1. No category

Addendum to Model Specification Testing of
Time Series Regressions
Herman J. Bierens
June 7, 2015
Abstract
In this addendum to
Bierens, H. J. (1984): ”Model Specification Testing of Time Series Regressions”, Journal of Econometrics 26, 323-353
I will derive the limiting null distribution of the proposed Weighted Integrated Condition Moment (WICM) test involved and show how to derive
upper bounds of the critical values of a standardized version of the WICM
test. Moreover, I will show how to approximate the critical values of the
WICM test via a bootstrap method. Furthermore, I will give a more formal
proof of the consistency of the WICM test. Also, I will solve the problem
how to standardize the conditioning variables before applying a bounded
transformation such that all the asymptotic results carry over. This was
one of the unsolved problems in Bierens (1984).
1. Introduction
In this addendum to Bierens (1984) [B84 hereafter] I will derive the limiting null
distribution of the test involved and show how to approximate the latter via a
bootstrap method, for possibly nonlinear strictly stationary time series regression
models. As in B84, I will focus on the nonlinear ARX model
Yt = f (Yt−1 , Yt−2 , ..., Yt−p , Xt−1 , Xt−2 , ..., Xt−q , θ0 ) + Ut
= ft−1 (θ0 ) + Ut , say,
(1.1)
(1.2)
where (1.2) is merely a short-hand notation for (1.1), and
Assumption 1.1.
(a) Zt = (Yt , Xt0 )0 ∈ R × Rs is a strictly stationary vector time series process, with
E[Yt2 ] < ∞ and Xt is a vector of exogenous variables;
(b) θ0 is a parameter vector contained in a given compact parameter space Θ ⊂ Rm ;
(c) f (v, θ) is a given real function on Rp+q.s × Θ which for each v ∈ Rp+q.s is
continuous in θ ∈ Θ, and for each θ ∈ Θ Borel measurable in v ∈ Rp+q.s , hence
for each θ ∈ Θ the sequence ft−1 (θ) is a well-defined strictly stationary time series
process;
(d) supθ∈Θ E[ft−1 (θ)2 ] < ∞,
and Ut is the error term.
In general, time series regression models aim to represent the best one-step
ahead forecasting scheme, in the sense that given the entire past of the time series
involved up to time t − 1, the mean square forecast error is minimal. For the
model (1.1) this is the case if and only if the error process Ut is a martingale
difference sequence w.r.t. the σ-algebra
¡
¢
t
F−∞
= σ {Zt−j }∞
j=0
generated by the sequence {Zt−j }∞
j=0 , i.e.,
£
¤
t
t−1
and E Ut |F−∞
Ut is measurable F−∞
= 0 a.s.
(1.3)
More formally, using the short-hand notation (1.2), the null hypothesis to be
tested is that
¡ £
¢
¤
t−1
= ft−1 (θ0 ) = 1, (1.4)
H0 : There exists a θ0 ∈ Θ such that Pr E Yt |F−∞
which is equivalent to the hypothesis that Ut = Yt − ft−1 (θ0 ) is a martingale
difference process, against the alternative hypothesis that H0 is false, i.e.,
¡ £
¢
¤
t−1
(1.5)
H1 : For all θ ∈ Θ, Pr E Yt |F−∞
= ft−1 (θ) < 1.
Note that if we define θ0 as
£
¤
θ0 = arg min E (Yt − ft−1 (θ))2
(1.6)
θ∈Θ
n h¡
i
h
io
£
¢2
¡ £
¤¢
¤
t−1 2
t−1
= arg min E Yt − E Yt |F−∞
+ E E Yt |F−∞
− ft−1 (θ)
θ∈Θ
h¡ £
¢2 i
¤
t−1
= arg min E E Yt |F−∞ − ft−1 (θ)
θ∈Θ
2
regardless whether H0 is true or not, then H0 and H1 become
¡ £
¢
¤
t−1
= ft−1 (θ0 ) = 1,
H0 : Pr E Yt |F−∞
¡ £
¢
¤
t−1
= ft−1 (θ0 ) < 1,
H1 : Pr E Yt |F−∞
(1.7)
(1.8)
Lemma 1.1. Under Assumption 1.1,
£
£
¤
¤
t−1
t−1
E Yt |F−∞
= lim E Yt |Ft−k
a.s.,
(1.9)
respectively.
To determine which one of the two hypotheses is true, we need to condition
on the one-sided infinite vector time series {Zt−j }∞
j=1 , which in practice is not
possible because Zt is usually only observed from a particular time t0 onwards.
Part of the solution to this problem is the following lemma.
k→∞
¡
¢
t−1
where Ft−k
= σ {Zt−j }kj=1 is the σ-algebra generated by Zt−1 , Zt−2 , ..., Zt−k .
Consequently, under H1 and with θ0 defined by (1.6) there exists an k0 ∈ N such
that for all t,
¡ £
¢
¤
t−1
sup Pr E Yt |Ft−k
(1.10)
= ft−1 (θ0 ) < 1.
k≥k0
Proof. The result (1.9) is well-known. See for example Theorem 9.4.8 in Chung
(1974) or Theorem 3.12 in Bierens (2004). As to part (1.10), let
¯ £
¯
¯ £
¯
¤
¤
t−1
t−1
Vt = ¯E Yt |F−∞
− ft−1 (θ0 )¯ , Vt,k = ¯E Yt |Ft−k
− ft−1 (θ0 )¯ .
d
Then by (1.9), limk→∞ Vt,k = Vt a.s., which implies that Vt,k → Vt for k → ∞ and
thus, by definition of convergence in distribution,
lim Pr [Vt,k > ε] = Pr [Vt > ε]
k→∞
(1.11)
for all continuity points ε of the distribution function of Vt . Moreover, under the
alternative hypothesis (1.8), Pr [Vt > 0] > 0, which implies that there exists a
continuity point ε > 0 of the distribution function of Vt such that Pr [Vt > ε] > 0
as well.1 It follows now from (1.11) that
lim inf Pr [Vt,k > 0] ≥ lim Pr [Vt,k > ε] = Pr [Vt > ε] > 0,
k→∞
1
k→∞
Because Pr [Vt > 1/n] ↑ Pr [Vt > 0] monotonically as n → ∞.
3
hence
¡ £
¢
¤
t−1
lim sup Pr E Yt |Ft−k
= ft−1 (θ0 ) = lim sup Pr [Vt,k = 0] < 1.
k→∞
(1.12)
k→∞
Since by the strict stationarity condition in Assumption 1.1 the probabilities in
(1.12) do not depend on t, the result (1.10) follows.
Now let Φ : Rs+1 → Rs+1 be a bounded Borel measurable one-to-one mapping
with Borel measurable inverse. Then
¡
¢
¡
¢
t−1
= σ {Zt−j }kj=1 = σ {Φ(Zt−j )}kj=1 ,
Ft−k
hence by Theorem 1 in Bierens (1982) and its generalization in Bierens (2015a,
Theorem 2.1),
¡ £
¢
¤
t−1
(1.13)
= ft−1 (θ0 ) < 1
H1 (k) : Pr E Yt |Ft−k
implies that the set
½
Sk =
τ = (τ10 , τ20 , ....τk0 )0 ∈ R(s+1)k :
"
à k
!#
)
X
E (Yt − ft−1 (θ0 )) exp i.
τj0 Φ(Zt−j )
=0
j=1
has Lebesgue measure zero and is nowhere dense. Consequently, denoting
"
à k
!#
X
τj0 Φ(Zt−j )
(1.14)
ψk (τ1 , τ2 , ....τk ) = E (Yt − ft−1 (θ0 )) exp i.
j=1
the following result holds.
Lemma 1.2. Assume that the conditions of Lemma 1.1 hold. Let Υ be a compact subset of Rs+1 with positive Lebesgue measure and let µ be an absolutely
continuous probability measure on Υ. Then
½
Z
> 0 under H1 (k),
2
|ψk (τ1 , τ2 , ....τk )| dµ(τ1 )dµ(τ2 )...dµ(τk )
= 0 under H0 .
Υk
4
Given that Zt is observed for 1 − t0 ≤ t ≤ n, where t0 = max(p, q), the
empirical counter-part of ψk (τ1 , τ2 , ....τk ) defined in (1.14) is
⎛
⎞
min(k,t−t0 )
n
X
X
e k,n (τ1 , τ2 , ....τk ) = 1
(Yt − ft−1 (b
θn )) exp ⎝i.
τj0 Φ(Zt−j )⎠ (1.15)
ψ
n t=1
j=1
for n > k + t0 ,where
1X
b
θn = arg min
(Yt − ft−1 (θ))2
θ∈Θ n
t=1
n
is the NLLS estimator of θ0 .
However, for the time being I will assume all the lagged Zt ’s are observed, so
that
à k
!
n
X
X
1
b k,n (τ1 , τ2 , ....τk ) =
(Yt − ft−1 (b
θn )) exp i.
τj0 Φ(Zt−j )
(1.16)
ψ
n t=1
j=1
will be treated as the empirical counter-part of ψk (τ1 , τ2 , ....τk ). In section 7 I will
set forth conditions under which the asymptotic results below on the basis of
(1.16) are the same as for (1.15).
In order to keep this addendum in tune with the addendums to Bierens
(1982), Bierens and Ploberger (1997) and Bierens and Wang (2012), i.e., Bierens
√ b
(2014,2015a,b), I will from this point onwards denote nψ
k,n (τ1 , τ2 , ....τk ) by
c
Wk,n (τ1 , τ2 , ....τk ). Thus,
à k
!
n
X
X
1
ck,n (τ1 , τ2 , ....τk ) = √
(Yt − ft−1 (b
θn )) exp i.
τj0 Φ(Zt−j ) .
(1.17)
W
n t=1
j=1
In this addendum to B84 I will set forth further conditions such that under
H0 the following results hold.
• For each k ∈ N,
bn,k def.
B
=
d
def.
→ Bk =
Z
k
ZΥ
Υk
¯2
¯
¯
¯c
¯Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk )
(1.18)
|Wk (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk ),
where Wk (τ1 , τ2 , ....τk ) is a complex-valued zero-mean Gaussian process on
Υk ;
5
• for any positive sequence γk satisfying
Ln of n satisfying limn→∞ Ln = ∞,
Ln
X
k=1
d
bn,k →
γk B
P∞
k=1
∞
X
γk < ∞ and any subsequence
γk Bk ;
k=1
P
• the critical values of ∞
k=1 γk Bk can be approximated by a bootstrap method,
similar to Bierens (2015a).
Moreover, under H1 the following results hold.
• For each k ∈ N,
p
def.
bn,k /n →
ηk =
B
Z
Υk
|ψk (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk )
(1.19)
where ηk > 0 for all but a finite number of k’s,
• and
The statistic
n
X
1X
p
bn,k →
γk B
γk ηk > 0
n k=1
k=1
L
Tbn =
∞
Ln
X
k=1
bn,k
γk B
will be referred to as the test statistic of the Weighted Integrated Conditional
Moment (WICM) test.
2. Nonlinear least squares
2.1. Consistency
In B84 the consistency of the NLLS estimator is b
θn derived under data heterogeneity. However, under the current strict stationarity condition in Assumption
1.1 the following conditions suffice.
Assumption 2.1.
6
(a) The vector time series process Zt in Assumption 1.1 (a) has a vanishing memt
ory, in the sense that the sets in its remote σ-algebra F−∞ = ∩t F−∞
have either
probability zero or one;
(b) E[supθ∈Θ ft−1 (θ)2 ] < ∞;
(c) The solution θ0 of (1.6) is unique.
See Bierens (2008, Ch. 7) for the motivation of the vanishing memory concept.
In particular this concept plays a key-role in the following uniform weak law of
large numbers (UWLLN).
See Bierens (2008, Ch. 7) for the motivation of the vanishing memory concept.
In particular this concept plays a key-role in the following uniform weak law of
large numbers (UWLLN).
Lemma 2.1. Let Vt ∈ Rk be a strictly stationary time series process with vanishing memory, and let Θ be a compact subset of a Euclidean space. Let g(v, θ) be a
real or complex-valued function on Rk × Θ satisfying the following conditions.
(1) For each θ ∈ Θ, g(v, θ) is Borel measurable in v ∈ Rk ;
(2) For each v ∈ Rk , g(v, θ) is continuous in θ ∈ Θ;
(3) E [supθ∈Θ |g(V1 , θ)|]¯ <P
∞.
¯
1
¯
Then p limn→∞ supθ∈Θ n nt=1 g(Vt , θ) − E[g(V1 , θ)]¯ = 0.
Proof. Similar to Theorem 7.8(a) in Bierens (2004, p. 187).
This lemma mimics the uniform strong law of large number of Jennrich (1969),
with Kolmogorov’s strong law of large numbers replaced by the following weak
version for time series.
Lemma 2.2. Let Vt ∈ Rk be a strictly stationary time series process with vanishing memory, with E[||Vt ||] < ∞. Then
1X
Vt = E[V1 ].
p lim
n→∞ n
t=1
n
Proof. See Bierens (2004, Theorem 7.4, p. 184).
7
0
0
Denoting Vt = (Zt0 , Zt−1
, ..., Zt−max(p,q)
)0 , it follows from Assumptions 1.1 and
2.1(a) that Vt is strictly stationary with vanishing memory.2 Next, let
g(Vt , θ) = (Yt − ft−1 (θ))2
and note that by Assumptions 1.1 and 2.1(b),
¸
¸
∙
∙
2
E sup |g(V1 , θ)| = E sup (Yt − ft−1 (θ))
θ∈Θ
θ∈Θ
¸
∙
2
2
≤ 2.E[Yt ] + 2.E sup ft−1 (θ) < ∞.
θ∈Θ
Therefore, it follows from Lemma 2.1 that under Assumption 1.1 and parts (a)
and (b) of Assumption 2.1,
¯
¯
¯b
¯
(2.1)
p lim sup ¯Q
(θ)
−
Q(θ)
¯ = 0,
n
n→∞ θ∈Θ
where
X
£
¤
bn (θ) = 1
Q
(Yt − ft−1 (θ))2 , Q(θ) = E (Yt − ft−1 (θ))2 .
n t=1
n
(2.2)
It is now a standard nonlinear regression exercise3 to verify that
Lemma 2.1. Under Assumptions 1.1 and 2.1 the NLLS estimator b
θn satisfies
θn = θ0
p lim b
n→∞
regardless whether H0 is true or not.
Of course, if H0 is not true then Ut = Yt − ft−1 (θ0 ) is no longer a martingale
difference process.
2.2. Asymptotic normality
In addition to Assumptions 1.1
2.1 the following conditions are sufficient for
√ and
b
the asymptotic normality of n(θn − θ0 ).
¡
¢
¡
¢
t
∞
Because F−∞
= σ {Zt−j }∞
j=0 = σ {Vt−j }j=0 .
3
See for example Bierens (2004, Ch. 6).
2
8
Assumption 2.2. Under H0 the following conditions hold.
(a) Θ is convex and θ0 is an interior point of Θ;
(b) ft−1 (θ) is a.s. twice continuously differentiable in the components θ1 , θ2 , ..., θm
of θ;
(c) for i1 , i2 = 1, 2, ..., m,
i
h
¯
2
¯
E (Yt − ft−1 (θ)) . |(∂/∂θi1 )ft−1 (θ)| . |(∂/∂θi2 )ft−1 (θ)| θ=θ0 < ∞, (2.3)
∙
¸
E sup |(∂/∂θi1 )ft−1 (θ)| . |(∂/∂θi2 )ft−1 (θ)| < ∞, (2.4)
θ∈Θ
¸
∙
E sup |Yt − ft−1 (θ)|. |(∂/∂θi1 )(∂/∂θi2 )ft−1 (θ)| < ∞; (2.5)
θ∈Θ
(d) The matrix
is nonsingular.
£
¤
A2 = E ((∂/∂θ0 )ft−1 (θ)) ((∂/∂θ)ft−1 (θ))|θ=θ0
Denoting
∇ft−1 (θ) = (∂/∂θ0 )ft−1 (θ)
bn (θ) and the
it follows then from the first-order conditions for a minimum of Q
mean value theorem that, with probability converging to 1,
√
1 X
e2,n n(b
0= √
Ut .∇ft−1 (θ0 ) − A
θn − θ0 )
n j=1
n
where
e2,n
A
⎞
((∂/∂θ1 )ft−1 (θ)) (∇ft−1 (θ))0 |θ=eθ1
1
⎜
⎟
..
=
⎝
⎠
.
n j=1
0
((∂/∂θm )ft−1 (θ)) (∇ft−1 (θ)) |θ=eθm
⎛
⎞
(Yt − ft−1 (θ)) ((∂/∂θ1 )(∇ft−1 (θ))0 )|θ=eθ1
n
1 X⎜
⎟
..
−
⎝
⎠,
.
n j=1
0
(Yt − ft−1 (θ)) ((∂/∂θm )(∇ft−1 (θ)) )|θ=eθm
n
X
⎛
9
with the e
θi ’s mean values satisfying
θn − θ0 ||, i = 1, 2, ..., m.
||e
θi − θ0 || ≤ ||b
(2.6)
Since by Assumption 2.2(c) and Lemma 2.1,
and
¯ n
¯1 X
¯
p lim sup ¯
(((∂/∂θi1 )ft−1 (θ)) ((∂/∂θi2 )ft−1 (θ))
n→∞ θ∈Θ ¯ n
j=1
¶¯
¯
− E [((∂/∂θi1 )ft−1 (θ)) ((∂/∂θi2 )ft−1 (θ))] ¯¯ = 0
¯ n
¯1 X
¯
p lim sup ¯
((Yt − ft−1 (θ)) ((∂/∂θii )(∂/∂θi2 )ft−1 (θ))
n→∞ θ∈Θ ¯ n
j=1
¶¯
¯
− E [(Yt − ft−1 (θ)) ((∂/∂θi )(∂/∂θi2 )ft−1 (θ))] ¯¯ = 0
e2,n = A2 ,
it follows straightforwardly from (2.6) and Lemma 2.1 that p limn→∞ A
hence by Assumption 2.2(d),
−1
e−1
p lim A
2,n = A2 .
n→∞
Finally, note that Ut .∇ft−1 (θ0 ) is a vector-valued martingale difference process,
so that by the martingale difference central limit theorem of McLeish (1974) and
condition (2.3),
1 X
d
√
Ut .∇ft−1 (θ0 ) → Nm [0, A1 ], where
n t=1
£
¤
A1 = E Ut2 .(∇ft−1 (θ0 ))(∇ft−1 (θ0 ))0 .
n
Summarizing, the following asymptotic normality result has been shown to
hold.
Lemma 2.2. Under H0 and Assumptions 1.1, 2.1 and 2.2,
n
√
1 X
b
n(θn − θ0 ) = √
Ut .A−1
2 ∇ft−1 (θ0 ) + op (1)
n j=1
£
¤
d
−1
→ Nm 0, A−1
.
2 A1 A2
10
3. The limiting null distribution of the WICM test
I will first show that
bn,k ≤ 2.B
b1,n,k +
Lemma 3.1. Under H0 and Assumptions 1.1, 2.1 and 2.2, B
b2,n,k = Op (1), hence for any positive
b1,n,k ] = E[U12 ] and supk∈N B
b2,n,k , where E[B
2B
P∞
sequence γk satisfying
k=1 γk < ∞ and any subsequence Ln of n such that
limn→∞ Ln = ∞,
∞
Ln
X
X
bn,k −
bn,k = op (1).
γk B
γk B
(3.1)
k=1
k=1
Proof. It follows from (1.17) that
ck,n (τ1 , τ2 , ....τk )
W
à k
!
n
X
1 X
Ut exp i.
τj0 Φ(Zt−j )
=√
n t=1
j=1
à k
!
n ³
´
X
X
1
ft−1 (b
θn ) − ft−1 (θ0 ) exp i.
τj0 Φ(Zt−j )
−√
n t=1
j=1
so that
Z
¯
¯2
¯c
¯
¯Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk )
k
Υ
à k
!
Z Ã
n
X
1 X
√
=
Ut cos
τj0 Φ(Zt−j )
n
Υk
t=1
j=1
à k
!!2
n
´
X
1 X³
θn ) − ft−1 (θ0 ) cos
τj0 Φ(Zt−j )
ft−1 (b
−√
n t=1
j=1
bn,k =
B
×dµ(τ1 )dµ(τ2 )...dµ(τk )
à k
!
Z Ã
n
X
1 X
√
+
Ut sin
τj0 Φ(Zt−j )
n
k
Υ
t=1
j=1
à k
!!2
n
´
X
1 X³
ft−1 (b
θn ) − ft−1 (θ0 ) sin
τj0 Φ(Zt−j )
−√
n t=1
j=1
11
(3.2)
×dµ(τ1 )dµ(τ2 )...dµ(τk )
à k
!!2
Z Ã
n
X
1 X
√
Ut cos
τj0 Φ(Zt−j )
dµ(τ1 )dµ(τ2 )...dµ(τk )
≤ 2
n
k
Υ
t=1
j=1
à k
!!2
Z Ã
n
´
X
1 X³
√
ft−1 (b
θn ) − ft−1 (θ0 ) cos
τj0 Φ(Zt−j )
+2
n
k
Υ
t=1
j=1
×dµ(τ1 )dµ(τ2 )...dµ(τk )
à k
!!2
Z Ã
n
X
1 X
√
Ut sin
τj0 Φ(Zt−j )
dµ(τ1 )dµ(τ2 )...dµ(τk )
+2
n
k
Υ
t=1
j=1
à k
!!2
Z Ã
n
´
X
1 X³
√
ft−1 (b
θn ) − ft−1 (θ0 ) sin
τj0 Φ(Zt−j )
+2
n
k
Υ
t=1
j=1
where
b1,n,k
B
b2,n,k
B
b1,n,k + 2B
b2,n,k
= 2.B
¯
à k
!¯2
n
¯ 1 X
¯
X
¯
¯
0
=
Ut exp i.
τj Φ(Zt−j ) ¯ dµ(τ1 )dµ(τ2 )...dµ(τk ) (3.3)
¯√
¯
Υk ¯ n t=1
j=1
à k
!!2
Z Ã
n
X
1 X
√
Ut cos
τj0 Φ(Zt−j )
dµ(τ1 )dµ(τ2 )...dµ(τk )
=
n
k
Υ
t=1
j=1
à k
!!2
Z Ã
n
X
1 X
√
Ut sin
τj0 Φ(Zt−j )
dµ(τ1 )dµ(τ2 )...dµ(τk )
+
n
k
Υ
t=1
j=1
à k
!¯2
Z ¯¯
n ³
¯
´
X
¯ 1 X
¯
ft−1 (b
=
θn ) − ft−1 (θ0 ) exp i.
τj0 Φ(Zt−j ) ¯
¯√
¯
Υk ¯ n t=1
Z
j=1
×dµ(τ1 )dµ(τ2 )...dµ(τk )
à k
!!2
Z Ã
n
´
X
1 X³
√
ft−1 (b
θn ) − ft−1 (θ0 ) cos
τj0 Φ(Zt−j )
=
n
k
Υ
t=1
j=1
×dµ(τ1 )dµ(τ2 )...dµ(τk )
à k
!!
Z Ã
n
´
X
1 X³
0
√
+
ft−1 (b
θn ) − ft−1 (θ0 ) sin
τj Φ(Zt−j )
n t=1
Υk
j=1
12
(3.4)
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Obviously,
b1,n,k ] = E[Ut2 ] = E[U12 ].
E[B
(3.5)
b2,n,k = Op (1), observe that by the mean value theorem,
To prove that supk∈N B
à k
!
n
´
X
1 X³
√
ft−1 (b
θn ) − ft−1 (θ0 ) cos
τj0 Φ(Zt−j )
n t=1
j=1
à k
!
n
X
X
√
1
0
0
θn − θ0 )
∇ft−1 (e
θn (τ )) cos
τj Φ(Zt−j )
(3.6)
= n(b
n t=1
j=1
where e
θn (τ ) is a mean value such that
θn − θ0 ||.
||e
θn (τ ) − θ0 || ≤ ||b
hence
¯
à k
!¯
n ³
¯ 1 X
¯
´
X
¯
¯
0
b
ft−1 (θn ) − ft−1 (θ0 ) cos
τj Φ(Zt−j ) ¯
¯√
¯ n t=1
¯
j=1
√
1
≤ || n(b
θn − θ0 )||.
n
and thus,
Z
Υk
Ã
n
X
sup ||∇ft−1 (θ)||
t=1 θ∈Θ
à k
!!2
n
´
X
1 X³
√
ft−1 (b
θn ) − ft−1 (θ0 ) cos
τj0 Φ(Zt−j )
n t=1
j=1
× dµ(τ1 )dµ(τ2 )...dµ(τk )
n
√
1X
b
sup ||∇ft−1 (θ)||
≤ || n(θn − θ0 )||.
n t=1 θ∈Θ
= Op (1), uniformly in k.
√ b
The latter follows from the fact that by Lemma 2.2, || P
n(θn −θ0 )|| = Op (1) and by
1
2
condition (2.4), E[supθ∈Θ ||∇ft−1 (θ)|| ] < ∞, hence n nt=1 supθ∈Θ ||∇ft−1 (θ)|| =
Op (1).
13
Similarly,
Z
Υk
Ã
à k
!!2
n
´
X
1 X³
√
ft−1 (b
θn ) − ft−1 (θ0 ) sin
τj0 Φ(Zt−j )
n t=1
j=1
× dµ(τ1 )dµ(τ2 )...dµ(τk )
n
√
1X
θn − θ0 )||.
sup ||∇ft−1 (θ)||
≤ || n(b
n t=1 θ∈Θ
= Op (1), uniformly in k.
hence
n
√
1X
b2,n,k ≤ 2|| n(b
θn − θ0 )||.
sup ||∇ft−1 (θ)|| = Op (1).
(3.7)
sup B
n t=1 θ∈Θ
k∈N
P
The P
result (3.1) follows now trivially from the fact that ∞
k=1 γk < ∞ implies
∞
lim`→∞ k=`+1 γk = 0.
Next, denote
"
Ã
bk (τ1 , τ2 , ....τk ) = E ∇ft−1 (θ0 ) exp i
Ã
φk,t−1 (τ1 , τ2 , ....τk ) = exp i
k
X
k
X
j=1
!
!#
τj0 Φ(Zt−j )
,
(3.8)
τj0 Φ(Zt−j )
j=1
−bk (τ1 , τ2 , ....τk )0 A−1
(3.9)
2 ∇ft−1 (θ0 ),
n
X
1
Ut .φk,t−1 (τ1 , τ2 , ....τk ),
(3.10)
Wk,n (τ1 , τ2 , ....τk ) = √
n t=1
Z
|Wk,n (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk ). (3.11)
Bk,n =
Υk
Then
Lemma 3.2. Under H0 and Assumptions 1.1, 2.1 and 2.2, and for fixed k ∈ N,
sup
(τ1 ,τ2 ,....τk )∈Υk
¯
¯
¯c
¯
¯Wk,n (τ1 , τ2 , ....τk ) − Wk,n (τ1 , τ2 , ....τk )¯ = op (1),
14
(3.12)
hence
bn,k = Bk,n + op (1).
B
(3.13)
Proof. Observe from (3.6) that
¯
à k
!
n ³
¯ 1 X
´
X
¯
f (b
θ ) − ft−1 (θ0 ) cos
τj0 Φ(Zt−j )
¯√
¯ n t=1 t−1 n
j=1
à k
!¯
n
¯
X
X
√
1
¯
0
0
b
∇ft−1 (θ0 ) cos
τj Φ(Zt−j ) ¯
− n(θn − θ0 )
¯
n t=1
j=1
n
√
1X
b
≤ || n(θn − θ0 )||
sup
||∇ft−1 (θ) − ∇ft−1 (θ0 )||
n t=1 ||θ−θ0 ||≤||bθn −θ0 ||
= op (1)
√
because || n(b
θn − θ0 )|| = Op (1) and
1X
sup
||∇ft−1 (θ) − ∇ft−1 (θ0 )|| = op (1).
n t=1 ||θ−θ0 ||≤||bθn −θ0 ||
(3.14)
n
(3.15)
To prove the latter, observe that for an arbitrary small ε > 0,
#
" n
1X
sup
||∇ft−1 (θ) − ∇ft−1 (θ0 )||
E
n t=1 ||θ−θ0 ||≤||bθn −θ0 ||
"
#
³
´
≤E
sup ||∇ft−1 (θ) − ∇ft−1 (θ0 )|| × I ||b
θn − θ0 || ≤ ε
||θ−θ0 ||≤ε
∙
³
´¸
b
+ 2E sup ||∇ft−1 (θ)|| × I ||θn − θ0 || > ε
θ∈Θ
"
#
≤E
sup
||θ−θ0 ||≤ε
||∇ft−1 (θ) − ∇ft−1 (θ0 )||
s ∙
¸r h
i
Pr ||b
θn − θ0 || > ε
+ 2 E sup ||∇ft−1 (θ)||2
=E
"
θ∈Θ
sup
||θ−θ0 ||≤ε
#
||∇ft−1 (θ) − ∇ft−1 (θ0 )|| + o(1),
15
(3.16)
where the o(1) term is due to the facts that by Lemma 2.1, Pr[||b
θn − θ0 || > ε] → 0
2
and that by condition (2.4), E[supθ∈Θ ||∇ft−1 (θ)|| ] < ∞. Moreover, since
"
#
∙
¸
E
sup
||θ−θ0 ||≤ε
||∇ft−1 (θ) − ∇ft−1 (θ0 )||
≤ 2E sup ||∇ft−1 (θ)||
θ∈Θ
s ∙
¸
2
E sup ||∇ft−1 (θ)|| < ∞
≤
θ∈Θ
and by a.s. continuity of ∇ft−1 (θ),
lim
sup
ε↓0 ||θ−θ0 ||≤ε
||∇ft−1 (θ) − ∇ft−1 (θ0 )|| = 0 a.s.,
it follows from the dominated convergence theorem that
"
#
lim E
ε↓0
sup
||θ−θ0 ||≤ε
||∇ft−1 (θ) − ∇ft−1 (θ0 )|| = 0.
(3.17)
It follows now straightforwardly from (3.16) and (3.17) that (3.15) holds.
Next, observe from Lemma 2.1 and Assumption 2.1(b) that for fixed k,
° n
°
à k
!
°1 X
°
X
°
°
sup
∇ft−1 (θ0 ) cos
τj0 Φ(Zt−j ) − Re[bk (τ1 , τ2 , ....τk )]° = op (1)
°
°
°
n
k
(τ1 ,τ2 ,....τk )∈Υ
t=1
j=1
and thus,
à k
!
n
X
X
√
1
n(b
θn − θ0 )0
∇ft−1 (θ0 ) cos
τj0 Φ(Zt−j )
n t=1
j=1
√ b
= n(θn − θ0 )0 Re[bk (τ1 , τ2 , ....τk )] + op (1)
n
X
0 −1 1
Ut .∇ft−1 (θ0 ) + op (1)
= Re[bk (τ1 , τ2 , ....τk )] A2 √
n j=1
uniformly on Υk , where the latter equality follows from Lemma 2.2. Hence by
(3.2) and (3.14),
"
à k
!#
n
h
i
X
X
1
0
ck,n (τ1 , τ2 , ....τk ) = Re √
Re W
Ut exp i.
τj Φ(Zt−j )
n t=1
j=1
16
0
− Re[bk (τ1 , τ2 , ....τk )]
"
1
A−1
2 √
n
1
= Re √
n
n
X
n
X
Ut .∇ft−1 (θ0 )) + op (1)
j=1
#
Ut φk,t−1 (τ1 , τ2 , ....τk ) + op (1)
t=1
= Re [Wk,n (τ1 , τ2 , ....τk )] + op (1)
uniformly on Υk . The same applies to the Im[.] case. This proves (3.12).
The proof that the latter implies (3.13) is not hard4 and is therefore left to
the reader.
Next, I will set forth conditions such that supk∈N E[Bk,n ] < ∞, as follows.
Observe from (3.8), (3.9) and (3.10) that
¤
£
E |Wk,n (τ1 , τ2 , ....τk )|2
i
h
= E Ut2 φk,t−1 (τ1 , τ2 , ....τk )φk,t−1 (τ1 , τ2 , ....τk )
" Ã
à k
!
!
X
= E Ut2 exp i
τj0 Φ(Zt−j ) − bk (τ1 , τ2 , ....τk )0 A−1
2 ∇ft−1 (θ0 )
Ã
Ã
× exp −i
=
E[Ut2 ]
j=1
k
X
!
τj0 Φ(Zt−j )
j=1
"
!#
− (∇ft−1 (θ0 ))0 A−1
2 bk (τ1 , τ2 , ....τk )
Ã
2
− bk (τ1 , τ2 , ....τk )0 A−1
2 E Ut ∇ft−1 (θ0 ) exp −i
"
Ã
− E Ut2 (∇ft−1 (θ0 ))0 exp i
0
+ bk (τ1 , τ2 , ....τk ) A−1
2 E
2
= E[U1 ]
4
k
X
j=1
!#
τj0 Φ(Zt−j )
k
X
j=1
!#
τj0 Φ(Zt−j )
A−1
2 bk (τ1 , τ2 , ....τk )
£ 2
¤
Ut ∇ft−1 (θ0 )∇ft−1 (θ0 ))0 A−1
2 bk (τ1 , τ2 , ....τk )
Denote the left-hand side of (3.12) by dk,n , and verify that
q
¯
¯
¯e
bk,n .
bk,n ¯¯ ≤ d2k,n + 4.dk,n B
¯Bk,n − B
17
− bk (τ1 , τ2 , ....τk )0 A−1
2 ck (τ1 , τ2 , ....τk )
− ck (τ1 , τ2 , ....τk )0 A−1
2 bk (τ1 , τ2 , ....τk )
−1
+ bk (τ1 , τ2 , ....τk )0 A−1
2 A1 A2 bk (τ1 , τ2 , ....τk ),
where the bar denotes the complex-conjugate, and
"
Ã
ck (τ1 , τ2 , ....τk ) = E Ut2 ∇ft−1 (θ0 ) exp i
k
X
!#
τj0 Φ(Zt−j )
j=1
.
It is an easy complex calculus exercise to verify that
bk (τ1 , τ2 , ....τk )0 A−1
2 ck (τ1 , τ2 , ....τk )
+ ck (τ1 , τ2 , ....τk )0 A−1
2 bk (τ1 , τ2 , ....τk )
= 2 Re [bk (τ1 , τ2 , ....τk )0 ] A−1
2 Re [ck (τ1 , τ2 , ....τk )]
0
+ 2 Im [bk (τ1 , τ2 , ....τk ) ] A−1
2 Im [ck (τ1 , τ2 , ....τk )]
and
−1
bk (τ1 , τ2 , ....τk )0 A−1
2 A1 A2 bk (τ1 , τ2 , ....τk )
−1
= Re [bk (τ1 , τ2 , ....τk )0 ] A−1
2 A1 A2 Re [bk (τ1 , τ2 , ....τk )]
−1
+ Im [bk (τ1 , τ2 , ....τk )0 ] A−1
2 A1 A2 Im [bk (τ1 , τ2 , ....τk )] ,
hence
¤
£
E |Wk,n (τ1 , τ2 , ....τk )|2 ≤ E[U12 ]
+ 2 Re [bk (τ1 , τ2 , ....τk )0 ] A−1
2 Re [ck (τ1 , τ2 , ....τk )]
+ 2 Im [bk (τ1 , τ2 , ....τk )0 ] A−1
2 Im [ck (τ1 , τ2 , ....τk )]
0
−1
+ Re [bk (τ1 , τ2 , ....τk ) ] A2 A1 A−1
2 Re [bk (τ1 , τ2 , ....τk )]
0
−1
+ Im [bk (τ1 , τ2 , ....τk ) ] A2 A1 A−1
2 Im [bk (τ1 , τ2 , ....τk )]
(3.18)
Let λ1 , ..., λm be the eigenvalues of A−1
2 , with corresponding orthonormal eigenvectors q1 , ..., qm . Then
¯
¯
¯
¯Re [bk (τ1 , τ2 , ....τk )0 ] A−1
2 Re [ck (τ1 , τ2 , ....τk )]
m
X
≤
λi |qi0 Re [qi0 bk (τ1 , τ2 , ....τk )]| . |Re [qi0 ck (τ1 , τ2 , ....τk )]|
i=1
18
=
m
X
i=1
¯ "
à k
!#¯
¯
¯
X
¯
¯
0
0
λi ¯E (∇ft−1 (θ0 )) qi cos
τj Φ(Zt−j ) ¯
¯
¯
j=1
¯ "
à k
!#¯
¯
¯
X
¯
¯
× ¯E Ut2 (∇ft−1 (θ0 ))0 qi cos
τj0 Φ(Zt−j ) ¯
¯
¯
j=1
≤
≤
≤
m
X
i=1
m
X
i=1
m
X
£
¤
λi E [|(∇ft−1 (θ0 ))0 qi |] E Ut2 |(∇ft−1 (θ0 ))0 qi |
£
¤
λi E [||∇ft−1 (θ0 )||] E Ut2 ||∇ft−1 (θ0 )||
λi
i=1
p
£
¤
E [||∇ft−1 (θ0 )||2 ]E Ut2 ||∇ft−1 (θ0 )||
m
X
p
£
¤
=
λi trace(A2 )E Ut2 ||∇ft−1 (θ0 )||
i=1
and similarly
p
£
¤
= trace(A2 ) trace(A2 )E Ut2 ||∇ft−1 (θ0 )||
Im [bk (τ1 , τ2 , ....τk )0 ] A−1
Im [ck (τ1 , τ2 , ....τk )]
p 2
£
¤
≤ trace(A2 ) trace(A2 )E Ut2 ||∇ft−1 (θ0 )||
−1
Moreover, with λ∗1 , ..., λ∗m the eigenvalues of A−1
2 A1 A2 we have similarly,
−1
Re [bk (τ1 , τ2 , ....τk )0 ] A−1
2 A1 A2 Re [bk (τ1 , τ2 , ....τk )]
m
X
λ∗i (E [||∇ft−1 (θ0 )||])2
≤
i=1
≤
=
m
X
i=1
m
X
£
¤
λ∗i E ||∇ft−1 (θ0 )||2
λ∗i .trace(A2 )
i=1
−1
= trace(A−1
2 A1 A2 ).trace(A2 )
and
−1
Im [bk (τ1 , τ2 , ....τk )0 ] A−1
2 A1 A2 Im [bk (τ1 , τ2 , ....τk )]
−1
≤ trace(A−1
2 A1 A2 ).trace(A2 ).
19
Hence,
Z
£
¤
E |Wk,n (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk )
Υk
p
£
¤
≤ E[U12 ] + 4.trace(A2 ) trace(A2 )E Ut2 ||∇ft−1 (θ0 )||
−1
+2.trace(A−1
2 A1 A2 ).trace(A2 ).
E [Bk,n ] =
(3.19)
Thus, if
Assumption 3.1. E [Ut2 ||∇ft−1 (θ0 )||] < ∞
then
Lemma 3.3. Under H0 and Assumptions 1.1, 2.1, 2.2 and 3.1, supk∈N E[Bk,n ] <
∞.
Lemmas 3.1-3.3 now yield the following corollary.
Lemma 3.4. Under H0 and Assumptions 1.1, 2.1, 2.2 and 3.1,
∞
X
k=1
bk,n =
γk B
∞
X
γk Bk,n + op (1)
k=1
for any positive sequence γk satisfying
P∞
k=1
γk < ∞.
Proof. For any K ∈ N,
¯
¯∞
∞
¯
¯X
X
¯
¯
bk,n −
γk B
γk Bk,n ¯ ≤ R0,n,K + R1,n,K + R2,n,K
¯
¯
¯
k=1
k=1
where
R0,n,K
¯
¯K
K
¯
¯X
X
¯
¯
b
= ¯
γk Bk,n −
γk Bk,n ¯
¯
¯
k=1
∞
X
R1,n,K = 2
R2,n,K = 2
k=1
b1,k,n +
γk B
k=K+1
à ∞
X
γk
k=K+1
20
!
∞
X
k=K+1
b2,k,n .
sup B
k∈N
γk Bk,n ,
Let ε > 0 and δ > 0 be arbitrary. By Chebyshev’s inequality for first moments
and Lemmas 3.1 and 3.3 there exists a K1 (δ, ε) ∈ N such that
Pr [R1,n,K > δ/2] ≤ 2E[R1,n,K ]/δ
P
P∞
b
2 ∞
k=K+1 γk E[B1,k,n ] +
k=K+1 γk E[Bk,n ]
= 2
δ
à ∞
!µ
¶
X
−1
4E[U1 ] + 2 sup E[Bk,n ]
γk
≤ δ
k=K+1
k∈N
< ε
for all K ≥ K1 (δ, ε). Moreover, by Lemma 3.1 there exists a K2 (δ, ε) ∈ N such
that
¸
∙
δ
b
<ε
Pr [R2,n,K > δ/2] = Pr sup B2,k,n > P∞
4 k=K+1 γk
k∈N
for all K ≥ K2 (δ, ε). Using the easy inequality
Pr[R1 + R2 > δ] ≤ Pr[R1 > δ/2] + Pr[R2 > δ/2]
(3.20)
for nonnegative random variables R1 and R2 , it follows now that for a fixed K0 ≥
max{K1 (δ, ε), K2 (δ, ε)},
Pr [R1,n,K0 + R2,n,K0 > δ] < 2ε
Finally, it follows from Lemma 3.2 that p limn→∞ R0,n,K0 = 0, so that there
exists an n0 ∈ N such that for all n ≥ n0 ,
Pr [R2,n,K0 > δ] < ε.
Hence by (3.20),
¯
#
"¯ ∞
∞
¯
¯X
X
¯
¯
bk,n −
γk B
γk Bk,n ¯ > 2δ
Pr ¯
¯
¯
k=1
k=1
≤ Pr [R2,n,K0 + R1,n,K0 + R2,n,K0 > 2δ]
≤ Pr [R2,n,K0 + R1,n,K0 + R2,n,K0 > 2δ]
≤ Pr [R2,n,K0 > δ] + Pr [R1,n,K0 + R2,n,K0 > δ]
< 3ε,
21
¯P
¯
P∞
¯ ∞
¯
b
which implies that p limn→∞ ¯ k=1 γk Bk,n − k=1 γk Bk,n ¯ = 0.
The next step is to prove that for each k ∈ N, Wk,n (τ1 , τ2 , ....τk ) converges
weakly to a zero-mean complex-valued Gaussian process Wk (τ1 , τ2 , ....τk ) on Υk ,
denoted by Wk,n ⇒ Wk , by showing that Wk,n (τ1 , τ2 , ....τk ) is tight on Υk and its finite distributions converge to the corresponding finite distribution of Wk (τ1 , τ2 , ....τk ).
See Bierens (2015a) and Billingsley (1968) for these concepts. However, it follows
from Bierens and Ploberger (1997, Lemma A.1) that Re[Wk,n (τ1 , τ2 , ....τk )] and
Im[Wk,n (τ1 , τ2 , ....τk )] are tight on Υk , hence so is Wk,n (τ1 , τ2 , ....τk ). Now it suffices to verify that the finite distributions of Wk,n (τ1 , τ2 , ....τk ) converge to the
corresponding finite distribution of Wk (τ1 , τ2 , ....τk ), which can be done similar to
Bierens (2015a), replacing the reference to standard central limit theorem for the
i.i.d. case with a reference to the martingale difference central limit theorem of
McLeish (1974). Thus,
Lemma 3.5. Under H0 and Assumptions 1.1, 2.1 and 2.2, and for each k ∈ N,
Wk,n (τ1 , τ2 , ...., τk ) ⇒ Wk (τ1 , τ2 , ...., τk ) on Υk , where the latter is a zero-mean
complex-valued Gaussian process on Υk with covariance function
Γk ((τ1,1 , τ1,2 , ....τ1,k ), (τ2,1 , τ2,2 , ....τ2,k ))
h
i
= E Wk (τ1,1 , τ1,2 , ....τ1,k )Wk (τ2,1 , τ2,2 , ....τ2,k )
h
i
2
= E Ut φk,t−1 (τ1,1 , τ1,2 , ....τ1,k )φk,t−1 (τ2,1 , τ2,2 , ....τ2,k ) .
Consequently, by the continuous mapping theorem,
Z
d
Bk,n → Bk =
|Wk (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk )
Υk
Note that
Wk,n (τ1 , τ2 , ...., τk ) = Wk+1,n (τ1 , τ2 , ...., τk , 0)
and similarly,
Wk (τ1 , τ2 , ...., τk ) = Wk+1 (τ1 , τ2 , ...., τk , 0).
Hence
⎛
⎜
⎜
⎜
⎝
W1,n (τ1 )
W2,n (τ1 , τ2 )
..
.
Wk,n (τ1 , τ2 , τ3 , ...., τk )
⎛
⎞
⎟ ⎜
⎟ ⎜
⎟=⎜
⎠ ⎝
22
Wk,n (τ1 , 0, 0, ...., 0)
Wk,n (τ1 , τ2 , 0, ...., 0)
..
.
Wk,n (τ1 , τ2 , τ3 , ...., τk )
⎞
⎟
⎟
⎟
⎠
(3.21)
⎛
⎜
⎜
⇒⎜
⎝
⎞
Wk (τ1 , 0, 0, ...., 0)
Wk (τ1 , τ2 , 0, ...., 0)
..
.
Wk (τ1 , τ2 , τ3 , ...., τk )
⎛
⎟ ⎜
⎟ ⎜
⎟=⎜
⎠ ⎝
W1 (τ1 )
W2 (τ1 , τ2 )
..
.
Wk (τ1 , τ2 , τ3 , ...., τk )
⎞
⎟
⎟
⎟
⎠
(3.22)
where the weak convergence result involved is not hard to verify. It follows now
from the continuous mapping theorem that
Lemma 3.6. Under the conditions of Lemma 3.5,
d
(B1,n , B2,n , ..., Bk,n )0 → (B1 , B2 , ..., Bk )0 .
It remains to show that
Assumptions 1.1, 2.1, 2.2 and 3.1, and for any
Lemma 3.7. Under H0 and P
positive sequence γk satisfying ∞
k=1 γk < ∞,
∞
X
k=1
d
γk Bk,n →
∞
X
γk Bk .
k=1
Proof. It follows from Lemma 3.5 that
Z
£
¤
E |Wk (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk )
E[Bk ] =
k
ZΥ
h
i
=
E Wk (τ1 , τ2 , ....τk )Wk (τ1 , τ2 , ....τk )
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Z
h
i
2
E Ut φk,t−1 (τ1 , τ2 , ....τk )φk,t−1 (τ1 , τ2 , ....τk )
=
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk )
hence, similar to Lemma 3.3 we have
sup E[Bk ] < ∞.
(3.23)
k∈N
Moreover, Lemma 3.6 implies that for each ` ∈ N,
`
X
k=1
d
γk Bk,n →
23
`
X
k=1
γk Bk .
(3.24)
P∞Let x, x − δ and x + δ, with δ > 0, be continuity points of the distribution of
k=1 γk Bk , Then
"∞
#
X
lim sup Pr
γk Bk,n ≤ x
n→∞
k=1
≤ lim Pr
n→∞
= Pr
" `
X
" `
X
k=1
+ Pr
k=1
k=1
≤ Pr
≤ Pr
k=1
"∞
X
k=1
γk Bk,n ≤ x = Pr
γk Bk ≤ x and
" `
X
"∞
X
#
∞
X
k=`+1
γk Bk ≤ x and
#
k=1
#
γk Bk > δ
k=`+1
"
γk Bk ≤ x + δ + δ −1
∞
X
#
γk Bk > δ
k=`+1
∞
X
Ã
#
γk Bk ≤ x
γk Bk ≤ δ
∞
X
γk Bk ≤ x + δ + Pr
#
" `
X
γk
k=`+1
!
#
sup E[Bk ],
k∈N
where first equality follows from (3.24) and the last inequality follows from (3.23)
and Chebyshev’s inequality for first moments. Hence, letting ` → ∞ first and
then letting δ ↓ 0 it follows that
"∞
#
"∞
#
X
X
lim sup Pr
γk Bk,n ≤ x ≤ Pr
γk Bk ≤ x .
(3.25)
n→∞
k=1
Moreover,
#
"∞
X
γk Bk,n ≤ x
Pr
k=1
= Pr
" `
X
k=1
+ Pr
γk Bk,n ≤ x −
" `
X
k=1
k=1
∞
X
γk Bk,n and
k=`+1
γk Bk,n ≤ x −
∞
X
k=`+1
24
∞
X
k=`+1
γk Bk,n and
γk Bk,n ≤ δ
∞
X
k=`+1
#
γk Bk,n > δ
#
≥ Pr
≥ Pr
= Pr
" `
X
k=1
" `
X
k=1
" `
X
k=1
− Pr
≥ Pr
Hence
lim inf Pr
n→∞
"∞
X
k=1
= Pr
k=1
≥ Pr
"∞
X
k=1
#
∞
X
k=`+1
#
γk Bk,n ≤ x − δ − Pr
#
" `
X
k=1
∞
X
k=`+1
γk Bk,n ≤ x − δ and
γk Bk,n ≤ x
≥ lim inf Pr
" `
X
γk Bk,n ≤ x − δ
k=1
k=1
γk Bk,n and
k=`+1
γk Bk,n ≤ x − δ and
" `
X
" `
X
n→∞
γk Bk,n ≤ x −
∞
X
γk Bk,n ≤ δ
∞
X
γk Bk,n > δ
"
∞
X
k=`+1
n→∞
γk Bk ≤ x − δ − lim sup Pr
n→∞
γk Bk ≤ x − δ − lim sup Pr
n→∞
"
∞
X
k=`+1
∞
X
"
#
γk Bk,n > δ
γk Bk,n ≤ x − δ − lim sup Pr
#
#
k=`+1
#
#
γk Bk,n ≤ δ
"
#
∞
X
γk Bk,n > δ
k=`+1
γk Bk,n > δ
γk Bk,n > δ
k=`+1
#
#
By Lemma 3.1, (3.20) and Chebyshev’s inequality for first moments,
" ∞
#
X
Pr
γk Bk,n > δ
k=`+1
"
≤ Pr 2
≤ Pr
"
∞
X
k=`+1
∞
X
k=`+1
e1,k,n + 2
γk B
∞
X
k=`+1
#
e2,k,n > δ
γk B
e1,k,n > δ/4 + Pr
γk B
25
"
∞
X
k=`+1
#
#
#
e2,k,n > δ/4
γk B
#
≤ 4.δ−1
Ã
∞
X
γk
k=`+1
!
"
e2,k,n >
E[U12 ] + Pr sup B
k∈N
δ
4.
¡P∞
k=`+1 γk
£P∞
¤
hence lim`→∞ lim supn→∞ Pr
γ
B
>
δ
= 0 and thus
k
k,n
k=`+1
"∞
#
"∞
#
X
X
γk Bk,n ≤ x ≥ Pr
γk Bk ≤ x − δ .
lim inf Pr
n→∞
k=1
¢ ,
k=1
Letting δ ↓ 0 it follows now that
"∞
#
"∞
#
X
X
lim inf Pr
γk Bk,n ≤ x ≥ Pr
γk Bk ≤ x
n→∞
#
k=1
(3.26)
k=1
Combining (3.25) and (3.26) yields
"∞
#
"∞
#
X
X
lim Pr
γk Bk,n ≤ x = Pr
γk Bk ≤ x ,
n→∞
k=1
k=1
which proves the lemma.
Summarizing, the following theorem has been proved.
P
Theorem 3.1. Let γk be an a priory chosen positive sequence satisfying ∞
k=1 γk <
∞, and let Ln < n be a subsequence of n satisfying limn→∞ Ln = ∞. The test
statistic of the WICM test takes the form
Tbn =
Ln
X
k=1
bn,k ,
γk B
(3.27)
bn,k are defined by (1.18). Under H0 and Assumpwhere the ICM test statistics B
tions 1.1, 2.1, 2.2 and 3.1,
∞
X
d
γk Bk ,
(3.28)
Tbn → T =
k=1
where each Bk takes the form
Z
|Wk (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk ),
Bk =
Υk
with Wk (τ1 , τ2 , ....τk ) a complex-valued zero-mean Gaussian process on Υk with
covariance function given in (3.21).
26
4. Consistency of the WICM test
In this section I will show that under H1 the WICM test statistic Tbn in Theorem
3.1 converges in probability to infinity, so that this test is consistent. This will be
done in the following three steps.
Lemma 4.1. Denote
à k
!
n
X
1X
ψk,n (τ1 , τ2 , ....τk ) =
Ut exp i.
τj0 Φ(Zt−j ) ,
n t=1
j=1
Z
ηk,n =
|ψk,n (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk ).
(4.1)
(4.2)
Υk
Under H1 and Assumptions 1.1 and 2.1,
sup
sup
k∈N (τ1 ,τ2 ,....τk )∈Υk
|ψk,n (τ1 , τ2 , ....τk ) − ψk,n (τ1 , τ2 , ....τk )| = op (1),
hence
¯
¯
¯b
¯
sup ¯Bk,n /n − ηk,n ¯ = op (1).
(4.3)
(4.4)
k∈N
Moreover, for each k ∈ N,
sup
(τ1 ,τ2 ,....τk )∈Υk
|ψk,n (τ1 , τ2 , ....τk ) − ψk (τ1 , τ2 , ....τk )| = op (1),
(4.5)
where ψk (τ1 , τ2 , ....τk ) is defined by (1.14), hence
ηk,n = ηk + op (1),
where ηk is defined in (1.19).
Proof. Observe from (3.2) that under H1 ,
b k,n (τ1 , τ2 , ....τk )
ψ
à k
!
n
X
X
1
(Yt − ft−1 (b
θn )) exp i.
τj0 Φ(Zt−j )
=
n t=1
j=1
à k
!
n
X
1X
Ut exp i.
τj0 Φ(Zt−j )
=
n t=1
j=1
à k
!
n
³
´
X
1X
0
ft−1 (b
θn ) − ft−1 (θ0 ) exp i.
τj Φ(Zt−j ) ,
−
n t=1
j=1
27
(4.6)
where now Ut = Yt − ft−1 (θ0 ) is no longer a martingale difference process. Hence,
¯
¯
¯b
¯
sup
sup
¯ψ k,n (τ1 , τ2 , ....τk ) − ψk,n (τ1 , τ2 , ....τk )¯
k∈N (τ1 ,τ2 ,....τk )∈Υk
n
¯
1 X ¯¯
¯
≤
θn ) − ft−1 (θ0 )¯ .
¯ft−1 (b
n t=1
Since by Lemma 2.1,
¯ n
¯
¯1 X
¯
¯
¯
sup ¯
|ft−1 (θ) − ft−1 (θ0 )| − E [|ft−1 (θ) − ft−1 (θ0 )|]¯ = op (1)
¯
θ∈Θ ¯ n t=1
p
and by Lemma 2.1, b
θn → θ0 , it follows from the continuity of E[|ft−1 (θ)−ft−1 (θ0 )|]
that
n
¯
1 X ¯¯
¯
b
¯ft−1 (θn ) − ft−1 (θ0 )¯ = E[|ft−1 (θ) − ft−1 (θ0 )|]|θ=bθn + op (1)
n t=1
= op (1),
which proves (4.3). The latter implies trivially the result (4.4).
The result (4.5) follows straightforwardly from Lemma 2.1, and (4.6) follows
straightforwardly from (4.5).
Lemma 4.2. Under H1 and Assumptions 1.1 and 2.1,
bk,n /n = Op (1), sup ηk,n = Op (1), sup ηk = O(1).
sup B
k∈N
k∈N
k∈N
Proof. It follows from (4.1) that
|ψk,n (τ1 , τ2 , ....τk )|2 ≤
Ã
!2
n
n
1X
1X 2
|Ut | ≤
U
n t=1
n t=1 t
hence by (4.2),
1X 2
bn (θ0 ) = Q(θ0 ) + op (1) = E[U12 ] + op (1)
U =Q
n t=1 t
n
sup ηk,n ≤
k∈N
28
(4.7)
where the equalities follow from (2.1) and (2.2). Obviously, (4.4) and (4.7) imply
bk,n /n = Op (1).
that supk∈N B
Similarly, it follows from (1.14) that
|ψk (τ1 , τ2 , ....τk )|2 ≤ (E[|U1 |])2 ≤ E[U12 ]
hence supk∈N ηk ≤ E[U12 ] = O(1).
Using Lemmas 4.1 and 4.2, I will now prove that the WICM test is consistent.
Theorem 4.1. Under H1 and Assumptions 1.1 and 2.1, and with γk and Ln as
in Theorem 3.1,
Ln
∞
X
X
p
b
b
γk Bk,n /n →
γk ηk > 0.
(4.8)
Tn /n =
k=1
k=1
Proof. For fixed ` ≤ Ln ,
¯
¯L
∞
`
n
¯ X
¯X
X
¯
¯
bk,n /n − ηk |
b
γk Bk,n /n −
γk ηk ¯ ≤
γk |B
¯
¯
¯
k=1
k=1
à ∞ k=1
!µ
¶
X
b
+
sup Bk,n /n + sup ηk,n + sup ηk .
γk
k∈N
k=`+1
k∈N
(4.9)
k∈N
Let ε > 0 and δ > 0 be arbitrary. By Lemma 4.2 we can choose ` so large that
!µ
#
"Ã ∞
¶
X
bk,n /n + sup ηk,n + sup ηk > δ/2 < ε (4.10)
sup B
lim sup Pr
γk
n→∞
k∈N
k=`+1
k∈N
k∈N
and for this ` it follows from parts (4.5) and (4.6) of Lemma 4.1 that
p lim
n→∞
hence
lim Pr
n→∞
`
X
k=1
" `
X
k=1
bk,n /n − ηk | = 0,
γk |B
#
bk,n /n − ηk | > δ/2 = 0.
γk |B
29
(4.11)
It follows now from (3.20), (4.9), (4.10) and (4.11) that
¯
#
"¯ L
∞
n
¯
¯X
X
¯
¯
bk,n /n −
lim sup Pr ¯
γk B
γk ηk ¯ > δ < ε,
¯
¯
n→∞
k=1
k=1
which by the arbitrariness of ε implies that
¯
#
"¯ L
∞
n
¯
¯X
X
¯
¯
bk,n /n −
lim Pr ¯
γk B
γk ηk ¯ > δ = 0.
n→∞
¯
¯
k=1
k=1
This result and the result of Lemma 1.2 imply (4.8).
5. Upper bounds of the critical values
In Bierens (2015b, Theorem 5.15 ) it has been shown that for a zero mean complexvalued continuous Gaussian random function W on a compact subset Υ of a
Euclidean space,
Z
∞
X
2
|W (t)| dµ(t) =
λj ε2j ,
(5.1)
Υ
j=1
where µ is a probability measure on Υ, the εj ’s are independent standard normally
distributed random variables and the λj ’s are positive numbers satisfying
∞
X
j=1
λj =
Z
Υ
E[|W (t)|2 ]dµ(t) < ∞.
(5.2)
Moreover, it has been shown in Bierens (2015a, Theorem 5.1) that similar to
Bierens and Ploberger (1997, Theorem 7), (5.1) and (5.2) imply that for all y > 0,
¸
∙ R
n
£ 2
¤
|W (t)|2 dµ(t)
1X 2
2
Υ
>
y
≤
Pr
Pr R
χ
>
y
,
where
χ
=
sup
εj
(5.3)
1
1
2 ]dµ(t)
n
E[|W
(t)|
n≥1
Υ
j=1
5
3.1.
Clearly, these results apply to each Bk in Theorem
P∞
I will now show that (5.3) applies to T = k=1 γk Bk in Theorem 3.1 as well.
This theorem is a corected version of Lemma 4 in Bierens and Wang (2012).
30
Theorem 5.1. Under the conditions of Theorem 3.1, and for fixed K ∈ N,
TK =
K
X
γk Bk =
∞
X
ωK,j ε2K,j
(5.4)
j=1
k=1
P
where the ωK,j ’s are non-negative and satisfy ∞
j=1 ωK,j = E[TK ], and the εK,j ’s
are independent standard normally distributed random variables. Hence,
£
¤
(5.5)
Pr [TK /E[TK ] > y] ≤ Pr χ21 > y
for all y > 0, with χ21 the same as in (5.3). Moreover, letting K → ∞, it follows
that for all y > 0,
£
¤
Pr [T /E[T ] > y] ≤ Pr χ21 > y .
(5.6)
R
Proof. Recall that Bk = Υk |Wk (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk ), where Wk (.)
is a continuous complex-valued zero-mean Gaussian process on Υk .
Denote
Wk+ (τ1 , τ2 , ..., τk , τk+1 ) =
√
γk Wk (τ1 , τ2 , ....τk )ρk+1 (τk+1 ),
where ρk (τ ) is a sequence of continuous functions on Υ such that for k, m =
1, 2, 3, .....,
Z
Z
ρk (τ )ρm (τ )dµ(τ ) = I(k = m),
ρk (τ )dµ(τ ) = 0.
Υ
Υ
Such a sequence can always be constructed. For example, let {ρk (τ )}∞
k=0 with
ρ0 (τ ) ≡ 1 be the sequence of orthonormal polynomials on Υ. Then for 1 ≤ m <
k ≤ K,
Z
Wm+ (τ1 , τ2 , ..., τm+1 )Wk+ (τ1 , τ2 , ..., τk+1 )dµ(τ1 )dµ(τ2 )...dµ(τK+1 )
K+1
Υ
Z
√ √
= γm γk
Wm (τ1 , τ2 , ..., τm )ρm+1 (τm+1 )Wk (τ1 , τ2 , ..., τk )
Υk
Z
× dµ(τ1 )dµ(τ2 )...dµ(τk ) ρk+1 (τ )dµ(τ )
Υ
=0
31
whereas for m = k,
Z
Wk+ (τ1 , τ2 , ..., τk+1 )Wk+ (τ1 , τ2 , ..., τk+1 )dµ(τ1 )dµ(τ2 )...dµ(τK+1 )
K+1
Υ
Z
= γk
Wk (τ1 , τ2 , ....τm )Wk (τ1 , τ2 , ....τk )dµ(τ1 )dµ(τ2 )...dµ(τk )
Υk Z
×
ρk+1 (τ )2 dµ(τ )
Υ
Z
= γk
|Wk (τ1 , τ2 , ....τm )|2 dµ(τ1 )dµ(τ2 )...dµ(τk )
Υk
= γk Bk
Consequently, denoting
W K (τ1 , τ2 , ..., τK+1 ) =
K
X
Wk+ (τ1 , τ2 , ...., τk+1 ),
k=1
we have
Z
ΥK+1
2
|W K (τ1 , τ2 , ..., τK+1 )| dµ(τ1 )dµ(τ2 )...dµ(τK+1 ) =
K
X
γk Bk .
(5.7)
k=1
Note that for k < K, Wk (τ1 , τ2 , ...., τk ) = WK (τ1 , τ2 , ...., τk , 0, .., 0), hence
W K (τ1 , τ2 , ..., τK+1 ) =
K−1
X
√
γk WK (τ1 , τ2 , ...., τk , 0, .., 0)ρk+1 (τk+1 )
k=1
√
+ γK WK (τ1 , τ2 , ...., τK )ρK+1 (τK+1 )
is a continuous complex-valued zero mean Gaussian process on ΥK+1 because
WK (τ1 , τ2 , ...., τK ) is a continuous complex-valued zero mean Gaussian process on
functions ρk (τ ) are continuous and nonrandom. Therefore, similar to
ΥK and the P
(5.1), TK = K
k=1 γk Bk has the representation (5.4), so that similar to (5.2),
"
#
∙
¸
n
£
¤
1X 2
TK
Pr
> y ≤ Pr sup
εK,j > y = Pr χ21 > y .
(5.8)
E[TK ]
n∈N n
j=1
Finally, recall from (3.23) that supk∈N E[Bk ] < ∞, which implies that
plim TK = T and lim E[TK ] = E[T ]
K→∞
K→∞
32
d
and thus TK /E[TK ] → T /E[T ] as K → ∞. The latter is equivalent to
lim Pr [TK /E[TK ] ≤ y] = Pr [T /E[T ] ≤ y]
K→∞
(5.9)
in the continuity points y of the distribution function of T /E[T ], which proves
inequality (5.9) for all continuity points y of the distribution of T /E[T ].
However, in Theorem 5.3 at the end of this subsection it will be shown that
the distribution function of T is continuous on (0, ∞), and the same applies to
T /E[T ]. Consequently, inequality (5.9) holds for all y > 0.
P n
bn,k
To make these result operational for the WICM test statistic Tbn = Lk=1
γk B
bn,k of the covariance functions Γk of the Wk (τ1 , τ2 , ...., τk )’s such
we need estimates Γ
that
b Tbn ) def.
E(
=
p
→
∞
X
k=1
γk
Ln
X
k=1
Z
γk
Z
Υk
bn,k ((τ1 , τ2 , ...., τk ), (τ1 , τ2 , ...., τk )) dµ(τ1 )dµ(τ2 )...dµ(τk )
Γ
Γk ((τ1 , τ2 , ...., τk ), (τ1 , τ2 , ...., τk )) dµ(τ1 )dµ(τ2 )...dµ(τk ) = E[T ].
Υk
In Bierens (2015a, Section 5.1) I have shown how to derive closed form expresbn,k of Γk for the case that Υ is a
bn,k ’s and the consistent estimates Γ
sions of the B
hypercube centered around the zero vector and µ is the uniform probability measure on Υ. The formulas involved can easily adapted to the present case, which is
left to the reader.
b Tbn ).
Now the following results hold for the standardized WICM test Tbn /E(
Theorem 5.2. For α ∈ (0, 1), let c(α) be such that Pr[χ21 > c(α)] = α, and let
b Tbn ) = Op (1) under H1 .
b Tbn ) be a consistent estimator of E[T ] under H0 , and E(
E(
Under H0 and the conditions of Theorem 3.1,
h
i
b Tbn ) > c(α) ≤ α under H0 ,
lim Pr Tbn /E(
n→∞
h
i
b
b
b
lim Pr Tn /E(Tn ) > c(α) = 1 under H1 .
n→∞
33
Remarks.
b Tbn ) = Op (1) under H1 is not restrictive, because
• The condition that E(
b Tbn ) on a bootstrap version of Tbn
without loss of generality we may base E(
b Tbn ) = Op (1) under H1 . See (6.3) in the next section.
for which E(
• The values of c(α) for α = 0.01, α = 0.05 and α = 0.1 have been calculated
in Bierens and Ploberger (1997), i.e.,
c(0.01) = 6.81, c(0.05) = 4.26, c(0.10) = 3.23.
(5.10)
Finally, I will show that the limiting distribution of the WICM test is continuous.
Theorem 5.3. Let G(x) = Pr[T ≤ x], where T is the asymptotic null distribution
of the WICM test in Theorem 3.1. Under the conditions of Theorem 3.1, G(x) is
continuous on (0, ∞).
Proof. It follows from (5.7) and Theorem 6.5 in Bierens (2015a) that the distribution function
"K
#
X
GK (x) = Pr
γk Bk ≤ x
k=1
is continuous on (0, ∞). Moreover, note that
K
X
k=1
as K → ∞, hence
d
γk Bk →
∞
X
γk Bk
k=1
lim GK (x) = G(x)
K→∞
in the continuity points of G.
To prove that G(x) is continuous on (0, ∞), suppose that x0 > 0 is a discontinuity point of G, i.e.,
G(x0 ) − lim G(x0 − δ) = ε > 0.
δ↓0
34
Hence, for all δ > 0 such that x0 − δ is a continuity point of G,
ε ≤ G(x0 ) − G(x0 − δ) = G(x0 ) − lim GK (x0 − δ)
K→∞
(5.11)
Since GK is continuous, for each K there exists a δK > 0 such that GK (x0 ) −
GK (x0 − δK ) < ε/2, and since for δ ∈ (0, δK ], GK (x0 ) − GK (x0 − δ) < ε/2 as well,
we may without loss of generality assume that limK→∞ δK = 0. Then for arbitrary
δ > 0 and K so large that δ > δK , GK (x0 − δK ) ≥ GK (x0 − δ), hence
G(x0 ) − GK (x0 − δ) ≤
=
≤
≤
G(x0 ) − GK (x0 − δK )
G(x0 ) − GK (x0 ) + GK (x0 ) − GK (x0 − δK )
G(x0 ) − GK (x0 ) + ε/2
ε/2
where that last inequality follows from G(x0 ) ≤ GK (x0 ). Again, assuming that x0 −
δ is a continuity point of G, we have
G(x0 ) − G(x0 − δ) = G(x0 ) − lim GK (x0 − δ) ≤ ε/2.
K→∞
(5.12)
However, (5.11) and (5.12) contradict, which proves that G(x) does not have any
discontinuity points.
6. Bootstrap
In this section I will extend the bootstrap procedure in Bierens (2015a) to the
present case. For that purpose it is necessary to strengthen some of the conditions
in Assumptions 2.2 and the condition in Assumption 3.1 a little bit, as follows.
Assumption 6.1. Let θ0 be defined by (1.6). The following conditions hold,
regardless whether H0 is true or not.
(a) Θ is convex and θ0 is an interior point of Θ.
(b) ft−1 (θ) is a.s. twice continuously differentiable in the components θ1 , θ2 , ..., θm
of θ ∈ Θ;
(c) E [supθ∈Θ ||∇ft−1 (θ)||2 ] < ∞.
(d) E [supθ∈Θ (Yt − ft−1 (θ))2 .||∇ft−1 (θ)||2 ] < ∞.
(e) E [supθ∈Θ (Yt − ft−1 (θ))2 ||∇ft−1 (θ)||] < ∞.6
6
Cf. Assumption 3.1.
35
(f ) For i1 , i2 = 1, 2, ..., m, E[supθ∈Θ |Yt − ft−1 (θ)|.|(∂/∂θi1 )(∂/∂θi2 )ft−1 (θ)|] < ∞.
(g) There exists a sequence Mt of random variables such that for all θ1 , θ2 ∈ Θ,
||(Yt − ft−1 (θ1 ))∇ft−1 (θ1 ) − (Yt − ft−1 (θ2 ))∇ft−1 (θ2 )|| ≤ Mt .||θ1 − θ2 ||
and
1X
E[Mt2 ] < ∞.
lim sup
n
n→∞
t=1
n
(h) For all θ ∈ Θ the matrix A2 (θ) = E[(∇ft−1 (θ))(∇ft−1 (θ))0 ] is nonsingular.7
Except for part (g), this assumption encompasses the conditions in Assumptions
2.2 and 3.1, for ease of reference. Using the mean value theorem, Assumption 6.1(g) can be broken down in more primitive conditions. In particular,
2
0
denoting
p∇ ft−1 (θ) = (∂/∂θ)(∂/∂θ )ft−1 (θ), and defining the matrix norm as
||A|| = trace(AA0 ), it follows from the mean value theorem that
||(Yt − ft−1 (θ1 ))∇ft−1 (θ1 ) − (Yt − ft−1 (θ2 ))∇ft−1 (θ2 )||
≤ |Yt − ft−1 (θ1 )|.||∇ft−1 (θ1 ) − ∇ft−1 (θ2 )|| + |ft−1 (θ1 ) − ft−1 (θ2 )|.||∇ft−1 (θ2 )||
µ
¶
2
2
≤ sup |Yt − ft−1 (θ)|. sup ||∇ ft−1 (θ)|| + sup ||∇ft−1 (θ)|| ||θ1 − θ2 ||.
θ∈Θ
θ∈Θ
θ∈Θ
Theorem 6.1. Denote
1 X
= √
εi,t (Yt − ft−1 (θ0 ))φk,t−1 (τ1 , τ2 , ....τk ),
n t=1
Z
¯
¯ ∗
¯Wi,k,n (τ1 , τ2 , ....τk )¯2 dµ(τ1 )dµ(τ2 )...dµ(τk ),
=
n
∗
(τ1 , τ2 , ....τk )
Wi,k,n
∗
Bi,k,n
∗
Ti,n
=
Υk
L
n
X
∗
γk Bi,k,n
, i = 1, 2, ..., M,
k=1
where the εi,t ’s are independent random drawings from the standard normal distribution, M is the number of bootstraps, φk,t−1 (τ1 , τ2 , ....τk ) is defined by (3.9),
7
Note that the elements of A2 (θ) are continuous in θ and therefore det(A2 (θ)) is continous
in θ. Consequently, if we adopt the condition (d) in Assumption 2.2, i.e., det(A2 (θ0 )) 6= 0, then
we can choose Θ0 so small that det(A2 (θ)) 6= 0 for all θ ∈ Θ0 .
36
and γk and Ln are the same as in Theorem 3.1. Under Assumptions 1.1, 2.1 and
6.1 the following results hold.
(a) For each k ∈ N and i = 1, 2, ..., M,
∗
∗
(τ1 , τ2 , ....τk ) ⇒ Wi,k
(τ1 , τ2 , ....τk )
Wi,k,n
∗
(τ1 , τ2 , ....τk ) is a zero-mean complex-valued Gaussian process
on Υk , where Wi,k
k
on Υ with covariance function
Γ∗k ((τ1,1 , τ1,2 , ....τ1,k ), (τ2,1 , τ2,2 , ....τ2,k ))
i
h
∗
∗
(τ1,1 , τ1,2 , ....τ1,k )Wi.k
(τ2,1 , τ2,2 , ....τ2,k )
= E Wi.k
h
i
2
= E (Yt − ft−1 (θ0 )) φk,t−1 (τ1,1 , τ1,2 , ....τ1,k )φk,t−1 (τ2,1 , τ2,2 , ....τ2,k )
h¡
i
£
¤¢
t−1 2
= E Yt − E Yt |F−∞
φk,t−1 (τ1,1 , τ1,2 , ....τ1,k )φk,t−1 (τ2,1 , τ2,2 , ....τ2,k )
∙
£
¡
¤¢
t−1 2
+ E ft−1 (θ0 ) − E Yt |F−∞
¸
× φk,t−1 (τ1,1 , τ1,2 , ....τ1,k )φk,t−1 (τ2,1 , τ2,2 , ....τ2,k )
= Γk ((τ1,1 , τ1,2 , ....τ1,k ), (τ2,1 , τ2,2 , ....τ2,k ))
∙
£
¤¢
¡
t−1 2
+ E ft−1 (θ0 ) − E Yt |F−∞
¸
× φk,t−1 (τ1,1 , τ1,2 , ....τ1,k )φk,t−1 (τ2,1 , τ2,2 , ....τ2,k )
(6.1)
d
∗
∗
→ Bi,k
, and
where Γk is the covariance function (3.21). Hence, Bi,k,n
∞
X
d
Ti,n → Ti∗ =
where
∗
Bi,k
(b) Moreover,
=
Z
Υk
∗
γk Bi,k
k=1
¯
¯ ∗
¯Wi,k (τ1 , τ2 , ....τk )¯2 dµ(τ1 )dµ(τ2 )...dµ(τk ).
d
∗
∗
∗
∗ 0
(T1,n
, T2,n
, ..., TM,n
)0 → (T1∗ , T2∗ , ..., TM
),
37
(6.2)
∗
where T1∗ , T2∗ , ..., TM
are i.i.d.. Furthermore, under H0 ,
d
∗
∗
∗
(T1,n
, T2,n
, ..., TM,n
)0 → (T1 , T2 , ..., TM )0 ,
where T1 , T2 , ..., TM are i.i.d. as T in (3.28) and are independent of T .
Proof. Part (a) follows similar to Lemma 3.5 and Theorem 3.1. As to part (b), it
follows from the independence of εi,t ’s that, conditional on the data, T1 , T2 , ..., TM
are i.i.d. and are therefore i.i.d. unconditionally as well. Moreover, under H0 the
covariance function (6.1) is exactly the same as (3.21), hence Ti ∼ T. Furthermore,
since the εi,t ’s are drawn independently of the data, (T1 , T2 , ..., TM )0 is independent
of T .
∗
If it were possible to compute the Ti,n
’s then the asymptotic α × 100% crit∗
ical values c(α) of the WICM test can be approximated by sorting the Ti,n
’s in
∗
decreasing order and then use cn,M (α) = T[αM],n as the approximation of c(α),
where [αM] denotes the largest natural number less or equal to αM.
However, obviously the computation of the T i,n ’s is not possible. Therefore,
to make this bootstrap procedure feasible, we need to construct feasible bootstrap
∗
WICM statistics Tei,n such that Tei,n = Ti,n
+ op (1), as follows.
Denote for i = 1, 2, ..., M,
à k
!
n
X
X
1
bbn,k (τ1 , τ2 , ....τk |θ) =
∇ft−1 (θ) exp i
τj0 Φ(Zt−j ) ,
n t=1
j=1
X
b2,n (θ) = 1
∇ft−1 (θ)(∇ft−1 (θ))0
A
n t=1
à k
!
X
bk,n,t−1 (τ1 , τ2 , ....τk |θ) = exp i
τj0 Φ(Zt−j )
φ
n
j=1
b−1
−bbn,k (τ1 , τ2 , ....τk |θ)0 A
2,n (θ)∇ft−1 (θ),
n
X
bk,n,t−1 (τ1 , τ2 , ....τk |θ),
fi,k,n (τ1 , τ2 , ....τk |θ) = √1
εi,t (Yt − ft−1 (θ))φ
W
n t=1
Z ¯
¯2
¯
¯f
b
e
Bi,k,n =
¯Wi,k,n (τ1 , τ2 , ....τk |θn )¯ dµ(τ1 )dµ(τ2 )...dµ(τk ),
Tei,n =
Υk
L
n
X
k=1
ei,k,n ,
γk B
38
(6.3)
where the εi,t ’s are the same as in Theorem 6.1.
We can write
fi,k,n (τ1 , τ2 , ....τk |b
θn)
W
à k
!
n
X
1 X
εi,t (Yt − ft−1 (θ0 )) exp i
τj0 Φ(Zt−j )
=√
n t=1
j=1
−b
an,k,i (θ, τ1 , τ2 , ....τk )
b−1 (b
− bbn,k (τ1 , τ2 , ....τk |b
θn )0 A
θn )dbn,k,i (b
θn , τ1 , τ2 , ....τk )
2,n
where
1 X
εi,t (ft−1 (θ) − ft−1 (θ0 ))
b
an,k,i (θ, τ1 , τ2 , ....τk ) = √
n t=1
!
à k
X
τj0 Φ(Zt−j ) ,
× exp i
n
j=1
n
1 X
b
εi,t Ψk,t (θ, τ1 , τ2 , ....τk ), with
dn,k,i (θ, τ1 , τ2 , ....τk ) = √
n t=1
à k
!
X
τj0 Φ(Zt−j ) .
Ψk,t (θ, τ1 , τ2 , ....τk ) = (Yt − ft−1 (θ))∇ft−1 (θ) exp i
j=1
Similarly,
∗
(τ1 , τ2 , ....τk )
Wi,k,n
à k
!
n
X
1 X
εi,t (Yt − ft−1 (θ0 )) exp i
τj0 Φ(Zt−j )
=√
n t=1
j=1
b
− bk (τ1 , τ2 , ....τk )0 A−1
2 dn,k,i (θ0 , τ1 , τ2 , ....τk ).
I will show first that
¯
¯
¯
¯f
∗
b
sup
¯Wi,k,n (τ1 , τ2 , ....τk |θn ) − Wi,k,n (τ1 , τ2 , ....τk )¯ = op (1),
(6.4)
(τ1 ,τ2 ,....τk )∈Υk
using the following three lemmas, and then that
∗
+ op (1).
Tei,n = Ti,n
39
(6.5)
Lemma 6.1. Under the conditions of Theorem 6.1 and for each k and i,
dbn,k,i (θ, τ1 , τ2 , ....τk ) is tight on Θ × Υk . Consequently,
°
°
°b
°
b
b
sup
(6.6)
°dn,k,i (θn , τ1 , τ2 , ....τk ) − dn,k,i (θ0 , τ1 , τ2 , ....τk )° = op (1),
(τ1 ,τ2 ,....τk )∈Υk
where here and in the sequel the norm ||.|| on Cm is defined as ||a + i.b|| =
√
a0 a + b0 b.
Proof. According to Lemma A.1 in Bierens and Ploberger (1997), in multivariate
form, Re[dbn,k,i (θ, τ1 , τ2 , ....τk )] is tight on Θ × Υk if for any pair
(θ1 , τ1,1 , τ1,2 , ....τ1,k ), (θ2 , τ2,1 , τ2,2 , ....τ2,k ) ∈ Θ × Υk ,
the following Lipschitz condition holds:
|| Re[Ψk,t (θ1 , τ1,1 , τ1,2 , ....τ1,k )] − Re[Ψk,t (θ2 , τ2,1 , τ2,2 , ....τ2,k )]|
v
u
k
X
u
t
2
≤ Kk,t ||θ1 − θ2 || +
||τ2,j − τ1,j ||2 ,
(6.7)
1X £ 2 2 ¤
E εi,t Kk,t < ∞,
n→∞ n
t=1
(6.8)
j=1
where Kk,t is a sequence of random variables such that
n
lim sup
and for an arbitrary point (θ, τ1 , τ2 , ....τk ) ∈ Θ × Υk ,
¤
1X £ 2
E εi,t || Re(Ψk,t (θ, τ1 , τ2 , ....τk ))||2 < ∞.
lim sup
n→∞ n
t=1
n
Condition (6.7) follows from
|| Re[Ψk,t (θ1 , τ1,1 , τ1,2 , ....τ1,k )] − Re[Ψk,t (θ2 , τ2,1 , τ2,2 , ....τ2,k )]||
°
à k
!
°
X
°
0
τ1,j
Φ(Zt−j )
= °(Yt − ft−1 (θ1 ))∇ft−1 (θ1 ) cos
°
j=1
40
(6.9)
− (Yt − ft−1 (θ2 ))∇ft−1 (θ2 ) cos
à k
X
!°
°
°
°
°
0
τ2,j
Φ(Zt−j )
j=1
≤ ||(Yt − ft−1 (θ1 ))∇ft−1 (θ1 ) − (Yt − ft−1 (θ2 ))∇ft−1 (θ2 )||
°
°
k
°
°
X
°
°
0
(τ2,j − τ1,j ) Φ(Zt−j )°
+ °(Yt − ft−1 (θ2 ))∇ft−1 (θ2 )
°
°
j=1
≤ Mt−1 .||θ1 − θ2 ||
à k
!
µ
¶
X
+ sup ||Φ(z)|| sup (|Yt − ft−1 (θ)|.||∇ft−1 (θ)||)
||τ2,j − τ1,j ||
z
θ∈Θ
j=1
v
u
k
X
u
t
2
||τ2,j − τ1,j ||2 ,
≤ Kk,t ||θ1 − θ2 || +
j=1
where the second inequality follows from Assumption 6.1(g), and the last inequality from
µ
µ
¶
¶
√
Kk,t = k + 1 Mt + sup ||Φ(z)|| sup (|Yt − ft−1 (θ)|.||∇ft−1 (θ)||)
z
θ∈Θ
and
||θ1 − θ2 || +
k
X
j=1
v
u
k
X
u
√
t
2
||τ2,j − τ1,j || ≤ k + 1 ||θ1 − θ2 || +
||τ2,j − τ1,j ||2 .
j=1
Moreover, condition (6.8) follows from
1X £ 2 2 ¤
1X £ 2 ¤
E εi,t Kk,t = lim sup
E Kk,t
lim sup
n→∞ n
n→∞ n
t=1
t=1
n
n
1X
≤ 2(k + 1) lim sup
E[Mt2 ]
n→∞ n
t=1
¸
∙
2
2
2
+ 2(k + 1) sup ||Φ(z)|| E sup(Yt − ft−1 (θ)) .||∇ft−1 (θ)||
n
z
θ∈Θ
< ∞,
where the last inequality is due to parts (d) and (g) of Assumption 6.1.
41
Finally, condition (6.9) follows from part (d) of Assumption 6.1 and the easy
inequality
¸
∙
£
¤
2
2
2
E || Re(Ψk,t (θ, τ1 , τ2 , ....τk ))|| ≤ E sup(Yt − ft−1 (θ)) ||∇ft−1 (θ)|| .
θ∈Θ
Thus, Re[dbn,k,i (θ, τ1 , τ2 , ....τk )] is tight on Θ × Υk . Obviously, the same holds for
Im[dbn,k,i (θ, τ1 , τ2 , ....τk )] and thus for dbn,k,i (θ, τ1 , τ2 , ....τk ) as well.
Using the martingale difference central limit theorem of McLeish (1974) it follows straightforwardly that the finite distributions of dbn,k,i (θ, τ1 , τ2 , ....τk ) converge
to multivariate zero-mean complex-valued normal distributions, hence
dbn,k,i (θ, τ1 , τ2 , ....τk ) ⇒ dk,i (θ, τ1 , τ2 , ....τk )
on Θ × Υk , where dk,i (θ, τ1 , τ2 , ....τk ) is a an m-dimensional zero-mean complexvalued Gaussian process on Θ × Υk . Similarly,
dbn,k,i (θ, τ1 , τ2 , ....τk ) − dbn,k,i (θ0 , τ1 , τ2 , ....τk )
k
⇒ dk,i (θ, τ1 , τ2 , ....τk ) − dk,i (θ0 , τ1 , τ2 , ....τk )
(6.10)
on Θ × Υ .
Now let Θ0 (ε) = {θ ∈ Rm : ||θ − θ0 || ≤ ε}, where ε > 0 is so small that
Θ0 (ε) ⊂ Θ. and denote
°
°
°b
°
b
b n,k,i (θ) =
d
sup
(θ,
τ
,
τ
,
....τ
)
−
d
(θ
,
τ
,
τ
,
....τ
)
∆
° n,k,i
1 2
k
n,k,i 0 1 2
k °,
(τ1 ,τ2 ,....τk )∈Υk
∆k,i (θ) =
sup
(τ1 ,τ2 ,....τk )∈Υk
kdk,i (θ, τ1 , τ2 , ....τk ) − dk,i (θ0 , τ1 , τ2 , ....τk )k .
Then by the continuous mapping theorem, (6.10) implies
d
b n,k,i (θ) →
sup ∆
sup ∆k,i (θ)
θ∈Θ0 (ε)
θ∈Θ0 (ε)
and thus for any δ > 0,
i
h
i
h
b
b
b
b
b
Pr ∆n,k,i (θn ) > δ = Pr ∆n,k,i (θn ) > δ and θn ∈ Θ0 (ε)
h
i
b n,k,i (b
+ Pr ∆
θn ) > δ and b
θn ∈
/ Θ0 (ε)
"
#
h
i
b n,k,i (θ) > δ + Pr ||b
≤ Pr sup ∆
θn − θ0 || > ε
θ∈Θ0 (ε)
→ Pr
"
sup ∆k,i (θ) > δ
θ∈Θ0 (ε)
42
#
Since limε↓0 supθ∈Θ0 (ε) ∆k,i (θ) = ∆k,i (θ0 ) = 0 a.s. and thus
"
#
lim Pr
ε↓0
sup ∆k,i (θ) > δ = 0,
θ∈Θ0 (ε)
h
i
b n,k,i (b
it follows now that for any δ > 0, limn→∞ Pr ∆
θn ) > δ = 0. The latter
proves (6.6).
Similarly,
Lemma 6.2. Under the conditions of Theorem 6.1 and for each k and i,
b
an,k,i (θ, τ1 , τ2 , ....τk ) is tight on Θ × Υk . Consequently,
¯
¯
¯
¯
b
an,k,i (θn , τ1 , τ2 , ....τk )¯ = op (1).
sup
¯b
(τ1 ,τ2 ,....τk )∈Υk
Moreover, it follows from Assumption 6.1(c) and Lemma 2.1 that
° h
i
³ h
i´°
°
°
b
b
sup
sup °Re bn,k (τ1 , τ2 , ....τk |θ) − E Re bn,k (τ1 , τ2 , ....τk |θ) °
(τ1 ,τ2 ,....τk )∈Υk θ∈Θ
= op (1),
sup
° h
i
³ h
i´°
°
°
b
b
sup °Im bn,k (τ1 , τ2 , ....τk |θ) − E Im bn,k (τ1 , τ2 , ....τk |θ) °
(τ1 ,τ2 ,....τk )∈Υk θ∈Θ
= op (1),
and
°
h
i°
°b
°
b
sup °A
2,n (θ) − E A2,n (θ) ° = op (1),
θ∈Θ
where in the latter case the matrix norm involved is defined by ||A|| =
hence by Lemma 2.1 and Assumption 6.1(h) it follows that
p
trace(AA0 ),
Lemma 6.3. Under the conditions of Theorem 6.1 and for each k,
°
°
° b b −1b
°
θn ) − A−1
(θ
)b
(τ
,
τ
,
....τ
)
sup
°A2,n (θn ) bn,k (τ1 , τ2 , ....τk |b
° = op (1).
0
k
1
2
k
2
(τ1 ,τ2 ,....τk )∈Υk
Combining the results of Lemmas 6.1-6.3 it follows that (6.4) holds, which in
its turn implies that,
43
ei,k,n − B ∗ | = op (1) for
Lemma 6.4. Under the conditions of Theorem 6.1, |B
i,k,n
each k ∈ N and each i = 1, 2, ..., M.
Proof. Denote
δi,k,n =
sup
(τ1 ,τ2 ,....τk )∈Υk
It is easy to verify that
¯
¯
¯f
¯
∗
θn ) − Wi,k,n
(τ1 , τ2 , ....τk )¯ .
¯Wi,k,n (τ1 , τ2 , ....τk |b
∗
ei,k,n − Bi,k,n
|
|B
¯¯
¯
¯2 ¯
¯¯
¯2 ¯
¯
∗
b
f
¯
¯
¯
¯
≤
sup
¯¯Wi,k,n (τ1 , τ2 , ....τk |θn )¯ − Wi,k,n (τ1 , τ2 , ....τk ) ¯
k
(τ1 ,τ2 ,....τk )∈Υ
¯
¯ ∗
2
¯Wi,k,n (τ1 , τ2 , ....τk )¯ .
≤ δi,k,n
+ 2δi,k,n
sup
(τ1 ,τ2 ,....τk )∈Υk
Since by part (a) of Theorem 6.1,
∗
∗
(τ1 , τ2 , ....τk ) ⇒ Wi,k
(τ1 , τ2 , ....τk )
Wi,k,n
on Υk , it follows from the continuous mapping theorem that
sup
(τ1 ,τ2 ,....τk )∈Υk
which implies that
¯ d
¯ ∗
¯Wi,k,n (τ1 , τ2 , ....τk )¯ →
sup
(τ1 ,τ2 ,....τk )∈Υk
sup
(τ1 ,τ2 ,....τk )∈Υk
¯
¯ ∗
¯Wi,k (τ1 , τ2 , ....τk )¯ ,
¯
¯ ∗
¯Wi,k,n (τ1 , τ2 , ....τk )¯ = Op (1).
The lemma now follows straightforwardly from the latter and δi,k,n = op (1).
Lemma 6.5. Under the conditions of Theorem 6.1,
Ln
X
k=1
hence
ei,k,n −
γk B
∞
X
∗
γk Bi,k,n
= op (1),
k=1
∗
∗
∗
, T2,n
, ..., TM,n
)0 + op (1),
(Te1,n , Te2,n , ..., TeM,n )0 = (T1,n
44
∗
where Ti,n
is defined in Theorem 6.1 and Tei,n in (6.3).
Proof. Note that similar to (3.18),
h
i
2
∞
f
E |Wi,k,n (τ1 , τ2 , ....τk |θ)| |F−∞
1X
(Yt − ft−1 (θ))2
n t=1
n
=
b−1
−2 Re[bbn,k (τ1 , τ2 , ....τk |θ)]0 A
cn,k (τ1 , τ2 , ....τk |θ)]
2,n (θ) Re[b
b−1
cn,k (τ1 , τ2 , ....τk |θ)]
−2 Im[bbn,k (τ1 , τ2 , ....τk |θ)]0 A
2,n (θ) Im[b
b
b−1
b
b−1
+ Re[bbn,k (τ1 , τ2 , ....τk |θ)]0 A
2,n (θ)A1,n (θ)A2,n (θ) Re[bn,k (τ1 , τ2 , ....τk |θ)]
b−1 (θ)A
b1,n (θ)A
b−1 (θ) Im[bbn,k (τ1 , τ2 , ....τk |θ)]
+ Im[bbn,k (τ1 , τ2 , ....τk |θ)]0 A
2,n
where
2,n
à k
!
n
X
1X
(Yt − ft−1 (θ))2 ∇ft−1 (θ) exp i
τj0 Φ(Zt−j ) ,
b
cn,k (τ1 , τ2 , ....τk |θ) =
n t=1
j=1
X
b1,n (θ) = 1
A
(Yt − ft−1 (θ))2 (∇ft−1 (θ))(∇ft−1 (θ))0 .
n t=1
n
Then similar to (3.19),
Z
i
h
∞
fi,k,n (τ1 , τ2 , ....τk |θ)|2 |F−∞
dµ(τ1 )dµ(τ2 )...dµ(τk )
E |W
Υk
q
n
1X
2
b
b2,n (θ))
(Yt − ft−1 (θ)) + 4.trace(A2,n (θ)) trace(A
≤
n t=1
q
n
X
b2,n (θ)) trace(A
b2,n (θ)) 1
+4.trace(A
(Yt − ft−1 (θ))2 ||∇ft−1 (θ)||
n
³
´ t=1³
´
−1
−1
b
b
b
b
+2.trace A2,n (θ)A1,n (θ)A2,n (θ) trace A2,n (θ)
bn (θ),
=R
h
i
∞
e
bn (b
say, hence E Bi,k,n |F−∞ ≤ R
θn ) and thus for K ∈ N,
¯
#
" ∞
∞
¯
X
X
¯
∞
b
e
b
γk Bi,k,n ¯ F−∞ ≤ Rn (θn )
γk .
E
¯
k=K
k=K
45
It is not hard (but ¯somewhat tedious)
to verify from Assumption 6.1 and
¯
¯b
¯
Lemma 2.1 that supθ∈Θ ¯Rn (θ) − R(θ)¯ = op (1) and thus by Lemma 2.1,
p
bn (b
θn ) → R(θ0 ),
R
where
p
R(θ) = E[(Yt − ft−1 (θ))2 ] + 4.trace(A2 (θ)) trace(A2 (θ))
p
£
¤
+4.trace(A2 (θ)) trace(A2 (θ))E (Yt − ft−1 (θ))2 ||∇ft−1 (θ)||
¢
¡
−1
(θ)A
(θ)A
(θ)
trace (A2 (θ))
+2.trace A−1
1
2
2
with A2 (θ) defined in Assumption 6.1(h) and
£
¤
A1 (θ) = E (Yt − ft−1 (θ))2 (∇ft−1 (θ))(∇ft−1 (θ))0 .
Now by Chebyshev’s inequality for conditional probabilities and first moments,
it follows that for arbitrary β > 0,
i
hP
¯
#
" ∞
∞
P∞
∞
e
¯
E
X
k=K γk Bi,k,n |F−∞
γk
¯
∞
ei,k,n > β.R
bn (b
≤ k=K ,
Pr
γk B
θn )¯ F−∞ ≤
b
b
¯
β
β.Rn (θn )
k=K
hence
P∞
¯
#!
¯
¯
∞
bn (b
ei,k,n > β.R
Pr
γk B
θn )¯ F−∞
¯
k=K
#
" ∞
X
bn (b
ei,k,n > β.R
γk B
θn )
= Pr
k=K γk
≥ E
β
Ã
= Pr
+ Pr
≥ Pr
+ Pr
"
∞
X
"k=K
∞
X
" k=K
∞
X
"k=K
∞
X
" k=K
∞
X
k=K
#
ei,k,n > β.R
bn (b
bn (b
γk B
θn ) and |R
θn ) − R(θ0 )| > 1
#
ei,k,n > β.R
bn (b
bn (b
γk B
θn ) and |R
θn ) − R(θ0 )| ≤ 1
#
ei,k,n > β.R
bn (b
bn (b
γk B
θn ) and |R
θn ) − R(θ0 )| > 1
#
ei,k,n > β.(R(θ0 ) + 1) and |R
bn (b
γk B
θn ) − R(θ0 )| ≤ 1
46
= Pr
+ Pr
− Pr
Since
lim sup Pr
n→∞
"
∞
X
" k=K
∞
X
"k=K
∞
X
∞
X
k=K
bn (b
bn (b
ei,k,n > β.R
γk B
θn ) and |R
θn ) − R(θ0 )| > 1
#
ei,k,n > β.(R(θ0 ) + 1)
γk B
#
bn (b
ei,k,n > β.(R(θ0 ) + 1) and |R
γk B
θn ) − R(θ0 )| >(6.11)
1
k=K
"
#
#
ei,k,n > β.R
bn (b
bn (b
γk B
θn ) and |R
θn ) − R(θ0 )| > 1
h
i
bn (b
≤ lim sup Pr |R
θn ) − R(θ0 )| > 1 = 0
n→∞
and
lim sup Pr
n→∞
"
∞
X
k=K
#
ei,k,n > β.(R(θ0 ) + 1) and |R
bn (b
γk B
θn ) − R(θ0 )| > 1
h
i
bn (b
≤ lim sup Pr |R
θn ) − R(θ0 )| > 1 = 0
n→∞
it follows now from (6.11) that for arbitrary β > 0,
" ∞
# P
∞
X
ei,k,n > β.(R(θ0 ) + 1) ≤ k=K γk
γk B
lim sup Pr
β
n→∞
k=K
For arbitraryPε > 0 and δ ∈ (0, 1), choose β = ε/(R(θ0 ) + 1) and let K0 (ε, δ) ∈
N be such that ∞
k=K0 (ε,δ) γk < δ.β. Then for K ≥ K0 (ε, δ),
" ∞
#
X
ei,k,n > ε < δ.
γk B
(6.12)
lim sup Pr
n→∞
k=K
∗
] ≤ R(θ0 ), so that K0 (ε, δ) can be chosen
Note that similar to (3.19), E[Bi,k,n
such that for K ≥ K0 (ε, δ),
" ∞
#
X
∗
γk Bi,k,n
> ε < δ.
(6.13)
lim sup Pr
n→∞
k=K
47
It follows now from (6.12), (6.13) and Lemma 6.4, similar to the proof of
P
P∞
P∞
∗
e
e
Lemma 3.4, that ∞
k=1 γk Bi,k,n −
k=1 γk Bi,k,n = op (1) and
k=Ln +1 γk Bi,k,n =
op (1).
Summarizing, the following results have been proved.
Theorem 6.2. Under the conditions and notations in Theorem 6.1 it follows
that
d
∗ 0
(Te1,n , Te2,n , ..., TeM,n )0 → (T1∗ , T2∗ , ..., TM
),
∗
are i.i.d.. Moreover, with Tbn the WICM test statistic, it
where T1∗ , T2∗ , ..., TM
follows that under H0 ,
d
(Tbn , Te1,n , Te2,n , ..., TeM,n )0 → (T, T1 , T2 , ..., TM )0 ,
where T, T1 , T2 , ..., TM are i.i.d.
Again, the asymptotic α × 100% critical values c0 (α) of the WICM test, i.e.,
Pr[T > c0 (α)] = α, can be approximated by sorting the Tei,n ’s in decreasing order
and then use e
cn,M (α) = Te[αM],n as the approximation of c0 (α). More precisely,
Theorem 6.3. Let e
cn,M (α) be the 1 − α quantile of the empirical distribution
function
M
´
1 X ³e
e
Gn,M (x) =
I Ti,n ≤ x ,
M i=1
i.e., e
cn,M (α) = arg minGe n,M (x)≥1−α x, and let c(α) be the 1 − α quantile of the
distribution functions G(x) = Pr[T ≤ x]. Then under H0 , the conditions of
Theorem 6.2 and for arbitrary δ > 0,
h
i
en,M (x) − G(x)| > δ ≤ 1
lim sup Pr |G
4δM
n→∞
hence
lim lim sup Pr[|e
cn,M (α) − c(α)| > δ] = 0.
M→∞
n→∞
Moreover, under H1 , Pr[Tbn > e
cn,M (α)] → 1 for any M ∈ N.
48
Proof. Similar to Theorems 6.2 and 6.3 in Bierens (2015a), using the fact that by
Theorem 5.3, G(x) is continuous on (0, ∞), so that for all α ∈ (0, 1), G(c(α)) =
1 − α.
Under H1 we have for arbitrary δ > 0
h
i
1
∗
e
lim sup Pr |Gn,M (x) − G (x)| > δ ≤
4δM
n→∞
where G∗ (x) = Pr[T1∗ ≤ x], and
cn,M (α) − c∗ (α)| > δ] = 0
lim lim sup Pr[|e
M→∞
n→∞
where c∗ (α) is the 1 − α quantile of G∗ (x), but these results are of no direct
used, except that they imply that under H1 , e
cn,M (α) = Op (1) and thus Pr[Tbn >
e
cn,M (α)] → 1, for any M ∈ N.
7. The initial value problem
The results in this addendum were based on the assumption that all lagged Zt ’s
are observed, which is obviously not realistic. In particular, given that Zt is
observed for 1 − t0 ≤ t ≤ n, where t0 ≥ max(p, q), the empirical counter-part of
ψk (τ1 , τ2 , ....τk ) defined in (1.14) is
⎛
⎞
min(k,t−t0 )
n
X
X
e k,n (τ1 , τ2 , ....τk ) = 1
ψ
(Yt − ft−1 (b
θn )) exp ⎝i.
τj0 Φ(Zt−j )⎠
n t=1
j=1
√
fk,n (τ1 , τ2 , ....τk )/ n
= W
whereas all the asymptotic results were based on
à k
!
n
X
X
1
b k,n (τ1 , τ2 , ....τk ) =
(Yt − ft−1 (b
θn )) exp i.
τj0 Φ(Zt−j )
ψ
n t=1
j=1
√
ck,n (τ1 , τ2 , ....τk )/ n
= W
As alluded to in the introduction, we need to set forth conditions under which
fk,n . The
ck,n are the same as for W
the asymptotic results above on the basis of W
condition involved is condition (7.1) in the following theorem.
49
Theorem 7.1. Denote
Z ¯
¯2
¯
¯f
e
Bn,k =
¯Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk ),
k
ZΥ ¯
¯2
¯c
¯
b
Bn,k =
¯Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk ).
Υk
If the sequence of weights γk is chosen such that
∞
X
k=1
√
k2 γk < ∞
(7.1)
then under H0 and the conditions and notations in Theorem 3.1,
Ln
X
k=1
whereas under H1 ,
Ln
X
k=1
en,k =
γk B
Ln
X
k=1
en,k /n =
γk B
√
bn,k + Op (1/ n),
γk B
Ln
X
k=1
bn,k /n + Op (1/n),
γk B
Proof. Observe that
¯
⎛
⎞¯
à k
!
¯
¯
min(k,t−t0 )
X
X
¯
¯
0
0
¯exp i.
τj Φ(Zt−j ) − exp ⎝i.
τj Φ(Zt−j )⎠¯¯
¯
¯
¯
j=1
j=1
¯
¯
¯
¯
k
X
¯
¯
0
¯
≤¯
τj Φ(Zt−j )¯¯
¯j=min(k,t−t0 )+1
¯
= 0 for t ≥ k + t0
≤ C(k + t0 − 1) for 1 ≤ t < k + t0
where C = supτ ∈Υ ||τ ||. supz∈Rs+1 ||Φ(z)||. Hence,
k+t0 −1
¯
¯
1 X
¯f
¯
c
|Yt − ft−1 (b
θn )|(k + t0 − 1)
¯Wk,n (τ1 , τ2 , ....τk ) − Wk,n (τ1 , τ2 , ....τk )¯ ≤ C. √
n t=1
50
and thus
¯
¯
¯f
2
2¯
c
¯|Wk,n (τ1 , τ2 , ....τk )| − |Wk,n (τ1 , τ2 , ....τk )| ¯
¯
¯2
¯f
¯
c
≤ ¯W
(τ
,
τ
,
....τ
)
−
W
(τ
,
τ
,
....τ
)
k,n 1 2
k
k,n 1 2
k ¯
¯
√ ¯¯
¯
ck,n (τ1 , τ2 , ....τk ) |
f
c
+ 2 2 ¯Wk,n (τ1 , τ2 , ....τk ) − Wk,n (τ1 , τ2 , ....τk )¯ × |W
Ã
!2
k+t0 −1
1 X
2
√
|Yt − ft−1 (b
θn )| (k + t0 − 1)2
≤C
n t=1
k+t0 −1
√
1 X
ck,n (τ1 , τ2 , ....τk )|
+ 2 2C √
|Yt − ft−1 (b
θn )|(k + t0 − 1) × |W
n t=1
2
2
where the
√ first inequality follow from the easy inequality ||z1 | − |z2 | | ≤ |z1 −
2
z2 | + 2 2|z1 − z2 |.|z2 | for complex numbers z1 and z2 . Consequently,
Ã
!2
k+t
0 −1
X
1
|Yt − ft−1 (b
θn )|
k + t0 − 1 t=1
q
k+t
0 −1
X
1 √
bn,k
+ √ 2 2C(k + t0 − 1)
|Yt − ft−1 (b
θn )| × B
n
t=1
¯
¯
¯e
bn,k ¯¯ ≤ 1 C 2 (k + t0 − 1)4
¯Bn,k − B
n
k+t
0 −1
X
1 2
1
4
C (k + t0 − 1)
≤
(Yt − ft−1 (b
θn ))2
n
k + t0 − 1 t=1
q
k+t
0 −1
X
1 √
b
bn,k
+ √ 2 2C(k + t0 − 1)
|Yt − ft−1 (θn )| × B
n
t=1
which implies that
¯
¯L
Ln
n
¯
¯X
X
¯
¯
e
b
γ
−
γ
B
B
¯
k n,k
k n,k ¯
¯
¯
k=1
k=1
!
k+t
0 −1
X
1
sup(Yt − ft−1 (θ))2
k + t0 − 1 t=1 θ∈Θ
Ã
!
k+t
Ln
0 −1
X
1 √ X
1
√
+ √ 2 2C
γk (k + t0 − 1)2
sup |Yt − ft−1 (θ)|
k
+
t
n
0 − 1 t=1 θ∈Θ
k=1
Ln
X
1
≤ C2
γk (k + t0 − 1)4
n k=1
Ã
51
v
u Ln
uX
bn,k
×t
γk B
k=1
Next, note that (7.1) implies
∞
X
k=1
k4 γk < ∞
(7.2)
Since by Assumption 1.1 and (7.1) and (7.2),
Ã
!#
"L
k+t
n
0 −1
X
X
1
γk (k + t0 − 1)4
sup(Yt − ft−1 (θ))2
E
k
+
t
0 − 1 t=1 θ∈Θ
k=1
¸X
∙
∞
2
γk (k + t0 − 1)4 < ∞
= E sup(Yt − ft−1 (θ))
θ∈Θ
k=1
and
"L
Ã
n
X
√
E
γk (k + t0 − 1)2
k=1
!#
k+t
0 −1
X
1
sup |Yt − ft−1 (θ)|
k + t0 − 1 t=1 θ∈Θ
¸X
∙
∞
√
γk (k + t0 − 1)2 < ∞,
= E sup |Yt − ft−1 (θ)|
θ∈Θ
k=1
PLn
d P∞
bn,k →
γk B
whereas by Theorem 3.1,
k=1 γk Bk under H0 , which implies
k=1
PLn
b
that k=1 γk Bn,k = Op (1), it follows now that under H0 ,
¯
¯L
Ln
n
¯
¯X
X
√
¯
¯
e
b
γk Bn,k −
γk Bn,k ¯ = Op (1/ n).
¯
¯
¯
k=1
k=1
Finally, the result under H1 follows easily from
¯L
¯
Ln
n
¯X
¯
X
¯
¯
e
b
γk Bn,k /n −
γk Bn,k /n¯
¯
¯
¯
k=1
k=1
Ã
!
k+t
Ln
0 −1
X
1 2X
1
≤ 2C
γk (k + t0 − 1)4
sup(Yt − ft−1 (θ))2
n
k
+
t
0 − 1 t=1 θ∈Θ
k=1
52
Ln
1 √ X
√
+ 2 2C
γk (k + t0 − 1)2
n
k=1
v
u Ln
uX
bn,k /n
×t
γk B
Ã
!
k+t
0 −1
X
1
sup |Yt − ft−1 (θ)|
k + t0 − 1 t=1 θ∈Θ
k=1
and Theorem 4.1.
8. Standardization of the conditioning variables
Another unresolved issue in B84 is how to standardize the conditioning lagged
variables in Zt before taking the bounded transformation Φ in order to preserve
enough variation in Φ(Zt ). See section 6 in B84.
To discuss this issue, let us for the time being assume that Zt = Yt , so that
the nonlinear ARX model (1.1) becomes a nonlinear AR model:
Yt = f (Yt−1 , Yt−2 , ..., Yt−p , θ0 ) + Ut = ft−1 (θ0 ) + Ut , say.
Moreover, as in B82, let us assume that
(8.1)
Φ(x) = arctan(x),
and note that
¡
¢−1
.
Φ0 (x) = 1 + x2
(8.2)
8.1. A wrong way to standardize
Following B82, one may be tempted to standardize each lagged Yt as Yen,t =
bn , where
σ
b−1
bn ) = α
b n Yt − β
n (Yt − µ
v
u
n
n
X
X
u
t
µ
bn = (1/(n + t0 )
Yt , σ
bn = (1/(n + t0 − 1))
(Yt − µ
bn )2 ,
t=1−t0
t=1−t0
bn = µ
α
b n = 1/b
σn, β
bn /b
σn,
and 1 − t0 is the time index of the first observed Yt . Then (1.17) becomes
!
à k
n
´
³
X
X
1
bn
fk,n (τ1 , τ2 , ....τk ) = √
W
.
(Yt − ft−1 (b
θn )) exp i.
τj Φ α
b n Yt−j − β
n t=1
j=1
(8.3)
53
Similarly, denoting
µ0 = E[Yt ], σ0 =
p
E[(Yt − µ)2 ], α0 = 1/σ0 , β0 = µ0 /σ0 ,
(8.4)
replace Φ(Yt−j ) in (1.17) by Φ (α0 Yt−j − β0 ) , i.e.,
à k
!
n
X
X
1
ck,n (τ1 , τ2 , ....τk ) = √
(Yt − ft−1 (b
θn )) exp i.
τj Φ (α0 Yt−j − β0 ) . (8.5)
W
n t=1
j=1
fk,n (τ1 , τ2 , ....τk ) and W
ck,n (τ1 , τ2 , ....τk ) are asNow the hope is that under H0 , W
fk,n (τ1 , τ2 , ....τk ) has
ymptotically equivalent, in the sense the WICM test based W
the same limiting null distribution as the (now infeasible) WICM test based on
ck,n (τ1 , τ2 , ....τk ).
W
bn = β0 + Op (n−1/2 ),
It can be shown that it does if α
b n = α0 + Op (n−1/2 ) and β
but it is unlikely that these conditions hold in the time series case. Moreover, even
if it does (which is doubtful), this particular standardization is not recommended
because it destroys the martingale difference structure of the first term in (8.3)
under the null hypothesis and therefore may lead to rejection of the true null
hypothesis in the case of small and medium size macro-economics data.
8.2. Martingale difference structure preserving standardization
Therefore, a better standardization procedure is the following. Denote
v
u
t
t
X
X
u
−1
t
−1
Yi , σ
bt = (t + t0 )
(Yi − µ
bt )2 , if t > 1 − t0 ,
µ
bt = (t + t0 )
i=1−t0
i=1−t0
µ
bt = 0, σ
bt = 1 if t ≥ 1 − t0 ,
bt = µ
σt, β
bt /b
σt,
α
b t = 1/b
bt , Y = α0 Yt − β0 ,
b t Yt − β
Ye t = α
t
(8.6)
where α0 and β0 are the same as in (8.4). Then
Lemma 8.1 Under Assumptions 1.1(a) and 2.1(a) and with Φ as in (8.1), it
follows that
¯ ³ ´
¯
¯
¯
(8.7)
p lim ¯Φ Ye t − Φ (Y t )¯ = 0.
t→∞
54
Consequently, for an arbitrary strictly stationary time series process Vt ∈ R satisfying E[|Vt |] < ∞ and each j ∈ N,
n
¯ ³
´
¡
¢¯¯i
1X h
¯
(8.8)
E |Vt |. ¯Φ Ye t−j − Φ Y t−j ¯ = 0,
lim
n→∞ n
t=1
which by Chebyshev’s inequality for first moments implies that
n
¯ ³
´
¡
¢¯¯
1X
¯
p lim
|Vt |. ¯Φ Ye t−j − Φ Y t−j ¯ = 0.
n→∞ n
t=1
Proof. Recall from Assumptions 1.1(a) and 2.1(a) that Yt and Yt2 are strictly
stationary with vanishing memory, and E[Yt2 ] < ∞, so that by Lemma 2.2,
p limt→∞ µ
bt = E[Y1 ] = µ0 , p limt→∞ σ
b2t = var(Y1 ) = σ02 , hence
b t = β0 .
b t = α0 , p lim β
p lim α
t→∞
t→∞
(8.9)
Moreover, with Φ as in (8.1), it follows from the mean value theorem and (8.2)
that
¯ ³ ´
¯
¯ ¯
¯
¯
¯ ¯e
bt − β0 |
e
αt − α0 |.|Yt | + |β
¯Φ Y t − Φ (Y t )¯ ≤ ¯Y t−j − Y t−j ¯ ≤ |b
so that (8.7) holds by (8.9) and the fact that by strict stationarity, |Yt | = Op (1).
As to (8.8), observe that for an arbitrary M > 0, and with I(.) the indicator
function,
n
¯ ³ ´
¯i
1X h
¯
¯
e
E |Vt |. ¯Φ Y t − Φ (Y t )¯
n t=1
n
¯ ³ ´
¯i
1X h
¯
¯
≤
E |Vt |.I (|Vt | ≤ M) ¯Φ Ye t − Φ (Y t )¯
n t=1
n
¯ ³ ´
¯i
1X h
¯
¯
+
E |Vt |.I (|Vt | > M) ¯Φ Ye t − Φ (Y t )¯
n t=1
n
¯i
1 X h¯¯ ³ e ´
¯
≤ M.
E ¯Φ Y t − Φ (Y t )¯ + π.E [|V1 |.I (|V1 | > M)] ,
n t=1
where the π follows from the fact that by (8.1), supx∈R |Φ(x)| = π/2. For an arbitrary ε > 0 we can choose M so large that π.E [|V1 |.I (|V1 | > M)] < ε. Moreover,
it follows from the bounded convergence theorem and (8.7) that
¯i
h¯ ³ ´
¯
¯
e
lim E ¯Φ Y t − Φ (Y t )¯ = 0,
t→∞
55
which trivially implies that
limn→∞ n1
¯i
h¯ ³ ´
¯
¯
e
t=1 E ¯Φ Y t − Φ (Y t )¯ . Hence,
Pn
#
n
¯ ³ ´
¯
1X
¯
¯
lim sup E
|Vt |. ¯Φ Ye t − Φ (Y t )¯ < ε,
n
n→∞
t=1
"
which, by the arbitrariness of ε > 0 implies that (8.8) holds.
Now (8.5) reads
à k
!
n
X
X
¡
¢
1
ck,n (τ1 , τ2 , ....τk ) = √
W
(Yt − ft−1 (b
θn )) exp i.
τj Φ Y t−j
n t=1
j=1
where
c(2) (τ1 , τ2 , ....τk )
c(1) (τ1 , τ2 , ....τk ) − W
= W
k,n
k,n
c(1) (τ1 , τ2 , ....τk )
W
k,n
à k
!
n
X
¡
¢
1 X
=√
Ut exp i.
τj Φ Y t−j ,
n t=1
j=1
c(2) (τ1 , τ2 , ....τk )
W
k,n
à k
!
n
´
X
¡
¢
1 X³
=√
ft−1 (b
θn ) − ft−1 (θ0 ) exp i.
τj Φ Y t−j .
n t=1
j=1
(8.10)
(8.11)
(8.12)
Similarly, (8.3) now becomes
à k
!
n
³
´
X
X
1
fk,n (τ1 , τ2 , ....τk ) = √
(Yt − ft−1 (b
θn )) exp i.
τj Φ Ye t−j
W
n t=1
j=1
where
f(2) (τ1 , τ2 , ....τk )
f(1) (τ1 , τ2 , ....τk ) − W
= W
k,n
k,n
à k
!
n
³
´
X
X
1
f(1) (τ1 , τ2 , ....τk ) = √
W
Ut exp i.
τj Φ Ye t−j
,
k,n
n t=1
j=1
(8.13)
(8.14)
f(2) (τ1 , τ2 , ....τk )
W
k,n
!
à k
n
´
´
³
X
1 X³
=√
ft−1 (b
. (8.15)
θn ) − ft−1 (θ0 ) exp i.
τj Φ Ye t−j
n t=1
j=1
56
8.2.1. The case H0
Using Lemma 8.1, I will show now that
Lemma 8.2. Under the conditions of Theorem 2.2, and with Φ the arctan function,
¯
¯
¯ f(2)
¯
(2)
c
sup
¯Wk,n (τ1 , τ2 , ....τk ) − Wk,n (τ1 , τ2 , ....τk )¯ = op (1).
(τ1 ,τ2 ,....τk )0 ∈Υk
Proof.
follows from the mean value theorem that ft−1 (b
θn ) − ft−1 (θ0 ) =
³
´It
0
b
θn − θ0 ∇ft−1 (e
θt ), where e
θt is a mean value satisfying ||e
θt − θ0 || ≤ ||b
θn − θ0 ||,
hence
¯ h
i
h
i¯
¯
f(2) (τ1 , τ2 , ....τk ) − Re W
c(2) (τ1 , τ2 , ....τk ) ¯¯
¯Re W
k,n
k,n
¯
n ³
¯ 1 X
´
¯
= ¯√
ft−1 (b
θn ) − ft−1 (θ0 )
¯ n t=1
!
à k
!!¯
à à k
¯
³
´
X
X
¡
¢
¯
e
τj Φ Y t−j
τj Φ Y t−j
− cos
× cos
¯
¯
j=1
j=1
¯
n ³
¯√ ³
´0 1 X
´
¯
b
≤ ¯ n θn − θ0
∇ft−1 (e
θt ) − ∇ft−1 (θ0 )
¯
n t=1
!
à k
!!¯
à à k
¯
´
³
X
X
¡
¢
¯
− cos
τj Φ Ye t−j
τj Φ Y t−j
× cos
¯
¯
j=1
j=1
¯
n
¯√ ³
´0 1 X
¯
+¯ n b
θn − θ0
∇ft−1 (θ0 )
¯
n t=1
à k
!
!!¯
à à k
¯
´
³
X
X
¢
¡
¯
e
− cos
τj Φ Y t−j
τj Φ Y t−j
× cos
¯
¯
j=1
j=1
n
°√ ³
´° 1 X
°
°
b
≤ 2 ° n θn − θ0 ° .
sup
k∇ft−1 (θ) − ∇ft−1 (θ0 )k
n t=1 ||θ−θ0 ||≤||bθn −θ0 ||
°√ ³
´°
°
°
θn − θ0 °
+ sup |τ | × ° n b
τ ∈Υ
57
×
k
n
X
1X
¯ ³
´
¡
¢¯¯
¯
e
||∇ft−1 (θ0 )||. ¯Φ Y t−j − Φ Y t−j ¯
n t=1
°√ ³
´°
°
°
Theorem 2.2 implies that ° n b
θn − θ0 ° = Op (1), and it follows from Lemma
¯ ³
´
¡
¢¯¯
Pk 1 Pn
¯
e
Φ
Y
−
Φ
Y
||∇f
(θ
)||.
8.1 that for fixed k,
¯
t−1 0
t−j
t−j ¯ = op (1).
j=1 n
t=1
Moreover, part
£ (2.4) of Assumption 2.2 implies ¤that for an arbitrary small ε >
0, κ(ε) = E sup||θ−θ0 ||≤ε k∇ft−1 (θ) − ∇ft−1 (θ0 )k < ∞, so that by Lemma 2.2,
P
p limn→∞ n1 nt=1 sup||θ−θ0 ||≤ε k∇ft−1 (θ) − ∇ft−1 (θ0 )k = κ(ε). It follows now easily
that
n
³
´
1X
b
sup
k∇ft−1 (θ) − ∇ft−1 (θ0 )k = p lim κ ||θn − θ0 ||
p lim
n→∞ n
n→∞
θn −θ0 ||
t=1 ||θ−θ0 ||≤||b
j=1
= κ(0) = 0.
(8.16)
¯ h
i
h
i¯
¯
f(2) (τ1 , τ2 , ....τk ) − Re W
c(2) (τ1 , τ2 , ....τk ) ¯¯ = op (1),
Thus, sup(τ1 ,τ2 ,....τk )0 ∈Υk ¯Re W
k,n
k,n
and the same applies to the Im[.] case.
Next, denote
Then
en,k
B
Z
¯
¯2
¯c
¯
¯Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk ),
k
ZΥ ¯
¯2
¯f
¯
=
¯Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk ).
bn,k =
B
Υk
en,k = B
bn,k + op (1)
Lemma 8.3. Under H0 and Assumptions 1.1, 2.1 and 2.2, B
for each k ∈ N.
Proof. Lemma 8.2 implies that
fk,n (τ1 , τ2 , ....τk ) − W
ck,n (τ1 , τ2 , ....τk ) = ∆Wk,n (τ1 , τ2 , ....τk ) + op (1),
W
uniformly in (τ1 , τ2 , ..., τk )0 ∈ Υk , where
∆Wk,n (τ1 , τ2 , ....τk )
Ã
à k
!!
!
à k
n
³
´
X
X
¢
¡
1 X
.
Ut exp i.
τj Φ Ye t−j )
− exp i.
τj Φ Y t−j
=√
n t=1
j=1
j=1
58
Hence
en,k =
B
Z
Υk
¯
¯2
¯c
¯
¯Wk,n (τ1 , τ2 , ....τk ) + ∆Wk,n (τ1 , τ2 , ....τk ) + op (1)¯
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Z ³ h
i´2
ck,n (τ1 , τ2 , ....τk )
Re W
dµ(τ1 )dµ(τ2 )...dµ(τk )
=
k
ΥZ
i´2
³ h
ck,n (τ1 , τ2 , ....τk )
+
dµ(τ1 )dµ(τ2 )...dµ(τk )
Im W
k
Υ
Z
+
(Re [∆Wk,n (τ1 , τ2 , ....τk )])2 dµ(τ1 )dµ(τ2 )...dµ(τk )
k
ZΥ
+
(Im [∆Wk,n (τ1 , τ2 , ....τk )])2 dµ(τ1 )dµ(τ2 )...dµ(τk )
k
ZΥ
h
i
ck,n (τ1 , τ2 , ....τk ) (Re [∆Wk,n (τ1 , τ2 , ....τk )])
+2
Re W
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Z
h
i
ck,n (τ1 , τ2 , ....τk ) (Im [∆Wk,n (τ1 , τ2 , ....τk )])
Im W
+2
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Z
h
i
c
Re Wk,n (τ1 , τ2 , ....τk ) dµ(τ1 )dµ(τ2 )...dµ(τk ) × op (1)
+2
k
ZΥ
h
i
ck,n (τ1 , τ2 , ....τk ) dµ(τ1 )dµ(τ2 )...dµ(τk ) × op (1)
+2
Im W
k
ZΥ
+2
Re [∆Wk,n (τ1 , τ2 , ....τk )] dµ(τ1 )dµ(τ2 )...dµ(τk ) × op (1)
k
Υ
Z
+2
Im [∆Wk,n (τ1 , τ2 , ....τk )] dµ(τ1 )dµ(τ2 )...dµ(τk ) × op (1)
Υk
+op (1)
and thus,
¯
¯
¯e
bn,k ¯¯
¯Bn,k − B
Z
≤
(Re [∆Wk,n (τ1 , τ2 , ....τk )])2 dµ(τ1 )dµ(τ2 )...dµ(τk )
k
ZΥ
+
(Im [∆Wk,n (τ1 , τ2 , ....τk )])2 dµ(τ1 )dµ(τ2 )...dµ(τk )
Υk
59
sZ
+2
Υk
×
sZ
+2
sZ
sZ
+2
(Re [∆Wk,n (τ1 , τ2 , ....τk )])2 dµ(τ1 )dµ(τ2 )...dµ(τk )
Υk
Υk
×
³ h
i´2
c
Re Wk,n (τ1 , τ2 , ....τk )
dµ(τ1 )dµ(τ2 )...dµ(τk )
³ h
i´2
ck,n (τ1 , τ2 , ....τk )
Im W
dµ(τ1 )dµ(τ2 )...dµ(τk )
sZ
Υk
sZ
+2
Υk
(Im [∆Wk,n (τ1 , τ2 , ....τk )])2 dµ(τ1 )dµ(τ2 )...dµ(τk )
Υk
³ h
i´2
ck,n (τ1 , τ2 , ....τk )
Re W
dµ(τ1 )dµ(τ2 )...dµ(τk ) × op (1)
³ h
i´2
ck,n (τ1 , τ2 , ....τk )
Im W
dµ(τ1 )dµ(τ2 )...dµ(τk ) × op (1)
sZ
(Re [∆Wk,n (τ1 , τ2 , ....τk )])2 dµ(τ1 )dµ(τ2 )...dµ(τk ) × op (1)
+2
k
Υ
Z
+2
(Im [∆Wk,n (τ1 , τ2 , ....τk )])2 dµ(τ1 )dµ(τ2 )...dµ(τk ) × op (1)
Υk
+op (1)
Z
|∆Wk,n (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk )
≤
Υk
Z
q
bn,k ×
+4 B
|∆Wk,n (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk )
Υk
q
bn,k × op (1)
+4 B
sZ
+4
|∆Wk,n (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk ) × op (1)
Υk
+op (1)
d
bn,k →
bn,k = Op (1), it suffices to prove
Since by Lemma 3.5, B
Bk , which implies B
that
Z
|∆Wk,n (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk ) = op (1),
(8.17)
Υk
60
as follows. Note that for (τ1 , τ2 , ....τk )0 ∈ Υk ,
£
¤
E |∆Wk,n (τ1 , τ2 , ....τk )|2
⎡ Ã Ã
!
à k
!!2 ⎤
n
k
³
´
X
X
X
¡
¢
1
⎦
=
E ⎣Ut2 cos
τj Φ Ye t−j )
− cos
τj Φ Y t−j
n t=1
j=1
j=1
⎡ Ã Ã
!!2 ⎤
!
à k
Z
n
k
³
´
X
X
X
¡
¢
1
⎦
E ⎣Ut2 sin
τj Φ Ye t−j )
− sin
τj Φ Y t−j
+
Υk n t=1
j=1
j=1
= 2E[Ut2 ]
"
à k
!
à k
!#
n
³
´
X
X
¡
¢
1X
−2
E Ut2 cos
τj Φ Ye t−j )
× cos
τj Φ Y t−j
n t=1
j=1
j=1
"
à k
!
à k
!#
Z
n
³
´
X
X
X
¡
¢
1
E Ut2 sin
τj Φ Ye t−j )
× sin
τj Φ Y t−j
−2
Υk n t=1
j=1
j=1
"
à k
!#
n
´
´
X
X ³ ³
¡
¢
1
E Ut2 cos
τj Φ Ye t−j − Φ Y t−j
= 2E[U12 ] − 2
n t=1
j=1
à k
" Ã
!!#
n
´
X ³ ³
¡
¢´
1X
2
E Ut 1 − cos
τj Φ Ye t−j − Φ Y t−j
= 2
n t=1
j=1
≤ 2
k
X
j=1
|τj .|
≤ 2 sup |τ |
τ ∈Υ
= o(1)
n
¡
¢¯¯i
1 X h 2 ¯¯ ³ e ´
E Ut ¯Φ Y t−j − Φ Y t−j ¯
n t=1
k
n
X
1X
j=1
n
t=1
E
h
Ut2
¯ ³
´
¡
¢¯¯i
¯
e
¯Φ Y t−j − Φ Y t−j ¯
where the first inequality follows from the mean value theorem and the last equality is due to Lemma 8.1. Thus,
Z
¤
£
E |∆Wk,n (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk ) = o(1)
Υk
which implies (8.17).
61
It follows now from Theorem 3.1, Lemma 3.1 and Lemma 8.3 that the following
general result holds.
Theorem 8.1. Denote
à k
!
n
X
X
1
fk,n (τ1 , τ2 , ....τk ) = √
W
(Yt − ft−1 (b
θn )) exp i.
τj0 Φ(Zet−j ) ,
n t=1
j=1
Z ¯
¯2
¯
¯f
en,k =
B
¯Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk ),
Ten =
Υk
L
n
X
k=1
en,k ,
γk B
where each component Zei,t of Zet is a standardized version of component Zi,t of
Zt , similar to Ye t in (8.6) with Yt replaced by Zi,t . Similarly, denote
à k
!
n
X
X
1
ck,n (τ1 , τ2 , ....τk ) = √
W
(Yt − ft−1 (b
θn )) exp i.
τj0 Φ(Z t−j ) ,
n t=1
j=1
Z ¯
¯2
¯c
¯
bn,k =
B
¯Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk ),
Tbn =
Υk
L
n
X
k=1
bn,k .
γk B
where each component Z i,t of Z t is a standardized version of component Zi,t of
Zt , similar to Y t in (8.6) with Yt replaced by Zi,t . Moreover, let all the components
of the one-to one mapping Φ be arctan(.) functions. Then under the conditions
of Theorem 3.1,
∞
X
d
b
e
γk Bk = T,
Tn = Tn + op (1) →
k=1
where for each k the random variable Bk represents the limiting distribution of
bn,k . Therefore, all the previous result for the WICM test statistic Tbn under H0
B
carry over to Ten .
Proof. Recall from Theorem 3.1 that
Ln
∞
X
X
d
b
b
Tn =
γk Bn,k →
γk Bk = T
k=1
k=1
62
(8.18)
en,k , i.e., B
en,k ≤ 2.B
e1,n,k + 2B
e2,n,k , where
and note that Lemma 3.1 carries over to B
e1,n,k ] = E[U12 ] and supk∈N B
e2,n,k = Op (1). Of course, Lemma 3.1 is directly
E[B
b1,n,k and B
b2,n,k as in Lemma
bn,k ’s as well. Then with B
applicable to the current B
3.1, it follows that for each K ∈ N,
∞
∞
¯
¯
³
´
X
X
¯e
¯
e
b
b
γk ¯Bn,k − Bn,k ¯ ≤ 2
γk B1,n,k + B1,n,k
k=K+1
k=K+1
∞
X
+2
=
k=K+1
∞
X
k=K+1
³
´
b
e
γk sup B2,n,k + B2,n,k
k∈N
γk × Op (1)
(8.19)
whereas by Lemma 8.3,
K
X
k=1
¯
¯
¯e
¯
b
γk ¯B
−
B
n,k
n,k ¯ = op (1).
(8.20)
The result (8.19) implies that for arbitrary ε ∈ (0, 1) and δ > 0 we can choose K
so large that
#
" ∞
¯
¯
X
¯e
¯
bn,k ¯ > δ/2 < ε/2,
lim sup Pr
γk ¯Bn,k − B
n→∞
k=K+1
whereas (8.20) implies that for this K,
#
"K
¯
X ¯¯
¯
en,k − B
bn,k ¯ > δ/2 < ε/2.
lim sup Pr
γk ¯B
n→∞
k=1
It follows now from the inequality (3.20) that
#
"∞
¯
X ¯¯
¯
en,k − B
bn,k ¯ > δ < ε,
lim sup Pr
γk ¯B
n→∞
hence
Ln
X
k=1
k=1
∞
¯
¯ X
¯
¯
¯e
¯
¯e
¯
b
b
γk ¯Bn,k − Bn,k ¯ ≤
γk ¯Bn,k − Bn,k ¯ = op (1).
k=1
The theorem under review now follows from (8.18) and (8.21).
63
(8.21)
8.2.2. The case H1
Also the consistency result in Theorem 4.1 carries over to the WICM test statistic
Ten .
Theorem 8.2. Under H1 and Assumptions 3.1 and 4.1, and under the notations
in Theorems 4.1 and 8.1,
Ten /n =
Proof. Denote
∞
X
γk ηk > 0.
k=1
Z
Υk
Z
¯
¯2
¯ c(2)
¯
¯Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk ),
k
ZΥ ¯
¯2
¯ f(2
¯
=
¯Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk ),
b (2) =
B
n,k
e (2)
B
n,k
k=1
p
ek,n /n = Tbn /n + op (1) →
γk B
¯
¯2
¯ c(1)
¯
¯Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk ),
k
ZΥ ¯
¯2
¯
¯ f(1)
=
¯Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk ),
b (1) =
B
n,k
e (1)
B
n,k
Ln
X
Υk
en,k be the same as in Theorem 8.1. Moreover, denote
bn,k and B
and let B
"
à k
!#
X
¡
¢
τj Φ Y t−j
ψk (τ1 , τ2 , ....τk ) = E Ut exp i.
ηk =
Z
Υk
j=1
|ψk (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk )
I will prove the theorem under review by showing that
b (1) = Op (1), sup B
b (1) = Op (1),
sup B
k∈N
n,k
k∈N
n,k
(8.22)
e (2) = op (1), sup B
b (2) = op (1),
sup B
(8.23)
n,k
n,k
k∈N
k∈N
¯
¯
¯
¯
¯e
¯b
e (1) ¯
b (1) ¯
sup ¯B
n,k /n − Bn,k /n¯ = op (1), sup ¯Bn,k /n − Bn,k /n¯ = op (1), (8.24)
k∈N
¯
¯k∈N
¯
¯ e (1)
(1)
b /n¯ = op (1), for each k ∈ N,
(8.25)
¯Bn,k /n − B
n,k
p
b (1) /n →
ηk , for each k ∈ N.
B
n,k
64
(8.26)
Then it not hard to verify that Theorem 8.2 holds.
Proof of (8.22). It is easy to verify from (8.11) and (8.14) that
¯
¯
¯
c(1) (τ1 , τ2 , ....τk )¯¯ ≤ C1,n ,
supk∈N sup(τ1 ,τ2 ,....τk )0 ∈Υk ¯n−1/2 W
k,n
¯
¯
¯ −1/2 f(1)
¯
Wk,n (τ1 , τ2 , ....τk )¯ ≤ C1,n ,
supk∈N sup(τ1 ,τ2 ,....τk )0 ∈Υk ¯n
where
C1,n
Hence,
n
√ 1X
= 2
|Ut | = Op (1).
n t=1
2
2
b (1) ≤ C1,n
b (1) ≤ C1,n
= Op (1), sup B
= Op (1).
sup B
n,k
n,k
k∈N
(8.27)
k∈N
Proof of (8.23). Again, it is easy to verify from (8.12) and (8.15) that
¯
¯
¯ −1/2 c(2)
¯
supk∈N sup(τ1 ,τ2 ,....τk )0 ∈Υk ¯n
Wk,n (τ1 , τ2 , ....τk )¯ ≤ C2,n ,
¯
¯
¯
f(2) (τ1 , τ2 , ....τk )¯¯ ≤ C2,n ,
supk∈N sup(τ1 ,τ2 ,....τk )0 ∈Υk ¯n−1/2 W
k,n
where
C2,n =
n ¯
¯
√ 1X
¯
¯
b
f
2
(
θ
)
−
f
(θ
)
¯ t−1 n
t−1 0 ¯
n t=1
n
√ 1X
≤
2
sup
|ft−1 (θ) − ft−1 (θ0 )|
n t=1 ||θ−θ0 ||≤||bθn −θ0 ||
= op (1).
p
The latter follows from ||b
θn − θ0 || → 0 and (8.16). Hence
2
2
e (2) ≤ C2,n
b (2) ≤ C2,n
= op (1), sup B
= op (1).
sup B
n,k
n,k
k∈N
(8.28)
k∈N
Proof of (8.24). Observe from (8.3) that
Z ¯
¯2
¯
¯ −1/2 f
e
Bn,k /n =
Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk )
¯n
k
ZΥ ³ h
i
h
i´2
f(1) (τ1 , τ2 , ....τk ) − Re n−1/2 W
f(2) (τ1 , τ2 , ....τk )
=
Re n−1/2 W
k,n
k,n
Υk
65
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Z ³ h
i
h
i´2
f(1) (τ1 , τ2 , ....τk ) − Im n−1/2 W
f(2) (τ1 , τ2 , ....τk )
Im n−1/2 W
+
k,n
k,n
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Z ¯
¯2
¯ −1/2 f(2)
¯
(1)
e
= Bn,k /n +
Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk )
¯n
Υk
Z
h
i
h
i
−1/2 f(1)
−1/2 f(2)
Wk,n (τ1 , τ2 , ....τk ) × Re n
Wk,n (τ1 , τ2 , ....τk )
−2
Re n
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Z
h
i
h
i
f(1) (τ1 , τ2 , ....τk ) × Im n−1/2 W
f(2) (τ1 , τ2 , ....τk )
Im n−1/2 W
−2
k,n
k,n
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk ),
hence
¯
¯
¯e
e (2) /n
e (1) /n¯¯ ≤ B
¯Bn,k /n − B
n,k
n,k
Z ¯ h
i¯ ¯ h
i¯
¯
f(1) (τ1 , τ2 , ....τk ) ¯¯ × ¯¯Re n−1/2 W
f(2) (τ1 , τ2 , ....τk ) ¯¯
+2
¯Re n−1/2 W
k,n
k,n
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Z ¯ h
i¯ ¯ h
i¯
¯
¯ ¯
¯
−1/2 f(1)
−1/2 f (2)
+2
Wk,n (τ1 , τ2 , ....τk ) ¯ × ¯Im n
Wk,n (τ1 , τ2 , ....τk ) ¯
¯Im n
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk )
q
q
e (2) /n + 4 B
e (1) /n B
e (2) /n
≤B
≤
n,k
2
C2,n
n,k
n,k
+ 4.C1,n .C2,n = op (1),
where the latter follows from (3.19) and (8.28). Obviously, the last inequality also
bn,k /n − B
b (1) /n. This proves (8.24).
holds for B
n,k
Proof of (8.25). Denote
and
(1)
f(1) (τ1 , τ2 , ....τk ) − W
c(1) (τ1 , τ2 , ....τk ).
∆Wk,n (τ1 , τ2 , ....τk ) = W
k,n
k,n
∆
Bk,n
=
Z
Υk
¯
¯2
¯
¯
(1)
¯∆Wk,n (τ1 , τ2 , ....τk )¯ dµ(τ1 )dµ(τ2 )...dµ(τk ).
It is not hard to verify from (8.11) and (8.14) and the mean value theorem that
¯
¯
¯ −1/2
¯
(1)
sup
∆Wk,n (τ1 , τ2 , ....τk )¯ ≤ Dk,n ,
¯n
(τ1 ,τ2 ,....τk )0 ∈Υk
66
where
Dk,n
k
n
¯ ³
´
X
√
¡
¢¯¯
1X
¯
e
= 2 sup |τ |
|Ut |. ¯Φ Y t−j − Φ Y t−j ¯ = op (1),
n t=1
τ ∈Υ
j=1
where the latter follows from Lemma 8.1. Hence
∆
2
Bk,n
/n ≤ Dk,n
= op (1).
(8.29)
Then
e (1) /n
B
n,k
Z ¯
¯2
¯ −1/2 c(1)
¯
(1)
−1/2
∆Wk,n (τ1 , τ2 , ....τk )¯
=
Wk,n (τ1 , τ2 , ....τk ) + n
¯n
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Z ³ h
i
h
i´2
c(1) (τ1 , τ2 , ....τk ) + Re n−1/2 ∆W (1) (τ1 , τ2 , ....τk )
Re n−1/2 W
=
k,n
k,n
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Z ³ h
i
h
i´2
c(1) (τ1 , τ2 , ....τk ) + Im n−1/2 ∆W (1) (τ1 , τ2 , ....τk )
Im n−1/2 W
+
k,n
k,n
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Z ³ h
i´2
−1/2 c (1)
Re n
dµ(τ1 )dµ(τ2 )...dµ(τk )
=
Wk,n (τ1 , τ2 , ....τk )
k
Υ
Z ³ h
i´2
−1/2 c(1)
+
Im n
dµ(τ1 )dµ(τ2 )...dµ(τk )
Wk,n (τ1 , τ2 , ....τk )
k
ZΥ ³ h
i´2
(1)
+
dµ(τ1 )dµ(τ2 )...dµ(τk )
Re n−1/2 ∆Wk,n (τ1 , τ2 , ....τk )
k
Υ
Z ³ h
i´
(1)
+
Im n−1/2 ∆Wk,n (τ1 , τ2 , ....τk ) dµ(τ1 )dµ(τ2 )...dµ(τk )
k
ZΥ ³ h
i´ ³ h
i´
(1)
−1/2
c(1) (τ1 , τ2 , ....τk )
+2
Re
n
∆W
(τ
,
τ
,
....τ
)
Re n−1/2 W
1
2
k
k,n
k,n
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Z ³ h
i´ ³ h
i´
(1)
−1/2 c(1)
−1/2
+2
Wk,n (τ1 , τ2 , ....τk )
Im n
∆Wk,n (τ1 , τ2 , ....τk )
Im n
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk )
∆
b (1) /n + Bk,n
/n
=B
n,k
67
+2
Z
Υk
³ h
i´ ³ h
i´
(1)
−1/2
c(1) (τ1 , τ2 , ....τk )
Re n−1/2 W
Re
n
∆W
(τ
,
τ
,
....τ
)
k
k,n
k,n 1 2
×dµ(τ1 )dµ(τ2 )...dµ(τk )
Z ³ h
i´ ³ h
i´
(1)
−1/2
c(1) (τ1 , τ2 , ....τk )
+2
Im
n
∆W
(τ
,
τ
,
....τ
)
Im n−1/2 W
1
2
k
k,n
k,n
Υk
×dµ(τ1 )dµ(τ2 )...dµ(τk ),
hence
¯
¯
¯ e (1)
∆
b (1) /n¯¯ ≤ Bk,n
/n
¯Bn,k /n − B
n,k
sZ
i´2
³ h
(1)
−1/2
c
Wk,n (τ1 , τ2 , ....τk )
dµ(τ1 )dµ(τ2 )...dµ(τk )
Re n
+2
Υk
Z ³ h
i´2
(1)
Re n−1/2 ∆Wk,n (τ1 , τ2 , ....τk )
dµ(τ1 )dµ(τ2 )...dµ(τk )
×
Υk
sZ
³ h
i´2
c(1) (τ1 , τ2 , ....τk )
+2
Im n−1/2 W
dµ(τ1 )dµ(τ2 )...dµ(τk )
k,n
Υk
Z ³ h
i´2
(1)
Im n−1/2 ∆Wk,n (τ1 , τ2 , ....τk )
dµ(τ1 )dµ(τ2 )...dµ(τk )
×
Υk
q
q
(1)
∆
2
∆
b
≤ Bk,n /n + 4 Bn,k /n Bk,n
/n ≤ Dk,n
+ 4C1,n Dk,n = op (1),
where the latter follows from (3.19) and (8.29).
Proof of (8.28). Next, denote
"
Ã
ψk (τ1 , τ2 , ....τk ) = E Ut exp i.
ηk =
Z
Υk
k
X
j=1
!#
¡
¢
τj Φ Y t−j
,
|ψk (τ1 , τ2 , ....τk )|2 dµ(τ1 )dµ(τ2 )...dµ(τk ),
and observe from Lemma 2.1 that under Assumptions 1.1 and 2.1,
¯
¯
¯ −1/2 c(1)
¯
sup
Wk,n (τ1 , τ2 , ....τk ) − ψk (τ1 , τ2 , ....τk )¯ = op (1)
¯n
(τ1 ,τ2 ,....τk )0 ∈Υk
p
b (1) /n →
ηk for each k ∈ N.
for each k ∈ N, which implies that B
n,k
68
9. Models with infinitely many lagged conditioning variables
So far the above results have been derived for nonlinear ARX models for which
the conditional expectation function ft−1 (θ0 ) depend on a finite number of lagged
dependent variables Yt and a finite number of lagged exogenous variables Xt .
t−1
],
However, since ft−1 (θ0 ) aims to represent the conditional expectation E[Yt |F−∞
which in general may depend on all lagged Yt ’s and lagged Xt ’s, it may be more
realistic to allow infinitely many lags in the specification of ft−1 (θ0 ). A convenient
way to do that is to specify the errors Ut as an ARMA process, for example an
ARMA(1,1) process Ut = κ0 Ut−1 +et −δ0 et−1 , where |κ0 | < 1, |δ0 | < 1 and δ0 6= κ0 ,
t−1
and now under H0 , E[et |F−∞
] = 0 a.s. Then model (1.2) becomes a nonlinear
ARMAX model,
Yt = ft−1 (θ0 ) + κ0 (Yt−1 − ft−2 (θ0 )) + et − δ0 et−1 ,
which by inverting the lag polynomial 1 − δ0 L can be written as an infinite-order
ARX model,
Yt = ft−1 (θ0 ) + (κ0 − δ0 )
= gt−1 (θ0 , δ0 , κ0 ) + et ,
∞
X
j=0
δ0j (Yt−1−j − ft−2−j (θ0 )) + et
say.
Previously it was assumed that Zt = (Yt , Xt0 )0 is observed for 1 − max(p, q) ≤
t ≤ n, where p is the maximum lag of Yt in ft−1 (θ0 ), and q is the maximum lag of
Xt in ft−1 (θ0 ). Adopting this assumption in the present case as well we need to
truncate gt−1 (θ0 , δ0 , κ0 ), for example by
g 0 (θ0 , δ0 , κ0 ) = f0 (θ0 ),
g t−1 (θ0 , δ0 , κ0 ) = ft−1 (θ0 ) + (κ0 − δ0 )
for t ≥ 2.
t−2
X
j=0
δ0j (Yt−1−j − ft−2−j (θ0 ))
Since gt−1 (θ, δ, κ)−gt−1 (θ, δ, κ) → 0 exponentially for t → ∞, using gt−1 (θ, δ, κ)
as a proxy for gt−1 (θ, δ, κ) yields, loosely speaking, asymptotically the same results
as if the gt−1 (θ, δ, κ)’s were used.
69
More generally, suppose that ft−1 (θ) depends on the entire sequence {Zt−j }∞
j=1 ,
>
and that we can specify a truncated version ft−1 (θ) of ft−1 (θ) that only depend on the observed lagged Yt and Xt . One can formulate conditions regard>
ing the speed of convergence to zero for t → ∞ of supθ∈Θ |ft−1 (θ) − ft−1
(θ)|,
0
0
>
0
supθ∈Θ ||(∂/∂θ )ft−1 (θ) − (∂/∂θ )ft−1 (θ)| and supθ∈Θ ||(∂/∂θ)(∂/∂θ )ft−1 (θ) − (∂/
>
∂θ)(∂/∂θ0 )ft−1
(θ)|| such that, after some modifications of the assumptions to take
>
into account that ft−1
(θ) is no longer strictly stationary, the above results carry
>
over, with ft−1 (θ) replaced by ft−1
(θ). This problem is related to the initial value
problem discussed in section 7. The details are left to the reader as an exercise.
10. Implementation
Choosing
Υ = Xs+1
i=1 [−c, c]
for some c > 0, the WICM test statistic can be written in slightly different but
asymptotically equivalent form as
¯
!¯2
à k
Z
n
Ln
¯
¯
X
X
X
1
¯
¯
(1)
0e
b
b
γk
τj Zt−j ¯
Ut exp i.
Tn =
¯√
(s+1).k
¯
Xi=1
[−c,c] ¯ n − Ln t=L +1
j=1
k=1
n
×dµ(τ1 )dµ(τ2 )...dµ(τk )
bt ’s are the NLLS residuals, the Zet ’s are vectors in Rs+1
where as before, the U
of bounded transformations of the standardized dependent and exogeneous variables, and Ln is a subsequence of the effective sample size n. Throughout it will
be assumed that µ is the uniform probability measure on Υ. The time shift accommodates the standardization procedure as well as the initial value issue.
(1) d
Recall that under H0 , Tbn → T, and that the actual WICM test is
Tbn = Tbn(1) /Tbn(2) ,
where under H0 , Tbn is a consistent estimator of E[T ], and under H1 , p limn→∞ Tbn ∈
(0, ∞).
(2)
(2)
Lemma 10.1. The statistic Tbn has the closed form expression
!Ã
!
ÃL
n
n
X
X
1
bt2
U
γk
Tbn(1) =
n
−
L
n t=L +1
k=1
(1)
n
70
1
+2
n − Ln
where
Pt1 ,t2
n
X
tX
1 −1
t1 =Ln +2 t2 =Ln +1
bt1 U
bt2 Pt1 ,t2
U
´´
³ ³
ei,t1 −j − Zei,t2 −j
k Y
s+1 sin c Z
Y
´
³
=
γk
e
e
c
Z
−
Z
j=1 i=1
k=1
i,t1 −j
i,t2 −j
Ln
X
(2)
with Zei,t component i of Zet . Moreover, the estimator Tbn has the closed form
expression
!
ÃL
n
i
h
i
h
X
−1 b
−1 b
b−1
b
b2n − 2.trace A
+
trace
A
Tbn(2) =
γk σ
A
C
A
C
1,n 2,n
1,n 2,n 1,n 1,n ,
k=1
where
σ
b2n =
b1,n =
A
b2,n =
A
b1,n =
C
b2,n =
C
n
X
1
b 2,
U
n − Ln t=L +1 t
n
1
n − Ln
1
n − Ln
n
X
t=Ln +1
n
X
∇ft (b
θn )∇ft (b
θn )0 ,
bt2 ∇ft (b
U
θn )∇ft (b
θn )0 ,
t=Ln +1
n
X
1
(n − Ln )2
1
(n − Ln )2
n
X
t1 =Ln +1 t2 =Ln +1
n
n
X
X
t1 =Ln +1 t2 =Ln +1
∇ft1 (b
θn )∇ft2 (b
θn )0 Pt1 ,t2 ,
bt2 ∇ft1 (b
θn )∇ft2 (b
θn )0 Pt1 ,t2 .
U
1
Proof. Similar to Lemma 5.2 in Bierens (2015a).
Finally, it is not too hard, but rather tedious, to show that
(1)
(1)
Lemma 10.2. A typical bootstrap version Ten of Tbn takes the form
!
ÃL
n
n
X
X
1
(1)
e
bt2
Tn =
γk
ε2t U
n
−
L
n t=L +1
k=1
n
71
1
+2
n − Ln
n
X
tX
1 −1
t1 =Ln +2 t2 =Ln +1
b−1
−2de0n A
en
1,n e
bt1 U
bt2 Pt1 ,t2
εt1 εt2 U
b−1
b b−1 en
+e
e0n A
1,n C1,n A1,n e
b1,n and C
b1,n are the same as before, and
where the εt ’s are i.i.d. N(0, 1), A
een
n
X
1
bt ∇ft (b
= √
εt U
θn ),
n − Ln t=L +1
n
den =
1
√
(n − Ln ) n − Ln
n
X
n
X
t1 =Ln +1 t2 =Ln +1
bt1 ∇ft2 (b
εt1 U
θn ).Pt1 ,t2 .
(1) d
(2)
Since Ten → T ∗ , where under H0 , T ∗ ∼ T, and in general p limn→∞ Tbn =
(1)
(2)
(1)
(2)
E[T ∗ ], we may use Ten = Ten /Tbn as a bootstrap version of Tbn = Tbn /Tbn .
11. A numerical example
11.1. Fitting an AR(1) model to an MA(1) process
In order to verify the finite sample performance of the WICM test, consider the
zero-mean Gaussian MA(1) process
Yt = Ut − ρUt−1 , where Ut ∼ i.i.d. N(0, 1), |ρ| < 1,
(11.1)
and suppose that it is incorrectly assumed that this is an AR(1) process i.e.,
£
¤
t−1
(11.2)
= θ1 Yt−1 + θ2 a.s.
H0 : E Yt |F−∞
¡
¢
¡
¢
t−1
∞
= σ {Yt−j }∞
for some θ = (θ1 , θ2 )0 , where F−∞
j=1 = σ {Ut−j }j=1 . Note that
£
2¤
E
(Y
θ0 = (θ0,1 , θ0,2 )0 = arg min
t − θ1 Yt−1 − θ2 )
(θ1 ,θ2 )0 ∈R2
£
2¤
= arg min
E
(U
t − (ρ + θ1 ) Ut−1 + ρθ1 Ut−2 − θ2 )
(θ1 ,θ2 )0 ∈R2
¡
¢
2
2 2
2
= arg min
+
2ρθ
1
+
ρ
1 + (1 + ρ )θ1 + θ2
(θ1 ,θ2 )0 ∈R2
¡
¢0
= −ρ/(1 + ρ2 ), 0 ,
72
hence, the error term of the AR(1) model involved is
ρ
Yt−1
1 + ρ2
ρ3
ρ2
= Ut −
U
−
Ut−2 .
t−1
1 + ρ2
1 + ρ2
Ut∗ = Yt +
It is easy to verify that
E [Ut∗ .Yt−m ] =
(
0 for m 6= 2,
for m = 2,
ρ2
− 1+ρ
2
so that Ut∗ is independent of all but one lagged Yt , with Yt−2 the exception. Moreover
⎧
⎨ −ρ3 /(1 + ρ2 ) for m = 1,
∗
−ρ2 /(1 + ρ2 ) for m = 2,
E [Ut .Ut−m ] =
⎩
0 for m ≥ 3,
hence by joint normality,
E [Ut∗ |Ut−1 , Ut−2 , ..., Ut−m ] = −
for m ≥ 2 and thus
¤
£
t−1
=
E Ut∗ |F−∞
ρ3
ρ2
U
−
Ut−2
t−1
1 + ρ2
1 + ρ2
lim E [Ut∗ |Ut−1 , Ut−2 , ..., Ut−m ]
m→∞
ρ2
(ρUt−1 + Ut−2 )
1 + ρ2
∞
X
ρ3
2
Yt−1 − ρ
ρj Yt−2−j .
= −
1 + ρ2
j=0
= −
The latter equality follows from inverting the MA(1) process (11.1), yielding Ut =
P
∞
j
j=0 ρ Yt−j .
Furthermore, note that
E
h¡ £
¤¢ i
t−1 2
E Ut∗ |F−∞
=
73
ρ4
1 + ρ2
(11.3)
The values of this expectation are are listed in the following table, for ρ = 0.2,
0.3, ..., 0.9.
ρ ρ4 /(1 + ρ2 ) ρ ρ4 /(1 + ρ2 )
0.2
0.0015
0.6
0.0953
0.3
0.0074
0.7
0.1611
(11.4)
0.4
0.0221
0.8
0.2498
0.5
0.0500
0.9
0.3625
Clearly, for low values of ρ it will be difficult to detect misspecification of the
AR(1) model in finite samples.
11.2. WICM test results
I have generated independently four time series Yt , t = 1, 2, ..., 500, according to
(11.1) for ρ = 0.8, ρ = 0.6, ρ = 0.4 and ρ = 0.2, respectively, and conducted for
each of these four time series the WICM test for the (false) null hypothesis (11.2).
The WICM test has been
√ implemented as in the previous section, with c = 5,
k
γk = (0.9) and Ln = [ n], where [x] denotes the largest integer ≤ x. Thus,
Ln = 22. Moreover, the bootrap sample size is M = 500.
(1)
(2)
In the following table, WICM is the test statistic Tbn = Tbn /Tbn as defined
in Lemma 10.1, b(0.01), b(0.05), b(0.10) are the bootstrap critical values for significance levels 1%, 5% and 10%, respectively, based on the bootstrap versions
(1)
(2)
(1)
Ten = Ten /Tbn of Tbn , with Ten defined in Lemma 10.2, and BPV is the bootstrap
p-value.
ρ WICM b(0.01) b(0.05) b(0.10) BPV
0.8 2.11
2.05
1.53
1.35 0.006
0.6 2.16
1.78
1.47
1.32 0.002
(11.5)
0.4 1.55
1.77
1.52
1.38 0.042
0.2 0.56
1.82
1.46
1.30 1.000
First note that the WICM test statistics are lower than each of the three the
upper bounds in (5.10). Thus, the tentative conclusion is that the false null hypothesis is not rejected. However, the results demonstrate that the upper bounds
(5.10) are way too conservative. They are only informative if they indicate rejection at the 1% or 5% significance levels, because then we may conclude without
conducting the bootstrap that the null hypothesis will be rejected on the basis of
the bootstrap results as well.
For ρ = 0.8 and ρ = 0.6 the false AR(1) null hypothesis is strongly rejected
and even for ρ = 0.4 the null hypothesis is rejected at the 5% significance level.
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In view of (11.3) and (11.4) the result for ρ = 0.2 is not surprising. In this case
we need a much longer time serie to detect the mispecification.
It would have been much more informative if I had done a complete Monte
Carlo analysis of the finite sample power of the WICM test for the above MA(1)AR(1) set up. However, each bootstrap round took about two hours computing
time, so such a Monte Carlo analysis with only 100 replications, for example, will
take about one month non-stop!
12. Concluding remarks
In this addendum to B84 I have derived the limiting distribution of the WICM test
and shown that similar to Bierens (2015a) the critical values can be approximated
by a bootstrap method, and that the same upper bounds of the critical values as
in the i.i.d. case apply to a standardized version of the WICM test. Moreover,
I have also solved the problem (see section 6 in B84) how to standardize the
conditioning lagged variables Zt before taking the bounded transformation Φ in
order to preserve enough variation in Φ(Zt ) (see section 6 in B84) while preserving
the asymptotic results as well.
References
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BP1997.PDF.
Bierens, H. J. (2015a): ”Addendum to Consistent Model Specification Tests”,
http://grizzly.econ.psu.edu/~hbierens/ADDENDUM_B1982.PDF.
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Bierens, H. J., and L. Wang (2012): ”Integrated Conditional Moment Tests
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