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In this article, we focus on the semi-parametric error-in-variables model with missing responses:
,
where y_{i}
are the response variables missing at random,
are design points, ζ_{i}
are the potential variables observed with measurement errors μ_{i}
, the unknown slope parameter ß
and nonparametric component g(·)
need to be estimated. Here we choose two different approaches to estimate ß
and g(·)
. Under appropriate conditions, we study the strong consistency for the proposed estimators.

Consider the following semi-parametric error-in-variables (EV) model

{ y i = ξ i β + g ( t i ) + ϵ i , x i = ξ i + μ i , (1.1)

where y i are the response variables, ( ξ i , t i ) are design points, ξ i are the potential variables observed with measurement errors μ i , E μ i = 0 , ϵ i are random errors with E ϵ i = 0 . β ∈ R is an unknown parameter that needs to be estimated. g ( ⋅ ) is a unknown function defined on close interval [ 0,1 ] , h ( ⋅ ) is a known function defined on [ 0,1 ] satisfying

ξ i = h ( t i ) + v i , (1.2)

where v i are also design points.

Model (1.1) and its special forms have gained much attention in recent years. When μ i ≡ 0 , ξ i are observed exactly, the model (1.1) reduces to the general semi-parametric model, which was first introduced by Engle et al. [

On the other hand, we often encounter incomplete data in the practical application of the models. In particular, some response variables may be missing, by design or by happenstance. For example, the responses y i may be very expensive to measure and only part of y i are available. Actually, missing of responses is very common in opinion polls, social-economic investigations, market research surveys and so on. Therefore, we focus our attention on the case that missing data occur only in the response variables. When ξ i can fully be observed, the model (1.1) reduces to the usual semi-parametric model which has been studied by many scholars in the literature: See Wang et al. [

To deal with missing data, one method is to impute a plausible value for each missing datum and then analyze the results as if they are complete. In regression problems, common imputation approaches include linear regression imputation by Healy and Westmacott [

In this paper, suppose we obtain a random sample of incomplete data { ( y i , δ i , x i , t i ) } from the model (1.1), where δ i = 0 if y i is missing, otherwise δ i = 1 . Throughout this paper, we assume that y i is missing at random. The assumption implies that δ i and y i are independent. That is, P ( δ i = 1 | y i ) = P ( δ i = 1 ) . This assumption is a common assumption for statistical analysis with missing data and is reasonable in many practical situations.

The paper is organized as follows. In Section 2, we list some assumptions. The main results are given in Section 3. Some preliminary lemmas are stated in Section 4. Proofs of the main results are provided in Sections 5.

In this section, we list some assumptions which will be used in the main results. Here a n = O ( b n ) means | a n | ≤ C | b n | for every n ≥ 1 , a n = o ( b n ) means a n / b n → 0 as n → ∞ , while a.s. is stand for almost sure.

(A0) Let { ϵ i ,1 ≤ i ≤ n } , { μ i ,1 ≤ i ≤ n } and { δ i ,1 ≤ i ≤ n } be sequences of independent random variables satisfying

i) E ϵ i = 0 , E μ i = 0 , E ϵ i 2 = 1 , E μ i 2 = Ξ μ 2 > 0 is known.

ii) sup i E | ϵ i | p < ∞ , sup i E | μ i | p < ∞ for some p > 4 .

iii) { ϵ i ,1 ≤ i ≤ n } , { μ i ,1 ≤ i ≤ n } , { δ i ,1 ≤ i ≤ n } are independent of each other.

(A1) Let { v ,1 ≤ i ≤ n } in (2) be a sequence satisfying

i) lim n → ∞ n − 1 ∑ i = 1 n v i 2 = Σ 0 , lim n → ∞ n − 1 ∑ i = 1 n δ i v i 2 = Σ 1 a .s . ( 0 < Σ 0 , Σ 1 < ∞ ) .

ii) lim n → ∞ sup n ( n log n ) − 1 ⋅ max 1 ≤ m ≤ n | ∑ i = 1 m v j i | < ∞ , where { j 1 , j 2 , ⋯ , j n } is a permutation of ( 1,2, ⋯ , n ) .

iii) max 1 ≤ i ≤ n | v i | = O ( 1 ) .

(A2) g ( ⋅ ) and h ( ⋅ ) are continuous functions satisfying the first-order Lipschitz condition on the close interval [ 0,1 ] .

(A3) Let W n j c ( t i ) ( 1 ≤ i , j ≤ n ) be weight functions defined on [0, 1] and satisfy

i) max 1 ≤ j ≤ n ∑ i = 1 n δ j W n j c ( t i ) = O ( 1 ) a.s.

ii) max 1 ≤ i ≤ n ∑ j = 1 n δ j W n j c ( t i ) I ( | t i − t j | > a ⋅ n − 1 / 4 ) = o ( n − 1 / 4 ) a.s. for any a > 0 .

iii) max 1 ≤ i , j ≤ n W n j c ( t i ) = o ( n − 1 / 2 log − 1 n ) a.s.

(A4) The probability weight functions W n j ( t i ) ( 1 ≤ i , j ≤ n ) are defined on [ 0,1 ] and satisfy

i) max 1 ≤ j ≤ n ∑ i = 1 n W n j ( t i ) = O ( 1 ) .

ii) max 1 ≤ i ≤ n ∑ j = 1 n W n j ( t i ) I ( | t i − t j | > a ⋅ n − 1 / 4 ) = o ( n − 1 / 4 ) , for any a > 0 .

iii) max 1 ≤ i , j ≤ n W n j ( t i ) = o ( n − 1 / 2 log − 1 n ) .

Remark 2.1. Conditions (A0)-(A4) are standard regularity conditions and used commonly in the literature, see Härdle et al. [

For model (1.1), we want to seek the estimators of β and g ( ⋅ ) . The most natural idea is to delete all the missing data. Therefore, one can get model δ i y i = δ i ξ i β + δ i g ( t i ) + δ i ϵ i . If ξ i can be observed, we can apply the least squares estimation method to estimate the parameter β . If the parameter β is known, using the complete data { ( δ i y i , δ i x i , δ i t i ) ,1 ≤ i ≤ n } , we can define the estimator of g ( ⋅ ) to be

g n * ( t , β ) = ∑ j = 1 n W n j c ( t ) ( δ j y j − δ j x j β ) ,

where W n j c ( t ) are weight functions satisfying (A3). On the other hand, under the condition of the semi-parametric EV model, Liang et al. [

S S ( β ) = ∑ i = 1 n δ i { [ y i − x i β − g n * ( t i , β ) ] 2 − Ξ μ 2 β 2 } = min !

Therefore, we can achieve the modified LSE of β as follow:

β ^ c = [ ∑ i = 1 n ( δ i x ˜ i c 2 − δ i Ξ μ 2 ) ] − 1 ∑ i = 1 n δ i x ˜ i c y ˜ i c , (3.1)

where x ˜ i c = x i − ∑ j = 1 n δ j W n j c ( t i ) x j , y ˜ i c = y i − ∑ j = 1 n δ j W n j c ( t i ) y j . We substitute (3.1) into g n * ( t , β ) , then

g ^ n c ( t ) = ∑ j = 1 n δ j W n j c ( t ) ( y j − x j β ^ c ) . (3.2)

Apparently, the estimators β ^ c and g ^ n c ( t ) are formed without taking all sample information into consideration. Hence, in order to make up for the missing data, we imply an imputation method from Wang and Sun [

U i [ I ] = δ i y i + ( 1 − δ i ) [ x i β ^ c + g ^ n c ( t i ) ] . (3.3)

Therefore, Using complete data { ( U i [ I ] , x i , t i ) ,1 ≤ i ≤ n } , similar to (3.1)-(3.2), one can get another estimators for β and g ( ⋅ ) , that is

β ^ I = [ ∑ i = 1 n ( x ˜ i 2 − δ i Ξ μ 2 ) ] − 1 ∑ i = 1 n x ˜ i U ˜ i [ I ] , (3.4)

g ^ n [ I ] ( t ) = ∑ j = 1 n W n j ( t ) ( U j [ I ] − x j β ^ I ) . (3.5)

where U ˜ i [ I ] = U i [ I ] − ∑ j = 1 n W n j ( t i ) U j [ I ] , x ˜ i = x i − ∑ j = 1 n W n j ( t i ) x j , W n j ( t ) are weight functions satisfying (A4).

Based on the estimators for β and g ( ⋅ ) , we have the following results.

Theorem 3.1 Suppose that (A0)-(A3) are satisfied. For every t ∈ [ 0,1 ] , we have

a) β ^ c → β a . s .

b) g ^ n c ( t ) → g ( t ) a . s .

Theorem 3.2 Suppose that (A0)-(A4) are satisfied. For every t ∈ [ 0,1 ] , we have

a) β ^ I → β a . s .

b) g ^ n [ I ] ( t ) → g ( t ) a . s .

In the sequel, let C , C 1 , ⋯ be some finite positive constants, whose values are unimportant and may change. Now, we introduce several lemmas, which will be used in the proof of the main results.

Lemma 4.1 (Baek ang Liang [

∑ i = 1 n a n i e i = o ( n − 1 / α ) a . s .

Lemma 4.2 (Hardle et al. [

max 1 ≤ i ≤ n | ∑ k = 1 n a k i V k | = O ( n − s log n ) a . s . for s = ( p 1 − p 2 ) / 2 .

Lemma 4.3

a) Let A ˜ i = A ( t i ) − ∑ j = 1 n W n j ( t i ) A ( t j ) , where A ( ⋅ ) = g ( ⋅ ) or h ( ⋅ ) . Let A ˜ i c = A ( t i ) − ∑ j = 1 n δ j W n j c ( t i ) A ( t j ) , where A ( ⋅ ) = g ( ⋅ ) or h ( ⋅ ) . Then, (A0)-(A4) imply that max 1 ≤ i ≤ n | A ˜ i | = o ( n − 1 / 4 ) and max 1 ≤ i ≤ n | A ˜ i c | = o ( n − 1 / 4 ) a . s .

b) (A0)-(A4) imply that n − 1 ∑ i = 1 n ξ ˜ i 2 → Σ 0 , ∑ i = 1 n | ξ ˜ i | ≤ C 1 n , n − 1 ∑ i = 1 n δ i ( ξ ˜ i c ) 2 → Σ 1 a . s . and ∑ i = 1 n | δ i ξ ˜ i c | ≤ C 2 n a . s .

c) (A0)-(A4) imply that max 1 ≤ i ≤ n | ξ ˜ i | = O ( 1 ) and max 1 ≤ i ≤ n | ξ ˜ i c | = O ( 1 ) a . s .

Lemma 4.4 Suppose that (A0)-(A4) are satisfied. Then one can deduce that

max 1 ≤ i ≤ n | g ^ n c ( t i ) − g ( t i ) | = o ( n − 1 4 ) a . s .

One can easily get Lemma 4.3 by (A0)-(A4). The proof Lemma 4.4 is analogous to the proof of Theorem 3.1(b).

Firstly, we introduce some notations, which will be used in the proofs below.

ξ ˜ i c = ξ i − ∑ j = 1 n δ j W n j c ( t i ) ξ j , μ ˜ i c = μ i − ∑ j = 1 n δ j W n j c ( t i ) μ j ,

g ˜ i c = g ( t i ) − ∑ j = 1 n δ j W n j c ( t i ) g ( t j ) , ϵ ˜ i c = ϵ i − ∑ j = 1 n δ j W n j c ( t i ) ϵ j ,

ξ ˜ i = ξ i − ∑ j = 1 n W n j ( t i ) ξ j , μ ˜ i = μ i − ∑ j = 1 n W n j ( t i ) μ j ,

g ˜ i = g ( t i ) − ∑ j = 1 n W n j ( t i ) g ( t j ) , ϵ ˜ i = ϵ i − ∑ j = 1 n W n j ( t i ) ϵ j , η i = ϵ i − μ i β ,

B 1 n 2 = ∑ i = 1 n δ i ξ ˜ i 2 , S n 2 = ∑ i = 1 n ξ ˜ i 2 , S 1 n 2 = ∑ i = 1 n ( δ i x ˜ i 2 − δ i Ξ μ 2 ) , S 2 n 2 = ∑ i = 1 n ( x ˜ i 2 − Ξ μ 2 ) .

Proof of Theorem 3.1(a). From (3.1), one can write that

β ^ c − β = S 1 n − 2 [ ∑ i = 1 n δ i ( ξ ˜ i c + μ ˜ i c ) ( y ˜ i c − ξ ˜ i c β − μ ˜ i c β ) + ∑ i = 1 n δ i Ξ μ 2 β ] = S 1 n − 2 { ∑ i = 1 n [ δ i ( ξ ˜ i c + μ ˜ i c ) ( ϵ ˜ i c − μ ˜ i c β ) + δ i Ξ μ 2 β ] + ∑ i = 1 n δ i ξ ˜ i c g ˜ i c + ∑ i = 1 n δ i μ ˜ i c g ˜ i c }

= S 1 n − 2 { ∑ i = 1 n δ i ξ ˜ i c ( ϵ i − μ i β ) + ∑ i = 1 n δ i μ i ϵ i − ∑ i = 1 n δ i ( μ 2 − Ξ μ 2 ) β + ∑ i = 1 n δ i ξ ˜ i c g ˜ i c + ∑ i = 1 n δ i μ ˜ i c g ˜ i c − ∑ i = 1 n ∑ j = 1 n δ i δ j W n j c ( t i ) ξ ˜ i c ϵ j − ∑ i = 1 n ∑ j = 1 n δ i δ j W n j c ( t i ) ϵ i μ j

− ∑ i = 1 n ∑ j = 1 n δ i δ j W n j c ( t i ) μ i ϵ j + ∑ i = 1 n ∑ j = 1 n δ i δ j W n j c ( t i ) ξ ˜ i c μ j β + 2 ∑ i = 1 n ∑ j = 1 n δ i δ j W n j c ( t i ) μ i μ j β + ∑ i = 1 n ∑ j = 1 n ∑ k = 1 n δ i δ j δ k W n j c ( t i ) W n k c ( t i ) μ j ϵ k − ∑ i = 1 n ∑ j = 1 n ∑ k = 1 n δ i δ j δ k W n j c ( t i ) W n k c ( t i ) μ j μ k β } : = S 1 n − 2 ∑ k = 1 12 A k n . (5.1)

Thus, to prove β ^ c → β a.s., we only need to verify that S 1 n − 2 ≤ C n − 1 a . s . and n − 1 A k n = o ( 1 ) a . s . for k = 1 , 2 , ⋯ , 12 .

Step 1. We prove S 1 n − 2 ≤ C n − 1 a . s . Note that

S 1 n 2 = ∑ i = 1 n [ δ i ( ξ ˜ i c + μ ˜ i c ) 2 − δ i Ξ μ 2 ] = ∑ i = 1 n δ i ( ξ ˜ i c ) 2 + ∑ i = 1 n δ i ( μ i 2 − Ξ μ 2 ) + ∑ i = 1 n δ i [ ∑ j = 1 n δ j W n j c ( t i ) μ j ] 2 + 2 ∑ i = 1 n δ i ξ ˜ i c μ i − 2 ∑ i = 1 n δ i ξ ˜ i c ∑ j = 1 n δ j W n j c ( t i ) μ j − 2 ∑ i = 1 n δ i μ i ∑ j = 1 n δ j W n j c ( t i ) μ j : = B 1 n + B 2 n + B 3 n + B 4 n + B 5 n + B 6 n .

By Lemma 4.3(a), we have n − 1 B 1 n → Σ 1 a.s. Hence, it suffices to verify that B k n = o ( B 1 n ) = o ( n ) a.s. for k = 2 , 3 , ⋯ , 6 . Applying (A0), taking r = p / 2 > 2 , p 1 = 1 / 2 , p 2 = 1 / 2 in Lemma 4.2, we can verity that

∑ i = 1 n ( ζ i − E ζ i ) = n 1 2 ⋅ ∑ i = 1 n n − 1 2 ( ζ i − E ζ i ) = O ( n 1 2 log n ) a . s . (5.2)

where { ζ i } is a sequence of independent random variables satisfying E ζ i = 0 and sup 1 ≤ i ≤ n E | ζ i | p / 2 < ∞ . Therefore, we obtain B 2 n = O ( n 1 / 2 log n ) = o ( n ) a . s . from (A0) and (5.2). On the other hand, taking α = 4 , p > 4 in Lemma 4.1, we have

max 1 ≤ i ≤ n | ∑ j = 1 n δ j W n j c ( t i ) ζ j | = o ( n − 1 4 ) a .s ., max 1 ≤ i ≤ n | ∑ j = 1 n W n j ( t i ) ζ j | = o ( n − 1 4 ) a .s . (5.3)

where { ζ i } is a sequence of independent random variables satisfying E ζ i = 0 and sup 1 ≤ i ≤ n E | ζ i | p < ∞ . By (A0) and Lemma 4.3, taking r = 4 , p 1 = 1 / 4 , p 2 = 3 / 4 in Lemma 4.2, one can also deduce that

| B 4 n | = 2 n 1 4 ⋅ | ∑ i = 1 n n − 1 4 δ i ξ ˜ i c μ i | = O ( n 1 2 log n ) = o ( n ) a .s . (5.4)

Note that, from Lemma 4.3(a), (5.2) and (5.3), we have

| B 3 n | ≤ ∑ i = 1 n | δ i | ⋅ max 1 ≤ i ≤ n | ∑ j = 1 n δ j W n j c ( t i ) μ j | 2 = o ( n 1 2 ) a .s . (5.5)

| B 5 n | ≤ 2 ∑ i = 1 n | δ i ξ ˜ i c | ⋅ max 1 ≤ i ≤ n | ∑ j = 1 n δ j W n j c ( t i ) μ j | = o ( n 3 4 ) a .s . (5.6)

| B 6 n | ≤ 2 [ ∑ i = 1 n ( | δ i μ i | − E | δ i μ i | ) + ∑ i = 1 n E | δ i μ i | ] ⋅ max 1 ≤ i ≤ n | ∑ j = 1 n δ j W n j c ( t i ) μ j | = o ( n 3 4 ) a .s . (5.7)

Therefore, for (5.2)-(5.7), one can deduce that S 1 n 2 = B 1 n + o ( n ) = B 1 n + o ( B 1 n ) a .s . , which yields that

lim n → ∞ B 1 n S 1 n 2 = lim n → ∞ B 1 n B 1 n + o ( B 1 n ) = 1 a .s .

Therefore, by the Lemma 4.3(b), we can get that S 1 n − 2 ≤ C n − 1 a .s .

Step 2. We verify that n − 1 A k n = o ( n − 1 / 4 ) a .s . for k = 1 , 2 , ⋯ , 12 . From (A0), we find out { η i = ϵ i − μ i β ,1 ≤ i ≤ n } is a sequences of independent random variables with E η i = 0 , sup i E | η i | p ≤ C sup i E | e i | p + C sup i E | μ i | p < ∞ , for some p > 4 . Similar to (4), we deduce that

n − 1 A 1 n = n − 1 ∑ i = 1 n δ i ξ ˜ i c η i = O ( n − 1 2 log n ) a .s .

Meanwhile, from (A0)-(A3), Lemma 4.3, (5.2) and (5.3), one can achieve that

S n − 2 A 2 n ≤ C n | ∑ i = 1 n δ i μ i ϵ i | = o ( 1 ) a .s .

S n − 2 A 3 n ≤ C n | ∑ i = 1 n δ i ( μ i 2 − Ξ μ 2 ) β | = o ( 1 ) a .s .

S n − 2 A 4 n ≤ C n | ∑ i = 1 n δ i ξ ˜ i c g ˜ i c | C n [ max 1 ≤ i ≤ n ∑ i = 1 n | δ i ξ ˜ i c | max 1 ≤ i ≤ n | g ˜ i c | ] = o ( 1 ) a .s .

S n − 2 A 5 n ≤ C n | ∑ i = 1 n δ i μ ˜ i c g ˜ i c | C n [ ∑ i = 1 n | μ i | ⋅ | δ i g ˜ i c | + ∑ i = 1 n | δ i g ˜ i c | ⋅ | ∑ j = 1 n δ j W n j c ( t i ) μ j | ] ≤ C n ⋅ max 1 ≤ i ≤ n ( ∑ i = 1 n ( | μ i | − E | μ i | ) + ∑ i = 1 n E | μ i | ) ⋅ max 1 ≤ i ≤ n | δ i g ˜ i c | + o ( n − 1 2 ) = o ( 1 ) a .s .

The proof of n − 1 A k n = o ( 1 ) a .s . for k = 6 , ⋯ , 12 is analogous. Thus, the proof of Theorem 3.1(a) is completed. ■

Proof of Theorem 3.1(b). From (3.2), for every t ∈ [ 0,1 ] , one can write that

g ^ n c ( t ) − g ( t ) = ∑ j = 1 n W n j c ( t ) δ j ( y j − x j β ^ c ) − g ( t ) = ∑ j = 1 n W n j c ( t ) δ j [ ξ j β + g ( t j ) + ϵ j − ( ξ j + μ j ) β ^ c ] − g ( t ) = ∑ j = 1 n W n j c ( t ) δ j ξ j ( β − β ^ c ) + ∑ j = 1 n W n j c ( t i ) δ j [ g ( t j ) − g ( t ) ] + ∑ j = 1 n W n j c ( t i ) δ j ϵ j + ∑ j = 1 n W n j c ( t i ) δ j μ j β + ∑ j = 1 n W n j c ( t i ) δ j μ j ( β ^ c − β ) : = F 1 n ( t ) + F 2 n ( t ) + F 3 n ( t ) + F 4 n ( t ) + F 5 n ( t ) .

Therefore, we only need to prove that F k n ( t ) → 0 a.s. for every t ∈ [ 0,1 ] and k = 1 , 2 , ⋯ , 5 . From (A0)-(A3), Theorem 3.1(a), Lemma 4.3, (2) and (3), for every t ∈ [ 0,1 ] and any a > 0 , one can get

F 1 n ( t ) ≤ | β − β ^ c | ⋅ max 1 ≤ j ≤ n | h ( t j ) + v j | ⋅ ∑ j = 1 n δ j W n j c ( t ) | = o ( n − 1 4 ) a .s .

F 2 n ( t ) ≤ ∑ j = 1 n δ j W n j c ( t ) ⋅ [ g ( t j ) − g ( t ) ] ⋅ I ( | t j − t | > a ⋅ n − 1 4 ) + ∑ j = 1 n δ j W n j c ( t ) ⋅ [ g ( t j ) − g ( t ) ] ⋅ I ( | t j − t | < a ⋅ n − 1 4 ) ≤ C ⋅ a ⋅ n − 1 4 = o ( 1 ) a .s .

F 3 n ( t ) ≤ | ∑ j = 1 n W n j c ( t ) δ j ϵ j | = o ( 1 ) a .s .

F 4 n ( t ) ≤ | ∑ j = 1 n W n j c ( t ) δ j μ j β | = o ( 1 ) a .s .

F 5 n ( t ) ≤ | β − β ^ c | ⋅ | ∑ j = 1 n W n j c ( t ) δ j μ j | = o ( 1 ) a .s .

Thus, the proof of Theorem 3.1(b) is completed. ■

Proof of Theorem 3.2(a). From (3.3)-(3.4), write that

β ^ I − β = S 2 n − 2 { ∑ i = 1 n ( ξ ˜ i + μ ˜ i ) [ δ i ( y i − ξ i β − μ i β ) + ( 1 − δ i ) ( ξ i + μ i ) ( β ^ c − β ) + ( 1 − δ i ) g ^ n c ( t i ) − ∑ j = 1 n W n j ( t i ) ( δ j ( y j − ξ j β − μ j β ) + ( 1 − δ j ) ( ξ j + μ j ) ( β ^ c − β ) + ( 1 − δ j ) g ^ n c ( t j ) ) ] + ∑ i = 1 n δ i Ξ μ 2 β }

= S 2 n − 2 { ∑ i = 1 n ( ξ ˜ i + μ ˜ i ) [ δ i ( ϵ i − μ i β ) + δ i ( g ( t i ) − g ^ n c ( t i ) ) + ( 1 − δ i ) ( ξ i + μ i ) ( β ^ c − β ) + g ^ n c ( t i ) − ∑ j = 1 n W n j ( t i ) ( δ j ( ϵ j − μ j β ) + δ j ( g ( t j ) − g ^ n c ( t j ) ) + ( 1 − δ j ) ( ξ j + μ j ) ( β ^ c − β ) + g ^ n c ( t j ) ) ] + ∑ i = 1 n δ i Ξ μ 2 β } = S 2 n − 2 { ∑ i = 1 n δ i ξ ˜ i ( ϵ i − μ i β ) + ∑ i = 1 n δ i ξ ˜ i ( g ( t i ) − g ^ n c ( t i ) )

+ ∑ i = 1 n ξ ˜ i ( ξ i + μ i ) ( 1 − δ i ) ( β ^ c − β ) − ∑ i = 1 n ∑ j = 1 n δ j W n j ( t i ) ξ ˜ i ϵ j + ∑ i = 1 n ∑ j = 1 n δ j W n j ( t i ) ξ ˜ i μ j β − ∑ i = 1 n ∑ j = 1 n δ j W n j ( t i ) ξ ˜ i ( g ( t j ) − g ^ n c ( t j ) ) − ∑ i = 1 n ∑ j = 1 n W n j ( t i ) ξ ˜ i ( 1 − δ j ) ( ξ j + μ j ) ( β ^ c − β ) + ∑ i = 1 n ξ ˜ i g ˜ i c + ∑ i = 1 n δ i μ i ϵ i

− ∑ i = 1 n ∑ j = 1 n δ i W n j ( t i ) ϵ i μ j − ∑ i = 1 n δ i ( μ i 2 − Ξ μ 2 ) β + ∑ i = 1 n ∑ j = 1 n δ i W n j ( t i ) μ i μ j β + ∑ i = 1 n δ i μ ˜ i ( g ( t i ) − g ^ n c ( t i ) ) + ∑ i = 1 n μ ˜ i ( 1 − δ i ) ( ξ i + μ i ) ( β ^ c − β )

− ∑ i = 1 n ∑ j = 1 n δ j W n j ( t i ) μ i ϵ j + ∑ i = 1 n ∑ j = 1 n ∑ k = 1 n δ j W n j ( t i ) W n k ( t i ) μ k ϵ j + ∑ i = 1 n ∑ j = 1 n δ j W n j ( t i ) μ i μ j β − ∑ i = 1 n ∑ j = 1 n ∑ k = 1 n δ j W n j ( t i ) W n k ( t i ) μ j μ k β − ∑ i = 1 n ∑ j = 1 n δ j W n j ( t i ) μ ˜ i ( g ( t j ) − g ^ n c ( t j ) ) − ∑ i = 1 n ∑ j = 1 n W n j ( t i ) ( 1 − δ j ) μ ˜ i ( ξ j + μ j ) ( β ^ c − β ) + ∑ i = 1 n μ ˜ i g ˜ i c : = S 2 n − 2 ∑ k = 1 21 D k n .

Using a similar approach as step 1 in the proof of Theorem 3.1(a), one can get S 2 n − 2 ≤ C n − 1 a .s .

Therefore, we only need to verify that n − 1 D k n = o ( 1 ) a . s . for k = 1 , 2 , ⋯ , 21 . From (A0)-(A4), Lemmas 4.2-4.4, Theorem 3.1(a), (5.2)-(5.4), we have

n − 1 D 1 n = n − 1 ∑ i = 1 n δ i ξ ˜ i ( ϵ i − μ i β ) = n − 1 ⋅ O ( n 1 2 log n ) = o ( 1 ) a .s .

n − 1 D 2 n ≤ n − 1 [ ∑ i = 1 n | δ i ξ ˜ i | ⋅ max 1 ≤ i ≤ n | g ( t i ) − g ^ n c ( t i ) | ] = o ( 1 ) a .s .

n − 1 D 3 n = n − 1 ∑ i = 1 n ξ ˜ i 2 ( 1 − δ i ) ( β ^ c − β ) + n − 1 ∑ i = 1 n ξ ˜ i ( 1 − δ i ) ∑ j = 1 n W n j ( t i ) ξ j ( β ^ c − β ) + n − 1 ∑ i = 1 n ξ ˜ i μ i ( 1 − δ i ) ( β ^ c − β ) = o ( 1 ) a .s .

n − 1 D 4 n ≤ n − 1 ∑ i = 1 n | ξ ˜ i | ⋅ max 1 ≤ i ≤ n | ∑ j = 1 n δ j W n j ( t i ) ϵ j | = o ( 1 ) a .s .

In the same way, from (A0)-(A4), Lemmas 4.2-4.4, (5.2) and (5.3), one can similarly deduce that n − 1 D k n = o ( n − 1 / 4 ) a .s . for k = 5 , 6 , ⋯ , 21 . Thus, the proof of Theorem 3.2(a) is completed. ■

Proof of Theorem 3.2(b). From (3.4), write that

g ^ n [ I ] ( t ) − g ( t ) = ∑ j = 1 n W n j ( t ) { δ j y j + ( 1 − δ j ) [ ( ξ j + μ j ) β ^ c + g ^ n c ( t j ) ] − ( ξ i + μ j ) β ^ I } − g ( t ) = ∑ j = 1 n W n j ( t ) { δ j [ ξ j β + g ( t j ) + ϵ j ] + ( 1 − δ j ) [ ( ξ j + μ j ) β ^ c + g ^ n c ( t j ) ] − ( ξ j + μ j ) β ^ I } − g (t)

= ∑ j = 1 n W n j ( t ) { δ j ξ j β + δ j g ( t j ) + δ j ϵ j + ξ j β ^ c + μ j β ^ c − δ j ξ j β ^ c − δ j μ j β ^ c + g ^ n c ( t j ) − δ j g ^ n c ( t j ) − ξ j β ^ I − μ j β ^ I } − g (t)

= ∑ j = 1 n W n j ( t ) δ j ξ j ( β − β ^ c ) + ∑ j = 1 n W n j ( t ) δ j [ g ( t j ) − g ^ n c ( t j ) ] + ∑ j = 1 n W n j ( t ) δ j ϵ j + ∑ j = 1 n W n j ( t ) ξ j ( β ^ c − β ) + ∑ j = 1 n W n j ( t ) μ j ( β ^ c − β ) − ∑ j = 1 n W n j ( t ) δ j μ j β ^ c

+ ∑ j = 1 n W n j ( t ) [ g ^ n c ( t j ) − g ( t j ) ] + ∑ j = 1 n W n j ( t ) [ g ( t j ) − g ( t ) ] + ∑ j = 1 n W n j ( t ) ξ j ( β − β ^ I ) + ∑ j = 1 n W n j ( t ) μ j ( β − β ^ I ) : = ∑ k = 1 10 G k n .

Therefore, we only need to prove that G k n ( t ) → 0 a.s. for every t ∈ [ 0,1 ] and k = 1 , 2 , ⋯ , 10 . From (A0)-(A4), Lemma 4.3-4.4, (5.2), (5.3), one can get

G 1 n ( t ) ≤ | β − β ^ c | ⋅ max 1 ≤ j ≤ n | δ j ξ j | ⋅ ∑ j = 1 n W n j ( t ) ≤ | β − β ^ c | ⋅ max 1 ≤ j ≤ n | ξ j | ⋅ ∑ j = 1 n W n j ( t ) = o ( 1 ) a .s .

G 2 n ( t ) ≤ max 1 ≤ j ≤ n | g ( t j ) − g ^ n c ( t j ) | ⋅ | ∑ j = 1 n W n j ( t ) δ j | = o ( 1 ) a .s .

Meanwhile, the proof of G k n ( t ) → 0 a .s . for every t ∈ [ 0,1 ] and k = 3 , ⋯ , 10 is analogous. Thus, the proof of Theorem 3.2(b) is completed. ■

The authors greatly appreciate the constructive comments and suggestions of the Editor and referee. This research was supported by the National Natural Science Foundation of China (11701368).

The authors declare no conflicts of interest regarding the publication of this paper.

Zhang, L.R. and Zhang, J.J. (2019) Strong Consistency of Estimators under Missing Responses. Journal of Applied Mathematics and Physics, 7, 93-103. https://doi.org/10.4236/jamp.2019.71008