0 Introduction

In this paper, we consider the following semiparametric regression model

Yi=Xi'β+g(Ti)+εi，

(1)

where {(X_i, Y_i, T_i), 1≤i≤n} are independent identically distributed sample, X_i=(X_i1, …, X_ip)', T_i∈[0, 1], 1≤i≤n, β is p×1 vector, g(t) is a unknown and smooth function on the ^{[0, 1]}. The {ε_i, 1≤i≤n} are independent identically distributed random variables with Eε_i=0, Eε₁²=σ₀²>0, where Eε_i and Eε_i² denote the expectation value of ε_i and ε_i², respectively.And it is independent on the {(X_i, T_i), 1≤i≤n}. The model (1) belongs to a sort of the semiparametric regression model and has extensive application in many practical problems. Recently, the model(1) has been discussed by many authors with complete data^{[1, 2, 3, 4]}. However, it is not surprising that little attention has been paid to model (1) with missing data. Limited studies in this direction have been discussed by reference ^[5], generalized the local linear estimation of ^[6], to the case with covariate missing at random.Reference ^[7] developed estimation theory for semiparametric regression analysis in the presence of missing response. Refercence^[8] discussed the generalized partially linear models with missing covariates.Reference^[9] discussed the partially linear models with missing responses at random. More references see ^{[10, 11, 12, 13]}. This paper focuses on establishing asymptotic normality and consistency of the two stage estimator in semiparametric regression model with missing response data.

In the semiparametric regression analysis setting up, the basic inference begins by considering the random sample

(Xi, Yi, Ti, δi),

(2)

for i=1, 2, …, n. Here all the X_i^' and T_i areobserved.If the response variable Y_i with companion X_i is observed, its associated indicator δ_i is set to be 1;otherwise, Y_i is missing and δ_i=0, for each i=1, 2, …, n.

By a purely semiparametric approach to discussing the missing data (2), the MAR assumption would require that there exists a chance mechanism denoted by p(X_i, T_i), such that

P(δ=1|Xi, Yi, Ti)=P(δ=1|Xi, Ti)=p(Xi, Ti)

(3)

holds almost surely. In practice, (3) is a common assumption for statistical analysis with missing data and is reasonable in many practical applications, see reference^[14].

1 The two stage estimator

In this section we define the estimators that we will analyze in this paper. We describe how to estimate the regression function.

Let α=Eg(T_i), e_i=g(T_i)-α+ε_i, i=1, …, n, the model (1) turn into following

Yi=α+XI'β+ei.

(4)

Where the e₁, …, e_n are independent identically distributed random variables with Ee₁=0 and 0<σ²=Ee₁²=Eε₁²+Var(g(T_i))=σ₀²+σ₁²<∞. The model (4) can be changed into the following form=

Yn=(1n, Xn)α β+en=1nα+Xnβ+en.

(5)

Where X_n=(X₁, …, X_n)', Y_n=(Y₁, …, Y_n)' and e_n=(e₁, …, e_n)', 1_n=(1, …, 1)'. Set S_n=X_n^'Q_nX_n, P_n=Q_nX_nS_n^-1X_n^'Q_n^' and d_n=1_n^'(Q_n-P_n)1_n=$\sum\limits_{i=1}^{n}{{}}$δ_i-1_n^'P_n1_n, where Q_n=Diag(δ₁, …, δ_n).

In order to obtain the solution of the following least squares problem (5), we have to find α and β to minimize

Wn=(Yn-Xnβ-1nα)'Qn(Yn-Xnβ-1nβ).

By optimization theory, we have that

$\left\{ \begin{align} & X_{n}^{'}{{Q}_{n}}{{X}_{n}}\beta +X_{n}^{'}{{Q}_{n}}{{1}_{n}}\alpha =X_{n}^{'}{{Q}_{n}}{{Y}_{n}}, \\ & 1_{n}^{'}{{Q}_{n}}{{1}_{n}}\alpha +1_{n}^{'}{{Q}_{n}}{{X}_{n}}\beta =1_{n}^{'}{{Q}_{n}}{{Y}_{n}}, \\ \end{align} \right.$

and thus

$\left\{ \begin{align} & \beta _{n}^{*}={{\left( X_{n}^{'}{{Q}_{n}}{{X}_{n}} \right)}^{-1}}X_{n}^{'}{{Q}_{n}}\left( {{Y}_{n}}-{{1}_{n}}\alpha _{n}^{*} \right), \\ & \alpha _{n}^{*}={{\left( 1_{n}^{'}{{Q}_{n}}{{1}_{n}} \right)}^{-1}}1_{n}^{'}{{Q}_{n}}\left( {{Y}_{n}}{{X}_{n}}\beta _{n}^{*} \right). \\ \end{align} \right.$

(7)

Joining the first equality into the second equality of (7), we obtain α_n^*=d_n^-1[1_n^'(Q_n-P_n)Y_n], and consequently

$\left\{ \begin{align} & \alpha _{n}^{*}=d_{n}^{-1}\left[ 1_{n}^{'}\left( {{Q}_{n}}-{{P}_{n}} \right){{Y}_{n}} \right], \\ & \beta _{n}^{*}=S_{n}^{-1}X_{n}^{'}{{Q}_{n}}{{Y}_{n}}-S_{n}^{-1}X_{n}^{'}{{Q}_{n}}{{1}_{n}}\alpha _{n}^{*}. \\ \end{align} \right.$

As a result, the first stage estimator β_n^* of β is obtained.Substituting β_n^* for β in the model(1), we have

Xi=Xi'βn*+g(Ti)+εi.

(8)

Now, we defined the nonparametric estimator of g(t) that

gn*(t)=$\sum\limits_{j=1}^{n}{{}}$ Wnj(t)(Yj-Xj'βn*)δj,

(9)

where W_nj=$\left[ K\frac{{{T}_{j}}-t}{{{h}_{n}}} \right]/\left[ \sum\limits_{r=1}^{n}{{}}K\left( \frac{{{T}_{r}}-t}{{{h}_{n}}} \right){{\delta }_{r}} \right]$, K(·) is a kernel function and h_n>0 is the bandwidth.Substituting g_n^* in the model (1), we have

Yi=Xi'β+gn*(Ti)+εi.

(10)

Using the generalized least squares for the model (10), we can find β to minimize

$\sum\limits_{i=1}^{n}{{}}$[Yi-Xi'β-gn*(Ti)]2,

(11)

and obtain estimator of β that

${\hat{\beta }}$n=Sn-1Xn'Qn(Yn-gn*(T)),

where g_n^*(T)=[g_n^*(T₁), g_n^*(T₂), …, g_n^*(T_n)]^T. The estimator of g(T) is as follows:

${\hat{g}}$n(t)=$\sum\limits_{j=1}^{n}{{}}$Wnj(t)(Yj-Xj'${\hat{\beta }}$n)δj.

So we now can obtain the estimation of θ=E(Y). The regression imputation estimator of θ can be denoted by

${\hat{\theta }}$I=$\frac{1}{n}$$\sum\limits_{i=1}^{n}{{}}${δiYi+(1-δi)(XiT${\hat{\beta }}$n+${\hat{g}}$n(Ti))｝.

(12)

Thus, we have the propensity score weighted estimator

${\hat{\theta }}$w=$\frac{1}{n}$$\sum\limits_{i=1}^{n}{{}}$$\left\{ \frac{{{\delta }_{i}}{{Y}_{i}}}{\hat{P}\left( {{X}_{i}}, {{T}_{i}} \right)}+\left( 1-\frac{{{\delta }_{i}}}{\hat{P}\left( {{X}_{i}}, {{T}_{i}} \right)} \right)X_{i}^{T}{{{\hat{\beta }}}_{n}}+{{{\hat{g}}}_{n}}\left( {{T}_{i}} \right) \right\}, $

(13)

where ${\hat{P}}$(x, t) is a high-dimensional kernel estimator of the propensity score defined by

$\hat{P}\left( x, t \right)=\left[ \sum\limits_{j=1}^{n}{{}}{{\delta }_{j}}W\left( \frac{x-{{X}_{j}}}{{{h}_{n}}}, \frac{t-{{T}_{j}}}{{{h}_{n}}} \right) \right]/\sum\limits_{j=1}^{n}{{}}W\left( \frac{x-{{X}_{j}}}{{{h}_{n}}}, \frac{t-{{T}_{j}}}{{{h}_{n}}} \right), $

(14)

with W(·, ·) is the weighting function and h_n is the bandwidth sequence.

2 The asymptotic properties and consistencies

We explore the asymptotic distribution and consistency of the all estimators. The following notation and assumptions are needed.

(ⅰ) The T₁, T₂…T_n are independent identically distributed random variables and the {T_i} is independent of the {e_i}.

(ⅱ) The rank(X_n)=p<n.

(ⅲ) $\underset{n>p}{\mathop{\sup }}\,\frac{1_{n}^{'}{{P}_{n}}{{1}_{n}}}{n}$<1.

(ⅳ) E[g(T₁)]²<∞.

(ⅴ) Existence 0<C₁≤C₂<∞ and have C₁I(‖u‖<2)≤K(u)≤C₂ I(‖u‖<2), u∈[0, 1].

(ⅵ) The probability density function of T_i is r(t) and

0<$\mathop {\inf }\limits_{0 \le t \le 1} $r(t)≤$\mathop {\sup }\limits_{0 \le t \le 1} $r(t)<∞.

(1)

In what follows the main results will be established for the asymptotic distribution and consistency of the semiparametric regression model.

Theorem 1  Under conditions (ⅰ)~(ⅴ), we have that

(1) α_n^*→α, a.s.

(2) β_n^*→β, a.s. if and only if s_n^-1→0.

(3) β_n^*$\xrightarrow{P}$β⇔β_n^*$\xrightarrow{{{L}_{r}}}$β(0<r≤2)⇔β_n^*→β, a.s.

Theorem 2  Under conditions (ⅰ)~(ⅴ), if $\underset{n\to \infty }{\mathop{\lim }}\, \underset{1\le k\le n}{\mathop{\max }}\, $X_k^'S_n^-1X_k=0 and $\underset{n\to \infty }{\mathop{\lim }}\, $nS_n^-1=p(x, t)Σ is symmetric and positive definite, where p(x, t)>0 for any x>0 and t>0, and Σ is a symmetric and positive definite matrix,

(1) then $\sqrt{n}$(β_n^*-β)$\xrightarrow{L}$N(0, σ²p(x, t)Σ),

(2) further, if the condition (ⅲ) is replaced by $\underset{n>p}{\mathop{\sup }}\, $1_n^'P_n 1_n=O(1), and g∈R is bounded when S_n^-1→0 and nS_n^-1→p(x, t)Σ, then

(ⅰ)${\hat{\beta }}$ _n→β, a.s.

(ⅱ) $\sqrt{n}$(_n-β)$\xrightarrow{L}$N(0, σ₀²p(x, t)Σ).

Theorem 3  Under conditions (ⅰ)~(ⅵ), suppose that T₁, T₂… T_n in the conditions (ⅰ) are independent identically distributed random variables and the probability density function r(t) is unknown substitute the condition of (ⅰ). In addition, inf(n^1-αh_n)>0, α∈(1/2, 1) and $\underset{n\to \infty }{\mathop{\lim }}\, \underset{1\le k\le n}{\mathop{\max }}\, $X_k^'S_n^-1X_k=0 when h_n→0 and [$\sqrt{n}$h_n]/[logn]→∞,

(1) then g_n^*(t)-g(t)→0, a.s. ∀t∈C_r∧{t, r(t)>0},

(2) further if g∈R is bounded when S_n^-1→0, then ${\hat{g}}$_n(t)-g(t)→0, a.s.

Theorem 4  Under conditions (i)~(vi), we have

$\sqrt{n}$(${\hat{\theta }}$-θ)$\xrightarrow{L}$N(0, σ02p(x, t)Σ).

(15)

3 Sketches of the proofs

In this section, we will give the proof of Theorem 1~3.The following lemmas are needed for our technical proofs.

Lemma 1  If h_n→0, nh_n^d/(n^1-1/rlogn)→∞, the kernel estimator of nonparametric regression under response missing is

$\hat{g}\left( x \right)=\sum\limits_{i=1}^{n}{{}}K\frac{{{X}_{i}}-x}{{{h}_{n}}}{{Y}_{i}}{{\delta }_{i}}/\sum\limits_{j=1}^{n}{{}}K\left( \frac{{{X}_{j}}-x}{{{h}_{n}}} \right){{\delta }_{j}}=\sum\limits_{i=1}^{n}{{}}W_{ni}^{'}\left( x \right){{Y}_{i}}, $

we have $\underset{n\to \infty }{\mathop{\lim }}\, \hat{g}$(x)=g(x), a.s.

Proof  Similar to the theorem in reference^[15].

Proof the Theorem 1  Similar to the Lemma 2.1 in reference ^[16].

Proof the Theorem 2

(1) $\sqrt{n}$(β_n^*-β)=$\sqrt{n}$[(S_n^-1X_n^'Q_nY_n-S_n^-1X_n^'Q_n1_nd_n^-11_n^'(Q_n-P_n)Y_n)-β]=
    $\sqrt{n}$[(S_n^-1X_n^'Q_n(1_nα+X_nβ+e_n)-
    S_n^-1X_n^'Q_n1_nd_n^-11_n^'(Q_n-P_n)(1_nα+X_nβ+e_n)-β]=
    $\sqrt{n}$[S_n^-1X_n^'Q_ne_n-S_n^-1X_n^'Q_n1_nd_n^-11_n^'(Q_n-P_n)X_nβ-
    S_n^-1X_n^'Q_n1_nd_n^-11_n^'(Q_n-P_n)e_n]=
    $\sqrt{n}$[S_n^-1X_n^'Q_n e_n-S_n^-1X_n^'Q_n1_nd_n^-11_n^'(Q_n-P_n)e_n]=
    $\sqrt{n}$S_n^-1X_n^'Q_n e_n-$\sqrt{n}$S_n^-1X_n^'Q_n1_n1_n^'(Q_n-P_n)e_n d_n^-1$\triangleq $J₁-J₂.

It is easy to prove J₂$\xrightarrow{P}$0 and J₁=$\sqrt{n}$S_n^-1X_n^'Q_ne_n. Thus, EJ₁=0, VarJ₁=Var($\sqrt{n}$S_n^-1X_n^'Q_nε_n)=σ²p(x, t)Σ.By Linderberg Theorem, we have

J1$\xrightarrow{L}$N(0, σ2p(x, t)Σ).

(2) Firstly, we prove the conclusion (ⅰ) of (2).

${\hat{\beta }}$n-β=Sn-1Xn'Qn(Yn-gn*(T))-β=    Sn-1Xn'Qn(Xnβ+g(T)+εn-gn*(T))-β=    Sn-1Xn'Qnεn-Sn-1Xn'Qn(gn*(T)-g(T)).

By Lemma 1 in reference^[17], in order to obtain the proof of (i), we only prove

Sn-1Xn'Qn(gn*(T)-g(T))→0, a.s.

By $\underset{n\to \infty }{\mathop{\lim }}\, $nS_n^-1=p(x, t)Σ>0, it is known that there is c₃>c₄>0 and c₃≥|S_n|/n≥c₄ such that

‖Sn-1Xn'Qn‖≤‖Sn-1‖‖Xn'‖≤‖Sn-1‖|$\sum\limits_{i, j}^{{}}{{}}$|Xij||≤${n \over {X_{nn}^2}}\sqrt {\sum\limits_{i,j}^n {} X_{ij}^2/n} \le \sqrt {{n \over {\left| {{S_n}} \right|}}} < c.$

According to the first conclusion of Theorem 3, g_n^*(T)-g(T)→0, a.s.Thus,

Sn-1Xn'Qn(gn*(T)-g(T))→0, a.s.

Now we prove conclusion (ⅱ) of (2).

$\sqrt{n}$(${\hat{\beta }}$n-β)=$\sqrt{n}$[Sn-1Xn'Qn(Yn-gn*(T))-β]=   $\sqrt{n}$Sn-1Xn'Qnεn-$\sqrt{n}$Sn-1Xn'Qn(gn*(T)-g(T)).

Because |X_n^'Q_n/n|≤$\sqrt{\left\| {{S}_{n}} \right\|/n}$≤c, nS_n^-1→ p(x, t)Σ is symmetric and positive definite, and g_n^*(T)→g(T), a.s.

$\sqrt{n}$Sn-1Xn'Qn(gn*(T)-g(T))=nSn-1Xn'Qn$\frac{1}{\sqrt{n}}$(gn*(T)-g(T))→0, a.s.

It is known that $\sqrt{n}$S_n^-1X_n^'Q_ne_n is i.i.d. random variable,

E($\sqrt{n}$Sn-1Xn'Qnen)=0.

and

Var$\sqrt{n}$Sn-1Xn'Qnεn=p(x, t)σ02Σ.

It follows from Linderberg theorem that

$\sqrt{n}$Sn-1Xn'Qnεn$\xrightarrow{L}$N(0, p(x, t)σ02Σ),

namely, $\sqrt{n}$(${\hat{\beta }}$_n-β)$\xrightarrow{L}$N(0, p(x, t)σ₀²Σ).This completes the proof of Theorem 2.

Theorem 3

(1) Let W_n(t)=(W_n1(t), …, W_nn(t))'=, where t_i∈C_f∧{T_i, f(t_i)>0}.Since

gn*(t)=$\sum\limits_{j=1}^{n}{{}}$Wnj(t)(Yj-Xj'βn*)δj=   $\sum\limits_{j=1}^{n}{{}}$Wnj(t)(Yj-Xj'β+Xj'β-Xj'βn*)δj=   Wn'(t)Qn(Yn-Xnβ)-Wn'(t)QnXn(βn*-β)$\triangleq $   J1-J2, J1=Wn'(t)Qn(Yn-Xnβ)=$\sum\limits_{j=1}^{n}{{}}$Wnj(t)(Yj-Xj'β)δj=   $\sum\limits_{j=1}^{n}{{}}$Wnj'(t)(g(tj)+εj)=$\sum\limits_{j=1}^{n}{{}}$Wnj'(t)kj,

where {k_j} is i.i.d. and 0<Ek₁²<∞, E(k₁|t₁=t)=g(t). By the Lemma 1, J₁→g(t), a.s.Thus, we only prove J₂→0, a.s.

J2=Wn'(t)QnXn(βn*-β)=   Wn'(t)QnXn[(Sn-1Xn'QnYn-Sn-1Xn'Qn1nαn*)-β]=   Wn'(t)Pnen-Wn'(t)Pn1n(αn*-α).

Let

bnk=$\sum\limits_{j=1}^{n}{{}}$Wnj(t)ajk(n)=$\sum\limits_{j=1}^{n}{{}}$K($\frac{{{t}_{j}}-t}{{{h}_{n}}}$)ajk(n)/$\sum\limits_{i=1}^{n}{{}}$K($\frac{{{t}_{j}}-t}{{{h}_{n}}}$)δi,

where P_n=(a_ij⁽ⁿ⁾).Now we prove W_n^'(t)P_ne_n→0, a.s.It is true that

$W_{n}^{'}\left( t \right){{P}_{n}}{{e}_{n}}=\sum\limits_{i=1}^{n}{{}}{{b}_{ni}}{{\varepsilon }_{i}}={{\left( {{f}_{n}}\left( t \right) \right)}^{-1}}\left\{ \sum\limits_{k=1}^{n}{{}}\left[ \sum\limits_{j=1}^{n}{{}}K\left( \frac{{{t}_{j}}-t}{{{h}_{n}}} \right)a_{jk}^{\left( n \right)} \right]{{\varepsilon }_{k}} \right\}\triangleq ={{\left( {{f}_{n}}\left( t \right) \right)}^{-1}}{{{\bar{u}}}_{n}}\left( t \right).$

It follows from Lemma 3 in reference^[9] that u_n(t)→0, a.s. Following Lemma 4 in reference ^[17], it is easy to know that f_n(t)→p(x, t)f(t), a.s. Thus,

Wn'(t)Pnen→0, a.s.

(16)

Within W_n^'(t)P_n1_n(α_n^*-α), let u_ik⁽ⁿ⁾=$\frac{{{A}_{ni}}\left( {{\delta }_{k}}-{{A}_{nk}} \right)}{{{d}_{n}}}$, where A_ni=$\sum\limits_{j=1}^{n}{{}}$a_ij⁽ⁿ⁾, P_n=(a_ij⁽ⁿ⁾). Then

Wn'(t)Pn1n(αn*-α)=Wn'(t)Pn1ndn-1[1n'(Qn-Pn)en]=    (fn(t))-1$\sum\limits_{k=1}^{n}{{}}$$\sum\limits_{j=1}^{n}{{}}$K$\left( \frac{{{t}_{j}}-t}{{{h}_{n}}} \right)$ujk(n)εknhn$\triangleq $    (fn(t))-1Vn(t).

By the condition (ⅱ), 0<d_n=1_n^'(Q_n-P_n)1_n=$\sum\limits_{k=1}^{n}{{}}$(δ_k-A_nk)=$\sum\limits_{k=1}^{n}{{}}$(δ_k-A_nk)².Since

$\sum\limits_{k=1}^{n}{{}}$Ank2=1n'Pn1n=$\sum\limits_{k=1}^{n}{{}}$Ank, $\sum\limits_{j, k=1}^{n}{{}}$(ujk(n))2=dn-2$\sum\limits_{j=1}^{n}{{}}$$\sum\limits_{k=1}^{n}{{}}$Anj2(δk-Ank)2= dn-1$\sum\limits_{j=1}^{n}{{}}$Anj2=dn-1(1n'Pn1n)=1n'Pn1n$\sum\limits_{j=1}^{n}{{}}$δj-1n'Pn1n= $\frac{1}{\sum\limits_{j=1}^{n}{{}}{{\delta }_{j}}/1_{n}^{'}{{P}_{n}}{{1}_{n}}-1}$≤c,

when n>p.The Cauchy-Schwarz inequality yields

$\sum\limits_{k=1}^{n}{{}}$$\sum\limits_{r=1}^{n}{{}}$ukk(n)urr(n)=dn-2$\sum\limits_{k=1}^{n}{{}}$$\sum\limits_{r=1}^{n}{{}}$Ank(1-Ank)Anr(1-Anr)≤<    dn-2[$\sum\limits_{k=1}^{n}{{}}$$\sum\limits_{r=1}^{n}{{}}$Ank2(1-Anr)2]1/2[$\sum\limits_{k=1}^{n}{{}}$$\sum\limits_{r=1}^{n}{{}}$Anr2(1-Ank)2]1/2≤c,

when n≥p.Therefore,

$\sum\limits_{k=1}^{n}{{}}$($\sum\limits_{j=1}^{n}{{}}$ujk(n))2≤n$\sum\limits_{k=1}^{n}{{}}$$\sum\limits_{j=1}^{n}{{}}$ (ujk(n))2≤cn.

Since the u_ik⁽ⁿ⁾ in the u_n(t) is the same function as the a_jk⁽ⁿ⁾ in the ${{{\bar{u}}}_{n}}$(t), ν_n(t)→0, a.s.can be obtained. According to the proof above and Lemma 4 in reference ^[17], we have

Wn'(t)Pn1n(αn*-α)→0, a.s.

(17)

(16) and (17) show J₂→0, a.s. This completes the proof (1) of Theorem 3.

(2) It is not difficult to obtain

${\hat{g}}$n(t)=$\sum\limits_{j=1}^{n}{{}}$Wnj(t)(Yj-Xj')δj= Wn'(t)Qn(Yn-Xnβ)-Wn'QnXn(${\hat{\beta }}$n-β), ${\hat{\beta }}$n-β=Sn-1Xn'Qn(Yn-gn*(T))-β= Sn-1Xn'Qnen-Sn-1Xn'Qn(gn*(T)-g(T)),

and

${\hat{g}}$n(t)=Wn'Qn(Yn-Xnβ)-Wn'(t)QnXnSn'Xn'Qnεn+   Wn'(t)QnXnSn-1Xn'Qn(gn*(T)-g(T))=   Wn'(t)Qn(Yn-Xnβ)-Wn'(t)Pnen+   Wn'(t)Pn(gn*(T)-g(T))=   I1-I2+I3.

(18)

It has been proved that

I1=Wn'(t)Qn(Yn-Xnβ)→ g(t), a.s.

and

Wn'(t)Pnen=$\sum\limits_{i=1}^{n}{{}}$bnjεi=    (fn(t))-1$\sum\limits_{k=1}^{n}{{}}$$\frac{1}{n{{h}_{n}}}$$\sum\limits_{j=1}^{n}{{}}$K$\left( \frac{{{t}_{j}}-t}{{{h}_{n}}} \right)$ajk(n)εk$\triangleq $    (fn(t))-1un'(t).

(19)

Using the same method of the proof for (1) of Theorem 3, it follows from Lemma 3 in reference ^[17] that

Wn'(t)Pnen→0, a.s.

(20)

By the conditions (Ⅴ) we know that

$\begin{align} & W_{n}^{'}\left( t \right){{P}_{n}}=\sum\limits_{j, k}^{{}}{{}}K\left( \frac{{{t}_{j}}-t}{{{h}_{n}}} \right)a_{jk}^{\left( n \right)}/\sum\limits_{i=1}^{n}{{}}\left( \frac{{{t}_{i}}-t}{{{h}_{n}}} \right){{\delta }_{i}}\le \\ & c\sum\limits_{j, k}^{{}}{{}}\left| a_{jk}^{\left( n \right)} \right|/\sum\limits_{i=1}^{n}{{}}K\left( \frac{{{t}_{i}}-t}{{{h}_{n}}} \right){{\delta }_{i}}\le \\ & c\sum\limits_{j, k}^{{}}{{}}\left| a_{jk}^{\left( n \right)} \right|n\le c\sum\limits_{k=1}^{{}}{{}}\frac{\left| {{A}_{nk}} \right|}{n}, \\ \end{align}$

and

1n'Pn1n=$\sum\limits_{j, k}^{{}}{{}}$ajk(n)=$\sum\limits_{k=1}^{n}{{}}$Ank=$\sum\limits_{k=1}^{n}{{}}$Ank2.

According to the conditions (ⅲ), it holds that $\sum\limits_{k=1}^{n}{{}}$($\frac{{{A}_{nk}}}{n}$)²≤c. Thus, when n>p, $\sum\limits_{k=1}^{n}{{}}$|$\frac{{{A}_{nk}}}{n}$|≤c.Namely, W_n^'(t)P_n≤c. From conclusion (1) of Theorem 3, we have g_n^*(T)-g(T)→0, a.s. Thus,

Wn'(t)Pn(gn*(T)-g(T))→0, a.s.

(21)

(18), (20) and (21) imply ${\hat{g}}$_n(t)→ g(t), a.s. This completes the proof of Theorem 3.