Question

The emissions benefit functions of ozone precursor emissions for the three major polluters in the area are given by

MEB1 = 5000 – 2 E1

MEB2 = 2500 – 2.5 E2

MEB3 = 1000 – 3 E3

where Ei is tons of emissions from firm i, and MEBi is the marginal benefit of being allowed to emit for firm i, in dollars per ton.

1. a. How many quotas will each firm buy (sell)? b. How much money will each firm spend (receive) because of the sale? c. How much will the total social cost of emissions abatement change as a result of the change to a transferable quota system?

2. If the damage function for sulfur is believed to be MED = 4000 + 4E, where MED is in dollars per ton,

   i. What is the optimal emissions tax and how much revenue will the government collect from the emissions tax?

   ii. What is the optimal emissions abatement subsidy (in dollars per ton of abatement from the unregulated level), and how large an expenditure will the government incur?

3. If the damage function for sulfur is grossly estimated to be MED = 10 + 0.1E, where MED is in dollars per ton,

   i. What is optimal emissions tax and how much revenue will the government collect from the emissions tax?

   ii. What is the optimal emissions abatement subsidy (in dollars per ton of abatement from the unregulated level), and how large an expenditure will the government incur?

4. Examine your answers to questions (1) through (3) and discuss:

   1. How the differences in MED in the two cases might affect your recommendation of whether to use a transferable allowance (quota) system versus a tax;

   2. How the differences in MED might affect your recommendation of whether to use a tax versus a subsidy.

Answer 1

x : Amount of nitrogen oxides(pollutant)

x follows normal distribution with mean = 1.6 and standard deviation =0.5

z : (x-)/ : (x-1.6) / 0.5 is a standard normal with  mean = 0 and standard deviation =1

a)

Probability that emission levels of a randomly selected car is between 1 and 2 = P(1<x<2) = P(x<2) - P(x<1)

P(x<2) :

z-score for 2 = (2-1.6) / 0.5 = 0.4/0.5 = 0.8

From standard normal table, P(z<0.8) = 0.7881

P(x<2) = P(z<0.8) = 0.7881

P(x<1)

z-score for 1= (1-1.6) / 0.5 = -0.6/0.5 = -1.2

From standard normal table, P(z<-1.2) = 0.1151

P(x<1) = P(z<-1.2) = 0.1151

P(1<x<2) = P(x<2) - P(x<1) = 0.7881 - 0.1151 = 0.673

Probability that emission levels of a randomly selected car is between 1 and 2 = P(1<x<2) = 0.673

b) P(x 2.5) = 1-P(x<2.5)

P(x<2.5)

Z-score for 2.5 = (2.5 - 1.6)/0.5 = 0.9/0.5 = 1.8

From Standard tables P(Z<1.8) = 0.9641

P(x<2.5) = P(Z<1.8) = 0.9641

P(x 2.5) = 1-P(x<2.5) = 1-0.9641= 0.0359

P(x 2.5) = 0.0359

c)

Let 'xw' be the emission level that constitutes the worst 10% of the vehicles.

Therefore ,

P(x xw) = 10/100 = 0.10

P(x xw) = 1- p(x<xw) = 0.10 ; p(x<xw) = 1-1.0 =0.90

let zw be the z-score for xw;

zw = (xw - 1.6)/0.5;  

xw = 1.6 + 0.5 xw

then p(z<zw) = 0.90

From standard normal tables, find the zw, such that p(z<zw) = 0.90

zw = 1.28

xw = 1.6 + 0.5 xw = 1.6 + 0.5 x 1.28 = 1.6 + 0.64 = 2.24

'xw' : the emission level that constitutes the worst 10% of the vehicles

The emission level that constitutes the worst 10% of the vehicles = 2.24