Multidimensional specification test based on non-stationary time series

Abstract

In the literature, most works of the specification tests focus on the problem with one-dimensional response or fixed multidimensional responses. In this paper, we develop a new specification test for the parametric models with non-stationary regressor under multidimensional setup, where the dimension of responses may tend to infinity, which fills a gap in the literature. The theoretical results about the asymptotic properties of the proposed test are studied and the optimal rate of the local departure under the alternative hypothesis is also given which ensures the models underpinning by the null and alternative hypotheses can be differentiated. Some simulation studies are done to evaluate the performance of the proposed test with the finite sample. Besides, a real data example based on the US aggregate consumers’ consumption data is employed to illustrate the performance. The results of simulation studies and real data analysis both demonstrate the efficiency of our proposed method.

Appendix

Here, the proofs of Theorems 1–3 are presented. In order to prove these theoretical results, some lemmas are needed to be shown firstly.

Lemma 1

Under Assumptions A-D, we have

• (1) $\sum \nolimits _{t=1}^n\dfrac{1}{d_t}\sim \dfrac{n^{\frac{\mu }{2}}}{p}$, $n\rightarrow \infty .$

• (2) $\sum \nolimits _{t=2}^n\sum \nolimits _{s=1}^{t-1}\dfrac{1}{d_{ts}}\dfrac{1}{d_s}<Cn^{\mu }$, $n\rightarrow \infty$, C is a constant.

• (3) $\sum \nolimits _{k=3}^n\sum \nolimits _{t=2}^{k-1}\sum \nolimits _{s=1}^{t-1}\dfrac{1}{d_{kt}}\dfrac{1}{d_{ts}}\dfrac{1}{d_s}<Cn^{\frac{3}{2}\mu }$, C is a constant.

Proof

• (1)
  $$\begin{aligned} \lim _{n\rightarrow \infty }\dfrac{\sum \nolimits _{t=1}^n\dfrac{1}{d_t}}{n^{\mu /2}/p}=\lim _{n\rightarrow \infty }\dfrac{\sum \nolimits _{t=1}^n\dfrac{1}{t^{1-\mu /2}log(t)}}{n^{\mu /2}/log(n)}=\lim _{n\rightarrow \infty }\dfrac{log(n)}{\frac{\mu }{2}log(n)-1}=\dfrac{2}{\mu }. \end{aligned}$$

  Since $\dfrac{2}{\mu }>0$ is a constant, we have $\sum \nolimits _{t=1}^n\dfrac{1}{d_t}\sim \dfrac{n^{\frac{\mu }{2}}}{p}$.

• (2)
  $$\begin{aligned}&\sum \limits _{t=2}^n\sum \limits _{s=1}^{t-1}\dfrac{1}{d_{ts}}\dfrac{1}{d_s}=C_4 \sum \limits _{t=2}^n\sum \limits _{s=1}^{t-1}\dfrac{1}{log(t)(t-s)^{1-\mu /2}}\dfrac{1}{log(s) s^{1-\mu /2}} \\&\quad<C_4 \sum \limits _{t=2}^n\sum \limits _{s=1}^{t-1}\dfrac{1}{(t-s)^{1-\mu /2}}\dfrac{1}{s^{1-\mu /2}}\sim C_4 \sum \limits _{s=1}^{n-1}\dfrac{1}{s^{1-\mu /2}}\displaystyle {\int _{s+1}^n (t-s)^{\mu /2-1}\mathrm {d}t} \\&\quad <\dfrac{2}{\mu }C_4 \sum \limits _{s=1}^{n-1} s^{\mu /2-1}n^{\mu /2}\sim Cn^{\mu }. \end{aligned}$$
• (3)
  $$\begin{aligned}&\sum \limits _{k=3}^n\sum \limits _{t=2}^{k-1}\sum \limits _{s=1}^{t-1}\dfrac{1}{d_{kt}}\dfrac{1}{d_{ts}}\dfrac{1}{d_s}\le C_5\sum \limits _{k=3}^n\sum \limits _{t=2}^{k-1}\sum \limits _{s=1}^{t-1}\dfrac{1}{(k-t)^{1-\mu /2}}\dfrac{1}{(t-s)^{1-\mu /2}}\dfrac{1}{s^{1-\mu /2}} \\&\quad<C_6n^{\mu /2}\sum \limits _{s=1}^{n-2}\dfrac{1}{s^{1-\mu /2}}\sum \limits _{t=s+1}^{n-1}\dfrac{1}{(t-s)^{1-\mu /2}}<Cn^{\frac{3}{2}\mu }. \end{aligned}$$

  $\square$

The following Lemma 2 shows some properties of the Hermite functions which which is proved in Dong and Gao (2018).

Lemma 2

Let ${\mathscr {T}}_k(x)=\dfrac{1}{k}\sum \nolimits _{i=0}^{k-1}{\mathscr {H}}_i^2(x)$ and ${\mathscr {T}}(x)=\dfrac{1}{2\pi }\sqrt{4-x^2}I\{|x|\le 2\}$, then ${\mathscr {T}}_k(x)$ converges to ${\mathscr {T}}(x)$ for any $x\in R$ as $k\rightarrow \infty$. Moreover, $\int _{-\infty }^{\infty }|{\mathscr {T}}_k(x)-{\mathscr {T}}(x)|\,\mathrm{d}x\rightarrow 0$ as $k\rightarrow \infty$.

Proof of Theorem 1

Under $H_0$, ${\hat{e}}_{t}^{(i)}=y_{t}^{(i)}-g(x_{t};{\hat{\theta }}^{(i)})=y_{t}^{(i)}-g(x_{t};\theta _{0}^{(i)})+g(x_{t};\theta _{0}^{(i)})-g(x_{t};{\hat{\theta }}^{(i)})$, denote ${\hat{g}}^{(i)}(x)=g(x;\theta _0^{(i)})-g(x;{\hat{\theta }}^{(i)})$, then ${\hat{e}}_{t}^{(i)}=e_t^{(i)}+{\hat{g}}^{(i)}(x_{t})$. The test statistic $T_{np}$ can be rewritten as

$$\begin{aligned} T_{np}&=\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n}Z(x_t)'Z(x_s)(e_t^{(i)}+{\hat{g}}^{(i)}(x_{t}))(e_s^{(i)}+{\hat{g}}^{(i)}(x_{s})) \\&=\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n}Z(x_t)'Z(x_s)e_{t}^{(i)}e_{s}^{(i)}+2\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n} Z(x_t)'Z(x_s)e_{t}^{(i)}{\hat{g}}^{(i)}(x_{s}) \\&\quad +\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n} Z(x_t)'Z(x_s){\hat{g}}^{(i)}(x_{t}){\hat{g}}^{(i)}(x_{s}) \\&=T_{np1}+T_{np2}+T_{np3}. \end{aligned}$$

We first consider the term $T_{np1}$,

$$\begin{aligned} T_{np1}&=\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n}Z(x_t)'Z(x_s)e_{t}^{(i)}e_{s}^{(i)} \\&=\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\Vert Z(x_t)\Vert ^2 \sigma _0+\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\Vert Z(x_t)\Vert ^2 ({e_{t}^{(i)}}^2-\sigma _0) \\&\quad +2\sum \limits _{i=1}^{p}\sum \limits _{t=2}^{n}\sum \limits _{s=1}^{t-1}Z(x_t)'Z(x_s)e_{t}^{(i)}e_{s}^{(i)}\\&=T_{np1}'+T_{np1}''+T_{np1}'''. \end{aligned}$$

Using the notations in Lemma 2, we have

$$\begin{aligned} \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np1}'&=\dfrac{1}{n^{\mu /2}K\sigma _0}\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\Vert Z(x_t)\Vert ^2 \sigma _0=\dfrac{1}{n^{\mu /2}K\sigma _0}p\sigma _0\sum \limits _{t=1}^{n}\Vert Z(x_t)\Vert ^2 \\&=\dfrac{p}{n^{\mu /2}K}\sum \limits _{t=1}^{n}\Vert Z(x_t)\Vert ^2 \\&=\dfrac{p}{n^{\mu /2}}\sum \limits _{t=1}^{n}{\mathscr {T}}(x_t)+\dfrac{p}{n^{\mu /2}}\sum \limits _{t=1}^{n}[{\mathscr {T}}_K(x_t)-{\mathscr {T}}(x_t)]. \end{aligned}$$

Let $h(n)=p^2\sim log^2(n)$, so h(n) is a slowly varying function. Then according to Corollary 2.1 in Wang and Phillips (2009) and $\int {\mathscr {T}}\,\mathrm{d}x=1$, we have

$$\begin{aligned} \dfrac{p}{n^{\mu /2}}\sum \limits _{t=1}^{n}{\mathscr {T}}(x_t)\xrightarrow {D}L_{W_{1-\mu /2}}(1,0). \end{aligned}$$

Noting that the standard normal density function is a bounded function, by Lemmas 1(1) and 2, we have

$$\begin{aligned}&E\left| \dfrac{p}{n^{\mu /2}}\sum \limits _{t=1}^n[{\mathscr {T}}_k(x_t)-{\mathscr {T}}(x_t)]\right| \le \dfrac{p}{n^{\mu /2}}\sum \limits _{t=1}^n E\left| {\mathscr {T}}_k(x_t)-{\mathscr {T}}(x_t)\right| \\&\quad \le \dfrac{O(1)p}{n^{\mu /2}}\sum \limits _{t=1}^n\dfrac{1}{d_t} \displaystyle {\int \left| {\mathscr {T}}_k(x)-{\mathscr {T}}(x)\right| \mathrm {d}x}\rightarrow 0. \end{aligned}$$

Hence, we have $\dfrac{1}{n^{\mu /2}K\sigma _0}T_{np1}'\xrightarrow {D}L_{W_{1-\mu /2}}(1,0)$.

For $T_{np1}''$, noting that $\Vert Z(x)\Vert ^2\le O(1)K$ uniformly for any x and $\int \Vert Z(x)\Vert ^2\,\mathrm{d}x=K$ by orthogonality and by Assumption B(b), we have

$$\begin{aligned}&E\left( \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np1}''\right) ^2 \\&\quad =E\left( \dfrac{1}{n^{\mu /2}K\sigma _0}\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\Vert Z(x_t)\Vert ^2 ({e_{t}^{(i)}}^2-\sigma _0)\right) ^2 \\&\quad =\dfrac{1}{n^{\mu }K^2\sigma _0^2}E\sum \limits _{i=1}^{p}\sum \limits _{j=1}^{p}\sum \limits _{t=1}^{n}\Vert Z(x_t)\Vert ^4 ({e_{t}^{(i)}}^2-\sigma _0)({e_{t}^{(j)}}^2-\sigma _0) \\&\quad \le \dfrac{O(1)}{n^{\mu }K}E\sum \limits _{i=1}^{p}\sum \limits _{j=1}^{p}\sum \limits _{t=1}^{n}\Vert Z(x_t)\Vert ^2 ({e_{t}^{(i)}}^2 {e_{t}^{(j)}}^2-{e_{t}^{(i)}}^2\sigma _0-{e_{t}^{(j)}}^2\sigma _0+\sigma _0^2) \\&\quad =\dfrac{O(1)}{n^{\mu }K}\sum \limits _{i=1}^{p}\sum \limits _{j=1}^{p}E({e_{t}^{(i)}}^2 {e_{t}^{(j)}}^2-\sigma _0^2) E\sum \limits _{t=1}^{n}\Vert Z(x_t)\Vert ^2\le \dfrac{O(1)p^2}{n^{\mu }K}\dfrac{n^{\mu /2}}{p}\displaystyle {\int \Vert Z(x)\Vert ^2\mathrm {d}x} \rightarrow 0. \end{aligned}$$

Now we move on to $T_{np1}'''$,

$$\begin{aligned}&E\left( \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np1}'''\right) ^2=E\left( \dfrac{2}{n^{\mu /2}K\sigma _0}\sum \limits _{i=1}^{p}\sum \limits _{t=2}^{n}\sum \limits _{s=1}^{t-1}Z(x_t)'Z(x_s)e_{t}^{(i)}e_{s}^{(i)}\right) ^2 \\&\quad =\dfrac{4}{n^{\mu }K^2\sigma _0^2}\sum \limits _{i=1}^{p}\sum \limits _{j=1}^{p}\sigma _{ij}^2 E\sum \limits _{t=2}^{n}\sum \limits _{s=1}^{t-1}|Z(x_t)'Z(x_s)|^2 \end{aligned}$$

Let ${\mathscr {F}}_s$ be the information flow of $\{u_1,\ldots ,u_s\}$. According to ${\int [Z(x)'Z(x_s)]^2\mathrm {d}x}=\Vert Z(x_s)\Vert ^2$, we have

$$\begin{aligned} E|Z(x_t)'Z(x_s)|^2&=E\displaystyle {\int \left[ Z\left( x_s+\dfrac{1}{d_{ts}}x\right) 'Z(x_s)\right] ^2\phi (x)\mathrm {d}x} \\&\le \dfrac{O(1)}{d_{ts}}E\displaystyle {\int \left[ Z(x)'Z(x_s)\right] ^2\mathrm {d}x}\le \dfrac{O(1)k}{d_{ts}d_{s}}. \end{aligned}$$

Thus, by Lemma 1(2), we have

$$\begin{aligned} E\left( \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np1}'''\right) ^2\le \dfrac{O(1)p^2}{n^{\mu }K^2}n^{\mu }K=\dfrac{O(1)p^2}{K}\rightarrow 0, \end{aligned}$$

and together with the last several results we yield

$$\begin{aligned} \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np1}\xrightarrow {D}L_{W_{1-\mu /2}}(1,0). \end{aligned}$$

For $T_{np3}$, we will show that $\dfrac{1}{n^{\mu /2}K\sigma _0}T_{np3}=o_p(1)$. Set $\xi _n\sim n^{\lambda _2}$, $n\rightarrow \infty$ where $\lambda _2$ satisfies $\lambda _0<\lambda _2<-\dfrac{\mu }{4}$. For any $\delta$, $\epsilon >0$,

$$\begin{aligned}&P\left( \dfrac{1}{n^{\mu /2}K\sigma _0}|T_{np3}|>\delta \right) \le P\left( \Vert {\hat{\theta }}^{(i)}-\theta _0^{(i)}\Vert>\xi _n\epsilon ,\exists i\right) \\&\quad +P\left( \dfrac{1}{n^{\mu /2}K\sigma _0}|T_{np3}|>\delta , \Vert {\hat{\theta }}^{(i)}-\theta _0^{(i)}\Vert \le \xi _n\epsilon ,\forall i\right) . \end{aligned}$$

For the first term of the right-hand side of the inequality, apply the Boole’s inequality,

$$\begin{aligned}P\left( \Vert {\hat{\theta }}^{(i)}-\theta _0^{(i)}\Vert>\xi _n\epsilon ,\exists i\right) \le p\cdot \sup _i P\left( \Vert {\hat{\theta }}^{(i)}-\theta _0^{(i)}\Vert >\xi _n\epsilon \right) =O(1)log(n)n^{\lambda _0-\lambda _2}=o(1).\end{aligned}$$

Then for the second term, according to the virtue of Markov’s inequality, we have

$$\begin{aligned}&P\left( \dfrac{1}{n^{\mu /2}K\sigma _0}|T_{np3}|>\delta , \Vert {\hat{\theta }}^{(i)}-\theta _0^{(i)}\Vert \le \xi _n\epsilon ,\forall i\right) \\&\quad \le \dfrac{1}{n^{\mu /2}K\sigma _0\delta }E\left[ |T_{np3}|\prod \limits _{i=1}^p I\{\Vert {\hat{\theta }}^{(i)}-\theta _0^{(i)}\Vert \le \xi _n\epsilon \}\right] \\&\quad =\dfrac{1}{n^{\mu /2}K\sigma _0\delta }E\vert \sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n} Z(x_t)'Z(x_s){\hat{g}}^{(i)}(x_{t}){\hat{g}}^{(i)}(x_{s})\prod \limits _{j=1}^p I\{\Vert {\hat{\theta }}^{(j)}-\theta _0^{(j)}\Vert \le \xi _n\epsilon \}\vert .&\end{aligned}$$

where $I(\cdot )$ stands for the indicator function.

Using Taylor expansion for $g(x;\theta )$ with respect to $\theta$ in a neighborhood of $\theta _0^{(i)}$, we have

$$\begin{aligned} g(x;{\hat{\theta }}^{(i)})=g(x;\theta _0^{(i)})+l_1(x;\theta _0^{(i)})'({\hat{\theta }}^{(i)}-\theta _0^{(i)})+\dfrac{1}{2}({\hat{\theta }}^{(i)}-\theta _0^{(i)})'l_2(x;{\bar{\theta }}^{(i)})({\hat{\theta }}^{(i)}-\theta _0^{(i)}) \end{aligned}$$

where ${\bar{\theta }}^{(i)}$ is on the line segment joining $\theta _0^{(i)}$ and ${\hat{\theta }}_0^{(i)}$. Therefore,

$$\begin{aligned}&{\hat{g}}^{(i)}(x_{t})=g(x;\theta _0^{(i)})-g(x;{\hat{\theta }}^{(i)})=l_1(x_t; \theta _0^{(i)})'(\theta _0^{(i)}-{\hat{\theta }}^{(i)}) \\&\quad -\dfrac{1}{2}({\hat{\theta }}^{(i)}-\theta _0^{(i)})'l_2(x_t;{\bar{\theta }}^{(i)})({\hat{\theta }}^{(i)}-\theta _0^{(i)}). \end{aligned}$$

Thus,

$$\begin{aligned}&\dfrac{1}{n^{\mu /2}K\sigma _0\delta }E\vert \sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n} Z(x_t)'Z(x_s){\hat{g}}^{(i)}(x_{t}){\hat{g}}^{(i)}(x_{s})\prod \limits _{j=1}^p I\{\Vert {\hat{\theta }}^{(j)}-\theta _0^{(j)}\Vert \le \xi _n\epsilon \}\vert \\&\quad \le \dfrac{O(1)}{n^{\mu /2}K}E\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n}\vert Z(x_t)'Z(x_s)l_1(x_t;\theta _0^{(i)})'(\theta _0^{(i)}-{\hat{\theta }}^{(i)})l_1(x_s;\theta _0^{(i)})'(\theta _0^{(i)}-{\hat{\theta }}^{(i)}) \\&\qquad \cdot \prod \limits _{j=1}^p I\{\Vert {\hat{\theta }}^{(j)}-\theta _0^{(j)}\Vert \le \xi _n\epsilon \}\vert +\dfrac{O(1)}{n^{\mu /2}K}E\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n}\vert Z(x_t)'Z(x_s)l_1(x_t;\theta _0^{(i)})'(\theta _0^{(i)}-{\hat{\theta }}^{(i)}) \\&\qquad \cdot ({\hat{\theta }}^{(i)}-\theta _0^{(i)})'l_2(x_s;{\bar{\theta }}^{(i)})({\hat{\theta }}^{(i)}-\theta _0^{(i)})\prod \limits _{j=1}^p I\{\Vert {\hat{\theta }}^{(j)}-\theta _0^{(j)}\Vert \le \xi _n\epsilon \}\vert \\&\qquad +\dfrac{O(1)}{n^{\mu /2}K}E\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n}\vert Z(x_t)'Z(x_s)({\hat{\theta }}^{(i)}-\theta _0^{(i)})'l_2(x_t;{\bar{\theta }}^{(i)})({\hat{\theta }}^{(i)}-\theta _0^{(i)}) \\&\qquad \cdot ({\hat{\theta }}^{(i)}-\theta _0^{(i)})'l_2(x_s;{\bar{\theta }}^{(i)})({\hat{\theta }}^{(i)}-\theta _0^{(i)})\prod \limits _{j=1}^p I\{\Vert {\hat{\theta }}^{(j)}-\theta _0^{(j)}\Vert \le \xi _n\epsilon \}\vert \\&\quad =T_{np3}'+T_{np3}''+T_{np3}'''. \end{aligned}$$

Since $\lambda _2<-\dfrac{\mu }{4}$ and $p\sim \log (n)$, $\xi _n^2 n^{\mu /2}p\sim n^{2\lambda _2+\mu /2}p=o(1)$. For $T_{np3}'$, due to the arbitrariness of $\epsilon$, by results (1) and (2) in Lemma 1, we have

$$\begin{aligned} T_{np3}'\le & {} \dfrac{O(1)\xi _n^2\epsilon ^2}{n^{\mu /2}K}E\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n} |Z(x_t)'Z(x_s)|\Vert l_1(x_{t};\theta _{0}^{(i)})\Vert \Vert l_1(x_{s};\theta _{0}^{(i)})\Vert \\\le & {} \dfrac{O(1)\xi _n^2\epsilon ^2}{n^{\mu /2}}E\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n} \Vert l_1(x_{t};\theta _{0}^{(i)})\Vert ^2 \\&\quad +\dfrac{O(1)\xi _n^2\epsilon ^2}{n^{\mu /2}}E\sum \limits _{i=1}^{p}\sum \limits _{t=2}^{n}\sum \limits _{s=1}^{t-1} \Vert l_1(x_{t};\theta _{0}^{(i)})\Vert \Vert l_1(x_{s};\theta _{0}^{(i)})\Vert \\\le & {} O(1)\xi ^2\epsilon ^2+O(1)\xi ^2\epsilon ^2 n^{\mu /2}p=o(1). \end{aligned}$$

Similarly, for $T_{np3}'''$, we have

$$\begin{aligned} T_{np3}'''&\le \dfrac{O(1)\xi _n^4\epsilon ^4}{n^{\mu /2}K}E\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n} |Z(x_t)'Z(x_s)|\Vert l_2(x_{t};{\bar{\theta }}^{(i)})\Vert \Vert l_2(x_{s};{\bar{\theta }}^{(i)})\Vert \\&\le O(1)\xi ^4\epsilon ^4+O(1)\xi ^4\epsilon ^4 n^{\mu /2}p=o(1).&\end{aligned}$$

Apply Cauchy–Schwartz inequality, we have

$$\begin{aligned} T_{np3}''\le 2\sqrt{T_{np3}'T_{np3}'''}=o(1). \end{aligned}$$

Thus,

$$\begin{aligned} \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np3}\xrightarrow {P} 0. \end{aligned}$$

Apply Cauchy–Schwartz inequality again, we have

$$\begin{aligned} \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np2}\xrightarrow {P} 0. \end{aligned}$$

Finally, we get

$$\begin{aligned} \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np}\xrightarrow {D}L_{W_{1-\mu /2}}(1,0). \end{aligned}$$

Then, the proof of Theorem 1 is finished. $\square$

Proof of Theorem 2

Under $H_1$ and Assumption D, we have $m^{(i)}(x_t)=g(x_t;\theta _0^{(i)})+\Delta _n^{(i)}(x_t)=g(x_t;\theta _0^{(i)})+\delta _n \Delta ^{(i)}(x_t)$. Thus

$$\begin{aligned} {\hat{e}}_{t}^{(i)}&=y_{t}^{(i)}- g(x_{t};{\hat{\theta }}^{(i)})=e_{t}^{(i)}+\delta _n \Delta ^{(i)}(x_t)+g(x_{t};\theta _{0}^{(i)})-g(x_{t};{\hat{\theta }}^{(i)}) \\&=e_{t}^{(i)}+\delta _n \Delta ^{(i)}(x_t)+{\hat{g}}^{(i)}(x_{t}). \end{aligned}$$

The test statistic $T_{np}$ can be rewritten as

$$\begin{aligned} T_{np}= & {} \sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n} Z(x_t)'Z(x_s)(e_{t}^{(i)}+\delta _n \Delta ^{(i)}(x_t))(e_{s}^{(i)}+\delta _n \Delta ^{(i)}(x_s)) \\&\quad +2\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n} Z(x_t)'Z(x_s)(e_{t}^{(i)}+\delta _n \Delta ^{(i)}(x_t)){\hat{g}}^{(i)}(x_{s}) \\&\quad +\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n} Z(x_t)'Z(x_s){\hat{g}}^{(i)}(x_{t}){\hat{g}}^{(i)}(x_{s})\\= & {} T_{np1}+T_{np2}+T_{np3}. \end{aligned}$$

It is already showed in the proof of Theorem 1 that $\dfrac{1}{n^{\mu /2}k\sigma _0}T_{np3}=o_p(1)$, and by Cauchy–Schwarz inequality, $|T_{np2}|\le 2\sqrt{T_{np1}T_{np3}}$. So we only need to show $\dfrac{1}{n^{\mu /2}k\sigma _0}T_{np1}\xrightarrow {p}\infty$ to finish the proof. Consider the term $T_{np1}$ firstly, we have

$$\begin{aligned} T_{np1}&=\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n} Z(x_t)'Z(x_s)e_{t}^{(i)}e_{s}^{(i)}+2\delta _n\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n} Z(x_t)'Z(x_s)\Delta ^{(i)}(x_t)e_{s}^{(i)} \\&\quad +\delta _n^2\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n}\sum \limits _{s=1}^{n} Z(x_t)'Z(x_s)\Delta ^{(i)}(x_t)\Delta ^{(i)}(x_s)=T_{np1}'+2T_{np1}''+T_{np1}'''. \end{aligned}$$

From the proof of Theorem 1, we know that $\dfrac{1}{n^{\mu /2}K\sigma _0}T_{np1}'\xrightarrow {D}L_{W_{1-\mu /2}}(1,0).$

For $T_{np1}'''$,

$$\begin{aligned} \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np1}'''=&\dfrac{\delta _n^2}{n^{\mu /2}K\sigma _0}\cdot \dfrac{n^{\mu }}{p^2}\sum \limits _{i=1}^{p}\sum \limits _{l=0}^{K-1}\left( \dfrac{p}{n^{\mu /2}}\sum \limits _{t=1}^n{\mathscr {H}}_l(x_t)\Delta ^{(i)}(x_t)\right) ^2 \\&\ge \dfrac{\delta _n^2 n^{\mu /2}}{p^2 K\sigma _0}\sum \limits _{i=1}^{p}\left( \dfrac{p}{n^{\mu /2}}\sum \limits _{t=1}^n{\mathscr {H}}_{l_i}(x_t)\Delta ^{(i)}(x_t)\right) ^2, \end{aligned}$$

where $l_i$ is the first integer such that $\int {\mathscr {H}}_{l_i}(x)\Delta ^{(i)}(x)\mathrm {d}x\ne 0$ under dimension i. Such $l_i$ must exist, otherwise, for any l, $\int {\mathscr {H}}_{l}(x)\Delta ^{(i)}(x)\mathrm {d}x=0$, which means $\Delta ^{(i)}(x)$ is orthogonal with every ${\mathscr {H}}_l(x)$, i.e., $\Delta ^{(i)}(x)$ is a zero function, that contradicts the assumption.

Apply Corollary 2.1 in Wang and Phillips (2009) again, as $n\rightarrow \infty$, we have

$$\begin{aligned} \dfrac{p}{n^{\mu /2}}\sum \limits _{t=1}^n{\mathscr {H}}_{l_i}(x_t)\Delta ^{(i)}(x_t)\xrightarrow {D}\int {\mathscr {H}}_{l_i}(x)\Delta ^{(i)}(x)\mathrm {d}xL_{W_{1-\mu /2}}(1,0). \end{aligned}$$

By continuous mapping theorem, we get

$$\begin{aligned} \left( \dfrac{p}{n^{\mu /2}}\sum \limits _{t=1}^n{\mathscr {H}}_{l_i}(x_t)\Delta ^{(i)}(x_t)\right) ^2\xrightarrow {D}\left( \int {\mathscr {H}}_{l_i}(x)\Delta ^{(i)}(x)\mathrm {d}xL_{W_{1-\mu /2}}(1,0)\right) ^2, \end{aligned}$$

as $n\rightarrow \infty$.

According to Assumption D, $\dfrac{\delta _n^2 n^{\mu /2}}{p^2 K\sigma _0}\rightarrow \infty$, thus

$$\begin{aligned} \dfrac{\delta _n^2 n^{\mu /2}}{p^2 K\sigma _0}\sum \limits _{i=1}^p\left( \int {\mathscr {H}}_{l_i}(x)\Delta ^{(i)}(x)\mathrm {d}xL_{W_{1-\mu /2}}(1,0)\right) ^2\xrightarrow {P}\infty . \end{aligned}$$

It follows that $\dfrac{1}{n^{\mu /2}K\sigma _0}T_{np1}'''=\varOmega _P\left( \dfrac{\delta _n^2 n^{\mu /2}}{p^2 K}\right) \rightarrow \infty .$

For $T_{np1}''$, it can be rewritten as

$$\begin{aligned} T_{np1}''&=\delta _n\sum \limits _{i=1}^{p}\sum \limits _{t=1}^{n} Z(x_t)'Z(x_s)\Delta ^{(i)}(x_t)e_{t}^{(i)}+\delta _n\sum \limits _{i=1}^p\mathop {\sum \limits ^n \sum \limits ^n}_{t>s} Z(x_t)'Z(x_s)\Delta ^{(i)}(x_t)e_{s}^{(i)} \\&\quad +\delta _n\sum \limits _{i=1}^p\mathop {\sum \limits ^n \sum \limits ^n}_{t<s} Z(x_t)'Z(x_s)\Delta ^{(i)}(x_t)e_{s}^{(i)} \\&=T_{np11}''+T_{np12}''+T_{np13}'' \end{aligned}$$

We first consider the term $T_{np11}''$,

$$\begin{aligned} E\left( \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np11}''\right) ^2&=\dfrac{\delta _n^2}{n^{\mu }K^2\sigma _0^2}\sum \limits _{i=1}^{p}\sum \limits _{j=1}^{p}\sum \limits _{t=1}^{n}\Vert Z(x_t)\Vert ^4\Delta ^{(i)}(x_t)\Delta ^{(j)}(x_t)\sigma _{ij} \\&\le \dfrac{O(1)\delta _n^2}{n^{\mu }K}\sum \limits _{i=1}^{p}\sum \limits _{j=1}^{p}\sum \limits _{t=1}^{n}\Vert Z(x_t)\Vert ^2\sigma _{ij}=\dfrac{O(1)\delta _n^2 p}{n^{\mu /2}}\rightarrow 0. \end{aligned}$$

Then, we move on to $T_{np12}''$,

$$\begin{aligned}&E\left( \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np12}''\right) ^2 \\&\quad =\dfrac{\delta _n^2}{n^{\mu }K^2\sigma _0^2}E\left( \sum \limits _{i=1}^{p}\sum \limits _{j=1}^{p}\sum \limits _{s=1}^{n-1}\sum \limits _{t_1=s+1}^{n}\sum \limits _{t_2=s+1}^{n} Z(x_{t_1})' Z(x_s) Z(x_{t_2})' Z(x_s)\Delta ^{(i)}(x_{t_1})\Delta ^{(j)}(x_{t_2})e_{s}^{(i)}e_{s}^{(j)}\right) \\&\quad =\dfrac{\delta _n^2}{n^{\mu }K^2\sigma _0^2}E\left( \sum \limits _{i=1}^{p}\sum \limits _{j=1}^{p}\sum \limits _{t=2}^{n}\sum \limits _{s=1}^{n-1} [Z(x_{t})' Z(x_s)]^2 \Delta ^{(i)}(x_{t})\Delta ^{(j)}(x_{t})\sigma _{ij}\right) \\&\qquad +\dfrac{\delta _n^2}{n^{\mu }K^2\sigma _0^2}E\left( 2\sum \limits _{i=1}^{p}\sum \limits _{j=1}^{p}\sum \limits _{t_1=t_2+1}^{n}\sum \limits _{t_2=s+1}^{n-1}\sum \limits _{s=1}^{n-2} Z(x_{t_1})' Z(x_s) Z(x_{t_2})' Z(x_s)\Delta ^{(i)}(x_{t_1})\Delta ^{(j)}(x_{t_2})\sigma _{ij}\right) \\&\quad =T_{np121}''+T_{np122}'' \end{aligned}$$

For $T_{np121}''$, by Lemma 1(2), $\dfrac{\delta _n^2}{n^{\mu }K^2\sigma _0^2}T_{np121}''\le \dfrac{O(1)\delta _n^2 p^2}{K}\rightarrow 0.$

For $T_{np122}$, by Lemma 1(3),

$$\begin{aligned}&\dfrac{\delta _n^2}{n^{\mu }K^2\sigma _0^2}T_{np122}''\le \dfrac{O(1)\delta _n^2 p^2}{n^{\mu }K^2}\sum \limits _{t_1=t_2+1}^{n}\sum \limits _{t_2=s+1}^{n-1}\sum \limits _{s=1}^{n-2} E| Z(x_{t_1})' Z(x_s) Z(x_{t_2})' Z(x_s) | \\&\quad =\dfrac{O(1)\delta _n^2 p^2}{n^{\mu }K^2}\sum \limits _{t_1=t_2+1}^{n}\sum \limits _{t_2=s+1}^{n-1}\sum \limits _{s=1}^{n-2} E| Z(x_{t_2}+x_{t_1}-x_{t_2})' Z(x_s) Z(x_{t_2})' Z(x_s) | \\&\quad \le \dfrac{O(1)\delta _n^2 p^2}{n^{\mu }K^2}\sum \limits _{t_1=t_2+1}^{n}\sum \limits _{t_2=s+1}^{n-1}\sum \limits _{s=1}^{n-2}\dfrac{1}{d_{t_1 t_2}}\dfrac{1}{d_{t_2 s}}\dfrac{1}{d_s}\iiint |Z(x)'Z(z)Z(y)'Z(z)|\mathrm {d}x\mathrm {d}y\mathrm {d}z \\&\quad \le \dfrac{O(1)\delta _n^2 p^2 n^{\mu /2}}{K^2}\iint |Z(x)'Z(z)|^2\mathrm {d}x\mathrm {d}z=\dfrac{O(1)\delta _n^2 p^2 n^{\mu /2}}{K^2}\int \Vert Z(z)\Vert ^2\mathrm {d}z=\dfrac{O(1)\delta _n^2 p^2 n^{\mu /2}}{K}. \end{aligned}$$

Thus, $\dfrac{1}{n^{\mu /2}K\sigma _0}T_{np12}''=O_p(\delta _n p n^{\mu /4} K^{-1/2}).$

Similarly, we can get

$$\begin{aligned} \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np13}''=O_p(\delta _n p n^{\mu /4} K^{-1/2}). \end{aligned}$$

Then

$$\begin{aligned} \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np1}''=O_p(\delta _n p n^{\mu /4} K^{-1/2}). \end{aligned}$$

As we just have proved earlier,

$$\begin{aligned} \dfrac{1}{n^{\mu /2}K\sigma _0}T_{np1}'''=\varOmega _P\left( \dfrac{\delta _n^2 n^{\mu /2}}{p^2 K}\right) , \end{aligned}$$

while

$$\begin{aligned} \dfrac{\delta _n p n^{\mu /4}K^{-1/2}}{\delta _n^2 n^{\mu /2}/(p^2K)}=\dfrac{p^3K^{1/2}}{\delta _n n^{\mu /4}}=\dfrac{p^3}{n^{\lambda /2}}\rightarrow 0, \end{aligned}$$

so the divergence rate of $\dfrac{1}{n^{\mu /2}K\sigma _0}T_{np1}'''$ dominates the divergence rate of $\dfrac{1}{n^{\mu /2}K\sigma _0}T_{np1}''$. Thus,

$$\begin{aligned} \dfrac{1}{n^{\frac{\mu }{2}}K\sigma _{0}}T_{np}\xrightarrow {P}\infty ,\quad K,n,p\rightarrow \infty . \end{aligned}$$

$\square$

Proof of Theorem 3

As ${\hat{\sigma }}$ is a consistent estimator for $\sigma _0$, this theorem is obvious due to Theorems 1 and 2. $\square$