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
Probability
Tail
A probability distribution is a collection of probabilities describing the odds of any particular event
1/6
1 2 3 4 5 6

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
Mean
We generally assume a Normal Distribution which can be characterized by a mean (average) and standard deviation (measure of dispersion)
Temperature

Without some math, we can’t find the probability of a specific outcome, but we can easily divide up the distribution
Probability
Mean-2SD
2.5% 13.5%
Mean -1SD
34%
Mean
34%
Mean+1SD
13.5%
Mean+2SD
Temperature
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)
23 41
Can’t we do a little better than this?
59 77 95
Temperature

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)
70 77 84 91 95 98
Conditioning on month gives us a more accurate probabilities!
Temperature

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
N


,

2

We could find this distribution by collecting temperature data for south bend
Sample Mean
(Average) x




1
N 

 i
N 

1 x i
 
Sample
Variance s 2 



1
N 

 i
N 

1
 x i
 x

2  
2
Note: Standard Deviation is the square root of the variance.

Probability distributions are scalable x

N y y

 kx
N
 k

,k
2 σ 2

3 X =
Mean = 1
Variance = 4
Std. Dev. = 2
Mean = 3
Variance = 36 (3*3*4)
Std. Dev. = 6

Probability distributions are additive x x y


 y
N
N



μ

N x y

,σ
,σ

2 x
2 y x


  y
,σ x
2  σ y
2 
2 cov xy

+
Mean = 1
Variance = 1
Std. Dev. = 1
COV = 2
Mean = 2
Variance = 9
Std. Dev. = 3
=
Mean = 3
Variance = 14 (1 + 9 + 2*2)
Std. Dev. = 3.7

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
Variance = 4
Std. Dev. = 2
Mean = $ 12,000
Variance = 4,000,000
Std. Dev. = $ 2,000

We could also use this to forecast:
Value = $20,000 - $1,000 (Age)
?
Value = $20,000 - $1,000 (6) = $14,000
Note: There is NO uncertainty in this prediction.

You believe that there is a relationship between age and value, but you don’t know what it is….
Note that while the true distribution of age is N(8,4), our collected sample
be N(8,4). This sampling error will create errors in our estimates!!

18000,00
16000,00
14000,00
12000,00
10000,00
8000,00
6000,00
4000,00
2000,00
0,00
0 a
2 4 6 8
Slope = b
10 12

Value = a + b * (Age) + error
14

Intercept
Age
12,354
- 854
653
80
Value = $12,354 - $854 * (Age) + error
18.9
-10.60
T-Stats bigger than 2 in absolute value are considered statistically significant!
Intercept (a)
Mean = $12,354
Std. Dev. = $653
Age (b)
Mean = -$854
Std. Dev. = $80

Regression Statistics
R Squared
Standard Error
0.36
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
Mean = $12,354
Std. Dev. = $653
Mean = $854
Std. Dev. = $ 80
Mean = $0
Std. Dev. = $2,250
StdDev

Var

X
2
Var

2 XCov
 

Var
 error

Cov
 
 
X Var
StdDev

653
2
(Recall, The Average Car age is 8 years)

 
80
2 
2
  
80
2 
2250
2 
$ 2 , 259


12 , 354

854 *
 

$ 7 , 230
StdDev

653
2 
 
80
2 
2
  
80
2 
2250
2 
$ 2 , 259
Value
Forecast Interval
+95%
-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
Pat
Buchanan’s
Votes
Observable
Demographics
Using Palm Beach data, forecast Pat Buchanan’s vote total for Palm
Beach
B

PB
F
 
PB

The Data: Demographic Data By County
County
Alachua
Baker
Black
(%)
21.8
16.8
Age 65
(%)
9.4
7.7
Hispanic
(%)
4.7
1.5
College
(%)
34.6
5.7
Income
(000s)
26.5
27.6
Buchanan
Votes
262
73
Total
Votes
84,966
8,128
What variables do you think should affect Pat Buchanan’s Vote total?
# of Buchanan votes
# of votes gained/lost for each percentage point increase in college educated population
V
 a
 bC
 
% of County that is college educated

Results Parameter
Value
Standard Error
T-Statistic a
5.35
58.5
.09
b
14.95
3.84
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
V

5 .
35

14 .
95 C
Plug in Values for
College % to get vote predictions
0 15
There is a 95% chance that the value for ‘b’ lies between
23 and 7
County
Alachua
Baker
College
(%)
34.6
5.7
Predicted
Votes
522
90
Actual
Votes
262
73
Error
260
17

Lets try something a little different…
County
Alachua
Baker
College (%) Buchanan
Votes
34.6
262
5.7
73
Log of Buchanan
Votes
5.57
4.29
Log of Buchanan votes
LN
 a
 bC
 
% of County that is college educated
Percentage increase/decease in votes for each percentage point increase in college educated population

Results Parameter
Value
Standard Error
T-Statistic a
3.45
.27
12.6
b
.09
.02
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%
LN

3 .
45

.
09 C
V
 e
LN
Plug in Values for
College % to get vote predictions
0 .09
There is a 95% chance that the value for ‘b’ lies between
.13 and .05
County
Alachua
Baker
College
(%)
34.6
5.7
Predicted
Votes
902
55
Actual
Votes
262
73
Error
640
-18

How about this…
County
Alachua
Baker
College (%) Buchanan
Votes
34.6
262
5.7
73
Log of College (%)
3.54
1.74
# of Buchanan votes
V
 a
 bLN
 
 
Log of % of County that is college educated
Gain/ Loss in votes for each percentage increase in college educated population

Results Parameter
Value
Standard Error
T-Statistic a
-424
139
-3.05
b
252
54
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
V
 
424

252 LN
0 .09
There is a 95% chance that the value for ‘b’ lies between
360 and 144
Plug in Values for
College % to get vote predictions
County
Alachua
Baker
College
(%)
34.6
5.7
Predicted
Votes
469
15
Actual
Votes
262
73
Error
207
-58

One More…
County College
(%)
Alachua 34.6
Baker 5.7
Buchanan
Votes
262
73
Log of College (%) Log of Buchanan
Votes
3.54
5.57
1.74
4.29
Log of Buchanan votes
LN
 a
 bLN
 
 
Log of % of County that is college educated
Percentage gain/Loss in votes for each percentage increase in college educated population

Results Parameter
Value
Standard Error
T-Statistic a
.71
.63
1.13
b
1.61
.24
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%
LN

.
71

1 .
61 LN
V
 e
LN
0 .09
There is a 95% chance that the value for ‘b’ lies between
2 and 1.13
Plug in Values for
College % to get vote predictions
County
Alachua
Baker
College
(%)
34.6
5.7
Predicted
Votes
624
34
Actual
Votes
262
73
Error
362
-39

It turns out the regression with the best fit looks like this.
County
Alachua
Baker
Black
(%)
21.8
16.8
Age 65
(%)
9.4
7.7
Hispanic
(%)
4.7
1.5
College
(%)
34.6
5.7
Income
(000s)
26.5
27.6
Buchanan
Votes
262
73
Total
Votes
84,966
8,128
LN
 a
1
 a
2
B
 a
2
A
65
 a
3
H
 a
4
C
 a
5
I
 
Error term
Buchanan Votes
Total Votes
*100
Parameters to be estimated

The Results:
Variable
Intercept
Black (%)
Age 65 (%)
Hispanic (%)
College (%)
Income (000s)
LN

2 .
146

.
0132
Now, we can make a forecast!

Coefficient
2.146
-.0132
-.0415
-.0349
-.0193
-.0658
.
0415
 
65

Standard Error
.396
.0057
.0057
.0050
.0068
.00113
.
0349 t - statistic
5.48
-2.88
-5.93
-6.08
-1.99
-4.58
R Squared = .73

.
0193

.
0658
 
County
Alachua
Baker
Black (%) Age 65 (%) Hispanic (%) College (%) Income
(000s)
21.8
9.4
4.7
34.6
26.5
16.8
7.7
1.5
5.7
27.6
Buchanan
Votes
262
73
Total
Votes
84,966
8,128
County
Alachua
Baker
Predicted
Votes
520
55
Actual
Votes
262
73
Error
258
-18

County Black
(%)
Palm Beach 21.8
Age 65
(%)
23.6
Hispanic
(%)
9.8
College
(%)
22.1
Income
(000s)
33.5
Buchanan
Votes
3,407
Total
Votes
431,621
LN

2 .
146

.
0132

.
0415
 
65

.
0349
LN
 
 
2 .
004
P
 e

2 .
004 
.
134 %

.
00134

431 , 621


578

.
0193

.
0658
 
This would be our prediction for Pat
Buchanan’s vote total!

LN
 
 
2 .
004
Probability
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
-2.004 – 2*(.2556)
= -2.5152
LN(%Votes)
-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 %
Probability
Next, lets convert the Logs to vote percentages e

2 .
5152 
.
08 % e

1 .
4928
% of Votes

.
22 %
There is a 95% chance that Buchanan’s vote percentage lies in this range


.
00134

431 , 621


578
Probability
Finally, we can convert to actual votes
3,407 votes turns out to be 7 standard deviations away from our forecast!!!

.
0008

431 , 621


348

.
0022

431 , 621

Votes

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”
Q
D

D

P , I ,...

Price
Quantity
Demanded
Price
Income
Other
“Demand
Shifters”
D
Quantity

Demand curves slope downwards – this reflects the negative relationship between price and quantity. Elasticity of Demand measures this effect quantitatively
Price
4
4
* 100

50 %
$6.00
$4.00

D

%

Q
%

P


50
50
 
1
1
D

I

$ 10

Quantity
2
2
2
* 100
 
50 %

For any fixed price, demand (typically) responds positively to increases in income. Income
Elasticity measures this effect quantitatively
Price

I

%

Q
%

I

100
100

1
$4.00
%

I


20

10
10
* 100

100 %
4
D

I

$ 20

D

I

$ 10

Quantity
2
%

Q

2
2
* 100

100 %

Cross price elasticity refers to the impact on demand of another price changing
Price
 p

%
%

Q
H

P
L

200
100

2
$4.00
%

P
L

2
2
* 100

100 %
6
D

P
L
D

P
L

$ 2


$ 4

Quantity
2
%

Q

2
2
* 100

200 %
A positive cross price elasticity refers to a substitute while a negative cross price elasticity refers to a compliment

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 Total Sales
Granger
Mishawaka
1.02
2.56
Average Income
(Thousands)
Competitor’s
Price
21.934
35.796
1.48
2.53
Advertising
Expenditures
(Thousands)
2.367
26.922
9,809
130,835
Lets begin by estimating a basic demand curve – quantity demanded is a linear function of price.
Q
 a
0
 a
1
P
Change in quantity demanded per $ change in price (to be estimated)

That is, we have estimated the following equation
Variable
Intercept
Price (X)
Regression Results
Coefficient
155,042
-46,087
Standard Error
18,133
7214
Regression Statistics
R Squared
Standard Error t Stat
8.55
-6.39
.17
48,074
Q

155 , 042

46 , 087 P
Every dollar increase in price lowers sales by 46,087 units.

Price of Pepsi $1.37
Q

155 , 042

46 , 087

1 .
37


91 , 903
P
We can now use this estimated demand curve along with price in South
Bend to estimate demand in South Bend $1.37
91,903
Q

We can get a better sense of magnitude if we convert the estimated coefficient to an elasticity
Q

155 , 042

46 , 087 P

Q
 p
 
46 , 087
 

Q
 p

 p
Q


 
46 , 087



1 .
37
91 , 903


P
$1.37
  
46 , 087



1 .
37
91 , 903


 
.
68
Q
P
Q

$ 1 .
37

155 , 042

46 , 087

1 .
37


91 , 903
91,903

As we did earlier, we can experiment with different functional forms by using logs
City Price Total Sales
Granger
Mishawaka
1.02
2.56
Average Income
(Thousands)
Competitor’s
Price
21.934
35.796
1.48
2.53
Advertising
Expenditures
(Thousands)
2.367
26.922
9,809
130,835
Adding logs changes the interpretation of the coefficients
Q
 a
0
 a
1
LN
Change in quantity demanded per percentage change in price
(to be estimated)

That is, we have estimated the following equation
Variable
Intercept
Price (X)
Regression Results
Coefficient
133,133
-103,973
Standard Error
14,892
16,407
Regression Statistics
R Squared
Standard Error t Stat
8.93
-6.33
.17
48,140
Q

133 , 133

103 , 973 LN
Every 1% increase in price lowers sales by 103,973 units.

Values For South Bend
Price of Pepsi
Log of Price
Q

133 , 133

103 , 973
 
P
We can now use this estimated demand curve along with price in South
Bend to estimate demand in South Bend
$1.37
.31
$1.37
100,402
Q

We can get a better sense of magnitude if we convert the estimated coefficient to an elasticity
Q

133 , 133

103 , 973 P

Q
%
 p
 
103 , 973
 

Q
%
 p


1
Q


 
103 , 973



1
100 , 402


P   
103 , 973



1
100 , 402


 
1 .
04
LN
Q

 

LN
133 , 133


1 .
37


.
31
103 , 973
 

100 , 402
$1.37
100,402
Q

As we did earlier, we can experiment with different functional forms by using logs
City Price Total Sales
Granger
Mishawaka
1.02
2.56
Average Income
(Thousands)
Competitor’s
Price
21.934
35.796
1.48
2.53
Advertising
Expenditures
(Thousands)
2.367
26.922
9,809
130,835
Adding logs changes the interpretation of the coefficients
LN
 a
0
 a
1
P
Percentage change in quantity demanded per $ change in price (to be estimated)

That is, we have estimated the following equation
Variable
Intercept
Price (X)
Regression Results
Coefficient
13
-1.22
Standard Error
.34
.13
Regression Statistics
R Squared
Standard Error
LN

13

1 .
22 P t Stat
38.1
-8.98
.28
.90
Every $1 increase in price lowers sales by 1.22%.

Values For South Bend
Price of Pepsi $1.37
LN

13

1 .
22

1 .
37


11 .
33
Q
 e
11 .
33 
83 , 283
P
We can now use this estimated demand curve along with price in South
Bend to estimate demand in South Bend $1.37
83,283
Q

We can get a better sense of magnitude if we convert the estimated coefficient to an elasticity
LN

13

1 .
22 P

%

Q
 p
 
1 .
22

%

Q
 p p
1
 
1 .
22
1 .
37
1
P   
1 .
22
1 .
37
1
 
1 .
67
LN

13

1 .
22

1 .
37


11 .
33
Q
 e
11 .
33 
83 , 283
$1.37
83,283
Q

As we did earlier, we can experiment with different functional forms by using logs
City Price Total Sales
Granger
Mishawaka
1.02
2.56
Average Income
(Thousands)
Competitor’s
Price
21.934
35.796
1.48
2.53
Advertising
Expenditures
(Thousands)
2.367
26.922
9,809
130,835
Adding logs changes the interpretation of the coefficients
LN
 a
0
 a
1
LN
Percentage change in quantity demanded per percentage change in price (to be estimated)

That is, we have estimated the following equation
Variable
Intercept
Price (X)
Regression Results
Coefficient
12.3
-2.60
Standard Error
.28
.31
Regression Statistics
R Squared
Standard Error
LN

12

2 .
6 LN t Stat
42.9
-8.21
.25
.93
Every 1% increase in price lowers sales by 2.6%.

Values For South Bend
Price of Pepsi
Log of Price
LN

12

2 .
6
Q
 e
11 .
19 
72 , 402
P
$1.37
.31

11 .
19
We can now use this estimated demand curve along with price in South
Bend to estimate demand in South Bend $1.37
72,402
Q

We can get a better sense of magnitude if we convert the estimated coefficient to an elasticity
LN

12

2 .
6 LN
%

Q
%
 p
 
2 .
6
 
P
LN

12

2 .
6
Q
 e
11 .
19

72 , 402

11 .
19
$1.37
72,402
  
2 .
6
Q

We can add as many variables as we want in whatever combination. The goal is to look for the best fit.
LN
 a
0
 a
1
P
 a
2
LN
 a
3
LN
  c
Variable
Intercept
Price
Log of Income
Log of Competitor’s Price
% 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
Coefficient
5.98
-1.29
1.46
2.00
Standard Error
1.29
.12
.34
.34
t Stat
4.63
-10.79
4.29
5.80
R Squared: .46

Values For South Bend
Price of Pepsi
Log of Income
Log of Competitor’s Price
$1.37
3.81
.80
Now we can make a prediction and calculate elasticities
LN

5 .
98
 
1 .
29

1 .
37


1 .
46

3 .
81


2 .
00
 

11 .
36
Q
 e
11 .
36 
87 , 142
P


I


Q

P

P
%

Q
%

I
  1

1 .
46

1 .
29

CP



%

Q
%

P c



2 .
00
1 .
37
1
 
1 .
76
$1.37
Q
87,142

We could use a cross sectional regression to forecast quantity demanded out into the future, but it would take a lot of information!
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
Time

Time Series estimation ignores the demand factors constant and estimates the variation in demand over 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

Time series estimation leaves the demand factors constant 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

2005:2
2005:3
2005:4
2006:1
2006:2
2006:3
2006:4
Time Period Quantity (millions of kilowatt hours)
2003:1 11
2003:2
2003:3
2003:4
15
12
14
2004:1
2004:2
2004:3
2004:4
2005:1
12
17
13
16
14
18
15
17
15
20
16
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 x t
 x
0
 bt
Estimated quarterly growth
(in kilowatt hours)
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

Variable
Intercept
Time Trend
Regression Results
Coefficient Standard Error
11.9
.953
.394
.099
t Stat
12.5
4.00
Regression Statistics
R Squared
Standard Error
Observations
.53
1.82
16 x t

11 .
9

Lets forecast electricity usage at the mean time period (t = 8)
.
394 t x
ˆ t

Var
11 .
9
  t


3 .
.
394
50

15 .
05

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
ˆ t

11 .
9
  t


.
394
47 .
7

41 .
85

One method of evaluating a forecast is to calculate the root mean squared error
2005:1
2005:2
2005:3
2005:4
2006:1
2006:2
2006:3
2006:4
Time Period Actual Predicted Error
2003:1 11 12.29
-1.29
2003:2
2003:3
15
12
12.68
13.08
2.31
-1.08
2003:4
2004:1
2004:2
2004:3
2004:4
14
12
17
13
16
13.47
13.87
14.26
14.66
15.05
.52
-1.87
2.73
-1.65
.94
15
20
16
19
14
18
15
17
15.44
15.84
16.23
16.63
17.02
17.41
17.81
18.20
-1.44
2.15
-1.23
.37
-2.02
2.58
-1.81
.79
Sum of squared forecast errors
RMSE

Number of
RMSE

Observations


A t n
1 .
70

F t

2

Lets take another look at the data…it seems that there is a regular pattern…
25
Q2
Q2
20
Q2
Q2
15
10
5
0
2003-1 2004-1 2005-1 2006-1
We are systematically under predicting usage in the second quarter

2005:2
2005:3
2005:4
2006:1
2006:2
2006:3
2006:4
Time Period Actual Predicted Ratio Adjusted
2003:1 11 12.29
.89
12.29(.87)=10.90
2003:2
2003:3
2003:4
2004:1
15
12
14
12
12.68
13.08
13.47
13.87
1.18
.91
1.03
.87
12.68(1.16) = 14.77
13.08(.91) = 11.86
13.47(1.04) = 14.04
13.87(.87) = 12.30
2004:2
2004:3
2004:4
2005:1
17
13
16
14
14.26
14.66
15.05
15.44
1.19
.88
1.06
.91
14.26(1.16) = 16.61
14.66(.91) = 13.29
15.05(1.04) = 15.68
15.44(.87) = 13.70
18
15
17
15
20
16
19
15.84
16.23
16.63
17.02
17.41
17.81
18.20
1.14
.92
1.02
.88
1.14
.89
1.04
15.84(1.16) = 18.45
16.23(.91) = 14.72
16.63(1.04) = 17.33
17.02(.87) = 15.10
17.41(1.16) = 20.28
17.81(.91) = 16.15
18.20(1.04) = 18.96
We can adjust for this seasonal component…
Average Ratios
•Q1 = .89
•Q2 = 1.16
•Q3 = .90
•Q4 = 1.04

Now, we have a pretty good fit!!
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
ˆ t

11 .
9

.
394

41 .
85

1 .
04


43 .
52

We could also account for seasonal variation by using dummy variables x t
 x
0
 b
0 t
 b
1
D
1
 b
2
D
2
 b
3
D
3
D i


1 ,

 if quarter
0 , else i
Note: we only need three quarter dummies. If the observation is from quarter 4, then
D
1 x t

 x
D
2
0

 b
0 t
D
3

0

Variable
Intercept
Time Trend
D1
D2
D3
Regression Results
Coefficient Standard Error
12.75
.226
.375
-2.375
1.75
-2.125
.0168
.219
.215
.213
Regression Statistics
R Squared
Standard Error
Observations
.99
.30
16
Note the much better fit!!
t Stat
56.38
22.2
-10.83
8.1
-9.93
x t

12 .
75

.
375 t

2 .
375 D
1

1 .
75 D
2

2 .
125 D
3

2004:3
2004:4
2005:1
2005:2
2005:3
2005:4
2006:1
2006:2
2006:3
2006:4
Time Period Actual Ratio Method Dummy
Variables
2003:1
2003:2
11
15
10.90
14.77
10.75
15.25
2003:3
2003:4
2004:1
2004:2
12
14
12
17
11.86
14.04
12.30
16.61
11.75
14.25
12.25
16.75
13
16
14
18
15
17
15
20
16
19
13.29
15.68
13.70
18.45
14.72
17.33
15.10
20.28
16.15
18.96
13.25
15.75
13.75
18.25
14.75
17.25
15.25
19.75
16.25
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 Ratio
2006-1

Recall our prediction for period 76 ( Year 2022 Q4)
70
60
50
40
30
20
10
0 x t

12 .
75

.
375
 

41 .
25

Recall, our trend line took the form… x t
 x
0
 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 t g % x t
 x
0 
1
100

If we wish to estimate this equation, we have a little work to do… x t
 x
0

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 x t
 ln x
0
 t ln

1
 g


Variable
Intercept
Time Trend
Regression Results
Coefficient Standard Error
2.49
.063
.026
.006
t Stat
39.6
4.06
Regression Statistics
R Squared
Standard Error
Observations
.54
.1197
16 ln x t

2 .
49

.
026 t
Lets forecast electricity usage at the mean time period (t = 8) ln
Var x
ˆ t
 x
ˆ t
2 .
49

.
026
 
 

.
0152

2 .
698
BE CAREFUL….THESE NUMBERS ARE LOGS !!!

ln
Var x
ˆ t
 x
ˆ t
2 .
49

.
026
 
 

.
0152

2 .
698
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
T = 8
2005-1 2006-1
RMSE

1 .
70

300
200
100
0
1
600
Errors is growth rates compound quickly!!
500
400
13 25 37 49 61 73 85 97 e
4 .
49

/

2

89 .
22
SD


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 p
 p
0
 bt
Price at time t
Price at April 1993 Number of months from April 1993
Variable
Intercept
Time Trend
Regression Results
Coefficient Standard Error
.67
.05
.010
.0004
t Stat
12.19
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 p t
 p
0
 b
0 t
 b
1
D
1
 b
2
D
2
 b
3
D
3
i

1 ,
0 ,
Variable
Intercept
Time Trend
D1
D2
D3
Regression Results
Coefficient Standard Error
.58
.07
.01
-.03
.15
.16
.0004
.075
.074
.075
t Stat
8.28
23.7
-.43
2.06
2.20
R Squared= .74

If we wanted to remove the seasonal component, we could by subtracting the seasonal dummy off each gas price
Date
1993 – 04
1993 - 07
1993 - 10
1994 - 01
1994 - 04
Price
Seasonalizing
1.05
1.06
1.06
.98
1.00
Regression coefficient
2 nd Quarter
3 rd Quarter
.15
.16
4 th Quarter
1 st Quarter
2 nd Quarter
0
-.03
.15
Seasonalized data
.90
90
1.06
1.01
.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
~ t
 p
0
 bt
Seasonalized Price Series
Variable
Intercept
Time Trend
Regression Results
Coefficient
.587
.010
Standard Error
.05
.0004
t Stat
11.06
23.92
p t
 p
0
 b
0 t
 b
1
D
1
 b
2
D
2
 b
3
D
3
Variable
Intercept
Time Trend
D1
D2
D3
Regression Results
Coefficient
.587
.010
0
0
0
Standard Error
.07
.0004
.075
.074
.075
t Stat
8.28
23.7
0
0
0

The regression we have in place gives us the trend plus the seasonal component of the data p t

.
58

.
01 t
 
.
03 D
1

.
15 D
2

.
16 D
3
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
Date
1993 - 04
1993 - 05
1993 - 06
1993 - 07
1993 - 08
Business Cycle Component
Actual Price
Predicted Price
(From regression)
1.050
1.071
.752
.763
1.075
1.064
1.048
773
.797
.807
Business Cycle
Component
.297
.308
.301
.267
.240

We can plot this and compare it with business cycle dates
Actual
Price p t
 p t
Predicted
Price

Date
1993 - 04
1993 - 05
1993 - 06
1993 - 07
1993 - 08
Variable
Intercept
Time Trend
D1
D2
D3
Data Breakdown
Actual Price
1.050
1.071
1.075
1.064
1.048
Trend
.58
.59
.60
.61
.62
Regression Results
Coefficient Standard Error
.58
.07
.01
-.03
.15
.16
.0004
.075
.074
.075
Seasonal
.15
.15
.15
.16
.16
Business Cycle
.320
.331
.325
.294
.268
t Stat
8.28
23.7
-.43
2.06
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
Percentage price impact of quarter I relative to quarter 4
Monthly growth rate ln p t
 p
0
 b
0 t
 b
1
D
1
 b
2
D
2
 b
3
D
3
i

1 ,
0 ,
Variable
Intercept
Time Trend
D1
D2
D3
Regression Results
Coefficient Standard Error
-.14
.03
.005
-.02
.06
.07
.0001
.032
.032
.032
t Stat
-4.64
29.9
-.59
2.07
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
Date
1993
– 04
1993 - 07
1993 - 10
1994 - 01
1994 - 04
Price
1.05
1.06
1.06
.98
1.00
Log of Price
Seasonalizing
Regression coefficient
.049
2 nd Quarter
.06
.062
.062
-.013
3 rd Quarter .07
4 th Quarter
0
1 st Quarter
-.02
.005
2 nd Quarter .06
Log of Seasonalized data
-.019
-.010
Seasonalized
Price
.98
.99
.062
.006
1.06
1.00
-.062
.94
Example: e

.
019 
.
98

The regression we have in place gives us the trend plus the seasonal component of the data ln p t
 
.
14

.
005 t

.
02 D
1

.
06 D
2

.
07 D
3
Predicted
Log of
Price
Trend
Seasonal
If we subtract our predicted price (from the regression) from the actual price, we will have isolated the business cycle and noise
Date
1993 - 04
1993 - 05
1993 - 06
1993 – 07
1993 - 08
Business Cycle Component
Actual Price
Predicted Log Price
(From regression)
1.050
1.071
-.069
-.063
1.075
1.064
1.048
-.057
-.047
-.041
e

.
069 
.
93
Predicted
Price
.93
.94
.94
.95
.96
Business Cycle
Component
.12
.13
.13
.11
.09

As you can see, very similar results
Actual
Price p t
 p t
Predicted
Price

In either case, we could make a forecast for gasoline prices next year. Lets say, April 2011.
Date
April 2011
Forecasting Data
Time Period
217
Quarter
2 p t

.
58

.
01
 

.
03
 

.
15
 

.
16
 

2 .
90
OR ln p t
 
.
14

.
005
 

.
02
 

.
06
 

.
07
 

1 .
005 e
1 .
005 
2 .
73

10
11
12
6
7
8
9
4
5
2
3
Quarter Market Share
1 20
22
23
24
18
23
19
17
22
23
18
23
Consider a new forecasting problem.
You are asked to forecast a company’s market share for the 13 th quarter.
15
10
5
0
30
25
20
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.
8
9
6
7
4
5
10
11
12
2
3
Quarter Market
Share
1 20
22
23
24
18
23
19
17
22
23
18
23
MA(3) MA(5)
21.67
23
21.67
21.67
20
19.67
19.33
20.67
21
21.4
22
21.4
20.2
19.8
20.8
19.8
A moving average of length N is equal to the average value over the previous N periods
MA
 
 

1 t t 
N
A t
N

The longer the moving average, the smoother the forecasts are…
30
25
20
15
10
5
0
1 2 3 4 5 6 7 8 9 10 11 12
Actual
MA(3)
MA(5)

Calculating forecasts is straightforward…
MA(3) MA(5)
9
10
11
12
5
6
7
8
3
4
1
2
Quarter Market
Share
20
22
23
24
18
23
19
17
22
23
18
23
21.67
23
21.67
21.67
20
19.67
19.33
20.67
21
21.4
22
21.4
20.2
19.8
20.8
19.8
MA(3)
23

18

23

21 .
33
3
MA(5)
23

18

23

22

17

20 .
6
5
So, how do we choose N??

9
10
11
12
4
5
6
7
8
2
3
Quarter Market
Share
1 20
22
23
24
18
23
19
17
22
23
18
23
RMSE

MA(3) Squared
Error
21.67
23
21.67
21.67
20
19.67
19.33
20.67
21
7.1289
4
Total = 78.3534
5.4289
25
1.7689
7.1289
9
5.4289
13.4689
78 .
3534

2 .
95
9
MA(5) Squared
Error
21.4
22
21.4
20.2
19.8
20.8
19.8
10.24
Total = 62.48
2.56
9
19.36
3.24
10.24
7.84
RMSE

62 .
48

2 .
99
7

Exponential smoothing involves a forecast equation that takes the following form
F t

1
 wA t


1
 w

F t w

Forecast for time t+1
Smoothing parameter t
Actual value at time
Forecast for time t
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
W=.3
W=.5
8
9
10
11
12
4
5
6
7
2
3
Quarter Market
Share
1 20
22
23
24
18
23
19
17
22
23
18
23
21.0
20.7
21.1
21.7
22.4
21.1
21.7
20.9
19.7
20.4
21.2
20.2
21.0
20.5
21.3
22.2
23.1
20.6
21.8
20.4
18.7
20.4
21.7
19.9
.
5
Usually, the initial forecast is chosen to equal the sample average
  
20 .
6


21 .
8

As was mentioned earlier, the smaller w will produce a smoother forecast
20
15
10
5
30
25
0
1 2 3 4 5
Actual
6 7 w=.3
8 9 w=.5
10 11 12

Calculating forecasts is straightforward…
8
9
10
11
12
4
5
6
7
2
3
Quarter Market
Share
1 20
22
23
24
18
23
19
17
22
23
18
23
W=.3
21.0
20.7
21.1
21.7
22.4
21.1
21.7
20.9
19.7
20.4
21.2
20.2
W=.5
21.0
20.5
21.3
22.2
23.1
20.6
21.8
20.4
18.7
20.4
21.7
19.9
W=.3
.
3
  
20 .
2


21 .
04
W=.5
.
5
   

21 .
45
So, how do we choose W??

9
10
11
12
4
5
6
7
8
2
3
Quarter Market
Share
1 20
22
23
24
18
23
19
17
22
23
18
23
RMSE

W = .3
21.0
20.7
21.1
21.7
22.4
21.1
21.7
20.9
19.7
20.4
21.2
20.2
7.84
Total = 87.19
Squared
Error
1
1.69
3.61
5.29
19.36
3.61
7.29
15.21
5.29
6.76
10.24
87 .
19

2 .
70
12
W=.5
21.0
20.5
21.3
22.2
23.1
20.6
21.8
20.4
18.7
20.4
21.7
19.9
9.61
Total = 101.5
Squared
Error
1
2.25
2.89
3.24
26.01
5.76
7.84
11.56
10.89
6.76
13.69
RMSE

101 .
5

2 .
91
12