1 Additional lemmas. Supplementary Material APPENDIX. that N 1 N } E } E { N } E + O(N 3 ), Proof. The results follow by straightforward calculation.

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1 1 Additional lemmas Supplementay Mateial APPENDIX Lemma A1. Let (T 1,1, T 2,1, T 3,1, T 4,1 ),..., (T 1,N, T 2,N, T 3,N, T 4,N ) be independent andom vectos of length 4 such that E(T,i ) = 0 (i = 1,..., N, = 1, 2, 3, 4). Assume that N 1 N i=1 E(T,i) 2 = O(1) fo = 1, 2, 3, 4. Then E Ẽ(T1 )Ẽ(T N 2) = N 2 E(T 1,i T 2,i ), i=1 E Ẽ(T1 )Ẽ(T2)Ẽ(T N 3) = N 3 E(T 1,i T 2,i T 3,i ), i=1 E Ẽ(T1 )Ẽ(T2)Ẽ(T3)Ẽ(T 4) = N 4 κ N i=1 E(T κ1,it κ2,i) N =1 E(T κ3,t κ4,) + O(N 3 ), whee κ = (κ 1, κ 2, κ 3, κ 4 ) and κ is taken ove κ = (1, 2, 3, 4), (1, 3, 2, 4), (1, 4, 2, 3). Poof. The esults follow by staightfowad calculation. Lemma A2. Let τ 1, τ 2, τ 3, τ 4 be vaiables taking values τ 1,i, τ 2,i, τ 3,i, τ 4,i (i = 1,..., N) on the finite population. Define T,i = (p 1 i R i 1)τ,i and T,i O = (p 1 i R i 1)τ,i O fo = 1, 2, 3, 4. Assume that (10) holds, lim inf N Ẽp(1 p) > 0, and Ẽ(τ 2 ) = O(1) fo = 1, 2, 3, 4. Then the following esults hold. (i) E Re Ẽ(T1 )Ẽ(T O 2) = E P ois Ẽ(T )Ẽ(T O 1 2 ) + o(n 1 ). (ii) Ẽ(T1 )Ẽ(T 2), Ẽ(T 3 )Ẽ(T O 4) = cov P ois Ẽ(T )Ẽ(T O 1 2 ), Ẽ(T3 O )Ẽ(T 4 O ) + o(n 2 ). (iii) Ẽ(T1 ), Ẽ(T 2 )Ẽ(T O 3) = cov P ois Ẽ(T 1 ), Ẽ(T2 O )Ẽ(T 3 O ) + σ 1 Ẽp(1 p)τ1 O O E P ois Ẽ(T )Ẽ(T O 2 3 ) + o(n 2 ). (iv) Ẽ(T1 ), Ẽ(T 2 )Ẽ(T3)Ẽ(T O 4) = cov P ois Ẽ(T 1 ), Ẽ(T2 O )Ẽ(T 3 O )Ẽ(T 4 O ) + o(n 2 ). Thoughout, E Re and ae computed unde eective sampling and E P ois and cov P ois ae computed unde Poisson sampling. 1

2 Poof. We show esult (iii) in detail. The othe esults can be shown similaly. By simple manipulation, [Ẽ(p 1 R 1)τ 1, Ẽ(p 1 R 1)τ 2 Ẽ(p 1 R 1)τ 3, = 1 E N 3 Re (R i π i )(R p )(R k p k ) p 1 i τ 1,i p 1 τ 2, p 1 k i,,k τ 3,k. By Haek (1981, Chapte 14), conside the egession pedicto ˆπ i p i )σ 1 (K n), whee K = N i=1 R i and n = N i=1 p i. Then = p i p i (1 E Re (Ri π i ) 2 = E P ois (Ri ˆπ i ) 2 + o(σ 1 ), E Re (R i π i )(R π ) = E P ois (R i ˆπ i )(R ˆπ ) + o(σ 1 ) (i ), whee σ o(σ 1 ) 0 as σ unifomly in i,. Futhemoe, conside the quadatic egession pedicto ˆπ iq = p i p i (1 p i )σ 1 (K n) p i (1 p i )( p p i )σ 1, whee p is defined as befoe in (18). Then E Re (Ri π i ) 3 = E P ois (Ri ˆπ iq ) 3 + o(σ 1 ) E Re (Ri π i ) 2 (R π ) = E P ois (Ri ˆπ iq ) 2 (R ˆπ Q ) + o(σ 1 ) (i ), E Re (R i π i )(R π )(R k π k ) = E P ois (R i ˆπ iq )(R ˆπ Q )(R k ˆπ kq ) + o(σ 2 ) (i k), whee σ o(σ 1 ) 0 and σ 2 o(σ 2 ) 0 as σ unifomly in i,, k. By using these elations and (18) and appoximating R p R ˆπ = (R ˆπ Q ) + (ˆπ Q ˆπ ), we find that [Ẽ(p 1 R 1)τ 1, Ẽ(p 1 R 1)τ 2 Ẽ(p 1 R 1)τ 3, ( 1 = 1 + o(1) N 3 i,,k [ 1 1 (1 p i )( p p i )τ 1,i Nσ N 2 i E P ois (R i ˆπ i )(R ˆπ )(R k ˆπ k ) p 1 i,k = (R π ) + (π p ) by τ 1,i p 1 τ 2, p 1 k τ 3,k E P ois (R ˆπ )(R k ˆπ k ) p 1 τ 2, p 1 k τ 3,k The esult follows because N 1 (R i ˆπ i )p 1 i τ 1,i = Ẽ(p 1 R 1)τ1 O, etc. The tem Ẽ(1 p)( p p)τ 1 equals Ẽ(1 p)( p p)τ O 1 because Ẽ(1 p)( p p)p = 0 and hence equals Ẽp(1 p)τ O 1 because Ẽ(1 p)τ O 1 = 0. ). 2

3 Lemma A3. In the setup of Lemma A2, define S,i = (π 1 i R i 1)τ,i fo = 1, 2, 3, 4. Unde the assumptions in Lemma A1, the following esults hold. (i) E Re Ẽ(S1 )Ẽ(S O 2) = E P ois Ẽ(T )Ẽ(T O 1 2 ) + o(n 1 ). (ii) Ẽ(S1 )Ẽ(S 2), Ẽ(S 3 )Ẽ(S O 4) = cov P ois Ẽ(T )Ẽ(T O 1 2 ), Ẽ(T3 O )Ẽ(T 4 O ) + o(n 2 ). (iii) Ẽ(S1 ), Ẽ(S 2 )Ẽ(S O 3) = cov P ois Ẽ(T 1 ), Ẽ(T2 O )Ẽ(T 3 O ) + σ 1 Ẽp(1 p)τ1 O O E P ois Ẽ(T )Ẽ(T O 2 3 ) + σ 1 Ẽp(1 p)τ2 O O E P ois Ẽ(T )Ẽ(T O 1 3 ) + σ 1 Ẽp(1 p)τ3 O O E P ois Ẽ(T )Ẽ(T O 1 2 ) + o(n 2 ). (iv) Ẽ(S1 ), Ẽ(S 2 )Ẽ(S3)Ẽ(S O 4) = cov P ois Ẽ(T 1 ), Ẽ(T2 O )Ẽ(T 3 O )Ẽ(T 4 O ) + o(n 2 ). Poof. We show esult (iii) in detail. By simple manipulation, [Ẽ(π 1 R 1)τ 1, Ẽ(π 1 R 1)τ 2 Ẽ(π 1 R 1)τ 3, = 1 E N 3 Re (R i π i )(R π )(R k π k ) π 1 i τ 1,i π 1 τ 2, π 1 k τ 3,k i,,k = 1 N 3 i,,k E Re (R i π i )(R p )(R k p k ) p 1 i i,,k τ 1,i p 1 τ 2, p 1 k τ 3,k + 1 E N 3 Re (R i π i )(p π )(R k p k ) p 1 i τ 1,i p 1 τ 2, p E N 3 Re (R i π i )(R π )(p k π k ) p 1 i i,,k τ 1,i p 1 k τ 3,k τ 2, p 1 k τ 3,k + o(n 2 ). Result (ii) follows by simila calculations to those fo Lemma A2 using the appoximation of p i π i by ˆπ i ˆπ iq. 2 Poofs Poof of Poposition 1. (i) Diect calculation shows that ˆα eg = g 1 + g 2 (ˆα eg ) Ẽ(eξT ) Ẽ 1 (ξξ T ) V 1 Ẽ(ξ). 3

4 Expansion (4) holds because Ẽ 1 (ξξ T ) V 1 = Ẽ 1 (ξξ T )Ẽ(ξξT ) V V 1 = V 1 Ẽ(ξξT ) V V 1 + o p (N 1/2 ) by the assumption E( ξ 4 ) <. the expansion of ˆα egc holds because Similaly, ˆα egc = g 1 + g 2 (ˆα egc ) + Ẽ(e 0)Ẽ(ξT )Ẽ 1 (ξξ T )Ẽ(ξ) Ẽ(e 0ξ T ) Ẽ 1 (ξξ T ) V 1 Ẽ(ξ). (ii) By Owen (2001, Chapte 11), max i ξ i = o p (N 1/2 ) and ˆλ = O p (N 1/2 ) povided that E( ξ 2 ) <. A Taylo expansion of (3) gives [ 0 = Ẽ(ξ) )ˆλ + Ẽ(ξ Ẽ(ξξT ξξ T )ˆλ ˆλ 1 ξ ξ k ξξ T Ẽ 2 (1 + λ ˆλ ˆλk ˆλ, T ξ) 4 whee λ is on the line oining 0 and ˆλ. By the Cauchy-Schwatz and tiangle inequalities, the diffeence between the last tem and k Ẽ(ξ ξ k ξξ T )ˆλ ˆλk ˆλ/2 is bounded in absolute value by [ 1 Ẽ ξ ξ k ξξ T max (1 + 2 λ T ξ i ) 4 1 ˆλ ˆλk ˆλ = o p (N 3/2 ) i k because max i λ T ξ i = o p (1) and E( ξ 4 ) <. By eaanging the expansion of (3) and solving fo ˆλ as in Newey and Smith (2004, Lemma A4), we find ˆλ = h 1 +h 2 (ˆλ)+ h 3 (ˆλ) + o p (N 3/2 ), whee h 1 = V 1 Ẽ(ξ) and h 2 (ˆλ) = V Ẽ(ξξT 1 ) V V 1 Ẽ(ξ) + V Ẽ(ξT 1 )V 1 E(ξ ξξ T ) V 1 Ẽ(ξ), h 3 (ˆλ) = V Ẽ(ξξT 1 ) V V Ẽ(ξξT 1 ) V V 1 Ẽ(ξ) V Ẽ(ξξT 1 ) V V Ẽ(ξT 1 )V 1 E(ξ ξξ T ) V 1 Ẽ(ξ) V [Ẽ(ξT 1 )V Ẽ(ξξT 1 ) V V 1 E(ξ ξξ T ) V 1 Ẽ(ξ) + V [Ẽ(ξT 1 )V Ẽ(ξT 1 )V 1 E(ξ ξξ T ) V 1 E(ξ ξξ T ) V 1 Ẽ(ξ) V Ẽ(ξT 1 )V 1 E(ξ ξξ T ) V Ẽ(ξξT 1 ) V V 1 Ẽ(ξ) + V Ẽ(ξT 1 )V 1 E(ξ ξξ T ) V ẼT 1 (ξ)v 1 E(ξ ξξ T ) V 1 Ẽ(ξ) + V [Ẽ(ξT 1 )V Ẽ(ξ 1 ξξt ) E(ξ ξξ T ) V 1 Ẽ(ξ) [ 1 2 V 1 E(ξ ξ k ξξ T ) V 1 Ẽ(ξ) V 1 Ẽ(ξ) V 1 Ẽ(ξ) k k Thoughout, each expession of the fom ẼT (ϕ T )UE(ψ ξξ T ), enclosed by culy backets, is defined as k Ẽ(ϕ k)u k E(ψ ξξ T ). Similaly, Ẽ T (ϕ T )UẼ(ψ ξξt ) is defined 4 k

5 as k Ẽ(ϕ k)u k Ẽ(ψ ξξ T ). Futhe, a Taylo expansion of ˆα lik gives [ ˆα lik = Ẽ(η) Ẽ(ηξT )ˆλ + ˆλ T Ẽ(ηξξ T )ˆλ 1 2 ˆλ ηξ ξξ T T Ẽ (1 + λ ˆλ T ξ) 4 ˆλ, whee λ is on the line oining 0 and ˆλ. The last tem is ˆλ T Ẽ(ηξ ξξ T )ˆλ ˆλ/2 + o p (N 3/2 ) similaly as in the expansion of (3). Wite h 2 = h 2 (ˆλ) and h 3 = h 3 (ˆλ). By substituting the expansion of ˆλ into that of ˆα lik, we find ˆα lik = Ẽ(η) E(ηξT )(h 1 + h 2 + h 3 ) Ẽ(ηξT ) E(ηξ T ) (h 1 + h 2 ) + h T 1E(ηξξ T )(h 1 + h 2 ) + h T 2E(ηξξ T )h 1 + h T 1 Ẽ(ηξξT ) E(ηξξ T ) 1 2 ht 1 E(ηξ ξξ T )(h 1 ) h 1 + O p (N 3/2 ). By eaanging the tems, we have g 2 (ˆα lik ) = E(ηξ T )h 2 Ẽ(ηξT ) E(ηξ T ) h 1 + h T 1E(ηξξ T )h 1, g 3 (ˆα lik ) = E(ηξ T )h 3 Ẽ(ηξT ) E(ηξ T ) h 2 + h T 1E(ηξξ T )h 2 + h T 2E(ηξξ T )h 1 + h T 1 Ẽ(ηξξT ) E(ηξξ T ) 1 2 ht 1 E(ηξ ξξ T )(h 1 ) h 1. Finally, by diect calculation, the g 2 tem is simplified to that in Poposition 1, and the g 3 tem is simplified to g 3 (ˆα lik ) = Ẽ(eξT )V Ẽ(ξξT 1 ) V V 1 Ẽ(ξ) Ẽ(eξT )V Ẽ(ξT 1 )V 1 E(ξ ξξ T ) V 1 Ẽ(ξ) Ẽ(ξT )V Ẽ(ξξT 1 ) V V 1 E(eξξ T )V 1 Ẽ(ξ) + Ẽ(ξT )V Ẽ(ξT 1 )V 1 E(ξ ξξ T ) V 1 E(eξξ T )V 1 Ẽ(ξ) Ẽ(ξT )V 1 E(eξξ T )V Ẽ(ξξT 1 ) V V 1 Ẽ(ξ) + Ẽ(ξT )V 1 E(eξξ T )V Ẽ(ξT 1 )V 1 E(ξ ξξ T ) V 1 Ẽ(ξ) + Ẽ(ξT )V Ẽ(eξξT 1 ) E(eξξ T ) V 1 Ẽ(ξ) 1 )V Ẽ(ξT 1 )V 1 E(eξ ξξ T ) V 1 Ẽ(ξ). 2Ẽ(ξT h 1 h 1 5

6 Poofs of Popositions 2 3. Poposition 2 follows by simple calculation. To show V(ˆα lik ), wite g 2 = g 2 (ˆα lik ), g 3 = g 3 (ˆα lik ), and the 8 tems of g 3 as g 31, g 32 and so on. By diect but tedious calculation using Lemma A1, va(g 2 ) = N 2 t(v 1 Σ eξ:eξ ) t(v 1 Σ eξξ V 1 Σ eξξ ), cov(g 1, g 21 ) = N 2 t(v 1 Σ e0 eξξ), cov(g 1, g 22 ) = N 2 t(v 1 Σ e0 ξξv 1 Σ eξξ ), cov(g 1, g 31 ) = N 2 t(v 1 Σ e0 ξξv 1 Σ eξξ ) + N 2 u, cov(g 1, g 32 ) = N 2 u, cov(g 1, g 33 ) = cov(g 1, g 35 ) = N 2 t(v 1 Σ e0 ξξv 1 Σ eξξ ), cov(g 1, g 37 ) = N 2 t(v 1 Σ e0 eξξ), up to tems of ode O(N 2 ) and cov(g 2, g 3l ) = o(n 2 ) fo l = 4, 6, 8, whee Σ eξ:eξ = E(e 2 ξξ T ), Σ eξξ = E(eξξ T ), and Σ e0 ξξ = E(e 0 ξξ T ) as defined in Popositions 2 3, Σ e0 eξξ = E(e 0 eξξ T ), and u = t E(e 0 eξ T )V 1 E(ξ ξξ T )V 1 = E(e 0 eξ T )V 1 E(ξξ T V 1 ξ). The fomula follows by substituting these expessions into the definition of V(ˆα lik ). To show V(ˆα eg,bc ), decompose the bias coection tem as (η ˆβ T ξ)ξ T Ẽ 1 (ξξ T )ξ N 1 Ẽ = N 1 Ẽ(eξ T V 1 ξ) + N Ẽ(eξT 1 V 1 ξ) E(eξ T V 1 ξ) N 1 Ẽ ( ˆβ β) T ξξ T V 1 ξ + N 1 Ẽ [eξ T Ẽ 1 (ξξ T ) V 1 ξ Substituting the fist-ode expansions fo ˆβ and Ẽ 1 (ξξ T ) above leads to thid-ode expansion (4) fo ˆα eg,bc, whee g 2 (ˆα eg,bc ) = Ẽ(eξT )V 1 Ẽ(ξ) + N 1 E(eξ T V 1 ξ), g 3 (ˆα eg,bc ) = Ẽ(eξT )V Ẽ(ξξT 1 ) V V 1 Ẽ(ξ) + N Ẽ(eξT 1 V 1 ξ) E(eξ T V 1 ξ). N 1 E(ξ T V 1 ξξ T )V 1 Ẽ(eξ) [ N 1 t V Ẽ(ξξT 1 ) V V 1 E(eξξ T ). 6

7 Wite g 2 = g 2 (ˆα eg,bc ), g 3 = g 3 (ˆα eg,bc ), and the 4 tems of g 3 as g 31, g 32 and so on. The fomula fo V(ˆα eg,bc ) follows by diect calculation: va(g 2 ) = N 2 t(v 1 Σ eξ:eξ ) + t(v 1 Σ eξξ V 1 Σ eξξ ), cov(g 1, g 2 ) = N 2 t(v 1 Σ e0 eξξ), cov(g 1, g 31 ) = N 2 t(v 1 Σ e0 ξξv 1 Σ eξξ ) + N 2 u, cov(g 1, g 32 ) = N 2 t(v 1 Σ e0 eξξ), cov(g 1, g 33 ) = N 2 u, cov(g 1, g 34 ) = N 2 t(v 1 Σ e0 ξξv 1 Σ eξξ ), up to tems of ode O(N 2 ). It is inteesting to mention that without bias coection, V(ˆα eg ) is equal to va(g 2 ) + 2cov(g 1, g 2 ) + 2cov(g 1, g 31 ). V(ˆα egc,bc ) can be poved by simila calculation. Poof of Poposition 4. Finally, the fomula fo (i) The expansion of α eg holds similaly to that of ˆα eg because α eg = g 1 +g 2 ( α eg ) Ẽ(εξT ) Ẽ 1 (ξζ T ) V 1 Ẽ(ξ) and Ẽ 1 (ξζ T ) V 1 = V 1 Ẽ(ξζ T ) V V 1 + o p (N 1/2 ). (ii) By Tan (2013), max i=1,...,n p 1 (X i ) 1X i = o p (N 1/2 ) and λ = O p (N 1/2 ). A Taylo expansion of (8) gives [ 0 = Ẽ(ξ) Ẽ(ζξT ) λ + Ẽ(ζ ξξ T ) λ λ 1 ζ ξ k ξξ T Ẽ 2 (1 + λ λ λk λ, T ξ) 4 k whee λ is on the line oining 0 and λ. The last tem is k Ẽ(ζ ξ k ξξ T )ˆλ ˆλk ˆλ/2+ o p (N 3/2 ) similaly as in the expansion of (3). By solving fo λ fom the expansion of (8), we find λ = h 1 + h 2 ( λ) + h 3 ( λ) + o p (N 3/2 ), whee h 2 ( λ) and h 3 ( λ) ae defined as h 2 (ˆλ) and h 3 (ˆλ) with ξξ T eplaced by ζξ T, ξ ξξ T by ζ ξξ T, and ξ ξ k ξξ T by ζ ξ k ξξ T thoughout. Futhe, a Taylo expansion of α lik gives α lik = Ẽ(η) Ẽ(ηξT ) λ + λ T Ẽ(ηξξ T ) λ 1 2 λ T [ ηξ ξξ T Ẽ (1 + λ λ T ξ) 4 λ, whee λ is on the line oining 0 and λ. The last tem is λ T Ẽ(ηξ ξξ T ) λ λ/2+ o p (N 3/2 ) similaly as in the expansion of (3). By substituting the expansion of λ into that of α lik and by simila calculation as in the poof of Poposition 1, we obtain 7

8 g 2 ( α lik ) as shown in Poposition 1 and g 3 ( α lik ) = Ẽ(εξT )V Ẽ(ζξT 1 ) V V 1 Ẽ(ξ) Ẽ(εξT )V Ẽ(ξT 1 )V 1 E(ζ ξξ T ) V 1 Ẽ(ξ) Ẽ(ξT )V Ẽ(ζξT 1 ) V V 1 E(εξξ T )V 1 Ẽ(ξ) + Ẽ(ξT )V Ẽ(ξT 1 )V 1 E(ζ ξξ T ) V 1 E(εξξ T )V 1 Ẽ(ξ) Ẽ(ξT )V 1 E(εξξ T )V Ẽ(ζξT 1 ) V V 1 Ẽ(ξ) + Ẽ(ξT )V 1 E(εξξ T )V Ẽ(ξT 1 )V 1 E(ζ ξξ T ) V 1 Ẽ(ξ) + Ẽ(ξT )V Ẽ(εξξT 1 ) E(εξξ T ) V 1 Ẽ(ξ) 1 )V Ẽ(ξT 1 )V 1 E(εξ ξξ T ) V 1 Ẽ(ξ). 2Ẽ(ξT Poofs of Popositions 5 6. Poposition 5 follows by simple calculation. To show V( α lik ), wite g 2 = g 2 ( α lik ), g 3 = g 3 ( α lik ), and the 8 tems of g 3 as g 31, g 32 and so on. By diect but tedious calculation using Lemma A1, va(g 2 ) = N 2 t(v 1 Σ εξ:εξ ) t(v 1 Σ εξξ V 1 Σ εξξ ), cov(g 1, g 21 ) = N 2 t(v 1 Σ e0 εξξ), cov(g 1, g 22 ) = N 2 t(v 1 Σ e0 ξξv 1 Σ εξξ ), cov(g 1, g 31 ) = N 2 t(v 1 Σ e0 ζξv 1 Σ εξξ ) + ν, cov(g 1, g 32 ) = N 2 ν, cov(g 1, g 33 ) = cov(g 1, g 35 ) = N 2 t(v 1 Σ e0 ζξv 1 Σ εξξ ), cov(g 1, g 37 ) = N 2 t(v 1 Σ e0 εξξ), up to tems of ode O(N 2 ) and cov(g 2, g 3l ) = o(n 2 ) fo l = 4, 6, 8, whee Σ εξ:εξ = E(ε 2 ξξ T ) and Σ εξξ = E(εξξ T ) as in Popositions 5 6, Σ e0 ξξ = E(e 0 ξξ T ) as in Poposition 2, Σ e0 ζξ = E(e 0 ζξ T ), Σ e0 εξξ = E(e 0 εξξ T ), and ν = t E(e 0 εξ T )V 1 E(ζ ξξ T )V 1 = E(e 0 εξ T )V 1 E(ζξ T V 1 ξ). The fomula fo V( α lik ) follows because Σ e0 ξξ = Σ e0 ζξ A. To show V( α eg,bc ), we 8

9 find that α eg,bc admits expansion (4) similaly to ˆα eg,bc, whee g 2 ( α eg,bc ) = Ẽ(εξT )V 1 Ẽ(ξ) + N 1 E(εξ T V 1 ξ), g 3 ( α eg,bc ) = Ẽ(εξT )V Ẽ(ζξT 1 ) V V 1 Ẽ(ξ) + N Ẽ(εξT 1 V 1 ξ) E(εξ T V 1 ξ) N 1 E(ξ T V 1 ξζ T )V 1 Ẽ(εξ) [ N 1 t V Ẽ(ζξT 1 ) V V 1 E(εξξ T ). Wite g 2 = g 2 ( α eg,bc ), g 3 = g 3 ( α eg,bc ), and the 4 tems of g 3 as g 31, g 32 and so on. The fomula fo V( α eg,bc ) follows by diect calculation: va(g 2 ) = N 2 t(v 1 Σ εξ:εξ ) + t(v 1 Σ εξξ V 1 Σ εξξ ), cov(g 1, g 2 ) = N 2 t(v 1 Σ e0 εξξ), cov(g 1, g 31 ) = N 2 t(v 1 Σ e0 ζξv 1 Σ eξξ ) + N 2 ν, cov(g 1, g 32 ) = N 2 t(v 1 Σ e0 εξξ), cov(g 1, g 33 ) = N 2 ν, cov(g 1, g 34 ) = N 2 t(v 1 Σ e0 ζξv 1 Σ eξξ ), up to tems of ode O(N 2 ). It is inteesting to mention that without bias coection, V( α eg ) is equal to va(g 2 ) + 2cov(g 1, g 2 ) + 2cov(g 1, g 31 ). Poof of Coollay 1. Assume that E(Y X) = β T X. Then A = 0 and Σ εξξ = 0. Theefoe, V( α eg,bc ) = V( α lik ) = N 2 t(v 1 Σ εξ:εξ ). By diect calculation, Σ εξ:εξ = E(ε 2 ξξ T ) = E (p 1 1) 2 p 1 (Y β T X) 2 XX T, Σ eξ:eξ = E(e 2 ξξ T ) = E (p 1 1) 2 (Y β T X) 2 XX T [ + E (p 1 1) (p 1 1)(Y β T X)X + (β T X)X 2, whee τ 2 = ττ T fo a column vecto τ. Then V(ˆα lik ) V( α lik ) = N 2 t[v 1 (Σ eξ:eξ Σ εξ:εξ ) Σ eξξ V 1 Σ eξξ. The matix in the culy backet equals E(β T X) 2 ξξ T E(β T X)ξξ T V 1 E(β T X)ξξ T, which is the vaiance matix of the esidual in poecting (β T X)ξ on ξ and hence nonnegative definite. Theefoe, V( α lik ) V(ˆα lik ). Poof of Poposition 7. The poof is simila to those of Popositions 1 and 4. The moment condition Ẽ( X 4 ) = O(1) is imposed fo ˆα eg and α eg to show that 9

10 Ẽ(ξξ T ) = V + O p (N 1/2 ) in g 3 (ˆα eg ) and Ẽ(ξζ T ) = V + O p (N 1/2 ) in g 3 ( α eg ) by Chebyshevs inequality. Similaly, the moment condition Ẽ( X 6 ) = O(1) is imposed fo ˆα lik and α lik to show that Ẽ(e ξξ T ) = E Ẽ(e ξξ T ) + O p (N 1/2 ) in g 3 (ˆα lik ) and Ẽ(ε ξξ T ) = E Ẽ(ε ξξ T ) + O p (N 1/2 ) in g 3 ( α lik ). Poof of Poposition 8. (i) The bias fomulas follow by simple calculation. To show V ( α lik ), wite g 2 = g 2 ( α lik ), g 3 = g 3 ( α lik ), and the 8 tems of g 3 as g 31, g 32 and so on. Let e = e Ȳ. By diect but tedious calculation using Lemma A1, va (g 2 ) = N 2 t(v 1 Ω ε ξ:ε ξ) t(v 1 Ω ε ξξv 1 cov (g 1, g 21 ) = N 2 t(v 1 Ω e ε ξξ) + N 2 t(v 1 B ), cov (g 1, g 22 ) = N 2 t(v 1 Ω e ξξv 1 Ω ε ξξ), cov (g 1, g 31 ) = N 2 t(v 1 Ω e ζξv 1 Ω εξξ) + N 2 ν, cov (g 1, g 32 ) = N 2 ν, cov (g 1, g 33 ) = cov (g 1, g 35 ) = N 2 t(v 1 cov (g 1, g 37 ) = N 2 t(v 1 Ω e ε ξξ), Ω e ζξv 1 Ω ε ξξ), Ω ε ξξ), up to tems of ode O(N 2 ) and cov (g 2, g 3l )=o(n 2 ) fo l = 4, 6, 8. Hee Ω ε ξ:ε ξ= Nva Ẽ(ε ξ) and Ω ε ξξ = E Ẽ(ε ξξ T ) as in Poposition 8, Ω e ζξ = E Ẽ(e ζξ T ) = Ω εξξ, Ω e ξξ = E Ẽ(e ξξ T ) = Ω εξξ A, Ω e εξξ = E Ẽ(e ε ξξ T ) = Ω εξ:εξ, and [ ν = t E Ẽ(e ε ξ T ) V 1 E Ẽ(ζ ξξt ) V 1 = E Ẽ(e ε ξ T ) V 1 E Ẽ(ζξT V 1 ξ). To show V (ˆα lik ), wite g 2 = g 2 (ˆα lik ), g 3 = g 3 (ˆα lik ), and the 8 tems of g 3 as g 31, g 32 and so on. By diect but tedious calculation using Lemma A1, va (g 2 ) = N 2 t(v 1 Ω e ξ:e ξ) t(v 1 Ω e ξξv 1 cov (g 1, g 21 ) = N 2 t(v 1 Ω e e ξξ) + N 2 t(v 1 B ), cov (g 1, g 22 ) = N 2 t(v 1 Ω e ξξv 1 Ω eξξ), cov (g 1, g 31 ) = N 2 t(v 1 Ω e ξξv 1 Ω e ξξ) + N 2 u, cov (g 1, g 32 ) = N 2 u, cov (g 1, g 33 ) = cov (g 1, g 35 ) = N 2 t(v 1 cov (g 1, g 37 ) = N 2 t(v 1 Ω e e ξξ), Ω e ξξv 1 Ω e ξξ), Ω eξξ), 10

11 up to tems of ode O(N 2 ) and cov (g 2, g 3l ) = o(n 2 ) fo l = 4, 6, 8. Hee Ω e ξ:e ξ = Nva Ẽ(e ξ) and Ω e ξξ = E Ẽ(e ξξ T ) as in Poposition 8, Ω e e ξξ = E Ẽ( e e ξξ T ), and [ u = t E Ẽ(e e ξ T ) V 1 E Ẽ(ξ ξξt ) V 1 = E Ẽ(e e ξ T ) V 1 E Ẽ(ξξT V 1 ξ). (ii) Expansion (4) can be poved fo ˆα eg,bc and α eg,bc similaly as in the missingdata poblem. The g 1, g 2, and g 3 tems ae defined as in the poofs of Popositions 3 and 6 except that e is eplaced thoughout by e, ε by ε, V by V, and so on. Then the second-ode vaiance can be calculated fo ˆα eg,bc and α eg,bc similaly in the missing-data poblem. The fomulas of va (g 2 ), cov (g 1, g 2 ), etc., ae simila to va(g 2 ), cov(g 1, g 2 ), etc., in the poofs of Popositions 3 and 6, but with subtle changes as in the calculation of V (ˆα lik ) and V ( α lik ). Poof of Coollay 2. All the finite population quantities such as V, β, Ω ε ξξ, and A involve means ove the finite population of polynomials of p 1 (X), Y, and X with degee of Y 2 and that of X 4. By Chow and Teiche (1997, Coollay 8.1.7), each of these finite population means conveges to the coesponding expectation unde L yx. Fo example, V conveges to V yx = E yx (p 1 1)XX T, β to β yx = V 1 yx E yx (p 1 1)XY, Ω ε ξξ to E yx (p 1 1) 2 (Y β T yxx)xx T, and A to E yx (p 1 1)(Y β T yxx)xx T, whee E yx ( ) denotes the expectation of a function of (Y, X) unde L yx. The fomulas of second-ode vaiances emain valid with the finite population means eplaced by the coesponding expectations. Assume that the conditional expectation of Y given X unde L yx, E yx (Y X), is of fom b T X fo a constant vecto b. Then E yx (Y X) = β T yxx and hence Ω εξξ = o(1) and A = o(1). Theefoe, V ( α eg,bc ) = V ( α lik ) = N 2 t(v 1 Ω εξ:εξ) + 2t(V 1 B ) + o(n 2 ). By diect calculation, Ω ε ξ:ε ξ = Nva Ẽ(ε ξ) = Ẽ (p 1 1) 3 (Y β T X) 2 XX T, [ Ω e ξ:e ξ = Nva Ẽ(e ξ) = Ẽ (p 1 1) (p 1 1)(Y β T X)X + (β T X)X 2. Then V (ˆα lik ) V ( α lik ) = N 2 t[v 1 (Ω eξ:eξ Ω εξ:εξ) Ω eξξv 1 Ω eξξ+o(n 2 ). The matix in the culy backet conveges to E yx (p 1 1)(β T yxx) 2 XX T E yx (p 1 1)(β T yxx)xx T V 1 yx E yx (p 1 1)(β T yxx)xx T, which is the expectation of the pod- 11

12 uct of the esidual and its tanspose in poecting (p 1 1) 1/2 (β T yxx)x on (p 1 1) 1/2 X and hence nonnegative definite. Theefoe, V ( α lik ) V (ˆα lik ). Poofs of Popositions 9 and 10. The poof of Poposition 9 is simila to that of Poposition 7. Fo Poposition 10, the bias fomulas follow by simple calculation using Lemma A2. To show V ( α lik ), wite g 2 = g 2 ( α lik ), g 3 = g 3 ( α lik ), and the 8 tems of g 3 as g 31, g 32, etc. Recall that ξ 0,i = R i p i and U 0 = Ẽp(1 p). By diect but tedious calculation using Lemma A2, va (g 2 ) = N 2 t( 1 Γ εξ 1 :ε ξ 1 ) t( 1 Γ εξ 1 ξ 1 1 Γ εξ 1 ξ 1 ), cov (g 1, g 21 ) = N 2 t( 1 Γ ε ξ 1 :ε ξ 1 ) + N 2 t( 1 D 1 ) N 2 ψt( 1 Γ ε ξ 1 ξ 1 ), cov (g 1, g 22 ) = N 2 t( 1 Γ e ξ 1 ξ 1 1 Γ ε ξ 1 ξ 1 ) + N 2 ψt( 1 Γ ε ξ 1 ξ 1 ), cov (g 1, g 31 ) = N 2 t( 1 Γ e ζ 1 ξ 1 1 Γ ε ξ 1 ξ 1 ) + t( 1 Γ e ζ 1 ξ 0 0 Γ T ε ξ 1 ξ 0 ) + N 2 ν, cov (g 1, g 32 ) = N 2 ν, cov (g 1, g 33 ) = cov (g 1, g 35 ) = N 2 t( 1 Γ e ζ 1 ξ 1 1 Γ ε ξ 1 ξ 1 ) + t( 1 Γ e ζ 1 ξ 0 0 Γ T ε ξ 1 ξ 0 ), cov (g 1, g 37 ) = N 2 t( 1 Γ e :ε ξ 1 ξ 1 ), up to tems of ode O(N 2 ) and cov (g 2, g 3l ) = o(n 2 ) fo l = 4, 6, 8. Hee Γ ε ξ 1 :ε ξ 1 = Nva Ẽ(ε ξ 1 ) = Ẽ[(p 1 1)(p 1 1)(Y β T X)Z O 2 +o(1), Γ ε ξ 1 ξ 1 = E P ois Ẽ(ε ξ 1 ξ T 1 ), and Γ ε ξ 1 ξ 0 = E P ois Ẽ(ε ξ 1 ξ T 0 ) as in Poposition 10, Γ e ζ 1 ξ 1 = E P ois Ẽ(e ζ 1 ξ T 1 ) = Γ ε ξ 1 ξ 1, Γ e ζ 1 ξ 0 = E P ois Ẽ(e ζ 1 ξ 0 ) = Γ ε ξ 1 ξ 0, Γ e ξ 1 ξ 1 = E P ois Ẽ( e ξ 1 ξ T 1 ) = Γ εξ 1 ξ 1 C 1, Γ e :ε ξ 1 ξ 1 = E P ois Ẽ(e ε ξ 1 ξ T 1 ) = Ẽ[(p 1 1) 3 (Y β T X) 2 ZZ T = Γ εξ 1 :ε ξ 1 + Q 1 + o(1), and ν = E P ois Ẽ(e ε ξ1 T ) U1 1 E P ois Ẽ(ζ1 ξ1 T U1 1 ξ 1 ) + E P ois Ẽ(e ε ξ 0 ) 0 E P ois Ẽ(Rξ T 1 1 ξ 1 ) To show V (ˆα lik ), wite g 2 = g 2 (ˆα lik ), g 3 = g 3 (ˆα lik ), and the 8 tems of g 3 as g 31, g 32,. 12

13 etc. By diect but tedious calculation using Lemma A2, va (g 2 ) = N 2 t( 1 Γ e ξ 1 :e ξ 1 ) t( 1 Γ e ξ 1 ξ 1 1 Γ e ξ 1 ξ 1 ), cov (g 1, g 21 ) = N 2 t( 1 Γ e :e ξ 1 :ξ 1 ) + N 2 t( 1 D 1 ) N 2 ψt( 1 Γ e ξ 1 ξ 1 ), cov (g 1, g 22 ) = N 2 t( 1 Γ e ξ 1 ξ 1 1 Γ e ξ 1 ξ 1 ) + N 2 ψt( 1 Γ e ξ 1 ξ 1 ), cov (g 1, g 31 ) = N 2 t( 1 Γ e ξ 1 ξ 1 1 Γ eξ 1 ξ 1 ) + t( 1 Γ e ξ 1 ξ 0 0 Γ T e ξ 1 ξ 0 ) + N 2 u, cov (g 1, g 32 ) = N 2 u, cov (g 1, g 33 ) = cov (g 1, g 35 ) = N 2 t( 1 Γ e ξ 1 ξ 1 1 Γ e ξ 1 ξ 1 ) + t( 1 Γ e ξ 1 ξ 0 0 Γ T e ξ 1 ξ 0 ), cov (g 1, g 37 ) = N 2 t( 1 Γ e :e ξ 1 ξ 1 ), up to tems of ode O(N 2 ) and cov (g 2, g 3l ) = o(n 2 ) fo l = 4, 6, 8. Hee Γ e ξ 1 :e ξ 1 = Nva Ẽ(e ξ 1 ) = Ẽ[(p 1 1)(p 1 1)(Y β T X)Z+(β T X)Z O 2 +o(1), Γ eξ 1 ξ 1 = E P ois Ẽ(e ξ 1 ξ T 1 ), and Γ eξ 1 ξ 0 = E P ois Ẽ(e ξ 1 ξ 0 ) as in Poposition 10, Γ e ξ 1 ξ 0 = E P ois Ẽ(e ξ 1 ξ 0 ) = Γ εξ 1 ξ 0 C 0, Γ e :e ξ 1 :ξ 1 = Γ εξ 1 :ε ξ 1 + Ẽ[(p 1 1)(p 1 2)(Y β T X)Z O (β T X)Z T O, Γ e :e ξ 1 ξ 1 = E P ois Ẽ(e e ξ 1 ξ T 1 ) = Γ e :ε ξ 1 ξ 1 +Ẽ(p 1 1)(p 1 2)(Y β T X)(β T X)ZZ T = Γ e :e ξ 1 :ξ 1 + Q 1 + K 1 + o(1), and u = E P ois Ẽ(e e ξ1 T ) U1 1 E P ois Ẽ(ξ1 ξ1 T U1 1 ξ 1 ) + E P ois Ẽ(e e ξ 0 ) 0 E P ois Ẽ(ξ0 ξ T 1 1 ξ 1 ) Finally, V (ˆα eg,bc ) and V ( α eg,bc ) can be shown similaly as in Poposition 8, but with subtle changes as in the foegoing calculation of V (ˆα lik ) and V ( α lik ). Poof of Coollay 3. The poof is simila to that of Coollay 2. Let U yx = E yx (p 1 1)ZZ T. Assume that E yx (Y X) is of fom b T X fo a constant vecto b. Then Γ ε ξ 1 ξ 1, Γ ε ξ 1 ξ 0, C 1, C 0, Q 1, K 1, and ψ ae o(1). Theefoe, V ( α eg,bc ) = V ( α lik ) = N 2 t( 1 Γ ε ξ 1 :ε ξ 1 ) + 2t( 1 D 1 ) + o(n 2 ). Moeove, V (ˆα lik ) V ( α lik ) = N 2 t[ 1 (Γ e ξ 1 :e ξ 1 Γ ε ξ 1 :ε ξ 1 ) Γ e ξ 1 ξ 1 1 Γ e ξ 1 ξ 1 + o(n 2 ). The matix in the culy backet conveges to E yx [(p 1 1)(β T yxx)z O 2 E yx (p 1 1)(β T yxx)zz T yx E yx (p 1 1)(β T yxx)zz T, which is the expectation of the poduct of the esidual and its tanspose in poecting (p 1 1) 1/2 (β T yxx)z O on (p 1 1) 1/2 Z and hence nonnegative definite. Theefoe, V ( α lik ) V (ˆα lik ).. 13

14 Poof of Poposition 11. The poof is simila to that of Poposition 10, by diect calculation using Lemma A3. Fo example, the tem cov (g 1, g 21 ) fo α lik,π involves the additional tem N 2 ϕ T 1 Γ εε ξ 1 by esult (iii) of Lemma A3. REFERENCES Chow, Y.-S. and Teiche, H. (1997) Pobability Theoy (3d edition), New Yok: Spinge. 14