ECON 251
Research Methods
11. Time Series Analysis
and Forecasting
1
Introduction
 Any variable that is measured over time in sequential order


is called a time series.
We analyze time series to detect patterns.
The patterns help in forecasting future values of the time
series.
Future expected value
The time series exhibit a
downward trend pattern.
2
Components of a Time Series
 A time series can consist of four components.
•
•
•
•
Long - term trend (T)
Cyclical effect (C)
Seasonal effect (S)
Random variation (R)
A trend is a long term relatively
smooth pattern or direction, that
persists usually for more than one
year.
3
Components of a Time Series
 A time series can consists of four components.
•
•
•
•
Long - term trend (T)
Cyclical effect (C)
Seasonal effect (S)
Random variation (R)
A cycle is a wavelike pattern
describing a long term behavior
(for more than one year).
6-88 12-88 6-89 12-89 6-90
Cycles are seldom regular, and
often appear in combination with
4
other components.
Components of a Time Series
 A time series can consists of four components.
•
•
•
•
Long - term trend (T)
Cyclical effect (C)
Seasonal effect (S)
Random variation (R)
The seasonal component of
the time-series exhibits a short
term (less than one year)
calendar repetitive behavior.
6-88 12-88 6-89 12-89 6-90
5
Components of a Time Series
 A time series can consists of four components.
•
•
•
•
Long - term trend (T)
Cyclical effect (C)
Seasonal effect (S)
Random variation (R)
Random variation comprises the
irregular unpredictable changes in the
time series. It tends to hide the other
(more predictable) components.
We try to remove random variation
thereby, identify the other components.
6
Time-series models
There are two commonly used time series models:
 The additive model
yt = Tt + Ct + St + Rt
 The multiplicative model
yt = Tt x Ct x St x Rt
7
Smoothing Techniques
 To produce a better forecast we need to determine which
components are present in a time series
 To identify the components present in the time series, we
need first to remove the random variation
 This can be easily done by smoothing techniques:
• moving averages
• exponential smoothing
8
Moving Averages
 A k-period moving average for time period t is the arithmetic


average of the time series values where the period t is the
center observation.
Example: A 3-period moving average for period t is
calculated by
(yt+1 + yt + yt-1)/3.
Example: A 5-period moving average for period t is
calculated by
(yt+2 + yt+1 + yt + yt-1 + yt-2)/5.
9
Example: Gasoline Sales
 To forecast future gasoline sales, the last four years


quarterly sales were recorded.
Calculate the three-quarter and five-quarter moving average.
Data
Period
1
2
3
4
5
6
Year/Quarter Gas Sales
1
1
39
2
37
3
61
4
58
2
1
18
2
56
10
Solution
3 period moving average 5 period moving average
(yt+1 + yt + yt-1)/3
(yt+2 + yt+1 + yt + yt-1 + yt-2)/5
• Solving by hand
Period
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Gas
Sales
39
37
61
58
18
56
82
27
41
69
49
66
54
42
90
66
3-period
moving Avg.
5-period
moving Avg.
52.00
45.67
44.00
52.00
55.00
50.00
45.67
53.00
61.33
56.33
54.00
62.00
66.00
46.00
55.00
48.20
44.80
55.00
53.60
50.40
55.80
56.00
60.20
63.60
*
45.67
*
*
*
42.60
*
* 11
Example: Gasoline Sales
Moving Average
100
Actual
50
3pr MA
15
13
11
9
7
5
3
1
0
Notice how the averaging process removes some of
the random variation.
There is some trend component present, as well as seasonality.
12
Example: Gasoline Sales
The 5-period moving average removes more
variation than the 3-period moving average. Too much smoothing
may eliminate patterns
of interest. Here, the
seasonality
component is removed
when using 5-period
moving average.
100
80
60
40
20
0
1 2
3 4
5 6
7 8
5-period moving average
9 10 11 12 13 14 15 16
3-period moving average
Too little smoothing
leaves much of the
variation, which
disguises the real
patterns.
13
Example – Armani’s Pizza Sales
 Below are the daily sales figures for Armani’s pizza.
Compute a 3-day moving average. (Armani TS.xls)
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Sunday
Week 1
35
42
56
46
67
51
39
Week 2
38
46
61
52
73
58
42
 The moving average that is associated with Monday of week
2 is?
14
Centered moving average
 With even number of observations included in the moving


average, the average is placed between the two periods in
the middle.
To place the moving average in an actual time period, we
need to center it.
Two consecutive moving averages are centered by taking
their average, and placing it in the middle between them.
15
Example – Armani’s Pizza Sales
 Calculate the 4-period moving average and center it, for the
data given below:
Period Time series Moving Avg.
1
15
2
27
(2.5)
19.0
3
20
(3.5)
21.5
4
14
(4.5)
17.5
5
25
6
11
Centerd Mov.Avg.
20.25
19.50
16
Example – Armani’s Pizza Sales
 Armani’s pizza is back. Compute a 2-day centered moving
average.
Week 1
Monday
35
Tuesday
42
Wednesday
56
Thursday
46
Friday
67
Saturday
51
Sunday
39
 The centered moving average that is associated with Friday
of week 1 is?
17
Components of a Time Series
 A time series can consist of four components.
• Long - term trend (T)
• Cyclical effect (C)
• Seasonal effect (S)
Random variation (R)
A trend is a long term relatively
smooth pattern or direction, that
persists usually for more than
one year.
18
Trend Analysis
 The trend component of a time series can be linear or non
linear.
It is easy to isolate the trend component by using linear
regression.
• For linear trend use the model y = b0 + b1t + e.
• For non-linear trend with one (major) change in slope use the
quadratic model y = b0 + b1t + b2t2 + e
19
Example – Pharmaceutical Sales (Identify trend)
 Annual sales for a pharmaceutical company are believed to


change linearly over time.
Based on the last 10 year sales records, measure the trend
component.
Start by renaming your years 1, 2, 3, etc.
Year
1990
1991
1992
1993
1994
Sales
18.0
19.4
18.0
19.9
19.3
Year
1995
1996
1997
1998
1999
Sales
21.1
23.5
23.2
20.4
24.4
20
Example – Pharmaceutical Sales (Identify trend)
 Solution
• Using Excel we have
yˆ  17 . 28  . 6254 (11 )  24 . 1599
30
25
20
Sales
Regression Statistics
Multiple R
0.829573
R Square
0.688192
Adj. R Sq
0.649216
Std Error
1.351968
Observations
10
Forecast
forFit
period
t Line
Plot11
15
yˆ  17 . 28  . 6254 t
10
5
0
ANOVA
0
df
Regression
Residual
Total
Intercept
t
5
SS
MS
F
Sig. F
1 32.27345 32.27345 17.65682 0.002988
8 14.62255 1.827818
9
46.896
Coeffs.
S.E.
t Stat
P-value
17.28 0.92357
18.71 6.88E-08
0.625455 0.148847 4.202002 0.002988
10
15
11
21
Example – Cigarette Consumption
 Determine the long-term trend of the number of cigarettes

smoked by Americans 18 years and older.
Data (Cigarettes.xls)
Consumption
5000
4000
3000
2000
1000
0
1950
1960
1970
1980
Year
1955
1956
1957
1958
1959
1960
1961
.
.
1990
Per Capita
Consumption
3675
3718
3762
3892
4066
4197
4284
.
.
2000
 Solution
• Cigarette consumption
increased between 1955
and 1963, but decreased
thereafter.
• A quadratic model seems
to fit this pattern.
22
Example – Cigarette Consumption
Regression Statistics
Multiple R
0.9821756
R Square
0.9646688
Adj. R Sq
0.962759
Std Error
97.98621
Observations
40
4500
4300
4100
3900
3700
3500
3300
3100
2900
2700
2500
The quadratic model fits
the data very well.
0
ANOVA
df
Regression
Residual
Total
Intercept
t
t-sq
10
20
2
yˆ  3 ,730 . 689  71 . 380 t  2 . 558 t
SS
MS
F
2 9699558.4 4849779.2 505.11707
37
355248 9601.2974
39 10054806
30
40
Sig. F
1.38332E-27
Coeffs
S.E.
t Stat
P-value
3730.6888 48.903765 76.28633 2.67E-42
71.379694 5.5009971 12.975774 2.429E-15
-2.558442 0.130116 -19.66278 3.518E-21
23
Example
 Forecast the trend components for the previous two
examples for periods 12, and 41 respectively.
• yˆ  17 . 28  . 6254 t
yˆ ( t  12 ) 
yˆ ( t  41 ) 
• yˆ  3, 730 . 689  71 . 380 t  2 . 558 t 2
yˆ ( t  12 ) 
yˆ ( t  41 ) 
24
Components of a Time Series
 A time series can consists of four components.
Long - term trend (T)
• Cyclical effect (C)
• Seasonal effect (S)
Random variation (R)
A cycle is a wavelike pattern
describing a long term behavior
(for more than one year).
6-88 12-88 6-89 12-89 6-90
Cycles are seldom regular, and
often appear in combination with
25
other components.
Measuring the Cyclical Effects
 Although often unpredictable, cycles need to be isolated.
 To identify cyclical variation we use the percentage of trend.
• Determine the trend line (by regression).
• Compute the trend value yˆ t for each period t.
• Calculate the percentage of trend by
y t
yˆ t   100
26
Example – Energy Demand
 Does the demand for energy in the US exhibit cyclic

behavior over time? Assume a linear trend and calculate the
percentage of trend.
Data (Energy Demand.xls)
• The collected data included annual energy consumption from
1970 through 1993.
 Find the values for trend and percentage of trend for years
1975 and 1976.
27
Example – Energy Demand
 Now consider the multiplicative model
y t  Tt  C t  R t
yt
yˆ t

Tt  C t  R t
Tt
(Assuming no seasonal effects).
 C t  Rt
The regression line represents trend.
No trend is observed,
but cyclicality and
randomness still exist.
Rate/Predicted Rate
110
105
100
95
90
85
1
3
5
7
9
11 13 15 17 19 21 23
28
(66.4/69.828)100
Period Line Fit Plot
90
yˆ  69 . 3  . 502 t
Consumption
85
80
75
Cnsptn
Pred. cnsmptn
% of trend
66.4
69.7
72.2
74.3
72.5
.
.
69.8283
70.3299
70.8314
71.3329
71.8344
.
.
95.0903
99.1044
101.9322
104.1595
100.9265
.
.
70
65
110.0000
60
0
5
10
15
105.0000
20
25
Period
100.0000
When groups of the
percentage of trend
alternate around 100%,
the cyclical effects
are present.
30
95.0000
90.0000
85.0000
1
3
5
7
9
11 13 15 17 19 21 23
29
Components of a Time Series
 A time series can consists of four components.
Long - term trend (T)
Cyclical effect (C)
• Seasonal effect (S)
Random variation (R)
The seasonal component of
the time-series exhibits a short
term (less than one year)
calendar repetitive behavior.
6-88 12-88 6-89 12-89 6-90
30
Measuring the Seasonal effects
 Seasonal variation may occur within a year or even within a

shorter time interval.
To measure the seasonal effects we can use two common
methods:
1. constructing seasonal indexes
2. using indicator variables
 We will do each in turn
31
Example – Hotel Occupancy
 Calculate the quarterly seasonal indexes for hotel

occupancy rate in order to measure seasonal variation.
Data (Hotel Occupancy.xls)
Year Quarter Rate Year Quarter Rate Year Quarter Rate
1991
1
0.561 1993
1
0.594 1995
1
0.665
2
0.702
2
0.738
2
0.835
3
0.800
3
0.729
3
0.873
4
0.568
4
0.600
4
0.670
1992
1
0.575 1994
1
0.622
2
0.738
2
0.708
3
0.868
3
0.806
4
0.605
4
0.632
32
Example – Hotel Occupancy
 Perform regression analysis for the model
y = b0 + b1t + e where t represents the chronological time,
and y represents the occupancy rate.
Rate
Time (t) Rate
1
0.561
2
0.702
3
0.800
4
0.568
5
0.575
6
0.738
7
0.868
8
0.605
.
.
.
.
yˆ  . 639368  . 005246 t
0
5
10
t
15
20
The regression line represents trend.
33
Example – Hotel Occupancy
 Now let’s consider the multiplicative model, and let’s
DETREND data
y t  Tt  S t  R t
(Assuming no cyclical effects)
Tt  S t  R t

 St  Rt
yˆ t
Tt
yt
The regression line represents trend.
Rate/Predicted rate
No trend is observed, but
seasonality and
randomness still exist.
1.5
1
0.5
0
1
3
5
7
9
11
13
15
17
19
34
Example – Hotel Occupancy
Rate/Predicted rate
0.870
1.080
1.221
0.860
0.864
1.100
1.284
0.888
0.865
1.067
1.046
0.854
0.879
0.993
1.122
0.874
0.913
1.138
1.181
0.900
To remove most of the random variation but leave the
seasonal effects, average the terms StRt for each season.
Rate/Predicted rate
1.5
1
0.5
0
1
3
5
7
9
11
13
15
17
19
Average ratio for quarter 1: (.870 + .864 + .865 + .879 + .913)/5 = .878
Average ratio for quarter 2: (1.080+1.100+1.067+.993+1.138)/5 = 1.076
Average ratio for quarter 3: (1.222+1.284+1.046+1.123+1.182)/5 = 1.171
Average ratio for quarter 4: (.861 +.888 + .854 + .874 + .900)/ 5 = .875
35
Example – Hotel Occupancy
 Interpreting the results
• The seasonal indexes tell us what is the ratio between
the time series value at a certain season, and the overall
seasonal average.
• In our problem:
7.6% above the
annual average
Annual average
occupancy (100%)
17.1% above the
117.1% annual average
12.5% below the
annual average
107.6%
12.2% below the
annual average
87.8%
87.5%
Quarter 1 Quarter 2 Quarter 3 Quarter 4 Quarter 1 Quarter 2 Quarter 3 Quarter 4
36
Creating Seasonal Indexes
 Normalizing the ratios:
• The sum of all the ratios must be 4, such that the average
ratio per season is equal to 1.
• If the sum of all the ratios is not 4, we need to normalize
(adjust) them proportionately. Suppose the sum of ratios
equaled 4.1. Then each ratio will be multiplied by 4/4.1
(Seasonal averaged ratio) (number of seasons)
Seasonal index =
Sum of averaged ratios
In our problem the sum of all the averaged ratios is equal to 4:
.878 + 1.076 + 1.171 + .875 = 4.0.
No normalization is needed. These ratios become the seasonal indexes.
37
Example
 Using a different example on quarterly GDP, assume that

the averaged ratios were:
Quarter 1:
.97
Quarter 2:
1.03
Quarter 3:
.87
Quarter 4:
1.00
Determine the seasonal indexes.
38
Deseasonalizing time series
Seasonally adjusted time series = Actual time series
Seasonal index
 By removing the seasonality, we can:
• compare data across different seasons
• identify changes in the other components of the time - series.
39
Example – Hotel Occupancy
 A new manager was hired in Q3 1994, when hotel

occupancy rate was 80.6%. In the next quarter, the
occupancy rate was 63.2%. Should the manager be fired
because of poor performance?
Method 1:
• compare occupancy rates in Q4 1994 (0.632) and Q3 1994
(0.806)
• the manager has underperformed  he should be fired
 Method 2:
• compare occupancy rates in Q4 1994 (0.632) and Q4 1993
(0.600)
• the manager has performed very well  he should get a raise
40
Example – Hotel Occupancy
 Method 3:
• compare deseasonalized occupancy rates in Q4 1994 and Q3
1994
Seasonally adjusted time series = Actual time series
Seasonal index
Seasonally adjusted
Actual occupancy rate in Q4 1994
=
occupancy rate in Q4
Seasonal index for Q4
1994
Recall:
Seasonally adjusted
occupancy rate in Q3
1994
the manager has indeed performed well  he should be maybe get a raise
41
Recomposing the time series
 Recomposition will recreate the time series, based only on
the trend and seasonal components – you can then deduce
the value of the random component.
yˆ t  Tˆ t  Sˆ t  (. 639  . 0052 t ) Sˆ t
In period #1 ( quarter 1):
yˆ 1  Tˆ1  Sˆ 1  (. 639  . 0052 (1))(. 878 )  . 566
In period #2 ( quarter 2):
yˆ 2  Tˆ 2  Sˆ 2  (. 639  . 0052 ( 2 ))( 1 . 076 )  . 699
Actual series
Smoothed series
0.9
0.8
0.7
0.6
The linear trend (regression) line
0.5
1
3
5
7
9
11
13
15
17
19
42
Example – Recomposing the Time Series
 Recompose the smoothed time series in the above example
for periods 3 and 4 assuming our initial seasonal index
figures of:
Quarter 1: 0.878
Quarter 3: 1.171
Quarter 2: 1.076
Quarter 4: 0.875
and equation of:
yˆ t  Tˆ t  Sˆ t  (. 639  . 0052 t ) Sˆ t
yˆ 3  ( 0 . 639  0 . 0052 * t ) * SI 3
 ( 0 . 639  0 . 0052 * 3 ) * 1 . 171
yˆ 4  ( 0 . 639  0 . 0052 * t ) * SI 4
 ( 0 . 639  0 . 0052 * 4 ) * 0 . 875
43
Seasonal Time Series with Indicator Variables
 We create a seasonal time series model by using indicator


variables to determine whether there is seasonal variation in
data.
Then, we can use this linear regression with indicator
variables to forecast seasonal effects.
If there are n seasons, we use n-1 indicator variables to
define the season of period t:
•
[Indicator variable n] = 1 if season n belongs to period t
•
[Indicator variable n] = 0 if season n does not
belong to period t
44
Example – Hotel Occupancy
 Let’s again build a model that would be able to forecast

quarterly occupancy rates at a luxurious hotel, but this time
we will use a different technique.
Data (Hotel Occupancy.xls)
Year Quarter Rate Year Quarter Rate Year Quarter Rate
1991
1
0.561 1993
1
0.594 1995
1
0.665
2
0.702
2
0.738
2
0.835
3
0.800
3
0.729
3
0.873
4
0.568
4
0.600
4
0.670
1992
1
0.575 1994
1
0.622
2
0.738
2
0.708
3
0.868
3
0.806
4
0.605
4
0.632
45
Example – Hotel Occupancy
 Use regression analysis and indicator variables to forecast
hotel occupancy rate in 1996
Data
Qi =
1 if quarter i occur at t
0 otherwise
Quarter 1 belongs to t = 1
y
0.561
0.702
0.800
0.568
0.575
0.738
0.868
0.605
.
.
Quarter 2 does not
belong to t = 1
Quarter 3 does not
belong to t = 1
Quarters 1, 2, 3, do not
belong to period t = 4.
t
1
2
3
4
5
6
7
8
.
.
Q1
1
0
0
0
1
0
0
0
.
.
Q2
0
1
0
0
0
1
0
0
.
.
Q3
0
0
1
0
0
0
1
0
.
.
46
 The regression model
There is insufficient
evidence to conclude
y = b0 + b1t + b2Q1 + b3Q2 + b4Q3 +e
that seasonality
Regression Statistics There is sufficient evidence to
causes occupancy
Multiple R 0.943022
conclude that trend is present. rate in quarter 1 be
R Square
0.88929
different than this of
Adj R Sq. 0.859767
Std. Error
0.03806
Good fit quarter 4!
Observations
20
ANOVA
df
Regression
Residual
Total
Intercept
t
Q1
Q2
Q3
SS
MS
F
4 0.174531 0.043633 30.12217
15 0.021728 0.001449
19 0.196259
Coeffs
0.55455
0.005037
0.003512
0.139275
0.205237
S.E.
0.024812
0.001504
0.02449
0.024258
0.024118
t Stat
22.35028
3.348434
0.143423
5.74134
8.509752
P-value
6.27E-13
0.004399
0.887865
3.9E-05
3.99E-07
Sig. F
5.2E-07
Seasonality
effects on
occupancy rate in
quarter 2 and 3
are different than
this of quarter 4!
47
 The estimated regression model
y = b0 + b1t + b2Q1 + b3Q2 + b4Q3 +e
Regression Statistics
Multiple R 0.943022
R Square
0.88929
Adj R Sq. 0.859767
Std. Error
0.03806
Observations
20
df
Intercept
t
Q1
Q2
Q3
yˆ 21  . 555  . 00504 ( 21)  . 0035 (1)  . 664
yˆ 22  . 555  . 00504 ( 22 )  . 139 (1)  . 805
yˆ 23  . 555  . 00504 ( 23 )  . 205 (1)  . 876
yˆ 24  . 555  . 00504 ( 24 )
ANOVA
Regression
Residual
Total
yˆ t  . 555  . 00504 t  . 0035 Q1  . 139 Q 2  . 205 Q 3
SS
MS
F
4 0.174531 0.043633 30.12217
15 0.021728 0.001449
19 0.196259
Coeffs
0.55455
0.005037
0.003512
0.139275
0.205237
S.E.
0.024812
0.001504
0.02449
0.024258
0.024118
t Stat
22.35028
3.348434
0.143423
5.74134
8.509752
P-value
6.27E-13
0.004399
0.887865
3.9E-05
3.99E-07
 . 675
Sig. F
5.2E-07
b4
b2
b0
b3
1 2
3 4
48
Time-Series Forecasting with Autoregressive (AR)
models
 Autocorrelation among the errors of the regression model
provides opportunity to produce accurate forecasts.
 Correlation between consecutive residuals leads to the following
autoregressive model:
yt = b0 + b1yt-1 + et
 This is type of model is termed a first-order autoregressive
model, and is denoted AR(1). A model which also includes both
yt-1 and yt-2 is a second order model AR(2) and so on.
 Selecting between them is done on the basis of the significance
of the highest order term. If you reject the highest order term, rerun the model without it. Apply the same criterion for the new
highest order term, and so on.
49
Example – Consumer Price Index (CPI)
 Forecast the increase in the Consumer Price Index (CPI) for

the year 1994, based on the data collected for the years
1951 -1993.
Data (CPI.xls)
t
1
2
3
4
.
.
43
Y (t)
7.9
2.2
0.8
0.5
.
.
3
50
Example – Consumer Price Index (CPI)
 Regress the Percent increase (yt) on the time variable (t)
 it does not lead to good results.
Y (t)
16
14
Yt = 2.14 + .098t
12
10
r2 = 15.0%
8
(+)
6
4
2
(-)
(-)
0
-2 0
10
20
30
t
40
50
The residuals
form a pattern
over time. Also,
this indicates the
potential
presence of first order
autocorrelation
between
consecutive residuals.
51
Example – Consumer Price Index (CPI)
 An autoregressive model appears to be a desirable
technique.
The increase in the CPI for periods 1, 2, 3,… are predictors of the increase
in the CPI for periods 2, 3, 4,…, respectively.
Y (t)
7.9
2.2
0.8
0.5
-0.4
.
.
Y (t-1)
7.9
2.2
0.8
0.5
.
.
Regression Statistics
Multiple R
0.797851
R Square
0.636566
Adj. R Sq.
0.62748
Std. Error
1.943366
Observations
42
df
Intercept
Y (t-1)
Forecast
for 1994 ( t  44 ) :
yˆ 44  . 8066  . 78705 y 43
 . 8066  . 78705 ( 3 )  3 . 16775
ANOVA
Regression
Residual
Total
yˆ t  . 8066  . 78705 y t 1
SS
MS
F
Sig. F
1 264.5979 264.5979 70.06111 2.51E-10
40 151.0669 3.776673
41 415.6648
Coeffs
S.E.
t Stat
P-value
0.806644 0.506089 1.59388 0.118835
0.787045 0.094029 8.370251 2.51E-10
52
16
Compare
14
linear trend model
Y (t)
12
10
8
6
4
2
autoregressive model
0
-2 0
10
20
30
40
50
Y (t)
t
-5
16
14
12
10
8
6
4
2
0
-2 0
5
Y (t-1)
10
15
53
Selecting Among Alternative Forecasting
Models
 There are many forecasting models available
* o
o
*
o o*
* Model 1
?
o* o*
o o*
* Model 2
Which model
performs better?
54
Model selection
 To choose a forecasting method, we can evaluate forecast

accuracy using the actual time series.
The two most commonly used measures of forecast
n
accuracy are:

• Mean Absolute Deviation
MAD  t  1
y t  Ft
n
• Sum of Squares for Forecast Error
n
SSE  SSFE 
 (y
t
 Ft )
2
t 1
 In practice, SSE is SSFE are used interchangeably.
55
Model selection
 Procedure for model selection.
1. Use some of the observations to develop several competing
forecasting models.
2. Run the models on the rest of the observations.
3. Calculate the accuracy of each model using both MAD and
SSE criterion.
4. Use the model which generates the lowest MAD value, unless
it is important to avoid (even a few) large errors - in this case
use best model as indicated by the lowest SSE.
56
Example – Model selection I
 Assume we have annual data from 1950 to 2001.
 We used data from 1950 to 1997 to develop three alternate

forecasting models (two different regression models and an
autoregressive model),
Use MAD and SSE to determine which model performed
best using the model forecasts versus the actual data for
1998-2001.
Year
1998
1999
2000
2001
Actual y
129
142
156
183
Forecast value
Reg 1
Reg 2
136
118
148
141
150
158
175
163
AR
130
146
170
180
57
Example – Model selection I
 Solution
Actual y
in 1998
• For model 1
MAD 
Forecast for y
in 1998
129 136  142 148  156 150  183 175
4
 6 . 75
SSE  (129  136 )  (142  148 ) 
2
2
 (156  150 )  (183  175 )  185
2
2
• Summary of results
MAD
SSE
Reg 1
6.75
185
Reg 2
8.5
526
AR
5.5
222
58
Example – Model selection II
 For the actual values and the forecast values of a time
series shown in the following table, calculate MAD and SSE.
Forecast Value Actual Value
Ft
173
186
192
211
223
MAD
SSE
yt
166
179
195
214
220
59