368 The Long Swings Puzzle
The shift dummyμ 01 is similarly decomposed:
μ 01 =αγ 01 +α⊥ 1 γ 11 +α⊥ 2 γ 21.
To investigate the effect of an unrestricted constant onxt, (8.26) is then inserted
in (8.21) using (8.23) and (8.24). The effect of cumulating the constant twice is
given by:
C 2
∑t
j= 1
∑j
i= 1
μ 0 =
∑t
j= 1
∑j
i= 1
β ̃⊥ 2 α′⊥ 2 (αγ 0 +α⊥ 1 γ 1 +α⊥ 2 γ 2 )
=β ̃⊥ 2 α′⊥ 2 α⊥ 2 γ 2 (t(t− 1 )/ 2 ), (8.27)
asα′⊥ 2 α= 0 andα′⊥ 2 α 1 ⊥= 0. Thus, anunrestricted constantterm in the VAR model
will allow for a quadratic trend inxtso we need to restrict theα⊥ 2 component of
μ 0 to avoid this. How to do this will be discussed below.
The effect of cumulating the constant term once is given by:
C 1
∑t
j= 1
μ 0 =(ω 0 α′+ω 1 α′⊥ 1 +ω 2 α′⊥ 2 )
∑t
j= 1
(αγ 0 +α⊥ 1 γ 1 +α⊥ 2 γ 2 )
=
⎡
⎢
⎢⎣(ω 0 α′αγ 0
︸︷︷︸
γ ̃ 0
+ω 1 α′⊥ 1 α⊥ 1 γ 1
︸ ︷︷ ︸
γ ̃ 1
+ω 2 α′⊥ 2 α⊥ 2 γ 2
︸ ︷︷ ︸
̃γ 2
)
⎤
⎥
⎥⎦t, (8.28)
asα′α⊥ 1 =0,α′α⊥ 2 = 0 andα′⊥ 1 α⊥ 2 = 0. Thus, there are three different linear
trends associated with theC 1 components of the constant term.
Most applications of theI( 2 )model are for nominal variables, implying that
linear trends in the data are a natural starting hypothesis (as average nominal
growth rates are generally non-zero). To achieve similarity in the rank test pro-
cedure (Nielsen and Rahbek, 2000), the model should allow for linear trends in
all directions consistent with the specification of trend-stationarity as a starting
hypothesis in (8.25). This means thatμ 1 t= 0 andμ 11 t91.1= 0 in (8.16), so the
vectorsμ 1 andμ 11 need to be decomposed similarly to the constant term and the
step dummy:
μ 1 =αρ 0 +α⊥ 1 ρ 1 +α⊥ 2 ρ 2 ,
and:
μ 11 =αρ 01 +α⊥ 1 ρ 11 +α⊥ 2 ρ 21.
We now focus on the linear trend term. The effect of cumulating this term twice
is given by:
C 2
∑t
j= 1
∑j
i= 1
μ 1 i=
∑t
j= 1
∑j
i= 1
β⊥ 2 α′⊥ 2 (αρ 0 +α⊥ 1 ρ 1 +α⊥ 2 ρ 2 )i
=
∑t
j= 1
∑j
i= 1
β⊥ 2 α′⊥ 2 α⊥ 2 ρ 2
︸︷︷ ︸
= 0
i. (8.29)