university of copenhagen department of economics

Econometrics II
Autoregressive Conditional
Heteroskedasticity (ARCH) Models
Morten Nyboe Tabor
university of copenhagen department of economics

Learning Outcomes

After completing this topic, you should be able to:

1 Give a precise definition and interpretation of the concept of
autoregressive conditional heteroskedasticity (ARCH).

2 Give an account of statistical models with ARCH and GARCH in financial

time series.

3 Explain the conditions for stationarity of ARCH and GARCH models.

4 Explain how to estimate the parameters in ARCH models using maximum

likelihood estimation.

5 Construct misspecification tests for no ARCH-effects.

6 Explain how ARCH models can be extended to allow for asymmetries and
explanatory variables in the conditional variance.
7 Construct forecasts of the conditional variance in ARCH and GARCH
models.

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Course Outline: ARCH Models

1 Introduction
2 ARCH Models
Definition
Properties
3 ARCH: Examples
4 Misspecification Test for No-ARCH
5 ARCH: Maximum Likelihood Estimation
6 GARCH Models
Definition
Example
7 Extensions to the Basic Model
Asymmetric ARCH Models and the News Impact Curve
(G)ARCH in Mean
8 GARCH with t-distributed innovations
9 Properties of the MLE

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1. Introduction
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Introduction

• Financial economists are typically interested in both the mean and the
variance. This reflects the trade-off between return and risk.
The conditional variance is a measure of ‘unexpected variation’ = risk.

• A stylized fact for financial time series is a non-constant variance

(volatility):
“...large changes tend to be followed by large changes, of either sign, and
small changes tend to be followed by small changes.”
Mandelbrot (1963).
Known as volatility clustering. ”Risk is time-varying.”

• ARCH and GARCH models are approaches to modelling this feature.

Specify equations for the conditional mean and the conditional variance.

• In a broader perspective, non-linear time series models (such as

ARCH/GARCH) are typically needed for the study of economic and
financial time series.

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Example I: Price of IBM Stock

Example I: Price of IBM Stock
(A) IBM stock, percent month-on-month (B) Squared returns

25
750

0 500

250
-25

1940 1960 1980 2000 1940 1960 1980 2000

(C) ACF - Returns (D) ACF - Squared returns

1.0 1.00

0.5 0.75

0.0 0.50

-0.5 0.25

-1.0 0.00
0 5 10 0 5 10
6 of 21

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2. ARCH Models
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ARCH Model Defined

• Consider an equation for the conditional mean:

yt = xt0 θ + t , t = 1, 2, ..., T . (∗)

Often xt contains lags of yt and dummies for special features of the

market.

• The ARCH model also specifies an equation for the conditional variance.
Consider the information set
It−1 = {t−1 , t−2 , ...} = {yt−1 , xt−1, yt−2 , xt−2 , ...}.
Assuming that E [t |It−1 ] = 0, we define σt2 ≡ E (2t | It−1 ).

• An ARCH(1) model uses

σt2 = $ + α2t−1 . (∗∗)

To ensure that σt2 > 0, we need $ > 0, α ≥ 0.

If 2t−1 is high, the variance of the next shock, t , is large.

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ARCH Model Defined

• The model:

yt = xt0 θ + t , t = 1, 2, ..., T .
E [t |It−1 ] = 0
E [2t |It−1 ] = $ + α2t−1 (= σt2 )

• If xt ∈ It−1 , Var(yt |It−1 ) = σt2 .

• Conditional on It−1 , t ∼ (0, σt2 ).

Alternatively, we may write the model as

t = σt zt , zt ∼ i.i.d.(0, 1),

with zt independent of It−1 .

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ARCH Model: Properties

• The error term is

t = σt zt , zt ∼ i.i.d.(0, 1),

with zt independent of It−1 . Suppose that zt ∼ N(0, 1).

• It holds that

E [t |It−1 ] = 0
E [2t |It−1 ] = $ + α2t−1 (= σt2 )

So t is conditionally heteroskedastic.
In fact, t |It−1 ∼ N(0, σt2 ), so the error term is conditionally normal.
• Moreover, E [t ] = 0, and if 0 ≤ α < 1,
$
E [2t ] = .
1−α
So t is unconditionally homoskedastic!
Typically, the unconditional distribution of t is heavy-tailed, i.e.
non-normal.

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ARCH Model: The Unconditional Distribution

p
Histogram of t = $ + α2t−1 zt , with zt ∼ i.i.d.N(0, 1), $ = 1, α = 0.7.
Note: E (2t ) = 1−α
$
= 3.33...

Density

0.3 x, α= 0.7 N(s=1.91)

0.2

0.1

-17.5 -15.0 -12.5 -10.0 -7.5 -5.0 -2.5 0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5
Density
0.02

x, α= 0.7 N(s=1.91)

0.01

-14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4

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ARCH Model: The Unconditional Distribution

... and we observe something similar in practice!

Density
0.075 ibm N(s=6.72)

0.050

0.025

-35 -30 -25 -20 -15 -10 -5 0 5 10 15 20 25 30 35

Density
0.015
ibm N(s=6.72)

0.010

0.005

-35.0 -32.5 -30.0 -27.5 -25.0 -22.5 -20.0 -17.5 -15.0 -12.5 -10.0

Econometrics II — ARCH Models — Slide 12/49

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ARCH Model Defined: ARCH model as an AR process for 2t

• It is useful to consider the decomposition

2t = E (2t | It−1 ) + vt

where vt = 2t − σt2 is the (uncorrelated) surprise in the squared

innovations.
Now use σt2 = 2t − vt to rewrite σt2 = $ + α2t−1 as

2t =

The squared innovation 2t follows an process.

• Generalizes to an ARCH(p) model:

σt2 = $ + α1 2t−1 + α2 2t−2 + ... + αp 2t−p ,

where 2t follows an AR(p).

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3. ARCH: Examples
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Example II: Danish Stock Market Index (KFX)

Example II: Danish Stock Market Index (KFX)
(A) KFX stock index (log) (B) Log return on KFX stock index
0.1

5.5

0.0
5.0

-0.1
4.5

1994 1996 1998 2000 2002 2004 1994 1996 1998 2000 2002 2004

(C) Squared returns (D) ACF- Squared returns

1.0

0.015
0.5

0.010
0.0

0.005 -0.5

-1.0
1994 1996 1998 2000 2002 2004 0 5 10
7 of 21

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Example III: Danish NEER, 1990-2005

Example III: Danish NEER, 1990-2005
(A) Nominal effective exchange rate (log) (B) Day-to-day change (log)
0.10
0.02

0.05
0.00

0.00 -0.02

-0.05 -0.04
0 600 1200 1800 2400 3000 3600 0 600 1200 1800 2400 3000 3600

(C) Squared change (D) ACF - Squared change

0.0003 1.00
Truncated; true
value is 0.0013
0.75
0.0002

0.50

0.0001
0.25

0.0000 0.00
0 600 1200 1800 2400 3000 3600 0 5 10
8 of 21

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4. Misspecification Test for No-ARCH
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Misspecification Test for No-ARCH

• Use the Breusch-Pagan LM test for heteroskedasticity.

• To test for ARCH of order p use the AR(p) implication and consider the
auxiliary regression model

2t = β0 + β1 2t−1 + β2 2t−2 + ... + βp 2t−p + error .

Under the null of no ARCH,

H0 : β1 = β2 = ... = βp = 0.

The hypothesis can be tested using the familiar statistic

T · R 2 → χ2 (p).

• The ARCH test does also have power against residual autocorrelation.
Always test for autocorrelation first.

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Example: IBM Stock Returns

• Consider a linear regression

yt = 1.149 + 0.076yt−1 + b
t .
(5.01) (2.26)

Note that yt−1 is borderline significant. R 2 = 0.00574 is very low.

There are no signs of autocorrelation in bt .

• To test for the presence of ARCH effects, consider the auxiliary regression:

2t = 25.804+0.153b
b 2t−1 +0.100b
2t−2 +0.048b
2t−3 +0.110b
2t−4 +0.017b
2t−5 +resid.
(6.74) (4.53) (2.93) (1.39) (3.18) (0.484)

We note that many of the lags are significant. The coefficient of

determination is R 2 = 0.06939 and the LM test statistic is

ξARCH = TR 2 = 882 · 0.06939 = 61.2,

which is highly significant in the asymptotic χ2 (5) distribution.

Clear signs of ARCH effects.

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5. ARCH: Maximum Likelihood Estimation
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Maximum Likelihood Estimation

• Consider again the model:

yt = xt0 θ + t , t = 1, 2, ..., T .
E [t |It−1 ] = 0
E [2t |It−1 ] = $ + α2t−1 (= σt2 )

• In the presence of ARCH, the OLS estimator for θ is consistent but

inefficient.
There exists a non-linear estimator that takes the ARCH structure into
account.

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Maximum Likelihood Estimation

• Consider the ARCH(1) case

yt = xt0 θ + t
σt2 = $ + α2t−1 .

• Assume conditional normality:

t = yt − xt0 θ = σt zt , zt ∼ N(0, 1).

We specify the normal likelihood contribution as

 
1 2t
Lt (θ, $, α | yt , xt , It−1 ) = p exp − ,
2πσt2 2σt2
 
1 (yt − xt0 θ)2
= p exp − 0
,
0
2π($ + α[yt−1 − xt−1 θ]2 ) 2($ + α[yt−1 − xt−1 θ]2 )

and maximize with respect to θ, $ and α.

Note: The likelihood equations cannot be solved analytically.
• Other (typically fat-tailed distributions) can also be used.
Often a t(v ) distribution where v is estimated.
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Example: IBM Stock Returns

• Using a normal likelihood, we estimate the AR(1)-ARCH(5) model:

yt = 1.198 + 0.102yt−1 + t
(6.12) (2.83)

σt2 = 24.838 + 0.1342t−1 + 0.0982t−2 + 0.1002t−3 + 0.0602t−4 + 0.0552t−5 .

(7.80) (2.61) (1.52) (2.13) (1.26) (1.67)

Log-likelihood is −2909.10.

• Many lags are borderline significant.

Difficult to precisely pin down the shape of the memory structure.
Estimated coefficients are often relatively unstable between models.

• Misspecification tests can be carried out by considering the standardized

residuals:
yt − xt0 θ̂
ẑt = .
σ̂t
By the model assumptions, zt ∼ i.i.d.N(0, 1), so we could test if ẑt is not
autocorrelated, have no-ARCH, and is normally distributed.

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6. GARCH Models
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GARCH Defined
• The popular GARCH(1,1) model is defined by

σt2 = $ + α1 2t−1 + β1 σt−1

2
.

For σt2 to be non-negative we require the coefficients to be non-negative.

• Using the definition σt2 = 2t − vt , we get that

which is an model for the squared innovation.

Stationarity requires that
• Generalizes to a GARCH(p,q) model:
p q
X X
σt2 =$+ αj 2t−j + 2
βj σt−j .
j=1 j=1

The GARCH model is equivalent to an (restricted) infinite ARCH model.

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GARCH(1,1) as an ARCH(∞)

Suppose that α + β < 1.

Then we can rewrite σt2 :

σt2 = $ + α2t−1 + βσt−1

2

= $ + α2t−1 + β($ + α2t−2 + βσt−2

2
)
= (1 + β)$ + α2t−1 + βα2t−2 + β 2 σt−2
2

..
. ! !
∞ ∞
X X
σt2 = βi $+ βi α 2t−1−i .
i=0 i=0

I.e. the GARCH(1,1) is an ARCH(∞) with a restricted parameter structure.

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Example: IBM Stock Returns

• Now estimate the GARCH(1,1) model:

yt = 1.179 + 0.104yt−1 + t
(6.00) (2.86)

σt2 = 2.932 + 0.0972t−1 + 0.837σt−1

2
.
(2.10) (3.12) (16.4)

Log-likelihood is −2901.02.

• Note that the lagged variance is large and significant.

• Note that α + β = 0.934 ≈ 1, which is often the case empirically.

α + β = 1 is known as the IGARCH model.
Could be due to misspecification.
A likelihood-ratio test can be carried out to test for IGARCH.
P5
• Note that α = 0.45 for the ARCH(5) and α + β = 0.93 for the
i=1 i
GARCH(1,1).
Very different models.

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Example:
Example: IBMReturns,
IBM Stock StockGARCH(1,1)
Returns, GARCH(1,1)
Residual and conditional confidence bands
Residual
95% confidence band
20

10

−10

−20

−30
1930 1940 1950 1960 1970 1980 1990 2000

15 of 21

Residual: ˆt .
Confidence bands: [−1.96σ̂t , 1.96σ̂t ].
Econometrics II — ARCH Models — Slide 28/49
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Forecasting σt2

Recall that σt2 = $ + α2t−1 + βσt−1

2
, t = 1, ..., T .

Forecasts of σt2 for t > T given IT are:

E [σT2 +1 |IT ] = E [$ + α2T + βσT2 |IT ] = $ + α2T + βσT2

E [σT2 +2 |IT ] = E [$ + α2T +1 + βσT2 +1 |IT ] = $ + (α + β)E [σT2 +1 |IT ]
..
.
E [σT2 +h |IT ] = $ + (α + β)E [σT2 +h−1 |IT ]
h−1
X
= (α + β)i $ + (α + β)h−1 (α2T + βσT2 ).
i=0

2 $
• If (α + β) < 1, then E [σT +h |IT ] → as h → ∞.
1−α−β
I.e. the forecast of the conditional variance tends to the unconditional
variance. (Similar to forecasting stationary AR models.)
2
• In practice we replace ($, α, β, T , σT ) with the estimated quantities
($̂, α̂, β̂, ˆT , σ̂T2 )

Econometrics II — ARCH Models — Slide 29/49

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Example: IBM Stock Returns

Example: IBM Stock Returns
Conditional standard deviation and forecast

14 Unconditional standard deviation

GARCH(1,1)
ARCH(5)
12

10

1990 1992 1994 1996 1998 2000 2002 2004

16 of 21

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7. Extensions to the Basic Model
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Threshold Asymmetric Model

• Sometimes we believe that negative shocks have a different impact.

• To model this we could suggest

σt2 = $ + α2t−1 + κ2t−1 I(t−1 < 0) + βσt−1

2
,

where I(t−1 < 0) = 1 if t−1 < 0 and zero otherwise.

• For the IBM stock return we get a threshold model given by

yt = 1.114 + 0.107yt−1 + t
(5.35) (3.07)

σt2 = 3.228 + 0.0642t−1 + 0.0632t−1 I(t−1 < 0) + 0.831σt−1

2
.
(2.20) (2.48) (1.16) (16.2)

Log-likelihood is −2899.308.
Little support for asymmetry.

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News Impact Curve

Plot of (t−1 , σt2 ). News Impact Curve
News Impact Curve
100
Basic GARCH model
90 Threshold model
80
70
60
50
40
30
20
10

−30 −25 −20 −15 −10 −5 0 5 10 15 20 25

18 of 21

Econometrics II — ARCH Models — Slide 33/49

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Asymmetric Model

• An alternative form of asymmetry is

σt2 = $ + α(t−1 − γ)2 + βσt−1

2
,

where zero is no-longer the neutral shock.

• For the IBM stock return we get a threshold model given by

yt = 1.125 + 0.107yt−1 + t
(5.32) (3.01)

σt2 = 2.988 + 0.096(t−1 − 1.332)2 + 0.832σt−1

2
.
(2.30) (3.58) (0.298) (17.0)

Log-likelihood is −2899.928.
Again little support for asymmetry.

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News Impact Curve

Plot of (t−1 , σt2 ). News Impact Curve
News Impact Curve

100 Basic GARCH model

Asymmetric model

80

60

40

20

−30 −25 −20 −15 −10 −5 0 5 10 15 20 25

20 of 21

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(G)ARCH in Mean

• Another interesting extension is where the volatility, measured by σt2 ,

affects the conditional mean, i.e.

yt = xt0 θ + δσt2 + t
or yt = xt0 θ + δσt + t .

This is often interpreted as the presence of a risk premium.

• We estimate the GARCH-in-mean model for the IBM stock return and
obtain

yt = 0.948 + 0.104yt−1 + 0.0060σt2 + t

(1.49) (2.92) (0.377)

σt2 = 2.894 + 0.0962t−1 2

+ 0.839σt−1 .
(2.24) (3.26) (16.9)

The log-likelihood of this specification is −2900.930.

No support for a risk premium.

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8. GARCH with t-distributed innovations
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GARCH with t-distributed innovations

Consider again the GARCH(1,1) model,

yt = xt0 θ + t
t = σt zt , zt ∼ IID(0, 1)
σt2 = $ + α2t−1 + βσt−1
2
.

Instead of assuming that zt is N(0, 1), we could assume a Student’s

t-distribution.
Student’s t-distribution
The Student’s t-distribution has density
 −( v +1
2 ) v +1


γ(v ) x2 Γ 2
f (x ) = √ 1+ , v > 0, γ(v ) := v

,
πv v Γ 2

where v is the degrees of freedom and Γ(·) is the so-called gamma function.
If X ∼ tv , then

E (X ) = 0, if v > 1,
2 v
E (X ) = , if v > 2.
v −2
Econometrics II — ARCH Models — Slide 38/49
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Student’s t-distribution

(Source: Wikipedia)

Note: v = +∞ corresponds to the standard normal distribution.

Econometrics II — ARCH Models — Slide 39/49
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GARCH with t-distributed innovations

With t = σt zt , and assuming that v > 2,

v
Var(t |It−1 ) = σt2
v −2
In order to preserve the interpretation of σt2 being the conditional variance, let
r
v −2
zt := z̃t , where z̃t ∼ tv ,
v

such that E (zt2 ) = 1.

We say that zt has a scaled Student’s t-distribution, denoted tv (0, 1).

It holds that zt has pdf

 (− v +1
2 )
γ(v ) z2
f (z) = 1+ .
(v − 2)
p
(v − 2)π

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Maximum Likelihood Estimation

We consider estimation with the likelihood function based on tv (0, 1):

T
X
L(θ, $, α, β, v ) = log Lt (θ, $, α, β, v ),
t=1
 −( v +1
2 )
γ(v ) 2t (θ)
Lt (θ, $, α, β, v ) = 1+ ,
(v − 2)σt2 ($, α, β)
p
(v − 2)πσt2 ($, α, β)
t (θ) = yt − xt0 θ,
σt2 ($, α, β) = $ + α2t−1 (θ) + βσt−1
2
($, α, β),

for some initial value σ02 ($, α, β) - typically, the sample variance of yt .

Note that v is a model parameter.

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Example: IBM Stock Returns

• Now estimate the GARCH(1,1) model based on the tv (0, 1) distribution:

yt = 1.226 + 0.071yt−1 + t
(6.17) (1.93)

σt2 = 2.536 + 0.0932t−1 + 0.850σt−1

2
.
(2.60) (3.88) (22.8)

Estimate of v is 8.41.
Log-likelihood is −2890.04.

• Note that the autoregressive coefficient is no longer significant.

• The estimates of $, α, and β are very similar to the N(0, 1) case.

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9. Properties of the MLE
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Introduction

• We consider an econometric model.

The model parameters are θ ∈ Θ ⊂ RK , where Θ denotes the parameter
space.

• Given a set of observations y1 , ..., yT , the log-likelihood function is

T
X
L(θ) = log Lt (θ),
t=1

where Lt (θ) denotes the likelihood contribution.

• The Maximum Likelihood Estimator (MLE) satisfies

θ̂ = arg max L(θ).

θ∈Θ

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Example: ARCH(1)
Econometric model:
yt = µ + t , t = 1, ..., T
t = σt zt , zt ∼ IIDN(0, 1)
σt2 = $ + α2t−1 ,
with some initial value 0 .
Parameters and parameter space:
θ = (µ, $, α)0 ,
Θ = R × (0, ∞) × [0, ∞).
Log-likelihood function:
T
X
L(θ) = log Lt (θ),
t=1
 
1 2 (θ)
Lt (θ) = p exp − t2 ,
2πσt2 (θ) 2σt (θ)
t (θ) = yt − µ,
σt2 (θ) = $ + α2t−1 (θ).
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Consistency

Assume:
1 The true DGP is contained in the statistical model.
Let θ0 denote the true parameters generating the data.

2 Identification: L(θ∗ ) 6= L(θ0 ) for all θ∗ 6= θ0 .

3 The data is stationary and weakly dependent such that a LLN applies.

Then (under some additional regularity conditions),

θ̂ → θ0 (in probability, as T → ∞),

i.e. the MLE is consistent.

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Asymptotic Normality
Suppose that θ0 is inside Θ, and consider a Taylor expansion of the score at θ0 :
∂L(θ̂) ∂L(θ0 ) ∂ 2 L(θ0 )
0 = = + (θ̂ − θ0 ) + RT ,
∂θ ∂θ ∂θ∂θ0
where RT denotes a remainder term.
Assume in addition:
1 The data is stationary and weakly dependent, such that a CLT applies to
the score:
1 ∂L(θ0 )
√ → N(0, J) (in distribution),
T ∂θ
where J is positive definite.
2 The data is stationary and weakly dependent, such that a LLN applies to
the Hessian:
1 ∂ 2 L(θ0 )
− → J (in probability).
T ∂θ∂θ0

3 The remainder term, RT , vanishes as T → ∞.

Then
√
T (θ̂ − θ0 ) → N(0, J −1 ).

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Asymptotic Normality

• Typically, the asymptotic covariance matrix, J, is unknown.

It can be estimated consistently based on

1 ∂ 2 L(θ̂)
ĴH = − ,
T ∂θ∂θ0
or
T
1 X ∂ log Lt (θ̂) ∂ log Lt (θ̂)
ĴS = .
T ∂θ ∂θ0
t=1

• More details are given in the MSc course Financial Econometrics A, with
particular emphasis on ARCH models.
Also some details in the MSc course Advanced Microeconometrics.

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The Likelihood Ratio (LR) Test

Consider a null hypothesis of interest, H0 , and an alternative, HA . E.g.,

H0 : R θ = q against HA : Rθ 6= q.
(J×K )

Let θe and θb denote the ML estimates under H0 and HA , respectively.

The likelihood ratio (LR) test is based on estimating the model under H0 and
under HA and look at the loss in likelihood, L(θb) − L(θe).
The test is only appropriate for nested models: H0 ⊂ HA
• The LR test statistic is given by
 
ξˆLR = −2 · L(θe) − L(θb) ,

where L(θe) and L(θb) are the two log-likelihood values.

Under the null, this is asymptotically distributed as ξˆLR → χ2 (J) .

Econometrics II — ARCH Models — Slide 49/49