Finance 30210: Managerial
Economics
Demand Estimation and Forecasting
What are the odds that a fair coin flip results in a head?
What are the odds that the toss of a fair die results in a 5?
What are the odds that tomorrow’s temperature is 95 degrees?
The answer to all these questions come from a probability distribution
Probability
1/2
Head
Tail
Probability
1/6
1
2
3
4
5
6
A probability
distribution is a
collection of
probabilities
describing the odds of
any particular event
The distribution for temperature in south bend is a bit more complicated because there
are so many possible outcomes, but the concept is the same
Probability
Standard Deviation
Temperature
Mean
We generally assume a Normal Distribution which can be characterized by a
mean (average) and standard deviation (measure of dispersion)
Without some math, we can’t find the probability of a specific outcome, but we
can easily divide up the distribution
Probability
Temperature
Mean-2SD
2.5%
Mean -1SD
13.5%
Mean
34%
Mean+1SD
34%
Mean+2SD
13.5%
2.5%
Annual Temperature in South Bend has a mean of 59 degrees and a standard deviation
of 18 degrees.
Probability
95 degrees is 2 standard deviations
to the right – there is a 2.5% chance
the temperature is 95 or greater
(97.5% chance it is cooler than 95)
Temperature
23
41
Can’t we do a little better than this?
59
77
95
Conditional distributions give us probabilities conditional on some observable information – the
temperature in South Bend conditional on the month of July has a mean of 84 with a standard
deviation of 7.
Probability
95 degrees falls a little more than
one standard deviation away (there
approximately a 16% chance that
the temperature is 95 or greater)
Temperature
70
77
84
91 95
98
Conditioning on month gives us a more accurate probabilities!
We know that there should be a “true” probability distribution that
governs the outcome of a coin toss (assuming a fair coin)
PrHeads  PrTails   .5
Suppose that we were to flip a coin over and over again and after
each flip, we calculate the percentage of heads & tails
# of Heads
Total Flips
(Sample Statistic)
.5
(True Probability)
That is, if we collect “enough” data, we can eventually learn the truth!
We can follow the same process for the temperature in South Bend
Temperature ~

N , 2

We could find this distribution by collecting temperature data for south bend
Sample Mean
(Average)
Sample
Variance
1N
x    xi  
 N  i 1
N
1


2
s 2     xi  x    2
 N  i 1
Note: Standard Deviation is the square root of the variance.
Some useful properties of probability distributions

x  N μ,σ 2
y  kx
Probability distributions
are scalable


y  N k,k 2σ 2

=
3X
Mean = 1
Mean = 3
Variance = 4
Variance = 36 (3*3*4)
Std. Dev. = 2
Std. Dev. = 6
 
y  N  ,σ 
x  y  N    ,σ
x  N μ x ,σ x2
Probability distributions
are additive
y
2
y
x
y
2
x
 σ y2  2 cov xy
=
+
Mean = 1
Mean = 2
Mean = 3
Variance = 1
Variance = 9
Variance = 14 (1 + 9 + 2*2)
Std. Dev. = 1
Std. Dev. = 3
Std. Dev. = 3.7
COV = 2

Suppose we know that the value of a car is determined by its age
Value = $20,000 - $1,000 (Age)
Car Age
Value
Mean = 8
Mean = $ 12,000
Variance = 4
Variance = 4,000,000
Std. Dev. = 2
Std. Dev. = $ 2,000
We could also use this to forecast:
Value = $20,000 - $1,000 (Age)
How much should a six
year old car be worth?
Value = $20,000 - $1,000 (6) = $14,000
Note: There is NO uncertainty in this
prediction.
Searching for the truth….
You believe that there is a relationship between age and value,
but you don’t know what it is….
1. Collect data on values and age
2. Estimate the relationship
between them
Note that while the true distribution of age is N(8,4), our
collected sample will not be N(8,4). This sampling error will
create errors in our estimates!!
18000.00
16000.00
Slope = b
14000.00
12000.00
10000.00
a
8000.00
6000.00
4000.00
2000.00
0.00
0
2
4
6
Value = a + b * (Age) + error
8
10
12
14

error  N 0,σ 2
We want to choose ‘a’ and ‘b’ to minimize the error!

Regression Results
Variable
Intercept
Age
Coefficients
Standard Error
t Stat
12,354
653
18.9
- 854
80
-10.60
We have our estimate of “the truth”
Value = $12,354 - $854 * (Age) + error
Intercept (a)
Age (b)
Mean = $12,354
Mean = -$854
Std. Dev. = $653
Std. Dev. = $80
T-Stats bigger
than 2 in
absolute value
are considered
statistically
significant!
Regression Statistics
R Squared
0.36
Standard Error
2250
Percentage of value
variance explained by
age
Error Term
Mean = 0
Std. Dev = $2,250
We can now forecast the value of a 6 year old car
6
Value = $12,354 - $854 * (Age) + error
Mean = $12,354
Mean = $854
Mean = $0
Std. Dev. = $653
Std. Dev. = $ 80
Std. Dev. = $2,250
StdDev  Var a   X 2Var b   2 XCova, b   Var error 
Cova, b   XVarb
 
(Recall, The Average Car age is 8 years)
StdDev  6532  6 2 80 2  26880 2  2250 2  $2,259
Value  12,354  854 * 6  $7,230
StdDev  653
2
 6 80
2
2
2
2



 2 6 8 80  2250
 $2,259
Value
+95%
Forecast Interval
-95%
Age  6
x 8
Age
Note that your forecast error will always be smallest at the sample mean! Also, your forecast
gets worse at an increasing rate as you depart from the mean
What are the odds that Pat Buchanan received 3,407 votes from Palm
Beach County in 2000?
The Strategy: Estimate a
relationship for Pat Buchanan’s
votes using every county EXCEPT
Palm Beach
“Are a function of”
B  F D
Pat
Buchanan’s
Votes
Observable
Demographics
Using Palm Beach data,
forecast Pat Buchanan’s
vote total for Palm
Beach
BPB  F DPB 
The Data: Demographic Data By County
County
Black
(%)
Age 65
(%)
Hispanic
(%)
College
(%)
Income
(000s)
Buchanan Total
Votes
Votes
Alachua
21.8
9.4
4.7
34.6
26.5
262
84,966
Baker
16.8
7.7
1.5
5.7
27.6
73
8,128
What variables do you think should affect Pat Buchanan’s Vote total?
# of Buchanan votes
V  a  bC  
% of County that is
college educated
# of votes gained/lost for each
percentage point increase in
college educated population
Results
Parameter
a
b
Value
5.35
14.95
Standard Error
58.5
3.84
T-Statistic
.09
3.89
R-Square = .19
19% of the variation in
Buchanan’s votes across
counties is explained by
college education
The distribution for
‘b’ has a mean of
15 and a standard
deviation of 4
Each percentage point increase in
college educated (i.e. from 10% to 11%)
raises Buchanan’s vote total by 15
0
15
There is a 95% chance that
the value for ‘b’ lies between
23 and 7
V  5.35  14.95C
Plug in Values for
College % to get
vote predictions
County
College
(%)
Predicted
Votes
Actual
Votes
Error
Alachua
34.6
522
262
260
Baker
5.7
90
73
17
Lets try something a little
different…
County
College (%)
Buchanan
Votes
Log of Buchanan
Votes
Alachua
34.6
262
5.57
Baker
5.7
73
4.29
Log of Buchanan votes
LN V   a  bC  
% of County that is
college educated
Percentage increase/decease in
votes for each percentage point
increase in college educated
population
Results
Parameter
a
b
Value
3.45
.09
Standard Error
.27
.02
T-Statistic
12.6
5.4
R-Square = .31
31% of the variation in
Buchanan’s votes across
counties is explained by
college education
The distribution for
‘b’ has a mean of
.09 and a
standard deviation
of .02
Each percentage point increase in
college educated (i.e. from 10% to 11%)
raises Buchanan’s vote total by .09%
0
.09
There is a 95% chance that
the value for ‘b’ lies between
.13 and .05
LN V   3.45  .09C
V e
LN V 
Plug in Values for
College % to get
vote predictions
County
College
(%)
Predicted
Votes
Actual
Votes
Error
Alachua
34.6
902
262
640
Baker
5.7
55
73
-18
How about this…
County
College (%)
Buchanan
Votes
Log of College (%)
Alachua
34.6
262
3.54
Baker
5.7
73
1.74
# of Buchanan votes
V  a  bLN C   
Log of % of County that
is college educated
Gain/ Loss in votes for each
percentage increase in college
educated population
Results
Parameter
a
b
Value
-424
252
Standard Error
139
54
T-Statistic
-3.05
4.6
R-Square = .25
25% of the variation in
Buchanan’s votes across
counties is explained by
college education
The distribution for
‘b’ has a mean of
252 and a
standard deviation
of 54
Each percentage increase in college
educated (i.e. from 30% to 30.3%) raises
Buchanan’s vote total by 252 votes
0
.09
There is a 95% chance that
the value for ‘b’ lies between
360 and 144
V  424  252LN C 
Plug in Values for
College % to get
vote predictions
County
College
(%)
Predicted
Votes
Actual
Votes
Error
Alachua
34.6
469
262
207
Baker
5.7
15
73
-58
One More…
County
College
(%)
Buchanan
Votes
Log of College (%) Log of Buchanan
Votes
Alachua
34.6
262
3.54
5.57
Baker
5.7
73
1.74
4.29
Log of Buchanan votes
LN V   a  bLN C   
Log of % of County that
is college educated
Percentage gain/Loss in votes for
each percentage increase in
college educated population
Results
Parameter
a
b
Value
.71
1.61
Standard Error
.63
.24
T-Statistic
1.13
6.53
R-Square = .40
40% of the variation in
Buchanan’s votes across
counties is explained by
college education
The distribution for
‘b’ has a mean of
1.61 and a
standard deviation
of .24
Each percentage increase in college
educated (i.e. from 30% to 30.3%) raises
Buchanan’s vote total by 1.61%
0
.09
There is a 95% chance that
the value for ‘b’ lies between
2 and 1.13
LN V   .71  1.61LN C 
V e
LN V 
Plug in Values for
College % to get
vote predictions
County
College
(%)
Predicted
Votes
Actual
Votes
Error
Alachua
34.6
624
262
362
Baker
5.7
34
73
-39
It turns out the regression with the best fit looks like this.
County
Black
(%)
Age 65
(%)
Hispanic
(%)
College
(%)
Income
(000s)
Buchanan Total
Votes
Votes
Alachua
21.8
9.4
4.7
34.6
26.5
262
84,966
Baker
16.8
7.7
1.5
5.7
27.6
73
8,128
LN P   a1  a2 B  a2 A65  a3 H  a4C  a5 I  
Buchanan Votes
Total Votes
Error term
*100
Parameters to be estimated
The Results:
Variable
Coefficient
Standard Error
t - statistic
Intercept
2.146
.396
5.48
Black (%)
-.0132
.0057
-2.88
Age 65 (%)
-.0415
.0057
-5.93
Hispanic (%)
-.0349
.0050
-6.08
College (%)
-.0193
.0068
-1.99
Income (000s)
-.0658
.00113
-4.58
R Squared = .73
LN P   2.146  .0132B   .0415 A65   .0349H   .0193C   .0658I 
Now, we can make a forecast!
County
Black (%)
Age 65 (%)
Hispanic (%)
College (%)
Income
(000s)
Buchanan
Votes
Total
Votes
Alachua
21.8
9.4
4.7
34.6
26.5
262
84,966
Baker
16.8
7.7
1.5
5.7
27.6
73
8,128
County
Predicted
Votes
Actual
Votes
Error
Alachua
520
262
258
Baker
55
73
-18
County
Black
(%)
Age 65
(%)
Hispanic
(%)
College
(%)
Income
(000s)
Buchanan Total
Votes
Votes
Palm Beach
21.8
23.6
9.8
22.1
33.5
3,407
431,621
LN P   2.146  .0132B   .0415 A65   .0349H   .0193C   .0658I 
LN P  2.004
P  e 2.004  .134%
.00134431,621  578
This would be our prediction for Pat
Buchanan’s vote total!
LN P  2.004
We know that the log of Buchanan’s vote percentage is
distributed normally with a mean of -2.004 and with a
standard deviation of .2556
Probability
LN(%Votes)
-2.004 – 2*(.2556)
= -2.5152
-2.004 + 2*(.2556)
= -1.4928
There is a 95% chance that the log of Buchanan’s vote
percentage lies in this range
P  e 2.004  .134%
Next, lets convert the Logs to vote percentages
Probability
% of Votes
e
2.5152
 .08%
e
1.4928
 .22%
There is a 95% chance that Buchanan’s vote percentage lies in
this range
.00134431,621  578
Finally, we can convert to actual votes
Probability
3,407 votes turns out to
be 7 standard
deviations away from
our forecast!!!
.0008431,621  348
Votes
.0022431,621  970
There is a 95% chance that Buchanan’s total vote lies in this
range
We know that the quantity of some good or service demanded should be related
to some basic variables
“ Is a function of”
QD  DP, I ,...
Price
Quantity
Demanded
Price
Income
D
Quantity
Other
“Demand
Shifters”
Demand Factors
Cross Sectional estimation holds the time period constant and estimates the variation in
demand resulting from variation in the demand factors
Time
t-1
t
t+1
For example: can we estimate demand for Pepsi in South Bend by looking at selected
statistics for South bend
Suppose that we have the following data for sales in 200 different Indiana cities
City
Price
Average Income
(Thousands)
Competitor’s
Price
Advertising
Expenditures
(Thousands)
Total Sales
Granger
1.02
21.934
1.48
2.367
9,809
Mishawaka
2.56
35.796
2.53
26.922
130,835
Lets begin by estimating a basic demand curve – quantity demanded is a linear
function of price.
Q  a0  a1P
Change in quantity demanded
per $ change in price (to be
estimated)
That is, we have estimated the following equation
Regression Results
Variable
Coefficient
Standard Error
t Stat
Intercept
155,042
18,133
8.55
Price (X)
-46,087
7214
-6.39
Regression Statistics
R Squared
.17
Standard Error
48,074
Q  155,042  46,087 P
Every dollar increase in price
lowers sales by 46,087 units.
Values For South Bend
Price of Pepsi
$1.37
Q  155,042  46,0871.37  91,903
P
 1.37 
  .68
 91,903 
  46,087
$1.37
Q
91,903
As we did earlier, we can experiment with different functional forms by using logs
City
Price
Average Income
(Thousands)
Competitor’s
Price
Advertising
Expenditures
(Thousands)
Total Sales
Granger
1.02
21.934
1.48
2.367
9,809
Mishawaka
2.56
35.796
2.53
26.922
130,835
Adding logs changes the interpretation of the coefficients
Q  a0  a1LN P 
Change in quantity demanded
per percentage change in price
(to be estimated)
That is, we have estimated the following equation
Regression Results
Variable
Coefficient
Standard Error
t Stat
Intercept
133,133
14,892
8.93
Price (X)
-103,973
16,407
-6.33
Regression Statistics
R Squared
.17
Standard Error
48,140
Q  133,133 103,973LN P
Every 1% increase in price
lowers sales by 103,973 units.
Values For South Bend
Price of Pepsi
$1.37
Log of Price
.31
Q  133,133 103,973.31
Q
 103,973
%p
P
Q  1 
1


   103,973


%p  Q 
 100,402 
$1.37
Q
100,402
As we did earlier, we can experiment with different functional forms by using logs
City
Price
Average Income
(Thousands)
Competitor’s
Price
Advertising
Expenditures
(Thousands)
Total Sales
Granger
1.02
21.934
1.48
2.367
9,809
Mishawaka
2.56
35.796
2.53
26.922
130,835
Adding logs changes the interpretation of the coefficients
LN Q   a0  a1P
Percentage change in quantity
demanded per $ change in
price (to be estimated)
That is, we have estimated the following equation
Regression Results
Variable
Coefficient
Standard Error
t Stat
Intercept
13
.34
38.1
Price (X)
-1.22
.13
-8.98
Regression Statistics
R Squared
.28
Standard Error
.90
LN Q  131.22P
Every $1 increase in price
lowers sales by 1.22%.
Values For South Bend
Price of Pepsi
$1.37
LN Q   13  1.221.37   11.33
Q  e11.33  83,283
We can now use this
estimated demand curve
along with price in South
Bend to estimate
demand in South Bend
% Q
 1.22
p

P
% Q  p 
 1.37 
   1.22

p  1 
 1 
$1.37
Q
83,283
As we did earlier, we can experiment with different functional forms by using logs
City
Price
Average Income
(Thousands)
Competitor’s
Price
Advertising
Expenditures
(Thousands)
Total Sales
Granger
1.02
21.934
1.48
2.367
9,809
Mishawaka
2.56
35.796
2.53
26.922
130,835
Adding logs changes the interpretation of the coefficients
LN Q   a0  a1LN P 
Percentage change in quantity
demanded per percentage
change in price (to be
estimated)
That is, we have estimated the following equation
Regression Results
Variable
Coefficient
Standard Error
t Stat
Intercept
12.3
.28
42.9
Price (X)
-2.60
.31
-8.21
Regression Statistics
R Squared
.25
Standard Error
.93
LN Q  12  2.6LN P
Every 1% increase in price
lowers sales by 2.6%.
Values For South Bend
Price of Pepsi
$1.37
Log of Price
.31
LN Q   12  2.6.31  11.19
Q  e11.19  72,402
P
%Q
 2.6  
%p
$1.37
Q
72,402
We can add as many variables as we want in whatever combination. The goal is
to look for the best fit.
LN Q   a0  a1P  a2 LN I   a3 LN Pc 
% change in
Sales per $
change in price
% change in Sales
per % change in
income
% change in
Sales per %
change in
competitor’s
price
Regression Results
Variable
Intercept
Coefficient
Standard Error
t Stat
5.98
1.29
4.63
-1.29
.12
-10.79
Log of Income
1.46
.34
4.29
Log of Competitor’s Price
2.00
.34
5.80
Price
R Squared: .46
Values For South Bend
Price of Pepsi
$1.37
Log of Income
3.81
Log of Competitor’s Price
Now we can make a
prediction and calculate
elasticities
.80
LN Q   5.98  1.291.37   1.463.81  2.00.80  11.36
Q  e11.36  87,142
P
 %Q  P 
 1.37 
   1.29
  1.76
 P  1 
 1 
 %Q 
I  
  1.46
 %I 
 %Q 
  2.00
 CP  
 %Pc 
 
$1.37
Q
87,142
Demand Factors
We could use a cross sectional regression to forecast quantity demanded out into the
future, but it would take a lot of information!
Time
t-1
t
Estimate a
demand curve
using data at
some point in
time
t+1
Use the estimated
demand curve and
forecasts of data to
forecast quantity
demanded
Demand Factors
Time Series estimation ignores the demand factors and estimates the variation
in demand over time
Time
t-1
t
t+1
For example: can we predict demand for Pepsi in South Bend next year by
looking at how demand varies across time
Time series estimation ignores the demand factors and looks at variations in demand
over time. Essentially, we want to separate demand changes into various frequencies
Trend: Long term movements in demand (i.e. demand
for movie tickets grows by an average of 6% per year)
Business Cycle: Movements in demand related to the
state of the economy (i.e. demand for movie tickets
grows by more than 6% during economic expansions
and less than 6% during recessions)
Seasonal: Movements in demand related to time of
year. (i.e. demand for movie tickets is highest in the
summer and around Christmas
Time Period
Quantity (millions of kilowatt
hours)
2003:1
11
2003:2
15
2003:3
12
2003:4
14
2004:1
12
2004:2
17
2004:3
13
2004:4
16
2005:1
14
2005:2
18
2005:3
15
2005:4
17
2006:1
15
2006:2
20
2006:3
16
2006:4
19
Suppose that you
work for a local
power company.
You have been
asked to forecast
energy demand for
the upcoming year.
You have data over
the previous 4
years:
First, let’s plot the data…what do you see?
25
20
15
10
5
0
2003-1
2004-1
2005-1
2006-1
This data seems to have a linear trend
A linear trend takes the following form:
Estimated value for
time zero
Estimated quarterly growth
(in millions of kilowatt hours)
xt  x0  bt
Forecasted value at time t
(note: time periods are
quarters and time zero is
2003:1)
Time period: t = 0 is 2003:1
and periods are quarters
Regression Results
Variable
Coefficient
Standard Error
t Stat
Intercept
11.9
.953
12.5
Time Trend
.394
.099
4.00
Regression Statistics
R Squared
Standard Error
Observations
.53
1.82
16
xt  11.9  .394t
Lets forecast electricity usage at the mean time period (t = 8)
xˆt  11.9  .3948  15.05
Var xˆt   3.50
Here’s a plot of our regression line with our error bands…again, note that the
forecast error will be lowest at the mean time period
25
20
15
10
5
0
2003-1
2004-1
2005-1
T=8
2006-1
We can use this linear trend model to predict as far out as we want, but note
that the error involved gets worse!
70
60
50
40
30
20
10
0
Sample
xˆt  11.9  .39476  41.85
Var xˆt   47.7
Lets take another look at the data…it seems that there is a regular pattern…
25
20
Q2
Q2
Q2
Q2
15
10
5
0
2003-1
2004-1
There appears to be a seasonal cycle…
2005-1
2006-1
One seasonal adjustment process is to adjust each quarter by the average of
actual to predicted
Average Ratios
Time Period
Actual
Predicted
Ratio
Adjusted
2003:1
11
12.29
.89
12.29(.87)=10.90
2003:2
15
12.68
1.18
12.68(1.16) = 14.77
2003:3
12
13.08
.91
13.08(.91) = 11.86
•Q2 = 1.16
2003:4
14
13.47
1.03
13.47(1.04) = 14.04
•Q3 = .91
2004:1
12
13.87
.87
13.87(.87) = 12.30
2004:2
17
14.26
1.19
14.26(1.16) = 16.61
2004:3
13
14.66
.88
14.66(.91) = 13.29
2004:4
16
15.05
1.06
15.05(1.04) = 15.68
2005:1
14
15.44
.91
15.44(.87) = 13.70
2005:2
18
15.84
1.14
15.84(1.16) = 18.45
2005:3
15
16.23
.92
16.23(.91) = 14.72
2005:4
17
16.63
1.02
16.63(1.04) = 17.33
2006:1
15
17.02
.88
17.02(.87) = 15.10
2006:2
20
17.41
1.14
17.41(1.16) = 20.28
2006:3
16
17.81
.89
17.81(.91) = 16.15
2006:4
19
18.20
1.04
18.20(1.04) = 18.96
•Q1 = .87
•Q4 = 1.04
For each observation:
•Calculate the ratio of
actual to predicted
•Average the ratios by
quarter
•Use the average ration to
adjust each predicted
value
With the seasonal adjustment, we don’t have any statistics to judge goodness of fit. One
method of evaluating a forecast is to calculate the root mean squared error
Time Period
Actual
Adjusted
Error
2003:1
11
10.90
-0.1
2003:2
15
14.77
-0.23
2003:3
12
11.86
-0.14
2003:4
14
14.04
0.04
2004:1
12
12.30
0.3
2004:2
17
16.61
-0.39
2004:3
13
13.29
0.29
2004:4
16
15.68
-0.32
2005:1
14
13.70
-0.3
2005:2
18
18.45
0.45
2005:3
15
14.72
-0.28
2005:4
17
17.33
0.33
2006:1
15
15.10
0.1
2006:2
20
20.28
0.28
2006:3
16
16.15
0.15
2006:4
19
18.96
-0.04
Sum of squared forecast
errors
 A  F 
2
RMSE 
t
n
Number of
Observations
RMSE  .26
t
Looks pretty good…
20
19
18
17
16
15
14
13
12
11
10
2003-1
2004-1
2005-1
2006-1
RMSE  .26
Recall our prediction for period 76 ( Year 2022 Q4)
70
60
50
40
30
20
10
0
xˆt  11.9  .39476  41.851.04  43.52
We could also account for seasonal variation by using dummy variables
xt  x0  b0t  b1D1  b2 D2  b3 D3
1, if quarter i
Di  
else
0,
Note: we only need three quarter dummies. If the observation is from quarter 4, then
D1  D2  D3  0
xt  x0  b0t
Regression Results
Variable
Coefficient
Intercept
Standard Error
t Stat
12.75
.226
56.38
.375
.0168
22.2
D1
-2.375
.219
-10.83
D2
1.75
.215
8.1
D3
-2.125
.213
-9.93
Time Trend
Regression Statistics
R Squared
.99
Standard Error
.30
Observations
16
Note the much better fit!!
xt  12.75  .375t  2.375D1  1.75D2  2.125D3
Time Period
Actual
Ratio Method
Dummy
Variables
2003:1
11
10.90
10.75
2003:2
15
14.77
15.25
2003:3
12
11.86
11.75
2003:4
14
14.04
14.25
2004:1
12
12.30
12.25
2004:2
17
16.61
16.75
2004:3
13
13.29
13.25
2004:4
16
15.68
15.75
2005:1
14
13.70
13.75
2005:2
18
18.45
18.25
2005:3
15
14.72
14.75
2005:4
17
17.33
17.25
2006:1
15
15.10
15.25
2006:2
20
20.28
19.75
2006:3
16
16.15
16.25
2006:4
19
18.96
18.75
Ratio Method
RMSE  .26
Dummy Variables
RMSE  .25
A plot confirms the similarity of the methods
20
19
18
17
16
15
14
13
12
11
10
2003-1
2004-1
2005-1
Dummy
2006-1
Ratio
Recall our prediction for period 76 ( Year 2022 Q4)
70
60
50
40
30
20
10
0
xt  12.75  .37576  41.25
Recall, our trend line took the form…
xt  x0  bt
This parameter is measuring quarterly change in
electricity demand in millions of kilowatt hours.
Often times, its more realistic to assume that demand grows by a constant
percentage rather that a constant quantity. For example, if we knew that
electricity demand grew by G% per quarter, then our forecasting equation would
take the form
 G% 
xt  x0 1 

 100 
t
If we wish to estimate this equation, we have a little work to do…
xt  x0 1  g 
t
Note: this growth rate is in
decimal form
If we convert our data to natural logs, we get the following linear relationship that can
be estimated
ln xt  ln x0  t ln 1  g 
Regression Results
Variable
Coefficient
Standard Error
t Stat
Intercept
2.49
.063
39.6
Time Trend
.026
.006
4.06
Regression Statistics
R Squared
Standard Error
Observations
.54
.1197
ln xt  2.49  .026t
16
Lets forecast electricity usage at the mean time period (t = 8)
ln xˆt  2.49  .0268  2.698
BE CAREFUL….THESE NUMBERS ARE LOGS !!!
Var xˆt   .0152
ln xˆt  2.49  .0268  2.698
Var xˆt   .0152
The natural log of forecasted demand is 2.698. Therefore, to get the actual demand
forecast, use the exponential function
e
2.698
 14.85
Likewise, with the error bands…a 95% confidence interval is +/- 2 SD
2.698  /  2 .0152  2.451,2.945
e
2.451
,e
2.945
  11.60,19.00
Again, here is a plot of our forecasts with the error bands
30
25
20
15
10
5
0
2003-1
2004-1
2005-1
T=8
2006-1
RMSE  1.70
Errors is growth rates compound quickly!!
600
500
400
300
200
100
0
1
13
25
37
49
61
73
85
97
e 4.49  89.22
 /  2SD  35.8,221.8
Let’s try one…suppose that we are interested in forecasting gasoline prices. We have the
following historical data. (the data is monthly from April 1993 – June 2010)
Does a linear (constant cents per gallon growth per year) look reasonable?
Let’s suppose we assume a linear trend. Then we are estimating the following
linear regression:
monthly growth in
dollars per gallon
pt  p0  bt
Price at time t
Price at April 1993
Number of months
from April 1993
Regression Results
Variable
Intercept
Time Trend
Coefficient
Standard Error
t Stat
.67
.05
12.19
.010
.0004
23.19
R Squared= .72
We can check for the presence of a seasonal cycle by adding seasonal dummy
variables:
dollars per gallon impact of quarter I relative
to quarter 4
1, if quarter i
Di  
 0, else
pt  p0  b0t  b1D1  b2 D2  b3 D3
Regression Results
Variable
Coefficient
Standard Error
t Stat
Intercept
.58
.07
8.28
Time Trend
.01
.0004
23.7
D1
-.03
.075
-.43
D2
.15
.074
2.06
D3
.16
.075
2.20
R Squared= .74
If we wanted to remove the seasonal component, we could by subtracting the
seasonal dummy off each gas price
Seasonalizing
Date
Regression
coefficient
Price
Seasonalized
data
1993 – 04
1.05
2nd Quarter
.15
.90
1993 - 07
1.06
3rd Quarter
.16
90
1993 - 10
1.06
4th Quarter
0
1.06
1994 - 01
.98
1st Quarter
-.03
1.01
1994 - 04
1.00
2nd Quarter
.15
.85
Note: Once the seasonal component has been removed, all that should be left is
trend, cycle, and noise. We could check this:
Seasonalized Price Series
Regression Results
~
pt  p0  bt
Variable
Coefficient
Standard Error
t Stat
Intercept
.587
.05
11.06
Time Trend
.010
.0004
23.92
Seasonalized Price Series
~
pt  p0  b0t  b1D1  b2 D2  b3 D3
Regression Results
Variable
Coefficient
Standard Error
t Stat
Intercept
.587
.07
8.28
Time Trend
.010
.0004
23.7
D1
0
.075
0
D2
0
.074
0
D3
0
.075
0
The regression we have in place gives us the trend plus the seasonal component
of the data
pt  .58  .01t  .03D1  .15D2  .16D3
Predicted
Trend
Seasonal
If we subtract our predicted price (from the regression) from the actual price, we will have isolated
the business cycle and noise
Business Cycle Component
Date
Actual Price
Predicted Price
(From
regression)
Business Cycle
Component
1993 - 04
1.050
.752
.297
1993 - 05
1.071
.763
.308
1993 - 06
1.075
773
.301
1993 - 07
1.064
.797
.267
1993 - 08
1.048
.807
.240
We can plot this and compare it with business cycle dates
Actual
Price
pt  pˆ t
Predicted
Price
Data Breakdown
Date
Actual Price
Trend
Seasonal
Business Cycle
1993 - 04
1.050
.58
.15
.320
1993 - 05
1.071
.59
.15
.331
1993 - 06
1.075
.60
.15
.325
1993 - 07
1.064
.61
.16
.294
1993 - 08
1.048
.62
.16
.268
Regression Results
Variable
Coefficient
Standard Error
t Stat
Intercept
.58
.07
8.28
Time Trend
.01
.0004
23.7
D1
-.03
.075
-.43
D2
.15
.074
2.06
D3
.16
.075
2.20
Perhaps an exponential trend would work better…
An exponential trend would indicate constant percentage growth rather than cents per gallon.
We already know that there is a seasonal component, so we can start with
dummy variables
Monthly growth rate
Percentage price impact of quarter I relative
to quarter 4
1, if quarter i
Di  
 0, else
ln pt  p0  b0t  b1D1  b2 D2  b3 D3
Regression Results
Variable
Coefficient
Standard Error
t Stat
Intercept
-.14
.03
-4.64
Time Trend
.005
.0001
29.9
D1
-.02
.032
-.59
D2
.06
.032
2.07
D3
.07
.032
2.19
R Squared= .81
If we wanted to remove the seasonal component, we could by subtracting the
seasonal dummy off each gas price, but now, the price is in logs
Seasonalizing
Date
Price
Log of Price
Regression
coefficient
Log of Seasonalized
data
Seasonalized
Price
1993 – 04
1.05
.049
2nd Quarter
.06
-.019
.98
1993 - 07
1.06
.062
3rd Quarter
.07
-.010
.99
1993 - 10
1.06
.062
0
.062
1.06
1994 - 01
.98
-.013
1st Quarter
-.02
.006
1.00
1994 - 04
1.00
.005
2nd Quarter
.06
-.062
.94
Example:
e .019  .98
4th Quarter
The regression we have in place gives us the trend plus the seasonal component
of the data
ln pt  .14  .005t  .02D1  .06D2  .07 D3
Predicted
Log of
Price
Seasonal
Trend
If we subtract our predicted price (from the regression) from the actual price, we will have isolated
the business cycle and noise
e .069  .93
Business Cycle Component
Date
Actual Price
Predicted Log Price
(From regression)
Predicted
Price
Business Cycle
Component
1993 - 04
1.050
-.069
.93
.12
1993 - 05
1.071
-.063
.94
.13
1993 - 06
1.075
-.057
.94
.13
1993 – 07
1.064
-.047
.95
.11
1993 - 08
1.048
-.041
.96
.09
As you can see, very similar results
Actual
Price
pt  pˆ t
Predicted
Price
In either case, we could make a forecast for gasoline prices next year. Lets
say, April 2011.
Forecasting Data
Date
Time Period
April 2011
217
Quarter
2
pt  .58  .01217  .030  .151  .160  2.90
OR
ln pt  .14  .005217   .020  .061  .070  1.005
e1.005  2.73
By the way, the actual price in April 2011 was $3.80
Quarter
Market Share
1
20
2
22
25
3
23
20
4
24
5
18
6
23
7
19
5
8
17
0
9
22
10
23
11
18
12
23
30
15
10
Consider a new forecasting problem. You
are asked to forecast a company’s market
share for the 13th quarter.
1
2
3
4
5
6
7
8
9
10
There doesn’t seem to be any
discernable trend here…
11
12
Smoothing techniques are often used when data exhibits no trend or seasonal/cyclical
component. They are used to filter out short term noise in the data.
Quarter
Market
Share
MA(3)
MA(5)
1
20
2
22
3
23
4
24
21.67
5
18
23
6
23
21.67
21.4
7
19
21.67
22
8
17
20
21.4
9
22
19.67
20.2
10
23
19.33
19.8
11
18
20.67
20.8
12
23
21
19.8
A moving average of length N is
equal to the average value over
the previous N periods
t 1
MAN  
A
tN
N
t
The longer the moving average, the smoother the forecasts are…
30
25
20
Actual
15
MA(3)
MA(5)
10
5
0
1
2
3
4
5
6
7
8
9
10
11
12
Calculating forecasts is straightforward…
MA(3)
Quarter
Market
Share
MA(3)
MA(5)
1
20
2
22
3
23
4
24
21.67
5
18
23
6
23
21.67
21.4
7
19
21.67
22
8
17
20
21.4
9
22
19.67
20.2
10
23
19.33
19.8
11
18
20.67
20.8
12
23
21
19.8
23  18  23
 21.33
3
MA(5)
23  18  23  22  17
 20.6
5
So, how do we choose N??
Quarter
Market
Share
MA(3)
Squared
Error
MA(5)
Squared
Error
1
20
2
22
3
23
4
24
21.67
5.4289
5
18
23
25
6
23
21.67
1.7689
21.4
2.56
7
19
21.67
7.1289
22
9
8
17
20
9
21.4
19.36
9
22
19.67
5.4289
20.2
3.24
10
23
19.33
13.4689
19.8
10.24
11
18
20.67
7.1289
20.8
7.84
12
23
21
4
19.8
10.24
Total = 78.3534
RMSE 
78.3534
 2.95
9
Total = 62.48
RMSE 
62.48
 2.99
7
Exponential smoothing involves a forecast equation that takes the following form
Ft 1  wAt  1  wFt
w  0,1
Forecast for time t
Forecast for time t+1
Actual value at time
t
Smoothing parameter
Note: when w = 1, your forecast is equal to the previous value. When w = 0, your
forecast is a constant.
For exponential smoothing, we need to choose a value for the weighting formula as
well as an initial forecast
Quarter
Market
Share
W=.3
W=.5
1
20
21.0
21.0
2
22
20.7
20.5
3
23
21.1
21.3
4
24
21.7
22.2
5
18
22.4
23.1
6
23
21.1
20.6
7
19
21.7
21.8
8
17
20.9
20.4
9
22
19.7
18.7
10
23
20.4
20.4
11
18
21.2
21.7
12
23
20.2
19.9
Usually, the initial forecast
is chosen to equal the
sample average
.523  .520.6  21.8
As was mentioned earlier, the smaller w will produce a smoother forecast
30
25
20
15
10
5
0
1
2
3
4
5
Actual
6
7
w=.3
8
9
w=.5
10
11
12
Calculating forecasts is straightforward…
W=.3
Quarter
Market
Share
W=.3
W=.5
1
20
21.0
21.0
2
22
20.7
20.5
3
23
21.1
21.3
4
24
21.7
22.2
5
18
22.4
23.1
6
23
21.1
20.6
7
19
21.7
21.8
8
17
20.9
20.4
9
22
19.7
18.7
10
23
20.4
20.4
11
18
21.2
21.7
12
23
20.2
19.9
.323  .720.2  21.04
W=.5
.523  .519.9  21.45
So, how do we choose W??
Quarter
Market
Share
W = .3
Squared
Error
W=.5
Squared
Error
1
20
21.0
1
21.0
1
2
22
20.7
1.69
20.5
2.25
3
23
21.1
3.61
21.3
2.89
4
24
21.7
5.29
22.2
3.24
5
18
22.4
19.36
23.1
26.01
6
23
21.1
3.61
20.6
5.76
7
19
21.7
7.29
21.8
7.84
8
17
20.9
15.21
20.4
11.56
9
22
19.7
5.29
18.7
10.89
10
23
20.4
6.76
20.4
6.76
11
18
21.2
10.24
21.7
13.69
12
23
20.2
7.84
19.9
9.61
Total = 87.19
RMSE 
87.19
 2.70
12
Total = 101.5
RMSE 
101.5
 2.91
12