aricol ICTAMI. 52KB Jun 04 2011 12:08:21 AM
ACTA UNIVERSITATIS APULENSIS
TESTING ON THE RECURRENCE OF COEFFICIENTS
IN THE LINEAR REGRESSIONAL MODEL
by
Nicoleta Breaz and Daniel Breaz
Abstract. The statistical tests on the coefficients of the linear regressional model are well
known. In this paper, we make three new tests, by means of which one can verify if the
coefficients are terms of a sequence given by a recurrencial relation namely geometrical and
aritmetical progression and a general linear reccurence of specified order.
Introduction.
Let the multiple linear regressional model
Y=
and a sample data
y T = ( y1 , y 2 ,..., y n ) ∈ n ,
(
x = x1 ,..., x p
)
x11
x 21
=
...
x n1
(
x12
x 22
...
xn 2
∑α
p
k =1
k
Xk +ε
(1)
... x1 p
... x 2 p
∈ M n , p (), n >> p
... ...
... x np
)
Denoting α T = α 1 ,α 2 ,..., α p ∈ p, ε T = (ε 1 , ε 2 ,..., ε n ) ∈ n, from (1) we obtain the
y = xα + ε
(2)
The principle of least squares leads to the fitting model
y = xa + e
matriceal form
(
)
with a T = a1 , a 2 ,..., a p ∈ p, e T = (e1 , e2 ,..., en ) ∈ n and
∑e
n
2
k
= min
If rang (x ) = p then the fitted coefficients are
k =1
( )
a = xT x xT y
We call the clasical regressional linear model the model (2) that satisfies the following
conditions.
i) E (ε ) = θ , θ T = (0,0,...,0) ∈ n
−1
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
ii) Var (ε ) = σ 2 I ,
iii) ε − N
which can be written as
(3)
(
y− N xα , σ 2 I
The following statistical tests are well known.
)
A. F-test on equality between the coefficients
It is testing the null hypothesis
H 0 : α 1 = ... = α q
versus the alternative H 1 :" H 0 − false" . For a fixed leavel ϕ , we find the value
f = f q −1;n − p ;1−ϕ
of a Fisher-Snedecor statistics with q − 1 and n − p degree of freedom, such that
P (F < f H 0 ) = 1 − ϕ
The null hypothesis is rejected if
f c > f where
n − p S 02 − S12
q − 1 S12
Here and elsewhere in this paper,we use the denotations
S 02 - the sum of squared residuals in the reduced model, obtained under H 0
fc =
S12 -the sum of squared residuals in the full model, obtained under H 1
B. F-test on signifiance of the coefficients
It is testing the null hypothesis
H 0 : α 1 = ... = α q = 0
versus the alternative H 1 :" there exists i0 from {1,2,..., q} such that α i0 ≠ 0" .
For a fixed level ϕ the test rejects the null hypothesis if f c > f where
and f = f q ;n − p ;1−ϕ
S 02 − S12
q
is deduced from
fc =
S12
n− p
P (F < f H 0 ) = 1 − ϕ
Remark. The statistical tests presented above are based on a lemma which provides a
Fisher-Snedecor (F) statistics. Using this lemma we make new tests that verifies some
recurrential connection between the coefficients.
16
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
Lemma. Let be the random vector ε T = (ε 1 ,..., ε n ) as
(
ε − N θ ;σ 2 I
and the matrix x ∈ M np (), A ∈ M pq () ,
If
( )
)
x0 = xA ∈ M nq ()
(
Q = I − x x T x x T and Q0 = I − x0 x0 x0
then the statistics
ε T Q0ε − ε T Qε
ε T Qε
F=
rank (Q0 ) − rank (Q ) rank (Q )
−1
T
)
−1
x0
T
has a Fisher-Snedecor distribution with rank (Q0 ) − rank (Q ) and rank (Q ) degrees of
freedom .
MAIN RESULTS
The following results are obtained on the classical regressional model defined by (3).
The aritmetical progression test
We consider the null hypothesis
H 0 : α 2 = α 1 + r , α 3 = α 1 + 2r ,..., α p = α 1 + ( p − 1)r , p > 2
versus the alternative H 1 : ” H 0 false “
Denoting
x TH 0 = x1 + x 2 + ... + x p , x 2 + 2 x3 + ... + ( p − 1)x p
(
we have the full model
and a reduced one
α HT 0 = (α 1 , r ) = (α 1 , α 2 − α 1 )
y = xα + ε (under H 1 )
y = x H 0 α H 0 + ε (under H 0 )
We note that for r = 0 is obtained the classical test A.
It can be proved the following proposition:
1.Proposition
We have
x H 0 = xA,
17
)
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
where A ∈ M p , 2
1
1
A=
...
1
2.Proposition
Let be the matrix
( )
Q = I − x xT x
−1
0
1
...
p − 1
(
x T and Q0 = I − x H 0 x TH 0 x H 0
and the random variable
S 02 = ε T Q0 ε = e0
The random variable given by
and S12 = ε T Qε = e
2
)
−1
x TH 0
2
S 02 − S12
S12
− F ( p − 2, n − p ),
p−2
n− p
has a Fisher-Snedecor distribution with p − 2 and n − p degrees of freedom.
Proof.
If in lemma we use q = 2 and matrix A from proposition 1., we obtain
F=
F=
ε T Q0ε − ε T Qε
rank (Q0 ) − rank (Q )
Moreover
ε T Qε
rank (Q )
− F (rank (Q0 ) − rank (Q ), rank (Q )),
rank (Q0 ) = n − 2
rank (Q0 ) − rank (Q ) = n − 2 − n + p = p − 2
so the result holds.
Testing
• ϕ → f f = f p − 2;n − p;1−ϕ : P(F < f H 0 ) = 1 − ϕ
S 02 − S12
S12
p−2
n− p
• if f c > f then the H 0 is rejected.
• fc =
The geometrical progression test
18
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
We consider the null hypothesis
H 0 : α 2 = α 1 r , α 3 = α 1 r 2 ,..., α p = α 1 r p −1 , p > 1
versus the alternative H 1 : ” H 0 false “
Denoting
(
)
x TH 0 = x1 + rx 2 + ... + r p −1 x p , r =
α3
α2
α HT 0 = α 1 .
we have the full model
and a reduced one
y = xα + ε (under H 1 )
y = x H 0 α H 0 + ε (under H 0 )
We note that for r = 0 is obtained the classical test B and for r = 1 is obtained the
clasical test A.
It can be proved the following proposition:
3.Proposition
We have
x H 0 = xA,
where A ∈ M p ,1
1
r
A = r2
...
p −1
r
4.Proposition
With simillar notations we have
S 2 − S12
S12
− F ( p − 1, n − p ),
F= 0
p −1
n− p
Proof.
In lemma we use q = 1 and matrix A from proposition 3.
Moreover
rang (Q0 ) = n − 1
19
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
rang (Q0 ) − rang (Q ) = n − 1 − n + p = p − 1
so the result holds.
Testing
• ϕ → f f = f p −1;n − p;1−ϕ : P(F < f H 0 ) = 1 − ϕ
S 02 − S12
S12
p −1
n− p
• if f c > f then the H 0 is rejected.
• fc =
Testing on the linear recurrence of p − 2 order, p > 2
We consider the null hypothesis
H 0 :"α 1 ,...,α p are terms of a sequence given by a linear recurence of p - 2 order
namely x n + p − 2 = k1 x n + k 2 x n +1 + ... + k p − 2 x n + p −3 + k p −1 "
versus the alternative H 1 : ” H 0 false “
The hypothesis H 0 can be written as
α p −1 = α 1 k1 + ... + α p − 2 k p − 2 + k p −1
α p = α 2 k1 + ... + α p − 2 k p −3 + k p − 2 α 1 k1 + ... + α p − 2 k p − 2 + k p −1 + k p −1
Denoting
x TH 0 = x1 + k1 x p −1 + k p − 2 k1 x p , x 2 + k 2 x p −1 + k1 x p + k p − 2 k 2 x p ,...,
(
(
)
(
)
x p − 2 + k p − 2 x p −1 + k p −3 x p + k p − 2 k p − 2 x p , x p −1 + x p + k p − 2 x p
α HT 0 = (α 1 , α 2 ,..., α p − 2 , k p −1 ) .
we have the full model
and a reduced one
y = xα + ε (under H 1 )
y = x H 0 α H 0 + ε (under H 0 )
It can be proved the following proposition:
5.Proposition
We have
x H 0 = xA,
where A ∈ M p , p −1
20
)
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
1
0
...
A=
0
k1
k1 k p − 2
0
1
...
0
k2
k1 + k 2 k p − 2
...
0
0
...
0
0
...
...
...
...
1
0
k p −2
...
1
... k p −3 + k p2 − 2 1 + k p − 2
6.Proposition
With simillar notations we have
S 2 − S12
S12
− F (1, n − p ),
F= 0
1
n− p
Proof.
In lemma we use q = p − 1 and matrix A from proposition 5.
Moreover
rang (Q0 ) = n − p + 1
rang (Q0 ) − rang (Q ) = n − p + 1 − n + p = 1
so the result holds.
Testing
• ϕ → f f = f1;n − p;1−ϕ : P(F < f H 0 ) = 1 − ϕ
S 02 − S12
S12
• fc =
1
n− p
• if f c > f then the H 0 is rejected.
7.Remark
The last test can be generalised for a linear recurrence of l order, 1 ≤ l ≤ p − 2 .
REFERENCES
1.Stapleton J.H.-Linear Statistical Models, A Willey-Interscience Publication, Series
in Probability and Statistics, New York,1995.
2.Saporta G.-Probabilités, analyse des données et statistique, Editions Technip, Paris,
1990
Authors:
21
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
Nicoleta Breaz- „1 Decembrie 1918” University of Alba Iulia, str. N. Iorga, No. 13,
Departament of Mathematics and Computer Science, email: nbreaz@lmm.uab.ro
Daniel Breaz- „1 Decembrie 1918” University of Alba Iulia, str. N. Iorga, No. 13,
Departament of Mathematics and Computer Science, email: dbreaz@lmm.uab.ro
22
TESTING ON THE RECURRENCE OF COEFFICIENTS
IN THE LINEAR REGRESSIONAL MODEL
by
Nicoleta Breaz and Daniel Breaz
Abstract. The statistical tests on the coefficients of the linear regressional model are well
known. In this paper, we make three new tests, by means of which one can verify if the
coefficients are terms of a sequence given by a recurrencial relation namely geometrical and
aritmetical progression and a general linear reccurence of specified order.
Introduction.
Let the multiple linear regressional model
Y=
and a sample data
y T = ( y1 , y 2 ,..., y n ) ∈ n ,
(
x = x1 ,..., x p
)
x11
x 21
=
...
x n1
(
x12
x 22
...
xn 2
∑α
p
k =1
k
Xk +ε
(1)
... x1 p
... x 2 p
∈ M n , p (), n >> p
... ...
... x np
)
Denoting α T = α 1 ,α 2 ,..., α p ∈ p, ε T = (ε 1 , ε 2 ,..., ε n ) ∈ n, from (1) we obtain the
y = xα + ε
(2)
The principle of least squares leads to the fitting model
y = xa + e
matriceal form
(
)
with a T = a1 , a 2 ,..., a p ∈ p, e T = (e1 , e2 ,..., en ) ∈ n and
∑e
n
2
k
= min
If rang (x ) = p then the fitted coefficients are
k =1
( )
a = xT x xT y
We call the clasical regressional linear model the model (2) that satisfies the following
conditions.
i) E (ε ) = θ , θ T = (0,0,...,0) ∈ n
−1
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
ii) Var (ε ) = σ 2 I ,
iii) ε − N
which can be written as
(3)
(
y− N xα , σ 2 I
The following statistical tests are well known.
)
A. F-test on equality between the coefficients
It is testing the null hypothesis
H 0 : α 1 = ... = α q
versus the alternative H 1 :" H 0 − false" . For a fixed leavel ϕ , we find the value
f = f q −1;n − p ;1−ϕ
of a Fisher-Snedecor statistics with q − 1 and n − p degree of freedom, such that
P (F < f H 0 ) = 1 − ϕ
The null hypothesis is rejected if
f c > f where
n − p S 02 − S12
q − 1 S12
Here and elsewhere in this paper,we use the denotations
S 02 - the sum of squared residuals in the reduced model, obtained under H 0
fc =
S12 -the sum of squared residuals in the full model, obtained under H 1
B. F-test on signifiance of the coefficients
It is testing the null hypothesis
H 0 : α 1 = ... = α q = 0
versus the alternative H 1 :" there exists i0 from {1,2,..., q} such that α i0 ≠ 0" .
For a fixed level ϕ the test rejects the null hypothesis if f c > f where
and f = f q ;n − p ;1−ϕ
S 02 − S12
q
is deduced from
fc =
S12
n− p
P (F < f H 0 ) = 1 − ϕ
Remark. The statistical tests presented above are based on a lemma which provides a
Fisher-Snedecor (F) statistics. Using this lemma we make new tests that verifies some
recurrential connection between the coefficients.
16
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
Lemma. Let be the random vector ε T = (ε 1 ,..., ε n ) as
(
ε − N θ ;σ 2 I
and the matrix x ∈ M np (), A ∈ M pq () ,
If
( )
)
x0 = xA ∈ M nq ()
(
Q = I − x x T x x T and Q0 = I − x0 x0 x0
then the statistics
ε T Q0ε − ε T Qε
ε T Qε
F=
rank (Q0 ) − rank (Q ) rank (Q )
−1
T
)
−1
x0
T
has a Fisher-Snedecor distribution with rank (Q0 ) − rank (Q ) and rank (Q ) degrees of
freedom .
MAIN RESULTS
The following results are obtained on the classical regressional model defined by (3).
The aritmetical progression test
We consider the null hypothesis
H 0 : α 2 = α 1 + r , α 3 = α 1 + 2r ,..., α p = α 1 + ( p − 1)r , p > 2
versus the alternative H 1 : ” H 0 false “
Denoting
x TH 0 = x1 + x 2 + ... + x p , x 2 + 2 x3 + ... + ( p − 1)x p
(
we have the full model
and a reduced one
α HT 0 = (α 1 , r ) = (α 1 , α 2 − α 1 )
y = xα + ε (under H 1 )
y = x H 0 α H 0 + ε (under H 0 )
We note that for r = 0 is obtained the classical test A.
It can be proved the following proposition:
1.Proposition
We have
x H 0 = xA,
17
)
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
where A ∈ M p , 2
1
1
A=
...
1
2.Proposition
Let be the matrix
( )
Q = I − x xT x
−1
0
1
...
p − 1
(
x T and Q0 = I − x H 0 x TH 0 x H 0
and the random variable
S 02 = ε T Q0 ε = e0
The random variable given by
and S12 = ε T Qε = e
2
)
−1
x TH 0
2
S 02 − S12
S12
− F ( p − 2, n − p ),
p−2
n− p
has a Fisher-Snedecor distribution with p − 2 and n − p degrees of freedom.
Proof.
If in lemma we use q = 2 and matrix A from proposition 1., we obtain
F=
F=
ε T Q0ε − ε T Qε
rank (Q0 ) − rank (Q )
Moreover
ε T Qε
rank (Q )
− F (rank (Q0 ) − rank (Q ), rank (Q )),
rank (Q0 ) = n − 2
rank (Q0 ) − rank (Q ) = n − 2 − n + p = p − 2
so the result holds.
Testing
• ϕ → f f = f p − 2;n − p;1−ϕ : P(F < f H 0 ) = 1 − ϕ
S 02 − S12
S12
p−2
n− p
• if f c > f then the H 0 is rejected.
• fc =
The geometrical progression test
18
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
We consider the null hypothesis
H 0 : α 2 = α 1 r , α 3 = α 1 r 2 ,..., α p = α 1 r p −1 , p > 1
versus the alternative H 1 : ” H 0 false “
Denoting
(
)
x TH 0 = x1 + rx 2 + ... + r p −1 x p , r =
α3
α2
α HT 0 = α 1 .
we have the full model
and a reduced one
y = xα + ε (under H 1 )
y = x H 0 α H 0 + ε (under H 0 )
We note that for r = 0 is obtained the classical test B and for r = 1 is obtained the
clasical test A.
It can be proved the following proposition:
3.Proposition
We have
x H 0 = xA,
where A ∈ M p ,1
1
r
A = r2
...
p −1
r
4.Proposition
With simillar notations we have
S 2 − S12
S12
− F ( p − 1, n − p ),
F= 0
p −1
n− p
Proof.
In lemma we use q = 1 and matrix A from proposition 3.
Moreover
rang (Q0 ) = n − 1
19
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
rang (Q0 ) − rang (Q ) = n − 1 − n + p = p − 1
so the result holds.
Testing
• ϕ → f f = f p −1;n − p;1−ϕ : P(F < f H 0 ) = 1 − ϕ
S 02 − S12
S12
p −1
n− p
• if f c > f then the H 0 is rejected.
• fc =
Testing on the linear recurrence of p − 2 order, p > 2
We consider the null hypothesis
H 0 :"α 1 ,...,α p are terms of a sequence given by a linear recurence of p - 2 order
namely x n + p − 2 = k1 x n + k 2 x n +1 + ... + k p − 2 x n + p −3 + k p −1 "
versus the alternative H 1 : ” H 0 false “
The hypothesis H 0 can be written as
α p −1 = α 1 k1 + ... + α p − 2 k p − 2 + k p −1
α p = α 2 k1 + ... + α p − 2 k p −3 + k p − 2 α 1 k1 + ... + α p − 2 k p − 2 + k p −1 + k p −1
Denoting
x TH 0 = x1 + k1 x p −1 + k p − 2 k1 x p , x 2 + k 2 x p −1 + k1 x p + k p − 2 k 2 x p ,...,
(
(
)
(
)
x p − 2 + k p − 2 x p −1 + k p −3 x p + k p − 2 k p − 2 x p , x p −1 + x p + k p − 2 x p
α HT 0 = (α 1 , α 2 ,..., α p − 2 , k p −1 ) .
we have the full model
and a reduced one
y = xα + ε (under H 1 )
y = x H 0 α H 0 + ε (under H 0 )
It can be proved the following proposition:
5.Proposition
We have
x H 0 = xA,
where A ∈ M p , p −1
20
)
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
1
0
...
A=
0
k1
k1 k p − 2
0
1
...
0
k2
k1 + k 2 k p − 2
...
0
0
...
0
0
...
...
...
...
1
0
k p −2
...
1
... k p −3 + k p2 − 2 1 + k p − 2
6.Proposition
With simillar notations we have
S 2 − S12
S12
− F (1, n − p ),
F= 0
1
n− p
Proof.
In lemma we use q = p − 1 and matrix A from proposition 5.
Moreover
rang (Q0 ) = n − p + 1
rang (Q0 ) − rang (Q ) = n − p + 1 − n + p = 1
so the result holds.
Testing
• ϕ → f f = f1;n − p;1−ϕ : P(F < f H 0 ) = 1 − ϕ
S 02 − S12
S12
• fc =
1
n− p
• if f c > f then the H 0 is rejected.
7.Remark
The last test can be generalised for a linear recurrence of l order, 1 ≤ l ≤ p − 2 .
REFERENCES
1.Stapleton J.H.-Linear Statistical Models, A Willey-Interscience Publication, Series
in Probability and Statistics, New York,1995.
2.Saporta G.-Probabilités, analyse des données et statistique, Editions Technip, Paris,
1990
Authors:
21
Nicoleta Breaz, Daniel Breaz Testing on the recurrence of coefficients
in the linear regressional model
Nicoleta Breaz- „1 Decembrie 1918” University of Alba Iulia, str. N. Iorga, No. 13,
Departament of Mathematics and Computer Science, email: nbreaz@lmm.uab.ro
Daniel Breaz- „1 Decembrie 1918” University of Alba Iulia, str. N. Iorga, No. 13,
Departament of Mathematics and Computer Science, email: dbreaz@lmm.uab.ro
22