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Appendix B: Proofs of Theorems 4.1, 4.2 and 4.5

Hereafter, for notational simplicity, we drop “*” from the symbols of underlying true values.

The following lemmas are used in the proofs of Theorems 4.1 and 4.2.

Lemma B.1 Under Assumptions A-D, (a) ˆPN−1ΛˆMΣˆee11

Proof of Lemma B.1. First, we consider (a). The left hand side is equal to Pˆ−1 1 which is bounded in norm by

C2Pˆ12[ 1 Moreover, by Corollary A.1(a), we have

1

together with Corollary A.1(b) and (B.1), we obtain (a).

Next, we consider (b). The left hand side can be written as Pˆ−1 1

which is bounded in norm by CPˆ−1[1

For part (c), the left hand side can be written as PˆN−1/2

which is bounded in norm by C2PˆN−1/22· by (A.5), this gives

1

Hence, expression in (B.2) is bounded by C2

which is further bounded by C2

The proofs of the remaining three parts are similar to those of the first three. The details are therefore omitted.

Lemma B.2 Under Assumptions A-D,

A≡(ˆΛ−Λ)MΣˆ−1eeMΛ ˆˆPN1 =Op(T−1/2) +Op(∥Λˆ−Λ∥2) +Op([1

Proof of Lemma B.2. Consider equation (A.16) in the proof of Proposition 4.1, we had shown A = op(1). So term AA is of a smaller order and hence negligible. With Lemma B.2 (a), (b) and (c), equation (A.16) can be simplified as

A+A =Op(T1/2) +op

([1 N

N i=1

σ2iσ2i)2]

1 2)

. (B.3)

By the identification condition, we know bothΛ(N1MΣee1M)ΛandΛˆ(N1MΣˆee1M)ˆΛare diagonal matrices, which implies

Ndg {

Λ(1

NMΣ−1eeM)Λ−Λˆ( 1

NMΣˆ−1eeM)ˆΛ }

= 0,

whereNdgdenotes the operator which sets the diagonal elements of its input to zeros. By adding and subtracting terms,

Ndg {

(ˆΛ−Λ)( 1

NMΣˆee1M)ˆΛ + ˆΛ(1

NMΣˆee1M)(ˆΛ−Λ) (B.4)

−(ˆΛ−Λ)( 1

NMΣˆee1M)(ˆΛ−Λ) + Λ[1

NM( ˆΣee1−Σee1)M]Λ }

= 0.

By Lemma A.2 (b), N1MΣˆ−1eeM = N1MΣ−1eeM+op(1) =R+op(1), where the last equation is due to Assumption C.3. So term(ˆΛ−Λ)(N1MΣˆee1M)(ˆΛ−Λ) =Op(∥Λˆ−Λ∥2). Given this result, together with Lemma A.2(a), we have

Ndg {

(ˆΛ−Λ)( 1

NMΣˆ−1eeM)ˆΛ + ˆΛ(1

NMΣˆ−1eeM)(ˆΛ−Λ) }

(B.5)

=Op(∥Λˆ−Λ∥2) +Op([1 N

N i=1

σ2iσ2i)2]1/2).

Notice that(ˆΛ−Λ)(N1MΣˆ−1eeM)ˆΛ = (ˆΛ−Λ)(N1MΣˆ−1eeM)ˆΛ ˆP−1Pˆ =AP, where the lastˆ inequality is due to the definition ofA. ByPˆ =P+op(1)from Corollary A.1 (a), we have

(ˆΛ−Λ)(1

NMΣˆ−1eeM)ˆΛ =AP +op(A).

According to the preceding result, we can rewrite (B.5) as Ndg{AP +P A}=Op(∥Λˆ −Λ∥2) +Op([1

N

N i=1

σi2σi2)2]1/2), (B.6) whereop(A) is discarded since it has an smaller order than other terms.

Now equation (B.3) has 12r(r+ 1)restrictions and equation (B.6) has 12r(r−1) restric-tions, ther×r matrixAcan be uniquely determined. Solving this linear equation system, we have

A=Op(T−1/2) +Op(∥Λˆ −Λ∥2) +Op([1 N

N i=1

σi2σi2)2]

1 2)

.

This completes the proof.

Proof of Theorem 4.1. We first consider the first order condition (3.4), which can be written as

diag{(Mzz−Σˆzz)−(Mzz−Σˆzz) ˆΣ−1eeMΛ ˆˆGΛˆMMΛ ˆˆGΛˆMΣˆ−1ee(Mzz−Σˆzz)}= 0, where “diag” denotes the diagonal operator and Gˆ = (Ir+ ˆΛMΣˆ−1eeMΛ)ˆ −1. By

Mzz =MΛΛM+ Σee+MΛ1 T

T t=1

ftet+ 1 T

T t=1

etftΛM+ 1 T

T t=1

(etet−Σee), with some algebra manipulations, we can further write the preceding equation as

ˆ

σi2σi2= 1 T

T t=1

(e2itσi2+ 2miΛ1 T

T t=1

fteit−2miΛ ˆˆGΛˆMΣˆee1MΛ1 T

T t=1

fteit

−2miΛ1 T

T t=1

ftetΣˆ−1eeMΛ ˆˆGΛˆmi−2miΛ ˆˆGΛˆMΣˆ−1ee 1 T

T t=1

[eteitE(eteit)] (B.7) +mi(ˆΛ−Λ)(ˆΛ−Λ)mi−2mi(ˆΛ−Λ)ˆΛmi+ 2mi(ˆΛ−Λ)ˆΛMΣˆ−1eeMΛ ˆˆGΛˆmi

+2miΛ(ˆΛ−Λ)MΣˆ−1eeMΛ ˆˆGΛˆmi+ 2σˆi2σi2 ˆ

σ2i miΛ ˆˆGΛˆmi.

By GˆPˆN = ˆPNGˆ =ING, we haveˆ Gˆ = (ING) ˆˆ PN−1 = ˆPN−1(ING). Then, the thirdˆ term on right hand side (ignoring the factor 2) is equal to

miΛ(Iˆ NG) ˆˆ PN−1ΛˆMΣˆ−1eeMΛ1 T

T t=1

fteit=miΛ(Iˆ NG)(Iˆ −A)1 T

T t=1

fteit (B.8) and the sum of the seventh and eighth terms is equal to−2mi(ˆΛ−Λ) ˆGΛˆmi. Define

ψ¨= 1 T

T t=1

ftetΣˆ−1eeMΛ ˆˆPN−1; ϕ¨= ˆPN−1ΛˆMΣˆ−1ee 1 T

T t=1

(etet−Σee) ˆΣ−1eeMΛ ˆˆPN−1. Now consider the sum of the fourth and ninth terms. ByGˆ = ˆPN1(ING), together withˆ the definitions ofψ, this term is equal to¨

−2miΛ1 T

T t=1

ftetΣˆee1MΛ ˆˆGΛˆmi+ 2miΛ(ˆΛ−Λ)MΣˆee1MΛ ˆˆGΛˆmi

=−2miΛ ¨ψ(ING)ˆˆ Λmi+ 2miΛA(ING)ˆˆ Λmi

= 2miΛ ¨ψGˆΛˆmi−2miΛAGˆΛˆmi−2miΛ ¨ψ(ˆΛ−Λ)mi+ 2miΛA(ˆΛ−Λ)mi +miΛ(A+Aψ¨−ψ¨mi.

Also, by (A.16), we have

A+A=AA+ ¨ϕ+ (IrA)ψ¨+ ¨ψ(IrA)PˆN−1ΛˆMΣˆ−1ee( ˆΣee−Σee) ˆΣ−1eeMΛ ˆˆPN−1, or equivalently

A+Aψ¨−ψ¨ =AA+ ¨ϕAψ¨−ψ¨APˆN1ΛˆMΣˆee1( ˆΣee−Σee) ˆΣee1MΛ ˆˆPN1.

Thus, it follows that

−2miΛ1 T

T t=1

ftetΣˆ−1eeMΛ ˆˆGΛˆmi+ 2miΛ(ˆΛ−Λ)MΣˆ−1eeMΛ ˆˆGΛˆmi (B.9)

= 2miΛ ¨ψGˆΛˆmi−2miΛAGˆΛˆmi−2miΛ ¨ψ(ˆΛ−Λ)mi+ 2miΛA(ˆΛ−Λ)mimiΛAmi

miΛ ¨ϕΛmi+ 2miΛAψΛ¨ mi+miΛ ˆPN−1ΛˆMΣˆ−1ee( ˆΣee−Σee) ˆΣ−1eeMΛ ˆˆPN−1Λmi. Using (B.8) and (B.9), we can rewrite (B.7) as

ˆ

σi2σi2 = 1 T

T t=1

(e2itσ2i)−2mi(ˆΛ−Λ)1 T

T t=1

fteit+ 2miΛ ˆˆG1 T

T t=1

fteit (B.10) + 2miΛAˆ 1

T

T t=1

fteit−2miΛ ˆˆGA1 T

T t=1

fteit+ 2miΛ ¨ψGˆΛˆmi

−2miΛAGˆΛˆmi−2miΛ ¨ψ(ˆΛ−Λ)mi+ 2miΛA(ˆΛ−Λ)mi

+miΛAmi−2miΛAψΛ¨ mi−2mi(ˆΛ−Λ) ˆGΛˆmi+ 2ˆσ2iσi2 ˆ

σi2 miΛ ˆˆGΛˆmi

+miΛ ¨ϕΛmimiΛ ˆPN−1ΛˆMΣˆ−1ee( ˆΣee−Σee) ˆΣ−1eeMΛ ˆˆPN−1Λmi

−2miΛ ˆˆGΛˆMΣˆ−1ee 1 T

T t=1

[eteitE(eteit)] +mi(ˆΛ−Λ)(ˆΛ−Λ)mi

=ai,1+ai,2+· · ·+ai,17, say.

By the Cauchy-Schwartz inequality, we have 1

N

N i=1

σi2σi2)2 ≤17 1 N

N i=1

(∥ai,12+· · ·+∥ai,172).

The first termN−1Ni=1a1i2 =Op(T−1) by E

[1 N

N i=1

1

T

T t=1

(e2itσ2i)2 ]

=O(T−1).

The second term is bounded in norm by 4C2∥Λˆ −Λ∥2 1

N

N i=1

1

T

T t=1

fteit2 =op(T1) byΛˆ−Λ =op(1)and

E [1

N

N i=1

1

T

T t=1

fteit

2

]

=O(T1).

Similarly, one can show that the 3rd, 4th, 5th, 6th, 8th, 11th and 14th terms are all op(T1). The 7th term is bounded in norm by

(4∥Λ∥2· ∥Λˆ∥2· ∥Gˆ∥2· ∥A2) 1 N

N i=1

mi4,

which isOp(N−2T−1) +Op(N−2Op(∥Λˆ−Λ∥4) +Op(N−2Op[N1 Ni=1σi2σi2)]byGˆ = Op(N−1),Λ = Λ+oˆ p(1)and Lemma B.2. This result can be simplified to N1 Ni=1ai,72 = op(T1) +op(∥Λˆ −Λ∥2) since Op(N2) ·Op[N1 Ni=1σi2σi2)] is of smaller order than

1 N

N

i=1σi2σi2)2. Similar to the 7th term, the 9th and 10th terms are both of the order op(T1) +op(∥Λˆ−Λ∥2). The 12th term isop(∥Λˆ−Λ∥2) byGˆ =Op(N1). The 13th term is of smaller order term than N1 Ni=1σ2iσi2) and therefore negligible. The 15th term is op(N1 Ni=1σ2iσi2)) by Lemma B.1 (f). The 16th term is Op(T−1). The last term is Op(∥Λˆ −Λ∥4). Given the above results, we have

1 N

N i=1

σi2σi2)2 =Op(T−1) +op(∥Λˆ−Λ∥2). (B.11) Next, we derive bounds for ∥Λˆ −Λ∥2. By equation (A.18), together with Lemma B.1(b), (d), (e) and (f) and Lemma B.2, we have

Λˆ−Λ =Op(T−1/2) +Op([1 N

N i=1

σi2σi2)2]1/2). (B.12) Substituting equation (B.12) into (B.11), we have N1 Ni=1σi2σi2)2 = Op(T1). This proves the second result of Theorem 4.1.

To prove the first result of Theorem 4.1, we need the following lemmas.

Lemma B.3 Under Assumptions A-D, we have (a) ˆPN−1ΛˆMΣˆee11

T

T t=1

(etet−Σee) ˆΣee1MΛ ˆˆPN−1

=Op(N1T1/2) +Op(N1/2T1) +Op(T3/2);

(b) ˆPN−1ΛˆMΣˆee11 T

T t=1

etft =Op(N1/2T1/2) +Op(T1);

(c) ˆPN1ΛˆMΣˆ−1ee( ˆΣee−Σee) ˆΣ−1eeMΛ ˆˆPN1=Op(N−1T−1/2);

(d) 1 T

T t=1

ftetΣˆee1MRˆN1 =Op(N1/2T1/2) +Op(T1);

(e) ˆPN1Λˆ(MΣˆ−1ee 1 T

T t=1

[etet−Σee] ˆΣ−1eeM)RˆN1

=Op(N−1T−1/2) +Op(N−1/2T−1) +Op(T−3/2);

(f) ˆPN−1ΛˆMΣˆ−1ee( ˆΣee−Σee) ˆΣ−1eeMRˆ−1N =Op(N−1T−1/2).

Proof of Lemma B.3. We first consider (a). We rewrite it as Pˆ−1Λˆ( 1

N2MΣˆ−1ee 1 T

T t=1

(etet−Σee) ˆΣ−1eeM)Λ ˆˆP−1.

Since we already know that ∥Pˆ−1∥ =Op(1) and ∥Λˆ∥ = Op(1), we only need to consider the term in the big parenthesis, which is

1

By the Cauchy-Schwarz inequality, one can show the first term is bounded in norm by C8(1 which isOp(T−3/2) by the second part of Theorem 4.1. The second term equals to

1 which is bounded in norm by

C4[ 1

Next, we consider (b). The left hand side of (b) is equivalent to Pˆ1Λˆ(1

Similarly to (a), it suffices to consider the term inside the parenthesis, which is 1

The first term isOp(N−1/2T−1/2). The second term is bounded in norm by C4[1

N

N i=1

σ2iσ2i)2]1/2[1 N

N i=1

1

T

T t=1

fteit

2]1/2

,

which isOp(T1) by the second part of Theorem 4.1. Hence. result (b) follows.

For part (c), the left hand side of (c) is equivalent to Pˆ1Λˆ( 1

N2MΣˆee1( ˆΣee−Σee) ˆΣee1M)Λ ˆˆP1. It suffices to consider the expression in the parenthesis:

1 N2

N i=1

mimiσˆ2iσ2i ˆ

σi4 ≤ 1 N

(1 N

N i=1

mi2)1/2(1 N

N i=1

mi2σ2iσ2i)2 ˆ σ8i

)1/2

,

which is Op(N−1T−1/2) by the second part of Theorem 4.1. This proves result (c). The proofs of results (d), (e) and (f) are similar to those of (a), (b) and (c). The details are therefore omitted.

Lemma B.4 Under Assumptions A-D, A≡(ˆΛ−Λ)MΣˆee1MΛ ˆˆPN1 =Op

( 1

N T )

+Op (1

T )

+Op(∥Λˆ−Λ∥2).

Proof of Lemma B.4. Consider equation (A.16). Using the results in Lemma B.3 and the fact that AA has an order smaller than that ofA and is therefore negligible, we have

A+A=Op ( 1

N T )

+Op (1

T )

. (B.13)

Now consider the term N1ΛM( ˆΣ−1ee −Σ−1ee)MΛ, which can be written as 1

NΛM( ˆΣee1−Σee1)MΛ =−Λ[1 N

N i=1

mimiσˆ2iσ2i ˆ σ2iσ2i

]Λ (B.14)

=−Λ[1 N

N i=1

mimiσˆ2iσ2i σi4

]Λ + Λ[ 1 N

N i=1

mimiσi2σi2)2 ˆ σ2iσ4i

]Λ.

The norm of the second expression on the right hand side of (B.14) is bounded by C 1

N

N i=1

σ2iσ2i)2 =Op(T1),

by the boundedness of miˆi2, σi2 by Assumptions C and D. Substituting (B.10) into the first expression on the right hand side of (B.14) and using the same arguments as we did at before (B.11), one can show that the first expression is Op(1

N T) +op(T1). Hence, we

have 1

NΛM( ˆΣee1−Σee1)MΛ =Op ( 1

N T )

+Op (1

T )

. (B.15)

Now consider (B.4). Using the same arguments as in the derivation of (B.6) except that the result for N1ΛM( ˆΣ−1ee −Σ−1ee)MΛis given by (B.15) instead of op([N1 Ni=1σ2iσ2i)2]1/2), we have

Ndg{AP +P A}=Op ( 1

N T )

+Op (1

T )

+Op(∥Λˆ −Λ∥2). (B.16) Solving the equation system (B.13) and (B.16), we have

A=Op ( 1

N T )

+Op (1

T )

+Op(∥Λˆ −Λ∥2), as asserted in this lemma. This proves Lemma B.4.

Proof of Theorem 4.1 (continued). Using the results in Lemma B.3 and Lemma B.4 and noticing that∥Λˆ−Λ∥2 is of smaller order thanΛˆ−Λ and therefore negligible, we have from (A.18)

Λˆ−Λ =Op ( 1

N T )

+Op (1

T )

,

as asserted by the first result of Theorem 4.1. This completes the proof of Theorem 4.1.

Corollary B.1 Under Assumptions A-D,

A≡(ˆΛ−Λ)MΣˆ−1eeMΛ ˆˆPN−1 =Op ( 1

N T )

+Op (1

T )

.

Corollary B.1 is a direct result of Lemma B.4 and Theorem 4.1.

Lemma B.5 Under Assumptions A-D, (a) 1

T

T t=1

ftetΣˆ−1eeMRˆ−1N = 1 T

T t=1

ftetΣ−1eeM R−1N +Op(N−1/2T−1) +Op(T−3/2);

(b) ˆPN−1ΛˆMΣˆ−1ee 1 T

T t=1

etft=PN−1ΛMΣ−1ee 1 T

T t=1

etft+Op(N−1/2T−1) +Op(T−3/2);

(c) 1

NM( ˆΣee1−Σee1)M =− 1 N T

N i=1

T t=1

1

σi4mimi(e2itσ2i) + 1 N T

N i=1

mimiκi,4σi4 σi4 +Op(N−1T−1/2) +Op(N−1/2T−1) +Op(T−3/2).

Proof of Lemma B.5. Equation (B.10) can be written as ˆ

σi2σi2 = 1 T

T t=1

(e2itσ2i) +Ri, (B.17) where

Ri =−2miΛ ˆˆGΛˆMΣˆee11 T

T t=1

[eteitE(eteit)] +Si

with

Si=−2mi(ˆΛ−Λ)1 T

T t=1

fteit+ 2miΛ ˆˆG1 T

T t=1

fteit+ 2miΛAˆ 1 T

T t=1

fteit−2miΛ ˆˆGA1 T

T t=1

fteit

+ 2miΛ ¨ψGˆΛˆmi−2miΛAGˆΛˆmi−2miΛ ¨ψ(ˆΛ−Λ)mi+ 2miΛA(ˆΛ−Λ)mi arguments in the derivation of (B.10), we have

1 which is bounded in norm by

C2∥Λˆ∥4· ∥GˆN2· 1 which, by the Cauchy-Schwarz inequality, is bounded by

2 1 The first expression isOp(N−1T−1). The second expression is bounded by

C10

Given the above result, we have 1

This result, together with (B.18), gives 1

Notice that

The second term can be written as 1

The second term of the above equation is bounded in norm by C5

We now consider (c). The left hand side of (c) is equal to

−1

We usei1 andi2to denote the two expressions on the right hand side of the above equation.

We first consideri1. Substituting (B.17) into this term, we obtain i1=−1

Consider the second expression. The(v, u) element of this expression (v, u= 1, . . . , k) is B.3(a). The third term is bounded by

C6 by (B.18). Hence, we have

i1 =− 1

Proceed to consideri2. By ˆ We analyze the three terms at right-hand-side of the above equation one by one. The second term is bounded in norm by

2C8

by (B.19). Finally, the first term can be written as 1 The first term of the above expression is equal to

1

The second term is bounded in norm by C10

Hence, we have

i2 = 1 N T

N i=1

κi,4σi4

σ6i mimi+Op ( 1

N T )

+Op ( 1

T3/2 )

.

Summarizing the results oni1 and i2, we have (c).

Proof of Theorem 4.2. We first derive the asymptotic behavior ofA. Consider equation (A.16), using Lemma B.3 (a) and (f), Lemma B.5 (b) and Lemma B.4, we have

A+A =η+η+Op(N−1T−1/2) +Op(N−1/2T−1) +Op(T−3/2), where

η= 1 N T

T t=1

ftetΣ−1eeMΛP−1.

Let vech(B) be the operation which stacks the elements on and below the diagonal of matrixB into a vector, for any square matrixB. Takingvechoperation on both sides, we get

vech(A+A) = vech(η+η) +Op(N1T1/2) +Op(N1/2T1) +Op(T3/2).

LetDr be ther-dimensional duplication matrix andD+r be its Moore-Penrose inverse. By the basic fact that vech(B+B) = 2D+rvec(B), for anyr×r matrixB, we have

2Dr+vec(A) = 2D+rvec(η) +Op(N−1T−1/2) +Op(N−1/2T−1) +Op(T−3/2). (B.21) Furthermore, define

ζ = Λ[ 1 N T

N i=1

T t=1

mimi

σi4 (e2itσ2i)]Λ, µ= Λ[ 1 N T

N i=1

κi,4σi4

σ6i mimi]Λ.

Proceed to consider equation (B.4). By Lemma B.5(c) and Λˆ −Λ = Op(N1/2T1/2) + Op(T−1) by Theorem 4.1, we have

Ndg{Λˆ(1

NMΣˆ−1eeM)(ˆΛ−Λ) + (ˆΛ−Λ)( 1

NMΣˆ−1eeM)ˆΛ}

=Ndg{ζµ}+Op(N1T1/2) +Op(N1/2T1) +Op(T3/2).

Using the same arguments in the derivation of (B.16), we have

Ndg(AP +P A) = Ndg(ζ−µ) +Op(N1T1/2) +Op(N1/2T1) +Op(T3/2).

Let veck(B) be the operation which stacks the elements below the diagonal of matrix B into a vector, for any square matrixB. Let Dbe the matrix such thatveck(B) =Dvec(B) for anyr×r matrixB. By the preceding equation,

veck(AP +P A) = veck(ζ−µ) +Op(N1T1/2) +Op(N1/2T1) +Op(T3/2),

or equivalently

Dvec(AP +P A) =Dvec(ζ−µ) +Op(N1T1/2) +Op(N1/2T1) +Op(T3/2).

Usingvec(ABC) = (CA)vec(B), we can rewrite the preceding equation as

D[(P⊗Ir)+(IrP)Kr]vec(A) =Dvec(ζ−µ)+Op(N1T1/2)+Op(N1/2T1)+Op(T3/2), (B.22) where Kr is the r-dimensional communication matrix such that Krvec(B) = vec(B) for any r×r matrixB. By (B.21) and (B.22), we have

The above result can be rewritten as

D1vec(A) =D2vec(η)+D3vec(ζ)−D3vec(µ)+Op( 1

Given the above three results, we can rewrite (B.24) as vec(A) =D1

Consider equation (A.18). Using the results of Lemma B.5 (a) and (b) and Lemma B.3

Taking vectorization operation on the both sides of (B.26), we have vec(ˆΛ−Λ) =[Kkr[(P−1Λ)⊗Λ] +R−1Ir] 1 Substituting (B.25) into (B.27),

vec(ˆΛ−Λ) =B1 1

Given the above results and by a Central Limit Theorem, we obtain as N, T → ∞ and N/T2 →0,

sqrtN T[vec(ˆΛ−Λ)− 1

T]−→d N(0,Ω), whereΩ = lim

N→∞N with

N =B1(R⊗Ir)B

1+B2[ 1 N

N i=1

κi,4σi4

σi8 (mimi)⊗(mimi)]B

2.

This completes the proof of Theorem 4.2.

Proof of Theorem 4.5. By the definition of fˆt= (ˆΛMΣˆ−1eeMΛ)ˆ −1ΛˆMΣˆ−1eezt and A, we have

fˆtft=−Aft+ ˆP−1 1

NΛˆMΣˆ−1eeet. From Corollary B.1, we know A = Op(1

N T) +Op(T1), then the first term of the above equation is Op(1

N T) +Op(T1). From Corollary A.1 (a)(b), we know Pˆ = P +op(1) and Pˆ =Op(1), and from Assumption C.3, we knowP= lim

N→∞P wherePis positive definite matrix. Consider the part N1ΛˆMΣˆ−1eeet, which can be rewritten as

1 N

N i=1

1 ˆ

σi2Λˆmieit= 1

NΛMΣ−1eeet,−1 N

N i=1

ˆ σi2σi2

ˆ

σi2σi2 Λmieit+ 1 N

N i=1

1 ˆ

σ2i(ˆΛ−Λ)mieit, wheremi is the transpose of the ith row of M. Usea1, a2, a3 to denote the three terms on the right hand side of the above equation. Terma2can be shown to beOp(1

N T)+Op(T31/2) by the equation (B.10). Terma3can be shown to beOp(1

N T)+Op(T1)by equation (A.18).

Then we have 1

NΛˆMΣˆ−1eeet= 1

NΛMΣ−1eeet+Op ( 1

N T )

+Op (1

T )

.

Therefore,

fˆtft=P−1 1

NΛMΣ−1eeet+Op ( 1

N T )

+Op (1

T )

Based on the above result, by a Central Limit Theorem, we obtain as N, T → ∞ and N/T2 →0,

N( ˆftft)−→d N(0, P−1).

This completes the proof of Theorem 4.5.

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