Nonparametric cointegration analysis



n

gn(x) = e "5ixteixt

1 e -0.5iχdy,(eix)t = 1
i         dx t= 1              i

( ʌ

ixd e,χ 1 eβ∙xx

dx     1 -eix

\          /


(       . ■                „                         \ A

1 -e         +e -°-i ix( 1-( x+1) einx)

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


_ cos(0.5x)-isin(0.5x_etnx^ + 2insin(0.5x)eιxx

-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

∑ expi (xt + У) =

t≡ 1


exp( i (y + 0.5 x ))-------1----exp( iKx )------

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




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