D:\687303163.doc
EXERCISE ON ESTIMATING A STRUCTURAL VAR IN EVIEWS
Introduction
The aim of this exercise is to describe the basic steps to estimate a structural VAR and
studying impulse response curves in EViews.
We use Australian data for our examples in this exercise. The dataset is quarterly data
spanning 21 years: 1980Q1 - 2001Q1
Preparations
Enter EViews and choose:
 File
 Open dataset
 And select Svardata.wf1 and click OK
1
D:\687303163.doc
1 Defining the model
We define a simple model of an economy (in this case Australia) for the estimate of the VAR, with four
endogenous variables1: Inflation (INF), Real exchange rate (REER), Interest rate (INTEREST), and
Output gap (GAP).
Q1. Before we estimate our model, what should we do to ensure that our estimates are not biased?
Answer: We use OLS to estimate the VARS so we need to ensure that all variables are stationary to
avoid the spurious regression problem associated with unit roots.
Check for stationarity or otherwise of the four variables (see Unit Root exercise). We assume that all
four variables are stationary I(0) variables for simplicity.
Q2. Estimate an unrestricted VAR with all four endogenous variables in their stationary form with a
constant.
Answer: Use Quick, Estimate VAR or type in the command window:
var var01.ls 1 5 INTEREST GAP INF REER,
this command specifies a VAR estimate with the name var01 of the four variables (interest in
difference note) and a lag length of 5.
Vector Autoregression Estimates
Date: 11/19/02 Time: 12:45
Sample(adjusted): 1981:2 2001:4
Included observations: 83 after adjusting endpoints
Standard errors in ( ) & t-statistics in [ ]
INTEREST(-1)
.
.
.
C
R-squared
Adj. R-squared
Sum sq. resids
S.E. equation
F-statistic
Log likelihood
Akaike AIC
Schwarz SC
Mean dependent
S.D. dependent
INTEREST
GAP
INF
REER
0.983846
(0.12698)
[ 7.74811]
0.001558
(0.00086)
[ 1.81260]
0.209413
(0.09527)
[ 2.19814]
0.622813
(0.40982)
[ 1.51972]
-0.325614
(1.57990)
[-0.20610]
0.013497
(0.01069)
[ 1.26221]
1.355519
(1.18535)
[ 1.14356]
12.57886
(5.09909)
[ 2.46688]
0.950040
0.933923
73.04422
1.085418
58.94917
-112.4692
3.216126
3.828121
9.580241
4.222534
0.809195
0.747645
0.003346
0.007346
13.14695
302.1585
-6.774905
-6.162909
-0.000263
0.014624
0.958091
0.944572
41.11679
0.814355
70.86924
-88.62130
2.641477
3.253473
5.088072
3.458975
0.959035
0.945821
760.8750
3.503167
72.57452
-209.7205
5.559530
6.171525
111.2133
15.05025
Determinant Residual Covariance
Log Likelihood (d.f. adjusted)
Akaike Information Criteria
Schwarz Criteria
1
0.000461
-152.2836
5.693581
8.141564
Variable names in dataset in parenthesis.
2
D:\687303163.doc
Q3. We've selected a lag length of 5 for this estimate but is that appropriate?
Answer: Adding more lags always improves the fit but it reduces the degrees of freedom and increases
the danger of overfitting. An objective and replicable way to decide between these competing
objectives is to maximise some weighted measures of these two parameters. This is how the Akaike
Information criterion (AIC) and the Schwarz-Bayesian criterion (SBC) work. These two statistics are
measures of the trade-off fit against loss of degrees of freedom so that the best lag length should
minimise2 both of them.
An alternative to minimising a weighted measure of the lag length and best fit, as above, is to
systematically test for the significance of each lag using a likelihood ratio test (discussed in Lutkepohl,
1991, section 4.3). Since a VAR of lag length n nests the same VAR of lag length (n-1), the log
likelihood difference multiplied by the number of observations less the number of regressors in the
VAR should be distributed as a Chi-sq distribution with 2k degree of freedom, 2(2k) i.e.
LR = (T-m) {logn-1- logn} ~ 2(2k)
Where T is the number of observations, m is the number of regressors , logn is the log likelihood
of the VAR with n lags. The idea is that for each lag length, if there is no improvement in the fit from
the inclusion of this last lag then the difference in errors should not be significantly different from
white noise.
For our example, we use a general-to-specific specification (although some authors might specify a
specific-to-general instead).
(i)
We start with a large lag length (say 10).
(ii)
For each lag, say n, note its log likelihood then calculate the log likelihood for a VAR of
lag (n-1)
(iii) Take the difference of the log likelihoods
(iv)
This difference in (iii) should be distributed as a Chi-sq distribution with eight degrees
of freedom 2(8) if it is superfluous.
Q4. Test for the number of lags using the AIC, SC, and the likelihood ratio test for the appropriate lag
length in the VAR
Answer: with the VAR open select VIEW/ LAG STRUCTURE/ LAG LENGTH CRITERIA then
enter 10 in the maximum lag specification.
The readout should be:
2
Some programs maximise the negative of these measures.
3
D:\687303163.doc
VAR Lag Order Selection Criteria
Endogenous variables: INTEREST GAP INF REER
Exogenous variables: C
Date: 11/19/02 Time: 12:34
Sample: 1980:1 2001:4
Included observations: 78
Lag
LogL
LR
FPE
AIC
SC
HQ
0
1
2
3
4
5
6
7
8
9
10
-459.7154
-169.8935
-152.9814
-137.4222
-125.4949
-95.16905
-80.29215
-60.13943
-51.09151
-30.89014
-19.16174
NA
542.4870
29.92147
25.93198
18.65545
44.32243*
20.21733
25.32009
10.43991
21.23733
11.12694
1.713665
0.001531
0.001501
0.001531
0.001726
0.001227*
0.001314
0.001249
0.001612
0.001606
0.002055
11.89014
4.869064
4.845676
4.856979
4.961408
4.594078
4.622876
4.516396*
4.694654
4.586927
4.696455
12.01099
5.473349*
5.933388
6.428118
7.015975
7.132072
7.644297
8.021245
8.682931
9.058631
9.651586
11.93852
5.110970*
5.281107
5.485935
5.783889
5.610084
5.832406
5.919451
6.291234
6.377032
6.680085
* indicates lag order selected by the criterion
LR: sequential modified LR test statistic (each test at 5% level)
FPE: Final prediction error
AIC: Akaike information criterion
SC: Schwarz information criterion
HQ: Hannan-Quinn information criterion
The correct lag length will depend on the criteria or measure we use. This is typical of these tests and
researchers often use the criterion most convenient for their needs. The SC criterion is generally more
conservative in terms of lag length than the AIC criterion.
Here we assume a lag length of 1 for convenience.
2 VAR identification
We have an underlying structural equation of the form:
Ayt  C ( L) yt  Bu t
(1)
where the stochastic error ut is normally distributed i.e. ut  N(0,I). Unfortunately we cannot estimate
this equation directly due to identification issues, but instead we have estimated an unrestricted VAR of
the form:
yt  A1C ( L) yt  A1 Bu t
(2)
Matrices A B and C are not separately observable. So how can we recover equation (1) from (2)? The
solution is to impose restrictions on our VAR to identify an underlying structure but kind of restrictions
are these?
Economic theory can sometime tell us something about the structure of the system we wish to estimate.
As economists, we must interpret these structure or assumption from theory into restrictions on the
VAR. Such restrictions can include for example:
1. Causal ordering of shock propagation e.g. Cholesky decomposition
2. Nominal variables to have no long run effect on real variables
3. Long run behaviour of variables e.g. real exchange rate is constant in the long run
4
D:\687303163.doc
There are two types of restrictions, one type imposes restrictions on the short-run behaviour of the
system, and the other type imposes restrictions on the long run behaviour of the system. EViews
allows either short run or long run identifying restrictions, but not both types in one VAR. Moreover, it
also assumes that matrix B is an invertable square matrix i.e. there are no identities in the system.
3 Imposing Short Run Restrictions
To impose short run restrictions in EViews we use equation (2):
yt  A1C ( L) yt  A1 Bu t
We can estimate the random stochastic residual A-1But from the residual et of the estimated unrestricted
VAR:
A 1 Bu t  et
(3)
Reformulating equation (3) we have A 1 Bu t ut ' B' A' 1  et et ' , and since u t u t '  I , we have:
A 1 BB ' A'1  et et '
(4)
Equation (4) says that if there are k variables, the symmetry property above imposes k(k+1)/2
restrictions on the 2k2 unknown elements in A and B. Hence an additional k(3k-1)/2 restrictions must
be imposed. EViews requires that such restriction schemes must be of the form:
Aet  Bu t
(5)
By imposing a structure on matrices A and B, we necessarily impose restrictions on the structural VAR
in equation (1). An example of this specification using the Cholesky decomposition identification
scheme is:
 1

a
A =  21
a
 31
a
 41
0
1
0
0
a32
a 42
1
a 43
0

0
,B=
0

1 
 a11

 0
 0

 0

0
0
a 22
0
0
0
a33
0
0 

0 
0 

a 44 
For our example, we have a VAR with four endogenous variables, therefore we require 22 = 4(3*4-1)/2
restrictions. We can impose this restriction in EViews in either matrix form or in text form. Imposing
restrictions in matrix form is relatively straightforward, so we explain the text form instead.3
3
Restriction using text form is also more flexible, as we can restrict values to be equal.
5
D:\687303163.doc
Q5. Impose the Cholesky decomposition. Assumes that shocks or innovations are propagated in the
order of INTEREST, GAP, INF and then REER,
Answer: To impose the restriction above in text format, we select Procs/Estimate
Structural Factorization from the VAR window menu. In the SVAR Options dialog, select
Text (or Matrix is appropriate).
Each endogenous variable has an associated variable number, in our example this is:
@e1 for INTEREST residuals
@e2 for GAP residuals
@e3 for INF residuals
@e4 for REER residuals
The identifying restriction is imposed in terms of variables e's which are the residuals from the VAR
estimates, and variables u's which are the fundamental or 'primitive' random (stochastic) errors in the
structural system.
Enter the following the text box:
@e1 = C(1)*@u1
@e2 = C(2)*@e1 + C(3)*@u2
@e3 = C(4)*@e1 + C(5)*@e2 + C(6)*@u3
@e4 = C(7)*@e1 + C(8)*@e2 + C(9)*@e3 + C(10)*@u4
To use the matrix form to impose a restriction, create two matrices with the following values (we call
them A and B here)
0
0
 1

0
 NA 1
A =
NA NA 1

 NA NA NA

0

0
,B=
0

1 
0
0 
 NA 0


0 
 0 NA 0
 0
0 NA 0 


 0
0
0 NA

Select the matrix specification in the structural factorisation described above and enter the names of the
matrices (A and B) as appropriate:
The output is:
6
D:\687303163.doc
Structural VAR Estimates
Sample(adjusted): 1981:2 2001:4
Included observations: 83 after adjusting endpoints
Estimation method: method of scoring (analytic derivatives)
Convergence achieved after 12 iterations
Structural VAR is just-identified
Model: Ae = Bu where E[ee']=I
Restriction Type: short-run pattern matrix
A=
1
0
C(1)
1
C(2)
C(4)
C(3)
C(5)
B=
C(7)
0
0
C(8)
0
0
0
0
0
0
1
C(6)
0
0
0
1
0
0
C(9)
0
0
0
0
C(10)
Coefficient
Std. Error
z-Statistic
Prob.
C(1)
C(2)
C(3)
C(4)
C(5)
C(6)
C(7)
C(8)
C(9)
C(10)
-0.001066
-0.091071
0.329808
1.055599
-48.42839
0.965941
1.085418
0.007255
0.808446
3.372914
0.000734
0.082789
0.347913
12.23200
51.03540
0.457947
0.084245
0.000563
0.062748
0.261789
-1.453529
-1.100032
0.947961
0.086298
-0.948918
2.109284
12.88410
12.88410
12.88410
12.88410
0.1461
0.2713
0.3431
0.9312
0.3427
0.0349
0.0000
0.0000
0.0000
0.0000
Log likelihood
-152.2836
0.000000
1.000000
1.055599
-48.42839
0.000000
0.000000
1.000000
0.965941
0.000000
0.000000
0.000000
1.000000
0.000000
0.007255
0.000000
0.000000
0.000000
0.000000
0.808446
0.000000
0.000000
0.000000
0.000000
3.372914
Estimated A matrix:
1.000000
-0.001066
-0.091071
0.329808
Estimated B matrix:
1.085418
0.000000
0.000000
0.000000
4 Imposing Long Run Restrictions
To impose long run restrictions, we return to the unrestricted VAR estimate:
yt  A1C ( L) yt  A1 Bu t
Rearrangement of this equation reveals equation (5) below:
[ I  A 1C ( L)] y t  A 1 Bu t
y t  [ I  A 1C ( L)] 1 A 1 Bu t
(6)
7
D:\687303163.doc
Equation (6) shows how the random (stochastic) shocks affect the long-run levels of the variables. If
we define a matrix C  [ I  A 1C ( L)] 1 A 1 B , the aggregate effect of a shock u's is given by matrix C.
Hence if we assume that the (long run) cumulative effect of a sub-shock ui on a variable yj is zero, then
column i and row j of matrix C should be zero. Knowing the values of matrix C tells us something
about matrices A and B.
Due to the number of restrictions required EViews also imposes the restriction that matrix A is the
identity matrix. It uses matrix C to estimate matrix B.
Long run imposition can be specified in matrix form (were matrix C is entered) or in text form in
EViews.
Going back to our example, we know that we must have at least 22 restrictions. Matrix A is assumed
to be the identity matrix. This is equivalent to 16 restrictions so a further 6 restrictions must be
imposed in matrix C.
We have the following variable numeration:
Q6. Impose the following (arbitrary) restriction to matrix C:
(i)
The two nominal variable (INTERST and INF) have no long run real effect, and
(ii)
INF shocks have no long run effect on other variables
(iii) Gap is unaffected by any shocks but its own
Answer: In matrix form this is:
 NA

 0
C= 
NA

 0

NA 0 NA

NA 0
0 
NA NA NA

NA 0 NA
In the text form restriction is:
@LR1(@U3)=0
@LR2(@U1)=0
@LR2(@U3)=0
@LR2(@U4)=0
@LR4(@U1)=0
@LR4(@U3)=0
The output is:
8
D:\687303163.doc
Structural VAR Estimates
Sample(adjusted): 1980:2 2001:4
Included observations: 87 after adjusting endpoints
Estimation method: method of scoring (analytic derivatives)
Convergence achieved after 13 iterations
Structural VAR is just-identified
Model: Ae = Bu where E[ee']=I
Restriction Type: long-run text form
Long-run response pattern:
C(1)
C(3)
0
C(4)
C(2)
C(5)
0
C(6)
0
0
C(7)
0
C(8)
0
C(9)
C(10)
Coefficient
Std. Error
z-Statistic
Prob.
C(1)
C(2)
C(3)
C(4)
C(5)
C(6)
C(7)
C(8)
C(9)
C(10)
10.96752
8.536423
-13.50216
-0.031006
-7.593409
-27.08645
5.160389
16.85658
14.40233
71.07902
0.831445
0.851401
2.386708
0.002351
1.964504
7.892281
0.391208
1.736553
1.528329
5.388486
13.19091
10.02632
-5.657231
-13.19091
-3.865306
-3.432018
13.19091
9.706919
9.423579
13.19091
0.0000
0.0000
0.0000
0.0000
0.0001
0.0006
0.0000
0.0000
0.0000
0.0000
Log likelihood
-200.6625
0.000000
1.000000
0.000000
0.000000
0.000000
0.000000
1.000000
0.000000
0.000000
0.000000
0.000000
1.000000
-0.022506
0.007326
-0.430771
0.327101
-0.298165
0.000116
0.470101
-3.389196
0.025184
0.002695
0.700787
1.608676
Estimated A matrix:
1.000000
0.000000
0.000000
0.000000
Estimated B matrix:
1.101966
0.002579
0.281968
-1.833790
5 Generating impulse responses
Q7. Generate impulse response curves for the identification scheme in our example.
Answer: Selecting VIEW/IMPULSE RESPONSES, on the IMPULSE DEFINITION tag select
STRUCTURAL DECOMPOSITION. This option shows the impulse response of each variable from
shocks to the underlying fundamental shocks (u's).
9
D:\687303163.doc
Response to Structural One S.D. Innovations
Response of INTEREST to Shock1
Response of INTEREST to Shock2
Response of INTEREST to Shock3
Response of INTEREST to Shock4
1.2
1.2
1.2
1.2
0.8
0.8
0.8
0.8
0.4
0.4
0.4
0.4
0.0
0.0
0.0
0.0
-0.4
-0.4
1
2
3
4
5
6
7
8
9
10
-0.4
1
Response of GAP to Shock1
2
3
4
5
6
7
8
9
10
-0.4
1
Response of GAP to Shock2
2
3
4
5
6
7
8
9
10
1
Response of GAP to Shock3
.008
.008
.008
.006
.006
.006
.006
.004
.004
.004
.004
.002
.002
.002
.002
.000
.000
.000
2
3
4
5
6
7
8
9
10
1
Response of INF to Shock1
2
3
4
5
6
7
8
9
10
Response of INF to Shock2
2
3
4
5
6
7
8
9
10
1
Response of INF to Shock3
.8
.8
.6
.6
.6
.4
.4
.4
.4
.2
.2
.2
.2
.0
.0
.0
.0
-.2
-.2
-.2
-.2
-.4
-.4
-.6
1
2
3
4
5
6
7
8
9
10
Response of REER to Shock1
2
3
4
5
6
7
8
9
10
Response of REER to Shock2
2
3
4
5
6
7
8
9
10
1
Response of REER to Shock3
3
3
2
2
2
2
1
1
1
1
0
0
0
0
-1
-1
-1
-1
-2
-2
-2
-2
-3
-3
-3
-4
2
3
4
5
6
7
8
9
10
2
3
4
5
6
7
8
9
10
9
10
2
3
4
5
6
7
8
9
10
2
3
4
5
6
7
8
9
10
-3
-4
1
8
Response of REER to Shock4
3
1
7
-.6
1
3
-4
6
-.4
-.6
1
5
Response of INF to Shock4
.8
.6
-.6
4
.000
1
.8
-.4
3
Response of GAP to Shock4
.008
1
2
-4
1
2
3
4
5
6
7
8
9
10
1
2
3
4
5
6
7
8
9
10
Can you go back and replicate this for the Cholesky decomposition? How does it compare with the
default Cholesky decomposition available in EViews? They should be the same.
6 Generating the ‘fundamental’ shocks
From the estimated structural VAR, how can we generate the fundamental shocks i.e. ut's in equation
(1), using EViews? To generate the fundamental shocks, we use equation Aet  Bu t . Where et is the
error or residual from the VAR regression which has been generated, and matrices A and B identified
in the estimated SVAR. The fundamental shocks are then simply: uˆt  B 1 Aet
Q8. Generate the fundamental shocks from the SVAR
Answer: Select PROCS\MAKE RESIDUALS. EViews will automatically generate series named
RESID?? in the same ordering as the VAR estimate. In this case we assume that they are RESID01,
RESID02, RESID03, RESID04. Change the names as appropriate.
10
D:\687303163.doc
Create two matrices call them matx_a and matx_b by typing in the command window
matrix(4,4) matx_a
matrix(4,4) matx_b
and enter the estimated co-efficient from the estimated SVAR i.e.
matx_a values
1.000000
0.000000
0.000000
0.000000
0.000000
1.000000
0.000000
0.000000
0.000000
0.000000
1.000000
0.000000
0.000000
0.000000
0.000000
1.000000
matx_b values
1.101966
0.002579
0.281968
-1.833790
-0.022506 -0.298165
0.007326 0.000116
-0.430771 0.470101
0.327101 -3.389196
0.025184
0.002695
0.700787
1.608676
Type in the command window:
group resgrp RESID01 RESID02 RESID03 RESID04
matrix resmatrx=@convert(resgrp)
matrix resfund=@transpose(@inverse(matx_b) * matx_a *
@transpose(resmatrx))
show resfund
Select View/graph/line to see a plot of the errors:
10
5
0
-5
-10
-15
-20
10
20
30
40
C1
C2
50
60
70
80
C3
C4
We can see that shock volatility was higher in the first half of the sample.
11
D:\687303163.doc
7 Estimating VARs in EViews 4.0
Estimating an unrestricted VAR is relatively straightforward in EViews 4.0, this was shown in question
Q1. Identifying the VAR to obtain a Structural VAR by imposing certain identifying assumption is
more involved and new techniques of doing this are available all the time. The question is then what
identifying assumption can be imposed in EViews 4.0 and what can’t we impose? EViews is fairly
flexible in that it only imposes two restrictions on the types of restrictions one can impose. This class of
restrictions are known as AB class of restrictions (Amisano and Giannini, 1997).
We begin with our original structural equation that we wish to identify:
Ayt  C ( L) yt  Bu t
and we list each restriction that EViews assumes below:
Restriction 1. Matrices A and B are square and invertible.
This means that the number of the fundamental shocks (ui’s) is equal to the number of endogenous
variables. This implicitly assumes each fundamental shock is identified uniquely with an endogenous
variable.
For a given number of endogenous variables in the system, there is no a priori reason to assume that
there are the same number fundamental shocks present. For instance it might more convenient to
assume that there are fewer number of fundamental shocks, although in practice most researchers
assume that the two are equal.
Restriction 2. Identifying restrictions can either be of the short run type characterised by equation (5)
in section 3, or long run type characterised by equation (6) in section 4, but not both.
By imposing a long-run restriction (in matrix C described above) we infer a restriction on the values of
matrices A and B, although such a restriction may not be simple. We often find that economic theory
infer combinations of short-run and long-run relationships amongst a set of given variables.
Combining short-run and long-run restrictions involve complex matrix algebra since we need to ensure
that such combination do not contradict each other.
8 References
Amisano,Gianni and Carlo Giannini (1997), Topics in structural VAR econometrics, Second edition.
Heidelberg and New York: Springer.
Blanchard, O and D, Quah, The Dynamic Effects of Aggregate Supply and Demand Disturbances,
American Economic Review (1989)
Lütkepohl, Helmut (1991). Introduction to Multiple Time Series Analysis, Springer-Verlag.
12