New efficient spline estimation for varying-coefficient models with two-step knot number selection

Abstract

One of the advantages for the varying-coefficient model is to allow the coefficients to vary as smooth functions of other variables and the coefficients functions can be estimated easily through a simple B-spline approximations method. This leads to a simple one-step estimation procedure. We show that such a one-step method cannot be optimal when some coefficient functions possess different degrees of smoothness. Under the regularity conditions, the consistency and asymptotic normality of the two step B-spline estimators are also derived. A few simulation studies show that the gain by the two-step procedure can be quite substantial. The methodology is illustrated by an AIDS data set.

Appendix: Proof of theorems

It will be convenient to introduce the following notation

$$\begin{aligned} \beta (u)= & {} {({\beta _1}(u),{\beta _2}(u), \ldots ,{\beta _p}(u))^T},\,U = {({u_1},{u_2} \ldots ,{u_n})^T}, \\ {s_j}(u)= & {} \pi _j^T(u){\alpha _j},\,s(u) = {({s_1}(u),{s_2}(u), \ldots ,{s_p}(u))^T}, \\ B(U)= & {} {({\beta ^T}({u_1}),{\beta ^T}({u_2}), \ldots ,{\beta ^T}({u_n}))^T},\,S(U) = {({s^T}({u_1}),{s^T}({u_2}), \ldots ,{s^T}({u_n}))^T}, \\ {\alpha _j}= & {} {({\alpha _{j1}},{\alpha _{j2}}, \ldots ,{\alpha _{j{N_j}}})^T},\,\alpha = {(\alpha _1^T,\alpha _2^T, \ldots ,\alpha {}_p^T)^T}, \\ \!\!X= & {} {({x_1},{x_2},\! \ldots ,\!{x_p})^T},\,{X_i} \!=\! ({x_{i1}},{x_{i2}}, \ldots ,{x_{ip}})^{T},\,{D_x} \!=\! diag(X_1^T,X_2^T, \ldots ,X_n^T), \\ \!\!\!\pi (u) \!= & {} \!\! {\left( {\begin{array}{*{20}{c}} {\pi _1^T(u)}&{}\quad 0&{}\quad \ldots &{}\quad 0\\ 0&{}\quad {\pi _2^T(u)}&{}\quad \ldots &{}\quad 0\\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ 0&{}\quad 0&{}\quad \ldots &{}\quad {\pi _p^T(u)}\end{array}} \right) },\, \pi _{j}(u)\!=\!(\pi _{j1}(u),\!\pi _{j2}(u),\!\ldots ,\!\pi _{jN_{j}}(u))^{T}, \end{aligned}$$

where $i=1,2,\ldots ,n$, $j=1,2,\ldots ,p$, then $D=D_x\cdot (\pi (u_1),\pi (u_2),\ldots ,\pi (u_n))^{T}$. $\lambda _{\max }^A$ and $\lambda _{\min }^A$ are, respectively , the maximum and minimum eigenvalue of A. $I_p$ is $p\times p$ unit matrix. $Q_n(x,u)$ is the empirical distribution of $(X_i,u_i)_{i=1}^{n}$. $Q_n(x|u)$ is the conditional empirical distribution, $Q_n(u)$ is the marginal empirical distribution.

Lemma 1

There exists constants $0<c'_{1}<c'_{2}<\infty$ (independent of n and $k_j$) such that

$$\begin{aligned} ({{c'}_1} + {o_p}(1))h{\left\| \alpha \right\| ^2} \le \int {{s^T}(u)} s(u){d{{Q_n}(u)}} \le ({{c'}_2} + {o_p}(1))h{\left\| \alpha \right\| ^2}. \end{aligned}$$

(A.1)

Proof

By the Lemma 6.1 of Zhou et al. (1998) , there exists the constants $0<c'_{1}<c'_{2}<\infty$ (independent of n and $k_j$) such that

$$\begin{aligned} ({{c'}_1} + {o_p}(1))h\sum \limits _{l = 1}^{{N_j}} {\alpha _{jl}^2} \le \int {s_j^2(u)} {d{{Q_n}(u)}} \le ({{c'}_2} + {o_p}(1))h\sum \limits _{l = 1}^{{N_j}} {\alpha _{jl}^2}. \end{aligned}$$

Hence

$$\begin{aligned} ({{c'}_1} + {o_p}(1))h\sum \limits _{j = 1}^p {\sum \limits _{l = 1}^{{N_j}} {\alpha _{jl}^2} } \le \int {\sum \limits _{j = 1}^p {s_j^2(u)} d{Q_n}(u)} \le ({{c'}_2} + {o_p}(1))h\sum \limits _{j = 1}^p {\sum \limits _{l = 1}^{{N_j}} {\alpha _{jl}^2} }, \end{aligned}$$

This is, (A.1) holds. $\square$

Lemma 2

If condition C3 holds, there exists the constants $0<c_1<c_2<\infty$ (independent of n and $k_j)$ such that

$$\begin{aligned} ({c_1} + {o_p}(1))nh \le \lambda _{\min }^{{D^T}D} \le \lambda _{\max }^{{D^T}D} \le ({c_2} + {o_p}(1))nh. \end{aligned}$$

(A.2)

Proof

By

$$\begin{aligned} \begin{array}{l} D\alpha = {D_x}\cdot {(\pi ({u_1}),\pi ({u_2}), \ldots ,\pi ({u_n}))^T} {(\alpha _1^T,\alpha _2^T, \ldots ,\alpha _p^T)^T}\\ \begin{array}{*{20}{c}} {}&{} = \end{array}{D_x}\cdot {({\pi ^T}({u_1}){\alpha _1}, \ldots ,{\pi ^T}({u_1}) {\alpha _P}, \ldots ,{\pi ^T}({u_n}){\alpha _1}, \ldots ,{\pi ^T}({u_n}) {\alpha _P})^T}\\ \begin{array}{*{20}{c}} {}&{} = \end{array}{D_x}\cdot {({s_1}({u_1}), \ldots ,{s_p}({u_1}), \ldots ,{s_1} ({u_n}), \ldots ,{s_p}({u_n}))^T}. \end{array} \end{aligned}$$

Hence

$$\begin{aligned} \displaystyle \begin{array}{l} {\alpha ^T}{D^T}D\alpha = \displaystyle \sum \limits _{i = 1}^n {{s^T}({u_i}) {X_i}X_i^Ts({u_i})} \\ \begin{array}{*{20}{c}} {}&{}{}&{}{} \end{array} = n\displaystyle \int {{s^T}(u)XX_{}^Ts(u)d{Q_n}(x,u)} \\ \begin{array}{*{20}{c}} {}&{}{}&{}{} \end{array} = n\displaystyle \int {{s^T}(u) \cdot \displaystyle \int {XX_{}^Td{Q_n}(x|u) \cdot s(u)d{Q_n}(u)} }. \end{array} \end{aligned}$$

(A.3)

By condition C3,

$$\begin{aligned} \displaystyle \int {{X^T}Xd{Q_n}(x|u)}{\mathop {\longrightarrow }\limits ^{p}}\displaystyle \int {X{X^T}d{Q_n}(x|u)} = G(u). \end{aligned}$$

From (A.3),

$$\begin{aligned} \begin{array}{l} {\alpha ^T}{D^T}D\alpha = n\displaystyle \int {{s^T}(u) \cdot \left[ G(u) + {o_p}(1)\right] \cdot s(u)d{Q_n}(u)} \\ \begin{array}{*{20}{c}} {}&{}{}&{}{}{} \end{array} = n\displaystyle \int {{s^T}(u)G(u)s(u)d{Q_n}(u)} + \left[ {n\displaystyle \int {{s^T}(u)s(u)d{Q_n}(u)} } \right] {o_p}(1). \end{array} \end{aligned}$$

(A.4)

From (3.2),

$$\begin{aligned} {m_3}\int {{s^T}(u)s(u)d{Q_n}(u) \le } \int {{s^T}(u)G(u)s(u)d{Q_n}(u)} \le {M_3}\int {{s^T}(u)s(u)d{Q_n}(u)}. \end{aligned}$$

By Lemma 1 and (A.4)

$$\begin{aligned} ({m_3} \!+\! {o_p}(1))({{c'}_1} \!+\! o_p(1))nh{\left\| \alpha \right\| ^2} \!\le \! {\alpha ^T}{D^T}D\alpha \le ({M_3} \!+\! {o_p}(1))({{c'}_2} \!+\! {o_p}(1))nh{\left\| \alpha \right\| ^2}. \end{aligned}$$

Let $c_1=c'_1m_3$, $c_2=c'_2M_3$, we have

$$\begin{aligned} ({c_1} + o_p(1))nh{\left\| \alpha \right\| ^2} \le {\alpha ^T}{D^T}D\alpha \le ({c_2} + {o_p}(1))nh{\left\| \alpha \right\| ^2}. \end{aligned}$$

Note that

$$\begin{aligned} \lambda _{\max }^{{D^T}D} = \mathop {\max }\limits _{\left\| \alpha \right\| = 1} {\alpha ^T}{D^T}D\alpha ,\ \lambda _{\min }^{{D^T}D} = \mathop {\min }\limits _{\left\| \alpha \right\| = 1} {\alpha ^T}{D^T}D\alpha , \end{aligned}$$

(A.2) holds.

By Lemma 2, we also know that

$$\begin{aligned} \left( {\frac{1}{{{c_2}}} + {o_p}(1)} \right) {(nh)^{ - 1}} \le \lambda _{\min }^{{{({D^T}D)}^{ - 1}}} \le \lambda _{\max }^{{{({D^T}D)}^{ - 1}}} \le \left( {\frac{1}{{{c_1}}} + {o_p}(1)} \right) {(nh)^{ - 1}}. \end{aligned}$$

(A.5)

$\square$

Lemma 3

If A and B are nonnegative matrices, then

$$\begin{aligned} \lambda _{\min }^Atr(AB) \le tr(AB) \le \lambda _{\max }^Atr(AB) \end{aligned}$$

Proof

The strategy to prove this lemma is similar to Lemma 6.5 of Zhou et al. (1998). Therefore, we omit the proof. $\square$

Lemma 4

If conditions C1 and C2 hold, then there exits a constant $M_5$ such that

$$\begin{aligned} \mathop {\sup }\limits _{u \in [a,b]} \left| {{\beta _j}(u) - \pi _j^T(u){\alpha _j}} \right| \le {M_5}{h^m}, \end{aligned}$$

(A.6)

$j=1,2,\ldots ,p$, where $\alpha _j$ is a $N_j \times 1$ vector depending on $\beta _j(u).$

Proof

Lemma 4 proof follows readily from Corollary 6.21 of Schumaker (1981). $\square$

Proof of Theorem 1

By (2.9), then

$$\begin{aligned} {E_\chi }({{\hat{\beta }}} (u)) = {\pi ^T }(u) \cdot {({D^T }D)^{ - 1}}{D^T }{D_x}\mathrm{B}(U), \end{aligned}$$

where $\chi =(X_i,u_i)^n_{i=1}$ and ${E_\chi }$ denotes the conditional expectation given $\chi$. So

$$\begin{aligned} \begin{array}{llll} &{}{E_\chi }({{\hat{\beta }}} (u)) - s(u)\\ &{}\quad = {\pi ^T}(u) \cdot {({D^T}D)^{ - 1}}{D^T}{D_x}\mathrm{B}(U) - {\pi ^T}(u) \cdot {({D^T}D)^{ - 1}}({D^T}D)\alpha \\ &{}\quad = {\pi ^T}(u) \cdot {({D^T}D)^{ - 1}}{D^T}{D_x}\mathrm{B}(U) - {\pi ^T}(u) \cdot {({D^T}D)^{ - 1}}{D^T}{D_x}S(U)\\ &{}\quad = {\pi ^T}(u) \cdot {({D^T}D)^{ - 1}} \cdot {D^T}{D_x}(B(U) - S(U)), \end{array} \end{aligned}$$

(A.7)

on the other hand,

$$\begin{aligned} \begin{array}{llllll} &{}{D^T}{D_x}(B(U) - S(U))\\ &{}\quad = (\pi ({u_1}),\pi ({u_{\mathrm{{2}}}}), \ldots ,\pi ({u_n}))D_x^T \cdot D{}_x(B(U) - S(U))\\ &{}\quad = \pi ({u_1}) \cdot {X_1}X_1^T \cdot (\beta ({u_1}) - s({u_1})) + \cdots + \pi ({u_n}) \cdot {X_n}X_n^T \cdot (\beta ({u_n}) - s({u_n}))\\ &{}\quad = n\displaystyle \frac{1}{n}\displaystyle \sum \limits _{i = 1}^n {\pi ({u_i}) \cdot {X_i}X_i^T \cdot (\beta ({u_i}) - s({u_i}))} \\ &{}\quad = n\displaystyle \int {\pi (u) \cdot X{X^T} \cdot (\beta (u) - s(u))} d{Q_n}(x,u)\\ &{}\quad = n\displaystyle \int {\pi (u) \cdot \displaystyle \int {X{X^T}} } d{Q_n}(x|u) \cdot (\beta (u) - s(u))d{Q_n}(u)\\ &{}\quad = n\displaystyle \int {\pi (u)} \cdot (G(u) + {o_p}(1)) \cdot (\beta (u) - s(u))d{Q_n}(u). \end{array} \end{aligned}$$

(A.8)

Let $\eta (u) = G(u)(\beta (u) - s(u)) = {({\eta _1}(u), \ldots ,{\eta _p}(u))^T }$ and ${\eta _i}(u) = \sum \nolimits _{j = 1}^p {{g_{ij}}} (u)({\beta _j}(u) - {s_j}(u)),\ i=1,2\ldots ,p$. Then, $\pi (u)G(u)(\beta (u) - s(u)) = \pi (u)\eta (u),$ by (3.1), the $((i-1)N+l)$th element of $\displaystyle \int {\pi (u)} \eta (u) d{Q_n}(u)$ is

$$\begin{aligned} \begin{array}{lll} &{}\displaystyle \int {{\pi _l}} (u){\eta _i}(u)d{Q_n}(u)\\ &{}\quad =\displaystyle \sum \limits _{j = 1}^p {\displaystyle \int {{\pi _l}(u){g_{ij}}(u)} ({\beta _j}(u) - {s_j}(u))} d{Q_n}(u)\\ &{}\quad \le {M_2}\displaystyle \sum \limits _{j = 1}^p {\displaystyle \int {{\pi _l}(u)} ({\beta _j}(u) - {s_j}(u))} d{Q_n}(u), \end{array} \end{aligned}$$

(A.9)

by Glivenko–Cantelli Theorem,

$$\begin{aligned} \sup _{a<<b}\left| Q_{n}(t)-Q(t)\right| =O_{p}\left( n^{-1 / 2}\right) , \end{aligned}$$

by $k_j=O(n^{\frac{1}{{2m + 1}}})$, $m>1$,

$$\begin{aligned} \mathop {\sup }\limits _{a \le u \le b} \left| {{Q_n}(u) - Q(u)} \right| = {\mathrm{{o}}_p}({k_j}^{ - 1}), \end{aligned}$$

(A.10)

by (A.10), condition C2 and Lemma 6.10 of Agarwal et al. (1980), for any $1\le j\le p$,

$$\begin{aligned} \mathop {\max }\limits _l \{ \displaystyle \int {{\pi _l}} (u)({\beta _j}(t) - {s_j}(t))d{Q_n}(u)\} = {o_p}({h^{m + 1}}), \end{aligned}$$

(A.11)

by (A.9),

$$\begin{aligned} \displaystyle \int {{\pi _l}} (u){\eta _i}(u)d{Q_n}(u) \le {o_p}({h^{m + 1}}). \end{aligned}$$

Let ${D^T}{D_x}(B(U) - S(U)) = W = {(W_1^T, \ldots ,W_p^T)^T},{W_i} = {({w_{i1}}, \ldots ,{w_{i{N_j}}})^T}.$

By (A.7)–(A.11),

$$\begin{aligned} {w_{il}} \le {o_p}(n{h^{m + 1}}), \end{aligned}$$

(A.12)

by Lemma 3

$$\begin{aligned} \begin{array}{lll} &{}{\left\| {{\pi ^T}(u) \cdot {{({D^T}D)}^{ - 1}} \cdot {D^\tau } {D_x}(B(U) - S(U))} \right\| ^2}\\ &{}\quad = tr({W^T}{({D^T}D)^{ - 1}}\pi (u) \cdot {\pi ^T}(u){({D^T}D)^{ - 1}}W)\\ &{}\quad \le {(\lambda _{\max }^{{{({D^T}D)}^{ - 1}}})^2}tr(\pi (u) \cdot {\pi ^T}(u) \cdot W{W^T})\\ &{}\quad = {\left( \lambda _{\max }^{{{({D^T}D)}^{ - 1}}}\right) ^2}tr({W^T}\pi (u) \cdot {\pi ^T}(u)W),\\ \end{array} \end{aligned}$$

(A.13)

by(2.5) and (A.12),

$$\begin{aligned} W_i^T\pi (u) = \sum \limits _l {{w_{il}}{\pi _l}(u)} \le {o_p}(n{h^{m + 1}})\sum {{\pi _l}} (u) = {o_p}(n{h^{m + 1}}). \end{aligned}$$

Hence

$$\begin{aligned} tr({W^T}\pi (u) \cdot {\pi ^T}(u)W) = \sum \limits _{j = 1}^p {{{(W_j^T\pi (u))}^2}} \le {o_p}({n^2}{h^{2(m + 1)}}), \end{aligned}$$

(A.13)

by (A.5), (A.7), (A.9) and (A.13),

$$\begin{aligned} \left\| {{E_\chi }({{\hat{\beta }}} (u) - s(u))} \right\| \le {o_p}({h^m}), \end{aligned}$$

(A.14)

by Lemma 4,

$$\begin{aligned} \left\| {s(u) - \beta (u)} \right\| = O({h^m}). \end{aligned}$$

Hence

$$\begin{aligned} \begin{aligned} \left\| {{E_\chi }({{\hat{\beta }}} (u) - \beta (u))} \right\|&= \left\| {{E_\chi }({{\hat{\beta }}} (u) - s(u) + s(u) - \beta (u))} \right\| \\&\le \left\| {{E_\chi }({{\hat{\beta }}} (u) - s(u)} \right\| + \left\| {s(u) - \beta (u))} \right\| = {O_p}({h^m}). \end{aligned} \end{aligned}$$

(A.15)

Now let us prove the equation

$$\begin{aligned} \left\| {Var({{\hat{\beta }}} (u))} \right\| = {o_p}\left( \frac{1}{{nh}}\right) , \end{aligned}$$

(A.16)

where $V\mathrm{{a}}{\mathrm{{r}}_\chi }({{\hat{\beta }}} (u)) = {E_\chi }(({{\hat{\beta }}} (u) - {E_\chi }{{\hat{\beta }}} (u)){({E_\chi } - {E_\chi }{{\hat{\beta }}} (u))^T})$.

By (2.9),

$$\begin{aligned} V\mathrm{{a}}{\mathrm{{r}}_\chi }({{\hat{\beta }}} (u)) = {\pi ^T}(u) \cdot {({D^T}D)^{ - 1}} \cdot \pi (u){\sigma ^2} = {\displaystyle \Sigma _\beta }(u){\sigma ^2}, \end{aligned}$$

where ${\Sigma _\beta }(u) = {\pi ^T }(u) \cdot {({D^T }D)^{ - 1}}\pi (u)$.

Let $c\in R^{p}$ be a constant vector. By Lemma 3,

$$\begin{aligned} \begin{aligned} {c^T}{\Sigma _\beta }(u)c&= tr({c^T}{\pi ^T}(u) \cdot {({D^T}D)^{ - 1}} \cdot \pi (u)c) =tr({({D^T}D)^{ - 1}} \cdot \pi (u)c{c^T}{\pi ^T}(u))\\&\le \lambda _{\max }^{{{({D^T}D)}^{ - 1}}}tr(\pi (u) \cdot c{c^T}{\pi ^T}(u)) =\lambda _{\max }^{{{({D^T}D)}^{ - 1}}}tr({c^T}\pi (u) \cdot \pi (u) \cdot c)\\&=\lambda _{\max }^{{{({D^T}D)}^{ - 1}}}tr({c^T}({\pi ^T}(u)\pi (u)) \cdot c) = \lambda _{\max }^{{{({D^T}D)}^{ - 1}}}{\left\| c \right\| ^2}{\left\| {\pi (u)} \right\| ^2}, \end{aligned}\nonumber \\ \end{aligned}$$

(A.17)

similarly,

$$\begin{aligned} {c^T}{\Sigma _\beta }(u)c \ge \lambda _{\min }^{{{({D^T}D)}^{ - 1}}}{\left\| c \right\| ^2}{\left\| {\pi (u)} \right\| ^2}. \end{aligned}$$

(A.18)

Let $u \in \left( {{{\pi '}_{{i_u}}},{{\pi '}_{{i_u} + 1}}} \right]$. By (2.5),

$$\begin{aligned}&\sum \limits _{l = 1}^{{N_j}} {\pi _{jl}^2(u)} = \sum \limits _{i = {i_u} - m}^{{i_u}} {\pi _{jl}^2(u)} \ge \frac{1}{{m + 1}}{\left( \sum \limits _{i = {i_u} - m}^{{i_u}} {{\pi _{jl}}(u)} \right) ^2} = \frac{1}{{m + 1}}, \end{aligned}$$

(A.19)

$$\begin{aligned}&\sum \limits _{j = 1}^{{N_j}} {\pi _{jl}^2(u)} \le \left( \sum \limits _{l = 1}^{{N_j}} {\pi _{jl}}(u)\right) ^{2} = 1, \end{aligned}$$

(A.20)

by (A.5) and (A.17)–(A.20),

$$\begin{aligned} \frac{\mathrm{{1}}}{{\mathrm{{m}} + \mathrm{{1}}}}\left( \frac{1}{{{c_2}}} + {o_p}(1)\right) {(nh)^{ - 1}}{\left\| c \right\| ^2} \le {c^T}{\Sigma _\beta }(u)c \le \left( \frac{1}{{{c_1}}} + {o_p}(1)\right) {(nh)^{ - 1}}{\left\| c \right\| ^2}.\nonumber \\ \end{aligned}$$

(A.21)

On the other hand, we have

$$\begin{aligned} \begin{aligned} {E_\chi }{\left\| {{{\hat{\beta }}} (u) - {E_\chi }{{\hat{\beta }}} (u)} \right\| ^2}&= tr({E_\chi }{({{\hat{\beta }}} (u) - {E_\chi }{{\hat{\beta }}} (u))^T} ({{\hat{\beta }}} (u) - {E_\chi }{{\hat{\beta }}} (u)))\\&=tr(Var({{\hat{\beta }}} (u)))= {O_p}\left( \frac{1}{{nh}}\right) . \end{aligned} \end{aligned}$$

Hence (A.16) holds.

By (A.15) and (A.16),

$$\begin{aligned} \begin{aligned} \left\| {{{\hat{\beta }}} (u) - \beta (u)} \right\|&= \left\| {{{\hat{\beta }}} (u) - {E_\chi }{{\hat{\beta }}} (u) + {E_\chi } {{\hat{\beta }}} (u) - \beta (u)} \right\| \\&\le \left\| {{{\hat{\beta }}} (u) - {E_\chi }{{\hat{\beta }}} (u)} \right\| + \left\| {{E_\chi }{{\hat{\beta }}} (u) - \beta (u)} \right\| \\&= {O_p}\left( \frac{1}{{\sqrt{nh} }} + {h^m}\right) ={O_p} \left( {n^{ - \frac{m}{{2m + 1}}}}\right) . \end{aligned} \end{aligned}$$

$\square$

Proof of Corollary 2

The Corollary 2 follows by the proof of Theorem 1. $\square$

Proof of Theorem 3

Let $c\in R^{p}$ be a constant vector. We have

$$\begin{aligned} \frac{{{c^T}({{\hat{\beta }}} (u) - \beta (u))}}{{\sqrt{{c^T}\sum \nolimits _\beta {(u)} } c}} = \frac{{{c^T}({{\hat{\beta }}} (u) - {E_\chi }{{\hat{\beta }}} (u) + {E_\chi }\beta (u) - \beta (u))}}{{\sqrt{{c^T}\sum \nolimits _\beta {(u)} } c}}, \end{aligned}$$

by (A.15), (A.21) and ${k_j} = O({n^{\frac{1}{{2m + 1}}}}),1 \le j \le p,$

$$\begin{aligned} \frac{{{c^T}({E_\chi }{{\hat{\beta }}} (u) - \beta (u))}}{{\sqrt{{c^T}\sum \nolimits _\beta {(u)} } c}} = \frac{{{O_p}({h^m})}}{{{O_p}({{(nh)}^{ - \frac{1}{2}}})}} = {o_p}(1), \end{aligned}$$

(A.22)

then, it suffices to show that

$$\begin{aligned} \frac{{{c^T}({{\hat{\beta }}} (u) - {E_\chi }\beta (u))}}{{\sqrt{{c^T}\sum \nolimits _\beta {(u)} } c}}\longrightarrow N(0,{\sigma ^2}). \end{aligned}$$

(A.23)

Noting that

$$\begin{aligned} {{\hat{\beta }}} (u) - {E_\chi }({{\hat{\beta }}} (u)) = {\pi ^T}(u) \cdot {({D^T}D)^{ - 1}}{D^T}\varepsilon = A\varepsilon , \end{aligned}$$

where $\varepsilon = {({\varepsilon _1},{\varepsilon _2}, \ldots ,{\varepsilon _n})^T }$, $A = {\pi ^T}(u) \cdot {({D^T}D)^{ - 1}}{D^T} = ({A_1},{A_2}, \ldots ,{A_n})$,

$$\begin{aligned} {A_i}= & {} {\pi ^T}(u) \cdot {({D^T}D)^{ - 1}}\pi ({u_i}) \cdot {X_i}, \\ {c^T}({{\hat{\beta }}} (u) - {E_\chi }{{\hat{\beta }}} (u))= & {} {c^T}A\varepsilon = \sum \limits _{i = 1}^n {{c^T}{A_i}{\varepsilon _i} = \sum \limits _{i = 1}^n {b{}_i{\varepsilon _i}} }, \end{aligned}$$

where ${b_i} = {c^\tau }{A_i}$. To check the required Lindeberg–Feller condition, it suffices to verify

$$\begin{aligned} \mathop {\max b_i^2}\limits _{1 \le i \le n} = o\left( \sum \limits _{i = 1}^n {b_i^2}\right) . \end{aligned}$$

By (A.21), $\sum \nolimits _{i = 1}^n {b_i^2} = {c^T}A{A^T}c = {c^T}{\pi ^T}(u){({D^T}D)^{ - 1}}\pi (u)c = c\sum \nolimits _\beta {(u)} c = O\left( \frac{1}{{nh}}\right) .$

On the other hand, By Lemma 3 and $\left| {{x_i}} \right| < {M_4}$,

$$\begin{aligned} \begin{aligned} b_i^2&= {c^T}{A_i}A_i^Tc = tr({c^T}{A_i}A_i^Tc)\\&\le {\left\| c \right\| ^2}tr({\pi ^T}(u) \cdot {({D^T}D)^{ - 1}}\pi ({u_i}) \cdot {X_i}X_i^T \cdot {\pi ^T}({u_i}) \cdot {({D^T}D)^{ - 1}}\pi (u))\\&\le {\left\| c \right\| ^2}{(\lambda _{\max }^{{{({D^T}D)}^{ - 1}}})^2}tr(\pi (u) {\pi ^T}(u)\pi ({u_i}) \cdot {X_i}X_i^T \cdot {\pi ^T}({u_i}))\\&\le {\left\| c \right\| ^2}{(\lambda _{\max }^{{{({D^T}D)}^{ - 1}}})^2}{p^2}tr({X_i}X_i^T) = {\left\| c \right\| ^2}{(\lambda _{\max }^{{{({D^T}D)}^{ - 1}}})^2}{p^2}(x_{i1}^2 + x_{i2}^2 + \cdots + x_{ip}^2)\\&= {p^2}{\left\| c \right\| ^2}{\left( \frac{1}{{nh}}\right) ^2}{\left\| {{X_i}} \right\| ^2}= O{\left( \frac{1}{{nh}}\right) ^2} \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} \frac{{\mathop {\max }\limits _i b_i^2}}{{\sum \nolimits _{i = 1}^n {b_i^2} }} = \frac{{O{{\left( \frac{1}{{nh}}\right) }^2}}}{{O\left( \frac{1}{{nh}}\right) }} = O\left( \frac{1}{{nh}}\right) = o(1). \end{aligned}$$

(A.23) holds. The proof is completed. $\square$