STAT 497
LECTURE NOTES 4
MODEL INDETIFICATION AND NONSTATIONARY TIME SERIES MODELS
1
MODEL IDENTIFICATION
• We have learned a large class of linear
parametric models for stationary time series
processes.
• Now, the question is how we can find out the
best suitable model for a given observed
series. How to choose the appropriate model
(on order of p and q).
2
MODEL IDENTIFICATION
• ACF and PACF show specific properties for
specific models. Hence, we can use them as a
criteria to identify the suitable model.
• Using the patterns of sample ACF and sample
PACF, we can identify the model.
3
MODEL SELECTION THROUGH CRITERIA
• Besides sACF and sPACF plots, we have also other
tools for model identification.
• With messy real data, sACF and sPACF plots
become complicated and harder to interpret.
• Don’t forget to choose the best model with as
few parameters as possible.
• It will be seen that many different models can fit
to the same data so that we should choose the
most appropriate (with less parameters) one and
the information criteria will help us to decide this.
4
MODEL SELECTION THROUGH CRITERIA
• The three well-known information criteria are
– Akaike’s information criterion (AIC) (Akaike, 1974)
– Schwarz’s Bayesian Criterion (SBC) (Schwarz, 1978).
Also known as Bayesian Information Criterion (BIC)
– Hannan-Quinn Criteria (HQIC) (Hannan&Quinn,
1979)
5
AIC
• Assume that a statistical model of M parameters
is fitted to data
AIC  2 lnmaximum likelihood  2M .
• For the ARMA model and n observations, the
log-likelihood function
n
1
2
ln L   ln 2 a  2 S  p , q ,  
2
2 a 
i.i.d

2 
 assuming at ~ N 0, a .


SS Re sidual
6
AIC
• Then, the maximized log-likelihood is
n
n
2
lnˆ L   ln ˆ a  1  ln 2 
2
2 

constant
AIC  n ln   2M
2
ˆa
Choose model (or the value of M) with
minimum AIC.
7
SBC
• The Bayesian information criterion (BIC) or
Schwarz Criterion (also SBC, SBIC) is a criterion
for model selection among a class of
parametric models with different numbers of
parameters.
• When estimating model parameters using
maximum likelihood estimation, it is possible to
increase the likelihood by adding additional
parameters, which may result in overfitting.
The BIC resolves this problem by introducing a
penalty term for the number of parameters in
the model.
8
SBC
• In SBC, the penalty for additional parameters
is stronger than that of the AIC.
SBC  n ln   M ln n
2
ˆa
• It has the most superior large sample
properties.
• It is consistent, unbiased and sufficient.
9
HQIC
• The Hannan-Quinn information criterion
(HQIC) is an alternative to AIC and SBC.
HQIC  n ln   2M lnln n 
2
ˆa
• It can be shown [see Hannan (1980)] that in
the case of common roots in the AR and MA
polynomials, the Hannan-Quinn and Schwarz
criteria still select the correct orders p and q
consistently.
10
THE INVERSE AUTOCORRELATION
FUNCTION
• The sample inverse autocorrelation function
(SIACF) plays much the same role in ARIMA
modeling as the sample partial
autocorrelation function (SPACF), but it
generally indicates subset and seasonal
autoregressive models better than the SPACF.
11
THE INVERSE AUTOCORRELATION
FUNCTION
• Additionally, the SIACF can be useful for detecting
over-differencing. If the data come from a
nonstationary or nearly nonstationary model, the
SIACF has the characteristics of a noninvertible
moving-average. Likewise, if the data come from
a model with a noninvertible moving average,
then the SIACF has nonstationary characteristics
and therefore decays slowly. In particular, if the
data have been over-differenced, the SIACF looks
like a SACF from a nonstationary process
12
THE INVERSE AUTOCORRELATION
FUNCTION
• Let Yt be generated by the ARMA(p, q) process
 p  B Yt   q  B at where at ~ WN 0, a2 .
• If (B) is invertible, then the model
 q B Zt   p  B at
is also a valid ARMA(q, p) model. This model
is sometimes referred to as the dual model.
The autocorrelation function (ACF) of this dual
model is called the inverse autocorrelation
function (IACF) of the original model.
13
THE INVERSE AUTOCORRELATION
FUNCTION
• Notice that if the original model is a pure
autoregressive model, then the IACF is an ACF that
corresponds to a pure moving-average model. Thus, it
cuts off sharply when the lag is greater than p; this
behavior is similar to the behavior of the partial
autocorrelation function (PACF).
• Under certain conditions, the sampling distribution of
the SIACF can be approximated by the sampling
distribution of the SACF of the dual model (Bhansali,
1980). In the plots generated by ARIMA, the
confidence limit marks (.) are located at 2n1/2. These
limits bound an approximate 95% confidence interval
for the hypothesis that the data are from a white noise
process.
14
EXAMPLE USING SIMULATED
SERIES 1
• Simulated 100 data from AR(1) where =0.5.
• SAS output
Autocorrelations
Lag
Covariance
Correlation
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1.498817
0.846806
0.333838
0.123482
0.039922
-0.110372
-0.162723
-0.301279
-0.405986
-0.318727
-0.178869
-0.162342
-0.180087
-0.132600
0.026849
0.175556
1.00000
0.56498
0.22273
0.08239
0.02664
-.07364
-.10857
-.20101
-.27087
-.21265
-.11934
-.10831
-.12015
-.08847
0.01791
0.11713
-1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1
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Std Error
0
0.100000
0.128000
0.131819
0.132333
0.132387
0.132796
0.133680
0.136670
0.141937
0.145088
0.146066
0.146867
0.147847
0.148375
0.148397
15
Inverse Autocorrelations
Lag
Correlation
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
-0.50606
0.09196
0.06683
-0.14221
0.16250
-0.07833
-0.02154
0.10714
-0.03611
0.03881
-0.04858
0.00989
0.09922
-0.09950
0.11284
-1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1
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Partial Autocorrelations
Lag
Correlation
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0.56498
-0.14170
0.02814
-0.01070
-0.11912
-0.00838
-0.17970
-0.11159
0.02214
-0.01280
-0.07174
-0.06860
-0.02706
0.07718
0.04869
-1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1
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16
EXAMPLE USING SIMULATED
SERIES 2
• Simulated 100 data from AR(1) where =0.5
and take a first order difference.
• SAS output
Autocorrelations
Lag
Covariance
Correlation
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1.301676
-0.133104
-0.296746
-0.131524
0.080946
-0.116677
0.080503
-0.016109
-0.176930
-0.055488
0.136477
0.022838
-0.067697
-0.117708
0.013985
0.0086790
1.00000
-.10226
-.22797
-.10104
0.06219
-.08964
0.06185
-.01238
-.13592
-.04263
0.10485
0.01754
-.05201
-.09043
0.01074
0.00667
-1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1
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Std Error
0
0.100504
0.101549
0.106593
0.107557
0.107919
0.108669
0.109024
0.109038
0.110736
0.110902
0.111898
0.111926
0.112170
0.112904
0.112914
17
Inverse Autocorrelations
Lag
Correlation
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0.58314
0.60399
0.56860
0.46544
0.51176
0.43134
0.40776
0.42360
0.36581
0.33397
0.28672
0.27159
0.26072
0.16769
0.17107
-1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1
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|*******
|*******
|******
|*****
|*****
|***.
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Partial Autocorrelations
Lag
Correlation
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
-0.10226
-0.24095
-0.16587
-0.03460
-0.16453
0.01299
-0.06425
-0.18066
-0.11338
-0.03592
-0.05754
-0.08183
-0.17169
-0.11056
-0.13018
-1 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 1
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18
THE EXTENDED SAMPLE
AUTOCORRELATION FUNCTION_ESACF
• The extended sample autocorrelation function
(ESACF) method can tentatively identify the
orders of a stationary or nonstationary ARMA
process based on iterated least squares
estimates of the autoregressive parameters.
Tsay and Tiao (1984) proposed the technique.
19
ESACF
• Consider ARMA(p, q) model
p
q
1  1B     p B Yt  0  1  1B    q B at
or
Yt  0  1Yt 1     pYt  p  at  1at 1     q at q
then
Z t  1  1B     p B p Yt
follows an MA(q) model
Zt  0  at  1at 1     q at q .
20
ESACF
• Given a stationary or nonstationary time series
Yt with mean corrected form Yt  Yt   with a
true autoregressive order of p+d and with a
true moving-average order of q, we can use the
ESACF method to estimate the unknown orders
and by analyzing the sample autocorrelation
functions associated with filtered series of the
form
m
 m, j 
 m, j 
ˆ
ˆ
Zt
  m, j   B Yt  Yt   i Yt  j
i 1
where ˆi ' s are the parameter estimates under the assumption
that the series is an ARMA m, j process.
21
ESACF
• It is known that OLS estimators for ARMA
process are not consistent so that an iterative
procedure is proposed to overcome this.
 m1, j 1
ˆ
 m, j 
 m1, j 1
 m, j 1 m1
ˆ
ˆ
ˆ
i
 i
 i 1
 m, j 1
ˆ
m
• The j-th lag of the sample autocorrelation
function of the filtered series is the extended
sample autocorrelation function, and it is
m 

denoted as ˆ j .
22
ESACF
ESACF TABLE
MA
AR
0
1
2
3
3
ˆ10 
ˆ11
ˆ1 2 
ˆ13
ˆ 20 
ˆ 21
ˆ 2 2 
ˆ 23
ˆ30 
ˆ 31
ˆ32 
ˆ33
ˆ 40 
ˆ 41
ˆ 4 2 
ˆ 43
…
…
…
…
…
0
1
2
…
…
…
…
…
…
23
ESACF
• For an ARMA(p,q) process, we have the
following convergence in probability, that is,
for m=1,2,… and j=1,2,…, we have
0, 0  m  p  j  q


m
ˆ

j

 X  0,otherwise
24
ESACF
• Thus, the asymptotic ESACF table for ARMA(1,1)
model becomes
MA
AR
0
1
2
3
4
…
0
X
X
X
X
X
…
1
X
0
0
0
0
…
2
X
X
0
0
0
…
3
X
X
X
0
0
…
4
X
X
X
X
0
…
…
…
…
…
…
…
…
25
ESACF
• In practice, we have finite samples and
ˆ jm  ,0  m  p  j  q may not be exactly zero.
However, we can use the Bartlett’s approximate
m 

formula for the asymptotic variance of ˆ j .
• The orders are tentatively identified by finding a
right (maximal) triangular pattern with vertices
located at (p+d, q) and (p+d, qmax) and in which
all elements are insignificant (based on
asymptotic normality of the autocorrelation
function). The vertex (p+d, q) identifies the order.
26
EXAMPLE (R CODE)
>
>
>
>
>
x=arima.sim(list(order = c(2,0,0), ar = c(-0.2,0.6)), n = 200)
par(mfrow=c(2,1))
par(mfrow=c(1,2))
acf(x)
pacf(x)
27
EXAMPLE (CONTD.)
• After Loading Package TSA in R:
> eacf(x)
AR/MA
0 1 2 3
0 x x x x
1 x x x x
2 o o o o
3 x o o o
4 x x o o
5 x o x o
6 x x o x
7 x x o x
4
x
x
o
o
o
o
o
o
5
x
o
o
o
o
o
o
o
6
x
o
o
o
o
o
o
o
7
x
o
o
o
o
o
o
o
8
x
o
o
o
o
o
o
o
9
x
o
o
o
o
o
o
o
10
x
o
o
o
o
o
o
o
11
x
o
o
o
o
o
o
o
12
x
o
o
o
o
o
o
o
13
x
o
o
o
o
o
o
o
28
MINIMUM INFORMATION CRITERION
MINIC TABLE
AR
0
1
2
3
…
0
SBC(0,0)
SBC(1,0)
SBC(2,0)
SBC(3,0)
…
1
SBC(0,1)
SBC(1,1)
SBC(2,1)
SBC(3,1)
…
MA
2
SBC(0,2)
SBC(1,2)
SBC(2,2)
SBC(3,2)
…
3
SBC(0,3)
SBC(1,3)
SBC(2,3)
SBC(3,3)
…
…
…
…
…
…
…
29
MINIC EXAMPLE
• Simulated 100 data from AR(1) where =0.5
• SAS Output
Minimum Information Criterion
Lags
AR
AR
AR
AR
AR
AR
0
1
2
3
4
5
MA 0
MA 1
MA 2
MA 3
MA 4
MA 5
0.366884
-0.03571
-0.0163
0.001216
0.037894
0.065179
0.074617
-0.00042
0.021657
0.034056
0.069766
0.099543
0.06748
0.038633
0.064698
0.080065
0.115222
0.143406
0.083827
0.027826
0.072834
0.118677
0.14586
0.185604
0.11816
0.064904
0.107481
0.152146
0.189454
0.230186
0.161974
0.097701
0.140204
0.183487
0.229528
0.272322
Error series model: AR(8)
Minimum Table Value: BIC(1,0) = -0.03571
30
NON-STATIONARY TIME SERIES MODELS
• Non-constant in mean
• Non-constant in variance
• Both
31
NON-STATIONARY TIME SERIES
MODELS
• Inspection of the ACF serves as a rough
indicator of whether a trend is present in a
series. A slow decay in ACF is indicative of a
large characteristic root; a true unit root
process, or a trend stationary process.
• Formal tests can help to determine whether a
system contains a trend and whether the
trend is deterministic or stochastic.
32
NON-STATIONARITY IN MEAN
• Deterministic trend
– Detrending
• Stochastic trend
– Differencing
33
DETERMINISTIC TREND
• A deterministic trend is when we say that the
series is trending because it is an explicit
function of time.
• Using a simple linear trend model, the
deterministic (global) trend can be estimated.
This way to proceed is very simple and
assumes the pattern represented by linear
trend remains fixed over the observed time
span of the series. A simple linear trend
model:
Yt    t  at
34
DETERMINISTIC TREND
• The parameter  measure the average change
in Yt from one period to the another:
Yt  Yt  Yt 1   t   t  1  at  at 1
E Yt   
• The sequence {Yt} will exhibit only temporary
departures from the trend line +t. This type
of model is called a trend stationary (TS)
model.
35
EXAMPLE
36
TREND STATIONARY
• If a series has a deterministic time trend, then
we simply regress Yt on an intercept and a
time trend (t=1,2,…,n) and save the residuals.
The residuals are detrended series. If Yt is
stochastic, we do not necessarily get
stationary series.
37
DETERMINISTIC TREND
• Many economic series exhibit “exponential
trend/growth”. They grow over time like an
exponential function over time instead of a
linear function.
• For such series, we want to work with the log
of the series:
lnYt     t  at
So the average growth rate is  :
E  lnYt   
38
DETERMINISTIC TREND
• Standard regression model can be used to
describe the phenomenon. If the deterministic
trend can be described by a k-th order
polynomial of time, the model of the process
Yt   0  1t   2t 2     k t k  at
where at ~ WN 0, .
2
a
• Estimate the parameters and obtain the
residuals. Residuals will give you the detrended
series.
39
DETERMINISTIC TREND
• This model has a short memory.
• If a shock hits a series, it goes back to
trend level in short time. Hence, the best
forecasts are not affected.
• Rarely model like this is useful in practice.
A more realistic model involves stochastic
(local) trend.
40
STOCHASTIC TREND
• A more modern approach is to consider trends
in time series as a variable. A variable trend
exists when a trend changes in an
unpredictable way. Therefore, it is considered
as stochastic.
41
STOCHASTIC TREND
• Recall the AR(1) model: Yt = c + Yt−1 + at.
• As long as || < 1, everything is fine (OLS is
consistent, t-stats are asymptotically normal, ...).
• Now consider the extreme case where  = 1, i.e.
Yt = c + Yt−1 + at.
• Where is the trend? No t term.
42
STOCHASTIC TREND
• Let us replace recursively the lag of Yt on the
right-hand side:
Yt  c  Yt 1  at
 c  c  Yt 2  at 1   at

t
 tc  Y0   ai
i 1
Deterministic trend
• This is what we call a “random walk with
drift”. If c = 0, it is a“random walk”.
43
STOCHASTIC TREND
• Each ai shock represents shift in the intercept.
Since all values of {ai} have a coefficient of unity,
the effect of each shock on the intercept term is
permanent.
• In the time series literature, such a sequence is
said to have a stochastic trend since each ai shock
imparts a permanent and random change in the
conditional mean of the series. To be able to
define this situation, we use Autoregressive
Integrated Moving Average (ARIMA) models.
44
DETERMINISTIC VS STOCHASTIC TREND
• They might appear similar since they both lead to
growth over time but they are quite different.
• To see why, suppose that through any policies, you
got a bigger Yt because the noise at is big. What
will happen next period?
– With a deterministic trend, Yt+1 = c +(t+1)+at+1.
The noise at is not affecting Yt+1. Your policy had a
one period impact.
– With a stochastic trend, Yt+1 = c + Yt + at+1 = c +
(c + Yt−1 + at) + at+1. The noise at is affecting Yt+1.
In fact, the policy will have a permanent impact.
45
DETERMINISTIC VS STOCHASTIC TREND
Conclusions:
– When dealing with trending series, we are always
interested in knowing whether the growth is a
deterministic or stochastic trend.
– There are also economic time series that do not grow
over time (e.g., interest rates) but we will need to check
if they have a behavior ”similar” to stochastic trends ( =
1 instead of || < a, while c = 0).
– A deterministic trend refers to the long-term trend that is
not affected by short term fluctuations in the series.
Some of the occurrences are random and may have a
permanent effect of the trend. Therefore the trend must
contain a deterministic and a stochastic component.
46
DETERMINISTIC TREND EXAMPLE
Simulate data from let’s say AR(1):
>x=arima.sim(list(order = c(1,0,0), ar = 0.6), n = 100)
Simulate data with deterministic trend
>y=2+time(x)*2+x
>plot(y)
47
DETERMINISTIC TREND EXAMPLE
> reg=lm(y~time(y))
> summary(reg)
Call:
lm(formula = y ~ time(y))
Residuals:
Min
1Q
Median
-2.74091 -0.77746 -0.09465
3Q
0.83162
Max
3.27567
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.179968
0.250772
8.693 8.25e-14 ***
time(y)
1.995380
0.004311 462.839 < 2e-16 ***
--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.244 on 98 degrees of freedom
Multiple R-squared: 0.9995,
Adjusted R-squared: 0.9995
F-statistic: 2.142e+05 on 1 and 98 DF, p-value: < 2.2e-16
48
DETERMINISTIC TREND EXAMPLE
> plot(y=rstudent(reg),x=as.vector(time(y)), ylab='Standardized
Residuals',xlab='Time',type='o')
49
DETERMINISTIC TREND EXAMPLE
> z=rstudent(reg) De-trended series
> par(mfrow=c(1,2))
> acf(z)
> pacf(z)
AR(1)
50
STOCHASTIC TREND EXAMPLE
Simulate data from ARIMA(0,1,1):
> x=arima.sim(list(order = c(0,1,1), ma = -0.7), n = 200)
> plot(x)
> acf(x)
> pacf(x)
51
AUTOREGRESSIVE INTEGRATED
MOVING AVERAGE (ARIMA) PROCESSES
• Consider an ARIMA(p,d,q) process
 p  B 1  B  Yt   0   q  B at
d
where  p  B   1  1B     p B and
p
 q  B   1  1B     q B q share no common roots
and at ~ WN 0, .
2
a
52
ARIMA MODELS
• When d=0, 0 is related to the mean of the
process.
0   1  1     p .
• When d>0, 0 is a deterministic trend term.
– Non-stationary in mean:
 p B 1  B Yt  0   q B at
– Non-stationary in level and slope:
 p  B 1  B  Yt   0   q  B at
2
53
RANDOM WALK PROCESS
• A random walk is defined as a process where
the current value of a variable is composed of
the past value plus an error term defined as a
white noise (a normal variable with zero mean
and variance one).
• ARIMA(0,1,0) PROCESS
Yt  Yt 1  at  Yt  1  B Yt  at
where at ~ WN 0, a2 .
54
RANDOM WALK PROCESS
•
•
•
•
Behavior of stock market.
Brownian motion.
Movement of a drunken men.
It is a limiting process of AR(1).
55
RANDOM WALK PROCESS
• The implication of a process of this type is that the best
prediction of Y for next period is the current value, or
in other words the process does not allow to predict
the change (YtYt-1). That is, the change of Y is
absolutely random.
• It can be shown that the mean of a random walk
process is constant but its variance is not. Therefore a
random walk process is nonstationary, and its variance
increases with t.
• In practice, the presence of a random walk process
makes the forecast process very simple since all the
future values of Yt+s for s > 0, is simply Yt.
56
RANDOM WALK PROCESS
57
RANDOM WALK PROCESS
58
RANDOM WALK WITH DRIFT
• Change in Yt is partially deterministic and
partially stochastic.
Yt  Yt 1   0  at

Yt
• It can also be written as
Yt  Y0 
t 0

t
  ai
i
1
deterministic
trend
Pure model of a trend
(no stationary component)
stochastic
trend
59
RANDOM WALK WITH DRIFT
E Yt   Y0  t0
After t periods, the cumulative change in Yt is t0.
E Yt  s Yt   Yt  s 0  not flat
Each ai shock has a permanent effect on the
mean of Yt.
60
RANDOM WALK WITH DRIFT
61
ARIMA(0,1,1) OR IMA(1,1) PROCESS
• Consider a process
1  B Yt  1  B at
where at ~ WN 0, .
2
a
• Letting
Wt  1  B Yt
Wt  1  B at  stationary
62
ARIMA(0,1,1) OR IMA(1,1) PROCESS
• Characterized by the sample ACF of the
original series failing to die out and by the
sample ACF of the first differenced series
shows the pattern of MA(1).
• IF:

j
Yt    1    Yt  j  at where   1   .
j 1

E Yt Yt 1, Yt 2 ,    1   
j 1
j 1
Yt  j
63
Exponentially decreasing. Weighted MA of its past values.
ARIMA(0,1,1) OR IMA(1,1) PROCESS
E Yt 1 Yt , Yt 1 ,  Yt  1   EYt Yt 1 , Yt 2 ,
where  is the smoothing constant in the
method of exponential smoothing.
64
REMOVING THE TREND
• Shocks to a stationary time series are
temporary over time. The series revert to its
long-run mean.
• A series containing a trend will not revert to a
long-run mean. The usual methods for
eliminating the trend are detrending and
differencing.
65
DETRENDING
• Detrending is used to remove deterministic
trend.
• Regress Yt on time and save the residuals.
• Then, check whether residuals are stationary.
66
DIFFERENCING
• Differencing is used for removing the
stochastic trend.
• d-th difference of ARIMA(p,d,q) model is
stationary. A series containing unit roots can
be made stationary by differencing.
• ARIMA(p,d,q)  d unit roots
Integrated of order d, I(d)
Yt ~ I d 
67
DIFFERENCING
• Random Walk:
Yt  Yt 1  at
Yt  at
Non-stationary
Stationary
68
DIFFERENCING
• Differencing always makes us to loose
observation.
• 1st regular difference: d=1
1  B Yt  Yt  Yt 1  Yt
• 2nd regular difference: d=2
1  B  Y   Y  1  2B  B Y  Y  2Y
2
2
t
2
t
t
t
t 1
 Yt 2
Yt  Yt 2 is not the 2nd difference
69
DIFFERENCING
Yt
Yt
2Yt
YtYt-2
3
*
*
*
8
83=5
*
*
5
58=3
35=8
53=2
9
95=4
4(3)=7
98=1
70
KPSS TEST
• To be able to test whether we have a
deterministic trend vs stochastic trend, we are
using KPSS (Kwiatkowski, Phillips, Schmidt and
Shin) Test (1992).
H 0 : Yt ~ I 0  level or trend  stationary
H1 : Yt ~ I 1  difference stationary
71
KPSS TEST
STEP 1: Regress Yt on a constant and trend and
construct the OLS residuals e=(e1,e2,…,en)’.
STEP 2: Obtain the partial sum of the residuals.
t
St   ei
i 1
STEP 3: Obtain the test statistic
St
KPSS  n  2
t 1ˆ
2
n
where ˆ is the estimate of the long-run variance
of the residuals.
2
72
KPSS TEST
• STEP 4: Reject H0 when KPSS is large, because
that is the evidence that the series wander
from its mean.
• Asymptotic distribution of the test statistic
uses the standard Brownian bridge.
• It is the most powerful unit root test but if
there is a volatility shift it cannot catch this
type non-stationarity.
73
DETERMINISTIC TREND EXAMPLE
kpss.test(x,null=c("Level"))
KPSS Test for Level Stationarity
data: x
KPSS Level = 3.4175, Truncation lag parameter = 2, p-value = 0.01
Warning message:
In kpss.test(x, null = c("Level")) : p-value smaller than printed
p-value
> kpss.test(x,null=c("Trend"))
KPSS Test for Trend Stationarity
data: x
KPSS Trend = 0.0435, Truncation lag parameter = 2, p-value = 0.1
Warning message:
In kpss.test(x, null = c("Trend")) : p-value greater than printed
p-value
Here, we have deterministic trend or trend stationary process. Hence, we
need de-trending to work with stationary series.
74
STOCHASTIC TREND EXAMPLE
> kpss.test(x, null = "Level")
KPSS Test for Level Stationarity
data: x
KPSS Level = 3.993, Truncation lag parameter = 3, p-value = 0.01
Warning message:
In kpss.test(x, null = "Level") : p-value smaller than printed pvalue
> kpss.test(x, null = "Trend")
KPSS Test for Trend Stationarity
data: x
KPSS Trend = 0.6846, Truncation lag parameter = 3, p-value = 0.01
Warning message:
In kpss.test(x, null = "Trend") : p-value smaller than printed pvalue
Here, we have stochastic trend or difference stationary process. Hence, we
need differencing to work with stationary series.
75
PROBLEM
• When an inappropriate method is used to
eliminate the trend, we may create other
problems like non-invertibility.
• E.g.
  B Yt   0  1t   t  Trend stationary
where the roots of   B   0 are outside the
unit circle and  t    B at .
76
PROBLEM
• But if we misjudge the series as difference
stationary, we need to take a difference.
Actually, detrending should be applied. Then,
the first difference:
 B Yt  1  1  B  t
Now, we create a non-invertible unit root
process in the MA component.
77
PROBLEM
• To overcome this, look at the inverse sample
autocorrelation function. If it has the same
ACF pattern of non-stationary process (that is,
slow decaying behavior), this means that we
over-differenced the series.
• Go back and de-trend the series instead of
differencing.
• There are also smoothing filters to eliminate
the trend (Decomposition Methods).
78
NON-STATIONARITY IN VARIANCE
• Stationarity in mean
Stationarity in variance
• Non-stationarity in mean
Non-stationarity in
variance
• If the mean function is time dependent,
1. The variance, Var(Yt) is time dependent.
2. Var(Yt) is unbounded as t.
3. Autocovariance and autocorrelation functions are
also time dependent.
4. If t is large wrt Y0, then k  1.
79
VARIANCE STABILIZING
TRANSFORMATION
• The variance of a non-stationary process
changes as its level changes
VarYt   c. f t 
for some positive constant c and a function f.
• Find a function T so that the transformed
series T(Yt) has a constant variance.
The Delta Method
80
VARIANCE STABILIZING
TRANSFORMATION
• Generally, we use the power function
T Yt  
Yt   1

(Box and Cox, 1964)

1
0.5
0
Transformation
0.5
(Yt)0.5
Yt (no transformation)
1
1/Yt
1/(Yt)0.5
ln Yt
81
VARIANCE STABILIZING
TRANSFORMATION
• Variance stabilizing transformation is only for
positive series. If your series has negative values,
then you need to add each value with a positive
number so that all the values in the series are
positive. Now, you can search for any need for
transformation.
• It should be performed before any other analysis
such as differencing.
• Not only stabilize the variance but also improves
the approximation of the distribution by Normal
distribution.
82
TRANSFORMATION
install(TSA)
library(TSA)
oil=ts(read.table('c:/oil.txt',header=T), start=1996, frequency=12)
BoxCox.ar(y=oil)
83