SAS/ETS 9.22 User''''s Guide 218 ppsx

This can be formulated as the hypothesis that the cointegrated relation contains only mt and yt through mt yt.. The test of weak exogeneity of y2t for the parameters .˛1; ˇ/ determines w

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Figure 32.56 Parameter Estimation with the ECTREND Option

The VARMAX Procedure

Parameter Alpha * Beta' Estimates

AR Coefficients of Differenced Lag

Model Parameter Estimates

Standard Equation Parameter Estimate Error t Value Pr > |t| Variable

AR2_1_1 -0.72759 0.04623 -15.74 0.0001 D_y1(t-1) AR2_1_2 -0.77463 0.04978 -15.56 0.0001 D_y2(t-1)

AR2_2_1 0.38982 0.04955 7.87 0.0001 D_y1(t-1) AR2_2_2 -0.55173 0.05336 -10.34 0.0001 D_y2(t-1)

Figure 32.56can be reported as follows:

yt D

0:48015 0:98126 3:24543 0:12538 0:25624 0:84748

2

4

y1;t 1

y2;t 1

1

3

5

C

0:72759 0:77463 0:38982 0:55173

yt 1C t

The keyword “EC” in the “Model Parameter Estimates” table means that the ECTREND option is used for fitting the model

For fitting Case 3,

proc varmax data=simul2;

model y1 y2 / p=2 ecm=(rank=1 normalize=y1)

print=(estimates);

run;

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Figure 32.57 Parameter Estimation without the ECTREND Option

Parameter Alpha * Beta' Estimates

AR Coefficients of Differenced Lag

Model Parameter Estimates

Standard Equation Parameter Estimate Error t Value Pr > |t| Variable

D_y1 CONST1 -2.60825 1.32398 -1.97 0.0518 1

AR2_1_1 -0.74052 0.05060 -14.63 0.0001 D_y1(t-1) AR2_1_2 -0.76305 0.05352 -14.26 0.0001 D_y2(t-1) D_y2 CONST2 3.43005 1.39587 2.46 0.0159 1

AR2_2_1 0.34820 0.05335 6.53 0.0001 D_y1(t-1) AR2_2_2 -0.51194 0.05643 -9.07 0.0001 D_y2(t-1)

Figure 32.57can be reported as follows:

yt D

0:46421 0:95103 0:17535 0:35293

yt 1C

0:74052 0:76305 0:34820 0:51194

yt 1

C

2:60825 3:43005

C t

Test for the Linear Restriction on the Parameters

Consider the example with the variables mt log real money, yt log real income, itd deposit interest rate, and itb bond interest rate It seems a natural hypothesis that in the long-run relation, money and income have equal coefficients with opposite signs This can be formulated as the hypothesis that the cointegrated relation contains only mt and yt through mt yt For the analysis, you can express these restrictions in the parameterization of H such that ˇD H, where H is a known k s matrix

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and is the s r.r s < k/ parameter matrix to be estimated For this example, H is given by

H D

2

6 6 4

1 0 0

0 1 0

0 0 1

3

7 7 5

RestrictionH0W ˇ D H

When the linear restriction ˇD H is given, it implies that the same restrictions are imposed on all cointegrating vectors You obtain the maximum likelihood estimator of ˇ by reduced rank regression

of yt on H yt 1corrected for yt 1; : : : ; yt pC1; Dt/, solving the following equation

jH0S11H H0S10S001S01Hj D 0

for the eigenvalues 1 > 1> > s > 0 and eigenvectors v1; : : : ; vs/, Sij given in the preceding section Then choose O D v1; : : : ; vr/ that corresponds to the r largest eigenvalues, and the Oˇ is

H O

The test statistic for H0W ˇ D H is given by

T

r

X

i D1

logf.1 i/=.1 i/g! d 2r.k s/

If the series has no deterministic trend, the constant term should be restricted by ˛0?ı0 D 0 as in Case 2 Then H is given by

H D

2

6 6 6 6 4

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

3

7 7 7 7 5

The following statements test that 2 ˇ1C ˇ2 D 0:

model y1 y2 / p=2 ecm=(rank=1 normalize=y1);

cointeg rank=1 h=(1,-2);

run;

Figure 32.58shows the results of testing H0W 2ˇ1C ˇ2 D 0 The input H matrix is H D 1 2/0 The adjustment coefficient is reestimated under the restriction, and the test indicates that you cannot reject the null hypothesis

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Figure 32.58 Testing of Linear Restriction (H= Option)

Beta Under Restriction

Alpha Under Restriction

Hypothesis Test

Restricted Index Eigenvalue Eigenvalue DF Chi-Square Pr > ChiSq

Test for the Weak Exogeneity and Restrictions of Alpha

Consider a vector error correction model:

yt D ˛ˇ0yt 1C

p 1

X

i D1

ˆiyt iC ADt C t

Divide the process yt into y01t; y02t/0with dimension k1and k2and the † into

†D

†11 †12

†21 †22

Similarly, the parameters can be decomposed as follows:

˛D

˛1

˛2

ˆi D

ˆ1i

ˆ2i

AD

A1

A2

Then the VECM(p) form can be rewritten by using the decomposed parameters and processes:

y1t

y2t

D

˛1

˛2

ˇ0yt 1C

p 1

X

i D1

ˆ1i

ˆ2i

yt iC

A1

A2

Dt C

1t

2t

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The conditional model for y1t given y2t is

y1t D !y2tC ˛1 !˛2/ˇ0yt 1C

p 1

X

i D1

.ˆ1i !ˆ2i/yt i

C.A1 !A2/DtC 1t !2t

and the marginal model of y2t is

y2t D ˛2ˇ0yt 1C

p 1

X

i D1

ˆ2iyt iC A2Dt C 2t

where ! D †12†221

The test of weak exogeneity of y2t for the parameters ˛1; ˇ/ determines whether ˛2 D 0 Weak exogeneity means that there is no information about ˇ in the marginal model or that the variables

y2tdo not react to a disequilibrium

RestrictionH0W ˛ D J

Consider the null hypothesis H0W ˛ D J , where J is a k m matrix with r m < k

From the previous residual regression equation

R0t D ˛ˇ0R1t C Ot D J ˇ0R1tC Ot

you can obtain

N

J0R0t D ˇ0R1t C NJ0Ot

J?0R0t D J?0 Ot

where NJ D J.J0J / 1and J?is orthogonal to J such that J?0J D 0

Define

†JJ? D NJ0†J? and †J?J? D J?0†J?

and let !D †JJ?†J1

? J? Then NJ0R0t can be written as N

J0R0t D ˇ0R1t C !J?0R0t C NJ0Ot !J?0 Ot

Using the marginal distribution of J?0R0tand the conditional distribution of NJ0R0t, the new residuals are computed as

Q

RJ t D JN0R0t SJJ?SJ1

? J?J?0R0t

Q

R1t D R1t S1J?SJ1J J?0R0t

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SJJ? D NJ0S00J?; SJ?J? D J?0S00J?; and SJ?1 D J?0S01

In terms of QRJ t and QR1t, the MLE of ˇ is computed by using the reduced rank regression Let

Sij:J? D T1

T

X

t D1

Q

Ri tRQjt0 ; for i; j D 1; J

Under the null hypothesis H0W ˛ D J , the MLE Qˇ is computed by solving the equation

jS11:J? S1J:J?SJJ:J1

?SJ1:J?j D 0

Then Qˇ D v1; : : : ; vr/, where the eigenvectors correspond to the r largest eigenvalues The likelihood ratio test for H0W ˛ D J is

T

r

X

i D1

logf.1 i/=.1 i/g! d 2r.k m/

The test of weak exogeneity of y2tis a special case of the test ˛D J , considering J D Ik1; 0/0 Consider the previous example with four variables ( mt; yt; itb; itd ) If rD 1, you formulate the weak exogeneity of (yt; itb; itd) for mt as J D Œ1; 0; 0; 00and the weak exogeneity of itd for (mt; yt; itb)

as J D ŒI3; 00

The following statements test the weak exogeneity of other variables, assuming r D 1:

cointeg rank=1 exogeneity;

run;

cointeg rank=1 j=exogeneity;

run;

Figure 32.59shows that each variable is not the weak exogeneity of other variable

Figure 32.59 Testing of Weak Exogeneity (EXOGENEITY Option)

Testing Weak Exogeneity of Each Variables

Variable DF Chi-Square Pr > ChiSq

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Forecasting of the VECM

Consider the cointegrated moving-average representation of the differenced process of yt

yt D ı C ‰.B/t

Assume that y0D 0 The linear process yt can be written as

yt D ıt C

t

X

i D1

t i

X

j D0

‰ji

Therefore, for any l > 0,

yt Cl D ı.t C l/ C

t

X

i D1

t Cl i

X

j D0

‰ji C

l

X

i D1

l i

X

j D0

‰jt Ci

The l -step-ahead forecast is derived from the preceding equation:

yt Cljt D t C l/ C

t

X

i D1

t Cl i

X

j D0

‰ji

Note that

lim

l!1ˇ0yt Cljt D 0

since liml!1Pt Cl i

j D0 ‰j D ‰.1/ and ˇ0‰.1/ D 0 The long-run forecast of the cointegrated system shows that the cointegrated relationship holds, although there might exist some deviations from the equilibrium status in the short-run The covariance matrix of the predict error et Cljt D

yt Cl yt Cljt is

†.l/D

l

X

i D1

Œ

l i

X

j D0

‰j/†

l i

X

j D0

‰0j/

When the linear process is represented as a VECM(p) model, you can obtain

yt D …yt 1C

p 1

X

j D1

ˆjyt j C ı C t

The transition equation is defined as

zt D F zt 1C et

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where zt D y0t 1; y0t; y0t 1; ; y0t pC2/0is a state vector and the transition matrix is

F D

2

6 6 6 6 6 4

… …C ˆ1/ ˆ2 ˆp 1

::

: ::: ::: : :: :::

3

7 7 7 7 7 5

where 0 is a k k zero matrix The observation equation can be written

yt D ıt C H zt

where H D ŒIk; Ik; 0; : : : ; 0

The l -step-ahead forecast is computed as

yt Cljt D ı.t C l/ C HFlzt

Cointegration with Exogenous Variables

The error correction model with exogenous variables can be written as follows:

yt D ˛ˇ0yt 1C

p 1

X

i D1

ˆiyt iC ADt C

s

X

i D0

‚ixt iC t

The following statements demonstrate how to fit VECMX(p; s), where pD 2 and s D 1 from the P=2 and XLAG=1 options:

model y1 y2 = x1 / p=2 xlag=1 ecm=(rank=1);

run;

The following statements demonstrate how to BVECMX(2,1):

model y1 y2 = x1 / p=2 xlag=1 ecm=(rank=1)

prior=(lambda=0.9 theta=0.1);

run;

I(2) Model

The VARX(p,s) model can be written in the error correction form:

yt D ˛ˇ0yt 1C

p 1

X

i D1

ˆiyt iC ADt C

s

X

i D0

‚ixt iC t

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Let ˆD Ik Pp 1i D1 ˆi.

If ˛ and ˇ have full-rank r , and rank.˛0?ˆˇ?/D k r, then yt is an I.1/ process

If the condition rank.˛0?ˆˇ?/ D k r fails and ˛0?ˆˇ? has reduced-rank ˛0?ˆˇ? D 0 where and are k r/ s matrices with s k r, then ˛? and ˇ?are defined as k k r/ matrices of full rank such that ˛0˛?D 0 and ˇ0ˇ?D 0

If and have full-rank s, then the process yt is I.2/, which has the implication of I.2/ model for the moving-average representation

yt D B0C B1t C C2

t

X

j D1

j

X

i D1

iC C1

t

X

i D1

iC C0.B/t

The matrices C1, C2, and C0.B/ are determined by the cointegration properties of the process, and

B0and B1are determined by the initial values For details, see Johansen (1995a)

The implication of the I.2/ model for the autoregressive representation is given by

2yt D …yt 1 ˆyt 1C

p 2

X

i D1

‰i2yt i C ADt C

s

X

i D0

‚ixt i C t

where ‰i D Pp 1

j DiC1ˆi and ˆD Ik Pp 1i D1 ˆi

Test for I(2)

The I.2/ cointegrated model is given by the following parameter restrictions:

Hr;sW … D ˛ˇ0and ˛0?ˆˇ?D 0

where and are k r/ s matrices with 0 s k r Let Hr0represent the I.1/ model where

˛ and ˇ have full-rank r, let Hr;s0 represent the I.2/ model where and have full-rank s, and let

Hr;srepresent the I.2/ model where and have rank s The following table shows the relation between the I.1/ models and the I.2/ models

Table 32.2 Relation between theI.1/andI.2/Models

0 H00 H01 H0;k 1 H0k D H 0

: :

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Johansen (1995a) proposed the two-step procedure to analyze the I.2/ model In the first step, the values of r; ˛; ˇ/ are estimated using the reduced rank regression analysis, performing the regression analysis 2yt, yt 1, and yt 1on 2yt 1; : : : ; 2yt pC2; and Dt This gives residuals R0t, R1t, and R2t, and residual product moment matrices

Mij D 1

T

X

t D1

Ri tR0jt for i; j D 0; 1; 2

Perform the reduced rank regression analysis 2yt on yt 1 corrected for yt 1,

2yt 1; : : : ; 2yt pC2; and Dt, and solve the eigenvalue problem of the equation

jM22:1 M20:1M00:11 M02:1j D 0

where Mij:1 D Mij Mi1M111M1j for i; j D 0; 2

In the second step, if r; ˛; ˇ/ are known, the values of s; ; / are determined using the reduced rank regression analysis, regressing O˛0?2yton Oˇ0?yt 1corrected for 2yt 1; : : : ; 2yt pC2; Dt, and O

ˇ0yt 1

The reduced rank regression analysis reduces to the solution of an eigenvalue problem for the equation

jMˇ?ˇ?:ˇ Mˇ?˛?:ˇM˛1

? ˛?:ˇM˛?ˇ?:ˇj D 0

where

Mˇ?ˇ?:ˇ D ˇ?0 M11 M11ˇ.ˇ0M11ˇ/ 1ˇ0M11/ˇ?

Mˇ0

? ˛?:ˇ D M˛?ˇ?:ˇ D N˛0?.M01 M01ˇ.ˇ0M11ˇ/ 1ˇ0M11/ˇ?

M˛?˛?:ˇ D N˛0?.M00 M01ˇ.ˇ0M11ˇ/ 1ˇ0M10/N˛?

where N˛ D ˛.˛0˛/ 1

The solution gives eigenvalues 1 > 1 > > s> 0 and eigenvectors v1; : : : ; vs/ Then, the ML estimators are

O D v1; : : : ; vs/

O D M˛?ˇ?:ˇO

The likelihood ratio test for the reduced rank model Hr;swith rank s in the model Hr;k r D Hr0

is given by

Qr;s D T

k r

X

i DsC1

log.1 i/; sD 0; : : : ; k r 1

The following statements compute the rank test to test for cointegrated order 2:

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