UNIT ROOT TEST UNDER INNOVATION OUTLIER CONTAMINATION
SMALL SAMPLE CASE
Lynda Atil, Hocine Fellag and Karima Nouali Department of Mathematics, Faculty of Sciences
University of Tizi Ouzou Tizi–Ouzou 15000, Algeria
Abstract
The two sided unit root test of a first-order autoregressive model in the presence of an innovation outlier is considered. In this paper, we present three tests; two are usual and one is new. We give formulas computing the size and the power of the three tests when an innovation outlier (IO) occurs at a specified time, say k. Using a comparative study, we show that the new statistic performs better under contamination. A Small sample case is considered only.
Keywords: autoregressive process, Dickey-Fuller test, innovation outlier, power, size.
2000 Mathematics Subject Classification: Primary 62F11;
Secondary 62M10.
1. Introduction Consider a time series {x
t} which follows the model
(1) (1 − ρB)x
t=
t, t = . . . , −1, 0, 1, . . . , n ,
where {
t}
t=1,...,nis a sequence of independent normally distributed random
variables with mean 0 and variance 1, whereas B denotes the backshift
operator such that Bx
t= x
t−1. We assume that x
0= 0 with probability 1.
Suppose that all what we observe is the segment of observations (2) x
1, x
2, . . . , x
nand we want to test the hypothesis
(3) H
0: ρ = 1 vs H
1: ρ 6= 1
at the significance level α.
Various authors have treated the problem of unit root test. Dickey and Fuller (1979)wrote a pioneer paper where they proposed their well known Dikey-Fuller statistic. Phillips (1987) and Phillips and Perron (1988) sug- gested a criterion for correction of the bias in Dickey-Fuller statistic. For more details, see Diebold (1988), Perron (1989) and Sims and Uhlig (1991) and Fuller (1996, Chap. 10). Also, many authors studied the effect of out- liers on unit root tests. Franses and Haldrup (1994) showed that, in the case of Dickey-Fuller tests, there is over-rejection of the unit root hypoth- esis when additive outliers occur. Outliers on unit root tests in AR(1) are investigated by Shin et al. (1996). Maddala and Rao (1997) show that, when n goes to infinity, the impacts of finite additive outliers will go to zero. Vogelsang (1999) proposed two robust procedures to detect outliers and adjust the observations.
According to (3), we consider tests of the form Reject H
0if ˆ ρ
2> c,
where ˆ ρ is a suitable statistic (estimator of ρ) and c a constant.
Three statistics are proposed. The first is based on Dickey-Fuller statis- tic defined by
(4) T
DF= n(ˆ ρ
LS− 1) ,
where
(5) ρ ˆ
LS=
"
nX
t=2
x
tx
t−1# "
nX
t=2
x
2t−1#
−1.
Note that ˆ ρ
LSis the well known least squares estimator of ρ.
The second is of the form
(6) T
SY M= −(n − 2)
1/2(1 + ˆ ρ
S)
−1/2(1 − ˆ ρ
S)
1/2, where ˆ ρ
Sis the simple symmetrical estimator of ρ defined by
(7) ρ ˆ
S=
n
X
t=2
x
t−1x
t1
2
x
21+ x
2n+
n−1
X
t=2
x
2t.
T
SY Mis the corresponding t-statistic given by Fuller (1996).
Finally, we propose a new statistic given by the following formula
(8) T
M ED= x
1/M ED{x
2, x
3, . . . , x
n} ,
where M ED{x
2, x
3, . . . , x
n} means the median of x
2, x
3, . . . , x
n.
In this paper, T
M EDis called a median statistic. We will show that the size of the usual tests changes slightly if an innovation outlier occurs.
However, when the test statistic T
M EDis used, the size of the test is stable under contamination.
2. The Dickey-Fuller and symmetrical statistics under contamination
Before studying the given statistics under contamination, note that, using easy computations, we can write
P
T
DF2> c
= 1 − P (1 − √
c/n < ˆ ρ
LS< 1 + √ c/n) and
P
T
SY M2> c
= 1 − P (ˆ ρ
S> (n − 2 − c)/(n − 2 + c)) .
Also, we remark that ˆ ρ
LSand ˆ ρ
Scan be written as a ratio of two quadratic forms. Indeed, let X = (x
1, x
2, . . . , x
n)
Tbe the vector of observations. Then we have
ˆ
ρ
LS= (X
T.M
2.X)
−1.(X
T.M
1.X) with
M
1=
0 1/2 0 . . . 0 1/2 0 1/2 . . . 0 0 1/2 0 . .. 0 .. . . .. ... ... 1/2
0 0 0 1/2 0
and M
2=
1 0 0 . . . 0 0 1 0 . . . 0 0 0 . .. ... 0 .. . . .. ... 1 0
0 0 0 0 0
.
Also, we have ˆ ρ
S= (X
T.D
2.X)
−1.(X
T.D
1.X) with D
1= M
1and
D
2=
1/2 0 0 . . . 0 0 1 0 . . . 0 0 0 . .. ... 0 .. . . .. ... 1 0
0 0 0 0 1/2
.
Both ˆ ρ
LSand ˆ ρ
Sare of the form T = (X
T.R
2.X)
−1.(X
T.R
1.X), where R
1and R
2are symmetric. Assume that, at a position k ∈]1, n[, an outlier of magnitude ∆ occurs. Hence, instead of the segment (2), we observe the following observations z
1, z
2, . . . , z
n, where
z
t= x
t∀t < k ; z
k= x
k+ ∆ and z
t= ρz
t−1+
t∀t > k.
The process {z
t} generated by the contaminant is called the innovation out- lier model (IO) introduced by Fox (1972). Assume that Z = (z
1, z
2, . . . , z
n)
T. Then, under IO contamination, we observe T
∗= (Z
T.R
2.Z)
−1.(Z
T.R
1.Z) instead of T = (X
T.R
2.X)
−1.(X
T.R
1.X).
Proposition 1. For a given ρ = ρ
0,
P
IO(T
∗> c) = 0.5 + 1 π
Z
∞ 0sin f
∗(u, ∆) ug
∗(u, ∆) du, where
(9) f
∗(u, ∆) = 1 2
n
X
i=1
Arctan(λ
iu) + ∆
2u 2
n
X
i=1
λ
iQ
2k,i1 + λ
2iu
2and
(10) g
∗(u, ∆) =
n
Y
i=1
1 + λ
2iu
21/4. Exp
( ∆
2u
22
n
X
i=1
λ
2iQ
2k,i1 + λ
2iu
2) ,
where λ
1, λ
2, . . . , λ
nare the eigenvalues of the matrix B = A
T.(R
1−c.R
2).A.
Q
k,iis the (k, i) element of the orthogonal matrix Q containing the normal- ized eigenvectors of B. The matrix A is defined by
A =
1 0 0 . . . 0
ρ
01 0 . . . 0 ρ
20ρ
01 . . . .. . .. . .. . . . . . .. 0 ρ
n−10ρ
n−20. . . ρ
01
an n × n-matrix generated by the coefficient ρ
0.
P roof. First, one can write Z = AV with V
T= (v
1, v
2, . . . , v
n) such that v
t= ε
t∀t 6= k and v
k= ε
k+ ∆,
P
IO(T
∗> c) = P
IO(V
T.B.V > 0) with B = A
T.(R
1− c.R
2).A . Due to the fact that B is a nonsingular symmetric matrix with real and distinct eigenvalues λ
1, λ
2, . . . , λ
n, we can find an orthogonal matrix Q = (Q
ij)
i,j=1,...,n(Q
−1= Q
T) such that
Λ = Q
−1.B.Q
is a diagonal matrix with diagonal elements λ
1, λ
2, . . . , λ
n.
P
IO(T
∗> 0) = P
IO(V
T.B.V > 0)
= P
IO(V
T.Q.Λ.Q
−1.V > 0)
= P
IO((Q
TV )
T.Λ.(Q
T.V ) > 0)
= P
IO
λ
1 nX
i=1
Q
i1v
i!
2+ . . . +
n
X
i=1
Q
inv
i!
2> 0
= P
IO(λ
1Y
1+ . . . + λ
nY
n> 0) ,
where Y
j= P
ni=1
Q
ijv
i(j = 1, . . . , n) are independent and distributed as N (Q
kj∆, 1).
Then
P
IO(V
T.B.V > 0) = P
IO nX
i=1
λ
jχ
21(β
j) > 0
!
= P
IO(S > 0),
where χ
21(β
j), j = 1, 2, . . . , n are independent random variables distributed according to chi-square with one degree of freedom and the non-centrality parameter β
j= Q
2kj∆
2.
According to Imhoff (1961) theorem, since T
∗is of the form T
∗= P
ni=1
λ
jχ
2hj
(δ
j2), we can write
P
IO(T
∗> c) = 0.5 + 1 π
Z
∞ 0sin f
∗(u, ∆) ug
∗(u, ∆) du,
where
f
∗(u, ∆) = 1 2
n
X
i=1
h
i. h
Arctan(λ
iu) + δ
i2λ
iu(1 + λ
2iu
2)
−1i
and
g
∗(u, ∆) =
n
Y
i=1
(1 + λ
2iu
2)
hi/4. Exp ( 1
2
n
X
i=1
δ
jλ
iu 1 + λ
2iu
2) .
In our case, h
i= 1 and δ
i= Q
ki∆, ∀i = 1, 2, . . . , n. Hence
f
∗(u, ∆) = 1 2
n
X
i=1
Arctan(λ
iu) + ∆
2u 2
n
X
i=1
λ
iQ
2k,i1 + λ
2iu
2and
g
∗(u, ∆) =
n
Y
i=1
(1 + λ
2iu
2)
1/4. Exp
( ∆
2u
22
n
X
i=1
λ
2iQ
2k,i1 + λ
2iu
2) .
This completes the proof.
To obtain the power of the test, we just have to write P
IOT
DF∗ 2> c
= 1 − P
IO1 − √
c/n <
Z
T.M
2.Z
−1.
Z
T.M
1.Z
< 1 + √ c/n and
P
IOT
SY M∗ 2> c
= 1 − P
IOZ
T.D
2.Z
−1.
Z
T.D
1.Z
>
n − 2 − c /
n − 2 + c
with T
DF∗and T
SY M∗are the statistics under contamination. The result can be obtained by applying the above proposition.
3. The Median statistic under contamination Recall that the median statistic is defined here by the formula
T
M ED= x
1/M ED{x
2, x
3, . . . , x
n}.
Note that P
T
M ED2> c
= P
− 1/ √
c ≤ 1/T
M ED≥ 1/ √ c
,
where c is a critical value such that P (T
M ED2> c) = α. Our aim is to study the behavior of this probability with respect to ∆ when we observe T
M EDin the contaminated model.
Since analytical treatments are rather complicated , let us give the exact formula of the power test when n = 3. Even if a series of length 3 has no practical sense, our aim is to illustrate mathematically the impact of an IO outlier on the T
M ED-test only.
Proposition 2. If n = 3 and an IO outlier of magnitude ∆ occurs at a specified position k, then for a given ρ = ρ
0P
IOT
M ED∗ 2> c
= P
IO(k
1< Y
1/Y
2< k
2),
where T
M ED∗is the statistic T
M EDobserved in the contaminated model , Y
1and Y
2are independent random variables distributed according to N (m
1, 1) and N (m
2, 1) respectively with m
1and m
2given by
(m
1, m
2) =
(0, 2∆) if k = 1 ,
(ρ
0∆/2γ, 0) if k = 2 ,
(∆/2γ, 0) if k = 3 , where γ = p1 + (1 + ρ
0)
2/2 and
k
1= − 1 γ
1
√ c + ρ
0+ ρ
202
and k
2= 1 γ
1
√ c − ρ
0+ ρ
202
.
P roof. First, observe that x
1=
1and P
IO(T
M ED∗ 2> c) = P
IO(−1/ √
c ≤ 1/T
M ED∗≤ 1/ √ c).
Then, using easy calculations, we can write:
If k = 1, then z
1= x
1+ ∆, z
2= x
2+ ρ
0∆ , z
3= x
3+ ρ
20∆ and 1
T
M ED∗= z
2+ z
32z
1= ρ
0+ ρ
202 + (1 + ρ
0)
2+
32(
1+ δ) = ρ
0+ ρ
202 + γ. Y
1Y
2, where Y
1is N (0, 1) and Y
2is N (∆, 1).
If k = 2, then z
1= x
1, z
2= x
2+ ∆ , z
3= x
3+ ρ∆ and 1
T
M ED∗= z
2+ z
32z
1= ρ
0+ ρ
202 + (1 + ρ
0)
2+
3+ ρ
0∆ 2
1= ρ
0+ ρ
202 + γ. Y
1Y
2,
where Y
1is N (ρ
0∆/2γ, 1) and Y
2is N (0, 1).
If k = 3, then z
1= x
1, z
2= x
2, z
3= x
3+ ∆ and
1
T
M ED∗= z
2+ z
32z
1= ρ
0+ ρ
202 + (1 + ρ
0)
2+
3+ ∆
2
1= ρ
0+ ρ
202 + γ. Y
1Y
2, where Y
1is N (∆/2γ, 1) and Y
2is N (0, 1).
Note that, in the three cases, 1/T
M ED∗is written in the form
1
T
M ED∗= ρ
0+ ρ
202 + γ. Y
1Y
2.
So we obtain
P
IOT
M ED∗ 2> c
= P
IO(−1/ √
c ≤ 1/T
M ED∗≥ 1/ √ c)
= P
IOk
1< Y
1Y
2< k
2with
k
1= − 1 γ
1
√ c + ρ
0+ ρ
202
and k
2= 1 γ
1
√ c − ρ
0+ ρ
202
.
This completes the proof of the proposition.
We can add that, following Cabuk and Springer (1990), the probability
density function of Y
1/Y
2for n = 3 is given by the formula:
h(y) = 1
2π Exp 1
2 ((m
1y + m
2)
2/(y
2+ 1) − (m
21+ m
22))
.
2
y
2+ 1 Exp(−(m
1y + m
2)
2/2(y
2+ 1))
+ h √
2π(m
1y + m
2)/(y
2+ 1)
3/2i h 2Φ
(m
1y + m
2)/ p
y
2+ 1
−1 io
−∞ < y < ∞, where Φ(u) =
21πR
u−∞
e
−x2/2dx. Using the formula given above for n = 3 the power of the test is given by
(11) P
IO(T
M ED∗ 2> c) = Z
k2k1