Variance stabilization and simple GARCH models Erik Lindstrm - PowerPoint PPT Presentation

Variance stabilization and simple GARCH models Erik Lindström

Simulation, GBM Standard model in math. finance, the GBM (1) Solution: 2 (2) d S t = µ S t d t + σ S t d W t (( ) ) µ − σ 2 S t = S 0 exp t + σ W t Problem: Estimate ˜ µ = µ − σ 2 2 or µ .

Data Showing 5 independent realizations Figure: 25 20 15 Index level 10 5 0 0 5 10 15 20 25 30 time

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

Histograms OLS WLS 600 300 400 200 200 100 0 0 -0.2 0 0.2 0.4 -0.1 0 0.1 0.2 Transformed OLS MLE 300 300 200 200 100 100 0 0 -0.1 0 0.1 0.2 -0.1 0 0.1 0.2

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

Classical airline passenger data Time Series Plot:AirlinePassengers 700 600 500 AirlinePassengers 400 300 200 100 01-Jan-1949 01-Jan-1954 01-Jan-1959

Taking logarithms Time Series Plot:AirlinePassengers 600 550 500 450 400 350 AirlinePassengers 300 250 200 150 01-Jan-1949 01-Jan-1954 01-Jan-1959

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

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

Dependence structures Dependence on the OMXS30. 1.2 1.2 1 1 0.8 0.8 Autocorrelation, abs returns Autocorrelation, returns 0.6 0.6 0.4 0.4 0.2 0.2 0 0 −0.2 −0.2 0 5 10 15 20 25 30 0 50 100 150 200 250 300 lag lag

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

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

Next, we compute Cov( r t r t h ) as t r 2 t (white noise!). It then follows that t is thus a ………process (with ARCH, solution 2 h ) is harder. Introduce t r 2 t 2 r 2 t p t t i 1 i r 2 t i t The r 2 t 0 Computing Cov( r 2 h z t E t z t t h z t h E E t z t t h t 1 E t t h z t h E z t t 1 heteroscedastic noise). ◮ 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.

t r 2 t (white noise!). It then follows that t is thus a ………process (with ARCH, solution 2 The r 2 t i t i r 2 1 i p t t r 2 t 2 t r 2 t h ) is harder. Introduce t Computing Cov( r 2 heteroscedastic noise). ◮ 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 .

ARCH, solution r 2 The r 2 p heteroscedastic noise). ◮ 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 . t , r 2 ◮ Computing Cov( r 2 t − h ) is harder. Introduce ν t = r 2 t − σ 2 t (white noise!). It then follows that ∑ t = σ 2 t + ν t = ω + α i r 2 t − i + ν t . i = 1 t is thus a ………process (with

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

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

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

Estimation of GARCH(1,1) on OMXS30 logreturns ω = 1 . 9 · 10 − 6 , α 1 = 0 . 0775 β 1 = 0 . 9152 OMXS30 logreturns Extimated GARCH(1,1) vol 0.1 0.04 0.035 0.05 0.03 0.025 0 0.02 0.015 −0.05 0.01 2000 2010 2000 2010 OMXS30 normalised logreturns NORMPLOT OMXS30 normalised logreturns 4 0.999 0.997 0.99 0.98 2 0.95 0.90 Probability 0.75 0 0.50 0.25 −2 0.10 0.05 0.02 0.01 0.003 −4 0.001 2000 2010 −4 −2 0 2 4 Data

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 t = α r 2 t − 1 + ( 1 − α ) σ 2 σ 2 t − 1

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

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