Intercept Homogeneity Test for Fixed Effect Models under Cross-sectional Dependence: Some Insights

Appendix

Proof of Theorem 1

Here we will use maximum eigenvalue norm.

Result 1:β^FE is consistent under both H_0A and H_0B.

Proof:

β^FE=(X′MX)1X′MY=(X′MX)−1X′M(D¯θ+Xβ+ϵ)=β+(X′MX)−1X′MD¯θ+(X′MX)−1X′Mϵ.

from Assumption 1, we claim that plim(X′MX)/NT=R, i.e we claim that the second order moment of MX exists. Now note that, ||X′MX||=||X′MM′X||≤||M|| ||X′X||=||X′X||. Hence the matrix R is a finite matrix.

Note that, (X′MX)⁻¹X′MD̅=0, where D̅=[11111…1]′=1_NT. This is because D̅ is a linear combination of the columns of D and hence belongs to the column space of D and we know that MD=0.

Hence,

β^FE=β+(X′MX)−1X′Mϵ=β+[(X′MX)/NT]−1(X′Mϵ/NT).

Now we take plim on both sides and we have

plimβ^FE=β+plim[(X′MX)/NT]−1plim(X′Mϵ/NT)=β+[plim(X′MX)/NT]−1plim(X′Mϵ/NT)=β+R−1plim(X′Mϵ/NT).

Now limE(X′Mϵ/NT)=0 by Assumption 2. Again, by Assumption 5,

||V(X′Mϵ/NT)||=(1/N2T2)||E[X′M′IT⊗ΩMX]||≤1/N2T2E||X′M′IT⊗ΩMX||≤1/N2T2||Ω|| E||X′X||.

Hence lim(V(X′Mϵ/NT))→0 as T→∞, both under weak and strong dependence. For weak dependence, lim(V(X′Mϵ/NT))→0 as either T and / or N→∞.

So we have plimβ^FE=β under both the Null.

Result 2:β^FE is consistent under both H_1A and H_1B.

Proof:

β^FE=(X′MX)−1X′MY=(X′MX)−1X′M(Dμ+Xβ+ϵ)=β+(X′MX)−1X′MDμ+(X′MX)−1X′Mϵ.

Here (X′MX)⁻¹X′MD=0, since MD=0 and plim(X′MX)⁻¹X′Mϵ=0, by similar arguments as stated in the previous proof both under weak and strong dependence. For weak dependence, lim(V(X′Mϵ/NT))→0 as either T and/or N→∞.

Hence plimβ^FE=β under the alternative hypotheses.

Result 3:β^ols is consistent under both H_0A and H_0B.

Proof:

β^ols=(X′M¯X)−1X′M¯Y=(X′M¯X)−1X′M¯(D¯θ+Xβ+ϵ)=β+(X′M¯X)−1X′M¯D¯θ+(X′M¯X)−1X′M¯ϵ.

Now it can be easily seen from the form of M̅=I_NT–D̅(D̅′D̅)⁻¹D̅′ that M̅D̅=0.

Now, β^ols=β+(X′M¯X)−1X′M¯ϵ. It is also easy to see from Assumption 1 that W=plim[(X′M̅X)/NT] is a finite matrix.

plimβ^ols=β+plim(X′M¯X)−1X′M¯ϵ=β+plim(X′M¯X/NT)−1(X′M¯ϵ/NT)=β+W−1plim(X′M¯ϵ/NT).

Now limE(X′M̅ε/NT)=0, by Assumption 2. Again, by Assumption 5,

||V(X′M¯ϵ/NT)||=1/N2T2||EX(X′M¯′IT⊗ΩM¯X)||≤1/N2T2EX||X′M¯′IT⊗ΩM¯X||≤1/N2T2||Ω|| EX||X′X||.

Hence lim(V(X′M̅ϵ/NT))→0 as T→∞, both under weak and strong dependence. For weak dependence, lim(V(X′M̅ϵ/NT))→0 as either T and / or N→∞.

Hence plim(X′M̅ϵ/NT)=0. So we have plimβ^ols=β under both the Null.

Result 4:β^ols is inconsistent under both H_1A and H_1B.

Proof:

β^ols=(X′M¯X)−1X′M¯Y=(X′M¯X)−1X′M¯(Dμ+Xβ+ϵ)=β+(X′M¯X)−1X′M¯Dμ+(X′M¯X)−1X′M¯ϵ=β+(X′M¯X/NT)−1(X′M¯Dμ/NT)+(X′M¯X/NT)−1(X′M¯ϵ/NT).

plimβ^ols=β+plim[(X′M¯X/NT)−1(X′M¯Dμ/NT)]+plim[(X′M¯X/NT)−1(X′M¯ϵ/NT)]=β+W−1plim(X′M¯Dμ/NT)+W−1plim(X′M¯ϵ/NT).

Now plim(X′M̅ϵ/NT)=0 as we have already seen in Result 1. But plim(X′M¯Dμ/NT)=plim1N∑(X¯i−X¯¯)′(μi−μ¯)≠0, hence the proof.

Proof of Theorem 2

We will apply conditional Liapounov CLT. In both the cases, conditioning is on {X_t}. Define a matrix based on fourth moments and cross moments whose dimension is N²×N². It is given by VF=E((εtε′t)⊗(εtε′t)). Note, trace(VF)≥λmax(VF)≥∑i,jΩij2≥λmax(Ω2)=λmax2(Ω). Hence for strong dependence O(λ_max(V_F))=O(N²), whereas, for weak dependence, in general, O(λ_max(V_F))≥O(N).

Thus,

∑EX(l′dtεtε′td′tl)2≤∑(l′dtd′tl)2λmax(VF)

where

∑(l′dtd′tl)2=∑O(1NT2)2,=O(1N2T3).

For strong dependence, if the eigenvector corresponding to the largest eigenvalue of Ω belongs to the row-space of (∑d′tdt) then

(∑l′dtΩd′tl)2={(∑l′dtd′tl)O(λmax(Ω)}2,=O(λmax(Ω)NT)2,

and hence,

∑EX(l′dtεtε′td′tl)2(∑l′dtΩd′tl)2=O(1T).

For weak dependence,

(∑l′dtΩd′tl)2≥{(∑l′dtd′tl)(λmin(Ω)}2,=O(1NT)2,

since λ_min(Ω)=O(1).

Therefore,

∑EX(l′dtεtε′td′tl)2(∑l′dtΩd′tl)2=O(NT),

as λ_max(V_F)=O(N).

Proof of Theorem 3

To prove this theorem, let us first concentrate on ∑EX||dt(et−ϵt)||2.

∑EX||dt(et−ϵt)||2≤∑{||dtd′t|| EX||et−ϵt||2}.

Let us denote, for notational simplicity, X˜t as a N×k+1 matrix of all regressors including the intercept variable. Similarly, ψ is the k+1×1 vector of regression parameter including the intercept. First let us concentrate on the term ||e_t–ε_t||².

||et−ϵt||2=[{X˜t(ψ^−ψ)}′{X˜t(ψ^−ψ)}],={(ψ^−ψ)′(X˜′tX˜t)(ψ^−ψ)},≤{(ψ^−ψ)′(ψ^−ψ)}||(X˜′tX˜t)||e,=||(ψ^−ψ)||2||(X˜′tX˜t)||e,

where, ψ^ is the estimator of intercept and the β^FE. Note that ||dtd′t||=Op(1NT2), ||(X˜′tX˜t)||e=Op(N). For strong dependence, ||(ψ^−ψ)||2=Op(1T); and that for weak dependence is, Op(1NT), by Theorem 1.

Hence, for weak dependence,

∑EX||dt(et−ϵt)||2≤Op(1NT2)

and, for strong dependence,

∑EX||dt(et−ϵt)||2≤Op(1T2).

Thus, if ∑dtεtε′td′t converges so is ∑dtete′td′t and to the same matrix.

Now let us concentrate on EX|l′(∑dtϵtϵ′td′t)l−l′Vll′Vl|r for r≤2. Note that EX|l′(∑dtϵtϵ′td′t)l−l′Vll′Vl|r=1(l′Vl)rEX|∑(l′dt(ϵtϵ′t−Ω)d′tl)|r. By Burkholder inequality,

EX|l′(∑dtϵtϵ′td′t)l−l′Vll′Vl|r≤Cr(l′Vl)rEX(∑{l′dt(ϵtϵ′t−Ω)d′tl}2)r/2,≤Cr(l′Vl)r(EX∑{l′dt(ϵtϵ′t−Ω)d′tl}2)r/2,   as r/2≤1,=Cr(l′Vl)r(∑Var{l′dtϵtϵ′td′tl})r/2,≤Cr(l′Vl)r(∑EX{l′dtϵtϵ′td′tl}2)r/2,

where C_r is independent of T. For r=2, we have already shown in Theorem 2 that

Cr(l′Vl)r(∑EX{l′dtϵtϵ′td′tl}2)r/2={O(1T),for strong dependenceO(NT),for weak dependence.

Hence ∑dtϵtϵ′td′t→V in probability, conditionally on {X}, in the sense that,

l′(∑dtϵtϵ′td′t)l−l′Vll′Vl→0   in probability.

Therefore,

l′(∑dtete′td′t)l−l′Vll′Vl=l′V¯ll′Vl−1→0   in probability,

uniformly in l(l′l=1)

which implies ||V^−1/2V̅V^−1/2–I||_e→0 in probability.

Again,

||V||=||∑dtΩd′t||≤∑||dtd′t|| ||Ω||≤{Op(1NT) for weak dependence,Op(1T) for strong dependence.

Also,

EX||V¯||=EX||∑dtete′td′t|| ≤ ∑||dtd′t|| EX||ete′t||≤ ∑||dtd′t||{EX{||(ψ^−ψ)||2}||(X˜′tX˜t)||e+||Ω||}≤ {Op(1NT) for weak dependence,Op(1T) for strong dependence.

Hence by Theorem 2,

∥(β^FE−β^ols)′(V−1−V¯−1)(β^FE−β^ols)∥ =∥(β^FE−β^ols)′V−1/2(I−V1/2V¯−1V1/2)V−1/2(β^FE−β^ols)∥ ≤∥I−V1/2V¯−1V1/2∥e(β^FE−β^ols)′V−1(β^FE−β^ols) =∥V−1/2V¯V−1/2−I∥e×Op(1) =op(1)   for both weak and strong dependence.

Hence, MD and EMD have same limiting distribution.

Proof of Theorem 4

Let δμ=plim[(X′M¯X/NT)−1(X′M¯Dμ/NT)]=W−1plim 1N∑(X¯i−X¯¯)′(μi−μ¯) and δ^μ=(X′M¯X)−1X′M¯Dμ. Under Alternative, by Result 3 of Theorem 1, β^FE=β+(X′MX)−1X′Mϵ, and by Result 4 of Theorem 1, β^ols=β+δ^μ+(X′M¯X)−1X′M¯ϵ. Therefore, under Alternative,

β^FE−[β^ols−δ^μ]=∑t=1Tdtϵ.t,

as in the notation of equation (6).

Hence, (β^FE−β^ols+δ^μ)′V−1(β^FE−β^ols+δ^μ) is asymptotically χk2 as in Theorem 2.

Thus, as in Theorem 3, (β^FE−β^ols+δ^μ)′V¯−1(β^FE−β^ols+δ^μ) is asymptotically χk2. Therefore, (β^FE−β^ols)′V¯−1(β^FE−β^ols) is asymptotically non-central χk2 with the non-centrality parameter as δ′μV−1δμ.