Modal regression for fixed effects panel data

Abstract

Most research on panel data focuses on mean or quantile regression, while there is not much research about regression methods based on the mode. In this paper, we propose a new model named fixed effects modal regression for panel data in which we model how the conditional mode of the response variable depends on the covariates and employ a kernel-based objective function to simplify the computation. The proposed modal regression can complement the mean and quantile regressions and provide better central tendency measure and prediction performance when the data are skewed. We present a linear dummy modal regression method and a pseudo-demodeing two-step method to estimate the proposed modal regression. The computations can be easily implemented using a modified modal–expectation–maximization algorithm. We investigate the asymptotic properties of the modal estimators under some mild regularity conditions when the number of individuals, N, and the number of time periods, T, go to infinity. The optimal bandwidths with order $(NT)^{-1/7}$ are obtained by minimizing the asymptotic weighted mean squared errors. Monte Carlo simulations and two real data analyses of a public capital productivity study and a carbon dioxide emissions study are presented to demonstrate the finite sample performance of the newly proposed modal regression.

Appendices

A Appendix-Tables

See Tables 1, 2, 3, 4, 5, 6

Table 1 Monte Carlo results: DGP 1-Case 1

Table 2 Monte Carlo results: DGP 1-Case 2

Table 3 Monte Carlo results: DGP 1-Case 3

Table 4 Monte Carlo results: DGP 2-more skewed

Table 5 Monte Carlo results: DGP 2-less skewed

Table 6 The results of estimates of (17)

Table 7 The results of estimates of (18)-individual effects

Table 8 The Results of Estimates of (18)-Time Effects

B Appendix-Proofs

1.1 Proof of Theorem 2.1

Recall that $Y_{it}={X}_{it}^T {\beta }+Z^T_{\mu ,i}\mu +{v}_{it}$. We define $X^{*}_{it}=({X}_{it}^T, Z^T_{\mu ,i})^T$, $\theta =(\beta ^T,\mu ^T)^T$, and $\theta _0=(\beta _0^T,\mu _0^T)^T$, where $\theta _{0}$ is the true parameter value. Let $\delta _{NT}=h_0^{2}+\sqrt{\left( NTh_0^{3}\right) ^{-1}}$, then it is sufficient to show that for any given $\eta ,$ there exists a large number constant a such that

$$\begin{aligned} P\left\{ \sup _{\Vert c\Vert =a} Q_{N T}\left( \theta _{0}+\delta _{NT} c\right) <Q_{N T}\left( \theta _{0}\right) \right\} \ge 1-\eta , \end{aligned}$$

(A.1)

where $\Vert \cdot \Vert$ represents the Euclidean distance, and $\theta _{0}$ is the true parameter value.

Applying the Taylor expansion, it follows

$$\begin{aligned}&Q_{N T}\left( \theta _{0}+\delta _{NT} c\right) -Q_{N T}\left( \theta _{0}\right) \nonumber \\&\quad =\frac{1}{NTh_0}\sum ^N_{i=1}\sum ^T_{t=1} \left[ \phi \left( \frac{v_{it}-\delta _{NT} c^{T} X^*_{i t}}{h_0} \right) -\phi \left( \frac{v_{it}}{h_0} \right) \right] \nonumber \\&\quad = \frac{1}{NTh_0}\sum ^N_{i=1}\sum ^T_{t=1}\left[ -\phi ^{(1)} \left( \frac{v_{it}}{h_0} \right) \left( \frac{\delta _{NT} c^{T} X^*_{i t}}{h_0} \right) +\frac{1}{2}\phi ^{(2)} \left( \frac{v_{it}}{h_0} \right) \left( \frac{\delta _{NT} c^{T} X^*_{i t}}{h_0} \right) ^{2} \right. \nonumber \\&\left. \qquad -\frac{1}{6}\phi ^{(3)} \left( \frac{v^*_{it}}{h_0} \right) \left( \frac{\delta _{NT} c^{T} X^*_{i t}}{h_0} \right) ^{3} \right] , \end{aligned}$$

(A.2)

where $v^*_{it}$ is between $v_{it}$ and $v_{it}-\delta _{NT} c^{T} X^*_{i t}$. Based on the result $T_{N T}={\mathbb {E}}\left( T_{N T}\right) +O_{p}(\sqrt{{\text {Var}}\left( T_{N T}\right) }),$ we consider each part of above Taylor expansion.

For the first part, which is $I_1=\frac{1}{NTh_0}\sum ^N_{i=1}\sum ^T_{t=1}\left( -\phi ^{(1)} \left( \frac{v_{it}}{h_0} \right) \left( \frac{\delta _{NT} c^{T} X^*_{i t}}{h_0} \right) \right)$, we can calculate it directly to achieve

$$\begin{aligned} {\mathbb {E}} \left( I_1 \right) =&\frac{-\delta _{NT}}{h^2_0} \iint \phi ^{(1)} \left( \frac{v}{h_0} \right) f_{v}(v \mid X^*) \left( c^{T} X^* \right) dv d F(X^*) \nonumber \\ =&O_p\left( \delta _{NT} a h^2_0 \right) . \end{aligned}$$

(A.3)

Meanwhile, we know that

$$\begin{aligned} {\mathbb {E}} \left( I_1 \right) ^2 =&\frac{\delta ^2_{NT}}{NTh^4_0} \iint \phi ^{(1)2} \left( \frac{v}{h_0} \right) f_{v}(v \mid X^*) \left( {c^{T} X^*}\right) ^2 dv dF(X^*) \nonumber \\ =&O_p\left( \delta ^2_{NT} a^2 (NTh^3_0)^{-1} \right) . \end{aligned}$$

(A.4)

Combing (A.3) and (A.4) obtains

$$\begin{aligned} I_1=O_p\left( \delta _{NT} a h^2_0 \right) +O_p\left( \sqrt{\delta ^2_{NT} a^2 (NTh^3_0)^{-1}} \right) =O_p\left( \delta ^2_{NT} a \right) . \end{aligned}$$

(A.5)

For the second part, which is $I_2=\frac{1}{NTh_0}\sum ^N_{i=1}\sum ^T_{t=1}\left( \frac{1}{2}\phi ^{(2)} \left( \frac{v_{it}}{h_0} \right) \left( \frac{\delta _{NT} c^{T} X^*_{i t}}{h_0} \right) ^{2} \right)$, we can prove

$$\begin{aligned} {\mathbb {E}} \left( I_2 \right) =&\frac{\delta ^2_{NT}}{2h^3_0} \iint \phi ^{(2)} \left( \frac{v}{h_0} \right) f_{v}(v \mid X^*) \left( {c^{T} X^*}\right) ^2 dv dF(X^*)\nonumber \\ =&O_p\left( \delta ^2_{NT} a^2 \right) , \end{aligned}$$

(A.6)

which indicates that the second part will dominate the first part when we choose a big enough.

For the third part, which is $I_3=\frac{1}{NTh_0}\sum ^N_{i=1}\sum ^T_{t=1}\left( -\frac{1}{6}\phi ^{(3)} \left( \frac{v^*_{it}}{h_0} \right) \left( \frac{\delta _{NT} c^{T} X^*_{i t}}{h_0} \right) ^{3} \right)$, as $v^*_{it}$ is between $v_{it}$ and $v_{it}-\delta _{NT} c^{T} X^*_{i t}$, after some direct calculations we can obtain

$$\begin{aligned} {\mathbb {E}} \left( I_3 \right) \approx&\frac{-\delta ^3_{NT}}{6h^4_0} \iint \phi ^{(3)} \left( \frac{v}{h_0} \right) f_{v}(v \mid X^*) \left( {c^{T} X^*}\right) ^3 dv F(X^*) \nonumber \\ =&O_p\left( \delta ^3_{NT} \right) , \end{aligned}$$

(A.7)

which indicates that the second part dominates the third part with the assumption $NTh^5_0 \rightarrow \infty$.

Based on these, we can choose a bigger enough such that the second term dominates the other two terms with probability $1-\eta$. Because the second term is negative, thus $P\left\{ \sup _{\Vert c\Vert =a} Q_{N T}\left( \theta _{0}+\delta _{T} c\right) <Q_{N T}\left( \theta _{0}\right) \right\} \ge 1-\eta$ holds. Hence, with the probability approaching 1, there exists a local maximizer $\hat{\theta }$ such that

$$\begin{aligned} \Vert \hat{\theta }-\theta _0 \Vert = O_p(h_0^{2}+\sqrt{\left( NTh_0^{3}\right) ^{-1}}). \end{aligned}$$

(A.8)

$\square$

1.2 Proof of Theorem 2.2

Recall that

$$\begin{aligned} Q_{NT}({\beta }, {\mu })=\frac{1}{NTh_0}\sum ^N_{i=1}\sum ^T_{t=1}\phi \left( \frac{{Y}_{it}-{X}_{it}^T {\beta }-Z^T_{\mu ,i}{\mu }}{h_0} \right) . \end{aligned}$$

Because $(\hat{\beta }, \hat{\mu })$ maximize $Q_{NT}({\beta }, {\mu })$, we can take the derivative of $Q_{NT}({\beta }, {\mu })$ respect to $\beta$ and $\mu$ to obtain

$$\begin{aligned} \frac{\partial Q_{NT}({\beta }, {\mu })}{\partial \beta } \Bigg |_{(\beta =\hat{\beta }, \mu =\hat{\mu })}=-\frac{1}{NTh_0}\sum ^N_{i=1}\sum ^T_{t=1}\phi ^{(1)} \left( \frac{{Y}_{it}-{X}_{it}^T \hat{\beta }-Z^T_{\mu ,i}\hat{\mu }}{h_0} \right) \left( \frac{{X}_{it}}{h_0} \right) =0. \nonumber \\ \end{aligned}$$

(A.9)

$$\begin{aligned} \frac{\partial Q_{NT}({\beta }, {\mu })}{\partial \mu } \Bigg |_{(\beta =\hat{\beta }, \mu =\hat{\mu })}=-\frac{1}{NTh_0}\sum ^N_{i=1}\sum ^T_{t=1}\phi ^{(1)} \left( \frac{{Y}_{it}-{X}_{it}^T \hat{\beta }-Z^T_{\mu ,i}\hat{\mu }}{h_0} \right) \left( \frac{Z_{\mu ,i}}{h_0} \right) =0. \nonumber \\ \end{aligned}$$

(A.10)

By applying Taylor expansion for (A.9) and (A.10), we can achieve

$$\begin{aligned}&\frac{1}{NTh^2_0}\sum ^N_{i=1}\sum ^T_{t=1} \left[ -{\phi ^{(1)} \left( \frac{v_{it}}{h_0} \right) X_{it}+\phi ^{(2)} \left( \frac{v_{it}}{h_0} \right) X_{it} \{ -X^T_{it}(\hat{\beta }-\beta _0)h^{-1}_0-Z^T_{\mu ,i}(\hat{\mu }-\mu _0)h_0^{-1}\}}\right. \nonumber \\&\quad \left. {+\frac{1}{2}\phi ^{(3)} \left( \frac{v^*_{it}}{h_0} \right) X_{it} \{ -X^T_{it}(\hat{\beta }-\beta _0)h^{-1}_0-Z^T_{\mu ,i}(\hat{\mu }-\mu _0)h_0^{-1}\}^2} \right] +o_p(\hat{\beta }-\beta _0)=0. \end{aligned}$$

(A.11)

$$\begin{aligned}&\frac{1}{NTh^2_0}\sum ^N_{i=1}\sum ^T_{t=1} \left[ -{\phi ^{(1)} \left( \frac{v_{it}}{h_0} \right) Z_{\mu ,i}+\phi ^{(2)} \left( \frac{v_{it}}{h_0} \right) Z_{\mu ,i} \{ -X^T_{it}(\hat{\beta }-\beta _0)h^{-1}_0-Z^T_{\mu ,i}(\hat{\mu }-\mu _0)h_0^{-1}\}}\right. \nonumber \\&\quad \left. {+\frac{1}{2}\phi ^{(3)} \left( \frac{v^*_{it}}{h_0} \right) Z_{\mu ,i} \{ -X^T_{it}(\hat{\beta }-\beta _0)h^{-1}_0-Z^T_{\mu ,i}(\hat{\mu }-\mu _0)h_0^{-1}\}^2 } \right] +o_p(\hat{\mu }-\mu _0)=0, \end{aligned}$$

(A.12)

where $v^*_{it}$ is between $v_{it}$ and ${v}_{it}-{X}_{it}^T (\hat{\beta }-\beta _0)-Z^T_{\mu ,i}(\hat{\mu }-\mu _0)$.

We focus on (A.12) firstly. Considering $-\frac{1}{NTh^2_0}\sum ^N_{i=1}\sum ^T_{t=1} \left( \phi ^{(1)}\left( \frac{v_{it}}{h_0} \right) Z_{\mu ,i} \right)$, we get

$$\begin{aligned}&{\mathbb {E}} \left( -\frac{1}{NTh^2_0}\sum ^N_{i=1}\sum ^T_{t=1} \left( \phi ^{(1)}\left( \frac{v_{it}}{h_0} \right) Z_{\mu ,i} \right) \right) \nonumber \\&\quad = \frac{-1}{h^2_0} \iint \phi ^{(1)} \left( \frac{v}{h_0} \right) f_{v}(v \mid X, Z) Z \ dv d F(X) \nonumber \\&\quad = \frac{1}{h_0} \iint \phi \left( \tau \right) \tau f_{v}(\tau h_0 \mid X, Z) Z \ d\tau d F(X) \nonumber \\&\quad = \frac{h^2_0}{2} {\mathbb {E}} \left( Z f^{(3)}_{v}(0 \mid X, Z) \right) . \end{aligned}$$

(A.13)

Considering $\frac{1}{NTh^3_0}\sum ^N_{i=1}\sum ^T_{t=1} \left( \phi ^{(2)}\left( \frac{v_{it}}{h_0} \right) \left( Z_{\mu ,i} X^{T}_{it} \right) \right)$, we achieve

$$\begin{aligned}&{\mathbb {E}} \left( \frac{1}{NTh^3_0}\sum ^N_{i=1}\sum ^T_{t=1}\left( \phi ^{(2)}\left( \frac{v_{it}}{h_0} \right) Z_{\mu ,i}X^{T}_{it} \right) \right) \nonumber \\&\quad = \frac{1}{h^3_0} \iint \phi ^{(2)} \left( \frac{v}{h_0} \right) f_{v}(v \mid X, Z) \left( ZX^{T}\right) dv d F(X) \nonumber \\&\quad = \frac{1}{h^2_0} \iint \phi \left( \tau \right) (\tau ^2-1) f_{v}(\tau h_0 \mid X, Z) \left( ZX^{T} \right) d\tau d F(X)\nonumber \\&\quad = {\mathbb {E}} \left( ZX^{T} f^{(2)}_{v}(0 \mid X, Z) \right) . \end{aligned}$$

(A.14)

Considering $\frac{1}{NTh_0}\sum ^N_{i=1}\sum ^T_{t=1} \left( \phi ^{(2)}\left( \frac{v_{it}}{h_0} \right) \left( \frac{Z_{\mu ,i}}{h_0}\frac{Z^{T}_{\mu ,i}}{h_0} \right) \right)$, we can obtain

$$\begin{aligned}&{\mathbb {E}} \left( \frac{1}{NTh^3_0}\sum ^N_{i=1}\sum ^T_{t=1}\phi ^{(2)}\left( \frac{v_{it}}{h_0} \right) \left( Z_{\mu ,i}Z^{T}_{\mu ,i} \right) \right) \nonumber \\&\quad = \frac{1}{h^3_0} \iint \phi ^{(2)} \left( \frac{v}{h_0} \right) f_{v}(v \mid X, Z) \left( ZZ^{T}\right) dv d F(X) \nonumber \\&\quad = \frac{1}{h^2_0} \iint \phi \left( \tau \right) (\tau ^2-1) f_{v}(\tau h_0 \mid X, Z) \left( ZZ^{T} \right) d\tau d F(X) \nonumber \\&\quad = {\mathbb {E}} \left( ZZ^{T} f^{(2)}_{v}(0 \mid X, Z) \right) . \end{aligned}$$

(A.15)

Then, it follows that

$$\begin{aligned} \hat{\mu }-\mu _0=\left( \Phi +o_p(1)\right) ^{-1} \{-\Psi (\hat{\beta }-\beta _0)+o_p(1)\}, \end{aligned}$$

(A.16)

where $\Phi =\lim _{N \rightarrow \infty } (1/N) \sum ^N_{i=1} {\mathbb {E}} \left( Z_{\mu ,i}Z^{T}_{\mu ,i} f^{(2)}_{v}(0 \mid X_{it}, Z_{\mu ,i}) \right)$ and $\Psi =\lim _{N,T \rightarrow \infty } (1/(NT))\sum ^N_{i=1}$ $\sum ^T_{t=1} {\mathbb {E}}\left( Z_{\mu ,i}X^{T}_{it} f^{(2)}_{v}(0 \mid X_{it}, Z_{\mu ,i}) \right)$.

Substituting (A.16) into (A.11), we can have

$$\begin{aligned}&\frac{1}{NTh^2_0}\sum ^N_{i=1}\sum ^T_{t=1} \left[ -\phi ^{(1)} \left( \frac{v_{it}}{h_0} \right) X_{it}{+}\phi ^{(2)} \left( \frac{v_{it}}{h_0} \right) X_{it} \{ -X^T_{it}(\hat{\beta }{-}\beta _0)h^{-1}_0{+}Z^T_{\mu ,i}(\Phi ^{-1}\Psi (\hat{\beta }-\beta _0))h_0^{-1}\}\right. \nonumber \\&\quad \left. {+\frac{1}{2}\phi ^{(3)} \left( \frac{v^*_{it}}{h_0} \right) X_{it} \{ -X^T_{it}(\hat{\beta }-\beta _0)h^{-1}_0+Z^T_{\mu ,i}(\Phi ^{-1}\Psi (\hat{\beta }-\beta _0))h_0^{-1}\}^2} \right] +o_p(\hat{\beta }-\beta _0)=0. \end{aligned}$$

(A.17)

Define

$$\begin{aligned} &M_{NT}=\frac{-1}{NTh^2_0}\sum ^N_{i=1}\sum ^T_{t=1} \left( \phi ^{(1)}\left( \frac{v_{it}}{h_0} \right) {X_{it}}\right) ,\\&J_{NT}=\frac{1}{NTh^2_0}\sum ^N_{i=1}\sum ^T_{t=1}\left( \phi ^{(2)}\left( \frac{v_{it}}{h_0} \right) {X_{it}}(X^T_{it}-Z^T_{\mu ,i}\Phi ^{-1}\Psi )h^{-1}_0\right) , \end{aligned}$$

we then get

$$\begin{aligned} \hat{\beta }-\beta =J^{-1}_{NT}M_{NT}(1+o_p(1)). \end{aligned}$$

(A.18)

With some calculations, we can obtain

$$\begin{aligned}&{\mathbb {E}} \left( \frac{-1}{NTh^2_0}\sum ^N_{i=1}\sum ^T_{t=1}\phi ^{(1)}\left( \frac{v_{it}}{h_0} \right) {X_{it}} \right) \nonumber \\&\quad = \frac{1}{h^2_0} \iint \phi ^{(1)} \left( \frac{v}{h_0} \right) f_{v}(v \mid X,Z) X \ dv d F(X) \nonumber \\&\quad = \frac{1}{h_0} \iint \phi \left( \tau \right) \tau f_{v}(\tau h_0 \mid X,Z) X \ d\tau d F(X) \nonumber \\&\quad = \frac{h^2_0}{2} {\mathbb {E}} \left( X f^{(3)}_{v}(0 \mid X,Z) \right) . \end{aligned}$$

(A.19)

$$\begin{aligned}&{\mathbb {E}} \left( \frac{1}{NTh^3_0}\sum ^N_{i=1}\sum ^T_{t=1}\phi ^{(2)}\left( \frac{v_{it}}{h_0} \right) \left( X_{it}(X^T_{it}-Z^T_{\mu ,i}\Phi ^{-1}\Psi \right) \right) \nonumber \\&\quad = \frac{1}{h^3_0} \iint \phi ^{(2)} \left( \frac{v}{h_0} \right) f_{v}(v \mid X, Z) \left( X(X^T-Z^T\Phi ^{-1}\Psi ) \right) dv d F(X) \nonumber \\&\quad = \frac{1}{h^2_0} \iint \phi \left( \tau \right) (\tau ^2-1) f_{v}(\tau h_0 \mid X, Z) \left( X(X^T-Z^T\Phi ^{-1}\Psi ) \right) d\tau d F(X) \nonumber \\&\quad = {\mathbb {E}} \left( X(X^T-Z^T\Phi ^{-1}\Psi ) f^{(2)}_{v}(0 \mid X, Z) \right) . \end{aligned}$$

(A.20)

Meanwhile, based on above calculations, we have

$$\begin{aligned}&{\text {Cov}}(M_{NT}) \nonumber \\&\quad ={\mathbb {E}} \Bigg \{ \left( \frac{1}{NTh_0}\sum ^N_{i=1}\sum ^T_{t=1} \left( \phi ^{(1)}\left( \frac{v_{it}}{h_0} \right) \frac{X_{it}}{h_0} \right) \right) \left( \frac{1}{NTh_0}\sum ^N_{i=1}\sum ^T_{t=1} \left( \phi ^{(1)}\left( \frac{v_{it}}{h_0} \right) \frac{X_{it}}{h_0} \right) \right) ^T \Bigg \} \{1+o_p(1) \}\nonumber \\&\quad = \frac{1}{NTh^4_0} \iint \phi ^{(1)2} \left( \frac{v}{h_0} \right) f_{v}(v \mid X, Z) X X^{T} dv d F(X) \{1+o_p(1) \}\nonumber \\&\quad =\frac{1}{NTh^3_0} v_2 {L}\{1+o_p(1) \}, \end{aligned}$$

(A.21)

where $v_2=\int \phi ^{2} \left( \tau \right) \tau ^2 d \tau$ and ${L}=\lim _{N,T \rightarrow \infty } (1/(NT))\sum ^N_{i=1}$ $\sum ^T_{t=1} {\mathbb {E}} \Big ( X_{it}X^{T}_{it} f_{v}(0 \mid X_{it}, Z_{\mu ,i}) \Big )$.

To show Theorem 2.2, it is sufficient to show the asymptotic normality for $M^*_{NT}=\sqrt{NTh^3_0}M_{NT}$, where we prove that for any unit vector $d \in {\mathbb {R}}^q$,

$$\begin{aligned} \{d^T {\text {Cov}}(M^*_{NT}) d\}^{-1/2} \{d^T M^*_{NT}-d^T {\mathbb {E}}(M^*_{NT}) \} {\mathop {\rightarrow }\limits ^{d}} N(0, 1). \end{aligned}$$

(A.22)

Then, we check Lyapunov’s condition. Let $\xi _i=-1/\sqrt{NTh_0}\phi ^{(1)}\left( \frac{v_{it}}{h_0} \right) d^T X_{it}$, we need to prove $NT {\mathbb {E}} |\xi _1 |^3 \rightarrow 0$. As $\left( d^T X_{it} \right) ^2 \le \Vert d\Vert ^2 \Vert X_{it} \Vert ^2$ and $\phi ^{(1)}(.)$ is bounded, we have

$$\begin{aligned} NT {\mathbb {E}} |\xi _1 |^3 \le O((NT)^{-1/2} h^{-3/2}_0) \rightarrow 0. \end{aligned}$$

(A.23)

Thus, the asymptotic normality for $M^*_{NT}$ holds with

$$\begin{aligned} \sqrt{NTh^3_0} \left( M_{NT}-h^2_0M/2 \right) {\mathop {\rightarrow }\limits ^{d}} N(0, {v_2}{L}). \end{aligned}$$

(A.24)

According to Slutsky’s Theorem, we obtain Theorem 2.2. $\square$

1.3 Proof of Theorem 3.1

The main proof steps here are similar with those of Proof of Theorem 2.1. We briefly outline the proof. Recall that

$$\begin{aligned} Q_{NT}(\gamma _1, \beta )&=\frac{1}{NTh_1}\sum ^N_{i=1}\sum ^T_{t=1}\phi \left( \frac{{\hat{Y}}_{it}-\gamma _1-{{X}}^T_{it} {\beta }}{h_1} \right) \nonumber \\&=\frac{1}{NTh_1}\sum ^N_{i=1}\sum ^T_{t=1}\phi \left( \frac{{Y}_{it}-\alpha _{i0}+\alpha _{i0}-\hat{\alpha }_i-{\tilde{X}}^{T}_{it} {\theta }}{h_1} \right) , \end{aligned}$$

(A.25)

where ${\tilde{X}}^{T}_{it}=(1,{X}_{it}^T)$ and $\theta =(\gamma _1, \beta ^T)^T$. Define $\delta _{NT}=h_1^{2}+\sqrt{\left( NTh_1^{3}\right) ^{-1}}$, it is sufficient to show that for any given $\eta ,$ there exists a large number constant a such that $P\left\{ \sup _{\Vert c\Vert =a} Q_{N T}\left( \theta _{0}+\delta _{NT} c\right) <Q_{N T}\left( \theta _{0}\right) \right\} \ge 1-\eta$, where $\Vert \cdot \Vert$ represents the Euclidean distance, and $\theta _{0}$ is the true parameter value. Applying the Taylor expansion, it follows

$$\begin{aligned}&Q_{N T}\left( \theta _{0}+\delta _{NT} c\right) -Q_{N T}\left( \theta _{0}\right) \nonumber \\ {}&\quad = \frac{1}{NTh_1}\sum ^N_{i=1}\sum ^T_{t=1} \left[ \phi \left( \frac{v_{it}+\alpha _{i0}-\hat{\alpha }_i-\delta _{NT} c^{T} {\tilde{X}}_{it}}{h_1} \right) -\phi \left( \frac{v_{it}+\alpha _{i0}-\hat{\alpha }_i}{h_1} \right) \right] \nonumber \\ {}&\quad = \frac{1}{NTh_1}\sum ^N_{i=1}\sum ^T_{t=1} \left[ {-\phi ^{(1)} \left( \frac{v_{it}+\alpha _{i0}-\hat{\alpha }_i}{h_1} \right) \left( \frac{\delta _{NT} c^{T} {\tilde{X}}_{it}}{h_1} \right) }\right. \nonumber \\ {}&\qquad \left. {+\frac{1}{2}\phi ^{(2)} \left( \frac{v_{it}+\alpha _{i0}-\hat{\alpha }_i}{h_1} \right) \left( \frac{\delta _{NT} c^{T} {\tilde{X}}_{it}}{h_1} \right) ^{2} -\frac{1}{6}\phi ^{(3)} \left( \frac{v^*_{it}}{h_1} \right) \left( \frac{\delta _{NT} c^{T} {\tilde{X}}_{it}}{h_1} \right) ^{3}} \right] , \end{aligned}$$

(A.26)

where $v^*_{it}$ is between $v_{it}+\alpha _{i0}-\hat{\alpha }_i$ and $v_{it}+\alpha _{i0}-\hat{\alpha }_i-\delta _{NT} c^{T} {\tilde{X}}_{it}$. Following the same steps as the Proof of Theorem 2.1, with assumption that $\sqrt{T}h^2_1 \rightarrow \infty$ (i.e., $N^a/T \rightarrow 0$ for some $a>4/3$), one can obtain $\Vert \hat{\theta }-\theta _0 \Vert \le \delta _{NT}$.

$\square$

1.4 Proof of Theorem 3.2

Recall that ${\tilde{X}}^T_{it}=(1, {X}_{it}^T)$, $\theta _0 =(\gamma _{10}, \beta _0^T)^T$, and $\hat{\theta }=(\hat{\gamma }_1, \hat{\beta }^T)^T$. If $\hat{\theta }$ maximizes (11), it will satisfy the following equation

$$\begin{aligned} \frac{1}{NTh_1}\sum ^N_{i=1}\sum ^T_{t=1}\phi ^{(1)} \left( \frac{{Y}_{it}-\alpha _{i0}-(\hat{\alpha }_i-\alpha _{i0})-{\tilde{X}}^T_{it} \hat{\theta }}{h_1} \right) \left( \frac{-{\tilde{X}}_{it}}{h_1} \right) =0. \end{aligned}$$

(A.27)

Then, we can achieve

$$\begin{aligned}&\frac{1}{NTh_1}\sum ^N_{i=1}\sum ^T_{t=1}\phi ^{(1)} \left( \frac{v_{it}}{h_1} \right) \left( \frac{-{\tilde{X}}_{it}}{h_1} \right) \nonumber \\ {}&\quad +\frac{1}{NTh_1}\sum ^N_{i=1}\sum ^T_{t=1}\phi ^{(2)} \left( \frac{v_{it}}{h_1} \right) \left( \frac{{\tilde{X}}_{it}}{h_1} \right) \left( \frac{\alpha _{i0}-\hat{\alpha }_i-{\tilde{X}}^T_{it} (\hat{\theta }-\theta _0 )}{h_1} \right) \nonumber \\ {}&\quad +\frac{1}{2NTh_1}\sum ^N_{i=1}\sum ^T_{t=1}\phi ^{(3)} \left( \frac{v^*_{it}}{h_1} \right) \left( \frac{{\tilde{X}}_{it}}{h_1} \right) \left( \frac{\alpha _{i0}-\hat{\alpha }_i-{\tilde{X}}^T_{it} (\hat{\theta }-\theta _0 )}{h_1} \right) ^2 =0, \end{aligned}$$

(A.28)

where $v^*_{it}$ is between $v_{it}$ and $v_{it}+\alpha _{i0}-\hat{\alpha }_i-{\tilde{X}}^T_{it} (\hat{\theta }-\theta _0 )$. It can be shown that the third term on the left-hand side of (A.28) is dominated by the second term. With assumption that $\sqrt{T}h^2_1 \rightarrow \infty$ (i.e., $N^a/T \rightarrow 0$ for some $a>4/3$), we could then follow the same proof steps as those of Proof of Theorem 2.2 to achieve the results.

$\square$