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Non-parametric estimation of a time varying GARCH model

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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 />

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