A Efficient Estimation under Conditional Mean Restrictions

Recall the notation from the body of the text so that we have ρit(β)=yit−μit(β)∑s=1Tyis∑s=1Tμis(β),ρ_i(β)=[ρ_i1(β), …, ρ_iT(β)]′, Di=E(∂ρi∂β′(β0)|xi), Σ_i=Var(ρ_i(β₀)|x_i), pit(β)=μit(β)∑s=1Tμis(β),p_i(β)′=[p_i1(β), …, p_iT(β)], W_i(β)=diag(p_i(β)), μ_it=μ_it(β₀).

Introducing some new notation, let μ′i=[μi1, …, μiT],y′i=[yi1, …, yiT], Σ_y,_i=Var(y_i|x_i), h_i=E(c_i|x_i), ∂μit=∂μit∂β′(β0),∂μ′i=[(∂μi1)′, …, (∂μiT)′].

We will also use M^[k] to denote the k^th column vector (element) of the matrix (row vector) M.

Also let:

so that ρ_i=ρ_i(β₀) can be rewritten as:

Lemma 1: Σ_i=Var(ρ_i|x_i) is singular.

Proof. Since ρ_i=(I–P_i)y_i and P_i is a function of x_i, we simply have to show that (I–P_i) is singular.

First note that P_i is idempotent, so that I–P_i is idempotent as well.

Therefore:

rank(I−Pi)=trace(I−Pi)=T−trace(Pi)=T−1

since trace(Pi)=∑t=1Tμit∑t=1Tμit=1.

Hence I–P_i is singular.□

Lemma 2:Σi−=(Σy,i−1−Σy,i−1μi(μ′i Σy,i−1μi)−1μ′i Σy,i−1)is a symmetric generalized inverse of Σ_i.

Proof. We can rewrite Σ_i as:

Note that:

and:

P′i  Σi−=P′i  (Σy,i−1−Σy,i−1μi(μ′i Σy,i−1μi)−1μ′i Σy,i−1)=0

Therefore:

Therefore:

Note that:

Therefore:

So Σi−=(Σy,i−1−Σy,i−1μi(μ′i Σy,i−1μi)−1μ′i Σy,i−1) is indeed a generalized inverse of Σ_i. It is clearly symmetric as well.□

Lemma 3:The asymptotic variance ofβ^optis the same independently of which symmetric generalized inverse of Σ_i,Σi+,is used.

Proof. Let Σi± and Σi+¯ be two symmetric generalized inverses of Σ_i.

Since ρ_i=(I–P_i)y_i, we have, for any k=1, …, dim(β₀):

so that:

From this expression for D_i we can show that ΣiΣi±Di=ΣiΣi+¯Di=Di by showing that, for any particular choice of i, the linear system of equations in w:

is consistent.

Consistency of (A.19) follows from:

PiDi=hi(μi∑t=1T∂μit∑t=1Tμit−μi∑t=1T∂μit∑t=1Tμit)=0

and recalling ΣiΣi−=I−Pi from the previous lemma, so that:

ΣiΣi−Di=(I−Pi)Di=Di−0=Di

Therefore E(D′i Σi±Di)=E(D′i Σi±ΣiΣi+¯Di)=E(D′i Σi+¯Di). Hence the result of this lemma is proved.□

Proof of Proposition 1.

Proof.Chamberlain (1992: p. 581), showed that the asymptotic information bound for estimating β₀ from (1) is:

V0,β0−1=E(hi∂μ′i(Σy,i−1−Σy,i−1μi(μ′i Σy,i−1μi)−1μ′i Σy,i−1)∂μihi)=E(hi∂μ′i Σi−∂μihi)

Note that:

μ′i Σi−=μ′i(Σy,i−1−Σy,i−1μi(μ′i Σy,i−1μi)−1μ′i Σy,i−1)=0

Therefore:

Therefore we have:

Because Σi− is a symmetric generalized inverse of Σ_i as shown in Lemma 2, the estimator defined by (4) with Σi+=Σi− also has inverse asymptotic variance:

Hence from Lemma 3, for any choice of a symmetric generalized inverse of Σ_i, Σi+, we have the result:

Therefore, for any choice of Σi+,β^opt is asymptotically efficient for estimating β₀ consistently from (1). Since (1) implies (3), β^opt is also asymptotically efficient for estimating β₀ from (3).□

B Efficient Estimation under the Poisson Fixed Effects Assumptions

Lemma 4 provides a useful alternative characterization of Σi−.

Lemma 4:Σi−is the unique matrix S that satisfies:

Proof. In the previous section we have shown that Σi− satisfies (B.1)–(B.3). Since Σi− and Σ_i are symmetric:

This solution is unique since for any S, S˜ satisfying these requirements ^[2]:

□

Proof of Proposition 2

Proof. Under (8) and (9) we have Σy,i=hidiag(μi)+viμiμ′i where v_i=Var(c_i|x_i). Therefore:

where the last equality follows from μitμis−μispit∑r=1Tμir=0.

Define:

where J=[1…1…1…1] and, by an abuse of notation, (1μi)′=[1μi1, …, 1μiT]. We will show that Σi−=Xi.

Note that:

and:

Therefore:

Since both Σ_i and X_i are symmetric:

So in order to show that Σi−=Xi, there only remains to show that X_i satisfies (B.1) and (B.2).

For (B.1):

For (B.2):

where the second equality follows from the previous section of the appendix.

Therefore we have shown that in this case:

Therefore:

where j′=[1, …, 1] and, by an abuse of notation, (∂μiμi)′=[∂μ′i1μi1, …, ∂μ′iTμiT].

Note that:

so that:

Therefore:

Hence we have shown that under (1), (8) and (9):

D′iΣi−=(∂pi(β0)∂β′)′Wi(β0)−1

Hence the asymptotic variance of n(β^PFE−β0) is equal to V_opt and from Proposition 1 this is equal to V0,β0, the efficiency bound for estimating β₀ consistently from (1).□