Page 29 - Contributed Paper Session (CPS) - Volume 1

CPS658 Sagaren P.
′
′
′
′
Yields ∆ = + −1 − − + + Δ −1
1
1

1
0
1
− +. . . + −1 Δ −+1 − −1 Δ +
1 1
1

∗ ∗
= + Π +. . . + −1 Δ −+1 +

∗
′
′
with = −Π + ( − … − −1 ) and Π = α[ ; ] is K×(K+1) and
0
1
1

∗
= [ −1 ]
− 1
Note that in this model V is unristricted whilst the linear term can be absorbed
in the error correction mechanism.
Case 5:
There is a separate linear trend in the VECM(p) form the model is:
′
∆ = −1 + + t + Δ −1 +. . . + −1 Δ −+1 +
0

1
1

5. Estimations
Estimations for each of the five cases follows below.
Case 1: No deterministic components
Table 1
Co-integration Rank Test Using Trace
H0: H1: Drift in
Rank=r Rank>r Eigenvalu Trace Pr > ECM Drift in
e Trace Process
0 0 0,4360 19,175 0,0032 NOINT Constant
2,5655 0,1291
1 1 0,0847 3

Table 1 shows the results of the "Co-integration Rank Test Using Trace"
where no intercept term is assumed,for a 5% significance level H0: Rank=0 can
be rejected because p< 0.05 and H0:Rank=1 cannot be rejected because p>
0.05 therefore both series are Co-intergrated.
The model excludes all deterministic components in the data, implying no
growth and zero intercepts in every co-integrating relation. From the given
data it is evident that an intercept is needed to account for the initial level of
measurements . Thus, this option is clearly not justified.
0

−0.21684 0.22084 −0.62327 0.34321
∆ = [ ] −1 [ ] ∆ −1 +

−0.24048 0.24492 −0.39214 −0.2925
′
= [1 − 0.101846] = [ −0.21648 ]
−24048

−0.21684
′
′
Therefore α = [ ] [1 − 0.101846] and
−0.24048

= [1 − 0.101846] [ ]
1
2

= − 0.101846 1 = 1.101548
2

Case 2: No separate drift in the VECM but a constant enters the Co-integration
space.
18 | I S I W S C 2 0 1 9