Emissions Fees and marketable Permits. Kolstad - Chapter 9

Transcription

1 Kolstad - Chapter 9

2 Introduction Problems when using incentives to control pollution: 1 Space 2 Time 3 Imperfect Competition Space analyze two terms: 1 Source 2 Receptor

3 Space concentration of pollution at any receptor j: p j = f j (e 1, e 2,..., e I ) + Bj p j = a ij e i + Bj i De nition: Suppose a change in emissions from source i (4e i ) results in a change in pollution at receptor j (4p j ). The transfer coe cient between the source i and the receptor j is de ned as the ratio of the change in pollution at j to the change in emissions at i: a ij = 4p j 4e i

4 How much pollution do we want? E cient Pollution (equating marginal damage (Ambient Pollution) with marginal savings (Emissions)): MDE (e i ) = fd(p + 4p) D(p)g/4e i MD(p) 4p 4e i = MD(p)a i MC = MD(p)a i 8i = 1,..., I MC n (e n ) a n = MD(p) MC/a is the marginal cost per unit of ambient pollution This is the equimarginal principle modi ed for ambient pollution.

5 Emission Fees MC i (e i ) = t i 8i = 1,..., I Therefore from above: t n = MD(p)an 8 rm n All rms face the same emission fee per unit of ambient pollution. It must be multiplied by the transfer coe cient Ine ciencies of uniform emission fee:

6 Marketable Ambient Permits De nition: An ambient pollution permit for a receptor j gives the holder the right to emit at any location, provided the incremental pollution at receptor j does not exceed the permitted amount. Example: Two Firms EPA issues L 1 ambient permits to rm 1 and L 2 permits to rm 2. L = L 1 + L 2 L 1 + L 2 = l 1 + l 2 (where l 1 and l 2 is the number of permits after trade) a 1 e 1 = l 1 a 2 e 2 = l 2 Assume π is the price of the permits (unknown)

7 Marketable Ambient Permits Example: FOC : TC 1 (e 1 ) = C 1 (e 1 ) + π(l 1 L 1 ) = C 1 (e 1 ) + π(a 1 e 1 L 1 ) TC 2 (e 2 ) = C 2 (e 2 ) + π(l 2 L 2 ) = C 2 (e 2 ) + π(a 2 e 2 L 2 ) MTC 1 (e 1 ) = MC 1 (e 1 ) + πa 1 = 0 MTC 2 (e 2 ) = MC 2 (e 2 ) + πa 2 = 0 MC 1 (e 1 ) a 1 = MC 2(e 2 ) a 2 = π or MS 1 (e 1 ) a 1 = MS 2(e 2 ) a 2 = π (A) Marginal Savings = Permit Price! We can obtain values of e 1 and e 2 using (A) and L = a 1 e 1 + a 2 e 2

8 Marketable Ambient Permits How much will be emitted? Many permits a rm may hold, if the price of permits is higher than the marginal savings from emitting (Sell) If the permit price is lower than the rm s marginal savings from polluting, buying permits is easier than controlling emissions (buy)

9 Zonal Instruments Di erent zones have di erent fees (More Flexible) Time: Stock Pollutant (Pollutants that accumulate over time) De nition: Assume a pollutant accumulates in the env. according to the process: s(t) = δs(t 1) + e(t) where δ is he persistent rate of the pollutant and s the stock of Assume that pollution damage depends only on s(t). Then if δ > 0 the pollutant is a stock pollutant. If δ = 0, the pollutant is a ow pollutant. (1 δ) is the fraction of stock that is cleansed out of the env. in one period.

10 Zonal Instruments There is no sharp line between ow and stock pollutants. E ciency is more complex. The Net Cost of pollution emissions e t : NC = dnc de 1 = t=1 t=1 β t 1 fc t (e t ) + D t (s t ) β t 1 fc 0 t (e t ) + dd t(s t ) ds t ds t de 1 g = 0

11 Zonal Instruments Note that s t = e t + δe t 1 + δ 2 e t δ i e t i + δ t s o n ds t = δ t 1 de 1 dnc = MC 1 (e 1 ) + de 1 MS 1 (e 1 ) = t=1 t=1 β t 1 δ t 1 MD t (s t ) β t 1 δ t 1 MD t (s t ) = 0 Marginal Savings from emitting a unit of pollution today equal to the sum of all mg. damages that may occur in the future. The discount factor diminishes future marginal damages

12 Implementing Marketable Permits Initial Permit Issuance: For the purpose of e ciency does not matter which rm initially receive marketable permits Dominant Firm: If there is a single large rm (or potentially several large rms), it can make a di erence how permits are initially allocated: TC (E ) = C (E ) P(L E )[L B E ] dtc (E ) de = C 0 (E ) + P 0 (L E )[L B E ] + P(L E ) dtc (E ) de = MC (E ) + MP(E )[L B E ] + P(E ) = 0 MS(E ) = MP(E )[L B E ] + P(E ) ) E E = MS = P(E ) Therefore L B = E ) E = E L B < E ) L B < E < E L B > E ) L B > E > E