Embed Size (px)
Citation preview
Demand Estimation and Forecasting
Finance 30210: Managerial Economics
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
Head Tail
1/2
Probability
1 6
1/6
Probability
2 3 4 5
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
Temperature
We generally assume a Normal Distribution which can be characterized by a mean (average) and standard deviation (measure of dispersion)
Mean
Standard Deviation
Probability
Temperature
Without some math, we can’t find the probability of a specific outcome, but we can easily divide up the distribution
Mean Mean+1SD Mean+2SDMean -1SDMean-2SD
2.5% 2.5%13.5% 34% 34% 13.5%
Annual Temperature in South Bend has a mean of 59 degrees and a standard deviation of 18 degrees.
Probability
Temperature59 77 954123
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)
Can’t we do a little better than this?
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
Temperature84 91 987770
95 degrees falls a little more than one standard deviation away (there approximately a 16% chance that the temperature is 95 or greater)
95
Conditioning on month gives us a more accurate probabilities!
5.PrPr TailsHeads
We know that there should be a “true” probability distribution that governs the outcome of a coin toss (assuming a fair coin)
Suppose that we were to flip a coin over and over again and after each flip, we calculate the percentage of heads & tails
FlipsTotal
Headsof
#5.
That is, if we collect “enough” data, we can eventually learn the truth!
(Sample Statistic) (True Probability)
We can follow the same process for the temperature in South Bend
Temperature ~ 2,N
We could find this distribution by collecting temperature data for south bend
N
iixN
x1
1
2
1
22 1
N
ii xx
Ns
Sample Mean
(Average)
Sample Variance
Note: Standard Deviation is the square root of the variance.
Mean = 1
Variance = 4
Std. Dev. = 2
Probability distributions are scalable
22
2
σ,kkNy
kxy
μ,σNx
3 X =
Mean = 3
Variance = 36 (3*3*4)
Std. Dev. = 6
Some useful properties of probability distributions
Mean = 1
Variance = 1
Std. Dev. = 1
Probability distributions are additive
xyyxyx
yy
xx
σ,σNyx
,σNy
,σμNx
cov222
2
2
+Mean = 2
Variance = 9
Std. Dev. = 3
COV = 2
=Mean = 3
Variance = 14 (1 + 9 + 2*2)
Std. Dev. = 3.7
Mean = 8
Variance = 4
Std. Dev. = 2
Mean = $ 12,000
Variance = 4,000,000
Std. Dev. = $ 2,000
Suppose we know that the value of a car is determined by its age
Value = $20,000 - $1,000 (Age)
Car Age Value
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!!
Value = a + b * (Age) + error 20,σNerror
We want to choose ‘a’ and ‘b’ to minimize the error!
0 2 4 6 8 10 12 140.00
2000.00
4000.00
6000.00
8000.00
10000.00
12000.00
14000.00
16000.00
18000.00
a
Slope = b
Regression Results
Variable Coefficients Standard Error t Stat
Intercept 12,354 653 18.9
Age - 854 80 -10.60
Value = $12,354 - $854 * (Age) + error
We have our estimate of “the truth”
Intercept (a)
Mean = $12,354
Std. Dev. = $653
Age (b)
Mean = -$854
Std. Dev. = $80
T-Stats bigger than 2 in absolute value are considered statistically significant!
Regression Statistics
R Squared 0.36
Standard Error 2250
Error Term
Mean = 0
Std. Dev = $2,250
Percentage of value variance explained by age
We can now forecast the value of a 6 year old car
Value = $12,354 - $854 * (Age) + error
6
Mean = $12,354
Std. Dev. = $653
Mean = $854
Std. Dev. = $ 80
Mean = $0
Std. Dev. = $2,250
errorVarbaXCovbVarXaVarStdDev ,22
bVarXbaCov , (Recall, The Average Car age is 8 years)
259,2$225080862806653 22222 StdDev
8x
+95%
-95%
Age
Value
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
6Age
Forecast Interval
259,2$225080862806653 22222 StdDev
230,7$6*854354,12 Value
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 Using Palm Beach data,
forecast Pat Buchanan’s vote total for Palm Beach
DFB
Pat Buchanan’s Votes
“Are a function of”
Observable Demographics
PBPB DFB
The Data: Demographic Data By County
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
What variables do you think should affect Pat Buchanan’s Vote total?
bCaV
% of County that is college educated
# of votes gained/lost for each percentage point increase in college educated population
# of Buchanan votes
Parameter a b
Value 5.35 14.95
Standard Error 58.5 3.84
T-Statistic .09 3.89
Results R-Square = .19
CV 95.1435.5
The distribution for ‘b’ has a mean of 15 and a standard deviation of 4
15
There is a 95% chance that the value for ‘b’ lies between 23 and 7
County College (%)
Predicted Votes
Actual Votes
Error
Alachua 34.6 522 262 260
Baker 5.7 90 73 17
0
Plug in Values for College % to get vote predictions
19% of the variation in Buchanan’s votes across counties is explained by college education
Each percentage point increase in college educated (i.e. from 10% to 11%) raises Buchanan’s vote total by 15
County College (%) Buchanan Votes
Log of Buchanan Votes
Alachua 34.6 262 5.57
Baker 5.7 73 4.29
Lets try something a little different…
bCaVLN
% of County that is college educated
Percentage increase/decease in votes for each percentage point increase in college educated population
Log of Buchanan votes
Parameter a b
Value 3.45 .09
Standard Error .27 .02
T-Statistic 12.6 5.4
Results R-Square = .31
CVLN 09.45.3
The distribution for ‘b’ has a mean of .09 and a standard deviation of .02
.09
There is a 95% chance that the value for ‘b’ lies between .13 and .05
County College (%)
Predicted Votes
Actual Votes
Error
Alachua 34.6 902 262 640
Baker 5.7 55 73 -18
0
Plug in Values for College % to get vote predictions
31% of the variation in Buchanan’s votes across counties is explained by college education
VLNeV
Each percentage point increase in college educated (i.e. from 10% to 11%) raises Buchanan’s vote total by .09%
County College (%) Buchanan Votes
Log of College (%)
Alachua 34.6 262 3.54
Baker 5.7 73 1.74
How about this…
CbLNaV
Log of % of County that is college educated
Gain/ Loss in votes for each percentage increase in college educated population
# of Buchanan votes
Parameter a b
Value -424 252
Standard Error 139 54
T-Statistic -3.05 4.6
Results R-Square = .25
CLNV 252424
The distribution for ‘b’ has a mean of 252 and a standard deviation of 54
.09
There is a 95% chance that the value for ‘b’ lies between 360 and 144
County College (%)
Predicted Votes
Actual Votes
Error
Alachua 34.6 469 262 207
Baker 5.7 15 73 -58
0
Plug in Values for College % to get vote predictions
25% of the variation in Buchanan’s votes across counties is explained by college education
Each percentage increase in college educated (i.e. from 30% to 30.3%) raises Buchanan’s vote total by 252 votes
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
One More…
CbLNaVLN
Log of % of County that is college educated
Percentage gain/Loss in votes for each percentage increase in college educated population
Log of Buchanan votes
Parameter a b
Value .71 1.61
Standard Error .63 .24
T-Statistic 1.13 6.53
Results R-Square = .40
CLNVLN 61.171.
The distribution for ‘b’ has a mean of 1.61 and a standard deviation of .24
.09
There is a 95% chance that the value for ‘b’ lies between 2 and 1.13
County College (%)
Predicted Votes
Actual Votes
Error
Alachua 34.6 624 262 362
Baker 5.7 34 73 -39
0
Plug in Values for College % to get vote predictions
40% of the variation in Buchanan’s votes across counties is explained by college education
Each percentage increase in college educated (i.e. from 30% to 30.3%) raises Buchanan’s vote total by 1.61%
VLNeV
It turns out the regression with the best fit looks like this.
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
IaCaHaAaBaaPLN 54365221
Parameters to be estimated
Error termBuchanan Votes
Total Votes*100
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
Now, we can make a forecast!
ICHABPLN 0658.0193.0349.0415.0132.146.2 65
R Squared = .73
County Predicted Votes
Actual Votes
Error
Alachua 520 262 258
Baker 55 73 -18
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 Black (%)
Age 65 (%)
Hispanic (%)
College (%)
Income (000s)
Buchanan Votes
Total Votes
Palm Beach 21.8 23.6 9.8 22.1 33.5 3,407 431,621
004.2PLN
%134.004.2 eP
578621,43100134. This would be our prediction for Pat Buchanan’s vote total!
ICHABPLN 0658.0193.0349.0415.0132.146.2 65
Probability
LN(%Votes)
There is a 95% chance that the log of Buchanan’s vote percentage lies in this range
-2.004 – 2*(.2556) -2.004 + 2*(.2556)= -2.5152 = -1.4928
004.2PLN 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
% of Votes
There is a 95% chance that Buchanan’s vote percentage lies in this range
%134.004.2 eP
%08.5152.2 e %22.4928.1 e
Next, lets convert the Logs to vote percentages
Probability
Votes
There is a 95% chance that Buchanan’s total vote lies in this range
348621,4310008. 970621,4310022.
Finally, we can convert to actual votes 578621,43100134.
3,407 votes turns out to be 7 standard deviations away from our forecast!!!
We know that the quantity of some good or service demanded should be related to some basic variables
Quantity
Price
D
,..., IPDQD
Quantity Demanded Price
Income
Other “Demand Shifters”
“ Is a function of”
Time
Dem
and
Fact
ors
t t+1t-1
Cross Sectional estimation holds the time period constant and estimates the variation in demand resulting from variation in the demand factors
For example: can we estimate demand for Pepsi in South Bend by looking at selected statistics for South bend
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
Suppose that we have the following data for sales in 200 different Indiana cities
Lets begin by estimating a basic demand curve – quantity demanded is a linear function of price.
PaaQ 10
Change in quantity demanded per $ change in price (to be estimated)
Regression Results
Variable Coefficient Standard Error t Stat
Intercept 155,042 18,133 8.55
Price (X) -46,087 7214 -6.39
PQ 087,46042,155
That is, we have estimated the following equation
Regression Statistics
R Squared .17
Standard Error 48,074
Every dollar increase in price lowers sales by 46,087 units.
Values For South BendPrice of Pepsi $1.37
903,9137.1087,46042,155 Q
P
Q
$1.37
91,903
68.903,91
37.1087,46
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
As we did earlier, we can experiment with different functional forms by using logs
Adding logs changes the interpretation of the coefficients
PLNaaQ 10
Change in quantity demanded per percentage change in price (to be estimated)
Regression Results
Variable Coefficient Standard Error t Stat
Intercept 133,133 14,892 8.93
Price (X) -103,973 16,407 -6.33
PLNQ 973,103133,133
That is, we have estimated the following equation
Regression Statistics
R Squared .17
Standard Error 48,140
Every 1% increase in price lowers sales by 103,973 units.
Values For South BendPrice of Pepsi $1.37
Log of Price .31
P
Q
$1.37
100,402
31.973,103133,133 Q
402,100
1973,103
1
%
973,103%
Qp
Q
p
Q
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
As we did earlier, we can experiment with different functional forms by using logs
PaaQLN 10
Percentage change in quantity demanded per $ change in price (to be estimated)
Adding logs changes the interpretation of the coefficients
Regression Results
Variable Coefficient Standard Error t Stat
Intercept 13 .34 38.1
Price (X) -1.22 .13 -8.98
PQLN 22.113
That is, we have estimated the following equation
Regression Statistics
R Squared .28
Standard Error .90
Every $1 increase in price lowers sales by 1.22%.
Values For South BendPrice of Pepsi $1.37
P
Q
$1.37
83,283
We can now use this estimated demand curve along with price in South Bend to estimate demand in South Bend
283,83
33.1137.122.11333.11
eQ
QLN
1
37.122.1
1
%
22.1%
p
p
Q
p
Q
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
As we did earlier, we can experiment with different functional forms by using logs
PLNaaQLN 10
Percentage change in quantity demanded per percentage change in price (to be estimated)
Adding logs changes the interpretation of the coefficients
Regression Results
Variable Coefficient Standard Error t Stat
Intercept 12.3 .28 42.9
Price (X) -2.60 .31 -8.21
PLNQLN 6.212
That is, we have estimated the following equation
Regression Statistics
R Squared .25
Standard Error .93
Every 1% increase in price lowers sales by 2.6%.
Values For South BendPrice of Pepsi $1.37
Log of Price .31
P
Q
$1.37
72,402
402,72
19.1131.6.21219.11
eQ
QLN
6.2%
%
p
Q
We can add as many variables as we want in whatever combination. The goal is to look for the best fit.
cPLNaILNaPaaQLN 3210
% 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 Coefficient Standard Error t Stat
Intercept 5.98 1.29 4.63
Price -1.29 .12 -10.79
Log of Income 1.46 .34 4.29
Log of Competitor’s Price 2.00 .34 5.80
R Squared: .46
Values For South BendPrice of Pepsi $1.37
Log of Income 3.81
Log of Competitor’s Price .80
P
Q
$1.37
87,142
142,87
36.1180.00.281.346.137.129.198.536.11
eQ
QLN
Now we can make a prediction and calculate elasticities
00.2%
%
46.1%
%
76.11
37.129.1
1
%
cCP
I
P
Q
I
Q
P
P
Q
Time
Dem
and
Fact
ors
t t+1t-1
We could use a cross sectional regression to forecast quantity demanded out into the future, but it would take a lot of information!
Estimate a demand curve using data at some point in time
Use the estimated demand curve and forecasts of data to forecast quantity demanded
Time
Dem
and
Fact
ors
t t+1t-1
Time Series estimation ignores the demand factors and estimates the variation in demand over time
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
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:
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
0
5
10
15
20
25
2003-1 2004-1 2005-1 2006-1
First, let’s plot the data…what do you see?
This data seems to have a linear trend
A linear trend takes the following form:
btxxt 0
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
Estimated value for time zero
Estimated quarterly growth (in millions of kilowatt hours)
Regression Results
Variable Coefficient Standard Error t Stat
Intercept 11.9 .953 12.5
Time Trend .394 .099 4.00
Regression Statistics
R Squared .53
Standard Error 1.82
Observations 16txt 394.9.11
Lets forecast electricity usage at the mean time period (t = 8)
50.3ˆ
05.158394.9.11ˆ
t
t
xVar
x
0
5
10
15
20
25
2003-1 2004-1 2005-1 2006-1
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
T = 8
0
10
20
30
40
50
60
70
Sample
We can use this linear trend model to predict as far out as we want, but note that the error involved gets worse!
7.47ˆ
85.4176394.9.11ˆ
t
t
xVar
x
0
5
10
15
20
25
2003-1 2004-1 2005-1 2006-1
Lets take another look at the data…it seems that there is a regular pattern…
Q2
Q2Q2
Q2
There appears to be a seasonal cycle…
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
2003:4 14 13.47 1.03 13.47(1.04) = 14.04
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
Average Ratios• Q1 = .87• Q2 = 1.16• Q3 = .91• Q4 = 1.04
One seasonal adjustment process is to adjust each quarter by the average of actual to predicted
For each observation:• Calculate the ratio of
actual to predicted• Average the ratios by
quarter• Use the average ration to
adjust each predicted value
Time Period Actual Adjusted Error
2003:1 11 10.90 -0.12003:2 15 14.77 -0.232003:3 12 11.86 -0.142003:4 14 14.04 0.042004:1 12 12.30 0.32004:2 17 16.61 -0.392004:3 13 13.29 0.292004:4 16 15.68 -0.322005:1 14 13.70 -0.32005:2 18 18.45 0.452005:3 15 14.72 -0.282005:4 17 17.33 0.332006:1 15 15.10 0.12006:2 20 20.28 0.282006:3 16 16.15 0.152006:4 19 18.96 -0.04
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
n
FARMSE tt
2
Number of Observations
Sum of squared forecast errors
26.RMSE
10
11
12
13
14
15
16
17
18
19
20
2003-1 2004-1 2005-1 2006-1
Looks pretty good…
26.RMSE
0
10
20
30
40
50
60
70
52.4304.185.4176394.9.11ˆ tx
Recall our prediction for period 76 ( Year 2022 Q4)
We could also account for seasonal variation by using dummy variables
33221100 DbDbDbtbxxt
else ,0
iquarter if ,1iD
Note: we only need three quarter dummies. If the observation is from quarter 4, then
tbxx
DDD
t 00
321 0
Regression Results
Variable Coefficient Standard Error t Stat
Intercept 12.75 .226 56.38
Time Trend .375 .0168 22.2
D1 -2.375 .219 -10.83
D2 1.75 .215 8.1
D3 -2.125 .213 -9.93
Regression Statistics
R Squared .99
Standard Error .30
Observations 16
321 125.275.1375.2375.75.12 DDDtxt
Note the much better fit!!
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
26.RMSE
Ratio Method
25.RMSE
Dummy Variables
10
11
12
13
14
15
16
17
18
19
20
2003-1 2004-1 2005-1 2006-1
Dummy Ratio
A plot confirms the similarity of the methods
0
10
20
30
40
50
60
70
Recall our prediction for period 76 ( Year 2022 Q4)
25.4176375.75.12 tx
btxxt 0
Recall, our trend line took the form…
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
t
Gxx
100
%10
tt gxx 10
If we wish to estimate this equation, we have a little work to do…
Note: this growth rate is in decimal form
gtxxt 1lnlnln 0
If we convert our data to natural logs, we get the following linear relationship that can be estimated
Regression Results
Variable Coefficient Standard Error t Stat
Intercept 2.49 .063 39.6
Time Trend .026 .006 4.06
Regression Statistics
R Squared .54
Standard Error .1197
Observations 16
txt 026.49.2ln
Lets forecast electricity usage at the mean time period (t = 8)
0152.ˆ
698.28026.49.2ˆln
t
t
xVar
xBE CAREFUL….THESE NUMBERS ARE LOGS !!!
0152.ˆ
698.28026.49.2ˆln
t
t
xVar
x
The natural log of forecasted demand is 2.698. Therefore, to get the actual demand forecast, use the exponential function
85.14698.2 e
Likewise, with the error bands…a 95% confidence interval is +/- 2 SD
945.2,451.20152.2/698.2
00.19,60.11, 945.2451.2 ee
0
5
10
15
20
25
30
2003-1 2004-1 2005-1 2006-1
Again, here is a plot of our forecasts with the error bands
T = 8 70.1RMSE
0
100
200
300
400
500
600
1 13 25 37 49 61 73 85 97
8.221,8.352/
22.8949.4
SD
e
Errors is growth rates compound quickly!!
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:
btppt 0
Price at time t Price at April 1993 Number of months from April 1993
monthly growth in dollars per gallon
Regression Results
Variable Coefficient Standard Error t Stat
Intercept .67 .05 12.19
Time Trend .010 .0004 23.19
R Squared= .72
We can check for the presence of a seasonal cycle by adding seasonal dummy variables:
33221100 DbDbDbtbppt
else
iquarterifDi ,0
,1
dollars per gallon impact of quarter I relative to quarter 4
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 PriceRegression
coefficientSeasonalized
data
1993 – 04 1.05 .15 .90
1993 - 07 1.06 .16 90
1993 - 10 1.06 0 1.06
1994 - 01 .98 -.03 1.01
1994 - 04 1.00 .15 .85
2nd Quarter
3rd Quarter
4th Quarter
1st Quarter
2nd Quarter
Note: Once the seasonal component has been removed, all that should be left is trend, cycle, and noise. We could check this:
btppt 0~
Seasonalized Price Series
Regression Results
Variable Coefficient Standard Error t Stat
Intercept .587 .05 11.06
Time Trend .010 .0004 23.92
33221100~ DbDbDbtbppt
Seasonalized Price Series
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
321 16.15.03.01.58. DDDtpt
The regression we have in place gives us the trend plus the seasonal component of the data
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
Predicted
We can plot this and compare it with business cycle dates
tt pp ˆActual Price
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
33221100ln DbDbDbtbppt
else
iquarterifDi ,0
,1
Percentage price impact of quarter I relative to quarter 4
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
Monthly growth rate
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 PriceRegression
coefficientLog of Seasonalized
dataSeasonalized
Price
1993 – 04 1.05 .049 .06 -.019 .98
1993 - 07 1.06 .062 .07 -.010 .99
1993 - 10 1.06 .062 0 .062 1.06
1994 - 01 .98 -.013 -.02 .006 1.00
1994 - 04 1.00 .005 .06 -.062 .94
2nd Quarter
3rd Quarter
4th Quarter
1st Quarter
2nd Quarter
98.019. e
Example:
321 07.06.02.005.14.ln DDDtpt
The regression we have in place gives us the trend plus the seasonal component of the data
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 PricePredicted Log Price
(From regression)Predicted
PriceBusiness 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
Predicted Log of Price
93.069. e
As you can see, very similar results
tt pp ˆActual Price
Predicted Price
73.2
005.1007.106.002.217005.14.ln005.1
e
pt
90.2016.115.003.21701.58. tp
In either case, we could make a forecast for gasoline prices next year. Lets say, April 2011.
Forecasting Data
Date Time Period Quarter
April 2011 217 2
OR
By the way, the actual price in April 2011 was $3.80
Quarter Market Share
1 20
2 22
3 23
4 24
5 18
6 23
7 19
8 17
9 22
10 23
11 18
12 23
Consider a new forecasting problem. You are asked to forecast a company’s market share for the 13th quarter.
0
5
10
15
20
25
30
1 2 3 4 5 6 7 8 9 10 11 12
There doesn’t seem to be any discernable trend here…
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
N
ANMA
t
Ntt
1
0
5
10
15
20
25
30
1 2 3 4 5 6 7 8 9 10 11 12
Actual
MA(3)
MA(5)
The longer the moving average, the smoother the forecasts are…
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
Calculating forecasts is straightforward…
MA(3)
33.213
231823
MA(5)
6.205
1722231823
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 Total = 62.48
95.29
3534.78RMSE 99.2
7
48.62RMSE
Exponential smoothing involves a forecast equation that takes the following form
ttt FwwAF 11
Forecast for time t+1
Actual value at time t
Forecast for time t
Smoothing parameter
Note: when w = 1, your forecast is equal to the previous value. When w = 0, your forecast is a constant.
1,0w
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
For exponential smoothing, we need to choose a value for the weighting formula as well as an initial forecast
Usually, the initial forecast is chosen to equal the sample average
8.216.205.235.
0
5
10
15
20
25
30
1 2 3 4 5 6 7 8 9 10 11 12
Actual w=.3 w=.5
As was mentioned earlier, the smaller w will produce a smoother forecast
Calculating forecasts is straightforward…
W=.3
04.212.207.233.
W=.5
45.219.195.235.
So, how do we choose W??
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
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 Total = 101.5
70.212
19.87RMSE 91.2
12
5.101RMSE
