volatility - jerry dwyerjerrydwyer.com/pdf/clemson/topic5volatilitypart1.pdf · february 2015....

Post on 20-Jul-2020

0 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

TRANSCRIPT

Volatility

Gerald P. Dwyer

Clemson University

February 2015

Outline

1 VolatilityCharacteristics of Time SeriesHeteroskedasticityExponentially Weighted Moving AverageUse of Shorter-term VarianceUse of intraday changes to estimate daily volatilityARCHProperties of ARCH(1)Generalizations of ARCH(1)GARCHTesting and EstimationSummary

Economic time series: Characteristic features (Stylizedfacts)

Trend in level (rgdp)

A high level of persistence (dlrgdp)

Volatility not constant over time (rgdp)

Series may have a random walk with drift component (rgdp)

Series share co-movements (m2sl and ambsl; exchange rates)

Financial time series: Characteristic features (Stylizedfacts)

Trend in level (CRSP and exchange rates)

A high level of persistence (dlcrsp and exchange rates)

Volatility not constant over time (dlcrsp and exchange rates)

Series may have a random walk with drift component (dlcrsp andexchange rates)

Series share co-movements (exchange rates)

Skewed (crsp)

High excess kurtosis (crsp)

CRSP daily index 1983 to 2013

4,000.0

400.0

40.0

4.0

0.430 40 50 60 70 80 90 00 10

VWINDD

CRSP daily returns 1983 to 2013

-.20

-.16

-.12

-.08

-.04

.00

.04

.08

.12

85 90 95 00 05 10

VWRETD

Absolute value of CRSP daily returns 1983 to 2013

.00

.04

.08

.12

.16

.20

85 90 95 00 05 10

ABS_VWRETD

Square of CRSP daily returns 1983 to 2013

.000

.005

.010

.015

.020

.025

.030

85 90 95 00 05 10

SQ_VWRETD

Exchange rate Eurozone and U.S. 1/4/1999 to 1/30/2015

0.8

0.9

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

99 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14

DEXUSEU

Change in log of exchange rate Eurozone and U.S.1/4/1999 to 1/30/2015

-.04

-.03

-.02

-.01

.00

.01

.02

.03

.04

.05

99 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14

DLDEXUSEU

Absolute value of change in log of exchange rate Eurozoneand U.S. 1/4/1999 to 1/30/2015

.00

.01

.02

.03

.04

.05

99 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14

DLDEXUSEU_ABS

Square of change in log of exchange rate Eurozone andU.S. 1/4/1999 to 1/30/2015

.0000

.0004

.0008

.0012

.0016

.0020

.0024

99 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14

DLDEXUSEU_SQ

Heteroskedasticity over time

These graphs suggest heteroskedasticity over time

I Time-varying volatility of returns

F Of interest in itself to characterize returnsF Matters for prices of options and some other financial instrumentsF Risk of holding assets – VaR

I Volatility clustering

These graphs are suggestive but don’t tell us too much

I Using individual observations on squared changes and absolute value toestimate variance and standard deviation as it changes

I Similar to using each individual observation to estimate mean as itchanges

I Can’t forecast anything going forward

Exponentially weighted moving average of variance

Exponentially weighted average assumes today’s variance forecast is aweighted average of variance yesterday and variance forecasted foryesterday

σ̂2t = (1− λ) (rt − r)2 + λσ̂2

t−1

And yesterday’s variance forecast is a weighted average of variance theday before that and variance forecasted for day before that, and so on

σ̂2t−1 = (1− λ) (rt−1 − r)2 + λσ̂2

t−2

λ = 0.94 daily frequency has been suggested

Use shorter-term returns to estimate variance over longerperiods

Use daily variance in the month to calculate variance for the month –realized volatility

I Let rmt be the monthly return in month tI Let rt,i be the daily return on day i in month tI Suppose that daily returns are serially uncorrelated and the daily

variance is constantI Then

rmt =n

∑i=1

rt,i

Var [rmt ] = nVar [rt,i ]

I and V̂ar [rt,i ] =∑ni=1(rt,i−rt )

2

n−1 where rt is the mean of the daily returnsI The estimated monthly variance thus is

(σ̂mt )2 = n

∑ni=1 (rt,i − rt)

2

n− 1

Daily variance to estimate monthly variance

The estimated monthly variance is simple to calculate

(σ̂mt )2 = n

∑ni=1 (rt,i − rt)

2

n− 1

This becomes more complicated if the daily returns are seriallycorrelated, but it’s still manageable

If daily log returns have high excess kurtosis and serial correlations,then this estimator may not be consistent

Garman-Klass estimator of daily variance

Use high, low, opening, and closing prices to estimate variance

I Can estimate daily variance just knowing opening, high, low andclosing prices

I Assume that price follows a random walkI Let ct be the logarithm of the closing price so rt = ct − ct−1I Conventional estimator is

σ̂2t = E

[(ct − ct−1)

2]

I Using only closing priceI High Ht , low Lt , and open Ot also often are available

F Lower case indicates logarithms

I Can estimate daily variance of price (not log price) from

σ̂2GK = 0.12

(ot − ct−1)2

f+ 0.88

0.5 (ht − lt)2 + 0.386 (ct − ot)

2

1− f

where f is the fraction of the day that the market is closedI Minimum variance unbiased estimator for a random walk with no drift

Yang and Zhang estimator

Use high, low, opening, and closing prices to estimate variance of logprices over a longer period

I Define

ot = lnOt − lnOt−1, ht = lnHt − lnOt−1`t = ln Lt − lnOt−1, ct = lnCt − lnOt−1

I Monthly variance based on n days of trading is

σ̂2YZ = σ̂o + k σ̂c + (1− k) σ̂rs

where σ̂o and σ̂c are is the estimated variances of ot and ct and

σ̂2rs =

1

n ∑ [ht (ht − ct) + `t (`t − ct)]

k =0.34

1.34 + (n+ 1) / (n− 1)

and k was chosen to minimize the variance of the estimator σ̂2YZ

Annualization

Volatilities commonly annualized

I Multiply variance by T (T is number of trading days per year)

Daily returns often not annualized

I Will want to annualize returns when comparing them to annualvolatility

I Why not annualize in general? Magnitudes would be ridiculousI A 1 percent return in one day is a 252 percentage point log return per

yearI A 2 percent return in one day is a 504 percentage point log return per

year

Serial correlation

Change in logarithm of value-weighted CRSP index measurescontinuously compounded return

Serial correlation of squared changes in logarithm of value-weightedCRSP index

Serial correlation of absolute values of change in logarithm ofvalue-weighted CRSP index

Simple way to deal with serial correlation

Suppose the mean equation is

yt = α0 + α1yt−1 + εt

Might think just to estimate a qth-order autoregression for thesquared residuals

ε̂2t = b0 + b1 ε̂2t−1 + ... + b1 ε̂2t−1−q + vt

Not as tractable as another setup

yt = α0 + α1yt−1 + εt E ε2t = 0, E ε2t = σ2t , E εtεs = 0∀t 6= s

σ2t = γ0 + γ1ε2t−1

Autoregressive conditional heteroskedasticity (ARCH)

Simple ARCH model

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

where yt is the variable being examined, σ2t is the variance of εt

conditional on past values of the squared innovations, ε2t−1I yt = α0 + α1yt−1 + εt is the mean equation for ytI σ2

t = γ0 + γ1ε2t−1 is the variance equation for ytI εt is the innovation in yt

Autoregressive conditional heteroskedasticity (ARCH) withAR(1)

Further explore AR(1) with ARCH(1) and its properties

Representation is

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

Autoregressive conditional heteroskedasticity (ARCH) withAR(1)

Further explore AR(1) with ARCH(1) and its properties

Representation is

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

Differences from textbook’s use of letters

I Lag coefficients in mean equation are α’s (not a)I γ’s are the coefficients in the variance equation (not α)

Properties of ARCH(1)

Representation is

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

Changing variance has no effect on the predictability of the series yt

I Both unconditional expected value and conditional expected value (i.e.forecast at t − 1 of yt)

Changing variance with deterministic equation σ2t = γ0 + γ1ε2t−1

implies there is predictability in the variance given recent values

In sum, predictable variance but the predictable variance does nothelp to predict the level of the series

Will show this by looking at conditional expectations

Conditional Expectation

Define conditional expectation at t − 1 as

Et−1 xt ≡ E [xt |Ft−1]

where Ft−1 is the set of all information available up through periodt − 1, which includes for example yt−1, εt−1, σt−1, and vt−1

Properties of ARCH(1)

Representation is

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

Want variance σ2t positive

I Sufficient to guarantee this by imposing γ0 > 0 and γ1 > 0

Properties of ARCH(1)

Representation is

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

Expected value of innovations εt

E εt = E σtvt = E

[(γ0 + γ1ε2t−1

) 12

]vt

= E

[(γ0 + γ1ε2t−1

) 12

]E vt

= E

[(γ0 + γ1ε2t−1

) 12

]· 0

= 0

Properties of ARCH(1)

Representation is

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

Conditional expectation of the innovations εt

Et−1 εt = Et−1 σtvt = Et−1(γ0 + γ1ε2t−1

) 12 vt

= Et−1

[(γ0 + γ1ε2t−1

) 12

]Et−1 vt

= Et−1

[(γ0 + γ1ε2t−1

) 12

]· 0

= 0

Properties of ARCH(1)

Representation is

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

Unconditional and conditional expected values

E εt = 0

Et−1 εt = 0

I Note that Et−1 εt = 0 implies

Et−1 yt = Et−1 [α0 + α1yt−1 + htvt ]

= α0 + α1 Et−1 yt−1 + Et−1 εt

= α0 + α1yt−1

Properties of ARCH(1)

Representation is

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

Variance of the innovations εt

E ε2t = E σ2t v

2t = E

(γ0 + γ1ε2t−1

)v2t

= E(γ0 + γ1ε2t−1

)E v2t

=(γ0 + γ1 E ε2t−1

)· 1

=(γ0 + γ1 E ε2t

)and therefore

E ε2t =γ0

1− γ1

I E ε2t > 0 if γ0 > 0 and 0 ≤ γ1 < 1

Properties of ARCH(1)

Representation is

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

Conditional variance of the innovations εt

Et−1 ε2t = Et−1 σ2t v

2t = Et−1

(γ0 + γ1ε2t−1

)v2t

= Et−1(γ0 + γ1ε2t−1

)Et−1 v

2t

=(γ0 + γ1 Et−1 ε2t−1

)· 1

= γ0 + γ1ε2t−1

and thereforeEt−1 ε2t = γ0 + γ1ε2t−1

Et−1 ε2t > 0 and stable difference equation if γ0 > 0 and |γ1| < 1

I But want γ1 > 0 for positive variance and therefore 0 ≤ γ1 < 1

Properties of ARCH(1)

Representation is

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

Unconditional and conditional expected values

E εt = 0

Et−1 εt = 0

Variance in ARCH model with γ0 > 0 and 0 ≤ γ1 < 1

E ε2t =γ0

1− γ1

Et−1 ε2t = γ0 + γ1ε2t−1

Properties of ARCH(1)

Representation is

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

Third moment

I If vt is symmetric, then σtvt is symmetric

Fourth moment

I Assume vt is normally distributedI Do we get fatter tails in εt than from the normal distribution?

Kurtosis

Kurtosis is the fourth moment of a distribution divided by thevariance squared

K =µ4

(σ2)2

I where µ4 is the fourth moment about the meanI Could also define conditional kurtosisI Kurtosis for a normal distribution is three

Excess kurtosis is kurtosis minus three

K e =µ4

(σ2)2− 3

Properties of ARCH(1)

Representation is

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

If vt is normally distributed, the unconditional kurtosis εt withnormally distributed vt is

K =E ε4t

Var [εt ]2= 3

1− γ21

1− 3γ21

E ε4t > 0 and Var [εt ]2 > 0 and therefore γ1 must satisfy(

1− 3γ21

)> 0 and therefore 0 ≤ γ2

1 <13

Therefore,E ε4t

Var [εt ]2> 3

I Fatter tails than for a normal distribution with 0 ≤ γ21 < 1

3

Properties of ARCH(1)

Representation is

yt = α0 + α1yt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

Third moment

I If vt is symmetric, then σtvt is symmetric

Fourth moment

I If vt is normally distributed, then εt has excess kurtosis

Generalizations of ARCH(1)

Generalizations – ARMA model for series would be

yt = α0 + α (L) yt−1 + β (L) εt−1 + εt εt = σtvt vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1

Fairly easy to see this point so we’ll just stick to a simple AR(1)

Generalizations of ARCH(1)

Generalizations – more lags in ARCH

yt = α0 + α1yt−1 + εt , εt = σtvt , vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1 + γ2ε2t−2 + ... + γpε2t−q

Sufficient condition to be sure σ2t is positive is that γi ≥ 0,

i = 0, ..., q

GARCH

ARCH models can require many lags

I Reduce lags in mean equations by using ARMA models

F In level equation, MA terms can substitute for several AR terms

I Including something like AR terms in ARCH equation can reducenumber of lags

GARCH

GARCH (Generalized ARCH) model

I ARCH Modelσ2t = γ0 + γ1ε2t−1 + ... + γqε2t−q

I Instead, try GARCH, here a GARCH(p,q) (order of lags often notconsistent across authors)

σ2t = γ0 + γ1ε2t−1 + ... + γqε2t−q + δ1σ2

t−1 + ... + δkσ2t−p

I Lag lengths are q for the part analogous to the moving average and pfor the part analogous to an autoregression

GARCH

GARCH for AR(1)

yt = α0 + α1yt−1 + εt , εt = σtvt , vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1 + ... + γqε2t−q + δ1σ2

t−1 + ... + δkσ2t−p

Sufficient conditions to be sure σ2t is positive are γi ≥ 0 and δi ≥ 0

with ∑max(p,q)i=1 (γi + δi ) < 1

Just as with an ARMA model, this can reflect more complicateddynamics with fewer parameters than only adding more laggedsquared innovations

Properties of GARCH models

σ2t = γ0 + γ1ε2t−1 + ... + γqε2t−q + δ1σ2

t−1 + ... + δpσ2t−p

Restrictions on parameters

γi > 0, δi > 0,max(q,p)

∑i=1

(γi + δi ) < 1

VarianceE[ε2t]=

γ0

1−∑max(q,p)i=1 (γi + δi )

Properties of GARCH models

σ2t = γ0 + γ1ε2t−1 + ... + γqε2t−q + δ1σ2

t−1 + ... + δpσ2t−p

Restrictions on parameters

γi > 0, δi > 0,max(q,p)

∑i=1

(γi + δi ) < 1

VarianceE[ε2t]=

γ0

1−∑max(q,p)i=1 (γi + δi )

Properties of GARCH models

σ2t = γ0 + γ1ε2t−1 + ... + γqε2t−q + δ1σ2

t−1 + ... + δpσ2t−p

Restrictions on parameters

γi > 0, δi > 0,max(q,p)

∑i=1

(γi + δi ) < 1

Kurtosis

For GARCH(1,1) with 1− (γ1 + δ1)2 − 2γ2

1 > 0, then

E[ε4t]

(E [ε2t ])2=

3[

1− (γ1 + δ1)2]

1− (γ1 + δ1)2 − 2γ2

1

> 3

Properties of GARCH models

σ2t = γ0 + γ1ε2t−1 + ... + γqε2t−q + δ1σ2

t−1 + ... + δpσ2t−p

Restrictions on parameters

γi > 0, δi > 0,max(q,p)

∑i=1

(γi + δi ) < 1

Kurtosis

For GARCH(1,1) with 1− (γ1 + δ1)2 − 2γ2

1 > 0, then

E[ε4t]

(E [ε2t ])2=

3[

1− (γ1 + δ1)2]

1− (γ1 + δ1)2 − 2γ2

1

> 3

Properties of GARCH models

σ2t = γ0 + γ1ε2t−1 + ... + γqε2t−q + δ1σ2

t−1 + ... + δpσ2t−p

Restrictions on parameters

γi > 0, δi > 0,max(q,p)

∑i=1

(γi + δi ) < 1

Kurtosis

For GARCH(1,1) with 1− (γ1 + δ1)2 − 2γ2

1 > 0, then

E[ε4t]

(E [ε2t ])2=

3[

1− (γ1 + δ1)2]

1− (γ1 + δ1)2 − 2γ2

1

> 3

Limitations of GARCH models

1 Symmetric effects of shocks. This is too restrictive for stock returns,where negative shocks have a larger effect on future variance thanpositive shocks

2 Returns, for example, tend to have some clusters of high and lowvariance, whereas GARCH models tend to predict slow decay to meanvariance from any current variance

3 Restrictive parametrizations, e.g. 0 ≤ γ21 <

13 for kurtosis to be well

defined for ARCH(1)

4 Deterministic equation for variance; no error term inσ2t = γ0 + γ1ε2t−1

5 Provides no evidence on source of changes in variance

Estimating a GARCH model

A simple GARCH model

yt = α0 + α1yt−1 + εt , εt = htvt , vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1 + δ1σ2

t−1

Steps in estimating a GARCH model

1 Estimate a model for the mean equation

2 Use the residuals of the mean equation to test for ARCH effects

3 Specify a variance model with ARCH effects if it seems warranted

4 Check the fitted model and refine as suggested by diagnostic statistics

Estimating mean equation

In general, there is no reason the mean equation can’t be ascomplicated as we like

yt = µt + εt

yt can be a complicated ARMA(p,q) or can have variables included

I yt is stationary in meanI yt may be first difference of original seriesI for example yt = Yt − Yt−1

Original estimates may be mis-specified if ignore conditionalheteroskedasticity of εt

Three tests for ARCH – first

McLeod-Li test – Box-Ljung test applied to squared residuals, ε̂2t , forsome pre-specified number of lags k

T (T + 2)k

∑l=1

ρ̂ (ε̂2t)2

l

T − l∼a χ2

k

where ρ(ε̂2t)l

is the lth serial correlation of the squared residuals andreject the null hypothesis that all of the first k autocorrelations arezero if p-value less than 0.05 for test statistic

Three tests for ARCH – second and third

Engle test based on a regression for the squared residuals

ε̂2t = γ0 + γ1 ε̂2t−1 + γ2 ε̂2t−2 + ... + γk ε̂2t−k + et (1)

where et is the error term in the regression for squared residuals

I Test whether γ1 = γ2 = ... = γk using

T · R2 ∼a χ2k

I where T is the number of observations and R2 is the R2 of equation(1)

I Can be very informative with just one lag

F-test for regression (1)

(SSR0 − SSR1) /kSSR1/ (T − k − 1)

∼ Fk,T−k−1

I where T is the number of observations in the estimated equation (1),SSR0 = ∑ ε̂2t and SSR1 = ∑ e2t

Estimation of GARCH model

Steps in estimating a GARCH model

1 Estimate a model for the mean equation – DISCUSSED

2 Use the residuals of the mean equation to test for ARCH effects –DISCUSSED

3 Specify a variance model with GARCH effects if it seems warranted

4 Check the fitted model and refine as suggested by diagnostic statistics

Estimation of GARCH model

Steps in estimating a GARCH model

1 Estimate a model for the mean equation – DISCUSSED

2 Use the residuals of the mean equation to test for ARCH effects –DISCUSSED

3 Specify a variance model with GARCH effects if it seems warranted

4 Check the fitted model and refine as suggested by diagnostic statistics

Estimation by maximum likelihood

Maximize likelihood function based on

yt = α0 + α1yt−1 + εt , εt = σtvt , vt ∼ iid (0, 1)

σ2t = γ0 + γ1ε2t−1 + δ1σ2

t−1

I Nonlinear and involves a latent variable σ2t which is not observed but is

estimated for each observationI Not as complicated as it could be because σ2

t is a deterministicfunction of ε2t−1 and past σ2

t−1I Still cannot be done by a linear algorithmI Nonlinear maximizationI Use computed information matrix as basis of variance-covariance

matrix and standard errors

Distributions for maximum likelihood

Distributions

I Normal distributionI t-distribution with a small number of degrees of freedomI Generalized error distribution

Quasi-maximum likelihood

Quasi-maximum likelihood estimation

I Consistent estimates of parameters under fairly general conditionsI Issue of correct standard errors of coefficients

Estimation by maximum likelihood with normal distribution

Likelihood function if normally distributed

yt = α0 + α1yt−1 + εt , εt = σtvt , vt ∼ IIDN (0, 1)

σ2t = γ0 + γ1ε2t−1 + δ1σ2

t−1

Maximize likelihood function with respect to parameters

L (v |θ) =T

∏t=1

1√2πσt

exp

(− v2t

2σ2t

)I This can be written

ln L (v |θ) = −T

2

T

∏t=1

ln (2π)− 1

2

T

∑t=1

ln(

σt2)− 1

2

T

∑t=1

v2tσ2t

I Not remotely the same as minimizingT

∑t=1

ε2t orT

∑t=1

v2t

Estimation by maximum likelihood with t distribution

Likelihood function if distributed Student’s t, which provides fattertails, is

L (v |θ) =T

∏t=1

Γ ((df + 1) /2)

Γ (df /2)√(df − 2)π

(1 +

v2tdf − 2

)−(df+1)/2

df > 2

where df represents the degrees of freedom in the distribution andΓ (x) is the gamma function Γ (x) =

∫ ∞0 zx−1e−zdy

Maximize likelihood function with respect to parameters

L (v |θ) =T

∏t=1

Γ ((df + 1) /2)

Γ (df /2)√(df − 2)π

(1 +

v2tdf − 2

)−(df+1)/2

df > 2

I df can be estimated or can be specified in advance – commonly 3 to 6

Practical issues with GARCH estimation

Nonlinear maximization (or minimization of -ln L (ε|θ))I Can be something of an art form to be sure have reached maximum

but generally not a problem with low-order GARCH and a well-specifiedmean equation

I Redundant parameters can create serious problems for convergenceI Generally speaking, it is hard to estimate more than a few lagsI Fairly common to use GARCH(1,1) for financial data

Summary I

Economic and financial time series have common characteristicsI Serial correlation of seriesI Serial correlation of volatility

Heteroskedasticity

Simple ways of dealing with heteroskedasticityI Use of higher frequency data to estimate changes in variance for lower

frequency data

ARCH and GARCHI Variance a deterministic function of squared innovations

Properties of ARCH and GARCHI Series can be have an unpredictable mean but predictable varianceI ARCH and GARCH can help to account for excess heteroskedasticityI ARCH and GARCH do not generate asymmetric distributions of series

from symmetric innovations

Estimate ARCH and GARCH by maximum likelihoodI All the desirable properties that go with maximum likelihood if

distribution of innovations correctly specified

Summary II

I Even if distribution of innovations not correctly specified, estimator stillconsistent under general conditions

I If only quasi-maximum likelihood, estimators of standard errors maywell be wrong

Testing for ARCH or GARCH in time seriesI This is important because nonlinear estimation with redundant

parameters can fail to converge without it being obvious why it fails toconverge

I Most common testsF McLeod-Li Q statistics on squared innovationsF Engle’s T · R2 test

top related