"Frontmatter". In: Analysis of Financial Time Series

134 NONLINEAR TIME SERIES4.1.3 Smooth Transition AR (STAR) ModelA criticism of the SETAR model is that its conditional mean equation is not continuous.The thresholds {γ j } are the discontinuity points of the conditional mean functionµ t . In response to this criticism, smooth TAR models have been proposed; see Chanand Tong (1986) and Teräsvirta (1994) and the references therein. A time series x t issaid to follow a two-regime STAR(p) model if it satisfiesx t = c 0 +p∑( ) (xt−d − φ 0,i x t−i + Fc 1 +si=1)p∑φ 1,i x t−i + a t , (4.13)where d is the delay parameter, and s are parameters representing the location andscale of model transition, and F(.) is a smooth transition function. In practice, F(.)often assumes one of three forms—namely, logistic, exponential, or a cumulativedistribution function. From Eq. (4.13), the conditional mean of a STAR model is aweighted linear combination between the following two equations:µ 1t = c 0 +p∑φ 0,i x t−i ,i=1µ 2t = (c 0 + c 1 ) +i=1p∑(φ 0,i + φ 1,i )x t−i .i=1The weights are determined in a continuous manner by F( x t−d−s). The prior twoequations also determine properties of a STAR model. For instance, a prerequisitefor the stationarity of a STAR model is that all zeros of both AR polynomials areoutside the unit circle. An advantage of the STAR model over the TAR model is thatthe conditional mean function is differentiable. However, experience shows that thetransition parameters and s of a STAR model are hard to estimate. In particular,most empirical studies show that standard errors of the estimates of and s are oftenquite large resulting in t ratios about 1.0; see Teräsvirta (1994). This uncertainty leadsto various complications in interpreting an estimated STAR model.Example 4.3. To illustrate the application of STAR models in financial timeseries analysis, we consider the monthly simple stock returns for Minnesota Miningand Manufacturing (3M) Company from February 1946 to December 1997. If ARCHmodels are entertained, we obtain the following ARCH(2) modelR t = 0.014 + a t , a t = σ t ɛ t , σ 2t = 0.003 + 0.108a 2 t−1 + 0.151a2 t−2 , (4.14)where standard errors of the estimates are 0.002, 0.0003, 0.045, and 0.058, respectively.As discussed before, such an ARCH model fails to show the asymmetricresponses of stock volatility to positive and negative prior shocks. The STAR modelprovides a simple alternative that may overcome this difficulty. Applying STAR mod-