1.
BUSINESS
STATISTICS
BEO1106
WEEK 11
BUSINESS AND LAW
Dr. Hubert Fernando and Dr. Sidney Lung
2012
WWW.VU.EDU.AU
1
2.
Business Statistics
TIME-SERIES ANALYSIS
Textbook Chapter 14
3.
TIME-SERIES ANALYSIS
•
According to classical time-series analysis an observed time series is
the combination of some pattern and random variations.
The aim is to separate them from each other in order to
a) describe the historical pattern in the data,
and to
b) prepare forecasts by projecting the revealed historical pattern
into the future.
•
The pattern itself is likely to contain some, or all, of the following three
components: trend, seasonal and cyclical.
3
4.
Trend: The long-term general change in the level of the data with a
duration of longer than a year.
Yt
It can be linear (straight line)
Yi = b 0 + b1X i
t
or non-linear (smooth curve),
Hourly earnings: Manufacturing: Major seven countries
Broad money: (sa): Sw
eden
1995=100
1995=100
120
140
100
120
100
80
80
60
60
40
40
20
0
20
Sep-70
Sep-80
Sep-90
Sep-00
0
Jan-61
Jan-71
Jan-81
Jan-91
Jan-01
4
5.
Seasonal variations:
Regular fluctuations within a
period of no longer than a year.
Yt
Seasonal variations are usually
associated with the four seasons of the
year.
t
Hungary: Commodity output: Cement
Australia: Retail turnover: Department stores
'000 tonnes
$m
600
2500
500
2000
400
1500
300
1000
200
500
100
0
Dec-82
Dec-85
Dec-88
Dec-91
Dec-94
Dec-97
Dec-00
0
Jan-83
Jan-86
Jan-89
Jan-92
Jan-95
Jan-98
Jan-01
5
6.
Cyclical variations:
Yt
Fluctuations around the long-term
trend, lasting longer than a year.
Peak
Beginning
trough
The time gap between the beginning
trough and ending trough is the length
of the cycle.
t
Ending
trough
Aus: Dw
elling units approved: Private: New houses
Cyclical variations are attributed to
business cycles; to the ups and downs
in the level of business activity.
Expenditure on GDP: Construction: United States: (sa)
Number
bln 96 USD
40000
900
800
35000
700
30000
600
25000
500
20000
15000
Dec-70
400
Dec-75
Dec-80
Dec-85
Dec-90
Dec-95
Dec-00
300
Dec-60
Dec-70
Dec-80
Dec-90
Dec-00
6
7.
• The random variations of the data comprise the deviations of the
observed time series from the underlying pattern.
When this irregular component is strong compared to the (quasi-)
regular components, it tends to hide the seasonal and cyclical
variations, and
it is difficult to be detached from the pattern.
However, if we manage to capture the trend, the seasonal and cyclical
variations, the remaining changes do not have any discernible pattern,
so they are totally unpredictable.
7
8.
The four components of a time series ,T: trend, S: seasonal, C: cyclical,
R: random) can be combined in different ways.
Additive:
Multiplicative:
Yi = Ti + S i + Ci + I i
Yi = Ti × S i × Ci × I i
135 = 60+12+36+27
135 = 60 x 1.2 x 1.5 x 0.25
e.g. If the trend is linear, these two models look as follows:
Yi = (b 0 + b1X i ) + S i + Ci + I i
Yi = (b 0 + b1X i ) × S i × Ci × I i
8
9.
Austria: Domestic demand: Retail sales: Volume
Australia: Retail turnover: Recreational goods
1995=100
$m
160
900
800
140
700
120
600
100
500
400
80
300
60
40
200
Dec-76
Dec-80
Dec-84
Dec-88
Dec-92
Dec-96
Dec-00
100
Jan-83
Jan-86
Jan-89
Jan-92
Jan-95
Jan-98
Jan-01
These time series have an increasing linear trend component.
the fluctuations around
this trend have the
same intensity.
the fluctuations around this
trend are more and more
intensive.
9
10.
SMOOTHING TECHNIQUES
Are used to reduce, the random fluctuations in a time series so as to more
clearly expose the existence of the other components.
Example1
The daily (Monday – Friday) sales figures during the last four weeks were
recorded in a medium-size merchandising firm.
70
60
Sales
50
40
30
20
10
week 1
Week 2
Week 3
Week 4
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Day
10
11.
Moving Average
Day
Sales
1
2
3
4
5
6
43
45
22
25
31
51
43 + 45 + 22 = 110
3-day moving average
3-day moving
sum
3-day moving
average
110.0
92.0
78.0
107.0
etc.
36.7
30.7
26.0
35.7
etc.
70
60
MA(3)
Sales
50
40
30
MA(5)
20
10
110
= 36.67
3
Longer the
period
stronger
the
smoothing
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Day
11
12.
Centered Moving Average
To calculate a 4-period moving average, MA(4) must be placed
between the second and third observations.
Since this makes interpretation and graphing difficult, we center the
moving averages which are the 2-period moving averages of the 4period moving averages.
4-day centered moving average CMA(4).
Day
1
2
3
4
5
6
7
Sales
43
45
22
25
31
51
41
MA(4)
33.8
30.8
32.3
37.0
40.0
etc.
CMA(4)
32.3
31.5
34.6
38.5
etc.
33.8 + 30.8
= 32.3
2
12
13.
Exponential Smoothing
Ei = wYi + (1 − w) Ei −1
where Ei : exponentially smoothed value for time period i ;
Ei-1 : exponentially smoothed value for time period i -1;
Yi : observed value for time period i ;
w : smoothing constant, 0 < w < 1.
a) Assuming that Y has been observed from i = 1, this formula can be
applied only from the second time period.
For i = 1 we set the smoothed value equal to the observed
value, i.e. E1 = Y1
b) The smoothing constant determines the strength of smoothing,
the larger the value of w the weaker the smoothing effect.
13
14.
Example 2:
Using the quarterly Australian unemployed persons (in thousands) data for
the years 1989-98,
a) Apply the exponential smoothing technique with W = 0.2 and W = 0.7.
w = 0.2
1989
1990
1
2
3
4
1
2
Y
E (W=0.2) E (W=0.7)
1735.6
1735.6
1735.6
1507.9
1690.1
1576.2
1450.2
1642.1
1488.0
1402.7
1594.2
1428.3
1689.9
1613.3
1611.4
1621.4
1615.0
1618.4
etc.
etc.
etc.
i
E1 = Y1 = 1735.6
E 2 = WY2 + (1 − W )E1
= 0.2 ×1507.9 + 0.8 ×1735.6 = 1690.1
E3 = WY3 + (1 − W)E 2
= 0.2 ×1450.2 + 0.8 ×1690.1 = 1642.1
14
15.
Plot the time series and the exponential smoothed values on the same
graph.
3500
If w = 0.7, Ei is quite similar to
Yi, i.e. there is very little
smoothing.
3000
2500
2000
However, if w = 0.2, Ei does
not have the fluctuations of
Yi, i.e. there is far more
smoothing.
1500
1000
500
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
0
1989
1990
1991
1992
Y
1993
1994
S (w=0.2)
1995
1996
1997
1998
S (w=0.7)
15
16.
HOW TO CAPTURE THE TREND, CYCLICAL
AND SEASONAL COMPONENTS?
If we decompose a time series into the trend, seasonal and cyclical
components, then we can construct a forecast by projecting these parts into
the future.
1) Trend analysis
The easiest way of isolating a long-term linear trend is by simple linear
regression, where the independent variable is the time variable i.
and i is equal to 1 for the first time period in the sample and increases
by one each period thereafter.
16
17.
Example 3:
The graph below shows Australian exports of footwear ($m) from 1988 through
2000.
Exports: 85: Footw ear: ANNUAL
$m
70
60
This time series has an upward trend,
which is linear.
50
40
30
20
10
1988
1990
1992
1994
1996
1998
2000
Estimate a linear trend line using Excel.
First you have to create a time variable Xi
and then regress fwexport on Xi.
ˆ
Yi = 15.308 + 4.505X i
year
1988
1989
1990
1991
1992
1993
i
1
2
3
4
5
etc.
fwexport
14
23
22
30
36
etc.
17
18.
80
70
60
50
40
30
ˆ
Yi = 15.308 + 4.505X i
20
10
0
1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000
fwexport
y-hat
Note: In the first year of the sample period i = 1 not 0.
b 0 = 15.308 In 1987 (i = 0) the trend value of footwear exports is 15.308 $m.
b1 = 4.505 Each year the value of footwear exports increases by 4.505 $m.
ˆ
In 1988 (i = 1) Y = 15.308 + 4.505 ×1 = 19.813 $m
1
ˆ
In 1999 (i = 12) Y12 = 15.308 + 4.505 ×12 = 69.368 $m
18
19.
2) Measuring the cyclical effect
Assume that the time series model is multiplicative and consists of only
two parts: the trend and the cyclical components so that
Ci =
Yi = Ti × Ci
Yi
Ti
Under these assumptions the cyclical effect can be measured by
expressing the actual data as the percentage of the trend:
Yi
× 100 %
ˆ
Y
i
Calculate and plot the percentage of trend.
year
1988
1989
1990
1991
1992
t
1
2
3
4
5
fwexport
14
23
22
30
etc.
Y-hat Y/Y-hat*100
19.81
70.66
24.32
94.58
28.82
76.32
33.33
90.01
etc.
etc.
14 / 19.81 * 100 ≈ 71
So in 1988 the actual exports of
footwear were about 29% below
the trend line.
19
20.
Boom
130
120
Expansion phase
110
100
90
Recession
80
70
Contraction phase
Yi
ˆ
Yi ×
100 %
60
1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000
We have assumed that the time series pattern does not have a
seasonal component and that the random variations are negligible.
In Example 3 the first of these assumptions is certainly satisfied since the
data is annual.
20
21.
3) Measuring the seasonal effect
Depending on the nature of the time series, the seasonal variations can
be captured in different ways.
i. Assume, for example, that the time series does not contain a
discernible cyclical component and can be described by the
following multiplicative model
Yi = Ti × S i × I i
Yi
= Si × Ii
Ti
This suggests that dividing the estimated trend component (y-hat)
into the time series we obtain an estimate for the product of the
seasonal and random variations.
ˆ
Seasonal factor: Yi / Yi
In order to remove the random variations from this ratio, we
average the seasonal factors for each season and adjust these
averages to ensure that they add up to the number of seasons.
21
22.
ii.
When the time series model is multiplicative and has all four parts,
Yi = Ti × Ci × S i × I i
the data is first divided by (centered) moving averages, which are
supposed to capture the trend and cyclical components,
CMA i = Ti × Ci
Yi
Yi
=
= Si × I i
CMAi Ti × Ci
Then the seasonal factors and indices are calculated from this
Yi
CMA i
ratio:
and the trend and cyclical components are estimated from the
centered moving averages, instead of the original data.
Note: The order of the centered moving average must be equal to the number of
seasons. For example, we use 4-quarter CMA if the data is quarterly and
seasonality has 4 phases a year, and we use 12-month CMA if the data is
monthly and seasonality has 12 phases a year.
22
23.
Example 4:
The graph below shows retail turnover for households goods ($m) for Australia
from the second quarter of 1982 through the fourth quarter of 2000.
Retail turnover: Original: Household good retailing: QUARTERLY
$m
5000
4500
This time series has an upward linear
trend and quarterly seasonal variations.
It probably has some cyclical variations
too, but this third component seems to be
less significant than the other two.
4000
3500
3000
2500
2000
1500
Dec-82
Dec-85
Dec-88
Dec-91
Dec-94
Dec-97
Dec-00
a) Estimate a linear trend line with Excel.
ˆ
Yi = 1589.189 + 36.604X i
23
24.
ˆ
Yi = 1589.189 + 36.604 X i
b) Calculate the seasonal factors and the seasonal indices.
quarter
Jun-82
Sep-82
Dec-82
Mar-83
Jun-83
t
1
2
3
4
5
retail
1553.2
1601.9
2052.2
1666.0
1680.4
Y-hat
Y/Y-hat
1625.8
0.955 1553.2 / 1625.8 = 0.955
1662.4
0.964
1699.0
1.208
1735.6
0.960
1772.2
0.948
In order to find the seasonal indices the
seasonal factors (Y/Y-hat) have to be grouped,
averaged and, if necessary, adjusted.
24
25.
Year
1982
1983
1984
etc.
0.929 ×
4.000
= 0.930
3.997
Index
0.960
0.948
etc.
1998
1999
2000
Sum
Average
Q1
Q2
0.955
0.948
0.890
etc.
Q3
0.964
0.962
0.905
etc.
Q4
1.208
1.240
1.163
etc.
0.914
0.909
0.973
0.908
0.922
1.043
0.909
0.971
0.990
1.031
1.129
1.144
16.728
0.929
18.062
0.951
18.283
0.962
21.945
1.155
Total
3.997
0.930
0.951
0.963
1.156
4.000
S Mar = 93.0%
S Sep = 96.3%
S Jun = 95.1%
S Dec = 115.6%
These seasonal indices suggest that in the March, June and
September quarters retail turnover is expected to be 7.0, 4.9 and 3.7%
below its trend value, while in the December quarter retail turnover is
expected to be 15.6% above its trend value.
25
26.
c) De-seasonalising a time series.
quarter
Jun-82
Sep-82
Dec-82
Mar-83
Jun-83
Sep-83
t
1
2
3
4
5
6
retail
1553.2
1601.9
2052.2
1666.0
1680.4
etc.
Seasonal indices:
S Mar = 93.0% S Jun = 94.8% S Sep = 96.5% S Dec = 115.7%
Seasonal indices can be used to deseasonalise a time series, i.e. to
remove the seasonal variations from the data.
The seasonally adjusted data (in publications usually denoted as sa) is
obtained by dividing the observed, unadjusted data by the seasonal indices.
e.g. For the June quarter of 1982 the seasonally adjusted retail turnover is
1553.2 / 94.8 × 100 = 1638.2 $m
26
27.
INTRODUCTION TO FORECASTING
•
•
After having studied the historical pattern of a time series, if there is
reason to believe that the most important features of the variable do not
change in the future, we can project the revealed pattern into the future
in order to develop forecasts.
If a time series exhibits no (or hardly any) trend, cyclical and seasonal
variations, exponential smoothing can provide a useful forecast for one
period ahead:
Fi +1 = E i
Example 5:
(Refer Example 2)
We have applied exponential smoothing with W = 0.2 and W = 0.7 on quarterly
Australian unemployed persons (in thousands).
Since this time series does have some seasonal variations, exponential
smoothing cannot be expected to forecast unemployment reasonably well.
Nevertheless, just for illustration, let us forecast unemployment for the first
quarter of 1999.
27
28.
1998
1
2
3
4
unemployed E (W=0.7)
2461.4
2402.8
2210.9
2268.5
2221.3
2235.5
2102.6
2142.5
This is the smoothed value for the fourth
quarter of 1998, and thus the forecast for
the first quarter of 1999.
• If a time series exhibits a long-term (linear) trend and seasonal
variations, we can use regression analysis to develop forecasts in two
different ways.
We can forecast using the estimated trend and seasonal indices as:
Fi = Ti × S i = (b 0 + b1X i ) × S i
28
29.
Example 6:
(Refer Example 4)
Forecast retail turnover for households goods for the first quarter of 2001
applying the first approach can be implemented as follows.
Obtain the trend estimate from part a and the March seasonal
index from part b so that
ˆ
i = 76, S76 = SMar = 0.930 and Yi = 1589.189 + 36.604X i
ˆ
F76 = Y76 = (1589.189 + 36.604x76)x0.930 = 4064.8
We have predicted retail turnover for households goods for the first quarter
of 2001. Suppose we had another forecast value of 4203.4 for the same
data and the same time period using a different forecasting model. How
would we decide which forecast is more accurate?
29
30.
Measuring the forecast Error
•
How can we decide which forecasting model is the most accurate in a
given situation?
Forecast the variable of interest for a number of of time periods using
alternative models and evaluate some measure(s) of forecast accuracy
for each of these models.
Among a number of possible criteria that can be used for this purpose
a commonly used method is the,
Mean absolute deviation:
1 n
MAD = ∑ yt − Ft
n t =1
30
31.
Example 7:
Two forecasting models were used to predict the future values of a time series.
They are shown next, together with the actual values. For each model, calculate
MAD to determine which was more accurate.
Ft
et
Model 1 Model 2
Model 1 Model 2
yt
6.0
6.6
7.3
9.4
7.5
6.3
5.4
8.2
6.3
6.7
7.1
7.5
-1.5
0.3
1.9
1.2
| et |
Model 1 Model 2
-0.3
-0.1
0.2
1.9
1.5
0.3
1.9
1.2
Total:
6.0 – 7.5
0.3
0.1
0.2
1.9
4.9
2.5
| -1.5 |
Model 1 : MAD = 4.9/4=1.225
Model 2 : MAD = 2.5/4=0.625
Model 2 is the more accurate.
31
A particular slide catching your eye?
Clipping is a handy way to collect important slides you want to go back to later.
Be the first to comment