MPRA

By the similar arguments used to show ii) in the proof of Theorem 1, we can conclude
that π = 0.
Proof of Theorem 2
Since
̂θ
(k)
n
strongly converges to θ (k)
0 which belongs to the interior of the parameter
t=1
˜Q
(k)
n
space Θ (k) , the derivative of the criterion is equal to zero at
1 ∑ n ∂ 2 l kt (θ (k) )
t=1
. Applying a Taylor expansion for Q (k)
n ∂θ (k) ∂θ (k)′ n
theorem gives
0 = 1 n∑ l kt (θ (k)
0 )
(¯θ(k)
) (̂θ(k)
+ J
n ∂θ (k) kkn n n
¯θ
(k)
n
̂θ
(k)
n
✷
̂θ
(k)
n . Let J kkn (θ (k) ) =
at θ (k)
0 and the mean-value
)
− θ (k)
0 , (29)
where is between and θ (k)
0 . By the points ii) and iii) in Lemma 2 and the
(k)
consistency of ̂θ (¯θ(k)
)
n , the matrix J kkn n is a.s. invertible for suciently large n. Hence
multiplying by √ n and solving for √ )
(̂θ(k)
n − θ (k)
0 gives
Let . l t (θ) =
have
√ n
(̂θ(k)
n
n
)
− θ (k) op(1)
0 = −J −1
kkn
(¯θ(k) n
) 1 √n
n∑
t=1
l kt (θ (k)
0 )
∂θ (k) . (30)
(
)
∂l 1t (θ (1) )
, . . . , ∂l ′
mt(θ (m) )
. Collecting all these Taylor expansions, we
∂θ (1)′ ∂θ (m)′
√ n
(̂θn − θ 0
) op(1)
= J −1 1 √ n
Note that η ∗ t = D −1
0t H 1/2
0t η t . From (26), we then have
Then, we get
.
(
l t (θ 0 ) = −∆ t T m vec η ∗ t η ∗′
t
(
= −∆ t T m D −1
0t H 1/2
0t
√ n
(̂θn − θ 0
) op(1)
= −J −1 1 √ n
n∑
t=1
We now introduce the martingale dierence
ν t = vec(ε t ε ′ t) − vec(H 0t ) =
n∑ .
l t (θ 0 ).
t=1
)
− D −1
0t H 0t D −1
0t
) ⊗2
vec
(η t η ′ t − I m
)
.
( ) ⊗2 )
∆ t T m D −1
0t H 1/2
0t vec
(η t η ′ t − I m . (31)
(
)
H 1/2
0t ⊗ H 1/2
0t vec(η t η ′ t − I m ).
22