Non-parametric estimation of a time varying GARCH model

The Assumption 1 (i) here is similar in spirit to the stationarity condition for GARCH
(1,1) model discussed by Nelson (1991). This condition is required for the existence
of a well defined unique solution to the tvGARCH process. It is also sufficient for the
tvGARCH to be a short memory process. The Lipschitz continuity condition for the
parameters in Assumption 1 (ii) is required for the local stationarity of the tvGARCH
process. Similar condition is also assumed by Dahlhaus and Subba Rao (2006) for pa-
rameters of the tvARCH process. Notice that we do not make any assumption on the
density function of ǫt. Therefore, the methodology introduced in the paper will be useful
for analyzing data with heavy tailed distributions which is a common phenomenon in
financial time series.
Before proceeding further, we show in Proposition 2.1 that the tvGARCH process
possesses a well defined unique solution. In the Proposition 2.2, we derive the covariance
structure of the tvGARCH process and show that tvGARCH is a short memory process.
Proposition 2.1. Let the Assumption 1 (i) hold. Then the variance process (2) has
a well defined unique solution given by
¯σ 2 t = ω � �
t + n
∞� i� �
α
i=1 j=1
� �
t−j+1
v n
2 t−j + β � ��
t−j+1
ω n
� �
t−i , n
such that |σ 2 t − ¯σ 2 t | → 0 a.s., if σ 2 0 (starting point) is finite with probability one. Also,
inf
u ω(u)/(1 − inf
u β(u)) ≤ ¯σ 2 t < ∞ ∀ t a.s.
Proposition 2.2. Suppose that the Assumption 1 (i) is satisfied for the tvGARCH
process. Further assume that E|vt| 4 < ∞. Then for a fixed k ≥ 0 and 0 < δ < 1,
Cov(ǫ 2 t,ǫ 2 t+k) = O �
(1 − δ) k�
.
Now we define a stationary GARCH (1,1) process, which locally approximates the original
process (2) in the neighborhood of a fixed point (see Proposition 2.3). Let �ǫt(u0), u0 ∈
(0, 1] be a process with E(�ǫt(u0)| � Ft−1) = 0 and E(�ǫ 2 t(u0)| � Ft−1) = �σ 2 t (u0) where � Ft−1 =
σ(�ǫt−1, �ǫt−2,...). Then {�ǫt(u0)} is said to follow a stationary GARCH process associated
with (2) at time point u0 if it satisfies,
�ǫt(u0) = �σt(u0)vt,
�σ 2 t (u0) = ω(u0) + α(u0)�ǫ 2 t−1(u0) + β(u0)�σ 2 t−1(u0).
6
(3)