1. Seminar Exercise 3 ECN 4910

2. Exercise 1 a) For the problem to make sense a > d. Social welfare is maximised by solving the following problem: max x  x (a – by)-(d + gy)dy = max x (a – d) x – ½(b + g)x 2  foc: (a – d) – (b + g)x = 0  x = (a – d)/(b + g)

3. Graphic Solution Make the area under a-bx minus the area under d-gx as large as possible a d a/ba/b (a – d)/(b + g) Gain from regulation

4. b) Coase theorem Important point. A set of property rights must be defined. Some point on the line [0, a/b] defines how much can be polluted. Let this point be x*. Then for every x* a net surplus is generated by moving from x* to (a – d)/(b + g). The gross gain to the winner is always larger than the gross loss to the loser so the winner can compensate the loser.

5. Graphic Solution a d (a – d)/(b + g) x* Net surplus always positive Loss to polluter

6. c) Pigouvian taxes/Subsidies The basic idea of this exercise is that the total compensation should be generated by taxes or subsidies. The polluter pays a tax or receives a subsidy. The total of this sum is used as compensation for participation

7. Graphic Solution with taxes The total tax is not enough to compensate the polluter a d (a – d)/(b + g) x* Net surplus always positive Loss to polluter +

8. Graphic Solution with Subsidy The total tax is not enough to compensate the polluter a d (a – d)/(b + g) x* Net surplus always positive Loss to polluter from subsidy Net gain to polluter

9. Exercise 2 First problem. Standard theory of the firm. Solve: Max x,y (px ⅓ y ⅓ – wx – ry) Solution: x = p 3 /(27rw 2 ), y=p 3 /(27r 2 w) These are the unconstrained levels. If y is required to be constrained at some level where y > p 3 /(27r 2 w). The marginal benefit of y is obviously zero. Inthe following it assumed that y is constrained below this level.

10. The effect of constraining y Two approaches. Direct approach. Solve Max x (px ⅓ y ⅓ – wx – ry) Find Solution: x* = Insert into px ⅓ y ⅓ – wx – ry and take the derivative with respect to y. Awful math!

11. Now recall the envelope theorem Let x be a vector and y be a parameter Consider the problem Max x (F(x,a)) subject to G(x,a) ≤ 0. Optimal solution may be written x*(a) The derivative dF(x*(a),a)/da = ∂L/∂a where L is the Lagrangian. THIS STUFF IS IMPORTANT. YOU CAN READ THE MATH IN ESSENTIAL MATHEMATICS FOR ECONOMISTS, K SYDSÆTER, section 14.2

12. The shadow price approach Solve Max x,Y (px ⅓ Y ⅓ – wx – rY) s.t. Y ≤ y Form Lagrangian: L= (px ⅓ Y ⅓ – wx – rY) – λ(Y – y) F.o.c: ∂L/∂x = ⅓px -⅔ Y ⅓ – w = 0 ∂L/∂Y = ⅓px ⅓ Y -⅔ – r – λ = 0 As y is now at or below the unconstrained level, the constraint is binding. Therefore y = Y and λ ≥ 0.

13. The math is still pretty bad Solution Y = y. x = and, tada... λ = λ(y) = By the envelope theorem the marginal benefit of y is given by λ.

14. Properties of λ(y) lim y→0 λ(y) = ∞ λ(y) = 0 → y ≥ y=p 3 /(27r 2 w) (Why?)

15. Total benefit Again: Two approaches Direct approach. We already have x* = Benefit from y is found by inserting x* into px ⅓ Y ⅓ – wx – rY Show-offs may find it by integrating λ(Y) from 0 to y