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					Cost and Benefit



  Avinash Kishore
  (avinash.kishore@gmail.com)

  Based on notes from Andrew Foss

  Economics 1661 / API-135

  February 11, 2011
  Review Section
Agenda


   Basic Econometrics
     – Bi-variate regression
     – Multivariate linear regression
     – Special cases

   Elasticity
   Social Welfare Cost
   Benefits
     – Aggregation of Demand Curves
     – Travelling Cost Method


                               Note: A good general reference on costs and
                               benefits is EPA, Guidelines for Preparing Economic
                               Analyses, September 2000 (link), Chapters 7 and 8    1
Measures of Central Tendency and Dispersion


   Measures of Central Tendency: Mean
     – Mean = ∑Xi /n
     – Mean takes all data points into account
     – Mean is sensitive to outliers. Outliers have a lot of weight on
       mean

   Measures of Dispersion
     – Variance = (Xi – mean)2/N
     – Standard Deviation =[(Xi – mu)2/N]1/2 = (Variance)1/2

   More dispersed data have higher variance
     – Like mean, standard deviation is also sensitive to outliers


                                                                         2
It is possible that two datasets have the same mean
   but different standard deviations




                                 From Wikipedia
                                                      3
OLS regression (2 Variables)       http://www.ats.ucla.edu/stat/Stata/examples/rwg/rwgstata2/rwgstata2.htm




                Y (wateruse) = 1201.124 + 47.54*income
                                                                                                             4
Bi-variate Regression Model


   Yi = β0 + β1Xi + εi
       – i = each observation; Y = Dependent Variable (water use); X =
         Independent Variable (income); εi = Error term
       – β0 = intercept. It tells us the predicted value of Y when X = 0.
       – β1 = The coefficient that tells us how Y changes for unit
         change in X.
     water81       Coef.   Std. Err.     t     P>t     [95% Conf. Interval]


     income    47.54869    4.652286    10.22   0.000   38.40798   56.6894
     _cons     1201.124    123.3245    9.74    0.000   958.8191   1443.43

       – N = 496; R-square = 0.1745

   3 Sources of Error Term:
       – Omitted variables, Measurement error and Chance or
         Randomness                                                            5
Multiple Regression


   More than one independent variables
      – Yi = β0 + β1Xi1 + β2Xi2 + β3Xi3 + εi

   Now, β1 is the change in value of Y for a unit change
    in X1 while holding constant (or controlling for) X2 and
    X3 (the marginal interpretation)
   Example:. reg water81 income educat peop81
         water81   Coef.         Std. Err.           t   P>t     [95% Conf.   Interval]


         income    32.05796    4.285087       7.48       0.000   23.63864     40.47729
         educat    -42.61099    17.23512      -2.47      0.014   -76.4745     -8.747484
         peop81    480.5194    31.73071      15.14       0.000   418.175      542.8639
         _cons     678.8889     245.195       2.77       0.006   197.1305     1160.647



  N = 496, R-squared = 0.38

                                                                                           6
The Coefficient and the Confidence Interval                                                                         (precision)



                                                    Student-t Distribution
                         0.1                         μ = 32.06, σ = 4.28
                        0.09
  Probability Density




                        0.08


                        0.07


                        0.06


                        0.05


                        0.04


                        0.03


                        0.02


                        0.01


                          0
                          14.88   18.304   21.728    25.152   28.576   32   35.424   38.848   42.272   45.696   49.12




                                                                                                                             7
Some Special Cases: binary independent variables


     Normally, continuous dependent and independent variables in OLS.
     But we can also have binary independents. Also Dummy Variables.
       – Βeta coefficient of a dummy variable is not interpreted as its slope.
     Let there be a dummy variable Female s.t. Female = 1 for female
      employees; 0 for male employees.
     Regression equation: Earnings = β0 + β1*Female + εi
       – Earnings = 16.99 – 3.45*Female + εi
     Interpretation: Constant, β0 = Mean earning of men = 16.99
  β1 = Difference in mean earnings of men and women. Here, β1 = -3.45. So,
      mean earning of women = 16.99 – 3.45 = 13.44.
     Multivariate Example:
       –   Earnings = β0 + β1 Years of Education + β2Female + εi

       – Earnings = 15.55 + 3.00* Years of Education –3.55*Female

   Show graphically                                                             8
Special Cases: Interaction Terms


   It is possible that both intercepts and slopes (returns per year of
    education in this case) are different for two groups.
   e.g. Earnning = β0 + β1Years of Education + β2Female + β3
    Female*Years of Education + εi
   So, for females: Earnings = β0 + β1Years of Education + β2*1 +
    β3*1*Years of Education+ εi
   = (β0 + β2) + (β1+ β3)*Years of Education+ εi
   for males: Earnings = β0 + β1Years of Education + β2*0 +
    β3*0* Years of Education + εi
                        = β0 + β1Years of Education + εi
   So, for females, Slope = (β1+ β3) and Intercept = β0 + β2
   For males, Slope = β1 and intercept = β0
                                                                          9
We can also capture non-linear relations through
linear regression

   Quadratic: If the dependent variable is a parabolic
    function of the independent, we can still capture the
    relationship by adding a squared term, e.g.
    Environmental Kuznets Curve hypothesis:
     – Pollution = β0 + β1GDPPC + β2 (GDPPC)2 + εi

   Categorical Dependent Variable (as in RUM)
     – We get estimates of probability
     – OLS is not appropriate in such cases
     – Logit or probit models




                                                            10
Cost and Benefits
Costs Estimation Methods and Elasticity


   Direct Compliance Cost Method
   Partial Equilibrium Analysis (behavioral response)
   General Equilibrium Analysis
   Want to know how consumers and firms will react to
    changes in prices for the good/service being regulated
   Depends on price elasticity of supply and demand
   Elasticity = % ΔQ/% ΔP = (ΔQ/Q)/ (ΔP/P) =
    (ΔQ/ΔP)*(P/Q) = (1/Slope)*(P/Q)
   Higher the elasticity, greater the behavioral response
    to regulation and greater the social welfare cost.
                                                             12
If demand is very responsive to change in price,
good is price elastic. If demand does not respond
much to price changes, demand is price inelastic




   Elasticity =   ∞               Elasticity = 0




                                                    13
Social Welfare Costs


     Illustration of social welfare costs
        – Losses in consumer and producer surplus from increased
          marginal cost across regulated industry


  Welfare Effects of Industry-Wide Increase in Marginal Cost
Price                                Price


              D                                    D           MC1 = S1
                                             CS1
        CS0             MC0 = S0                               MC0 = S0
                                       P1
  P0                                   P0    PS1
        PS0



                  Q0     Quantity                      Q1 Q0    Quantity   14
      Benefits: Aggregation of Demand Curves (Private Goods)
          Let’s say Q is a private good. Person 1 demand for Q is: Q1 = 100 – P
          Person 2 demand for Q is : Q2 = 100 – P
          So, what is the aggregate market demand for Q?
120                                                              120


100                                                              100


 80                                                              80


 60                                                              60


 40                                                              40


 20                                                              20


  0                                                                0
      0      20     40      60       80        100         120         0   20    40     60      80      100    120
                          120
                                                                                        Algebraically, it is
                          100
                                                                                        QT = 200 – 2P
                           80


                           60                                                           Note: The algebra and
                                                                                        the graph won’t be this
                           40
                                                                                        simple if demand
                           20
                                                                                        functions are different

                            0
                                 0        50         100          150      200    250                          15
 Benefits: Aggregation of Demand Curves (Public Goods)
          Now assume Q is a public good (non-rivalrous and non-excludable)

          Does that change how we aggregate demand curves?
120                                              120

100                                              100

 80                                               80

 60                                               60

 40                                               40

 20                                               20

  0                                                0
      0    20   40   60         80   100   120         0        20     40    60     80     100     120
                      250


                      200
                                                                                  Algebraically, it is
                      150
                                                                                  P = 200 – 2Q
                      100
                                                                                  You work with inverse
                       50                                                         demand curve

                        0
                            0        20     40   60        80        100    120                           16
One more example: public park provision


   There are 500 people in a locality
   Each has demand function: Q = 100 – P
     – P = $ price per unit area of park people are willing to pay for Q
       sq. yards of the park preserved
     – MC = $ 10,000/sq. yard

   How many sq yard of park area should be preserved?
     – P = 100 – Q
     – Aggregate demand: P =500[100 – Q]
     – or P = 50,000 – 500Q

   50,000 – 500Q = 10,000, Q* = 80 sq. yards
   What is the total benefit @ Q*? Net benefit = ?
                                                                           17
Travel Cost Method:
Example Problem

     Ruritania is a country with three cities and a beautiful
      park at its center
     Environmental economists in Ruritania have collected
      the following data on park visitors from the three cities
Origin        Population   Visitors                  Travel Cost
1             100,000      15,000                    $12.50
2             750,000      75,000                    $25.00
3             1,000,000    50,000                    $37.50

     The environmental economists want to estimate the
      recreational value (i.e., non-market use value) of the
      park to the people of Ruritania
                                                                   18
Travel Cost Method:
Example Problem

     First calculate the visitation rate for each origin
Origin         Population   Visitors     Visit Rate    Travel Cost
1              100,000      15,000       0.15          $12.50
2              750,000      75,000       0.10          $25.00
3              1,000,000    50,000       0.05          $37.50




                                                                     19
    Travel Cost Method:
    Example Problem

                     Next plot the relationship between visitation rate and
                         travel cost and express it in an equation



                   $50
                                      Origin 3
                   $40
Travel Cost (TC)




                                                                                TC = - (50 / 0.20) * R + 50
                                                    Origin 2
                   $30
                                                                                     = -250 * R + 50
                   $20
                                                                   Origin 1     or
                   $10
                                                                                R = - (0.20 / 50) * TC + 0.2
                    $0
                                                                                     = -0.004 * TC + 0.2
                      0.00     0.05          0.10           0.15         0.20
                                      Visitation Rate (R)
                                                                                                              20
    Travel Cost Method:
    Example Problem

                     Suppose a fee were charged to enter the park
                     Calculate the relationship between Origin 1’s visitation
                      rate and the hypothetical fee

                   $40
Hypothetical Fee




                   $30                                                  R = -0.004 * (TC + Fee) + 0.2

                   $20                                                  TC1 = 12.5

                   $10                                                  R1 = -0.004 * (12.5 + Fee) + 0.2

                    $0                                                       = -0.004 * Fee + 0.15
                      0.00   0.05          0.10           0.15   0.20
                                                                        or
                                    Visitation Rate (R)
                                                                        Fee = -250 * R1 + 37.5
                                        Origin 1                                                        21
    Travel Cost Method:
    Example Problem

                     Calculate the relationship between Origin 2’s visitation
                         rate and the hypothetical fee



                   $40
Hypothetical Fee




                   $30                                                    R = -0.004 * (TC + Fee) + 0.2

                   $20                                                    TC2 = 25

                   $10                                                    R2 = -0.004 * (25 + Fee) + 0.2

                    $0                                                         = -0.004 * Fee + 0.10
                      0.00     0.05          0.10           0.15   0.20
                                                                          or
                                      Visitation Rate (R)
                                                                          Fee = -250 * R2 + 25
                                          Origin 2                                                         22
    Travel Cost Method:
    Example Problem

                     Calculate the relationship between Origin 3’s visitation
                         rate and the hypothetical fee



                   $40
Hypothetical Fee




                   $30                                                    R = -0.004 * (TC + Fee) + 0.2

                   $20                                                    TC3 = 37.5

                   $10                                                    R3 = -0.004 * (37.5 + Fee) + 0.2

                    $0                                                         = -0.004 * Fee + 0.05
                      0.00     0.05          0.10           0.15   0.20
                                                                          or
                                      Visitation Rate (R)
                                                                          Fee = -250 * R3 + 12.5
                                          Origin 3                                                        23
    Travel Cost Method:
    Example Problem

                     Calculate the relationship between all three origins’
                         visitation rate and the hypothetical fee



                   $40
Hypothetical Fee




                   $30                                                      R = -0.004 * (TC + Fee) + 0.2

                   $20                                                      R1 = -0.004 * Fee + 0.15

                   $10                                                      R2 = -0.004 * Fee + 0.10

                    $0                                                      R3 = -0.004 * Fee + 0.05
                      0.00       0.05          0.10           0.15   0.20
                                        Visitation Rate (R)

                             Origin 1       Origin 2      Origin 3                                          24
                   Travel Cost Method:
                   Example Problem

                          Calculate the relationship between the number of
                             visitors from Origin 1 and the hypothetical fee

                   $40
                                                                            Per capita demand function:
Hypothetical Fee




                   $30                                                      R1 = -0.004 * Fee + 0.15

                   $20                                                      Population = 100,000

                   $10                                                      So, demand function is:
                                                                            = 100,000*(-0.004 * Fee + 0.15)
                    $0
                         0     30,000   60,000   90,000   120,000 150,000   = -400 * Fee + 15,000
                                          Visitors (Q)

                                           Origin 1


                                                                                                              25
    Travel Cost Method:
    Example Problem

                     Calculate the relationship between the number of visitors
                         from Origin 2 and the hypothetical fee


                                                                          Per capita demand function:
                   $40                                                    R1 = -0.004 * Fee + 0.10
Hypothetical Fee




                   $30                                                    Population = 75,000
                   $20
                                                                          So, total demand function is:
                   $10
                                                                          = 75,000*(-0.004 * Fee + 0.10)

                    $0                                                    Q2 = -3,000 * Fee + 75,000
                         0   30,000   60,000   90,000   120,000 150,000
                                        Visitors (Q)

                                         Origin 2                                                       26
                   Travel Cost Method:
                   Example Problem

                          Calculate the relationship between the number of
                             visitors from Origin 3 and the hypothetical fee


                   $40
                                                                            Per capita demand function:
Hypothetical Fee




                   $30                                                      R1 = -0.004 * Fee + 0.05
                   $20
                                                                            Population = 50,000
                   $10
                                                                            So, total demand is:
                    $0                                                      = 50,000*(-0.004 * Fee + 0.05)
                         0     30,000   60,000   90,000   120,000 150,000   Q3 = -4,000 * Fee + 50,000
                                          Visitors (Q)

                                           Origin 3
                                                                                                          27
      Travel Cost Method:
      Example Problem
                             – Aggregate curve (from horizontal summation) represents total
                               recreational demand as a function of hypothetic park fee
                             – 1st kink @ Fee = $25, Qagg = 5000
                             – 2nd kink @ Fee = $ 12.50, Qagg = 10,000 + 37,500 = 47,500
                             – X-intercept = 15,000 + 75,000 + 50,000 = 140,000

                   $40
                             –
Hypothetical Fee




                                                                                Q1 = -400 * Fee + 15,000
                   $30
                                                                                Q2 = -3,000 * Fee + 75,000
                   $20
                                                                                Q3 = -4,000 * Fee + 50,000
                   $10

                    $0
                         0       30,000   60,000   90,000     120,000 150,000
                                            Visitors (Q)

                     Origin 1        Origin 2      Origin 3      Aggregate                                   28
    Travel Cost Method:
    Example Problem

                     Calculate recreational value as area under the
                         aggregate curve
                             – The recreational value of the park is $1,531,250


                   $40
Hypothetical Fee




                             Area A
                   $30                                                        Area A = ½ * (37.5 - 25) * 5,000
                                 Area B                                              = 31,250
                   $20
                                             Area C                           Area B = ½ * (25 - 12.5) *
                   $10                                                                 (5,000 + 47,500)
                                                                                     = 328,125
                    $0
                         0      30,000    60,000   90,000   120,000 150,000   Area C = ½ * (12.5 - 0) *
                                                                                       (47,500 + 140,000)
                                            Visitors (Q)                             = 1,171,875

                                            Aggregate                         Total Area = 1,531,250         29
    Travel Cost Method:
    Example Problem

                     Calculate recreational value as area under each of the
                         origin-specific demand curves (alternative method)
                             – The recreational value of the park is $1,531,250


                   $40
Hypothetical Fee




                   $30                                                          Area 1 = ½ * 37.5 * 15,000
                                 Area 1                                                = 281,250
                   $20
                                     Area 2
                                                                                Area 2 = ½ * 25 * 75,000
                                        Area 3
                   $10                                                                 = 937,500

                    $0                                                          Area 3 = ½ * 12.5 * 50,000
                         0      30,000     60,000   90,000    120,000 150,000          = 312,500

                                             Visitors (Q)                       Total Area = 1,531,250

                                Origin 1      Origin 2       Origin 3                                        30
Travel Cost Method:
Example Problem

   Calculate the number of visitors and consumer surplus
    if no fee were charged to enter the park
     – Total park visitation is 15,000 + 75,000 + 50,000 = 140,000
     – CS is equal to recreational value: $1,531,250




                                                                     31
    Travel Cost Method:
    Example Problem

                     Calculate the number of visitors and consumer surplus
                         if a fee of $20 were charged to enter the park
                             – Total park visitation is 7,000 (Or. 1) + 15,000 (Or. 2) = 22,000
                             – CS is area below demand curve and above price: $98,750
                   $40
Hypothetical Fee




                             Area A
                   $30                                                       Area A = ½ * (37.5 - 25) * 5,000
                                Area D                                              = 31,250
                                             Fee = $20
                   $20
                                                                             Area D = ½ * (25 - 20) *
                   $10                                                                (5,000 + 22,000)
                                                                                    = 67,500
                    $0
                         0      30,000   60,000   90,000   120,000 150,000   Total Area = 98,750

                                           Visitors (Q)

                                           Aggregate                                                        32
Benefits: Mid-term 2005
     The City of Miami, Florida proposes to invest in a new water reservoir for its public water
      system, and estimates its cost. To justify this substantial expenditure of public funds, the
      Mayor explains that if the new reservoir is not constructed then the next most costly way to
      increase Miami’s water supply will be to invest in a desalinization plant, which will be even
      more expensive. Hence, the Mayor explains, the social benefits of building the new reservoir
      clearly exceed its social costs. There are no environmental or other externalities involved with
      either alternative. How would you assess this reasoning from an economic perspective?




       – This is the “avoided cost” method of evaluating benefits.
         It is incorrect because it ignores demand for the public good
         (water), i.e., water’s real benefits to the society.


                Avoided                          Willingness                        Willingness
                 Costs                             to Pay                            to Accept



                                                                                                         33
To Be Continued…


   Next time: More on benefit estimation methods




                                                    34

				
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