Katarina Juselius 373inflation rates have to react in a non-homogeneous manner if relative prices move
persistently apart.
Case 2 {r=1,s 1 =1,s 2 = 1 }is consistent with:
⎡
⎢
⎣p1,t
p2,t
s12,t⎤
⎥
⎦=⎡
⎢
⎣c
c
0⎤
⎥
⎦∑tj= 1∑ji= 1u1,i+⎡
⎢
⎣b 11 b 12
b 21 b 22
b 31 b 32⎤
⎥
⎦⎡
⎢⎢
⎢⎢
⎣∑j
i= 1u1,i∑j
i= 1u2,i⎤
⎥⎥
⎥⎥
⎦+⎡
⎢
⎣ε1,t
ε2,t
ε3,t⎤
⎥
⎦.In this case one would not expect to find a directly cointegrating relation, as
r−s 2 =0. This result is easily seen from the nominal-to-real transformed system:
⎡
⎢
⎣p1,t−p2,t
p1,t
s12,t⎤
⎥
⎦=⎡
⎢
⎣b 11 −b 21 b 12 −b 22
c 0
b 31 b 32⎤
⎥
⎦⎡
⎢
⎢⎢
⎢
⎣∑j
i= 1u1,i∑j
i= 1u2,i⎤
⎥
⎥⎥
⎥
⎦+⎡
⎢
⎣ε ̃1,t
ε ̃2,t
ε3,t⎤
⎥
⎦.It is now easy to see that stationarity ofppptcan only be achieved in the spe-
cial case whenb 11 −b 21 =b 31 andb 12 −b 22 =b 32. But in general, empirical
support forppptcan only be achieved by polynomial cointegration, i.e., in the
form of a dynamic long-run adjustment relation. For example, ifb 12 −b 22 =b 32
andc=−(b 11 −b 21 −b 31 )/ω, then
{
p1,t−p2,t−s12,t+ωp1,t}
∼I( 0 ). The lat-
ter can be interpreted as evidence of the following dynamic adjustment relation:
p1,t=− 1 /ω
{
p1,t−p2,t−s12,t}. In this case, either inflation rates or the currency
depreciation/appreciation rate have to move in an offsetting direction whenppp
has persistently deviated from its benchmark values.
Thus the outcome of testing rank indices in theI( 2 )model has strong implica-
tions for whether support for a stationary relation can be found or not.
8.9 AnI( 2 )analysis of prices and exchange rates
8.9.1 Determining the two rank indices
The number of stationary multi-cointegrating relations,r, and the number ofI( 1 )
trends,s 1 , among the common stochastic trends,p−r, can be determined by the
ML procedure in Nielsen and Rahbek (2007), where the trace test is calculated for
all possible combinations ofrands 1 so that the joint hypothesis (r,s 1 )can be
tested, as explained below.
Table 8.5 reports the ML tests of the joint hypothesis of (r,s 1 ), which corresponds
to the two reduced rank hypotheses in (8.4) and (8.5). The test procedure starts
with the most restricted model (r=0,s 1 =0,s 2 = 3 )in the upper left-hand corner,
continues to the end of the first row (r=0,s 1 =3,s 2 = 0 ), and proceeds similarly
row-wise from left to right until the first acceptance. Based on the tests, the first
acceptance is at (r=1,s 1 =1,s 2 = 1 ), which was also the preferred choice in
section 8.4. The last column of the table correponds to theI( 1 )trace test. When