13 Appendices
A Proof of Lemma 1
It is convenient to introduce ξ = a'1ζp1, so that a11 = a1ξ and a12 = a1ξ1. Because
the estimates of p,τ,β,δ are superconsistent we can find the asymptotic variance of
a11 and a12 by finding derivatives of a11 and a 2 with respect to a and ζ, and then
transform the asymptotic variance of (a, ¢) by an application of the δ- method.
It is convenient to express the matrix expansion
(a + u)1 = aɪ — aɪ ua ' + 0(∣u∣2)
in terms of differentials
(dal) = — aɪ (da)a '. (25)
As an example of the δ-method we can find the asymptotic variance of a1 from (16)
and (25) as
asVar(a1) = a(Ir; 0)Φ(Ir; 0)'a' 0 a'1Ωa1.
Applying (25) we find derivatives of ξ = a'1ζp1,ξ1, and a1 :
dξ = —a1(da)a 'ζ p1 + al1(dζ )p1,
dξ± = lp'ι(ζ 'ɑ(da)' — (dζ )')a12,
da 1 = —a(da)'a 1.
Hence from a12 = a1ξ 1, we get
( (da)' ʌ
I (<)' ) ',
da12 = (da±)ξ 1 + a1(dξ1) = — (Ip — a1ξp'1 ζ')a; a1ξp'1)
where we find
0 = aιξp'ι = a1a,1 ζpι(p1ζ 'aɪ a1CpJ~1p1,
which gives the expression (17). We next find
da11 = d(a 1)a1ζp1 + a 1d(a1)ζp1 + a 1a1 (dζ )p1, (26)
= —a(da)' a11 — a 1a1(da)a 'ζp1 + a 1a'1(dζ )p1.
In order to find the elements of the asymptotic variance of a 11 it is enough to have
an expression for the asymptotic variance of v'a 11u for any vectors v,u. We find from
(26)
vlda11 u = — u'a(da)'a11v — u'a 1a1(da)a 'ζ p1v + u'a 1al1(dζ )p1u
= — v'a'11(da)a 'u + u'a 1a1[(dζ ) — (da)a 'ζ ]p1u
= N + N2
Then Var(v'da11u) = Var(N1) + Var(N2) + 2Cov(N1, N2), which is the expression in
(18).
28
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