n2T (H) →trace(W*mV* ζ1J, (35)
m∖ s, m s,q r,ml
whereas if this null hypothesis is false, Tm(H) converges in distribution to the sum, T1,m(H),
say, of the s 1 solutions of (34), hence plimn→∞n2Tm(H) = ∞. Thus, denoting the critical value
of the trace test at the α×100% significance level by Mα,s,q-r,m, we reject the null if n 2 rTm (H) ≥
Mα,s,q-r,m .
6.3. The choice of "m"
A Monte Carlo simulation of the limiting distribution (35), based on 10,000
replications of the random vectors Y*k* and X*k for k = 1,...,m, with m = max(s,q-r),...,20,
reveals that Mα,s,q-r,m is decreasing in m for m ≥ q-r+s, and infinite for m < q-r+s, due to
(near-) singularity of the matrix (32). See the separate appendix to this paper. Using the
approximation
P ( n 2 Tm (H) ≥ Mαm ) ≈ PT1m (H) ≥ n 2Mas,j
and
Lemma 7.
PT1,m (H)
≥ n 2M
α, s, q-r, mj
n ∖M
α α, s, q-r, m
λ R (RaτrC(1)C(1)tR )
mιn∖ q r v∙ 7 v∙ q q-rj
where the right-hand side probability is an increasing function of m, and λmin(.) stands for the
minimum eigenvalue of the matrix involved,
it follows that in order to boost the power of the test we should choose m "large". On the
other hand, m should not be too large, as otherwise m acts as being dependent on the sample
size n, which may distort the size of the test. Since the critical values Mα,s,q-r,m hardly change
anymore for m > 2(q-r+s), and since we have to choose m ≥ q as otherwise the matrix Âm
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