2. Specification1
Answers to the exam questions in Advanced Macroeconometrics 2005
Question 2.1
The graphs of the level and differences of the variables can be seen in the appendix. The
different interest rates seem to follow the same pattern. They are definitely not stationary as they
are not at all mean-reverting in levels. However, in differences the interest rates seem to be very
mean-reverting and therefore stationary, even though there are some signs of ARCH in the
variables. Hence, all of the interest rates seem to be I(1). From the graphs of the levels, it looks
like the data contains a linear trend since they are all declining in this period. However, it is
highly unlikely that there is a linear trend in data, because then it would be possible to some
extent to predict the future interest rates, and this is in contradiction to the assumption of rational
behaviour in the financial markets. The constant term μ0 should be restricted to lie in the
cointegrating space. Using the notation in chapter 62 this is equivalent to γ0=0, meaning
that 0        0 . This is summed up by case 2 on page 112 where μ1=0, γ0=0 but β0≠0.
1
Question 2.2
ARCH in data means that the variances of the residuals will not be constant and this could make
the cointegration estimates inefficient. However, according to Rahbek et al.(2002) the
cointegration tests are quite robust against moderate ARCH (chapter 4 p. 83). Even though the
period mentioned in the text is not included in this sample, there still seems to be ARCH present
in the data (cf. appendix: the hypothesis of no ARCH is rejected).
Question 2.3
The unrestricted model (with no dummies) is estimated with a restricted constant (in CATS
terminology: dettrend=CImean). The standard misspecification test (see appendix) shows that
the residuals are not well behaved. The multivariate normality test strongly rejects the null
hypothesis of normality. There seems to be problems with autocorrelation, kurtosis and ARCH.
There seems to be moderate skewness in R3M and TB3, but not for the other interest rates . This
is good because skewness is much worse than kurtosis in relation to the VAR analysis. The
misspecification is, however, likely to be improved by including relevant dummies.
1
2
Only CATS has been used in this analysis.
All of the references made to other material is made to the syllabus, unless otherwise stated.
Looking at the standardized residuals most of the outliers (|std. res.| > 3) are present in the
short interest rates. Outliers in R1M are present on the following dates: 1984:10, 1987:12 (this
seems to be transitory as it is followed by relatively large negative residuals), 1992:7. In R3M
the outliers are at 1985:02, 1985:04 (as they are of opposite signs it is probably a transitory
effect of a shock, i.e. an additive outlier), 1987:12 (also transitory as in R1M). There are no
outliers in R1Y. In TB3 there are outliers on 1984:11 (but also a relatively large std. residual in
the month before), 1985:02, 1985:04 (the same applies as for R3M) and in 2001:09. For B10Y
there is only one outlier in 1987:05. Hence there are several outliers which seem to be due to
transitory effects of shocks or simple mistakes, i.e. they are likely to be additive outliers and
should therefore not be modelled, see question 2.4. There are however three outliers which can
be explained by economic events: 1984:10 (This is chosen this because there is a relatively large
outlier in this month in the TB3 residuals, hence the effect seem to start here rather than
1984:11. One can choose either date as long as one remembers to include an additional blip
dummy at the left-over date), 1992:7 and 200:09. To take account of possible changes in the risk
premium due to these events three shift dummies have been created. These are restricted to the
cointegrating space to allow for the possibility that the events may have had a permanent effect
on the mean of the CI relations, i.e. the risk premium.
The model program in CATS should now now look like:
smpl 1984:6 2002:1 *Remark: insert your sample period here
source C:\CATS\catsmain.src; *Remark: the directory of your CATS program
@cats(lags=2,season=0,dettrend=CImean,shift)
# R1m R3m R1y Tb3 B10y
# Ds8410 Ds9207 Ds0109
# Dp8411
Note that the output in the Appendix is calculated without adding the Dp8411 (only a few students had
in fact remembered that CATS automatically add one blip dummy to the model, but not two!)
From this estimation it is possible to test whether there is a shift in the risk premium. Unfortunately the
inclusion of the shift dummies does not improve the misspecification tests significantly.
Question 2.4
Additive outliers are not and should not be a part of the VAR dynamics. Hence they must
be corrected prior to the analysis or simply ignored. It is not advisable to dummy out additive
outliers as the dummies then will generate spurious innovational effects in the VAR model. The
side-effect of leaving the additive transitory outliers in the dataset can be some presence of
negative residual autocorrelation. I choose to leave the additive outliers in the data set. If the
2
outliers are of minor size, they will cumulate to zero in the cointegration space and not affect the
common stochastic trends.
Question 2.5
a) Since there are dummies included in the model the asymptotic tables reported by CATS
are no longer correct. Instead one can use SimBreaks to simulate new tables. Note that This was
not required, but if done the CATS command looks as follows:
@SimBreaks(dettrend=cimean,niter=2500,tper=400,prmax=5,nbreak=3,break=||0.0143,0.4571,0.9810||);
The tables are affected both by the number of shift dummies and of their position in the sample
period. The results of the trace test are shown in table 1 below. When the standard tables are
used none of the tests are accepted. But when the tables corrected for the shift dummies are used
the null hypothesis of at least p-r=2 unit roots is just accepted. Thus, correcting for shift
dummies shifts the tables to the right so that the probability of accepting the null hypothesis of a
unit root increases.
b) Additive outliers are not and should not be a part of the VAR dynamics. Hence they
must be corrected prior to the analysis or simply ignored. It is not advisable to dummy out
additive outliers as the dummies then will generate spurious innovational effects in the VAR
model. The side-effect of leaving the additive transitory outliers in the dataset can be some
presence of negative residual autocorrelation. I choose to leave the additive outliers in the data
set. If the outliers are of minor size, they will cumulate to zero in the cointegration space and not
affect the common stochastic trends.
Whether the sample is large enough for the tables to be a reasonably good approximation
depends not just on the sample size, but also on the information in the data. We have 210
observations for each variable in this data set so the sample is not very small. Furthermore, the
equilibrium errors seem to cross the mean line several times over the sample period which can
be seen from the subsequent cointegration analysis. Thus, the data are quite informative about
the hypothetical long-run relations, which is the reason why the Bartlett corrected trace tests are
quite close to the uncorrected ones. Applying the Bartlett correction has the effect of lowering
the trace statistics, thereby making it harder to reject the null hypothesis of p-r unit roots. This
means that the Bartlett correction makes the size of the test (type 1 error) more correct, but it
does not improve the power of the test (the ability to reject a false null hypothesis). This is can
be a problem as the power is unfortunately often low for roots not too far away from the unit
circle.
3
c) From the modulus of the five largest roots in the companion matrix in table1 it can be
seen that setting r=3 ensures that the remaining roots are relatively small (less than 0.75). If r is
set to 2 there is no improvement in the size of the remaining roots. Hence r=3 seems to be the
best choice also in this case.
From the t-values of the α-matrix in the unrestricted VAR (see appendix) we notice that all
α columns have t-values greater than 2.0. Since the last (p-r) relations are non-stationary we
should use the Dickey Fuller distributions rather than the t-statistics when evaluating these αcoefficients. However, the last two columns of the α-matrix have more significant coefficients
than the second and the third column, which clearly seems inconsistent with the theoretical
assumptions. This is likely to be consequence of the fact that our VAR model is not completely
well-specified.
Note: recursive testing was not requested in the exam. The recursive graphs of the trace
statistics can also be used to determine rank, see appendix 3. Here we can only use the R version,
which is cleaned for short run effects. The graphs should grow linearly for i=1,…,r since they
are functions of non-zero eigenvalues and be constant for i=r+1…p since they are functions of
zero eigenvalues. All of them actually seem to grow over the period, however the two lowest not
very much. Hence it is not clear from this test which rank to choose, but on the other hand the
evidence is not necessarily inconsistent with r=3.
From the cointegration relations shown in the appendix, it can be seen that at least the first
two relations are stationary. Hence the rank is at least 2. There is a lot of persistence in the fifth
relation, hence it looks non-stationary in spite of the relatively large t-values of its αcoefficients. It is more difficult to say whether the third and the fourth relations are stationary.
They generally seem mean-reverting but with some persistence in some periods. Thus, this
graphical inspection could suggest that r=2, but the evidence is far from clear cut.
Economic interpretation can also give a hint of what the rank should be although one
should be careful at such an early point in the analysis. The α-coefficients suggest that the first
relation could be a R1M relation. The picture is less clear with respect to the other relations.
Overall the test and the indicators seem to suggest that the system contains three cointegrating
relations and hence 2 common trends, possibly four cointegrating relations and one common
trend.
d) According to the expectations theory and the term structure of interest rates there should
just be 1 common trend driving the term structure. Hence the choice of r=3 is not consistent with
The baseline period was chosen to be 1984:8 – 1986:12. Due to the shift dummies the X version of the graphs
cannot be calculated until 2001:10 and is therefore of little value.
3
4
this theory as it implies 2 common driving trends rather than one.
Table 1:
r
p-r
Eigenv. Trace
Trace
Trace95
0
1
2
3
4
5
4
3
2
1
0.4916 268.3
0.2471 126.1
0.1565 66.5
0.0863 30.8
0.0549 11.84
256.9
121.2
63.4
29.4
11.2
Trace95
Shift corrected
Bartlett corrected
76.8
53.9
35.1
20.2
9.1
100.8
74.0
52.0
31.5
16.3
Modulus of five largest roots
r=5 r=4 r=3 r=2 r=1
0.93
0.86
0.86
0.70
0.34
1
1
1
1
0.90 1
1
1
0.72 0.73 1
1
0.70 0.73 0.73 1
0.44 0.41 0.45 0.42
Question 2.6
a) R6m is likely to be weakly exogenous. This is because the cointegration rank divides the
data into r relations which are adjusting and p-r relations are pushing the process. The latter can
be interpreted as the common trends and are as such weakly exogenous variables. Hence
including R6m and leaving rank unaltered means that p-r increases by one, indicating one more
weakly exogenous variable.
b) R6m has to be long-run excludable because if it were not, it would be cointegrating with
some of the other interest rates and hence r would have increased.
c) R6m is not stationary. If it was, then r would probably have increased, since the
cointegration rank classifies the eigenvectors into r stationary and p-r non-stationary directions.
d) R6m would not have a unit vector in α, because in this case R6m would be purely
adjusting and hence imply that r would have increased. Also, since R6m is weakly exogenous it
cannot have a unit vector in α as this is the opposite of being weakly exogenous.
Thus, R6m is both weakly exogenous and long-run excludable and will not add anything to
the long-run structure. However, it is still possible that R6m could explain some of the short-run
variation in the model.
3. Testing
Question 3.1
The results of the variable exclusion test are shown in the appendix. This is a LR test, χ2(v)distributes with v=rm=3*1=3 (rank times the number of restrictions). We have p1=9 variables in the
cointegration space (p=5, 3 shift dummies and 1 constant). The test has m=1 restrictions and s=p1m=8 free parameters. The design and the restrictions matrix for formulating a test of whether the
mean shift can be excluded looks like this (H is p1xs and R is p1xm):
5
10 0 0 0 0 0 0 


 010 0 0 0 0 0 
 0 010 0 0 0 0 


 0 0 010 0 0 0 
H   0 0 0 010 0 0 


 0 0 0 0 010 0 
 0 0 0 0 0 010 


 0 0 0 0 0 0 01 
00000000


R '   0 0 0 0 0 0 0 01
Regardless of the rank, the hypothesis of no constant term in the cointegration relations is
rejected. The tests whether the shift dummies can be excluded are similar to the above test. The
shift dummy Ds0109 can clearly be left out for all the different ranks. The hypothesis that
Ds9207 can be excluded is clearly rejected for all of the different rank possibilities. The shift
dummy in 1984:10 can be left out of the cointegration relations for r=3 at a 1% significance
level. Thus the model is re-specified without Ds0109 but with an unrestricted blip dummy
Dp0109 in the model.4
Question 3.2
The tests of weak exogeneity are shown in the appendix. The individual hypothesis corresponds
to a zero row in α. If it is accepted it defines a common driving trend as the cumulated sum of
the empirical shocks to the variable in question. The design and the restriction matrices testing
whether R1M is weakly exogenous are given by:
0000


10 0 0 
H   010 0  R '  10 0 0 0 


 0 010 
 0 0 01 


The only interest rate which is not weakly exogenous is the one month interest rate. All of the
others cannot be rejected as weakly exogenous almost regardless of the rank. However since the
weak exogeneity tests of single variables are not independent, the joint exogeneity hypothesis of
several individually acceptable weakly exogenous variables can be rejected (and often is).
It does not really make sense to test whether the 4 variables are jointly weakly exogenous,
4
The first relation is normalized on R1M, the second on on R1Y, and the third on TB3 because these were the most
significant variables in the unrestricted α matrix.
6
nor whether 3 are jointly weakly exogenous. The condition s ≥ r5 implies that the number of
non-zero rows in α must not be greater than r. A zero row in α means the variable does not
adjust to long run relations, but is rather a common driving trend. There can at most be p-r
common trends hence the number of zero restrictions can be at most be 2 in this case.
By testing different combinations of the 4 individually weakly exogenous variables (in
pairs) it appears that only the three month interest rate and the ten year bond rate could be
accepted as jointly weakly exogenous based on a test statistic χ2(rm=6) = 7.075 with a p-value
of 0.31. a In this case the restriction matrix looks like:
 010 0 0 
R'  

 0 0 0 01
Question 3.3
The tests of a unit vector in α for each of the variables under the assumption that r=3 is shown in
table 2.6 These are LR tests which are distributed asymptotically as χ2(v) where v=m(p-r) (here
m is the number of known α vectors). The results show that the hypothesis of a unit vector in α
is clearly accepted for the one month interest rate, the one year interest rate and three month
Treasury bill rate. The hypothesis is also barely accepted for the three month interest rate. This
is contrary to the result in question 3.2, where we found that the B10Y and the R3M was found
to be jointly weakly exogenous.
Table 2:
r
3
v
2
χ2(v)
5
.
9
9
R
1
1
.
M
3
R
5
(0.51)
3
5
.
M
3
R
4
(0.069)
1
1
.
Y
6
T
7
(0.43)
3
B
.
3
5
2
(0.17)
B
1
1
0
2
Y
.
4
(0.002)
When a variable has a unit vector in α it implies that the variable in question is exclusively
adjusting to the long-run relations and its unanticipated shocks have no permanent effect on the
variables of the system. Because a unit vector is the opposite of weak exogeneity (implying that
a variable is exclusively pushing) a variable cannot at the same time have a unit vector in α and
be weakly exogenous. Thus, the finding of four potentially exclusively adjusting variables is
clearly inconsistent with the assumption of r=3 and the previous finding of RM3 and B10Y
being weakly exogenous..The joint hypothesis of a unit vector in R1M, R1Y and TB3 is
accepted with a p-value of 0.31 based on a test value of 7.08 distributed as χ2(6),, which is
identical to the test of the two weakly exogenous variables above. Based on this we have
5
6
See p. 222
In CATS the menu “Some alpha vectors known” under I(1) is used to make the tests.
7
tentatively identified the pushing forces to be associated with B10Y and R3M and the pulling
forces with R1M, R1Y and TB3.
How can we explain the puzzling result that both a unit vector and zero row in α was borderline
acceptable for R3M at the same time? The explanation is that R3M is only borderline
significantly adjusting to the third cointegration vector, but not to the first two relations (see the
unrestricted α in the Appendix) explaining the weak exogeneity result. However, as it is only
adjusting to one cointegration relation with a borderline significant α coefficient the test of a
unit vector in α cannot be rejected. Thus you should think about a situation where the α matrix
contains a row and a column where all elements are zero except for the intersection element
(which is a borderline significant adjustment coefficient of R3M). In this case we will not be
able to individually reject the zero row and the unit vector in α. This explanation was not
required in the exam!
Question 3.4
This question could be interpreted as testing the homogeneity of a single relation or testing
overall homogeneity. The latter interpretation was actually the one I (KJ) had in mind when
writing the exam questions. However, both interpretations were equally accepted.
This is an answer to the first interpretation:
Long-run price homogeneity of the 5 interest rates means that the coefficients in β sum to zero.
The design and restriction matrices of long run price homogeneity between the five interest rates
are given by:
 1 0 0 00


 1  1 0 0 0 
 0 1 1 0 0
1 1 1 1 1 0 0 0 


0
0

1

1
0
 R   0 0 0 0 01 0 0 
H 


 0 0 0 1 0
 0 0 0 0 0 01 0 




 0 0 0 0 0
 0 0 0 0 0


 0 0 0 0 1
Here the two dummies are set to zero in order to get 3 restrictions such that it can be tested. The
homogeneity of the five interest rates alone only counts as one restriction. Thus it is not testable
as such as we can impose r-1=2 restrictions without changing the likelihood function. The
results are shown in the appendix7. The hypothesis is clearly accepted with a p-value of 0.66 and
7
The test is made in CATS using the “Restrictions on subsets of beta”.
8
a test statistic of only 0.187. The test is χ2(v) distributed, where v=m-(r-1)=3-2=1. There are 3
binding restrictions and 5 free parameters. The acceptance of the price homogeneity of the five
interest rates implies that the interest rates are cointegrating but the spreads are not in general
stationary because a linear combination between two interest rates only exceptionally can cancel
the two common trends in the data. However, combinations of the interest rate spreads would be
stationary in this case. With two common driving trends it would be quite likely that interest rate
spreads would be pairwise cointegrating.
This is an answer to the second interpretation:
Overall long-run homogeneity is rejected based on the standard LR test χ2 (3)=14.23, p.val. =
0.00, but accepted if we employ a Bartlett correction based on χ2 (3)=3.88, p.val. = 0.28
(Correction Factor: 3.67). In this case there are 3 binding restrictions, one for each relation, and
7 free parameter in each relation. In case we interpret the result to mean that there is long-run
homogeneity between the interest rates then the spreads would generally be I(1), but pairwise
cointegrating.
4. Identification of the long-run structure
Question 4.1
In order to estimate a just-identifying structure, we need to place two restrictions on each of the
β-vectors. Of course these cannot be the same restrictions on each β, because then the rank
conditions would not be satisfied and the structure would not be generically identified. In order to
have empirical identification the rank conditions must not be violated when setting an insignificant
β coefficient to zero.
The just identified structure can be chosen in many different ways. For example, it can
describe the long run reduced form, assuming that R1M, R1Y and TB3 are endogenous and that
R3M and B10Y are weakly exogenous. In this case the weakly exogenous variables and the
dummies and the constant should be included in each of the relations as described by the design
matrices:
10 00 0 0 
 0 0 00 0 0 
 0 0 00 0 0 






 010 0 0 0 
10 0 0 0 0 
10 0 0 0 0 
000000
 010 0 0 0 
000000






0 00 00 0 
0 00 00 0 
01 0 00 0 



H1 
H 
H 
 0 010 0 0  2  0 010 0 0  3  0 010 0 0 






 0 0 010 0 
 0 0 010 0 
 0 0 010 0 
 0 0 0010 
 0 0 0010 
 0 0 0010 






 0 0 0 0 01 
 0 0 0 0 01 
 0 0 0 0 01 
9
Thus, H1 picks up R1M, H2 R1Y and H3 picks up TB3. The estimation results is shown in the
appendix. From the α-matrix it can be seen that R1M adjusts to deviations from equilibrium in
all three relations and R1Y and TB3 adjusts only to their own relations respectively. The graphs
of all the relations look stationary, but there seems to be some persistence in the beginning of
the sample (not shown in the appendix). In appendix the rank conditions8 are found to be just
satisfied. Since insignificant coefficients can be set to zero without violating the rank conditions
this model is empirically identified.9
Question 4.2
Since the coefficient to the shift dummy in 1984:10 is not significantly different from zero in the
third relation, this can set to zero. The estimation is shown in the appendix. The coefficient to
the constant is now significant. From the appendix it can also be seen that the rank conditions
are fulfilled. Hence the restrictions are over-identifying with one degrees of freedom and it is
now possible to test the overidentifying restriction using an LR test which is asymptotically
r
χ2(v) distributed, where v    mi  r  1 . The restriction is accepted with a p-value of 0.57 and
i 1
a test statistics of 0.33 (v=1). However, the recursive graphs of β constancy clearly reject
constancy.10 The eigenvalue fluctuation test indicates that the problem of non-constancy is due
to the short interest rate relation (R1M). This suggests that the structure should be reconsidered,
especially with respect to the short interest rate.
Question 4.3
The long run homogeneity restrictions on all of the cointegration relations is not identifying by
itself because it only imposes one restriction on each vector and this is identical for all β. Hence
the rank conditions would not be satisfied even if we imposed an additional identifying zero
restriction on each of the β vectors.
5. The moving average representation
Question 5.1
The identification of the common trends is similar to the identification of the long run
8
The definition of the rank condition (12.5) is found on p. 238 in the syllabus
It is also economically identified if it in addition to being generically and empirically identified it can be given a
relevant economic interpretation
10
The reference period is 1993:01-2002:01 due to suspicion of problems in the beginning of the sample.
9
10
structure. We can impose (p-r-1)=1 restriction on the α┴ vector without changing the likelihood
function. However, the space spanned by α┴ and β̃┴ is uniquely determined, so that the estimated
C matrix is unique. The first step in just-identifying the common trends model is to normalise on
one of the variables in each vector, which in turn results in a zero restriction on that variable in
the other vector. So by normalising on a variable, we are actually imposing a zero restriction.
The choice of which to normalise on can be guided by the size of the coefficients. The estimates
are shown in appendix. From the unrestricted α┴ we note that the largest coefficient of α┴,1 picks
up the weakly exogenous B10Y, whereas the largest coefficient of α┴,2 picks up TB3 rather than
the weakly exogenous R3M. This is again a puzzling result which can be explained by the
borderline significance of R3M already discussed above. Even though the joint weak exogeneity
of R3M and B10Y could not be rejected it was not because R3M was a strongly pushing
variable in the system, but because its was generally quite insignificant in the system. Even
though it generally makes sense to normalise on weakly exogenous variable (because they should
in general have a non-zero coefficient in α┴) in this case it would probably be better to normalize
on TB3. This can be seen from the estimates of the C matrix where there are several significant
coefficients in the column of TB3, whereas essentially none in the column of R3M. If R3M had
been truly pushing we would have expected to find insignificant coefficients in the columns
corresponding to the variables R1M, R1Y and TB3, and significant coefficients in the columns of
B10Y and R3M. This is not the case and we interpret the result to mean that there is essentially
one weakly exogenous variable, B10Y, and another one, TB3, which is mostly pushing but also
adjusting to some extent.
Question 5.2
The C matrix contains useful information on the overall effects of the stochastic driving forces
in the system. The columns show how the cumulated residual from each VAR equation load into
the variables individually. Hence a column of insignificant coefficients means that the
corresponding variable has only exhibited transitory effects on the variables in the system, and
significant coefficients in a column means that the variable in question has affected some
variables of the system permanently. The rows of the C matrix shows the weights with which
each of the variables in the system have been affected by cumulated empirical shocks.
The appendix shows the estimated C matrix. As expected the coefficients of columns of the
empirical shocks of R1M and R1Y are insignificant, meaning that they only have transitory
effects on the variables of the system. But, as already discussed, the coefficients in the column
of the empirical shocks to R3M are also insignificant and the coefficients in the column of TB3
11
are instead significant. The last column of the matrix shows that TB3 and the B10Y are
permanently affected by shocks of the B10Y. Since B10Y is also affected by the cumulated
empirical shocks of the TB3, it does not seem to be strongly exogenous. If B10Y only had one
significant variable in its row and if this variable was the coefficient to the empirical shocks of
itself then we could conclude that B10Y was strongly exogenous. If a variable is strongly
exogenous, it is itself is a common trend (the rows of the Гi matrices associated with the weakly
exogenous variable have to be zero).
Question 5.3
According to the expectations hypothesis and the term structure the short interest rate is the
driving force of the system. All of the other interest rates should react to changes and expected
changes in the short interest rates and, thus, be exclusively adjusting. Hence there should only
be one weakly exogenous variable in this case and that should be the short interest rate, here
R1M (the central bank is more likely to have control over the over-night lending rate but that is
not included in this analysis). In such a case ̂  could be expressed by:
ˆ c '  10000
Thereby the C matrix would consist only of zero columns except for in the first column, which
would show how the variables of the systems were affected by the empirical shocks of the short
interest rate, i.e. in this case all interest rates would be driven by the same common stochastic
trend defined as the cumulated shocks to the short-term interest rate.
a 
a 
  t
xt   a    R1M ,i stationary components
  i 1
a 

a 

Based on the results of this analysis the theoretical formulation does not seem to be
empirically supported. Even though the results of the VAR analysis were not completely
unambiguous, they did not suggest that the short-term interest rate was weakly exogenous and
thereby the driving trend of the system. Rather they suggested that it was the long bond rate and
one of the three month interest rates which were the driving forces of the system. Hence
according to this analysis the expectations hypothesis and the term structure would have been
rejected.
As already discussed the empirical analyses produced some puzzling results which suggest
that the data may not contain enough information to decisively conclude about the direction of
the causal link between the different interest rates. This is not so surprising as financial data
12
often contain a lot of noise and the signal-to-noise ratio is often small (the explanatory power of
the model is not very high). Furthermore, the specification of the empirical model was not yet
perfect: The underlying assumption of multivariate normality was not fully satisfied, the data
have not been cleaned for additive dummies which probably would improve normality, and
there is evidence of parameter nonconstancy over the sample period. If the assumptions
underlying the VAR model are not satisfied the analysis might very well produce unreliable and
even incorrect results.
6. A small theoretical exercise
Alternative solution 1:
First the VAR model is defined by:
1
 0.03   0 
xt     0.5, 0.5  xt 1  

  Ds ,t   t
0
 0.02   0.01 
which corresponds to (6.39) in Juselius (2005):
xt   ' xt 1  0   s Ds,t   t ,
t
IN  0,  
which defines α,β,μ0 and φs and, thus,   and   :
 0.5 
 0.03 
 0.00 
 
 ,  '  1 1  , 0  
 ,s  

0
0.02




 0.01 
1
0 1
     ,    0 1  , C     '  

1
0 1
The MA represenation can now be found as:
t 1
 t 1

xt  C    i  0t   s  Ds ,i   C *  L    t  0   s Ds ,t   X 0
i 1
 i 1

t

1
1
 0.02   0.01  t 1
*
xt      2,i  
t  
  Ds ,i  C  L    t  0   s Ds ,t   X 0
 1  i 1
 0.02   0.01  i 1
where
t 1
 1  t 1
 0.02 
 0.01 
C   i      2,i , C 0  
 ,C s  

i 1
 1  i 1
 0.02 
 0.01 
It follows that:
1. both x1,t and x2,t contains a linear trend,
13
2. both x1,t and x2,t contains a broken linear trend with the slope coefficient 0.02 until the
break point and 0.01+0.02=0.03 thereafter,
3. both x1,t and x2,t contains a stochastic trend defined as the cumulated shocks to x2,t,
 t 1

  2,i 
t 1
i 1

because C  i   t 1


i 1
  2,i 
 i 1

Alternative solution 2:
It is also possible to first solve for x2,t, which can be re-written as:
x2,t  x2,t 1  0.02  0.01Dst   2,t
and then substitute for x2,t-1 recursively which yields the following MA expression for x2,t.
t
t
i 1
i 1
x2,t  x2,0  0.02t  0.01 Dsi    2,i
Now we subtract the two equations given in the text:
x1,t  x2,t  0.01  0.01Dst  0.5  x1,t 1  x2,t 1   1,t   2,t
x1,t  x2,t  0.01  0.01Dst  0.5  x1,t 1  x2,t 1   1,t   2,t
t
i
i
i
t 1
t 1
t 1
1
1
1
1
x1,t  x2,t     x1,0  x2,0   0.01    0.01   Dst i      1,t i   2,t i 
2
2
2
 


i 0 
i 0 
i 0  2 
Inserting the MA expression for x2,t derived above we get an MA expression for x1,t:
1
x1,t  x2,0   
2
t
1
x1,t  x2,0   
2
t
x
1,0
x
1,0
i
i
i
t 1
t
t 1
t
t 1
1
1
1
 x2,0   0.02t  0.01    0.01 Dsi  0.01   Dst i    2,i      1,t i   2,t i 
i 0  2 
i 1
i 0  2 
i 1
i 0  2 
i
t i
i
t 1
t
t 1 
t
t 1
1
1 
1
 x2,0   0.02t  0.01    0.01 Dsi  0.01 1     Dsi    2,i      1,t i   2,t i 
 2 
i 0  2 
i 1
i 1 
i 1
i 0  2 

Question 6.1
From the equations of x1,t and x2,t it is clear that both contain a linear trend, since both have “t”
in their equations.
Question 6.2
Both variables contain a broken trend since each of the equations contains the sum of the shift
dummy,  Dst . A shift in the mean of xt becomes a broken linear trend in xt. The slope of the
broken trend is the same in x1,t and x2,t and has a slope of 0.02 before the break and 0.03 after
the break.
Question 6.3
t
Both variables also contain a stochastic trend which is given by   2,i , i.e. it is generated by
i 1
14
shocks to the second variable. Hence the common trend is related to x2,t, which is therefore the
driving force of the system, whereas x1,t is exclusively adjusting.
Appendix
15