Nonparametric cointegration analysis

( ʌ

ixd _e,_χ 1 eβ∙x^x

dx     1 -_e^ix

\          /

(       . ■                „                         \ A

1 -e         ₊e -°-^{i ix}( 1-( x+1) e^inx)

e -0.5ix_e 0.5ix}²        e ~^o∙5ix-e 0'5ix

_ cos(0.5x)-isin(0.5x)ɑ__et_nx^ ₊ 2insin(0.5x)e^ιxx

-4sin²(0.5 x )                   -4sin²(0.5 x )

(A.54)

Thus,

cos(0.5x)(1 - cos(xx) - (2x— 1)sin(0.5x)sin(xx)

-4sin²(0.5 x )

i (2x— 1)sin(0.5x)cos(xx) - cos(0.5x)sin(xx) + sin(0.5x)

-4sin²(0.5 x )

gx^(x ) ⁺ gx⁽ “x ) =

cos(0.5x)(1 - cos(xx) - (2x-1)sin(0.5x)sin(xx)

-2sin²(0.5 x )

(A.55)

Since cos(2kπ) = 1 and sin(2kπ) = 0, the second equality in (38) follows. The proof of the first
equality goes similarly, and (39) is trivial. Q.E.D.

Proof of Lemma 9: Observe that

K

∑ exp^{i (xt +} У) =

t≡ 1

exp( i (y + 0.5 x ))-------¹----^{exp( i}K^x )------

exp(-0.5 ix) - exp(0.5ix)

_ (cos(y + 0.5x) + i sin(y + 0.5 x))(1 - cos(Kx) - i sin(Kx))
-2 i sin(0.5 x )

_ i cos(y + 0.5x)(1 - cos(Kx)) + sin(y + 0.5 x)sin(Kx)
2sin(0.5 x )

cos(y + 0.5x)sin(Kx) - sin(y + 0.5x)(1 - cos(Kx))

2sin(0.5 x )                       ’

(A.56)

hence

48