_i-v)/b_n)+o_p(1)\nonumber \\= & {} B_{n3}^{[1]}(v)+B_{n3}^{[2]}(v)+o_p(1). \end{aligned}$$

(A.21)

Note that $1/(nb^q_n)\sum _{i=1}^{n}{\dot{g}}_{\theta }(\xi _i,\theta _0)\lambda ((V_i-v)/b_n)=f_v(v){\dot{g}}_{\theta }(\xi (v),\theta )+o_p(1).$ Therefore by the result of Lemma 1, it yields

$$\begin{aligned} B_{n3}^{[2]}(v)= & {} f_v(v){\dot{g}}_{\theta }(\xi (v),\theta )^\top n^{1/2}b_n^{q/2}(\theta _0-{\hat{\theta }}_{n})+o_p(1) \\= & {} -f_v(v){\dot{g}}_{\theta }(\xi (v),\theta )^\top \varSigma _0^{-1}\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )]+o_p(1). \end{aligned}$$

By the result of Lemma 1 with $\rho =q/2,$ we have $\Vert \theta _0-{\hat{\theta }}_{n}\Vert _2=O_{p}(n^{-1/2}b_n^{-q/2}).$ Then with Condition C2, we have

$$\begin{aligned} |B_{n3}^{[1]}(v)|\le & {} c\sup _i \Vert {\hat{\xi }}_n(V_i) - \xi (V_i)\Vert _2 \Vert \theta _0-{\hat{\theta }}_{n})\Vert _2\left\{ n^{-1}b_n^{-q}\sum _{i=1}^{n}\lambda ((V_i-v)/b_n)\right\} , \end{aligned}$$

where c is a finite positive number. Note that $1/(nb^q_n)\sum _{i=1}^{n}\lambda ((V_i-v)/b_n)=f_v(v)+o_p(1)$ and $\sup _i \Vert {\hat{\xi }}_n(V_i) - \xi (V_i)\Vert _2 = o_p(1).$ Note that $f_v(v)$ is bound away from zero and infinity in Conditions C4. Then it yields $B_{n3}^{[1]}(v)=o_{p}(1).$ Combining the above discussions and (A.21), we have that

$$\begin{aligned} B_{n3}(v)= -f_v(v){\dot{g}}_{\theta }(\xi (v),\theta )^\top \varSigma _0^{-1}\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )]+o_p(1). \end{aligned}$$

(A.22)

Thus from (A.16)–(A.17), (A.20) and (A.22), it yields

$$\begin{aligned} B_n(v)= & {} n^{-1/2}b_n^{-q/2}\sum _{i=1}^{n}\left\{ \varepsilon _i+(\xi _i -{\tilde{\xi }}_i)^\top {\dot{g}}_{\xi }(\xi _i,\theta _0)\right\} \lambda ((V_i-v)/b_n)\\&\quad -\, f_v(v){\dot{g}}_{\theta }(\xi (v),\theta )^\top \varSigma _0^{-1}\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )] +f_v(v) \mathrm{E}(S(\xi )|V=v)+o_p(1). \end{aligned}$$

Note that the kernel function $\lambda (\cdot )$ is a real function, then the class $\{\{\varepsilon +(X-{\tilde{\xi }})^\top {\dot{g}}_{\xi }(\xi ,\theta _0)\}\lambda ((V-v)/b_n): v\in {\mathbb {R}}^q\}$ is a V–C class by the Lemma 28 of Pollard (2012) and Lemma 2.6.18 of van der Vaart and Wellner (1996). Thus we can prove that $B_n(v)$ converges to a Gauss process $\varUpsilon (v).$ Further by Theorem 1.11.1 of van der Vaart and Wellner (1996), we prove the result of $T_n^{\mathtt {WISD}}.$ $\square$

Proof of Theorem 5

Recall that $T_n^{\mathtt {U}} = n^{-1}b_n^{q/2}\sum _{i\ne j}(Y_i-g({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n}))(Y_j-g({\hat{\xi }}_n(V_j),{\hat{\theta }}_{n})) \lambda (V_{i,j}/b_n)/b_n^q.$ Let $\lambda _{b_n}(\cdot )= \lambda (\cdot /b_n)/b_n^q$ and ${\hat{\xi }}_i = {\hat{\xi }}_n(V_i).$ Then

$$\begin{aligned} T_n^{\mathtt {U}}= & {} \sum _{k=1}^3 D_{nk}, \end{aligned}$$

(A.23)

where

$$\begin{aligned} D_{n1}= & {} n^{-1}b_n^{q/2}\sum _{j\ne i}\{Y_i-g(\xi _i,\theta _0)\}\{Y_j-g(\xi _j,\theta _0)\} \lambda _{b_n}(V_{i,j}),\\ D_{n2}= & {} n^{-1}b_n^{q/2}\sum _{j\ne i}\{g(\xi _i,\theta _0)-g({\hat{\xi }}_i,{\hat{\theta }}_{n})\}\{g(\xi _j,{\theta }_{0})-g({\hat{\xi }}_j,{\hat{\theta }}_{n})\}\lambda _{b_n}(V_{i,j}),\\ D_{n3}= & {} 2n^{-1}b_n^{q/2}\sum _{j\ne i}\{Y_i-g(\xi _i,\theta _0)\}\{g(\xi _j,{\theta }_{0})-g({\hat{\xi }}_j,{\hat{\theta }}_{n})\} \lambda _{b_n}(V_{i,j}). \end{aligned}$$

For $D_{n1},$ under the local alternative models (3.1) with $\rho =q/4,$ it is shown that

$$\begin{aligned} D_{n1}= & {} n^{-1}b_n^{q/2}\sum _{j\ne i}\varepsilon _i\varepsilon _j \lambda _{b_n}(V_{i,j})+n^{-2}\sum _{j\ne i}S(\xi _i)S(\xi _j)\lambda _{b_n}(V_{i,j})\nonumber \\&\quad +\,n^{-3/2}b_n^{q/4}\sum _{j\ne i}\{S(\xi _i)\varepsilon _j+S(\xi _j)\varepsilon _i\} \lambda _{b_n}(V_{i,j})+o_p(1). \end{aligned}$$

(A.24)

Following the similar argument as Lemma 3.3b in Zheng (1996), we have

$$\begin{aligned} \frac{1}{n(n-1)}\sum _{j\ne i}S(\xi _j)\varepsilon _i \lambda _{b_n}(V_{i,j}) = O_p(n^{-1/2}). \end{aligned}$$

Then with Condition C6(ii),

$$\begin{aligned} n^{-3/2}b_n^{q/4}\sum _{j\ne i}\{S(\xi _i)\varepsilon _j+S(\xi _j)\varepsilon _i\} \lambda _{b_n}(V_{i,j}) = O_p(b_n^{q/4}) = o_p(1). \end{aligned}$$

Rewriting $n^{-2}\sum _{j\ne i}S(\xi _i)S(\xi _j)\lambda _{b_n}(V_{i,j})$ as $n^{-1}\sum _{i= 1}^n S(\xi _i)\frac{n-1}{n}\{(n-1)^{-1}\sum _{j= 1,j\ne i}^n$ $S(\xi _j)\lambda _{b_n}(V_{i,j})\},$ then it is equal to $n^{-1}\sum _{i= 1}^nS(\xi _i) E\{S(\xi )|V=V_i\}f_v(V_i)+ o_p(1)=\mathrm{E}(S^2(\xi )f_v(V))+o_p(1).$ Thus the expression (A.24) for $D_{n1}$ reduces to

$$\begin{aligned} D_{n1}= & {} n^{-1}b_n^{q/2}\sum _{j\ne i}\varepsilon _i\varepsilon _j \lambda _{b_n}(V_{i,j})+\mathrm{E}(S^2(\xi )f_v(V))+o_p(1). \end{aligned}$$

(A.25)

Next we turn to the term $D_{n2}$ in (A.23). Define

$$\begin{aligned} D_{n2}^*= & {} n^{-1}b_n^{q/2}\sum _{i\ne j}({\hat{\theta }}_{n}-\theta _0)^\top {\dot{g}}_{\theta }(\xi _i,\theta _0){\dot{g}}_{\theta }(\xi _j,\theta _0)^\top ({\hat{\theta }}_{n}-\theta _0) \lambda _{b_n}(V_{i,j})\nonumber \\&\quad +\,n^{-1}b_n^{q/2}\sum _{i\ne j}(\xi _i-{\hat{\xi }}_{i})^\top {\dot{g}}_{\xi }(\xi _i,\theta _0){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top (\xi _j-{\hat{\xi }}_{j}) \lambda _{b_n}(V_{i,j})\nonumber \\&\quad -\,2({\hat{\theta }}_{n}-\theta _0)^\top \{n^{-1}b_n^{q/2}\sum _{i\ne j} {\dot{g}}_{\theta }(\xi _i,\theta _0){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top (\xi _j-{\hat{\xi }}_{j}) \lambda _{b_n}(V_{i,j})\}\nonumber \\:= & {} D_{n2}^{*[1]}+D_{n2}^{*[2]}-2D_{n2}^{*[3]}. \end{aligned}$$

(A.26)

By Taylor expansion and Conditions C2, C4 and C5, we have $D_{n2} = D_{n2}^*+o_p(1).$ We only need to focus on $D_{n2}^*$ for $D_{n2}.$ By the result of Lemma 1 under the alternative hypothesis (3.1) with $\rho =q/4,$ we have $n^{1/2}b_n^{q/4}({\hat{\theta }}_{n}-\theta _0)=\varSigma _0^{-1}\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )]+o_p(1).$ Therefore, it follows that

$$\begin{aligned} D_{n2}^{*[1]}= & {} n^{1/2}b_n^{q/4}({\hat{\theta }}_{n}-\theta _0)^\top \left\{ n^{-2}\sum _{i\ne j}{\dot{g}}_{\theta }(\xi _i,\theta _0){\dot{g}}_{\theta }(\xi _j,\theta _0)^\top \lambda _{b_n}(V_{i,j})\right\} n^{1/2}b_n^{q/4}(\theta _0-{\hat{\theta }}_{n})\nonumber \\= & {} \mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )]^\top \varSigma _0^{-1} \mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0){\dot{g}}_{\theta }(\xi ,\theta _0)^\top f_v(V)]\nonumber \\&\quad \varSigma _0^{-1}\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )]+o_p(1). \end{aligned}$$

(A.27)

Recall that $\varDelta _{\theta }(\xi ,{\tilde{\xi }}) =({\tilde{\xi }}-\xi )^\top {\dot{g}}_{\xi }(\xi ,\theta ).$ By Lemma 4, we have

$$\begin{aligned} D_{n2}^{*[2]}= n^{-1}b_n^{q/2} \sum _{i\ne j} \varDelta _{\theta _0}(\xi _i,{\tilde{\xi }}_i) \varDelta _{\theta _0}(\xi _j,{\tilde{\xi }}_j)\lambda _{b_n}(V_{i,j}) +o_p(1). \end{aligned}$$

(A.28)

Let $n^{-3/2}b_n^{q/2}\sum _{i\ne j} {\dot{g}}_{\theta }(\xi _i,\theta _0){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top (\xi _j-{\hat{\xi }}_{j}) \lambda _{b_n}(V_{i,j})=:M_n.$ Then it yields

$$\begin{aligned} M_n= & {} n^{-1/2}b_n^{q/2} \sum _{j=1}^n \left\{ n^{-1}\sum _{i=1,i\ne j}^n{\dot{g}}_{\theta }(\xi _i,\theta _0) \lambda _{b_n}(V_{i,j})\right\} {\dot{g}}_{\xi }(\xi _j,\theta _0)^\top (\xi _j-{\hat{\xi }}_{j}) \\= & {} n^{-1/2}b_n^{q/2} \sum _{j=1}^n {\dot{g}}_{\theta }(\xi _j,\theta _0)f_v(V_j){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top (\xi _j-{\hat{\xi }}_{j})+o_p(1). \end{aligned}$$

By denoting ${\dot{g}}_{\theta }(\xi _j,\theta _0)f_v(V_j){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top$ as a new function, for example $V(\xi _j,\theta _0),$ similar to the proof of (A.7), it yields $M_n= n^{-1/2}b_n^{q/2} \sum _{j=1}^n {\dot{g}}_{\theta }(\xi _j,\theta _0)f_v(V_j){\dot{g}}_{\xi }(\xi _j,\theta _0)^\top$ $(\xi _j-{\hat{\xi }}_{j})+o_p(1).$ Further recalling that $n^{1/2}b_n^{q/4}(\theta _0-{\hat{\theta }}_{n})=O_p(1),$ we have

$$\begin{aligned} D_{n2}^{*[3]}=O_p\left( b_n^{q/2}n^{-1/2}b_n^{-q/4}\right) =O_p\left( n^{-1/2}b_n^{q/4}\right) =o_p(1). \end{aligned}$$

(A.29)

From (A.26), (A.27), (A.28), (A.29) and $D_{n2} = D_{n2}^*+o_p(1),$ we have

$$\begin{aligned} D_{n2}= & {} n^{-1}b_n^{q/2} \sum _{i\ne j} \varDelta _{\theta _0}(\xi _i,{\tilde{\xi }}_i) \varDelta _{\theta _0}(\xi _j,{\tilde{\xi }}_j)\lambda _{b_n}(V_{i,j})+\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )]^\top \varSigma _0^{-1} \nonumber \\&\quad \times \,\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0){\dot{g}}_{\theta }(\xi ,\theta _0)^\top f_v(V)]\varSigma _0^{-1}\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )] + o_p(1). \end{aligned}$$

(A.30)

Under (3.1) with $\rho =q/4,$ noticing that$n^{1/2}b_n^{q/4}(\theta _0-{\hat{\theta }}_{n})=O_p(1),$ we have

$$\begin{aligned} D_{n3}= & {} 2n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{j\ne i}\varepsilon _i\{g(\xi _j,{\theta }_{0})-g(\xi _j,{\hat{\theta }}_{n})+g(\xi _j,{\hat{\theta }}_{n})-g({\hat{\xi }}_j,{\hat{\theta }}_{n})\} \lambda _{b_n}(V_{i,j})\nonumber \\&\quad +\,2n^{-3/2}b_n^{q/4}\sum _{i=1}^n\sum _{j\ne i}S(\xi _i)\{g(\xi _j,{\theta }_{0})-g(\xi _j,{\hat{\theta }}_{n})\nonumber \\&\quad +\,g(\xi _j,{\hat{\theta }}_{n})-g({\hat{\xi }}_j,{\hat{\theta }}_{n})\} \lambda _{b_n}(V_{i,j})\nonumber \\= & {} 2n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{j\ne i}\varepsilon _i{\dot{g}}_{\theta }(\xi _j,{\theta }_{0})^\top (\theta _0-{\hat{\theta }}_{n}) \lambda _{b_n}(V_{i,j})\nonumber \\&\quad +\,2n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{j\ne i}\varepsilon _i{\dot{g}}_{\xi }(\xi _j,{\theta }_{0})^\top (\xi _j-{\hat{\xi }}_j) \lambda _{b_n}(V_{i,j})\nonumber \\&\quad +\,2n^{-3/2}b_n^{q/4}\sum _{i=1}^n\sum _{j\ne i}S(\xi _i){\dot{g}}_{\theta }(\xi _j,{\theta }_{0})^\top (\theta _0-{\hat{\theta }}_{n}) \lambda _{b_n}(V_{i,j})\nonumber \\&\quad +\,2n^{-3/2}b_n^{q/4}\sum _{i=1}^n\sum _{j\ne i}S(\xi _i){\dot{g}}_{\xi }(\xi _j,{\theta }_{0})^\top (\xi _j-{\hat{\xi }}_j) \lambda _{b_n}(V_{i,j})+o_p(1)\nonumber \\:= & {} D_{n3}^{[1]}+D_{n3}^{[2]}+D_{n3}^{[3]}+D_{n3}^{[4]}+o_p(1). \end{aligned}$$

(A.31)

We can prove that $n^{-3/2}\sum _{i=1}^n\sum _{j\ne i}\varepsilon _i {\dot{g}}_{\theta }(\xi _j,{\theta }_{0}) \lambda _{b_n}(V_{i,j})$ $= n^{-1/2}\sum _{i=1}^n\varepsilon _i\left\{ n^{-1}\sum _{j=1,j\ne i}^n {\dot{g}}_{\theta }(\xi _j,{\theta }_{0}) \lambda _{b_n}(V_{i,j})\right\} = n^{-1/2}\sum _{i=1}^n\varepsilon _i {\dot{g}}_{\theta }(\xi _i,{\theta }_{0})f_v(V_i)+o_p(1).$ Further by the fact that $n^{1/2}b_n^{q/4}(\theta -{\hat{\theta }}_{n})=O_p(1),$ we have that

$$\begin{aligned} D_{n3}^{[1]} = O_p\left( b_n^{q/4}\right) =o_p(1). \end{aligned}$$

(A.32)

For $D_{n3},$ we first consider the part $n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{j=1}^n\varepsilon _i{\dot{g}}_{\xi }(\xi _j,{\theta }_{0})^\top {\hat{\xi }}_j \lambda _{b_n}(V_{i,j})$:

$$\begin{aligned}&n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{j=1}^n\varepsilon _i{\dot{g}}_{\xi }(\xi _j,{\theta }_{0})^\top {\hat{\xi }}_j) \lambda _{b_n}(V_{i,j}) \nonumber \\&\quad = n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{j=1}^n\varepsilon _i{\dot{g}}_{\xi }(\xi _j,{\theta }_{0})^\top \nonumber \\&\qquad \times \,\left\{ n^{-1}h_n^{-q}\sum _{s=1}^n{\tilde{\xi }}_s k(V_{j,s}/h_n)/f_v(V_j)\right\} \lambda _{b_n}(V_{i,j})+o_p(1) \nonumber \\&\quad = n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{s=1}^n\varepsilon _i \left\{ n^{-1}h_n^{-q}\sum _{j=1}^n{\dot{g}}_{\xi }(\xi _j,{\theta }_{0})^\top \lambda _{b_n}(V_{i,j}) k(V_{j,s}/h_n)/f_v(V_j)\right\} {\tilde{\xi }}_s\nonumber \\&\qquad +\,o_p(1) \nonumber \\&\quad = n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{s=1}^n\varepsilon _i {\dot{g}}_{\xi }(\xi _s,{\theta }_{0})^\top {\tilde{\xi }}_s\lambda _{b_n}(V_{i,s})+o_p(1). \end{aligned}$$

(A.33)

Thus it is easy to obtain that

$$\begin{aligned} D_{n3}^{[2]}= & {} -2n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{j\ne i}\varepsilon _i\varDelta _{\theta _0}(\xi _j,{\tilde{\xi }}_j) \lambda _{b_n}(V_{i,j}) +o_p(1). \end{aligned}$$

(A.34)

By applying law of large numbers twice, it can be validated that

$$\begin{aligned} n^{-2}\sum _{i=1}^n\sum _{j\ne i}S(\xi _i){\dot{g}}_{\theta }(\xi _j,\theta _0) \lambda _b(V_i-V_j)=\mathrm{E}[S(\xi ){\dot{g}}_{\theta }(\xi ,\theta _0)f_v(V)]+o_p(1). \end{aligned}$$

By Lemma 1 with $\rho =q/4,$ $n^{1/2}b_n^{q/4}({\hat{\theta }}_{n}-\theta _0)=\varSigma _0^{-1}\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )]+o_p(1).$ Combining the above two results, it yields

$$\begin{aligned} D_{n3}^{[3]} = -2 \mathrm{E}[S(\xi ){\dot{g}}_{\theta }(\xi ,\theta _0)^\top f_v(V)]\varSigma _0^{-1}\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )]+o_p(1). \end{aligned}$$

(A.35)

For $D_{n3}^{[4]},$ we can validate that

$$\begin{aligned} D_{n3}^{[4]}= & {} 2n^{-1/2}b_n^{q/4}\sum _{s=1}^n{\dot{g}}_{\xi }(\xi _s,{\theta }_{0})^\top (\xi _s-{\tilde{\xi }}_s)\left\{ n^{-1}\sum _{i=1,i\ne s}^n S(\xi _i) \lambda _{b_n}(V_{i,s})\right\} + o_p(1)\nonumber \\= & {} 2n^{-1/2}b_n^{q/4}\sum _{s=1}^n{\dot{g}}_{\xi }(\xi _s,{\theta }_{0})^\top (\xi _s-{\tilde{\xi }}_s) E\{ S(\xi _i) |V=V_s\} + o_p(1)\nonumber \\= & {} O_p\left( b_n^{q/4}\right) = o_p(1). \end{aligned}$$

(A.36)

Therefore, with (A.31)–(A.36), we have

$$\begin{aligned} D_{n3}= & {} -2 \mathrm{E}[S(\xi ){\dot{g}}_{\theta }(\xi ,\theta _0)^\top f_v(V)]\varSigma _0^{-1}\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )]\nonumber \\&\quad -\,2n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{j\ne i}\varepsilon _i\varDelta _{\theta _0}(\xi _j,{\tilde{\xi }}_j) \lambda _{b_n}(V_{i,j}) + o_p(1). \end{aligned}$$

(A.37)

From (A.23), (A.24), (A.30) and (A.37), we can validate that

$$\begin{aligned} T_n^{\mathtt {U}}= & {} n^{-1}b_n^{q/2}\sum _{i=1}^n\sum _{j\ne i} \left\{ \varepsilon _i- \varDelta _{\theta _0}(\xi _i,{\tilde{\xi }}_i)\right\} \left\{ \varepsilon _j- \varDelta _{\theta _0}(\xi _j,{\tilde{\xi }}_j)\right\} \lambda _{b_n}(V_{i,j})\nonumber \\&\quad + \,\mathrm{E}[\mu (\xi )^2 f_v(V)] +o_p(1), \end{aligned}$$

(A.38)

where $\mu (\xi ) = S(\xi )-{\dot{g}}_{\theta }(\xi ,\theta _0)^\top \varSigma _0^{-1}\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )].$ Since the first term of (A.38) is a degenerated U-statistic, by similar method of Li and Wang (1998), we can validate that the conditions of Theorem 1 of Hall (1984) hold for $T_n^{\mathtt {U}}.$ Thus we have proved the asymptotic normality of $T_n^{\mathtt {U}}.$

Proof of Theorem 6

Recall that $\varphi _n(v)=n^{-1/2}\sum \nolimits _{i=1}^n\{ Y_i-g({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})\}{\mathbf {1}}(V_i\le v)$ and the EP based test $T^{\mathtt {EP}}_n=\int \{\varphi _n(v)\}^2dF_n(v).$ Then we have

$$\begin{aligned} \varphi _n(v)= & {} n^{-1/2}\sum \limits _{i=1}^n\{Y_i-g(\xi _i,\theta _0)\}{\mathbf {1}}(V_i\le v)\nonumber \\&\quad +\,n^{-1/2}\sum \limits _{i=1}^n\{g(\xi _i,\theta _0)-g({\hat{\xi }}_n(V_i),{\hat{\theta }}_{n})\}{\mathbf {1}}(V_i\le v)\nonumber \\=: & {} G_{n1}+ G_{n2}. \end{aligned}$$

(A.39)

Under the alternative hypothetical model with $\rho =0$ in (3.1), that is $Y=g(\xi ,\theta _0)+n^{-1/2}S(\xi )+\varepsilon ,$ we have

$$\begin{aligned} G_{n1}= & {} n^{-1/2}\sum \limits _{i=1}^n \varepsilon _i {\mathbf {1}}(V_i\le v)+n^{-1}\sum \limits _{i=1}^n S(\xi _i) {\mathbf {1}}(V_i\le v)\nonumber \\= & {} n^{-1/2}\sum \limits _{i=1}^n \varepsilon _i {\mathbf {1}}(V_i\le v)+\mathrm{E}[S(\xi ) {\mathbf {1}}(V\le v)] +o_p(1). \end{aligned}$$

(A.40)

For $G_{n2},$ by Taylor expansion and Condition C2, we have

$$\begin{aligned} G_{n2}= & {} n^{-1/2}\sum \limits _{i=1}^n {\dot{g}}_{\theta }(\xi _i,\theta _0)^\top (\theta _0-{\hat{\theta }}_{n} ){\mathbf {1}}(V_i\le v)\nonumber \\&\quad +\, n^{-1/2}\sum \limits _{i=1}^n {\dot{g}}_{\xi }(\xi _i,\theta _0)^\top (\xi _i-{\hat{\xi }}_n(V_i) ){\mathbf {1}}(V_i\le v)+o_p(1)\nonumber \\=: & {} G_{n2}^{[1]}+ G_{n2}^{[2]}+o_p(1). \end{aligned}$$

(A.41)

Lemma 1 shows that $n^{1/2}({\hat{\theta }}_{n}-\theta _0)=n^{-1/2}\varSigma _0^{-1} \sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,\theta _0)\left\{ \varepsilon _i- \varDelta _{\theta _0}(\xi _i,{\tilde{\xi }}_i)\right\} +\varSigma _0^{-1}\mathrm{E}\{{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )\}+o_p(1)$ under (3.1) with $\rho =0.$ Thus we have $n^{1/2}({\hat{\theta }}_{n}-\theta _0)=O_p(1).$ Further by law of large numbers, it yields

$$\begin{aligned} G_{n2}^{[1]}= & {} \left\{ n^{-1}\sum \limits _{i=1}^n {\dot{g}}_{\theta }(\xi _i,\theta _0)^\top {\mathbf {1}}(V_i\le v)\right\} n^{1/2}(\theta _0-{\hat{\theta }}_{n} )\nonumber \\= & {} -\,n^{-1/2}\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)^\top {\mathbf {1}}(V\le v)]\varSigma _0^{-1} \sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,\theta _0)\left\{ \varepsilon _i- \varDelta _{\theta _0}(\xi _i,{\tilde{\xi }}_i)\right\} \nonumber \\&\quad -\,\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)^\top {\mathbf {1}}(V\le v)]\varSigma _0^{-1}\mathrm{E}\{{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )\}+o_p(1). \end{aligned}$$

(A.42)

For $G_{n2}^{[2]},$ similar to prove (A.7), we can prove

$$\begin{aligned} G_{n2}^{[2]}= & {} -n^{-1/2}\sum \limits _{i=1}^n \varDelta _{\theta _0}(\xi _i,{\tilde{\xi }}_i){\mathbf {1}}(V_i\le v)+o_p(1). \end{aligned}$$

(A.43)

From the results (A.41)–(A.43), we have

$$\begin{aligned} G_{n2}= & {} -\mathrm{E}\{{\dot{g}}_{\theta }(\xi ,\theta _0)^\top {\mathbf {1}}(V\le v)\}\varSigma _0^{-1}\mathrm{E}\{{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )\}\nonumber \\&\quad -\,n^{-1/2}\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)^\top {\mathbf {1}}(V\le v)]\varSigma _0^{-1} \sum _{i=1}^n{\dot{g}}_{\theta }(\xi _i,\theta _0)\left\{ \varepsilon _i- \varDelta _{\theta _0}(\xi _i,{\tilde{\xi }}_i)\right\} \nonumber \\&\quad -\,n^{-1/2}\sum \limits _{i=1}^n \varDelta _{\theta _0}(\xi _i,{\tilde{\xi }}_i){\mathbf {1}}(V_i\le v)+o_p(1). \end{aligned}$$

(A.44)

From the definition of $\mathtt {IF}_{v}^{\mathtt {EP}}(\varepsilon , \xi , {\tilde{\xi }},V),$ (A.39), (A.40) and (A.44), we have

$$\begin{aligned} \varphi _n(v)= & {} n^{-1/2}\sum \limits _{i=1}^n \mathtt {IF}_{v}^{\mathtt {EP}}(\varepsilon _i, \xi _i, {\tilde{\xi }}_i, V_i)+\mathrm{E}[S(\xi ) {\mathbf {1}}(V\le v)]\\&\quad -\,\mathrm{E}[{\dot{g}}_{\theta }(\xi ,\theta _0)^\top {\mathbf {1}}(V\le v)]\varSigma _0^{-1}\mathrm{E}\{{\dot{g}}_{\theta }(\xi ,\theta _0)S(\xi )\}+o_p(1). \end{aligned}$$

Note that the function class $\{{\mathbf {1}}(V\le v): v \in R^q \}$ is a V–C class since ${\mathbf {1}}(V\le v)$ is the product of q univariate indicator functions. Further by Lemma 2.6.18 of van der Vaart and Wellner (1996), we can obtain that the function class $\{\mathtt {IF}_{v}^{\mathtt {EP}}(\varepsilon , \xi , {\tilde{\xi }}, V_i): v \in R^q \}$ is a V–C class. By Theorem 2.6.8 of van der Vaart and Wellner (1996), we can prove the weak convergence of $\mathtt {IF}_{v}^{\mathtt {EP}}(\varepsilon , \xi , {\tilde{\xi }},V).$ The result of $T^{\mathtt {EP}}_n$ can be proved from Theorem 1.11.1 of van der Vaart and Wellner (1996). $\square$