We now prove the asymptotic normality of 9) under standard assumption moment. Then the first element of the vector in()satisfies , and about Et. Suppose that Et is i.i. d with mean zero, variance a finite fourth (1/V⑦∑et-NO by the Lindenberg- Levy CLt. For the second element of the vector in( 9), observe that((t/T)Et is a martin- gale difference sequence that satisfies the definition on p 13 of Ch. 4. Specifically, its variance Is 2=E(t/)2=a2(t2/T2 where (1/∑=(1/∑→2/3 Furthermore, to apply clt of a martingale difference sequence, we need to show that(1/)Et[(t/T)Et]2+02/3 as the condition(iii)on page 26 of Ch. 4 To prove this, notices that E(/m∑m2-(1m∑q)=E(/n∑m2-(/m∑m2 E((1/m)∑(t/)2(=2-a2) (1/)2∑(t/m)4E(e2-a2 t=1 E(e2-02)2(1/m∑+)→0 according to(6) and fourth moment of Et exist by assumption (1/)∑(t/m)2-(1/m)∑2m0 t=1 t=1 which also imply thatWe now prove the asymptotic normality of (9) under standard assumption about εt . Suppose that εt is i.i.d. with mean zero, variance σ 2 , and finite fourth moment. Then the first element of the vector in (9) satisfies (1/ √ T) Xεt L−→ N(0, σ 2 ) by the Linderberg-Le´vy CLT. For the second element of the vector in (9), observe that {(t/T)εt} is a martin￾gale difference sequence that satisfies the definition on p.13 of Ch. 4. Specifically, its variance is σ 2 t = E[(t/T)εt ] 2 = σ 2 · (t 2 /T2 ), where (1/T) X T t=1 σ 2 t = σ 2 (1/T3 ) X T t=1 t 2 → σ 2 /3. Furthermore, to apply CLT of a martingale difference sequence, we need to show that (1/T) PT t=1[(t/T)εt ] 2 p −→ σ 2/3 as the condition (iii) on page 26 of Ch. 4. To prove this, notices that E (1/T) X T t=1 [(t/T)εt ] 2 − (1/T) X T t=1 σ 2 t !2 = E (1/T) X T t=1 [(t/T)εt ] 2 − (1/T) X T t=1 (t/T) 2σ 2 !2 = E (1/T) X T t=1 (t/T) 2 (ε 2 t − σ 2 ) !2 = (1/T) 2X T t=1 (t/T) 4E(ε 2 t − σ 2 ) 2 = E(ε 2 t − σ 2 ) 2 1/T6X T t=1 t 4 ! → 0, according to (6) and fourth moment of εt exist by assumption. This imply that (1/T) X T t=1 [(t/T)εt ] 2 − (1/T) X T t=1 σ 2 t m.s −→ 0, which also imply that (1/T) X T t=1 [(t/T)εt ] 2 p −→ σ 2 /3. 10