Non-parametric estimation of a time varying GARCH model
Non-parametric estimation of a time varying GARCH model
Non-parametric estimation of a time varying GARCH model
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
Pro<strong>of</strong>. Let Ft−1 = σ(ǫ2 t−1,ǫ2 t−2,...). Then<br />
�<br />
n�<br />
V ar (uk − u0)<br />
k=p+1<br />
iKh(uk − u0)(v2 k − 1)σ2 k[1,ǫ2 k−1,...,ǫ 2 k−p] ⊤<br />
�<br />
�<br />
n�<br />
= E (uk − u0)<br />
k=p+1<br />
2iK2 h(uk − u0)V ar �<br />
(v2 k − 1)σ2 k[1,ǫ2 k−1,...,ǫ 2 k−p] ⊤ �<br />
|Fk−1<br />
�<br />
= �<br />
n�<br />
E (uk − u0)<br />
k=p+1<br />
2iK2 h(uk − u0)V ar(v2 k) �<br />
σ4 k[1,ǫ2 k−1,...,ǫ 2 k−p] ⊤ [1,ǫ2 k−1,...,ǫ 2 k−p] ��<br />
= nh2i−1ν2iV ar(v2 t )Ω(1 + oP(1)), (using Lemma A.2(ii))<br />
Pro<strong>of</strong> <strong>of</strong> Theorem 4.1. Let us denote β1 = [α00,α01,...,α0d,...,αp0,...,αpd] ⊤ . Using<br />
Taylor’s series expansion, we can write,<br />
�<br />
Y1 = X1 α0(u0),α (1)<br />
0 (u0),... α(d)<br />
+<br />
⎡<br />
1 ⎢<br />
(d + 1)!<br />
⎣<br />
+<br />
α (d+1)<br />
0 (ζ0(p+1))(up+1 − u0) d+1<br />
.<br />
α (d+1)<br />
0 (ζ0(n))(un − u0) d+1<br />
⎡<br />
p� 1 ⎢<br />
(d + 1)!<br />
⎣ .<br />
j=1<br />
0 (u0)<br />
,α1(u0),...,αp(u0),... d! α(d)<br />
�⊤ p (u0)<br />
d!<br />
⎤<br />
⎥<br />
⎦<br />
α (d+1)<br />
j (ζj(p+1))(up+1 − u0) d+1ǫ2 p+1−j<br />
α (d+1)<br />
j (ζj(n))(un − u0) d+1ǫ2 n−j<br />
⎤<br />
⎥<br />
⎦ +σ2 ∗ (v 2 − en−p)<br />
where σ 2 = [σ 2 p+1,σ 2 p+2,...,σ 2 n] ⊤ , v 2 = [v 2 p+1,v 2 p+2,...,v 2 n] ⊤ , ∗ denotes the component<br />
wise product 3 <strong>of</strong> vectors and ζjk, j = 0, 1,...,p, k = p + 1,...,n are between uk and u0.<br />
Multiplying both sides by (X ⊤ 1 W1X1) −1 X ⊤ 1 W1,<br />
⎡<br />
⎢<br />
× ⎢<br />
⎣ .<br />
ˆβ1(u0) = β1(u0) +<br />
⎡<br />
⎢<br />
× ⎢<br />
⎣ .<br />
α (d+1)<br />
0 (ζ0(p+1))(up+1 − u0) d+1<br />
α (d+1)<br />
0 (ζ0(n))(un − u0) d+1<br />
α (d+1)<br />
j (ζj(p+1))(up+1 − u0) d+1ǫ2 p+1−j<br />
α (d+1)<br />
j (ζj(n))(un − u0) d+1ǫ2 n−j<br />
1<br />
(d + 1)! (X⊤ 1 W1X1) −1 X ⊤ 1 W1<br />
⎤<br />
⎤<br />
⎥<br />
⎦ +<br />
1<br />
(d + 1)!<br />
Now it is not difficult to show using Lemma A.2 (i) that<br />
⎡<br />
⎤<br />
X ⊤ ⎢<br />
1 W1<br />
⎢<br />
⎣<br />
α (d+1)<br />
0 (ζ0(p+1))(up+1 − u0) d+1<br />
.<br />
α (d+1)<br />
0 (ζ0(n))(un − u0) d+1<br />
p�<br />
(X<br />
j=1<br />
⊤ 1 W1X1) −1 X ⊤ 1 W1<br />
⎥<br />
⎦ + (X⊤ 1 W1X1) −1 X ⊤ 1 W1(σ 2 ∗ (v 2 − en−p)). (12)<br />
3 Let x = [x1,x2,...,xp] ⊤ and y = [y1,y2,...,yp] ⊤ , then x ∗ y = [x1y1,x2y2,...,xpyp] ⊤ .<br />
22<br />
⎥<br />
⎦