Conclusion
In this paper, we first derive the condition under which it is cost-effective for a regulator to induce perfect compliance in an emissions control program. This condition depends on the cost of monitoring and sanctioning firms, as well as on the structure of the penalty for violations. It is not instrument-dependent. If the condition is met, the regulator has to induce perfect compliance independently of whether it is implementing emission standards or transferable permits. Because we assume that the regulator's monitoring and sanctioning costs are firm-specific, the condition itself is firm-specific. In other words, it is possible that cost-effectiveness calls the regulator to induce some firms to comply with the legislation while at the same time letting others violate it. This cannot happen when one assumes that the regulator's monitoring and sanctioning costs are the same for all firms. In this case, the regulator either has to induce compliance on all firms or allow all firms to violate.
Second, we characterize the total-cost-minimizing design of a program that caps aggregate emissions of a given pollutant from a set of heterogeneous firms based on emissions standards when it is cost-effective to induce perfect compliance and when it is not. We then allow the regulator to choose the optimality of inducing compliance or not. Doing this, we find that the total cost-effective design of such a program is one in which standards are firm-specific and perfectly enforced.
Third, we compare the costs of such an optimally designed program with that of an optimally designed program based on a perfectly competitive emission permits market, which also calls for perfect enforcement according to Stranlund (2007). This comparison allows us to conclude that the total costs of the latter are always larger than the costs of the former, except when the regulator's cost of monitoring a firm's emissions are the same for all firms or the marginal penalty for violations is constant. Moreover, when it is cost-effective to allow violations, tradable permits minimize costs only under even more special conditions.
In deriving the above results, we assume that the regulator has perfect information on the firms' abatement costs. This assumption is, of course, unrealistic. For this reason, we also derive the condition under which it is expected-cost-effective for a regulator to induce perfect compliance in an emissions control program based on emission standards when it has imperfect information on abatement costs. This condition is different from the corresponding condition under complete information, depending on covariance and expectation terms that capture the fact that the regulator is uncertain with respect to the firms' reactions to the different pairs of emission standards and monitoring probability. Quite differently, in the case of tradable permits, the regulator could surmount the informational problem (and therefore the uncertainty of whether it has to induce perfect compliance or not in this program) using a constant marginal penalty tied to the observed price of the permits. On the contrary, the regulator cannot surmount the informational problem in the case of tradable permits if it uses an increasing marginal penalty. The policy recommendation that emerges from these results is clear: when capping emissions from a set of sources whose abatement costs are not perfectly known, environmental regulators should use tradable permits and perfectly enforce them with a constant marginal penalty tied to the permit price if they want to minimize the total cost of the program.
Appendix
Proof of Proposition 1 If    from (FOC 4) we know that    Because we have also that    we can re-write the first-order conditions (FOC 1) and (FOC 2) of the regulator's problem as:

Re-arranging the expressions and dividing:

From the firm's optimal choice of emissions, we know that:

From where,

  and

Because a cost-minimizing regulator that wants to achieve    will set    such that    in order not to waste monitoring resources, we can write:

or

From where, using    

Dividing both sides of equation (cond 1) by    , we obtain   for all    We have proven that when a cost-minimizing regulator induces perfect compliance, this condition is met. The reverse is also true. Assume to the contrary that    holds but    If    we know from (FOC 4) that    and 

From these and the firm's optimal choice of emissions:

After substituting for the functional form of    operating, and rearranging, we can write:

which is a contradiction. Hence, when    is met, it is cost-effective for the regulator to allow firm    to comply with the emission standard, Q.E.D.
Proof of Proposition 3 In order to prove Proposition 3, we need first to answer a previous question: what is the cost-minimizing structure of the fine when it is optimum to induce compliance and when it is not?
If the optimal policy is going to induce compliance for all   , condition (con1') requires that    for all    We also know from Section 3 that, in this case, the characterization of the cost-effective design of a program based on standards calls    From this, we can conclude that the regulator must choose as high a linear component    of the fine structure as possible because this will decrease the optimum level of the inspection probability   and thereby the monitoring costs. Conceptually, this calls for    because this will make the monitoring costs equal to zero, but in the real world, there may be limits to the upper value of   . If we call    the highest possible value of    any value of      will still make it cost-effective to induce compliance for every firm and will not have an effect on the minimum costs of the program, namely   
Therefore, if the optimal policy induces compliance for all   , the cost-minimizing shape of the fine must be such that the linear component    is set as high as possible. The value of the progressive component    is irrelevant in equilibrium as long as    where    is the chosen level of   .
If the regulator is going to allow non-compliance, how does it have to choose    and    in order to minimize the costs of a program that produces    In other words, can the regulator decrease the costs of the program by altering the fine structure (the value of    and    once the optimal standards, inspections, probabilities, and emissions have been chosen? Notice that to choose the appropriate fine structure, the regulator should optimize the values of    and  ,  keeping violations and fines constant. If    changing    and    so as to keep    constant requires    But with    firms, it is impossible to move    and    such that    for all    Keeping    constant for all    requires firm-specific fine parameters. We assume that this is the case, and we then show that the optimal design of the program calls for a uniform fine structure.
If the fine structure is firm-specific, we have    and    for each    Now we ask how to choose    and    in order to minimize the costs of a program that produces    when it is optimal to allow violations. Following Arguedas (2008), we ask ourselves whether we can decrease the costs of a program that allows a certain level of violation for each firm by changing the fine structure (changing the values of    and   ) while choosing    In order to answer this question, we evaluate the Lagrangian of the regulator's problem at    when    and    and change    and    such that    that is   

Dividing both sides by   and substituting    for    we obtain:

We know that    and    Substituting:

And after some operations, we obtain:

This means that the regulator can decrease the costs of a program that allows a violation    for each firm by decreasing    (and increasing    accordingly so as to keep the equilibrium fine constant).
Now, decreasing    has a limit and this limit is    Under a negative value of    , a (sufficiently small) level of violation it will always exist to make the fine negative. But a negative fine violates our assumption that    for all levels of violations. On the other hand, there is no theoretical maximum value for    In theory, this value is infinite and, therefore, it is not firm-specific. Therefore, the cost-minimizing design of a program based on standards calls for a uniform penalty structure for all firms:    for all    The regulator always decreases monitoring costs by increasing    for the same level of violation. This is true for all firms and, therefore, the regulator must set    as high as possible for all firms. Because we are considering the case in which the regulator allows non-compliance, condition    for all    must hold. And because we have just said that the cost-minimizing shape of the penalty function requires    for all    the above condition only requires    In conclusion, if the optimal policy allows non-compliance, the best shape of the penalty function is one in which the linear component    and the progressive component are set "as high as possible" for all firms.
Having determined what is the cost-minimizing structure of the fine when it is optimum to induce compliance and when it is not, we now prove Proposition 3. Following Arguedas (2008), we assume that it is optimum to allow non-compliance, and call the optimal policy    with    for all    (with    as high as possible following the results above),    , and    Now consider an alternative policy    such that    and    for all    and    for all    with    By construction, this policy induces compliance because    for all    Moreover,    is cheaper than    in terms because abatement costs are the same under both programs (   for all    monitoring costs are the same under both programs (   for all    but there are no sanctioning costs under policy    because there are no violations, Q.E.D.
Proof of Proposition 6 The Lagrange of the regulator's problem can be written as:

with    and    being the    multipliers. The    necessary Kuhn-Tucker conditions for positive levels of the standard and the auditing probability are:

We assume that these conditions are necessary and sufficient to characterize the optimal solution of the problem.
If    from (Proof8'KT4) we know that    Using the linearity of    and operating, we can re-write the Kuhn-Tucker conditions (Proof8'KT1) and (Proof8'KT2) as:

and

Dividing both expressions, we obtain:

From the firm's optimal choice of emissions, we know that:

if the firm violates the standard (   and 

if    From where:

  and

if   , and 

if   . From these analyses, we can write

where    is the maximum value of    for which the firm is compliant with   , given    That is    is such that    Therefore,    can be interpreted as the probability that the firm    complies with the standard    and    the probability that the firm    violates the standard    Incorporating the assumption of a quadratic fine structure so that    for all    we can rewrite the above expression as:

Similarly, we can also write:

Using these two expressions we can write:

Now, in order to compare this expression with (E(de/ds)/E(de/dpi)), we need to evaluate (31) at    Assume that    is a discrete random variable. Then:

where    are all the possible values of    ordered from the lowest to the largest, and    is the associated probability of each of these values. It is easy to see that    for all   with    denoting the value of the monitoring probability that makes firm    comply if    In other words,    if and only if    i.e., the plant is monitored as if it had the highest possible marginal abatement costs. The proof of this result is quite intuitive. Assume that the regulator monitors firm    with a monitoring probability    such that    for all possible values of    except for    In this case,       for all possible values of    and    Given this,    Therefore,    requires   
Assuming a cost-minimizing regulator, this will set   . But note that if the regulator sets    , it will not only induce    but it will also induce    (perfect compliance with certainty). Therefore, we can write (E(de/ds)/E(de/dpi)2) as:

and (E(de/ds)/E(de/dpi)) as:

Combining these two equations, we obtain:

or, after operating and using   

Q.E.D.
The sign of    Without loss of generality, we assume that    may take only two possible values: low    and high   . We also assume that that    is linear. Then, we can write:

Assuming    as we do, and recalling that    the first term of the above expression is positive. With respect to the second one,    and rewriting

as

it is easy to see that this expression is also negative.
Therefore, we can conclude that    is positive. Q.E.D.
The sign of    Similarly to what we assume in the above proof, we assume that    may take only two possible values: low    and high    and that    is linear. Then, we can write:

Assuming    as we do throughout, and recalling that    the first term of the above expression is positive. With respect to the second one,    and rewriting

as

Therefore, we can conclude that    is positive. Q.E.D.
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