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Chapter 8
Modelling volatility and correlation
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1. An Excursion into Non-linearity Land
• Our “traditional” structural model could be something like:
yt = β 1 + β 2 x2t + ... + β k xkt + ut, or y = X β + u.
We also assume ut ∼ N(0,σ2).
• Motivation: the linear structural (and time series) models cannotexplain a number of important features common to muchfinancial data:
- leptokurtosis: returns have dist. with higher peak and fatter tail
- volatility clustering or volatility pooling (bunches of volatility,autocorrelation) where volatility is variance of returns
- leverage effects (asymmetry): higher volatility following a
price fall than following a price rise of the same magnitude
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Non-linear Models: A Definition
• Campbell, Lo and MacKinlay (1997) define a non-linear
data generating process as one that can be written yt = f (ut , ut -1, ut -2, …)
where ut is an iid error term and f is a non-linear function.
• They also give a slightly more specific definition as
yt = g(ut -1, ut -2, …)+ ut σ2(ut -1, ut -2, …)
where g is a function of past error terms only and σ2 is a
variance term.
• Models with nonlinear g(•) are “non-linear in mean”, while
those with nonlinear σ2
(•) are “non-linear in variance”.
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Types of non-linear models
• The linear paradigm is a useful one. Many apparently non-linearrelationships can be made linear by a suitable transformation. Onthe other hand, it is likely that many relationships in finance areintrinsically non-linear.
• There are many types of non-linear models, e.g.
- ARCH / GARCH: useful for modeling and forecasting volatility- switching models: series following different processes at different
points in time
….
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4. Implied Volatility Model
• Implied Volatility is the market forecast of the volatility ofunderlying asset returns over the lifetime of the option.
• It is can be used as an input for all pricing models for
financial options
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Heteroscedasticity Revisited
• An example of a structural model is
with ut ∼ N(0, ).
• The assumption that the variance of the errors is constant is known as
homoscedasticity, i.e. Var (ut ) = .
• What if the variance of the errors is not constant?
- heteroscedasticity
- would imply that standard error estimates could be wrong.
• Is the variance of the errors likely to be constant over time? Not forfinancial data.
σ u2
σ u2
t = β 1 + β 2 x2t + β 3 x3t + β 4 x4t + u t
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7. Autoregressive Conditionally Heteroscedastic(ARCH) Models
• So use a model which does not assume that the variance is constant.
• Definition of the conditional variance of ut := Var(u
t ⏐ u
t -1 , u
t -2 ,...) = E[(u
t -E(u
t ))2⏐ u
t -1 , u
t -2,...]
We usually assume that E(ut ) = 0
so = Var(ut ⏐ u
t -1 , u
t -2 ,...) = E[u
t 2⏐ u
t -1 , u
t -2,...].
• What could the current value of the variance of the errors plausiblydepend upon?
– Previous squared error terms.
• This leads to the AutoRegressive Conditionally Heteroscedastic Modelfor the variance of the errors:
= α 0
+ α 1
• This is known as an ARCH(1) model.
σ t 2
σ t 2
σ t 2
ut −12
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Autoregressive Conditionally Heteroscedastic(ARCH) Models (cont’d)
• The full model would be
yt = β 1 + β 2 x2t + ... + β k xkt + ut , ut ∼ N(0, )where = α
0+ α
1
• We can easily extend this to the general case where the error variancedepends on q lags of squared errors:
= α 0 + α 1 +α 2 +...+α q• This is an ARCH(q) model.
• Instead of calling the variance , in the literature it is usually called ht ,
so the model is yt = β 1 + β 2 x2t + ... + β k xkt + ut , u
t ∼ N(0,h
t )
where ht = α
0+ α
1+α
2+...+α
q
σ t 2
σ t 2
σ t 2
ut −12
ut q−
2
ut q−2
σ t 2
2
1−t u2
2−t u
2
1−t u2
2−t u
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Another Way of Writing ARCH Models
• For illustration, consider an ARCH(1). Instead of the above, we can
write
yt = β 1 + β 2 x2t + ... + β k xkt + ut , ut = v
t σ
t
, vt ∼ N(0,1)
• The two are different ways of expressing exactly the same model. The
first form is easier to understand while the second form is required forsimulating from an ARCH model, for example.
σ α α t t u= + −0 1 12
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Testing for “ARCH Effects”
1. First, run any postulated linear regression of the form given in the equation
above, e.g. yt = β 1 + β 2 x2t + ... + β k xkt + ut
saving the residuals, .
2. Then square the residuals, and regress them on q own lags to test for ARCH
of order q, i.e. run the regression
where vt is an error term.
Obtain R2 from this regression.
3. The test statistic is defined as TR2 (the number of observations multiplied by the coefficient of multiple correlation) from the last regression, and isdistributed as a χ 2(q).
t u
t qt qt t t vuuuu +++++= −−−22
22
2
110
2 ˆ...ˆˆˆ γ γ γ γ
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Testing for “ARCH Effects” (cont’d)
4. The null and alternative hypotheses are
H0
: γ 1
= 0 and γ 2
= 0 and γ 3
= 0 and ... and γ q
= 0
H1
: γ 1
≠ 0 or γ 2 ≠ 0 or γ
3 ≠ 0 or ... or γ
q ≠ 0.
If the value of the test statistic is greater than the critical value from the χ 2 distribution, then reject the null hypothesis (proof of autocorrelation
of volatility).
• Note: that the ARCH test is also sometimes applied directly to returnsinstead of the residuals from Stage 1 above.
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Problems with ARCH(q) Models
• How do we decide on q?• The required value of q might be very large
• Non-negativity constraints might be violated.
– When we estimate an ARCH model, we require α i >0 ∀i=1,2,...,q (since variance cannot be negative)
• A natural extension of an ARCH(q) model which gets
around some of these problems is a GARCH model.
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Generalised ARCH (GARCH) Models (cont’d)
• Now substituting into (2) for σ t -22
• An infinite number of successive substitutions would yield
• So the GARCH(1,1) model can be written as an infinite order ARCH model.
• We can again extend the GARCH(1,1) model to a GARCH( p,q):
σ t 2
=α 0 + α 12
1−t u +α 0 β + α 1 β 2
2−t u + β 2
(α 0 + α 12
3−t u + βσ t -32
)
σ t 2 = α 0 + α 1
2
1−t u +α 0 β + α 1 β 2
2−t u +α 0 β 2 + α 1 β
2 2
3−t u + β 3σ t -3
2
σ t 2 = α 0 (1+ β + β
2) + α 1
2
1−t u (1+ β L+ β 2 L
2 ) + β
3σ t -3
2
σ t 2 = α 0 (1+ β + β
2+...) + α 1
2
1−t u (1+ β L+ β 2 L2
+...) + β ∞ σ 02
σ t 2
= α 0+α 12
1−t u +α 22
2−t u +...+α q2
qt u − + β 1σ t-12+ β 2σ t-2
2+...+ β pσ t-p
2
σ t 2 = ∑ ∑
= =
−− ++q
i
p
j
jt jit iu1 1
22
0 σ β α α
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Generalised ARCH (GARCH) Models (cont’d)
• But in general a GARCH(1,1) model will be sufficient tocapture the volatility clustering in the data.
• Why is GARCH Better than ARCH?
- more parsimonious - avoids overfitting
- then less likely to breech non-negativity constraints
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The Unconditional Variance under the GARCHSpecification
• The unconditional variance of ut is given by
when
• is termed “non-stationarity” in variance
• is termed Intergrated GARCH
• For stationarity in variance, the conditional variance forecasts will
converge to the long-term average of variance• For non-stationarity in variance, the conditional variance forecasts will
go to infinity
• For IGARCH, the conditional variance forecasts will not converge to
their unconditional value as the horizon increases.
Var(ut ) =)(1 1
0
β α α
+−
β α +1 < 1
β α +1 ≥ 1
β α +1 = 1
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Estimation of ARCH / GARCH Models (cont’d)
• The steps involved in actually estimating an ARCH or GARCH model
are as follows
1. Specify the appropriate equations for the mean and the variance - e.g. an
AR(1)- GARCH(1,1) model:
2. Specify the log-likelihood function to maximise:
3. The computer will maximise the function and give parameter values and
their standard errors
t = μ + φ yt-1 + ut , ut ∼ N(0,σ t 2
)σ t
2 = α 0 + α 1
2
1−t u + βσ t-12
∑∑ =−
=−−−−−=
T
t
t t t
T
t
t y yT
L1
221
1
2 /)(2
1)log(
2
1)2log(
2σ φ μ σ π
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Example: Parameter Estimation
using Maximum Likelihood for a linear model
• Consider the bivariate regression case with homoscedastic errors for
simplicity:
• Assuming that ut ∼ N(0,σ 2), then yt ∼ N( , σ 2) so that the
probability density function for a normally distributed random variable
with this mean and variance is given by(1)
• Successive values of yt would trace out the familiar bell-shaped curve.
• Assuming that ut are iid, then y
t will also be iid.
t t t u x y ++= 21 β
t x21 β β +
⎭⎬⎫
⎩⎨⎧ −−
−=+2
2
212
21
)(
2
1exp
2
1),(
σ
β β
π σ σ β β t t
t t
x y x y f
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Parameter Estimation using Maximum Likelihood(cont’d)
• Then the joint pdf for all the y’s can be expressed as a product of theindividual density functions
(2)
• Substituting into equation (2) for every yt from equation (1),
(3)⎭⎬⎫
⎩⎨⎧ −−
−=+ ∑=
T
t
t t
T T t T
x y x y y y f
12
2
212
2121
)(
2
1exp
)2(
1),,...,,(
σ
β β
π σ σ β β
∏=
+=
+
++=+
T
t t t
T
t T
X y f
X y f
X y f X y f X y y y f
1
2
21
2
421
2
2212
2
1211
2
2121
),(
),(
)...,(),(),,...,,(
σ β β
σ β β
σ β β σ β β σ β β
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Parameter Estimation using Maximum Likelihood(cont’d)
• The typical situation we have is that the xt and y
t are given and we want to
estimate β 1
, β 2
, σ 2. If this is the case, then f(•) is known as the likelihoodfunction, denoted LF ( β
1, β
2, σ 2), so we write
(4)
• Maximum likelihood estimation involves choosing parameter values ( β 1, β
2 ,σ 2) that maximise this function.
• We want to differentiate (4) w.r.t. β 1, β
2 ,σ 2, but (4) is a product containingT terms.
⎭⎬
⎫
⎩⎨
⎧ −−−=
∑=
T
t
t t
T T
x y LF
1 2
2
212
21
)(
2
1exp
)2(
1),,(
σ
β β
π σ σ β β
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• Since , we can take logs of (4).
• Then, using the various laws for transforming functions containinglogarithms, we obtain the log-likelihood function, LLF :
• which is equivalent to
(5)
• Differentiating (5) w.r.t. β 1, β
2 ,σ 2, we obtain
(6)
max ( ) maxlog( ( )) x x
f x f x=
Parameter Estimation using Maximum Likelihood(cont’d)
∑=
−−−−−=
T
t
t t x yT T LLF
12
2
21 )(
2
1)2log(
2
log
σ
β β π σ
∑=
−−−−−=
T
t
t t x yT T LLF
12
2
212 )(
2
1)2log(
2log
2 σ
β β π σ
∑ −−−−=
2
21
1
1.2).(
2
1
σ
β β
∂β
∂ t t x y LLF
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(7)
(8)
• Setting (6)-(8) to zero to minimise the functions, and putting hats above
the parameters to denote the maximum likelihood estimators,
• From (6),
(9)
Parameter Estimation using Maximum Likelihood(cont’d)
∑ −−+−=
4
2
21
22
)(
2
11
2 σ
β β
σ ∂σ
∂ t t x yT LLF ∑
−−−−=
2
21
2
.2).(
2
1
σ
β β
∂β
∂ t t t x x y LLF
∑ =−− 0)ˆˆ( 21 t t x y β β
∑ ∑ ∑ =−− 0ˆˆ21 t t x y β β
∑ ∑ =−− 0ˆˆ21 t t xT y β β
∑ ∑ =−− 01ˆˆ1
21 t t xT
yT
β β
x y 21ˆˆ β β −=
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• From (7),
(10)
• From (8),
Parameter Estimation using Maximum Likelihood(cont’d)
∑ =−− 0)ˆˆ( 21 t t t x x y β β
∑ ∑∑ =−− 0ˆˆ 221 t t t t x x x y β β
∑ ∑∑ =−− 0ˆˆ 2
21 t t t t x x x y β β
∑ ∑∑ −−= t t t t x x y x y x )ˆ(ˆ2
2
2 β β
∑∑ −−= 2
2
2
2ˆˆ xT y xT x y x t t t β β
∑∑ −=− y xT x y xT x t t t )(ˆ 22
2 β
)(ˆ
222 xT x
y xT x y
t
t t
∑∑ −−= β
∑ −−= 2
2142)ˆˆ(
ˆ
1
ˆt t x y
T β β
σ σ
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• Rearranging,
(11)
• How do these formulae compare with the OLS estimators?
(9) & (10) are identical to OLS(11) is different. The OLS estimator was
• Therefore the ML estimator of the variance of the disturbances is biased,although it is consistent.
• Q: But how to use ML in estimating heteroscedastic models?
σ 2 21= ∑
T ut
σ 2 21=
− ∑T k
ut
Parameter Estimation using Maximum Likelihood(cont’d)
∑ −−= 2
21
2 )ˆˆ(1
ˆt t x y
T β β σ
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Estimation of GARCH Models Using
Maximum Likelihood
• Now we have yt = μ + φ y
t-1 + ut
, ut ∼ N(0, )
• Unfortunately, the LLF for a model with time-varying variances cannot bemaximised analytically, except in the simplest cases. So a numerical procedureis used to maximise the log-likelihood function. A potential problem: localoptima or multimodalities in the likelihood surface.
• The way we do the optimisation is:1. Set up LLF.
2. Use regression to get initial guesses for the mean parameters.
3. Choose some initial guesses for the conditional variance parameters.
4. Specify a convergence criterion - either by criterion or by value.
σ t 2
σ t 2 = α 0 + α 1 21−t u + βσ t-12
∑∑=
−=
−−−−−=T
t
t t t
T
t
t y yT
L1
22
1
1
2/)(
2
1)log(
2
1)2log(
2σ φ μ σ π
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Non-Normality and Maximum Likelihood
• Recall that the conditional normality assumption for ut
is essential in
specifying LF.• We can test for normality using the following representation
ut = v
t σ
t v
t ∼ N(0,1)
• The sample counterpart is
• Are the normal? Typically are still leptokurtic, although less so thanthe . Is this a problem? Not really, as we can use the ML with a robustvariance/covariance estimator. ML with robust standard errors is called Quasi-Maximum Likelihood or QML.
σ α α α σ t t t u= + +− −0 1 1
2
2 1
2 vu
t t
t
=σ
t
t t
uv
σ ˆ
ˆˆ =
t v t v
t u
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10. Extensions to the Basic GARCH Model
• Since the GARCH model was developed, a huge number of extensions
and variants have been proposed. Three of the most important
examples are EGARCH, GJR, and GARCH-M models.
• Problems with GARCH( p,q) Models:
- Non-negativity constraints may still be violated
- GARCH models cannot account for leverage effects (asymmetry)
• Possible solutions: the exponential GARCH (EGARCH) model or theGJR model, which are asymmetric GARCH models.
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13. The EGARCH Model
• Suggested by Nelson (1991). The variance equation is given by
• Advantages of the model
- Since we model the log(σ t 2), then even if the parameters are negative, σ
t 2
will be positive.
- We can account for the leverage effect: if the relationship between
volatility and returns is negative, γ will be negative.
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡−+++=
−
−
−
−−
π σ α
σ γ σ β ω σ
2)log()log(
2
1
1
2
1
12
1
2
t
t
t
t
t t
uu
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An Example of the use of a GJR Model
• Using monthly S&P 500 returns, December 1979- June 1998
• Estimating a GJR model, we obtain the following results, with t-ratios:
)198.3(
172.0=t y
)772.5()999.14()437.0()372.16(
604.0498.0015.0243.1 1
2
1
2
1
2
1
2
−−−− +++= t t t t t I uu σ σ
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News Impact Curves
The news impact curve plots the next period volatility (ht ) that would arise from various
positive and negative values of ut -1, given an estimated model.
News Impact Curves for S&P 500 Returns using Coefficients from GARCH and GJR
Model Estimates:
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Value of Lagged Shock
V a l u e o f C o n d i t i o n a l V a r i a n c e
GARCH
GJR
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16. GARCH-in Mean
• Most models in finance suppose that higher risk is compensated by a higher
return. So let the return of a security be partly determined by its risk?
• Engle, Lilien and Robins (1987) suggested the ARCH-M specification. AGARCH-M model would be
• δ can be interpreted as a sort of risk premium.
• It is possible to combine all or some of these models together to get morecomplex “hybrid” models - e.g. an ARMA-EGARCH(1,1)-M model.
t = μ + δσ t-1+ ut , ut ∼ N(0,σ t 2
)σ t
2 = α 0 + α 1
2
1−t u + βσ t -12
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17. What Use Are GARCH-type Models?
• GARCH can model the volatility clustering effect since the conditional
variance is autoregressive. Such models can be used to forecast volatility.
• We could show that
Var ( yt ⏐ y
t -1 , yt -2, ...) = Var (u
t ⏐ u
t -1 , ut -2, ...)
• So modelling σ t 2 will give us models and forecasts for y
t as well.
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Forecasting Variances using GARCH Models
• Producing conditional variance forecasts from GARCH models uses avery similar approach to producing forecasts from ARMA models.
• It is again an exercise in iterating with the conditional expectationsoperator.
• Consider the following GARCH(1,1) model:
, ut ∼ N(0,σ
t 2),
• What is needed is to generate all forecasts of σ T +12
⏐ΩT , σ T +22
⏐ΩT , ...,σ T +s
2 ⏐ΩT
where ΩT
denotes all information available up to andincluding observation T .
• Adding one to each of the time subscripts of the above conditionalvariance equation, and then two, and then three would yield the
following equationsσ
T +12 = α
0+ α
1u
T 2 + βσ
T 2 ,
σ T +2
2 = α 0 + α 1 uT +1
2 + β σ T +1
2 ,
σ T +3
2 = α 0 + α
1 uT +2
2 + β σ T +2
2
t t u y += 2
1
2
110
2
−− ++= t t t u βσ α α σ
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Forecasting Variancesusing GARCH Models (Cont’d)
• Let be the one step ahead forecast for σ 2 made at time T . This is
easy to calculate since, at time T , the values of all the terms on theRHS are known.
• would be obtained by taking the conditional expectation of the
first equation at the bottom of slide 36:
• Given, how is , the 2-step ahead forecast for σ 2 made at time T ,
calculated? Taking the conditional expectation of the second equation
at the bottom of slide 36:
= α 0 + α
1E( ⏐ ΩT
) + β
• where E( ⏐ ΩT ) is the expectation, made at time T , of , which is
the squared disturbance term.
2
,1
f
T σ
2
,1
f
T σ
2,1
f T σ = α 0 + α 1 2
T u + βσ T 2
2
,1
f
T σ 2
,2
f
T σ
2,2
f T σ 2
1+T u 2,1
f T σ
2
1+T u2
1+T u
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Forecasting Variancesusing GARCH Models (Cont’d)
• We can write
E(uT +1
2 | Ωt ) = σ
T +1
2
• But σ T +1
2 is not known at time T , so it is replaced with the forecast forit, , so that the 2-step ahead forecast is given by
= α 0 + α
1+ β
= α 0 + (α
1+ β )
• By similar arguments, the 3-step ahead forecast will be given by
= ET (α
0 + α 1
+ βσ T +2
2)
= α 0 + (α 1+ β )
= α 0
+ (α 1
+ β )[ α 0
+ (α 1
+ β ) ]
= α 0 + α 0(α 1+ β ) + (α 1+ β )2
• Any s-step ahead forecast (s ≥ 2) would be produced by
2
,1
f
T σ 2
,2
f
T σ 2
,1
f
T σ 2
,1
f
T σ 2
,2
f
T
σ 2
,1
f
T σ
2
,3
f
T σ 2
,2
f
T σ 2
,1
f
T σ 2
,1
f
T σ
f
T
ss
i
i f
T s hh ,1
1
1
1
1
1
10, )()( −−
=
− +++= ∑ β α β α α
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22. What Use Are Volatility Forecasts?
1. Option pricing
C = f(S, X, σ2, T, r f )
2. Conditional betas
3. Dynamic hedge ratiosThe Hedge Ratio - the size of the futures position to the size of the
underlying exposure, i.e. the number of futures contracts to buy or sell per
unit of the spot good.
β σ
σ i t
im t
m t
,,
,
=2
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What Use Are Volatility Forecasts? (Cont’d)
• What is the optimal value of the hedge ratio?
• Assuming that the objective of hedging is to minimise the variance of thehedged portfolio, the optimal hedge ratio will be given by
where h = hedge ratio
p = correlation coefficient between change in spot price (∆S ) andchange in futures price (∆F )
σ S = standard deviation of ∆S
σ F = standard deviation of ∆F
• What if the standard deviations and correlation are changing over time?
Use
h ps
F
= σ
σ
t F
t s
t t ph,
,
σ
σ =
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Likelihood Ratio Tests
• Estimate under the null hypothesis and under the alternative.
• Then compare the maximised values of the LLF.• So we estimate the unconstrained model and achieve a given maximised
value of the LLF, denoted Lu
• Then estimate the model imposing the constraint(s) and get a new value ofthe LLF denoted L
r .
• Which will be bigger?
• Lr ≤ L
ucomparable to RRSS ≥ URSS
• The LR test statistic is given by
LR = -2( Lr
- Lu) ∼ χ2(m)
where m = number of restrictions
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Likelihood Ratio Tests (cont’d)
• Example: We estimate a GARCH model and obtain a maximised LLF of
66.85. We are interested in testing whether β = 0 in the following equation.
yt = μ + φ y
t-1 + ut
, ut ∼ N(0, )
= α 0 + α
1+ β
• We estimate the model imposing the restriction and observe the maximisedLLF falls to 64.54. Can we accept the restriction?
• LR = -2(64.54-66.85) = 4.62.
• The test follows a χ2
(1) = 3.84 at 5%, so reject the null.• Denoting the maximised value of the LLF by unconstrained ML as L( )
and the constrained optimum as . Then we can illustrate the 3 testing procedures in the following diagram:
σ t 2
σ t 2 ut −1
2
L(~
)θ
2
1−t σ
θ
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Hypothesis Testing under Maximum Likelihood
• The vertical distance forms the basis of the LR test.
• The Wald test is based on a comparison of the horizontal distance.
• The LM test compares the slopes of the curve at A and B.
• We know at the unrestricted MLE, L( ), the slope of the curve is zero.
• But is it “significantly steep” at ?
• This formulation of the test is usually easiest to estimate.
L(~
)θ
θ
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VECH and Diagonal VECH
• In the case of the VECH, the conditional variances and covariances wouldeach depend upon lagged values of all of the variances and covariances
and on lags of the squares of both error terms and their cross products.• In matrix form, it would be written
• Writing out all of the elements gives the 3 equations as
• Such a model would be hard to estimate. The diagonal VECH is much
simpler and is specified, in the 2 variable case, as follows:
112212111012
1222
2
121022
1112
2
111011
−−−
−−
−−
++=++=++=
t t t t
t t t
t t t
huuh
huh
huh
γ γ γ
β β β
α α α
( ) ( ) ( )111 −−− +Ξ′Ξ+= t t t t H VECH BVECH AC H VECH ( )t t t H N ,0~1−Ξ ψ
1123312232111312133
2
232
2
1313112
1122312222111212123
2
222
2
1212122
11213122121111121132212
21111111
−−−
−−−
−−−
++++++=++++++=++++++=
t t t t t t t t
t t t t t t t t
t t t t t t t t
hbhbhbuuauauach
hbhbhbuuauauach
hbhbhbuuauauach
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BEKK and Model Estimation for M-GARCH
• Neither the VECH nor the diagonal VECH ensure a positive definite variance-
covariance matrix.• An alternative approach is the BEKK model (Engle & Kroner, 1995).
• In matrix form, the BEKK model is
• Model estimation for all classes of multivariate GARCH model is again
performed using maximum likelihood with the following LLF :
where N is the number of variables in the system (assumed 2 above), θ is a
vector containing all of the parameters to be estimated, and T is the number of
observations.
B B A H AW W H t t t t 111 −−− Ξ′Ξ′+′+′=
( ) ( )∑=
− ΞΞ+−−=T
t
t t t t H H TN
1
1'log
2
12log
2
π θ
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28. An Example: Estimating a Time-Varying Hedge
Ratio for FTSE Stock Index Returns(Brooks, Henry and Persand, 2002).
• Data comprises 3580 daily observations on the FTSE 100 stock index and
stock index futures contract spanning the period 1 January 1985 - 9 April 1999.
• Several competing models for determining the optimal hedge ratio are
constructed. Define the hedge ratio as β .
– No hedge ( β =0)
– Naïve hedge ( β =1) – Multivariate GARCH hedges:
• Symmetric BEKK
• Asymmetric BEKK
In both cases, estimating the OHR involves forming a 1-step aheadforecast and computing
t
t F
t CF
t h
hOHR Ω−=
+
++
1,
1,
1
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OHR Results
In Sample
Unhedged
β = 0
Naïve Hedge
β = 1
Symmetric Time
Varying
Hedge
t F
t FC
t h
h
,
,= β
Asymmetric
Time Varying
Hedge
t F
t FC
t h
h
,
,= β
Return 0.0389
{2.3713}
-0.0003
{-0.0351}
0.0061
{0.9562}
0.0060
{0.9580}Variance 0.8286 0.1718 0.1240 0.1211
Out of Sample
Unhedged
β = 0
Naïve Hedge
β = 1
Symmetric Time
VaryingHedge
t F
t FC
t h
h
,
,= β
Asymmetric
Time VaryingHedge
t F
t FC
t h
h
,
,= β
Return 0.0819
{1.4958}
-0.0004
{0.0216}
0.0120
{0.7761}
0.0140
{0.9083}Variance 1.4972 0.1696 0.1186 0.1188
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Plot of the OHR from Multivariate GARCH
Conclusions
- OHR is time-varying and less
than 1
- M-GARCH OHR provides a
better hedge, both in-sample
and out-of-sample.
- No role in calculating OHR for
asymmetries (not much
improvement)
Symmetric BEKK
Asymmetric BEKK
Time Varying Hedge Ratios
500 1000 1500 2000 2500 30000.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
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Homework
• Exercises: 1abcefgh, 2abc, 3ab