Nonparametric cointegration analysis

∞∞

Σ ^α2k⁺ Σ^{β 2}kk

4π² -j1________^j-1_____,

У α2k ₊ A βi
ʌ' 2 2          2 2

j 1 j j 1 j

which we are going to minimize subject to the conditions in Lemma 6, as follows: First,
choose for some large natural number N and all j,k > N, α_j,k =β_j,k = 0. Denote θ_k =
(α1,k,β1,k,α2,k,β2,k,...,αN,k,βN,k)^T, J = diag(1,1,2,2,...,N,N). Then
hence the unconstrained minimum of γ_k² corresponds to the minimum solution of the
generalized eigenvalue problem det(I - λJ^-2) = 0. Taking k = 1, this minimum eigenvalue is 1
(twice), with corresponding normalized eigenvectors (1,0,0,...,0)^T and (0,1,0,...,0)^T. Thus, the
unconstrained optimal solution satisfies α_j,1 = β_j,1 = 0 forj > 1, and then the conditions in
Lemma 6 imply that also β_1,1 = 0, whereas without loss of generality we may take α_1,1 =1.
Next, for k = 2 the conditions in Lemma 6, except condition (31), imply that the optimal
solution corresponds to the minimum eigenvalue for which the corresponding eigenvector is
orthogonal to (1,0,0,...,)^T and (0,1,0,...,0)^T. Clearly, this minimum eigenvalue is k² = 4, and the
corresponding normalized eigenvectors are (0,0,1,0,...,0)^T and (0,0,0,1,...,0)^T. Thus, α_j,2 = β_j,2 =
0 forj ≠ 2, and condition (31) implies that also β_2,2 = 0, whereas again we may choose α_2,2 =
1. Continuing this argument shows that the optimal solution for F_k is the one in Lemma 3,
provided that the stronger conditions (6) and (7) also hold. The latter has already been
established in Lemma 3. Since N was chosen arbitrary, it follows by induction that:

Theorem 3. The choice F_k(x) = cos(2kπx) for the weight functions is optimal in the sense that
then for any fixed positive integer m the lower bound (25) of the power of the lambda-min
test is maximal.

17

<<   14   15   16   17   18   19   20   >>

More intriguing information