aricol ICTAMI. 52KB Jun 04 2011 12:08:21 AM

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

∑

₌ +

= p

k

k kX

Y

1

ε

α (1)

and a sample data

(

)

∈

= n

T

y y y

y ₁, ₂,..., n ,

(

)

n p

np n

n

p p

p M

x x

x

x x

x

x x

x

x x

x _,

2 1

2 22

21

1 12

11

1

... ... ... ... ...

... ...

,..., ∈

   

 

   

  =

= (),n>> p

Denoting =

(

p

)

∈

T α α α

α ₁, ₂,..., p, =

(

n

)

∈

T ε ε ε

ε ₁, ₂,..., n, from (1) we obtain the matriceal form

ε α+ =x

y (2)

The principle of least squares leads to the fitting model

e xa

y= +

with aT =

(

a1,a2,...,a_p

)

∈



p_, =

(

)

∈

n T

e e e

e ₁, ₂,..., n and min

1

2 =

∑

n₌ k

k

e

If rang

( )

x = p then the fitted coefficients are

( )

x x x y a= T −1 T

We call the clasical regressional linear model the model (2) that satisfies the following conditions.

ii)Var

( )

ε =σ2I ,

iii)ε

−

N (3)

which can be written as

y

−

N

(

xα,σ2I

)

The following statistical tests are well known.

A. F-test on equality between the coefficients

It is testing the null hypothesis

q

H₀ :α₁ =...=α

versus the alternative H₁:"H₀−false". For a fixed leavel ϕ, we find the value

ϕ

− − − = fq 1;n p;1

f

of a Fisher-Snedecor statistics with q−1 and n−p degree of freedom, such that

(

F < f H₀

)

=1−ϕ

P

The null hypothesis is rejected if fc > f where

2 1

2 1 2 0

1 S

S S q

p n fc

− − − =

Here and elsewhere in this paper,we use the denotations

2 0

S - the sum of squared residuals in the reduced model, obtained under H₀

2 1

S -the sum of squared residuals in the full model, obtained under H₁ B. F-test on signifiance of the coefficients

It is testing the null hypothesis

0 ...

: ₁

0 = = q =

H α α

versus the alternative :"thereexists from

{

1,2,...,

}

such that 0"

0

0

1 i q i ≠

H α .

For a fixed level ϕ the test rejects the null hypothesis if fc > f where

p n

S q

S S fc

− −

= 12

2 1 2

0

and f = fq_;n₋p_;1−ϕ is deduced from

(

F < f H₀

)

=1−ϕ

P

Lemma. Let be the random vector

(

n

)

T ε ε

ε = ₁,..., as

ε

−

N

(

θ;σ2I

)

and the matrix x∈M_np(), A∈M_pq() ,

nq

M xA

x₀ = ∈ ()

If

( )

T T

x x x x I

Q= − −1 and Q₀ =I −x₀

( )

x₀Tx₀ −1x₀T

then the statistics

( )

( )

rank

( )

Q

Q Q

rank Q

rank

Q Q

F

T T

T ε ε ε ε ε

ε

− − =

0 0

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

( )

1 , 2

,..., 2 ,

: _{2 1 3 1 1}

0 = +r = + r = + p− r p>

H α α α α αp α

versus the alternative H₁: ”H₀false “ Denoting

( )

(

p p

)

T

H x x x x x p x

x _{1 2} ... , ₂ 2 ₃ ... 1

0 = + + + + + + −

( ) (

1, 1, 2 1

)

0 α α α α αTH = r = − we have the full model

ε α + =x

y (underH₁) and a reduced one

ε α + =xH0 H0

y (under H₀)

We note that for r =0 is obtained the classical test A. It can be proved the following proposition:

1.Proposition

We have

,

0 xA

where A∈Mp_,2

   

 

   

 

− =

1 1

... ...

1 1

0 1

p A

2.Proposition

Let be the matrix

( )

T T

x x x x I

Q= − −1 and

(

H

)

TH

T H

H x x x

x I Q

0 0 0 0

1 0

− −

= and the random variable

2 0 0 2

0 Q e

S =εT ε = and S₁2 =εTQε = e 2

The random variable given by

p n

S p

S S F

− −

−

= 12

2 1 2 0

2

−

F

(

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

( )

( )

rank

( )

Q

Q Q

rank Q

rank

Q Q

F

T T

T ε ε ε ε ε

ε

− − =

0

0

−

(

( )

( )

_,

( )

)

_,

0 rank Q rank Q

Q rank

F −

Moreover

( )

Q₀ =n−2

rank

( )

Q₀ −rank

( )

Q =n−2−n+ p= p−2

rank

so the result holds.

Testing

•

ϕ → f f = fp_−2;n₋p_;1−ϕ:P

(

F < f H0

)

=1−ϕ

•

p n

S p

S S fc

− −

−

= 12

2 1 2 0

2

•

if fc > f then the H0 is rejected.

We consider the null hypothesis

1 1 2

1 3 1 2

0: = , = ,..., = −

p

p r

r r

H α α α α α α , 1p>

versus the alternative H₁: ”H₀false “ Denoting

(

)

2 3 1

2

1 ... ,

0 α

α

= +

+ +

= _{x rx r} − _{x r}

x p

p T

H

1

0 α αT =

H .

we have the full model

ε α + =x

y (underH₁) and a reduced one

ε α + =xH0 H0

y (under H0)

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

,

0 xA

xH =

where A∈Mp,1

     

 

     

 

=

−1 2

... 1

p

r r

r

A

4.Proposition

With simillar notations we have

p n

S p

S S F

− −

−

= 12

2 1 2 0

1

−

F

(

p−1,n−p

)

,

Proof.

In lemma we use q=1 and matrix A from proposition 3. Moreover

( )

Q0 =n−1

( )

Q₀ −rang

( )

Q =n−1−n+p= p−1

rang

so the result holds.

Testing

•

ϕ → f f = fp₋1;n₋p;1_−ϕ:P

(

F < f H0

)

=1−ϕ

•

p n S p S S fc − − −

= 12

2 1 2 0

1

•

if fc > f then the H0 is rejected.

Testing on the linear recurrence of p−2 order, p>2 We consider the null hypothesis

" ... namely order 2 -p of recurence linear a by given sequence a of terms are ,..., :" 1 3 2 1 2 1 2 1 0 − − + − + −

+p = n + n + + p n p + p n p k x k x k x k x

H α α

versus the alternative H₁: ”H₀false “

The hypothesis H₀ can be written as

(

)

(

)

   + + + + + + + = + + + = − − − − − − − − − − − 1 1 2 2 1 1 2 3 2 1 2 1 2 2 1 1 1 ... ... ... p p p p p p p p p p p p k k k k k k k k k k α α α α α α α α Denoting

(

_{1 1 1 2 1} , _{2 2 1 1 2 2} ,...,

0 p p p p p p p

T

H x k x k k x x k x k x k k x

x = + ₋ + ₋ + ₋ + + ₋

x_p−2 +k_p−2x_p−1 +k_p−3x_p +k_p−2k_p−2x_p, x_p−1+x_p +k_p−2x_p

)

(

1, 2,..., 2, 1

)

0 = p− p−

T

H α α α k

α .

we have the full model

ε α + =x

y (underH₁) and a reduced one

ε α + =xH0 H0

y (under H₀)

It can be proved the following proposition:

5.Proposition

We have

,

0 xA

xH =

       

 

       

 

+ +

+ =

− −

− −

−

−

2 2

2 3 2

2 1 2 1

2 2

1

1 ...

1 ...

0 1

... 0

0

... ...

... ...

...

0 0

... 1

0

0 0

... 0

1

p p

p p

p

p

k k

k k

k k k k

k k

k

A

6.Proposition

With simillar notations we have

p n

S S S F

− −

= 12

2 1 2 0

1

−

F

(

1,n−p

)

,

Proof.

In lemma we use q= p−1 and matrix A from proposition 5. Moreover

( )

Q₀ =n−p+1

rang

( )

Q0 −rang

( )

Q =n− p+1−n+ p=1

rang

so the result holds.

Testing

•

ϕ → f f = f_{1;n−p;1−ϕ}:P

(

F < f H₀

)

=1−ϕ

•

p n

S S S fc

− −

= 12

2 1 2 0

1

•

if f_c > f then the H₀ 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

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