1) A sample of 10 observations is selected from a normal populatio.docx

1) A sample of 10 observations is selected from a normal
population for which the population standard deviation is known
to be 6. The sample mean is 23. (Round your answers to 3
decimal places.)
A) The standard error of the mean is ________
B) The 99 percent confidence interval for the
population mean is between _______and ________
2) The owner of Britten's Egg Farm wants to estimate the mean
number of eggs laid per chicken. A sample of 20 chickens
shows they laid an average of 20 eggs per month with a standard
deviation of 2.63 eggs per month (Round your answer to 3
decimal places.)
A) What is the best estimate of this value?
B) For a 99 percent confidence interval, the value of t is
______
C) The 99 percent confidence interval for the population mean
is _______to ________
3) As a condition of employment, Fashion Industries applicants
must pass a drug test. Of the last 230 applicants 26 failed the
test.
A) Develop a 90 percent confidence interval for the proportion
of applicants that fail the test. (Round your answers to 3
decimal places.)
For the applicants the confidence interval is between _______
and _______
B) Would it be reasonable to conclude that more than 11 percent
of the applicants are now failing the test? Yes or No
C) In addition to the testing of applicants, Fashion Industries
randomly tests its employees throughout the year. Last year in
the 520 random tests conducted, 22 employees failed the test.

Would it be reasonable to conclude that less than 6 percent of
the employees are not able to pass the random drug test? Yes or
No
4) A sample of 48 observations is selected from a normal
population. The sample mean is 22, and the population standard
deviation is 6.
Conduct the following test of hypothesis using the .05
significance level.
H0 : μ ≤ 21
H1 : μ > 21
A)
Is this a one- or two-tailed test?
B)
What is the decision rule? (Round your answer to 2 decimal
places.)
H0 and H1 when z >
C)
What is the value of the test statistic? (Round your answer to 2
decimal places.)
Value of the test statistic
D)
What is your decision regarding H0?

There is evidence to conclude that the population mean is
greater than 21.
E)
What is the p-value? (Round your answer to 4 decimal places.)
5) Most air travelers now use e-tickets. Electronic ticketing
allows passengers to not worry about a paper ticket, and it costs
the airline companies less to handle than paper ticketing.
However, in recent times the airlines have received complaints
from passengers regarding their e-tickets, particularly when
connecting flights and a change of airlines were involved. To
investigate the problem an independent watchdog agency
contacted a random sample of 20 airports and collected
information on the number of complaints the airport had with e-
tickets for the month of March. The information is reported
below.
14
14
16
12
12
14
13
16
15
14

12
15
15
14
13
13
12
13
10
13
At the .05 significance level can the watchdog agency conclude
the mean number of complaints per airport is less than 15 per
month? Conduct a test of hypothesis and interpret the results.
Click here for the Excel Data File
(a)
What assumption is necessary before conducting a test of
hypothesis?
(b)
What is the decision rule? (Negative amount should be indicated
by a minus sign. Round your answer to 3 decimal places.)
Reject H0 : μ ≥ 15 and accept H1 : μ < 15 when the test
statistic is .
The value of the test statistic is . (Negative amount should be
indicated by a minus sign. Round your answer to 2 decimal
places.)

(c)
What is your decision regarding H0 ?
6)
The liquid chlorine added to swimming pools to combat algae
has a relatively short shelf life before it loses its effectiveness.
Records indicate that the mean shelf life of a 5-gallon jug of
chlorine is 2,160 hours (90 days). As an experiment, Holdlonger
was added to the chlorine to find whether it would increase the
shelf life. A sample of nine jugs of chlorine had these shelf
lives (in hours):
2,159
2,170
2,180
2,179
2,160
2,167
2,171
2,181
2,185
At the .025 level, has Holdlonger increased the shelf life of the
chlorine?
Click here for the Excel Data File
A)
What is the decision rule? (Round your answer to 3 decimal
places.)
Reject H0 : μ ≤ 2,160 and accept H1 : μ > 2,160 when the test
statistic is
.

B)
The value of the test statistic is . (Round your answer to 2
decimal places.)
C)
What is your decision regarding H0 ?
D) The p-value is
7)
The following hypotheses are given.
H0 : π ≤ 0.82
H1 : π > 0.82
A sample of 100 observations revealed that p=0.93. At the 0.02
significance level, can the null hypothesis be rejected?
A)
State the decision rule. (Round your answer to 2 decimal
places.)
H0 and H1 if z >
B)
Compute the value of the test statistic. (Round your answer to 2
decimal places.)
C)
What is your decision regarding the null hypothesis?
H0.

8) Tina Dennis is the comptroller for Meek Industries. She
believes that the current cashflow problem at Meek is due to the
slow collection of accounts receivable. She believes that more
than 60 percent of the accounts are in arrears more than three
months. A random sample of 200 accounts showed that 140
were more than three months old. At the .01 significance level,
can she conclude that more than 60 percent of the accounts are
in arrears for more than three months?
A)
State the null hypothesis and the alternate hypothesis. (Round
your answers to 2 decimal places.)
H0: π ≤
H1: π >
B)
State the decision rule for .01 significance level. (Round your
answer to 2 decimal places.)
Reject H0 if z is >
C)
decimal places.)
z value is

D)
At the .01 significance level, can she conclude that more than
60 percent of the accounts are in arrears for more than three
months?
H0. Ms. Dennis is in concluding that more than 60% of the
accounts are more than 3 months old.
9) The cost of weddings in the United States has skyrocketed in
recent years. As a result many couples are opting to have their
weddings in the Caribbean. A Caribbean vacation resort
recently advertised in Bride Magazine that the cost of a
Caribbean wedding was less than $10,000. Listed below is a
total cost in $000 for a sample of 8 Caribbean weddings. At the
.005 significance level is it reasonable to conclude the mean
wedding cost is less than $10,000 as advertised?
9.5
9.1
10.8
9.7
8.5
9.4
8.6
9.4
A)
State the null hypothesis and the alternate hypothesis. Use a
.005 level of significance. (Enter your answers in thousands of
dollars.)
H0: μ ≥
H1: μ <

B)
State the decision rule for .005 significance level. (Round your
answer to 3 decimal places.)
Reject H0 if t <
C)
decimal places.)
D)
At the .005 significance level is it reasonable to conclude the
mean wedding cost is less than $10,000 as advertised?
H0. The cost is than $10,000.
(Click to select)
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Lecture Note of Bus 41202, Spring 2012:
More Volatility Models. Mr. Ruey Tsay
The GARCH-M model
rt = μ + cσ
2
t + at, at = σt�t, σ
2
t = α0 + α1a

2
t−1 + β1σ
2
t−1
where c is referred to as risk premium, which is expected to be
posi-
tive.
Example: A GARCH(1,1)-M model for the monthly excess
returns
of S&P 500 index from January 1926 to December 1991. For
numer-
ical stability, I use percentage returns. The fitted model is
rt = 0.743 + 0.048σ
2
t + at, σ
2
t = 0.812 + 0.123a
2
t−1 + .854σ
2
t−1.
Std err of risk premium is 0.141 so that the estimate is not
statistically

significant at the usual 5% level.
R demonstration
> source("garchM.R")
> mn=garchM(sp5*100)
[1] 2399.822
0: 2380.0229: 0.422452 0.00561297 0.806149 0.121976
0.854361
3: 2379.9525: 0.427596 0.00652832 0.806107 0.121466
0.855756
6: 2378.4838: 0.606807 0.0375055 0.801889 0.126193 0.846833
9: 2377.9587: 0.673997 0.00166937 0.798191 0.125517
0.851915
12: 2377.8474: 0.692209 0.0444464 0.802574 0.122278
0.854141
15: 2377.7922: 0.742796 0.0480959 0.812187 0.122530
0.853507
Maximized log-likehood: 2377.792
Coefficient(s):
Estimate Std. Error t value Pr(>|t|)
mu 0.7427956 0.1540336 4.82229 1.4192e-06 ***
gamma 0.0480959 0.1408765 0.34140 0.7327991
omega 0.8121873 0.2858128 2.84168 0.0044877 **

alpha 0.1225297 0.0220596 5.55449 2.7843e-08 ***
beta 0.8535072 0.0219079 38.95885 < 2.22e-16 ***
---
1
Remarks: This R script is relatively slow. It takes longer time
due
to its use of recursive loop in evaluating likelihood function.
The EGARCH model
Asymmetry in responses to past positive and negative returns:
g(�t) = θ�t + γ[|�t|− E(|�t|)],
with E[g(�t)] = 0.
To see asymmetry of g(�t), rewrite it as
g(�t) =
⎧ ⎪ ⎪ ⎨
⎪ ⎪ ⎩
(θ + γ)�t − γE(|�t|) if �t ≥ 0,
(θ − γ)�t − γE(|�t|) if �t < 0.
An EGARCH(m, s) model:
at = σt�t, ln(σ

2
t ) = α0 +
1 + β1B + · · · + βs−1Bs−1
1 − α1B −···− αmBm
g(�t−1).
Some features of EGARCH models:
• uses log trans. to relax the positiveness constraint
• asymmetric responses
Consider an EGARCH(1,1) model
at = σt�t, (1 − αB) ln(σ2t ) = (1 − α)α0 + g(�t−1),
Under normality, E(|�t|) =
√
2/π and the model becomes
(1 − αB) ln(σ2t ) =
⎧ ⎪ ⎪ ⎨
⎪ ⎪ ⎩
α∗ + (θ + γ)�t−1 if �t−1 ≥ 0,
α∗ + (θ − γ)�t−1 if �t−1 < 0
2
where α∗ = (1 − α)α0 −
√
2
π
γ.

This is a nonlinear fun. similar to that of the threshold AR
model of
Tong (1978, 1990).
Specifically, we have
σ2t = σ
2α
t−1 exp(α∗ )
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
exp[(θ + γ)
at−1√
σ2t−1
] if at−1 ≥ 0,
exp[(θ − γ) at−1√
σ2t−1
] if at−1 < 0.
The coefs (θ + γ) & (θ − γ) show the asymmetry in response to
positive and negative at−1. The model is, therefore, nonlinear if
θ �= 0. Thus, θ is referred to as the leverage parameter.
Focus on the function g(�t−1). The leverage parameter θ shows
the
effect of the sign of at−1 whereas γ denotes the magnitude
effect.
See Nelson (1991) for an exmaple of EGARCH model.

Another example: Monthly log returns of IBM stock from Jan-
uary 1926 to December 1997 for 864 observations.
For textbook, an AR(1)-EGARCH(1,1) is obtained (RATS
program):
rt = 0.0105 + 0.092rt−1 + at, at = σt�t
ln(σ2t ) = −5.496 +
g(�t−1)
1 − .856B,
g(�t−1) = −.0795�t−1 + .2647[|�t−1|−
√
2/π],
Model checking:
For ãt: Q(10) = 6.31(0.71) and Q(20) = 21.4(0.32)
For ã2t: Q(10) = 4.13(0.90) and Q(20) = 15.93(0.66)
3
Discussion:
Using
√
2/π ≈ 0.7979 ≈ 0.8, we obtain
ln(σ2t ) = −1.0 + 0.856 ln(σ2t−1) +

⎧ ⎪ ⎪ ⎨
⎪ ⎪ ⎩
0.1852�t−1 if �t−1 ≥ 0
−0.3442�t−1 if �t−1 < 0.
Taking anti-log transformation, we have
σ2t = σ
2×0.856
t−1 e
−1.001 ×
⎧ ⎪ ⎪ ⎨
⎪ ⎪ ⎩
e0.1852�t−1 if �t−1 ≥ 0
e−0.3442�t−1 if �t−1 < 0.
For a standardized shock with magnitude 2, (i.e. two standard
devi-
ations), we have
σ2t (�t−1 = −2)
σ2t (�t−1 = 2)
=
exp[−0.3442 × (−2)]
exp(0.1852 × 2) = e
0.318 = 1.374.
Therefore, the impact of a negative shock of size two-standard
de-
viations is about 37.4% higher than that of a positive shock of
the

same size.
Forecasting: some recursive formula available
Another parameterization of EGARCH models
ln(σ2t ) = α0 + α1
|at−1| + γ1at−1
σt−1
+ β1 ln(σ
2
t−1),
where γ1 denotes the leverage effect.
Below, I re-analyze the IBM log returns by extending the data
to
December 2009. The sample size is 1008.
The fitted model is
rt = 0.012 + at, at = σt�t
4
ln(σ2t ) = −0.611 +
0.231|at−1|− 0.250at−1
σt−1
+ 0.92 ln(σ2t−1).

Since EGARCH and TGARCH (below) share similar objective
and
the latter is easier to estimate. We shall use TGARCH model.
The Threshold GARCH (TGARCH) or GJR Model A
TGARCH(s, m) or GJR(s, m) model is defined as
rt = μt + at, at = σt�tσ
2
t = α0 +
s∑
i=1
(αi + γiNt−i)a
2
t−i +
m∑
j=1
βjσ
2
t−j,
where Nt−i is an indicator variable such that
Nt−i =
⎧ ⎪ ⎪ ⎨
⎪ ⎪ ⎩
1 if at−i < 0,

0 otherwise.
One expects γi to be positive so that prior negative returns have
higher impact on the volatility.
The Asymmetric Power ARCH (APARCH) Model.
This model was introduced by Ding, Engle and Granger (1993)
as a
general class of volatility models. The basic form is
rt = μt + at, at = σt�t, �t ∼ D(0, 1)
σδt = ω +
s∑
i=1
αi(|at−i|− γiat−i)δ +
m∑
j=1
βjσ
δ
t−j
where δ is a non-negative real number. In particular, δ = 2 gives
rise
to the TGARCH model and δ = 0 corresponds to using log(σt).
Theoretically, one can use any power δ to obtain a model. In
practice,
two things deserve further consideration. First, δ will also affect

the
5
specification of the mean equation, i.e., model for μt. Second, it
is
hard to interpret δ, except for some special values such as 0, 1,
2.
In R, one can fix the value of δ a priori using the subcommand
include.delta=F, delta = 2.
Here I pre-fix δ = 2. Thus, we can use APARCH model to
estimate
TGARCH model. Consider the percentage log returns of
monthly
IBM stock from 1926 to 2009.
R demonstration
> da=read.table("m-ibm2609.txt",header=T)
> head(da)
date ibm
1 19260130 -0.010381
.....

6 19260630 0.068493
> ibm=log(da$ibm+1)*100
>
m1=garchFit(~aparch(1,1),data=ibm,trace=F,delta=2,include.del
ta=F)
> summary(m1)
Title:
GARCH Modelling
Call:
garchFit(formula = ~aparch(1, 1), data = ibm, delta = 2,
include.delta = F,
trace = F)
Mean and Variance Equation:
data ~ aparch(1, 1)
[data = ibm]
Conditional Distribution: norm
Coefficient(s):
mu omega alpha1 gamma1 beta1
1.18659 4.33663 0.10767 0.22732 0.79468
Std. Errors: based on Hessian

Error Analysis:
mu 1.18659 0.20019 5.927 3.08e-09 ***
omega 4.33663 1.34161 3.232 0.00123 **
6
alpha1 0.10767 0.02548 4.225 2.39e-05 ***
gamma1 0.22732 0.10018 2.269 0.02326 *
beta1 0.79468 0.04554 17.449 < 2e-16 ***
---
Log Likelihood:
-3329.177 normalized: -3.302755
Standardised Residuals Tests:
Statistic p-Value
Jarque-Bera Test R Chi^2 67.07416 2.775558e-15
Shapiro-Wilk Test R W 0.9870142 8.597234e-08
Ljung-Box Test R Q(10) 16.90603 0.07646942
Ljung-Box Test R Q(15) 24.19033 0.06193099

Ljung-Box Test R Q(20) 31.89097 0.04447407
Ljung-Box Test R^2 Q(10) 4.591691 0.9167342
Ljung-Box Test R^2 Q(15) 11.98464 0.6801912
Ljung-Box Test R^2 Q(20) 14.79531 0.7879979
LM Arch Test R TR^2 7.162971 0.8466584
Information Criterion Statistics:
AIC BIC SIC HQIC
6.615430 6.639814 6.615381 6.624694
> plot(m1) <= shows normal distribution is not a good fit.
>
>
m1=garchFit(~aparch(1,1),data=ibm,trace=F,delta=2,include.del
ta=F,cond.dist="std")
> summary(m1)
Title:
GARCH Modelling
Call:
garchFit(formula = ~aparch(1, 1), data = ibm, delta = 2,
cond.dist = "std",

include.delta = F, trace = F)
data ~ aparch(1, 1)
[data = ibm]
Conditional Distribution: std
Coefficient(s):
mu omega alpha1 gamma1 beta1 shape
1.20476 3.98975 0.10468 0.22366 0.80711 6.67329
Std. Errors: based on Hessian
Error Analysis:
7
mu 1.20476 0.18715 6.437 1.22e-10 ***
omega 3.98975 1.45331 2.745 0.006046 **
alpha1 0.10468 0.02793 3.747 0.000179 ***
gamma1 0.22366 0.11595 1.929 0.053738 .
beta1 0.80711 0.04825 16.727 < 2e-16 ***

shape 6.67329 1.32779 5.026 5.01e-07 ***
---
Log Likelihood:
-3310.21 normalized: -3.283938
Statistic p-Value
Ljung-Box Test R Q(10) 16.91352 0.07629962
Ljung-Box Test R Q(15) 24.08691 0.06363224
Ljung-Box Test R Q(20) 31.75305 0.04600187
Ljung-Box Test R^2 Q(10) 4.553248 0.9189583
Ljung-Box Test R^2 Q(15) 11.66891 0.7038973
Ljung-Box Test R^2 Q(20) 14.18533 0.8209764
LM Arch Test R TR^2 6.771675 0.872326
AIC BIC SIC HQIC
6.579782 6.609042 6.579711 6.590898

> plot(m1)
Make a plot selection (or 0 to exit):
1: Time Series
2: Conditional SD
3: Series with 2 Conditional SD Superimposed
4: ACF of Observations
5: ACF of Squared Observations
6: Cross Correlation
7: Residuals
8: Conditional SDs
9: Standardized Residuals
10: ACF of Standardized Residuals
11: ACF of Squared Standardized Residuals
12: Cross Correlation between r^2 and r
13: QQ-Plot of Standardized Residuals
Selection: 13
For the percentage log returns of IBM stock from 1926 to 2009,
the
8

fitted GJR model is
rt = 1.20 + at, at = σt�t, �t ∼ t∗ 6.67
σ2t = 3.99 + 0.105(|at−1|− 0.224at−1)2 + .807σ2t−1,
where all estimates are significant, and model checking
indicates that
the fitted model is adequate.
Note that, we can obtain the model for the log returns as
rt = 0.012 + at, at = σt�t
σ2t = 3.99 × 10−4 + 0.105(|at−1|− 0.224at−1)2 + .807σ2t−1.
The sample variance of the IBM log returns is about 0.005 and
the
empirical 2.5% percentile of the data is about −0.130. If we use
these
two quantities for σ2t−1 and at−1, respectively, then we have
σ2t (−)
σ2t (+)
=
0.0004 + 0.105(0.130 + 0.224 × 0.130)2 + 0.807 × 0.005
0.0004 + 0.105(0.130 − 0.224 × 0.130)2 + 0.807 × 0.005
= 1.849.
In this particular case, the negative prior return has about 85%

higher
impact on the conditional variance.
Stochastic volatility model
A (simple) SV model is
at = σt�t, (1 − α1B −···− αmBm) ln(σ2t ) = α0 + vt
where �t’s are iid N(0, 1), vt’s are iid N(0, σ
2
v), {�t} and {vt} are
independent.
9
�3 �2 �1 0 1 2 3
�
2
0
2
4
qnorm � QQ Plot
Theoretical Quantiles
S
a

m
p
le
Q
u
a
n
til
e
s
Figure 1: Normal probability plot for TGARCH(1,1) model
fitted to monthly percentage log
returns of IBM stock from 1926 to 2009
10
�4 �2 0 2 4
�
2
0
2
4
qstd � QQ Plot

S
a
m
p
le
Q
u
a
n
til
e
s
Figure 2: QQ plot for TGARCH(1,1) model fitted to monthly
percentage log returns of IBM
stock from 1926 to 2009.
11
Long-memory SV model
A simple LMSV is
at = σt�t, σt = σ exp(ut/2), (1 − B)dut = ηt
where σ > 0, �t’s are iid N(0, 1), ηt’s are iid N(0, σ
2
η) and indepen-
dent of �t, and 0 < d < 0.5.

The model says
ln(a2t) = ln(σ
2) + ut + ln(�
2
t)
= [ln(σ2) + E(ln �2t)] + ut + [ln(�
2
t) − E(ln �2t)]
≡ μ + ut + et.
Thus, the ln(a2t) series is a Gaussian long-memory signal plus a
non-
Gaussian white noise; see Breidt, Crato and de Lima (1998).
Application
see Examples 3.4 & 3.5
12
Estimation: Conditional MLE or Quasi MLE
Special Note: In this course, we estimate volatility models using
the R package fGarch with garchFit command. The program is
easy to use and allows for several types of innovational
distributions:

The default is Gaussian (norm), standardized Student-t
distribution
(std), generalized error distribution (ged), skew normal
distribution
(snorm), skew Student-t (sstd), skew generalized error
distribution
(sged), and standardized inverse normal distribution (snig).
Except
for the inverse normal distribution, other distribution functions
are
discussed in the textbook. Readers should check the book for
details
about the density functions and their parameters.
Example: Monthly log returns of Intel stock
R demonstration: The Use fGarch package.
> library(fGarch)
> da=read.table("m-intc7303.txt",header=T)
> head(da)
date rtn
1 19730131 0.01005
.....

6 19730629 0.13333
> intc=log(da$rtn+1) <== log returns
> acf(intc)
> acf(intc^2)
> pacf(intc^2)
> Box.test(intc^2,lag=10,type=’Ljung’)
Box-Ljung test
data: intc^2
X-squared = 59.7216, df = 10, p-value = 4.091e-09
> m1=garchFit(~garch(3,0),data=intc,trace=F) <== trace=F
reduces the amount of output.
> summary(m1)
Title:
GARCH Modelling
Call:
garchFit(formula = ~garch(3, 0), data = intc, trace = F)
7

data ~ garch(3, 0)
[data = intc]
Conditional Distribution:
norm
Coefficient(s):
mu omega alpha1 alpha2 alpha3
0.016572 0.012043 0.208649 0.071837 0.049045
Std. Errors:
based on Hessian
Error Analysis:
mu 0.016572 0.006423 2.580 0.00988 **
omega 0.012043 0.001579 7.627 2.4e-14 ***
alpha1 0.208649 0.129177 1.615 0.10626
alpha2 0.071837 0.048551 1.480 0.13897
alpha3 0.049045 0.048847 1.004 0.31536
---

Statistic p-Value
Jarque-Bera Test R Chi^2 169.7731 0
Ljung-Box Test R Q(10) 10.97025 0.3598405
Ljung-Box Test R Q(15) 19.59024 0.1882211
Ljung-Box Test R Q(20) 20.82192 0.40768
Ljung-Box Test R^2 Q(10) 5.376602 0.864644
Ljung-Box Test R^2 Q(15) 22.73460 0.08993976
Ljung-Box Test R^2 Q(20) 23.70577 0.255481
LM Arch Test R TR^2 20.48506 0.05844884
AIC BIC SIC HQIC
-1.228111 -1.175437 -1.228466 -1.207193
> m1=garchFit(~garch(1,0),data=intc,trace=F)
> summary(m1)
Title:
GARCH Modelling

Call:
garchFit(formula = ~garch(1, 0), data = intc, trace = F)
8
data ~ garch(1, 0)
[data = intc]
Conditional Distribution:
norm
Coefficient(s):
mu omega alpha1
0.016570 0.012490 0.363447
Std. Errors:
based on Hessian
Error Analysis:
mu 0.016570 0.006161 2.689 0.00716 **
omega 0.012490 0.001549 8.061 6.66e-16 ***

alpha1 0.363447 0.131598 2.762 0.00575 **
---
Log Likelihood:
230.2423 normalized: 0.6189309
Statistic p-Value
Ljung-Box Test R Q(10) 13.72604 0.1858587 <=== Meaning?
Ljung-Box Test R Q(15) 22.31714 0.09975386 <====
implication?
Ljung-Box Test R Q(20) 23.88257 0.2475594
Ljung-Box Test R^2 Q(10) 12.50025 0.2529700
Ljung-Box Test R^2 Q(15) 30.11276 0.01152131
Ljung-Box Test R^2 Q(20) 31.46404 0.04935483
LM Arch Test R TR^2 22.036 0.0371183
AIC BIC SIC HQIC
-1.221733 -1.190129 -1.221861 -1.209182

> plot(m1)
1: Time Series
2: Conditional SD
9
7: Residuals
8: Conditional SDs
Selection: 13

1: Time Series
2: Conditional SD
7: Residuals
8: Conditional SDs
Selection: 0
The fitted ARCH(1) model is
rt = 0.0176 + at, at = σt�t, �t ∼ N(0, 1)
σ2t = 0.0125 + 0.363σ

2
t−1.
Model checking statistics indicate that there are some higher
order
dependence in the volatility, e.g., see Q(15) for the squared
standard-
ized residuals. It turns out that a GARCH(1,1) model fares
better
for the data.
Next, consider Student-t innovations.
R demonstration
> m2=garchFit(~garch(1,0),data=intc,cond.dist="std",trace=F)
10
�3 �2 �1 0 1 2 3
�4
�2
0
2
qnorm � QQ Plot
Sa
m

ple
Q
ua
nt
ile
s
Figure 3: QQ-plot for standardized residuals of an ARCH(1)
model with Gaussian innova-
tions for monthly log returns of INTC stock: 1973 to 2003.
> summary(m2)
Title:
GARCH Modelling
Call:
garchFit(formula = ~garch(1, 0), data = intc, cond.dist = "std",
trace = F)
data ~ garch(1, 0)
[data = intc]
Conditional Distribution: <====== Standardized Student-t.
std

Coefficient(s):
mu omega alpha1 shape
0.021571 0.013424 0.259867 5.985979
Std. Errors:
based on Hessian
11
Error Analysis:
mu 0.021571 0.006054 3.563 0.000366 ***
omega 0.013424 0.001968 6.820 9.09e-12 ***
alpha1 0.259867 0.119901 2.167 0.030209 *
shape 5.985979 1.660030 3.606 0.000311 *** <== Estimate of
degrees of freedom
---
Log Likelihood:
242.9678 normalized: 0.6531391
Statistic p-Value

Ljung-Box Test R Q(10) 14.31288 0.1591926
Ljung-Box Test R Q(15) 23.34043 0.07717449
Ljung-Box Test R Q(20) 24.87286 0.2063387
Ljung-Box Test R^2 Q(10) 15.35917 0.1195054
Ljung-Box Test R^2 Q(15) 33.96318 0.003446127
Ljung-Box Test R^2 Q(20) 35.46828 0.01774746
LM Arch Test R TR^2 24.11517 0.01961957
AIC BIC SIC HQIC
-1.284773 -1.242634 -1.285001 -1.268039
> plot(m2)
1: Time Series
2: Conditional SD

7: Residuals
8: Conditional SDs
Selection: 13 <== The plot shows that the model needs further
improvements.
12
> predict(m2,5) <===== Prediction
meanForecast meanError standardDeviation
1 0.02157100 0.1207911 0.1207911
2 0.02157100 0.1312069 0.1312069
3 0.02157100 0.1337810 0.1337810

4 0.02157100 0.1344418 0.1344418
5 0.02157100 0.1346130 0.1346130
The fitted model with Student-t innovations is
rt = 0.0216 + at, at = σt�t, � ∼ t5.99
σ2t = 0.0134 + 0.260a
2
t−1.
We use t5.99 to denote the standardized Student-t distribution
with
5.99 d.f.
Comparison with normal innovations:
• Using a heavy-tailed dist for �t reduces the ARCH effect.
• The difference between the models is small for this particular
instance.
You may try other distributions for �t.
GARCH Model
at = σt�t,
σ2t = α0 +
m∑
i=1
αia
2

t−i +
s∑
j=1
βjσ
2
t−j
where {�t} is defined as before, α0 > 0, αi ≥ 0, βj ≥ 0, and
∑max(m,s)
i=1 (αi + βi) < 1.
Re-parameterization:
Let ηt = a
2
t − σ2t . {ηt} un-correlated series.
The GARCH model becomes
a2t = α0 +
max(m,s)∑
i=1
(αi + βi)a
2
t−i + ηt −
s∑
j=1
βjηt−j.

13
This is an ARMA form for the squared series a2t .
Use it to understand properties of GARCH models, e.g. moment
equations, forecasting, etc.
Focus on a GARCH(1,1) model
σ2t = α0 + α1a
2
t−1 + β1σ
2
t−1,
• Weak stationarity: 0 ≤ α1, β1 ≤ 1, (α1 + β1) < 1.
• Volatility clusters
• Heavy tails: if 1− 2α21 − (α1 + β1)2 > 0, then
E(a4t )
[E(a2t )]
2
=
3[1− (α1 + β1)2]
1− (α1 + β1)2 − 2α21
> 3.
• For 1-step ahead forecast,
σ2h(1) = α0 + α1a

2
h + β1σ
2
h.
For multi-step ahead forecasts, use a2t = σ
2
t �
2
t and rewrite the
model as
σ2t+1 = α0 + (α1 + β1)σ
2
t + α1σ
2
t (�
2
t − 1).
2-step ahead volatility forecast
σ2h(2) = α0 + (α1 + β1)σ
2
h(1).
In general, we have
σ2h( ) = α0 + (α1 + β1)σ
2

h( − 1), > 1.
This result is exactly the same as that of an ARMA(1,1) model
with AR polynomial 1− (α1 + β1)B.
14
Example: Monthly excess returns of S&P 500 index starting
from
1926 for 792 observations.
The fitted of a Gaussian AR(3) model
r ̃t = rt − 0.0062
r ̃t = .089r ̃t−1 − .024r ̃t−2 − .123r ̃t−3 + .007 + at,
σ̂2a = 0.00333.
For the GARCH effects, use a GARCH(1,1) model, we have
A joint estimation:
rt = 0.032rt−1 − 0.030rt−2 − 0.011rt−3 + 0.0077 + at
σ2t = 7.98× 10−5 + .853σ2t−1 + 0.124a2t−1.
Implied unconditional variance of at is
0.0000798
1− 0.853− 0.1243 = 0.00352
close to the expected value. All AR coefficients are statistically
in-

significant.
A simplified model:
rt = 0.00745 + at, σ
2
t = 8.06× 10−5 + .854σ2t−1 + .122a2t−1.
Model checking:
For ãt: Q(10) = 11.22(0.34) and Q(20) = 24.30(0.23).
For ã2t : Q(10) = 9.92(0.45) and Q(20) = 16.75(0.67).
Forecast: 1-step ahead forecast:
σ2h(1) = 0.00008 + 0.854σ
2
h + 0.122a
2
h
Horizon 1 2 3 4 5 ∞
Return .0074 .0074 .0074 .0074 .0074 .0074
Volatility .054 .054 .054 .054 .054 .059
15
R demonstration:
> sp5=scan("sp500.txt")

Read 792 items
> pacf(sp5)
> m1=arima(sp5,order=c(3,0,0))
> m1
Call:
arima(x = sp5, order = c(3, 0, 0))
Coefficients:
ar1 ar2 ar3 intercept
0.0890 -0.0238 -0.1229 0.0062
s.e. 0.0353 0.0355 0.0353 0.0019
sigma^2 estimated as 0.00333: log likelihood = 1135.25, aic=-
2260.5
> m2=garchFit(~arma(3,0)+garch(1,1),data=sp5,trace=F)
> summary(m2)
Title:
GARCH Modelling
Call:
garchFit(formula = ~arma(3,0)+garch(1, 1), data = sp5, trace =
F)

data ~ arma(3, 0) + garch(1, 1)
[data = sp5]
Error Analysis:
mu 7.708e-03 1.607e-03 4.798 1.61e-06 ***
ar1 3.197e-02 3.837e-02 0.833 0.40473
ar2 -3.026e-02 3.841e-02 -0.788 0.43076
ar3 -1.065e-02 3.756e-02 -0.284 0.77677
omega 7.975e-05 2.810e-05 2.838 0.00454 **
alpha1 1.242e-01 2.247e-02 5.529 3.22e-08 ***
beta1 8.530e-01 2.183e-02 39.075 < 2e-16 ***
---
Log Likelihood:
1272.179 normalized: 1.606287
Statistic p-Value

Ljung-Box Test R Q(10) 11.56744 0.315048
16
Ljung-Box Test R Q(15) 17.78747 0.2740039
Ljung-Box Test R Q(20) 24.11916 0.2372256
Ljung-Box Test R^2 Q(10) 10.31614 0.4132089
Ljung-Box Test R^2 Q(15) 14.22819 0.5082978
Ljung-Box Test R^2 Q(20) 16.79404 0.6663038
LM Arch Test R TR^2 13.34305 0.3446075
AIC BIC SIC HQIC
-3.194897 -3.153581 -3.195051 -3.179018
> m2=garchFit(~garch(1,1),data=sp5,trace=F)
> summary(m2)
Title: GARCH Modelling
Call:

garchFit(formula = ~garch(1, 1), data = sp5, trace = F)
data ~ garch(1, 1)
[data = sp5]
Error Analysis:
mu 7.450e-03 1.538e-03 4.845 1.27e-06 ***
omega 8.061e-05 2.833e-05 2.845 0.00444 **
alpha1 1.220e-01 2.202e-02 5.540 3.02e-08 ***
beta1 8.544e-01 2.175e-02 39.276 < 2e-16 ***
---
Log Likelihood:
1269.455 normalized: 1.602848
Statistic p-Value

Ljung-Box Test R Q(10) 11.22050 0.340599
Ljung-Box Test R Q(15) 17.99703 0.262822
Ljung-Box Test R Q(20) 24.29896 0.2295768
Ljung-Box Test R^2 Q(10) 9.920157 0.4475259
Ljung-Box Test R^2 Q(15) 14.21124 0.509572
Ljung-Box Test R^2 Q(20) 16.75081 0.6690903
LM Arch Test R TR^2 13.04872 0.3655092
17
AIC BIC SIC HQIC
-3.195594 -3.171985 -3.195645 -3.186520
> plot(m2)
1: Time Series
2: Conditional SD

7: Residuals
8: Conditional SDs
Selection: 3
> predict(m2,6)
meanForecast meanError standardDeviation
1 0.007449721 0.05377242 0.05377242
2 0.007449721 0.05388567 0.05388567
3 0.007449721 0.05399601 0.05399601
4 0.007449721 0.05410353 0.05410353
5 0.007449721 0.05420829 0.05420829
6 0.007449721 0.05431038 0.05431038

Turn to Student-t innovation. (R output omitted.)
Estimation of degrees of freedom:
rt = 0.0085 + at, at = σt�t, �t ∼ t7
σ2t = .000125 + .113a
2
t−1 + .842σ
2
t−1,
where the estimated degrees of freedom is 7.00.
Forecasting evaluation
Not easy to do; see Andersen and Bollerslev (1998).
IGARCH model
18
0 200 400 600 800
�0
.2
0.
0
0.
2

0.
4
Series with 2 Conditional SD Superimposed
Index
x
Figure 4: Monthly S&P 500 excess returns and fitted volatility
An IGARCH(1,1) model:
at = σt�t, σ
2
t = α0 + β1σ
2
t−1 + (1− β1)a2t−1.
For the monthly excess returns of the S&P 500 index, we have
rt = .007 + at, σ
2
t = .0001 + .806σ
2
t−1 + .194a
2
t−1
For an IGARCH(1,1) model,
σ2h( ) = σ

2
h(1) + ( − 1)α0, ≥ 1,
where h is the forecast origin.
Effect of σ2h(1) on future volatilities is persistent, and the
volatility
forecasts form a straight line with slope α0. See Nelson (1990)
for
more info.
Special case: α0 = 0.
used in RiskMetrics to VaR calculation.
19
Example: An IGARCH(1,1) model for the monthly excess
returns
of S&P500 index from 1926 to 1991 is given below via R.
rt = 0.0074 + at, at = σt�t
σ2t = 5.11× 10−5 + .143a2t−1 + .857σ2t−1.
R demonstration: Using R script Igarch.R.
> source("Igarch.R")
> m4=Igarch(sp5)
Maximized log-likehood: -1268.205

Coefficient(s):
mu 7.41587e-03 1.52545e-03 4.86144 1.1653e-06 ***
omega 5.10855e-05 1.74923e-05 2.92046 0.0034952 **
beta 8.57124e-01 2.14420e-02 39.97404 < 2.22e-16 ***
---
20
HW Set 6 Data Sets/djao2.tsm
3621.63 1941.50 Sep 10, 93
3634.21 1938.30 DJ AO daily closing values
3615.76 1912.90
3633.65 1903.60
3630.85 1902.60
3613.25 1925.50
3575.80 1924.10
3537.24 1925.20
3547.02 1919.30

3539.75 1928.60
3543.11 1946.50
3567.70 1943.00
3566.02 1942.30
3566.30 1951.40
3555.12 1964.40
3581.11 1972.70
3577.76 1977.00
3587.26 1998.50
3598.99 2018.80
3583.63 2022.50
3584.74 2026.20
3593.41 2039.80
3593.13 2028.00
3603.19 2038.60
3621.63 2062.00
3629.73 2074.10
3642.31 2085.50

3635.32 2075.50
3645.10 2051.70
3636.16 2060.40
3649.30 2061.40
3673.61 2046.90
3672.49 2055.70
3664.66 2068.30
3687.86 2076.30
3680.59 2112.20
3692.61 2132.40
3697.64 2125.30
3661.87 2108.40
3624.98 2101.60
3643.43 2079.90
3647.90 2054.20
3640.07 2050.80
3663.55 2042.90
3662.43 2052.40

3684.51 2074.00
3677.52 2082.90
3710.77 2083.80
3704.35 2104.30
3685.34 2108.00
3694.01 2083.20
3670.25 2049.30
3674.17 2009.60
3687.58 2032.40
3687.58 2042.00
3683.95 2043.10
3677.80 2010.30
3683.95 2009.40
3697.08 2005.40
3702.11 2047.30
3704.07 2047.40
3710.21 2053.70
3718.88 2073.90

3734.53 2096.00
3729.78 2095.70
3740.67 2084.90
3764.43 2094.50
3742.63 2086.60
3716.92 2069.90
3726.14 2074.80
3751.57 2080.20
3755.21 2076.00
3745.15 2067.00
3762.19 2053.20
3757.72 2068.80
3757.72 2089.20
3792.93 2089.20
3793.77 2089.20
3794.33 2126.90
3775.88 2154.50
3754.09 2173.60

3756.60 2173.60
3783.90 2174.30
3798.82 2193.40
3803.88 2200.30
3820.77 2186.00
3865.51 2198.60
3850.31 2206.70
3848.63 2195.60
3842.43 2177.50
3867.20 2206.40
3870.29 2238.20
3870.29 2232.10
3884.37 2248.20
3891.96 2266.20
3914.48 2250.30
3912.79 2224.50
3895.34 2221.90
3908.00 2221.90

3926.30 2250.70
3945.43 2259.90
3978.36 2310.80
3964.01 2310.10
3975.54 2312.10
3967.66 2340.60
3871.42 2332.80
3906.32 2281.10
3906.03 2305.40
3931.92 2270.90
3895.34 2234.30
3894.78 2241.40
3904.06 2238.60
3928.27 2234.00
3937.27 2249.00
3922.64 2240.90
3887.46 2223.20
3887.46 2178.50

3911.66 2202.50
3891.68 2218.90
3839.90 2197.00
3838.78 2148.80
3832.02 2180.10
3809.23 2181.70
3831.74 2154.00
3824.42 2151.40
3832.30 2116.80
3856.22 2144.70
3851.72 2171.70
3853.41 2146.80
3830.62 2155.10
3862.70 2153.10
3862.98 2179.30
3849.59 2172.50
3848.15 2173.50
3865.14 2164.40

3895.65 2163.50
3864.85 2140.50
3862.55 2140.80
3869.46 2180.90
3821.09 2169.80
3774.73 2151.60
3762.35 2108.90
3699.02 2100.80
3626.75 2092.40
3635.96 2053.10
3635.96 2053.10
3593.35 2053.10
3675.41 2050.00
3679.73 2084.10
3693.26 2087.40
3674.26 2082.00
3688.83 2076.00
3681.69 2095.10

3661.47 2111.20
3663.25 2095.00
3661.47 2080.60
3620.42 2095.90
3619.82 2061.40
3598.71 2046.60
3652.54 2029.60
3648.68 2042.50
3705.78 2042.50
3699.54 2069.40
3699.54 2059.70
3668.31 2069.10
3681.69 2066.10
3701.02 2047.90
3714.41 2044.20
3697.75 2018.40
3695.97 1988.10
3669.50 2004.30

3629.04 2009.30
3656.41 2008.20
3629.04 2034.60
3652.84 2041.40
3659.68 2070.00
3671.50 2110.90
3720.61 2096.00
3732.89 2107.80
3758.98 2093.70
3766.35 2103.90
3742.41 2121.00
3745.17 2132.40
3755.30 2105.90
3753.46 2096.90
3757.14 2102.20
3757.14 2091.80
3758.37 2081.80
3760.83 2097.20

3758.99 2077.00
3772.22 2078.60
3768.52 2072.50
3755.91 2070.20
3749.45 2079.70
3753.14 2076.70
3773.45 2069.40
3783.12 2069.40
3814.83 2076.60
3790.41 2074.40
3811.34 2056.00
3776.78 2051.20
3741.90 2024.40
3707.97 1993.60
3724.77 2010.90
3699.09 2022.50
3636.94 2017.90
3685.50 1957.40

3669.64 1974.40
3667.05 1975.10
3624.96 1989.10
3646.65 1965.80
3646.65 1987.10
3652.48 2003.40
3674.50 1991.20
3688.42 1962.20
3709.14 1964.90
3702.99 1961.20
3702.66 1972.90
3704.28 1978.60
3739.25 2007.70
3753.81 2058.00
3755.43 2072.30
3748.31 2077.40
3727.27 2078.60
3732.45 2049.20

3735.04 2052.50
3741.84 2048.30
3735.68 2041.30
3720.47 2041.70
3730.83 2042.10
3764.50 2061.50
3798.17 2082.10
3796.22 2086.90
3792.16 2072.30
3765.79 2083.50
3747.02 2091.90
3753.81 2081.10
3755.76 2086.80
3766.76 2076.50
3750.90 2062.80
3768.71 2051.90
3760.29 2055.70
3784.57 2040.00

3776.48 2059.50
3755.43 2066.80
3755.11 2061.30
3751.22 2063.60
3775.83 2051.60
3846.73 2061.10
3829.89 2077.80
3881.05 2077.20 Aug 26, 94
�
HW Set 6 Data Sets/IBM2012.tsm
-0.796816965
2.004075088
2.459140314
-3.265596297
0.40241503
2.036730282
1.242252
-2.06192872

-0.813012608
0
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HW Set 6 Data Sets/m-INTC7308.txt
date rtn
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19730228 -0.139303
19730330 0.069364
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1) A sample of 10 observations is selected from a normal populatio.docx