Nonparametric cointegration analysis



JJexp(2 i π j 1 x )exp(2 i π j 2 y )min(x, y ) dxdy
x

= exp(2 i π j 1 x)ʃy exp(2 i π j2y)dydx - x exp(2 i πj 1 x)exp(2 i π j2y)dydx

x exp(2 iπ (j'i +j2)x)       ʃ exp(2 iπ (j'i +j2)x)       ʃ exp(2 iπ ji x)

______________dx ʃ I__________LJ___dx + I_________1__dx

(A.42)


2 i π j 2              2      (2 i j 2)2           2 (2 i π j 2)2

ʃx exp(2 iπ (j' 1 +j2) x) ʃxexp(2 iπ j 1 x)

- I____________LJ___dx + I___________1__dx

2       2 i π j 2             2     2 i π j 2

= _    1     + 1(j' 1+j 2=0)

4 π 2j 12      4 π2j 2

and

x

exp(2 i π j 1 x)exp(2 i π j2y)dydx ʃ ʃ
0


exp(2 iπ (j' 1 +j2) x)

2 i π j 2


dx -


exp(2 i π j 1 x )

2 i π j 2


dx


I(j-1÷j 2=0)
2
i π.j,


(A.43)


It follows now from (A.38) and (A.42) that

ʃʃ':(x)Fm(y)min(x,у)dxdy = 2 Σ
jj                                 4π2 j0

(                      V


j2



Λj


4π2


Ij-1


cc




j-i


A

Ï

c....


Ï

c_..„


(A.44)


ʌ Vλ β j,kβ j,
2^t    2 2

j=1     j


V


λ.1


A
. A
β
jm


and it follows from (A.38) and (A.43) that

ʃFk(x )fFm ) dydx = ɪ Σ cjk-j
0      ∙0                  2 i π j≠0    j


« jkβ j


1 τL « ∙t
= ⅛∑ —


4π


Moreover,


45


A
j « β t
Γλ j, mtJ,k


j=1


(A.45)




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