MPRA

In the representation of vec(H 0t ) obtained from (2), we replace vec(H 0t ) by vec(ε t ε ′ t)−ν t .
Then, we get
vec (ε t ε ′ t − E(ε t ε ′ t)) = ( ) (
A ⊗2
0 + B ⊗2
0 vec εt−1 ε ′ t−1 − E(ε t−1 ε ′ t−1) )
+ C ⊗2
0 vec ( x t−1 x ′ t−1 − Ex t−1 x ′ t−1
) ( )
+ νt − B ⊗2
0 ν t−1 .
Note that under assumption A5, the matrix I m 2 − A ⊗2
0 − B ⊗2
0 is inversible. Taking the
average of the two sides of the equality for t = 1, . . . , n gives
̂γ ε,n − γ ε,0 =L m (I m 2 − A ⊗2
0 − B ⊗2
0 ) −1 (I m 2 − B ⊗2
0 ) 1 n∑
n
+ L m (I m 2 − A ⊗2
0 − B ⊗2
0 ) −1 C ⊗2
0 D r
(̂γx,n − γ x,0
)
+ op (1), a.s.
t=1
ν t
We then have
⎛
√ ) ⎞
n
(̂θn − θ 0 √ ⎜ n (̂γεn − γ
⎝
ε ) ⎟
⎠
√ n (̂γxn − γ x )
o p(1)
=
⎛
⎞
−J −1 0 0
⎜ 0 N
⎝
1 N 2 ⎟
⎠
0 0 I r(r+1)/2
⎛
1
√ ⎝
n
The arguments for establishing the limiting distribution of
∑ n
t=1 Υ 0tvec ( η t η ′ t − I m
)
∑ n
t=1 vech(x tx ′ t − E(x t x t ))
⎛
1
√ ⎝
n
∑ n
t=1 Υ 0tvec ( η t η ′ t − I m
)
⎞
⎠ .
∑ n
t=1 vech(x tx ′ t − E(x t x t ))
⎞
⎠
being very similar to Lemma 6 in Thieu (2016), I just give a sketch of proof. For any
s > 0, the k-th individual volatility in (4) can be written σkt 2 = σ2 kts + σ2 kts , where
⎧ (
s∑ ⎨
m
) 2 (
σ 2 kts = b 2j
k
⎩ ω ∑
r∑
) ⎫
2⎬ kk + a kl ε l,t−j−1 + c ks x s,t−j−1
⎭
j=0
l=1
s=1
⎧ (
∞∑ ⎨
m
) 2 (
σ 2 kts = b 2j
k
⎩ ω ∑
r∑
) ⎫
2⎬ kk + a kl ε l,t−j−1 + c ks x s,t−j−1
⎭ .
s=1
j=s+1
l=1
Then we can write Υ 0t vec(η t η ′ t − I m ) = Y t,s + Y ∗ t,s, where
⎛ ( ) ( ) ⎞
Y t,s = ⎝ ∆ tsT m D −1
0tsH 1/2
0t,s ⊗ D −1
0tsH 1/2
0t,s
⎠ vec(η
H 1/2
0t,s ⊗ H 1/2
t η ′ t − I m ),
0t,s
with D 0ts = diag(σ 1ts , . . . , σ mts ) and ∆ ts = diag(∆ 1ts , . . . ∆ mts ), ∆ kts = 1
for k = 1, . . . , m and Y ∗ t,s is stationary and centered process satisfying
(∥ )
∥∥∥∥ n∑
lim lim sup P n −1/2 Y ∗ t,s
s→∞ n→∞
∥ > ɛ = 0.
23
t=1
σ 2 kts
∂σ 2 kts (θ(k) 0 )
∂θ (k) ,