Variance stabilization and simple GARCH models. Erik Lindström

Transcription

1 Variance stabilization and simple GARCH models Erik Lindström

2 Simulation, GBM Standard model in math. finance, the GBM ds t = µs t dt + σs t dw t (1) Solution: S t = S 0 exp ) ) ((µ σ2 t + σw t 2 (2) Problem: Estimate µ = µ σ2 2 or µ.

3 Data Showing 5 independent realizations Index level time Figure:

4 4 alternatives OLS WLS OLS on transformed data MLE Derive estimators on the black board.

5 Histograms 600 OLS 300 WLS Transformed OLS 300 MLE

6 Finding a transformation Several strategies Box-Cox transformations Doss transform (SDEs)

7 Classical airline passenger data 700 Time Series Plot:AirlinePassengers 600 AirlinePassengers Jan Jan Jan-1959

8 Taking logarithms AirlinePassengers Time Series Plot:AirlinePassengers Jan Jan Jan-1959

9 Time series models Let r t be a stochastic process. µ t = E[r t F t 1 ] is the conditional mean modeled by an AR, ARMA, SETAR, STAR etc. model. Having a correctly specified model for the conditional mean allows us to model the conditional variance. I will for the rest of the lecture assume that r t is the zero mean returns. σ 2 t = V[r t F t 1 ] is modeled using a dynamic variance model.

10 Why are we interested in (conditional) variances? Several financial applications: Mean variance portfolio optimization VaR and ES calculations Conservative estimate of quantiles via the Chebyshev inequality P ( X µ > a) σ2 a 2 (3)

11 Dependence structures Dependence on the OMXS Autocorrelation, returns Autocorrelation, abs returns lag lag

12 e ARCH family ARCH (1982), Bank of Sweden (2003) GARCH (1986) EGARCH (1991) Special cases (IGARCH, A-GARCH, GJR-GARCH, EWMA) FIGARCH (1996) SW-GARCH GARCH in Mean (1987)

13 ARCH The ARCH (Auto Regressive Conditional Heteroscedasticity) model The (mean free) model is given by r t = σ t z t, The conditional variance is given by σ 2 t = ω + p α i r 2 t i i=1 Easy to estimate as σ 2 t F t 1! Q : Compute cov(r t, r t h ) and cov(r 2 t, r2 t h ) for this model. Hint: Use properties of expectations

14 ARCH, solution We have that E[r t ] = E[E[σ t z t F t 1 ]] = E[σ t E[z t F t 1 ]] = 0.

15 ARCH, solution We have that E[r t ] = E[E[σ t z t F t 1 ]] = E[σ t E[z t F t 1 ]] = 0. Next, we compute Cov(r t, r t h )) as E[σ t z t σ t h z t h ] = E[E[σ t z t σ t h z t h F t 1 ]] = E[σ t σ t h z t h E[z t F t 1 ]] = 0.

16 ARCH, solution We have that E[r t ] = E[E[σ t z t F t 1 ]] = E[σ t E[z t F t 1 ]] = 0. Next, we compute Cov(r t, r t h )) as E[σ t z t σ t h z t h ] = E[E[σ t z t σ t h z t h F t 1 ]] = E[σ t σ t h z t h E[z t F t 1 ]] = 0. Computing Cov(r 2 t, r2 t h ) is harder. Introduce ν t = r 2 t σ2 t (white noise!). It then follows that p r 2 t = σt 2 + ν t = ω + α i r 2 t i + ν t. i=1 The r 2 t is thus a process (with heteroscedastic noise).

17 ARCH, limitations Large number of lags are needed to fit data. The model is rather restrictive, as the parameters must be bounded if moments should be finite Exercise: Compute the restrictions for the ARCH(1) model to have finite variance.

18 GARCH (Generalized ARCH) Is the most common dynamics variance model. The conditional variance is given by p q σt 2 = ω + α i r 2 t i + β j σt j 2 i=1 j=1 A GARCH(1,1) is often sufficent! Conditions on the parameters.

19 GARCH Cov(r t, r t h )= 0 as in the ARCH model. Computing Cov(r 2 t, r2 t h ) is similar to the ARCH model. Reintroducing ν t = r 2 t σ2 t gives (assume p = q) p p r 2 t = σt 2 + ν t = ω + α i r 2 t i + β j σt j 2 + ν t = ω + = ω + i=1 j=1 p p α i r 2 t i + β j (r 2 t j ν t j) + ν t i=1 j=1 p p (α i + β i )r 2 t i + β j ν t j + ν t i=1 j=1 The r 2 t is thus a process (with heteroscedastic noise).

20 Estimation of GARCH(1,1) on OMXS30 logreturns ω = , α 1 = β 1 = OMXS30 logreturns Extimated GARCH(1,1) vol OMXS30 normalised logreturns Probability NORMPLOT OMXS30 normalised logreturns Data

21 GARCH, special cases An IGARCH (integrated GARCH) is a GARCH where α i + β i = 1 and ω > 0. The EWMA(exponentially weighted moving average) process is a process where α + β = 1 and ω = 0, i.e. the volatility is given by σ 2 t = αr 2 t 1 + (1 α)σ2 t 1

22 EGARCH (Exponential GARCH) The conditional variance is given by p q log σt 2 = ω + α i f(r t i ) + β j log σt j 2 i=1 j=1 log σ 2 may be negative! Thus no (fewer) restrictions on the parameters.

23 Variations Several improvements can be applied to any of the models. Bad news tend to increase the variance more than good news. We can replace r 2 t i by (rt i + γ) 2 (Type I) ( rt i + cr 2 t i ) (Type II) Replace αi with (α i + α i 1 {rt i <0}) (GJR, Glosten-Jagannathan-Runkle). Distributions Stationarity problems.

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