Hindawi Publishing Corporation
Mathematical Problems in Engineering
Volume 2012, Article ID 862398, 19 pages
doi:10.1155/2012/862398
Research Article
Hypothesis Testing in Generalized
Linear Models with Functional Coefficient
Autoregressive Processes
Lei Song,1 Hongchang Hu,1 and Xiaosheng Cheng2
1
2
School of Mathematics and Statistics, Hubei Normal University, Huangshi 435002, China
Department of Mathematics, Huizhou University, Huizhou 516007, China
Correspondence should be addressed to Hongchang Hu, retutome@163.com
Received 28 January 2012; Accepted 25 March 2012
Academic Editor: Ming Li
Copyright q 2012 Lei Song et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
The paper studies the hypothesis testing in generalized linear models with functional coefficient
autoregressive FCA processes. The quasi-maximum likelihood QML estimators are given,
which extend those estimators of Hu 2010 and Maller 2003. Asymptotic chi-squares
distributions of pseudo likelihood ratio LR statistics are investigated.
1. Introduction
Consider the following generalized linear model:
yt g xtT β εt ,
t 1, 2, . . . , n,
1.1
where β is d-dimensional unknown parameter, {εt , t 1, 2, . . . , n} are functional coefficient
autoregressive processes given by
ε1 η1 ,
εt ft θεt−1 ηt ,
t 2, 3, . . . , n,
1.2
where {ηt , t 1, 2, . . . , n} are independent and identically distributed random variable errors
with zero mean and finite variance σ 2 , θ is a one-dimensional unknown parameter, and
ft θ is a real valued function defined on a compact set Θ which contains the true value θ0 as
2
Mathematical Problems in Engineering
an inner point and is a subset of R1 . The values of θ0 and σ 2 are unknown. g· is a known
continuous differentiable function.
Model 1.1 includes many special cases, such as an ordinary regression model
when ft θ ≡ 0, gτ τ; see 1–7
, an ordinary generalized regression model when
ft θ ≡ 0; see 8–13
, a linear regression model with constant coefficient autoregressive
processes when ft θ θ, gτ τ; see 14–16
, time-dependent and function coefficient
autoregressive processes when gτ 0; see 17
, constant coefficient autoregressive processes when ft θ θ, gτ 0; see 18–20
, time-dependent or time-varying autoregressive processes when ft θ at , gτ 0; see 21–23
, and a linear regression model with
functional coefficient autoregressive processes when gτ τ; see 24
. Many authors have
discussed some special cases of models 1.1 and 1.2 see 1–24
. However, few people
investigate the model 1.1 with 1.2. This paper studies the model 1.1 with 1.2. The
organization of this paper is as follows. In Section 2, some estimators are given by the quasimaximum likelihood method. In Section 3, the main results are investigated. The proofs of
the main results are presented in Section 4, with the conclusions and some open problems in
Section 5.
2. The Quasi-Maximum Likelihood Estimate
Write the “true” model as
yt g xtT β0 et ,
e 1 η1 ,
t 1, 2, . . . , n,
et ft θ0 et−1 ηt ,
2.1
t 2, 3, . . . , n,
where g τ dgτ/dτ /
0, ft θ dft θ/dθ /
0. Define
we have
et j−1
t−1 j0
−1
i0 ft−i θ0 2.2
1, and by 2.2,
2.3
ft−i θ0 ηt−j .
i0
Thus et is measurable with respect to the σ−field H generated by η1 , η2 , . . . , ηt , and
Eet 0,
Varet σ02
j−1
t−1 j0
2
ft−i
θ0 .
2.4
i0
Assume at first that the ηt are i.i.d. N0, σ 2 , we get the log-likelihood of y2 , . . . , yn
conditional on y1 given by
n − 1 ln σ 2
Φn ln Ln −
−
2
n t2
εt − ft θεt−1
2σ 2
2
−
n − 1 ln 2π
.
2
2.5
Mathematical Problems in Engineering
3
At this stage we drop the normality assumption, but still maximize 2.5 to obtain QML
estimators, denoted by σn2 , βn , θn . The estimating equations for unknown parameters in 2.5
may be written as
n
2
n−1
1 ∂Φn
−
εt − ft θεt−1 ,
4
∂σ 2
2σ 2
2σ t2
2.6
n
∂Φn
1
2 ft θ εt − ft θεt−1 εt−1 ,
∂θ
σ t2
n
∂Φn
1
T
2
β xt−1 .
εt − ft θεt−1 · g xtT β xt − ft θg xt−1
∂βd×1 σ t2
2.7
Thus, σn2 , βn , θn satisfy the following estimation equations
σn2 n n 2
1 εt − ft θn εt−1 ,
n − 1 t2
2.8
εt − ft θn εt−1 ft θn εt−1 0,
2.9
T εt − ft θn εt−1 g xtT βn xt − ft θn g xt−1
βn xt−1 0,
2.10
t2
n t2
where
εt yt − g xtT βn .
2.11
Remark 2.1. If gxtT β xtT β, then the above equations become the same as Hu’s see 24
.
If ft θ θ, gxtT β xtT β, then the above equations become the same as Maller’s see 15
.
Thus we extend those QML estimators of Hu 24
and Maller 15
.
For ease of exposition, we will introduce the following notations, which will be used
T
later in the paper. Let d 1 × 1− vector ϕ βT , θ . Define
∂Φn
∂Φn ∂Φn
σ2
,
Sn ϕ σ 2
,
∂ϕ
∂β ∂θ
∂2 Φn
.
Fn ϕ −σ 2
∂ϕ∂ϕT
2.12
By 2.7, we have
⎛
⎜
Fn ϕ ⎜
⎝
Xn ϕ, ω
∗
⎞
U
n ft 2 θ
t2
ft θft θ
2
εt−1
−
ft θεt εt−1
⎟
⎟,
⎠
2.13
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Mathematical Problems in Engineering
where the ∗ indicates that the elements are filled in by symmetry,
Xn ϕ, ω −σ
U
n 2
∂2 Φn
∂β∂βT
,
T
T
ft θεt−1 g xtT β xt ft θεt g xt−1
β xt−1 − 2ft θft θεt−1 g xt−1
β xt−1 ,
t2
n T
∂2 Φn
1
T
T
T
T
g
x
x
g
x
x
−
β
x
−
f
β
x
β
x
−
f
β
x
θg
θg
t
t
t−1
t
t
t−1
t
t
t−1
t−1
σ 2 t2
∂β∂βT
n
T T
1
T
T
g
x
x
.
−
f
β
x
x
−
f
β
x
x
ε
θε
θg
t
t
t−1
t
t
t−1
t
t
t−1
t−1
σ 2 t2
2.14
Because {et−1 } and {ηt } are mutually independent, we have
⎛
Dn E Fn ϕ0
⎜
⎜
⎝
Xn ϕ0
0
⎞
0
n
2
ft 2 θ0 Eet−1
⎟
⎟
⎠
Xn ϕ0
0
0
Δθ0 , σ0 ,
2.15
t2
where
n T
T
T
g xtT β0 xt − ft θ0 g xt−1
β0 xt−1 g xtT β0 xt − ft θ0 g xt−1
β0 xt−1 ,
Xn ϕ0 t2
Δθ0 , σ0 n
2
2
ft θ0 Eet−1
t2
σ02
n
t2
2
ft θ0 j−1
t−2 j0
2
ft−i
θ
On.
i0
2.16
By 2.8 2.7 and Eηt 0, we have
σ02 E
n
T
∂Φn Eηt g xtT β0 xt − ft θ0 g xt−1
β0 xt−1 0,
∂β ββ0
t2
n
∂Φn 2
ft θ0 E ηt et−1 0.
σ0 E
∂θ θθ0
t2
2.17
3. Statement of Main Results
In the section pseudo likelihood ratio LR statistics for various hypothesis tests of interest
are derived. We consider the following hypothesis:
H1 : g·, f· are continuous functions, and f · /
0, σ02 > 0.
3.1
Mathematical Problems in Engineering
5
2
be the
When the parameter space is restricted by a hypothesis H0j , j 1, 2, . . . , let βjn , θjn , σjn
2
corresponding QML estimators of β, θ, σ , and let
jn −2Φn βjn , θjn , σ 2
L
jn
3.2
be minus twice the log-likelihood, evaluated at the fitted parameters. Also let
n −2Φn βn , θn , σn2 ,
L
jn − L
n
djn L
3.3
be the “deviance” statistic for testing H0j against H1 . From 2.5 and 2.8,
n n − 1 ln σn2 n − 11 ln 2π
L
3.4
jn n − 1 ln σ 2 n − 11 ln 2π.
L
jn
3.5
and similarly
In order to obtain our results, we give some sufficient conditions as follows.
A1 Xn nt2 xt xtT is positive definite for sufficiently large n and
lim max xtT Xn−1 xt O n−α ,
n → ∞ 1≤t≤n
∀α ∈
1
,1 ,
2
lim sup |λ|max Xn−1/2 Zn Xn−T/2 < 1, 3.6
n→∞
T
where Zn 1/2 nt2 xt xt−1
xt−1 xtT and |λ|max · denotes the maximum in absolute value
of the eigenvalues of a symmetric matrix.
A2 There is a constant α > 0 such that
j−1
t
j1
t−j−1
n
max
ft−i θ0 ≤ γ.
1≤j≤n
i0
tj1
2
ft−i
θ
≤ α,
i0
3.7
A3 ft θ dft θ/dθ /
0 and ft θ dft θ/dθ exist and are bounded, and g· is
≤
twice continuously differentiable, 0 < m ≤ maxu |g u| ≤ M < ∞, 0 < m
< ∞.
maxu |g u| ≤ M
Theorem 3.1. Assume 2.1, 2.2 and (A1)–(A3).
1 Suppose H01 : ft θ θ and gu is a continuous function, σ02 > 0 holds. Then
D
d1n −→ χ21 ,
n −→ ∞.
3.8
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Mathematical Problems in Engineering
2 Suppose H02 : ft θ θ, gu u, σ02 > 0 holds. Then
D
d2n −→ χ21 ,
n −→ ∞.
3.9
3 Suppose H03 : ft θ θ, gu eu /1 eu , σ02 > 0 holds. Then
D
d3n −→ χ21 ,
n −→ ∞.
3.10
4. Proof of Theorem
To prove Theorem 3.1, we first introduce the following lemmas.
Lemma 4.1. Suppose that (A1)–(A3) hold. Then, for all A > 0,
P
→ 0,
sup Dn−1/2 Fn ϕ Dn−T/2 − Φn −
n → ∞,
ϕ∈Nn A
4.1
where
Φn diag Id ,
n
2
ft 2 θ0 et−1
Δn θ0 , σ0 t2
,
T Nn A ϕ ∈ Rd1 : ϕ − ϕ0 Dn ϕ − ϕ0 ≤ A2 .
4.2
4.3
Proof. Similar to proof of Lemma 4.1 in Hu 24
, here we omit.
Lemma 4.2. Suppose that (A1)–(A3) hold. Then ϕn → ϕ0 , σn2 → σ02 and
Xn β∗ , β∗∗ , θn −→ Xn ϕ0 ,
4.4
where β∗ , β∗∗ are on the line of β0 and βn .
Proof. Similar to proof of Theorem 3.1 in Hu 24
, we easily prove that ϕn → ϕ0 , and σn2 →
σ02 . Since 4.4 is easily proved, here we omit the proof 4.4.
Proof of Theorem 3.1. Note that Sn ϕn 0 and Fn ϕn are nonsingular. By Taylor’s expansion,
we have
0 Sn ϕn Sn ϕ0 − Fn ϕn ϕn − ϕ0 ,
4.5
Mathematical Problems in Engineering
7
where ϕn aϕn 1 − aϕ0 for some 0 ≤ a ≤ 1. Since ϕn ∈ Nn A, also ϕn ∈ Nn A. By 4.1,
we have
n DnT/2 .
Fn ϕn Dn1/2 Φn A
4.6
P
n is a symmetric matrix with A
n −
Thus A
→ 0. By 4.5 and 4.6, we have
−1
n Dn−1/2 Sn ϕ0 .
DnT/2 ϕn − ϕ0 DnT/2 Fn−1 ϕn Sn ϕ0 Φn A
β
β
θ
4.7
θ
Let Sn ϕ, Fn ϕ denote Sn ϕ, Sn ϕ, and Fn ϕ, Fn ϕ, respectively. By 4.7, we have
β θ Φn DnT/2 βn − β0 , θn − θ0 Dn−1/2 Sn ϕ0 , Sn ϕ0 oP 1.
4.8
Note that
Φn DnT/2
⎛ T/2 ⎞
Xn ϕ0
0
⎜
⎟
⎜
⎟
n
2
2
⎜
⎟,
f
θ
e
0
t2 t
t−1 ⎠
⎝
0
Δn θ0 , σ0 ⎛
Dn−1/2
⎜
⎝
−1/2 Xn
ϕ0
0
0
1
⎞
4.9
⎟
⎠.
Δn θ0 , σ0 By 2.15, 4.2 and 4.8, we get
β XnT/2 ϕ0 βn − β0 Xn−1/2 ϕ0 Sn ϕ0 oP 1
n
T
ηt g xtT β0 xt − ft θ0 g xt−1
β0 xt−1 oP 1,
Xn−1/2 ϕ0
t2
4.10
n
2
θ 2
ft θ0 et−1
θn − θ0 Sn ϕ0 oP
Δn θ0 , σ0 t2
n
ft θ0 ηt et−1 oP
4.11
Δn θ0 , σ0 .
t2
Note that
εt yt − g xtT β g xtT β∗ xtT β0 − β et
4.12
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Mathematical Problems in Engineering
By 2.1, 2.11 and 4.12, we have
T
T
β0 − βn et − ft θn et−1 .
β∗∗ xt−1
εt − ft θn εt−1 g xtT β∗ xtT − ft θn g xt−1
4.13
By 4.13 and 2.10, we have
n n 2 T
T
εt − ft θn εt−1 εt − ft θn εt−1
g xtT β∗ xtT − ft θn g xt−1
β∗∗ xt−1
t2
t2
× β0 − βn et − ft θn et−1
n T
T
εt − ft θn εt−1 g xtT β∗ xtT − ft θn g xt−1
β0 − βn
β∗∗ xt−1
t2
n εt − ft θn εt−1 et − ft θn et−1
t2
n εt − ft θn εt−1 et − ft θn et−1 .
t2
4.14
By 4.13, we have
T
T
g xtT β∗ xtT − ft θn g xt−1
β0 − βn εt − ft θn εt−1 − et − ft θn et−1 .
β∗∗ xt−1
4.15
By 4.15, we have
n 2
T
T
g xtT β∗ xtT − ft θn g xt−1
β0 − βn
β∗∗ xt−1
t2
n n 2 2
εt − ft θn εt−1 et − ft θn et−1
t2
−2
t2
n εt − ft θn εt−1 et − ft θn et−1
t2
n n 2 2
et − ft θn et−1 −
εt − ft θn εt−1 .
t2
t2
4.16
Mathematical Problems in Engineering
9
By 4.14 and 4.16, we have
n n 2 2
εt − ft θn εt−1 et − ft θn et−1
t2
t2
n 2
T
T
g xtT β∗ xtT − ft θn g xt−1
β0 − βn
−
β∗∗ xt−1
.
4.17
t2
By 4.15, we have
n 2
εt − ft θn εt−1
t2
n n 2 2
T
T
et − ft θn et−1 g xtT β∗ xtT − ft θn g xt−1
β0 − βn
β∗∗ xt−1
t2
2
t2
n T
T
et − ft θn et−1
g xtT β∗ xtT − ft θn g xt−1
β0 − βn .
β∗∗ xt−1
t2
4.18
Thus, by 4.17 and 4.18, we have
n 2
T
T
g xtT β∗ xtT − ft θn g xt−1
β0 − βn
β∗∗ xt−1
t2
n T
T
et − ft θn et−1
g xtT β∗ xtT − ft θn g xt−1
β0 − βn
0.
β∗∗ xt−1
t2
4.19
Since ηt et − ft θ0 et−1 , we have
n n 2 2
et − ft θn et−1 ηt ft θ0 et−1 − ft θn et−1
t2
t2
n
n 2
2
ft θ0 − ft θn
et−1
2 ft θ0 − ft θn ηt et−1 .
ηt2 t2
t1
Thus, by 4.17, 4.20 and mean value theorem, we have
σn2 n − 1
n 2
εt − ft θn εt−1
t2
n
t1
ηt2 n 2
2
ft θ0 − ft θn
et−1
2 ft θ0 − ft θn ηt et−1
t2
4.20
10
Mathematical Problems in Engineering
−
n 2
T
T
g xtT β∗ xtT − ft θn g xt−1
β0 − βn
β∗∗ xt−1
t2
n
n
n
2 2
2
ηt2 θ0 − θn
ft θ et−1
2 θ0 − θn
ft θ et−1 ηt
t2
t1
−
t2
n 2
T
T
g xtT β∗ xtT − ft θn g xt−1
β0 − βn
β∗∗ xt−1
,
t2
4.21
where θ aθ0 1 − aθn for some 0 ≤ a ≤ 1.
It is easy to know that
βn − β0
T
Xn ϕ0 βn − β0
n
ηt Xn−1/2
2
T
β0 xt−1
ϕ0 g xtT β0 xt − ft θ0 g xt−1
4.22
op 1.
t2
By Lemma 4.2 and 4.22, we have
σn2 n − 1
n
n
n
2 2
2
ηt2 θ0 − θn
ft θ et−1
2 θ0 − θn
ft θ et−1 ηt
t2
t1
−
n
ηt Xn−1/2
β , β , θn
∗
∗∗
t2
2
T
g xtT β0 xt − ft θ0 g xt−1
β0 xt−1
oP 1
t2
n
n
n
2 2
2
ηt2 θ0 − θn
ft θ et−1
2 θ0 − θn
ft θ et−1 ηt
t2
t1
−
n
ηt Xn−1/2
t2
2
T
β0 xt−1
ϕ0 g xtT β0 xt − ft θ0 g xt−1
t2
oP 1.
4.23
Hence, by 4.11, we have
θn − θ0 n
t2 ft θ0 ηt et−1
n 2
2
t2 ft θ0 et−1
n
t2 ft θ0 ηt et−1
n 2
2
t2 ft θ0 et−1
Δn θ0 , σ0 oP n
2
2
t2 ft θ0 et−1
⎛
⎞
⎜
oP ⎝ n
t2
1
2
ft 2 θ0 et−1
⎟
⎠.
4.24
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11
By 4.24, we have
θ0 − θn
n
2 ft
2
n
2
2 θ0 − θn
ft θ et−1 ηt
θ et−1
t2
t2
⎛
⎞⎞2
n
f
e
θ
η
1
0
t
t−1
⎜
⎟⎟ 2 2
⎜
t
⎝ t2
o
f
θ et−1
⎝
⎠
⎠
P
t
n
n 2
2
2
2
t2
t2 ft θ0 et−1
f
θ
e
0
t2 t
t−1
⎛
n
⎛
⎞⎞
n
f
e
θ
η
1
⎜
⎟⎟ ⎜
t 0 t t−1
2⎝ t2
o
ft θ et−1 ηt oP 1
⎝
⎠
⎠
P
2
n
2
n
2
2
t2
t2 ft θ0 et−1
t2 ft θ0 et−1
⎛
⎛
⎜
⎝
n
⎛
n
t2 ft θ0 ηt et−1
n 2
2
t2 ft θ0 et−1
⎞⎞2
⎜
oP ⎝ 1
n
t2
2
ft 2 θ0 et−1
⎛
⎛
n
2 2
⎟⎟ ft θ0 o1 et−1
⎠⎠
t2
4.25
⎞⎞
n ft θ0 ηt et−1
1
⎜
⎟⎟
⎜
2⎝ t2
oP ⎝ ⎠⎠
n
2
2
2
n
2
f
t2 t θ0 et−1
f
θ
e
0 t−1
t2 t
·
n
ft θ0 o1 et−1 ηt oP 1
t2
n
2
t2 ft θ0 ηt et−1
n 2
2
t2 ft θ0 et−1
2
2 nt2 ft θ0 ηt et−1
−
oP 1
n 2
2
t2 ft θ0 et−1
2
t2 ft θ0 ηt et−1
n 2
2
t2 ft θ0 et−1
n
−
oP 1.
By Lemma 4.2, we have
n −
1
σn2
n
−
t1
−
2
t2 ft θ0 ηt et−1
n 2
2
t2 ft θ0 et−1
n
ηt2
n
ηt Xn−1/2
β , β , θn
∗
∗∗
2
T
g xtT β0 xt − ft θ0 g xt−1
β0 xt−1
oP 1
t2
n
−
t1
−
2
t2 ft θ0 ηt et−1
n 2
2
t2 ft θ0 et−1
n
ηt2
n
t2
ηt Xn−1/2
2
T
β0 xt−1
ϕ0 g xtT β0 xt − ft θ0 g xt−1
oP 1.
4.26
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Mathematical Problems in Engineering
Now, we prove 3.8. By 4.12, we have
∗
xtT β0 − β1n et .
εt 1 yt − g xtT β1n g xtT β1n
4.27
T
T
β∗∗ xt−1
εt − ft θ0 εt−1 g xtT β∗ xtT − ft θ0 g xt−1
β0 − β ηt .
4.28
Note that
From 4.28, we have
∗
T ∗∗
T
εt 1 − θ1n εt−1 1 g xtT β1n
xtT − θ1n g xt−1
β0 − β1n ηt .
β1n xt−1
4.29
By 2.8 and 2.10, we have
0
n εt 1 − θ1n εt−1 1
∗
T ∗∗
g xtT β1n
xt − θ1n g xt−1
β1n xt−1
t2
n ∗
T ∗∗
T
∗
T ∗∗
g xtT β1n
xtT − θ1n g xt−1
β0 − β1n g xtT β1n
xt − θ1n g xt−1
β1n xt−1
β1n xt−1
t2
n
∗
T ∗∗
xt − θ1n g xt−1
ηt g xtT β1n
β1n xt−1
t2
n
T
∗ ∗∗ ∗
T ∗∗
β0 − β1n X1n β1n
xt − θ1n g xt−1
, β1n , θ1n ηt g xtT β1n
β1n xt−1 .
t2
4.30
From 4.30, we obtain that
n
−1 ∗ ∗∗ T β1n − β0 X1n
ηt g xtT β1n xt − θ1n g xt−1
β1n xt−1 .
β1n , β1n , θ1n
t2
By 4.29, 4.31 and Lemma 4.2, we have
2
σ1n
n − 1
n εt 1 − θ1n εt−1 1
2
t2
n
T
∗ ∗∗ ηt2 β0 − β1n X1n β1n
, β1n , θ1n β0 − β1n
t1
n
T ∗
T ∗∗
xt − θ1n g xt−1
ηt g xtT β1n
2 β0 − β1n
β1n xt−1
t2
4.31
Mathematical Problems in Engineering
n
ηt2
−
n
∗ ∗∗ , β1n , θ1n
β1n
2
∗
T ∗∗
g xtT β1n
xt − θ1n g xt−1
β1n xt−1
op 1
t2
t1
n
−1/2
ηt X1n
13
ηt2
−
n
2
T
g xtT β0 xt − θ0 g xt−1
β0 xt−1
−1/2 ηt X1n
ϕ0
op .
t2
t1
4.32
By 3.3–3.5, we have
d1n
2 2 σ1n
σ1n
n − 1
− 1 oP 1.
L1n − Ln n − 1 ln
σn2
σn2
4.33
Under the H01 , and by 4.26, 4.32 and 4.33, we have
2
− σn2
n − 1 σ1n
σn2
2
n
t2 ηt et−1
2 n
2
σn t2 et−1
oP 1
2
t2 ηt et−1
oP 1.
2
σ02 nt2 et−1
n
4.34
It is easily proven that
n
ηt et−1
t2
n 2 −→ N0, 1.
σ0
t2 et−1
4.35
Thus, by 4.33–4.35, we finish the proof of 3.8.
Next we prove 3.9. Under H02 : ft θ θ, gu u, and yt xtT β0 et , we have
εt 2 yt − xtT β2n xtT β0 − xtT β2n et xtT β0 − β2n et .
4.36
Hence
T
β0 − β2n et−1
εt 2 − θ2n εt−1 2 xtT β0 − β2n et − θ2n xt−1
T
β0 − β2n ηt .
xtT − θ2n xt−1
4.37
By 2.8, 2.10, we have
0
n εt 2 − θ2n εt−1 2
xt − θ2n xt−1
t2
n xtT
t2
T
− θ2n xt−1
β0 − β2n
n
xt − θ2n xt−1 ηt xt − θ2n xt−1 .
t2
4.38
14
Mathematical Problems in Engineering
From 4.38, we obtain,
n
−1 ηt xt − θ2n xt−1 .
β2n − β0 X2n
θ2n
4.39
t2
Thus, by 4.37, 4.39 and Lemma 4.2, we have
2
σ2n
n − 1
n εt 2 − θ2n εt−1 2
2
t2
n
n
T
T ηt2 β0 − β2n X2n θ2n β0 − β2n 2 β0 − β2n
ηt xt − θ2n xt−1
t2
t1
n
ηt2
−
ηt xt − θ2n xt−1
T
−1
X2n
θ2n
n
t2
t1
n
n
ηt2
−
n
ηt xt − θ2n xt−1
t2
2
−1/2
ηt X2n
θ0 xt
− θ0 xt−1 op 1.
t2
t1
4.40
By 3.3–3.5, we have
d2n
2 2 σ2n
σ2n
L2n − Ln n − 1 ln
n − 1
− 1 oP 1.
σn2
σn2
4.41
Under the H02 , by 4.26, 4.40, and 4.41, we obtain
2
− σn2
n − 1 σ2n
σn2
2
n
t2 ηt et−1
2 n
2
σn t2 et−1
4.42
2
n
t2 ηt et−1
2 n
2
σ0 t2 et−1
oP 1
oP 1.
Thus, by 4.35, 4.42, 3.9 holds.
Finally, we prove 3.10. Under H03 , we have
T ∗
εt 3 yt −
ext β3n
1e
∗
xtT β3n
T ∗
ext β3n
T ∗
1 ext β3n
T
2 xt β0 − β3n et .
4.43
Mathematical Problems in Engineering
15
Thus
ext β3n
T
3n et
β
x
−
β
εt 3 − θ3n εt−1 3 0
t
2
T ∗
1 ext β3n
T ∗
ext−1 β3n
T
3n − θ3n et−1
β
− θ3n x
−
β
0
t−1
T ∗∗ 2
1 ext−1 β3n
∗∗
T
⎛
4.44
⎞
e
⎜
T
T ⎟
3n e
3n ηt .
β
⎝
x
−
θ
x
−
β
⎠
0
t−1
2 t
T ∗∗ 2
T ∗
1 ext β3n
1 ext−1 β3n
T ∗∗
β3n
xt−1
∗
xtT β3n
By 2.8 and 2.10, we have
0
n t2
⎛
⎜
εt 3 − θ3n εt−1 3 ⎝ T ∗
ext β3n
1e
∗
xtT β3n
T
∗∗
2 xt − θ3n ext−1 β3n
1e
T ∗∗
β3n
xt−1
⎞
⎟
2 xt−1 ⎠
⎛
⎞
T ∗∗
∗
n
β3n
xt−1
xtT β3n
e
⎜
T
T ⎟
3n e
3n
β
x
−
θ
x
−
β
⎝
⎠
0
t
t−1
2
T ∗∗ 2
T ∗
t2
1 ext β3n
1 ext−1 β3n
⎛
⎞
T ∗∗
β3n
xt−1
∗
xtT β3n
e
e
⎜
⎟
× ⎝
2 xt − θ3n xt−1 ⎠
T ∗∗ 2
∗
β3n
xt−1
xtT β3n
1e
1e
n
⎛
⎞
T ∗∗
β3n
xt−1
∗
xtT β3n
4.45
e
e
⎜
⎟
ηt ⎝ 2 xt − θ3n xt−1 ⎠
T ∗∗ 2
T ∗
t2
1 ext β3n
1 ext−1 β3n
⎛
⎞
T ∗∗
∗
n
T
β3n
xt−1
xtT β3n
e
e
⎜
⎟
∗ ∗∗ β0 − β3n X3n β3n
, β3n , θ3n ηt ⎝ 2 xt − θ3n xt−1 ⎠.
T ∗∗ 2
T ∗
β
x
β
x
t2
1 e t 3n
1 e t−1 3n
From 4.45, we obtain
−1 ∗ ∗∗ β3n , β3n , θ3n
β3n − β0 X3n
n
⎛
∗
xtT β3n
T ∗∗
β3n
xt−1
⎞
e
e
⎜
⎟
ηt ⎝ 2 xt − θ3n xt−1 ⎠.
T ∗∗ 2
T ∗
β
x
t2
1 ext β3n
1 e t−1 3n
4.46
16
Mathematical Problems in Engineering
By 4.44, 4.46 and Lemma 4.2, we have
2
σ3n
n − 1
n εt 3 − θ3n εt−1 3
2
t2
n
T
∗ ∗∗ ηt2 β0 − β3n X3n β3n
, β3n , θ3n β0 − β3n
t1
⎛
⎞
T ∗∗
∗
n
T β3n
xt−1
xtT β3n
e
e
⎜
⎟
2 β0 − β3n
ηt ⎝ 2 xt − θ3n xt−1 ⎠
T ∗∗ 2
T ∗
β
x
β
x
t2
1 e t 3n
1 e t−1 3n
n
⎛
n
⎜
−1/2 ∗ ∗∗ ηt2 − ⎝
ηt X3n
β3n , β3n , θ3n
t2
t1
⎛
T ∗∗
β3n
xt−1
∗
xtT β3n
⎞⎞2
e
e
⎜
T
T ⎟⎟
×⎝ 2 xt − θ3n xt−1 ⎠⎠
T ∗∗ 2
T ∗
1 ext β3n
1 ext−1 β3n
⎛
⎞⎞2
T
T
n
x
β
x
β
0
0
e t
e t−1
⎟⎟
⎜
−1/2 ⎜
ηt2 − ⎝ ηt X3n
ϕ0 ⎝ 2 xt − θ0 2 xt−1 ⎠⎠ op 1.
T
T
t2
t1
1 e xt β0
1 ext−1 β0
n
⎛
4.47
By 3.3–3.5, we know that
d3n
2 2 σ3n
σ3n
n − 1
− 1 oP 1.
L3n − Ln n − 1 ln
σn2
σn2
4.48
Under the H03 , by 4.26, 4.47 and 4.48, we have
2
− σn2
n − 1 σ3n
σn2
n
2
t2 ηt et−1
2 n
2
σ0 t2 et−1
oP 1.
4.49
Thus, 3.10 follows from 4.48, 4.49, and 4.35. Therefore, we complete the proof of
Theorem 3.1.
5. Conclusions and Open Problems
In the paper, we consider the generalized linear mode with FCA processes, which includes
many special cases, such as an ordinary regression model, an ordinary generalized regression
model, a linear regression model with constant coefficient autoregressive processes, timedependent and function coefficient autoregressive processes, constant coefficient autoregressive processes, time-dependent or time-varying autoregressive processes, and a linear
Mathematical Problems in Engineering
17
regression model with functional coefficient autoregressive processes. And then we obtain
the QML estimators for some unknown parameters in the generalized linear mode model and
extend some estimators. At last, we use pseudo LR method to investigate three hypothesis
tests of interest and obtain the asymptotic chi-squares distributions of statistics.
However, several lines of future work remain open.
1 It is well known that a conventional time series can be regarded as the solution to a
differential equation of integer order with the excitation of white noise in mathematics, and a
fractal time series can be regarded as the solution to a differential equation of fractional order
with a white noise in the domain of stochastic processes see 25
. In the paper, {εt } is a
conventional nonlinear time series. We may investigate some hypothesis tests by pseudo LR
method when the {εt } is a fractal time series the idea is given by an anonymous reviewer.
In particular, we assume that
p
ap−i Dvi εt ηt ,
5.1
i0
where vp , vp−1 , . . . , v0 is strictly decreasing sequence of nonnegative numbers, ai is a constant
sequence, and Dv is the Riemann-Liouville integral operator of order v > 0 given by
1
D ht Γv
v
t
t − uv−1 hudu,
5.2
0
where Γ is the Gamma function, and ht is a piecewise continuous on 0, ∞ and integrable
on any finite subinterval of 0, ∞ See 25, 26
. Fractal time series may have a heavytailed probability distribution function and has been applied various fields of sciences and
technologies see 25, 27–32
. Thus it is very significant to investigate various regression
models with fractal time series errors, including regression model 1.1 with 5.1.
2 We maybe investigate the others hypothesis tests, for example:
H04 : ft θ 0, gu u, σ02 > 0;
H05 : ft θ θ, gu 0, σ02 > 0;
H06 : ft θ 0, gu eu /1 eu , σ02 > 0;
H07 : ft θ at and gu is a continuous function, σ02 > 0;
H08 : ft θ at , gu u, σ02 > 0;
H09 : ft θ at , gu eu /1 eu , σ02 > 0.
Acknowledgments
The authors would like to thank the anonymous referees for their valuable comments which
have led to this much improved version of the paper. The paper was supported by Scientific
Research Item of Department of Education, Hubei no. D20112503, Scientific Research Item
of Ministry of Education, China no. 209078, and Natural Science Foundation of China no.
11071022, 11101174.
18
Mathematical Problems in Engineering
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