Chapter 11 Audits, enforcement and moral hazard

Last lecture was about adverse selection Polluters could be of different types and the type was private information The source of the regulators problem was to regulate without knowing the type. This is only one of a large number of problems with enforcing environmental regulation Today’s topic is ”Hidden actions”

The problem with democracies and civilisation The economically optimal punishment for any crime is to hang people with probability zero. The punishment should be so severe that no- one wants to commit crimes. Alas, even the harshest punishment regimes do not deter all criminals. People commit crimes by mistake. In courts severe punishment may be counter- productive as judges and juries are human and prefer to let people go rather than impose extreme punishments.

The cost of hidden actions. A regulator can impose a standard but cannot observe emission levels. (I.e. can not tax or directly regulate emissions.) The cost to the firm of technology a and emission level f is given by C(a,f). Emissions are given by e(a, f). Damages are given by D(e). The firm wants to minimise C(a,f). The regulator wants to minimise D(e) + C(a,f). (Maximise respectively -C(a,f) and –(D(e) + C(a,f))

The optimal solution First order conditions aa –D’e’ a – C’ a = 0 ff –D’e’ f – C’ f = 0 Gives the solution that the regulator would choose if she could observe f.

The game Regulator choose a. Then the firm choose f. Stackelberg game solved by backwards induction. We first find the firm’s optimal choice of f as a function of a, and then find the regulator’s optimal a

The firms problem Maximising -C(a,f) implies C’ f (a, f) = 0. Gives firm’s f as a function h(a). The derivative of h depends on technology (The derivative C’’ fa (a, f) ) f C’ f (a, f)

The choice of a The regulator wishes to maximise –D(e(a,h(a)) – C(a, h(a)) First order condition is: aa = ( ff –D’e’ a – C’ a = (D’e’ f + C’ f )h’(a) ff > 0. D’e’ f + C’ f > 0. Therefore aa > 0. Too little a and vice versa. Therefore h’(a) > 0  –D’e’ a – C’ a > 0. Too little a and vice versa. D(e(a,f)) + C(a,f)

What to do? Create incentive structures These are very problem specific. We will og through a couple of examples

Unobservable emissions Two polluters emit pollution to a single receptor Only the total level of pollution at the receptor is observable Cost of pollution reduction is C i (e i ) i = 1,2. Cost of pollution is D(p) = D(a 1 e 1 + a 2 e 2 ). Optimal level of pollution is given by: –dC i (e i )/de i = a i D’(p*). i = 1,2.

Incentive compatible scheme We can’t impose tax on emissions as we can not monitor them. Let us try taxing pollution. Tax paid by polluter i is given by: T i = t(p – p*) Polluter i then faces the cost of pollution TC(e i ) = C i (e i ) + t(p – p*) = C i (e i ) + t(a 1 e 1 + a 2 e 2 – p*)

The outcome for polluters F.o.c for polluters: dC i (e i )/de i + ta i = 0 Optimal tax is then D’(p*) because that implies that dC i (e i )/de i =– D’(p*)a i which is the optimality condition Without observing actions we still can impose optimal taxes! Problematic approach if there are more than a few polluters

Midnight dumping Making polluters dump their waste in the right place. It is cheap to dump it in the mayor’s garden. It is expensive to have it safely disposed of. Waste is only observed when it is created and if it is safely disposed If waste is dumped illegally TC(w) = C i (w). If safely disposed TC(w) = C s (w) – sw. s is a waste disposal subsidy. Problem: With this scheme tha polluter has an incentive to create w. It becomes a waste factory.

Using both taxes and subsidies If waste is dumped illegally TC(w) = C i (w). If safely disposed TC(w) = C s (w) – sw + tw. t is a waste production tax. t and s must solve two problems: –Make waste production efficient given that the polluter disposes w safely –Ensure that the polluter actually deposits all waste safely.

This gives the regulator two constraints Make waste production efficient given that the polluter disposes w safely F.o.c : dC s (w s )/dw s – s + t = 0. Ensure that the polluter actually deposits all waste safely. C s (w) – sw + tw.  C i (w unregulated )

The regulators problem min s,t (C s (w) + D s (w) +  (sw - tw)) subject to the constraints.  is here the social cost of taxation. s must be taken from somewhere. t can be returned. A problem with two constraints and two variables. In general, both constraints will be binding so there really is not much to optimize.

Enforcement Here a polluter choose a pollution level. The emissions may not be observed, but can be revealed through an audit. The firm faces penalties for increasing e above some level s*. F(e) = 0 for e  s* F(e) = πf(e – s*) + D for e > s* Total cost of emission reductions T(e) = C(e) + F(e)

Two cases First case: relatively high D, f and π

Two cases Second case: Relatively low D, f and π

Auditing with more detail Assume there is a probability p of an accident where emissions are given by . The regulator wants the firm to produce something which requires emissions, but not too much emissions. The regulator does not know whether emissions are caused by accident or by non-compliance

A model With probability (1 – p) emissions are e, damages are D(e) and costs are C(e). With probability p, emissions are , damages are D(  ) and costs are C(e). Optimal emissions are defined by: min e (p(C(e)+D(e)) + (1– p)(C(e)+D(  ))) Foc: C’(e) = pD’(e)  optimal e = s*

Incentive structure The firm minimise: H(e) = p(F(e)+ C(e)) + (1 – p)(F(  )+ C(e)) if they comply. Incentive compatability requires H(s*) > H(  ) Participation constraint requires H(s*) > H(0) (Only one of these constraints will be binding) Efficency requires s* = argmin e H(e)

Regulators problem Given that e should be s*, the regulator chooses to minimise the cost of auditing. That is πk. By choosing large D and f, π can be chosen as small as one wants, that is some ε above 0.