# Note 5 ARMA

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```					EC 413                                                    FALL 2005
Lecture 5. Univariate time series modelling and forecasting
(Box-Jenkins Method)
Topics:
 Modeling & Forecasting Cycles

Diebold Ch 6 and 7
Univariate Time Series Models

Where we attempt to predict returns using ONE variable based on only
information contained in their past values.

Overview
Stationary ARMA (p, q) Model

What is the ARMA model?

yt = c + 1 yt-1 + 2 yt-2 + .. + p yt-p + 1ut-1 + 2ut-2 + ... + qut-q + ut
[ ] [--------------- AR ------------- ]   [--------- MA ------------------ ] [error]

.. ARMA(p, q) model

Add Independent variables Xt to have ARMA-X models.

yt = c +  Xt + 1 yt-1 + 1ut-1 + ut

What is AR model?
• An autoregressive model of order p, an AR(p) can be expressed as

yt = c + 1 yt-1 + 2 yt-2 + .. + p yt-p + ut

.. AR(p) model
2

What is MA model?

• Let ut (t = 1,2,3,...) be a white noise process, a sequence of
independently and identically distributed (iid) random variables with
E(ut) = 0 and Var(ut) = 2. The qth order MA model is given as a
function of past errors:
yt = c + ut + 1ut-1 + 2ut-2 + ... + qut-q
.. MA(q) model
What is a White Noise Process, ut?

• A white noise process is one with (virtually) no discernible structure.
It is an independently and identically distributed (iid) random
variables… purely random!

• Thus, the ACFs and PACFs (see below) are all zero (not significantly
different from 0). The plots of ACFs and PACFs will lie inside of the
confidence interval (see below).

Key Point of Estimating ARMA Models:

Allow for enough AR & MA terms so that the error term looks like a
white noise process.
ARIMA (p, d, q) Model
If yt is non-stationary, we take a first-difference of yt so that yt becomes
stationary.
yt= ytyt-1
   (d = 1 implies one time differencing. d = 1 is enough in most cases.)

yt = c + 1 yt-1 + .. + p yt-p + 1ut-1 + ... + qut-q + ut

.. ARIMA(p, 1, q) model

Note: When d = 0, we simply denote ARMA(p,q), instead of
ARIMA(p,0,q).
3

Important Remark: If the data is non-stationary (d = 1), the plot of ACFs
is relatively flat and decline slowly.

Unit root = non-stationary = needs first differencing
= integrated process = I(1) process
= needs ARIMA models

Note: In unusual cases, we use a second difference (d = 2), implying a
first difference of the first differenced series
yt= yt- yt-1  ytyt-1 yt-1yt-2) =  ytyt-1yt-2
 I(2) process

Two Main Questions:
 How many AR and MA terms are necessary?
 Add enough AR or MA terms until the residual looks likes a
white noise process.
 How can we tell that the residual term is a white noise process?
We look at the plot of the ACFs.
 People often adopt the model specification criteria such as
Akaike inf. criterion (AIC) and Schwarz Baysian inf. Criterion
(SIC or BIC or SBIC or SBC).
 Do we need to difference the data?
 Yes, if the data is non-stationary.
 How can you tell if the data is non-stationary?
 What do you mean by “stationary”?
 Box-Jenkins provided procedures to estimate an optimal ARIMA model.

Basic Technical Concepts

A Weakly “Stationary” Process
4

If a series satisfies the next three equations, it is said to be weakly or
covariance stationary: they do not depend on t.
1. E(yt) =  ,       t = 1,2,..., … Unconditional mean.
2. Var(yt) = 2 <                  … Unconditional variance
3. Cov(yt, yt-s) = s               … Auto-covariance

• So if the process is covariance stationary, (i) all the variances are the
same and (ii) all the covariances depend on the difference between t
and t-s.
Ex) Var(yt) = Var(yt-1) = Var(yt-10) = 2
Cov(yt, yt-5) = Cov(yt-1, yt-6) = Cov(yt-4, yt-9) = 5
(All these are the same!)
; We say that 5 is the autocovariance function with lag 5.

Auto-Correlation Function (ACF)
• The covariances, s, are known as autocovariances.
One can find 1, 2, 3, 4¸s, ….. and so on.
Here, 0 = Cov(yt, yt-0) = Var(yt) = 2
1 = Cov(yt, yt-1) = Cov(yt-1, yt-2) = ..

• However, the value of the autocovariances depend on the units of
measurement of yt.

• It is thus more convenient to use the autocorrelations which are the
autocovariances normalised by dividing by the variance:

s
s =         -1 s  s = 0,1,2, ...
0

s is the autocorrelation between yt and yt-s.
5

 If we plot s against s=0,1,2,... then we obtain the AutoCorrelation
Function (ACF) or called, correlogram.

1

0.9
acf

0.8                                               pacf

0.7

0.6
acf and pacf

0.5

0.4

0.3

0.2

0.1

0
1   2     3     4   5          6   7   8     9    10
-0.1
Lags

White Noise Process (ut)

• A white noise process is one with (virtually) no discernible structure.
It is an independently and identically distributed (iid) random
variables.

(i) E(ut)=0
(ii) Var(ut) = 2
(iii) s = 0 if s  0
That is, 0 = 2, 1 = 2 = …. = 0 (s > 0)

Thus the autocorrelation function (s) will be zero apart from a single
peak of 1 at s = 0.

• We have a distribution of s under the null that a time series is a white
noise.
s  approximately N(0,1/T) where T = sample size
This implies:
Var(s) = 1/T
SE(s) = 1/T
If so, the 95% confidence interval of s of a white noice process is
s = 0  1.96 * SE(s) = 0  1.96 * 1/T
6

For instance, if T = 100, C.I. = 0  1.96 * 1/100 = 

• Point: If a time series is a white noise, |^ | should NOT be bigger
s
than the confidence interval, 1.96 * 1/T for (almost) all values of
s.

Testing for a White Noise Process

Does your data look like a white noise process?
Does the residual look like a white noise process?

• We can use this to do significance tests for the autocorrelation
coefficients by constructing a confidence interval.

If ^ falls outside this region (1.96 * 1/T) for any value of s, then
s
we reject the null hypothesis that the true value of the coefficient at
lag s is zero, which implies a white noise process.

Ex) RATS example

1.00

0.75

0.50

0.25

0.00

-0.25

-0.50

-0.75

-1.00
2     4   6   8         10   12     14   16   18   20
ACF        PACF

Correlations of Series U
Autocorrelations
1: -0.1154773 -0.0722445 0.1479023 -0.0814163 -0.0977566 0.0595666
7: 0.0028219 0.0512692 -0.0584719 -0.0513372 0.0491615 -0.1019420
13: 0.1220512 0.0973237 -0.0764919 0.0347748 -0.0208870 0.0551283
19: 0.1662776 -0.1114049
7

Partial Autocorrelations (later on this)
1: -0.1154773 -0.0867362 0.1315508 -0.0565580 -0.0969105 0.0106180
7: 0.0179404 0.0821972 -0.0697137 -0.0695742 0.0229997 -0.0766431
13: 0.1335587 0.0833780 -0.0288299 -0.0014239 -0.0451313 0.1250012
19: 0.1921916 -0.0795414

• What if only one or a few ^ lie outside the confidence interval? Is
s
there any test for joint insignificance up to lag s?

-> Joint restriction Test

• We can also test the joint hypothesis that all m of the k correlation
coefficients are jointly equal to zero using the Q-statistic developed by
Box and Pierce:
m
Q(m) = T     k2
k=1
where T = sample size, m = maximum lag length

Ho: 1 = 2 = 3 = … = m = 0 (implying a white noise process)

The Q-statistic is asymptotically distributed as a chi-square
distribution ().

If Q(m) > m (m = degree of freedom), the reject the null
hypothesis and we say that the time series is NOT a white noice
process

If Q(m) < m, then we cannot reject the null of a white noise process
using lags up to m. We often choose m = 10, 20, or 30.

• However, the Box Pierce test has poor small sample properties, so a
variant has been developed, called the Ljung-Box statistic:
m
*
Q (m) = T(T+2)       k2/(T-k)
k=1
The Q-statistic is aso asymptotically distributed as a chi-square
distribution () and Q* can be used instead of Q.

This statistic is commonly used as a portmanteau (general) test of
linear dependence (autocorrelation) in time series.
8

Question (Exercise):

Suppose that a researcher had estimated the first 5 autocorrelation
coefficients using a series of length 100 observations, and found them
to be (from 1 to 5): 0.207, -0.013, 0.086, 0.005, -0.022. Test each of
the individual coefficient for significance, and use both the Box-
Pierce and Ljung-Box tests to establish whether they are jointly
significant.
Solution:

A coefficient would be significant if it lies outside (-0.196, +0.196) at
the 5% level, so only the first autocorrelation coefficient is significant.
(Here, 1.96 * 1/T = 0.196)

Next, the joint test is on:

Ho: 1 = 2 = … = 5 = 0 (implying a white noise process)
m
Q(5) = T    k2 = 5.09 and
k=1
m
Q*(5) = T(T+2)    k2/(T-k) = 5.26
k=1

Compared with a tabulated 2c with df 5 is 11.1 at the 5% level (see
2c chart), these are smaller than 11.1. So, we can say that the 5
coefficients are jointly insignificant. Then, the time series appears a
white noise process.

Note: In RATS, we can also examine p-values. (if p-value > 5%, the
null is not rejected. Thus, it is a white noise process. Otherwise, it is
not a white noise process).

Example) RATS reports:

Ljung-Box Q-Statistics
Q(20-0)=     18.0031. Significance Level 0.58720637
9

 Since the p-value> 0.05, we cannot reject the null that the
series is a white noise process using 20 lags.

• Most time series data are NOT a white noise process. The Box-
Jenkins method is to find a model in such as way that by adding AR
and MA terms the resulting residuals will look like a white noise
process.

If yt is not a white noise process, add AR and MA terms until ut looks
like a white noice process.

Yt = (p # of AR terms) + (q # of MA terms) + ut
If p and q are properly selected, ut will be a white noise process.
^
Point: After adding AR-MA terms, check if the residual (ut) is a
white noise process. (more on this, later)
^
(i) Use ACF plots of ut and check if any k lies outside the CI.
^
(ii) Use Q* statistic on ut.

SAS example (basics.sas              using basics.xls):

Proc import
datafile = "C:\Documents and Settings\jlee\
My Documents\EC413-1\Basics1.xls"
out = basics2 replace;
run;

Proc Arima data = basics2;
identify var = RATE;
title 'identify with Original Series';
Run;

Lag       Correlation

0            1.00000
10

1         0.97924
2         0.94808
3         0.92150
4         0.89886
5         0.87777
6         0.85502
7         0.83928
8         0.83106
9         0.81964
10         0.80072
11         0.77945
12         0.75884
13         0.74224
14         0.72398
15         0.69878
16         0.67720
17         0.65538
18         0.62905
19         0.59874
20         0.56914
21         0.54872
22         0.53439
23         0.52003
24         0.50575

Slowly declining ACFs RATE is possibly non-stationary.


To          Chi-                 Pr >
Lag        Square        DF      ChiSq

6       2264.38         6     <.0001
12       4045.20        12     <.0001
18       5366.10        18     <.0001
24       6210.75        24     <.0001

Since p-vlaues are less than 5% in all cases, we
can reject the null of a white noise at all lags.
11

RATS Example (ARMA1.prg)

calendar 1960 1 12
allocate 1999:9

open data basics.wks
data(format=wks,org=cols) / rate

graph(key=upleft)
# rate

* Compute and graph autocorrelations:
correlate(partial=PACF, QSTATS) rate / ACF
graph(key=below,style=bar,nodates,min=-1.0,max=1.0,number=1) 2
# ACF
# PACF

Results:
1.00

0.75

0.50

0.25

0.00

-0.25

-0.50

-0.75

-1.00
5     10   15         20     25   30   35   40
ACF        PACF

Ljung-Box Q-Statistics
Q(41-0)=    7378.5980. Significance Level 0.00000000

 Clearly, RATE is not a white noise process. (why?)
Perhaps, RATE is non-stationary. (why?)

SAS output
(SAS – Solutions - Analysis – Time Series Viewer)
12

Testing for a white noise and nonstationarity in SAS

SAS output provides other useful information (p-values):

White noise test; Null hypothesis = white noise

If p-value < 0.05, the null is rejected
 Not a white noise

If p-value > 0.05, the null is Not rejected
 a white noise

Unit root test: Null hypothesis = non-stationary (unit root)

If p-value < 0.05, the null is rejected
 stationary (no need to take a first difference)

If p-value > 0.05, the null is NOT rejected
 non-stationary (need to take a first difference)
13

Example:

White noise test: we               the null of a white noise.

Thus,

Unit root test: we                 the null of a unit root.

Thus,

Therefore, we conclude:
First, we need to take                                  .

Second, we need to add                                  . But, how?

Building ARMA Models - The Box Jenkins Approach

• Box and Jenkins (1970) were the first to approach the task of
estimating an ARMA model in a systematic manner. There are 3
steps to their approach:
14

1. Identification
2. Estimation
3. Model diagnostic checking
Step 1: (identification: use Unit root test and AIC/BIC)

- determining whether to difference the data or not.
- determining p and q values

Step 2: (BOXJENK)

- Estimation of the parameters
- Can be done using least squares or maximum likelihood
Step 3: (check on residuals)

- Model checking on residuals

 Box and Jenkins suggest 2 methods:

- Deliberate overfitting
- Residual diagnostics

• Identification would typically not be done using ACF and PACF’s.

• We want to form a parsimonious model.
- variance of estimators is inversely proportional to the number of
degrees of freedom.
- models which are profligate might be inclined to fit to data
specific features

• This gives motivation for using information criteria.

• The object is to choose the number of parameters which minimises the
information criterion. And the information criteria vary according to
how stiff the penalty term is.
15

• The two most popular criteria are Akaike’s (1974) information
criterion (AIC), Schwarz’s (1978) and Bayesian information criterion
(SIC or BIC).
^
AIC = ln( 2) + 2 k/T
^
SIC (BIC) = ln( 2) + (k/T) lnT

where k = p + q + 1, T = sample size.
Point: Choose a model with a LOWER values.
• Forecasting: based on the estimated ARMA models.

Practical Applications

(1) Variable: Interest rate (RATE) from the data file, basics.xls

1st Difference or not?

We have already examined that we need to take the first difference.
(The null of a unit root is not rejected.) Thus, we use d = 1.

2nd Using d = 1, find the optimal values of p and q.

Note: We usually add the constant term since the mean is not zero in
most cases after taking the first difference.

We make an AIC table.
[d = 1 is used in all cases, and the constant term is added.]

AIC table for RATE

AR terms (p)
0          1        2           3
MA        0     -543.97    -576.05 -593.05      -591.07
terms     1     -590.14    -593.38 -591.72      -593.51
(q)       2     -593.24    -594.21 -592.45        n/a
3     -591.44    -592.29 -607.41      -592.59

Therefore, we choose p = 2, q = 3 (while d = 1)  ARIMA(2,1,3)!
16

Note: One could use the SIC table, and the result can be different.

Parameter Estimates       22:23 Monday, October 24, 2005   14
RATE: RATE
ARIMA(2,1,3)

Std.
Model Parameter                       Estimate          Error      T       Prob>|T|

Intercept                              0.00440      0.0301        0.1461      0.8839
Moving Average, Lag 1                 -0.09664      0.0646       -1.4965      0.1352
Moving Average, Lag 2                 -0.82568      0.0539      -15.3144      <.0001
Moving Average, Lag 3                 -0.21649      0.0547       -3.9562      <.0001
Autoregressive, Lag 1                  0.23770      0.0407        5.8358      <.0001
Autoregressive, Lag 2                 -0.92626      0.0405      -22.8570      <.0001
Model Variance (sigma squared)         0.25191           .             .       .

Thus, the estimated equation is

Ratet = 0.0044 + 0.2377Ratet-1 - 0.9263Ratet-2 + ut – 0.097ut-1 – 0.826ut-2 – 0.216ut-3

We can use this regression for forecasting!

Undifferencing is needed:

After we obtain the predicted values of Ratet,
RateT+1 = RateT + RateT+1
RateT+2 = RateT+1 + RateT+2
RateT+3 = RateT+2 + RateT+3
And so on. SAS does this calculation automatically!

(2) Variable: M1 (Money Supply) from the data file, basics.xls

1st. For this, we want to take logarithm and use log(M1).

Why? We can stabilize the fluctuation of M1 by take log. (big values)

2nd Difference or not?

We now examine if the the null of a unit root is not rejected.
(solution-analysis-time series viewers-unit root tests)

Do ACFs decline slowly?
Do we reject the null of a white noise?
Do we reject the null of a unit root?
17

We need to take the first difference
Thus, we use d = 1.

The first difference of logged series = growth rates of M1
(Recall Lecture 3).

3rd Using d = 1, find the optimal values of p and q.

We make an AIC table.
[d = 1 is used in all cases, log transformation, and the constant term is

AIC table for log(M1)

AR terms (p)
0         1         2           3
MA        0
terms     1
(q)       2
3

Exercise: GNP82.XLS (or GNP82.DAT); quarterly data

(a) Import the data and create QTR (date variable),
starting from 1947, Q.1 (the data ends with 1988, Q3)

Then, export the data to gnp82new.xls (also including QTR).

(b) Find the optimal ARIMA model for GNP82.

Differencing? log? p & q values?

RATS example: ARMA3.PRG
Using bjident.src and bjfore.src in RATS
18

Exercise: ARMA Identification and Estimation

The files entitled Y1.PRN, Y2.PRN and Y3.PRN contain 1000 values of
the data that follow ARMA(p,q) models.

Repeat the following questions for EACH of the data file.

(a) Plot the data against time and obtain ACF, PACF and Q-statistics.
Does the series appear to be stationary? (Note that the series would
be non-stationary when the ACFs decay slowly.) Does the series
appear to be a white noise process? Why or why not?

(b) Estimate different ARMA models using various values of p and q
from 0 to 5, and identify the best ARMA(p,q) model. Then, provide
the estimation result of the best ARMA model. Also, provide
diagnostic check-ups for the the residuals of the model you estimated
by examining the ACF, PACF and Q-statistics. Does the residual
series appear to be a white noise process?

(c) Obtain the out-of-sample forecasts of the next 20 periods using the
estimated ARMA model.
19

TECHNICAL DETAILS

MA (Moving Average) Models

MA model

• Let ut (t=1,2,3,...) be a white noise process, a sequence of
independently and identically distributed (iid) random variables with
E(ut)=0 and Var(ut) = 2. The qth order MA model is given as:
yt =  + ut + 1ut-1 + 2ut-2 + ... + qut-q
.. MA(q) model

• The model is expressed in terms of past errors. We wish to estimate
the coefficients j, j=1,..,q, and use the model for forecasting.
• Its properties are

(i) Only q errors affect the current leve yt but higher order errors do
not affect yt.

(ii) It is a short memory model.
 For an MA(q) model, s = 0 for s > q.
Ex) MA(1) model
 yt =  + ut + 1ut-1

 Of course, 1 = ACF(1) is not 0 (it is not a white noise
process).
But, 2 = ACF(2) = 0. Also, 3 = 4 = .. = 0.

Ex) RATS example
Point: If k = k+1 = … = 0, then the process may follow an MA(k-1)
model.
20

Finding theoretical Variance and Covariances of MA models
Example 1)
Consider an MA(1) model, yt =  + ut + 1ut-1
(i) Mean:            E(yt)= + 0 + 0 = .
(ii) Variance:      0 = Var(yt) = Var( + ut + 1ut-1) = 0 + 2 + 12 2
= (1 + 12)2

(iii) Covariances: s = Cov(yt, yt-s)

1 = Cov(yt, yt-1) = Cov( + ut + 1ut-1,  + ut-1 + 1ut-2)
= E[yt - E(yt)][yt-1 - E(yt-1 )]
= E[(ut + 1ut-1)( ut-1 + 1ut-2)]
since yt - E(yt) = ( + ut + 1ut-1) -  = ut + 1ut-1
= 1  2
Thus, 1 = 1 / 0 = 1 2 / [(1 + 12)2] = 1/(1 + 12)

2 = Cov(yt, yt-2) = Cov( + ut + 1ut-1,  + ut-2 + 1ut-3)
= E[(ut + 1ut-1)( ut-2 + 1ut-3)]
= 0
3 = E[(ut + 1ut-1)( ut-3 + 1ut-4)] = 0
4 = 0…

Thus, s = 0, s = 2, 3, 4..

Point: If s = 0, s = 2, 3, 4.., then it may be an MA(1) model.

Ex: if 1  0, but if 2 = 3 = 4 = .. = 0, it is an MA(1) model.
Ex: if 1  0, 2  0,but if 3 = 4 = 5 = .. = 0, it is an MA(2) model.

Exercise
Find E(yt), 0, 1, 2, 3, ..
(a) yt = 3 + ut + 0.5ut-1
(b) yt = ut - 0.9ut-1
(c) yt = ut - 0.3ut-1 + 0.2ut-2
21

AR (Autoregressive) Models

AR model

• An autoregressive model of order p, an AR(p) can be expressed as

yt = c + 1 yt-1 + 2 yt-2 + .. + p yt-p + ut

where ut is a white noise process

• The model is expressed in terms of past values. We wish to estimate
the coefficients j, j=1,..,p, and use the model for forecasting.
• Its properties are

(i) All previous values will have cumulative effects on the current
level yt.

(ii) Thus, it is a long-run memory model.

o s = ACF(s) does not die out easily. It takes a longer time to
have ACF close to zero.
1

0.9
acf

0.8                                           pacf

0.7

0.6
acf and pacf

0.5

0.4

0.3

0.2

0.1

0
1   2   3   4   5          6   7   8     9    10
-0.1
Lags

o In theory, we say that there is a persistent effect over time.
o Many time series follow AR(p) models.
Ex) AR(1) model
o yt = c +  yt-1 + ut
 (Unconditional) Mean:
E(yt) = E(c +  yt-1 + ut) = c + E(yt-1 ) + 0
22

Letting E(yt) = E(yt-1 ) = , (assuming weakly
stationarity)
 = c +    = c / (1-).
 Variance

Var(yt) = Var(c +  yt-1 + ut) = 0 + 2Var(yt-1) + 2
Letting Var(yt) = Var(yt-1 ) = 0, (assuming weakly
stationarity)

0 = 20 + 2  0 = 2 / (1 - 2)
   Covariance (s)
From yt = c +  yt-1 + ut, let c = 0 for simplicity.
We can have: yt-1 =  yt-2 + ut-1
yt =  yt-1 + ut =  ( yt-2 + ut-1) + ut
=  (( yt-3 + ut-2) + ut-1) + ut
= (keep substituting..)
E(ytyt-1) = 
E(ytyt-2) = 2
E(ytyt-3) = 3
.. so on

Example)  = 0.99

1.00

0.75

0.50

0.25

0.00

-0.25

-0.50

-0.75

-1.00
5    10    15         20   25     30   35   40
ACF        PACF
23

Since || < 1 for stationary series, k  0 as k increases,
but it takes a while (k not small) to have k close to 0.

o Note: If  = 1 (as in the random walk model, yt = yt-1 + ut), the
unconditional mean does not exist, and k = 1 for any k. This is
the case of non-stationary data. If so, we need to do the first-
difference. yt = yt – yt-1. (more on this later.)

Partial Autocorrelation Functions (PACF)

• Measures the correlation between an observation k periods ago and
the current observation, after controlling for observations at
intermediate lags (i.e. all lags < k).
PACF(k) = ACF(k) after controlling the effects of (yt-1, .., yt-k+1)
yt, (yt-1, .. , yt-k+1), yt-k

At lag 1, the ACF(1) = PACF(1) always.

At lag 2, PACF(2) = (2-12) / (1-12).

For lags 3+, the formulae are more complex.

 PACF(k) can be found as the coefficient of yt-k in the regression:

Yt = 0 + 1yt-1 + 2yt-2 + ….. + k-1yt-k+1 + kyt-k + ut

 k = PACF(k)

 The PACF is useful for telling the maximum order of an AR process.

o For AR(q) models, PACF(q+s) = 0, s 
Ex) If PACF(2) = PACF(3) = .. 0, then it may be an AR(1) model.

Ex) If PACF(3) = PACF(4) = .. 0, then it may be an AR(2) model.

 Point: For an AR(p), the theoretical pacf will be zero after lag p.

• For an MA(q), the theoretical PACF will be geometrically declining.
24

Point:

1. For an AR(p), the theoretical PACF will be zero after lag p.

2. For an MA(q), the theoretical ACF will be zero after lag q.

Summary
 An AR process has
o a geometrically decaying ACF
o number of spikes of PACF = AR order
 An MA process has
o Number of spikes of ACF = MA order
o a geometrically decaying PACF

Stationarity Conditions for AR models
• The condition for stationarity of a general AR(p) model is that the
roots of 1 - 1 z - 2z2 - … - pzp = 0 all lie outside the unit circle.

Examples) Is yt stationary?
(a) yt = yt-1 + ut

1 - z   z 

The characteristic root is 1, so it is a unit root process (so non-
stationary)
(b) yt = 0.3yt-1 + 0.35yt-2 + ut

1 – 0.3 z z = 0 or (1-0.7z)(1+0.5z) = 0. hus,

zand z = 

The characteristic roots are 1.43 and 2. When any one of these lies
outside the unit circle (bigger than 1), the process is stationary
• MA models are already stationary. Why?
25

• A stationary AR(p) model can have an MA() representation.

Ex) yt = 0.5 yt-1 + ut
yt – 0.5yt-1 = ut or (1 – 0.5L)yt = ut
using an lag operator, Lyt = yt-1 (in genelra, Lkyt = yt-k)
1
yt = 1 - 0.5L ut = (1 + 0.5L+ 0.52L2 + 0.53 L3 + …)ut

= ut + 0.5ut-1 + 0.52ut-2 + 0.53 ut-3 +     .. MA() form

Point: Any stationalry AR model can be seen as an MA model. Any
(invertible) MA model can be seen as AR model. Thus, AR or MA
models are interchangeable.

ARMA (Autoregressive MA) Models

An ARMA(p, q) is expressed as:
yt = c + 1 yt-1 + 2 yt-2 + .. + p yt-p + ut + 1ut-1 + 2ut-2 + ... + qut-q

….. A Combination f AR and MA models.

 Either ACF or PACF cannot solely provide the information on the
maximum orders of p or q.

 In reality, it is rare to have pure AR or MA models. Thus, we resor to
the information criterion to find optimal values of p and q.
26

Exercise Questions

Part 1. Practical Applications

Q. 1. The file entitled US.PRN contains quarterly values of the U.S. GDP85 from 1960:1
to 1991:4.

(d) Plot the data against time and obtain ACF, PACF and Q-statistics. Does the series
appear to be stationary? (Note that the series would be non-stationary when the
ACFs decay slowly.)
(e) You must have concluded that the series appears non-stationary. To make it
stationary, first difference the logged series and call it as GDP_FD: GDP_FD =
log(GDP85)t = log(GDP85)t - log(GDP85)t-1
Then, plot this detrended data against time and obtain ACF, PACF and Q-
statistics. Does the series appear to be stationary? Does the series appear to be a
white noise process?
(f) Construct the AIC table using different values of p and q from 0 to 4, and identify
the best ARMA(p,q) model. Then, provide the estimation result of the best
ARMA model.
Then, provide diagnostic check-ups for the the residuals of the model you
estimated by examining the ACF, PACF and Q-statistics. Does the residual series
appear to be a white noise process?
(g) Obtain the out-of-sample forecasts of the next 10 quarters using the estimated
ARMA model.

Q. 2. Repeat Q. 1, using the data, TBILL, contained in the same file, US.PRN.

Part 2. Technical Details

1. Determine if each of the following time series is stationary.

(a) yt + 0.2 yt-1 – 0.48yt-2 = ut
(b) yt + 1.8 yt-1 + 0.81yt-2 = ut
(c) yt - 0.6 yt-2 = ut + 1.2ut-1

2. Find the ACF (s), for s = 0, 1, 2, 3, of each of the following models.

(a) yt = ut + 0.5ut-1
(b) yt = ut + 0.25ut-1 + 0.5ut-2
(c) yt = 0.4 yt-1 + ut
(d) yt - 0.6 yt-1 = ut + 0.4ut-1

3. Find the MA( representation, j=1,2,3, for each of the models in Q. 2 and the
corresponding impulse response function.

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