ma j-0 no v-0 ap r-0 ok t- ma r- 0 au g-9 feb -9 9 jul98 jan - 98 jun - 97 Time series Sales figures jan 98 - dec 01 403020100

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Transcript ma j-0 no v-0 ap r-0 ok t- ma r- 0 au g-9 feb -9 9 jul98 jan - 98 jun - 97 Time series Sales figures jan 98 - dec 01 403020100

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Time series
Sales figures jan 98 - dec 01
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2001-01-15
2000-01-15
1999-01-15
1998-01-15
1997-01-15
1996-01-15
1995-01-15
1994-01-15
1993-01-15
1992-01-15
1991-01-15
1990-01-15
1989-01-15
1988-01-15
1987-01-15
1986-01-15
1985-01-15
1984-01-15
1983-01-15
1982-01-15
1981-01-15
1980-01-15
Tot-P ug/l, Råån, Helsingborg 1980-2001
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800
700
600
500
400
300
200
100
0
Characteristics
• Non-independent observations (correlations
structure)
• Systematic variation within a year (seasonal
effects)
• Long-term increasing or decreasing level
(trend)
• Irregular variation of small magnitude
(noise)
Where can time series be found?
• Economic indicators: Sales figures,
employment statistics, stock market indices,
…
• Meteorological data: precipitation,
temperature,…
• Environmental monitoring: concentrations
of nutrients and pollutants in air masses,
rivers, marine basins,…
Time series analysis
• Purpose: Estimate different parts of a time
series in order to
– understand the historical pattern
– judge upon the current status
– make forecasts of the future development
• Methodologies:
Method
This course?
Time series regression
Yes
Classical decomposition
Yes
Exponential smoothing
Yes
ARIMA modelling (Box-Jenkins)
Yes
Non-parametric tests
No
Transfer function and intervention models
No
State space modelling
No
Spectral domain analysis
No
Time series regression?
Let
yt=(Observed) value of times series at time point t
and assume a year is divided into L seasons
Regession model (with linear trend):
yt=0+ 1t+j sj xj,t+t
where xj,t=1 if yt belongs to season j and 0 otherwise, j=1,…,L-1
and {t } are assumed to have zero mean and constant variance
(2 )
The parameters 0, 1, s1,…, s,L-1 are estimated by the Ordinary Least
Squares method:
(b0, b1, bs1, … ,bs,L-1)=argmin {(yt – (0+ 1t+j sj xj,t)2}
Advantages:
• Simple and robust method
• Easily interpreted components
• Normal inference (conf.intervals, hypothesis testing) directly
applicable
•Drawbacks:
•Fixed components in model (mathematical trend function and
constant seasonal components)
•No consideration to correlation between observations
Example: Sales figures
Sales figures January 1998 - December 2001
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month
jan-98
feb-98
mar-98
apr-98
maj-98
jun-98
jul-98
aug-98
sep-98
okt-98
nov-98
dec-98
20.33
20.96
23.06
24.48
25.47
28.81
30.32
29.56
30.01
26.78
23.75
24.06
jan-99
feb-99
mar-99
apr-99
maj-99
jun-99
jul-99
aug-99
sep-99
okt-99
nov-99
dec-99
23.58
24.61
27.28
27.69
29.99
30.87
32.09
34.53
30.85
30.24
27.86
24.67
jan-00
feb-00
mar-00
apr-00
maj-00
jun-00
jul-00
aug-00
sep-00
okt-00
nov-00
dec-00
26.09
26.66
29.61
32.12
34.01
32.98
36.38
35.90
36.42
34.04
31.29
28.50
jan-01
feb-01
mar-01
apr-01
maj-01
jun-01
jul-01
aug-01
sep-01
okt-01
nov-01
dec-01
28.43
29.92
33.44
34.56
34.22
38.91
41.31
38.89
40.90
38.27
32.02
29.78
Construct seasonal indicators: x1, x2, … , x12
January (1998-2001):
x1 = 1, x2 = 0, x3 = 0, …, x12 = 0
February (1998-2001):
x1 = 0, x2 = 1, x3 = 0, …, x12 = 0
etc.
x1 = 0, x2 = 0, x3 = 0, …, x12 = 1
December (1998-2001):
sales
time
x1
x2
x3
x4
x5
x6
x7
x8
x9
x10
x11
x12
20.33
1
1
0
0
0
0
0
0
0
0
0
0
0
20.96
2
0
1
0
0
0
0
0
0
0
0
0
0
23.06
3
0
0
1
0
0
0
0
0
0
0
0
0
24.48
4
0
0
0
1
0
0
0
0
0
0
0
0
I
I
I
I
I
I
I
I
I
I
I
I
I
I
32.02
47
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0
0
0
0
1
0
29.78
48
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0
0
0
0
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0
1
Use 11 indicators, e.g. x1 - x11 in the regression model
Regression Analysis: sales versus time, x1, ...
The regression equation is
sales = 18.9 + 0.263 time + 0.750 x1 + 1.42 x2 + 3.96 x3 + 5.07 x4 + 6.01 x5
+ 7.72 x6 + 9.59 x7 + 9.02 x8 + 8.58 x9 + 6.11 x10 + 2.24 x11
Predictor
Coef
SE Coef
T
P
Constant
18.8583
0.6467
29.16
0.000
time
0.26314
0.01169
22.51
0.000
x1
0.7495
0.7791
0.96
0.343
x2
1.4164
0.7772
1.82
0.077
x3
3.9632
0.7756
5.11
0.000
x4
5.0651
0.7741
6.54
0.000
x5
6.0120
0.7728
7.78
0.000
x6
7.7188
0.7716
10.00
0.000
x7
9.5882
0.7706
12.44
0.000
x8
9.0201
0.7698
11.72
0.000
x9
8.5819
0.7692
11.16
0.000
x10
6.1063
0.7688
7.94
0.000
x11
2.2406
0.7685
2.92
0.006
S = 1.087
R-Sq = 96.6%
R-Sq(adj) = 95.5%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
12
1179.818
98.318
83.26
0.000
Residual Error
35
41.331
1.181
Total
47
1221.150
Source
DF
Seq SS
time
1
683.542
x1
1
79.515
x2
1
72.040
x3
1
16.541
x4
1
4.873
x5
1
0.204
x6
1
10.320
x7
1
63.284
x8
1
72.664
x9
1
100.570
x10
1
66.226
x11
1
10.039
Unusual Observations
Obs
time
sales
Fit
SE Fit
Residual
St Resid
12
12.0
24.060
22.016
0.583
2.044
2.23R
21
21.0
30.850
32.966
0.548
-2.116
-2.25R
R denotes an observation with a large standardized residual
Predicted Values for New Observations
New Obs
1
Fit
SE Fit
32.502
0.647
95.0% CI
(
31.189,
95.0% PI
33.815)
(
29.934,
35.069)
Values of Predictors for New Observations
New Obs
time
x1
x2
x3
x4
x5
x6
1
49.0
1.00
0.000000
0.000000
0.000000
0.000000
0.000000
x7
x8
x9
x10
x11
0.000000
0.000000
0.000000
0.000000
0.000000
New Obs
1
Sales figures with predicted value
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month
What about serial correlation in data?
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Positive serial correlation:
Values follow a smooth pattern
Negative serial correlation:
Values show a “thorny” pattern
How to obtain it?
Use the residuals.
11
ˆ

et  yt  yˆ t  yt    0  ˆ1  t   ˆs , j  x j ,t  ; t  1,...,48
j 1


Residual plot from the regression analysis:
2
Smooth or thorny?
1
0
-1
-2
10
20
30
Month number (from jan 1998)
Durbin Watson test on residuals:
n
d
2
(
e

e
)
 t t 1
t 2
n
2
e
t
t 1
Thumb rule:
If d < 1 or d > 3, the conclusion is that residuals (and original data)
are correlated.
Use shape of figure (smooth or thorny) to decide if positive or
negative)
(More thorough rules for comparisons and decisions about positive or
negative correlations exist.)
Durbin-Watson statistic = 2.05
(Comes in the output )
Value > 1 and < 3  No significant serial correlation in residuals!
• Decompose – Analyse the observed time
series in its different components:
–
–
–
–
Trend part
Seasonal part
Cyclical part
Irregular part
(TR)
(SN)
(CL)
(IR)
Cyclical part: State-of-market in economic time series
In environmental series, usually together with
TR
• Multiplicative model:
yt=TRt·SNt ·CLt ·IRt
 Suitable for economic indicators
 Level is present in TRt or in
TCt=(TR∙CL)t
 SNt , IRt (and CLt) works as
indices
 Seasonal variation increases
with level of yt
• Additive model:
yt=TRt+SNt +CLt +IRt
 More suitable for environmental
data
 Requires constant seasonal
variation
 SNt , IRt (and CLt) vary around 0
Example 1: Sales data
Observed (blue) and deseasonalised (magenta)
Sales figures jan 98 - dec 01
Observed (blue) and theoretical trend (magenta)
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jul-
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20.00
15.00
10.00
02
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jul-
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5.00
jan
-97
-98
Observed (blue) with estimated trendline (black)
45.00
40.00
35.00
30.00
25.00
20.00
15.00
10.00
5.00
0.00
jun
jan
jun
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feb
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jul98
jan
- 98
jun
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-97
45.00
40.00
35.00
30.00
25.00
20.00
15.00
10.00
5.00
0.00
45.00
40.00
35.00
30.00
25.00
20.00
15.00
10.00
5.00
0.00
0.00
mar-97
jul-98
dec-99
apr-01
sep-02
Example 2:
Estimation of components, working scheme
1.
Seasonally adjustment/Deseasonalisation:
•
•
SNt usually has the largest amount of variation among the components.
The time series is deseasonalised by calculating centred and weighted Moving
Averages:
M
( L)
t

yt ( L / 2)  yt ( L / 21)  2  ...  yt  2  ...  yt ( L / 21)  2  yt ( L / 2)
L2
where L=Number of seasons within a year (L=2 for ½-year data, 4 for
quaerterly data och 12 för monthly data)
– Mt becomes a rough estimate of (TR∙CL)t .
– Rough seasonal components are obtained by
• yt/Mt in a multiplicative model
• yt – Mt in an additive model
– Mean values of the rough seasonal components are calculated for
eacj season separetly.  L means.
– The L means are adjusted to
• have an exact average of 1 (i.e. their sum equals L ) in a multiplicative
model.
• Have an exact average of 0 (i.e. their sum equals zero) in an additive
model.
– Final estimates of the seasonal components are set to these
adjusted means and are denoted:
sn1 ,, snL
– The time series is now deaseasonalised by
•
yt*  yt / snt
in a multiplicative model
•
yt*  yt  snt
in an additive model
where
snt
is one of
sn1 ,, snL
depending on which of the seasons t represents.
2.
Seasonally adjusted values are used to estimate the trend
component and occasionally the cyclical component.
If no cyclical component is present:
•
•
Apply simple linear regression on the seasonally adjusted values
Estimates trt of linear or quadratic trend component.
The residuals from the regression fit constitutes estimates, irt of
the irregular component
If cyclical component is present:
•
Estimate trend and cyclical component as a whole (do not split
them) by
tct 
yt*m  yt*( m1)    yt*  yt*1    yt* m
2  m 1
i.e. A non-weighted centred Moving Average with length 2m+1
caclulated over the seasonally adjusted values
– Common values for 2m+1: 3, 5, 7, 9, 11, 13
– Choice of m is based on properties of the final
estimate of IRt which is calculated as
*
ir

y
•
t
t /(tct )
•
irt  yt*  (tct )
in a multiplicative model
in an additive model
– m is chosen so to minimise the serial correlation
and the variance of irt .
– 2m+1 is called (number of) points of the
Moving Average.
Example, cont: Home sales data
Minitab can be used for decomposition by
StatTime seriesDecomposition
Val av modelltyp
Option to choose
between two
models
Time Series Decomposition
Data
Sold
Length
47,0000
NMissing
0
Trend Line Equation
Yt = 5,77613 + 4,30E-02*t
Seasonal Indices
Period
Index
1
-4,09028
2
-4,13194
3
0,909722
4
-1,09028
5
3,70139
6
0,618056
7
4,70139
MAPE:
8
4,70139
MAD:
0,9025
9
-1,96528
MSD:
1,6902
10
0,118056
11
-1,29861
12
-2,17361
Accuracy of Model
16,4122
Deseasonalised data have been stored in a column with head DESE1.
Moving Averages on these column can be calculated by
StatTime seriesMoving average
Choice of 2m+1
TC component with 2m +1 = 3 (blue)
MSD should be kept as small as possible
By saving residuals from the moving averages we can calculate MSD
and serial correlations for each choice of 2m+1.
2m+1
MSD
Corr(et,et-1)
3
1.817
-0.444
5
1.577
-0.473
7
1.564
-0.424
9
1.602
-0.396
11
1.542
-0.431
13
1.612
-0.405
A 7-points or 9-points moving average seems most reasonable.
Serial correlations are simply calculated by
StatTime seriesLag
and further
StatBasic statisticsCorrelation
Or manually in Session window:
MTB > lag ’RESI4’ c50
MTB > corr ’RESI4’ c50
Analysis with multiplicative model:
Time Series Decomposition
Data
Sold
Length
47,0000
NMissing
0
Trend Line Equation
Yt = 5,77613 + 4,30E-02*t
Seasonal Indices
Period
Index
1
0,425997
2
0,425278
3
1,14238
4
0,856404
5
1,52471
6
1,10138
MAPE:
7
1,65646
MAD:
0,9057
8
1,65053
MSD:
1,6388
9
0,670985
10
1,02048
11
0,825072
12
0,700325
Accuracy of Model
16,8643
additive
additive
additive
Classical decomposition, summary
Multiplicative model:
yt  TRt  SNt  CLt  IRt
Additive model:
yt  TRt  SNt  CLt  IRt
Deseasonalisation
• Estimate trend+cyclical component by a
centred moving average:
CMAt 
yt ( L / 2)  yt ( L / 21)  2  ...  yt  2  ...  yt ( L / 21)  2  yt ( L / 2)
L2
where L is the number of seasons (e.g. 12, 4, 2)
• Filter out seasonal and error (irregular)
components:
– Multiplicative model:
yt
snt  irt 
CMAt
-- Additive model:
snt  irt  yt  CMAt
Calculate monthly averages
Multiplicative model:
sn m 
1
nm

nm
( snl  irl )
Additive model:
sn m 
1
nm

for seasons m=1,…,L
nm
( snl  irl )
Normalise the monhtly means
Multiplicative model:
snm 
sn m
1
L l 1 sn l
L
L


L
l 1
sn l
Additive model:
snm  sn m 
1
L

L
l 1
sn l
Deseasonalise
Multiplicative model:
yt
dt 
snt
Additive model:
dt  yt  snt
where snt = snm for current month m
Fit trend function, detrend (deaseasonalised) data
trt  f (t )
Multiplicative model:
dt
clt  irt 
trt
Additive model:
clt  irt  dt  trt
Estimate cyclical component and separate from error
component
Multiplicative model:
clt 
irt 
(cl  ir )t k  (cl  ir )t ( k 1)  ...  (cl  ir )t  ...  (cl  ir )t  k
2  k 1
(cl  ir )t
clt
Additive model:
clt 
(cl  ir )t k  (cl  ir )t ( k 1)  ...  (cl  ir )t  ...  (cl  ir )t  k
irt  (cl  ir )t  clt
2  k 1