Hola Carlos. Estuve leyendo detenidamente el paper de Arguedas (2007). El mismo es perfectamente aplicable al caso que estamos tratando: regulador con información perfecta que quiere llegar a un límite agregado de emisiones E aplicando estándares, con dios tipos de multas (lineal y convexa en la violación) y cuyo objetivo es minimizar la suma de los costos de abatimiento y los costos esperados de control (monitoreo y sanción).

Lo que sí es que algunas de sus conclusiones respecto a la forma de la función de multa que minimiza los costos.

Te copio abajo el modelo y los resultados de Arguedas con nuestra notación y para nuestro caso (objetivo E en lugar de nivel eficiente de emisiones). También le agrego un grupo de fimras heterogéneas (Arguedas lo hace para una sola firma).

Por último, Arguedas hace que la función de multa tenga un parámetro fijo, F_{0}, el cual no depende de la violación de la siguiente forma,

F (e − s) = F_{0} + f(e − s)

donde, como siempre, f(e−s) > 0, f′(e−s) > 0 y f′′(e−s) > 0 para todo e > s y f(e − s) = 0 para todo e ≤ s. Arguedas llama a F_{0} the non-gravity component y a f(e − s) the gravity component.

Pero al final concluye que el mejor diseño de la multa en cualquier circunstancia es con F_{0} = 0, así que me parece que lo podemos ignorar tranquilamente.

1 The behavior of the Firm

Es idéntico a lo que tenemos planteado y ya sabemos.

Su lema 1.

Lemma 1:

e_{i}(π_{i}, s_{i}) = \{ \begin{matrix}s_{i} & si & c′_{i}(s_{i}) + π_{i}f′(0) ≥ 0 \\ e_{i}^{∗} > s_{i} & si & c′_{i}(s_{i}) + π_{i}f′(0) < 0 \end{matrix} \}

2 The Optimal Policy under given penalties

The regulator selects (π_{i}, s_{i}) for a given fine structure f(e_{i} − s_{i}) : The regulator minimizes in (π_{i}, s_{i})

∑_{i} c_{i}(e_{i}) + μ ∑_{i} π_{i} + β ∑_{i} π_{i}f(e_{i} − s_{i})  (1)

subject to e_{i} = e_{i}(π_{i}, s_{i}) e_{i} ≥ s_{i} s_{i} ≥ 0

∑_{i} e_{i} = E

where e_{i}(π_{i}, s_{i}) is the firm´ s best response characterized by Lemma 1. A sufficient condition for the Total-Cost-Effective policy to induce com-

pliance is the following:

Proposition 1 1: Let (s_{i}^{∗}, π_{i}^{∗}) be the solution to (1), then (s_{i}^{∗}, π_{i}^{∗}) induces compliance if

μ\frac{f′′(0)}{f′(0)} ≤ βf′(0)

Proof of proposition 1

The lagreangean of the problem is

L(π_{i}, s_{i}, λ_{1}^{i}, λ_{2}^{i}, λ_{3}^{i}, λ_{4}) = ∑_{i} c_{i}(e_{i}) + μ ∑_{i} π_{i} + β ∑_{i} π_{i}f(e_{i} − s_{i})

+ ∑_{i} λ_{1}^{i} [c′_{i}(e_{i}) + π_{i}f′(e_{i} − s_{i})]

+ ∑_{i} λ_{2}^{i} [−e_{i} + s_{i}] − ∑_{i} λ_{3}^{i}s_{i} + λ_{4} [E − ∑_{i} e_{i}]

For "analytical conveniencie2 Arguedas considers the pollution level as a choice varibale of the regulator. "The problem is mathematically equivalent to the one where the regulator chooses (π_{i}, s_{i}) knowing that the firm chooses e_{i}(π_{i}, s_{i}) as a response to the policy". (CARLOS YO COPIO EL PROCEDIMIENTO DE ARGUEDAS. SI DISCREPAS CON ALGO AVISA PORQUE A MI HAY COSAS QUE NO ME GUSTAN).

The n × 6 + 1 optimality conditions are:

\frac{∂L}{∂e_{i}} = c′_{i}(e_{i})+βπ_{i}f′(e_{i}−s_{i})+λ_{1}^{i} [c′′_{i}(e_{i}) + π_{i}f′′(e_{i} − s_{i})]−λ_{2}^{i}−λ_{4} = 0, i = 1, ...n

(2)

\frac{∂L}{∂π_{i}} = μ + βf(e_{i} − s_{i}) + λ_{1}^{i}f′(e_{i} − s_{i}) = 0, i = 1, ...n  (3)
\frac{∂L}{∂s_{i}} = −βπ_{i}f′(e_{i} − s_{i}) − λ_{1}^{i}π_{i}f′′(e_{i} − s_{i}) + λ_{2}^{i} − λ_{3}^{i} = 0, i = 1, ...n (4)
\frac{∂L}{∂λ_{1}^{i}} = c′_{i}(e_{i}) + π_{i}f′(e_{i} − s_{i}) = 0, i = 1, ...n  (5)
\frac{∂L}{∂λ_{2}^{i}} = −e_{i} + s_{i} ≤ 0; λ_{2}^{i} ≥ 0; (−e_{i} + s_{i}) λ_{2}^{i} = 0, i = 1, ...n  (6)
\frac{∂L}{∂λ_{3}^{i}} = −s_{i} ≤ 0; λ_{3}^{i} ≥ 0; λ_{3}^{i}s_{i} = 0, i = 1, ...n  (7)
\frac{∂L}{∂λ_{4}} = E − ∑_{i} e_{i} = 0  (8)

If (s_{i}^{∗}, π_{i}^{∗}) induces compliance, then e_{i}^{∗} = s_{i}^{∗}, λ_{2}^{i} ≥ 0 and λ_{3}^{i} ≥ 0 then by 4

−βπ_{i}f′(0) − λ_{1}^{i}π_{i}f′′(0) = −λ_{2}^{i} + λ_{3}^{i}

By 3

λ_{1}^{i}f′(0) = − [μ + βf(0)] = −μ
λ_{1}^{i} = −\frac{μ}{f′(0)}

And combining these two

−βπ_{i}f′(0) + \frac{μ}{f′(0)}π_{i}f′′(0) = −λ_{2}^{i} + λ_{3}^{i}

Since e_{i}^{∗} > 0, e_{i}^{∗} = s_{i}^{∗} implies s_{i}^{∗} > 0 and by 7 λ_{3}^{i} = 0. Then,

−βπ_{i}f′(0) + \frac{μ}{f′(0)}π_{i}f′′(0) = −λ_{2}^{i} ≤ 0

μ\frac{π_{i}f′′(0)}{f′(0)} ≤ βπ_{i}f′(0)

μ\frac{f′′(0)}{f′(0)} ≤ βf′(0)  (9)

QED Intuición: El regulador se encuentra en e_{i}^{∗} = s_{i} y se pregunta si existe alguna otra

combinación de s_{i} y π_{i} que induzca non-compliance y que alcance E a un costo menor. El regulador debe mover ambos s_{i} y π_{i}, porque de otra manera no logra E. Por ejemplo, si baja s_{i} solamente, e_{i} baja y por lo tanto E > ∑_{i} e_{i}. Lo que tendría que hacer también es bajar π_{i} para que∑_{i} e_{i} vuelva a subir. ¿Cuál es la correspondiente baja en π_{i} ante una baja marginal en s_{i}? Nosotros sabemos que ∂e_{i}/∂π_{i} = −f′/ (c′′_{i} + π_{i}f′′) < 0 y ∂e_{i}/∂s_{i} = π_{i}f′′/ (c′′_{i} + π_{i}f′′) > 0, por lo tanto la respuesta es

\frac{∂e_{i}/∂s_{i}}{∂e_{i}/∂π_{i}} = \frac{∂π_{i}}{∂s_{i}}_{e_{i}^{∗}=s_{i}} = \frac{−π_{i}f′′(0)}{f′(0)}

Cuando s_{i} baja marginalmente, los sanctioning costs aumentan en βπ_{i}f′(0). (\frac{∂e_{i}^{∗}}{∂s_{i}} > 0)

Cuando π_{i} baja en la cantidad necesaria mantener constante las emisiones , los monitoring costos bajan μ\frac{π_{i}f′′(0)}{f′(0)}. (\frac{∂e_{i}^{∗}}{∂π_{i}} < 0)

Si este ahorro marginal de costos por bajar π es menor o igual al incremento de costos marginales por bajar s_{i}, no conviene moverse. No conviene otra cosa que full-compliance.

Más importante, (9) nos dice que la forma de f afecta la optimalidad de inducir compliance o no. Aún más, si f es lineal (f′′(0) = 0), la condición se reduce a 0 ≤ βf′(0), la cual se cumple siempre.

Es decir, si f es lineal, siempre va a ser el caso que el regulador minimiza los costos totales del programa haciendo full compliance.

*************************************************************** Lo que queda sin responder es si los costos totales del programa (mínimos)

son menores con:

(1) f′′(.) = 0 y full compliance

(2) f′′(.) > 0 y full compliance (Prop. 1 se cumple)

(3) f′′(0) >0 y no full-compliance (Prop.1 no se cumple). ****************************************************************

Proposition 2: If the optimal policy (s_{i}^{∗}, π_{i}^{∗}) induces compliance, it is characterized by

c′_{i}(s_{i}^{∗}) − λ_{4} + μ\frac{dπ_{i} (s_{i}^{∗})}{ds_{i}} = 0

where π_{i}^{∗} = π_{i} (s_{i}^{∗}) = \frac{−c′_{i}(s_{i}^{∗})}{f′(0)}.

Proof of Proposition 2: Sustituyendo e_{i}^{∗} = s_{i}^{∗} and π_{i}^{∗} = π_{i} (s_{i}^{∗}) en las CPO del problema del

regulador, éstas quedan:

c′_{i}(s_{i}^{∗}) + βπ_{i}f′(0) + λ_{1}^{i} [c′′_{i}(s_{i}^{∗}) + π_{i}f′′(0)] − λ_{2}^{i} − λ_{4} = 0, i = 1, ...n (10)

μ + λ_{1}^{i}f′(0) = 0, i = 1, ...n  (11)
−βπ_{i}f′(0) − λ_{1}^{i}π_{i}f′′(0) + λ_{2}^{i} − λ_{3}^{i} = 0, i = 1, ...n  (12)
c′_{i}(s_{i}^{∗}) + π_{i}f′(0) = 0, i = 1, ...n  (13)
−e_{i} + s_{i} = 0; λ_{2}^{i} ≥ 0, i = 1, ...n  (14)
−s_{i} < 0; λ_{3}^{i} = 0, i = 1, ...n  (15)
E − ∑_{i} e_{i} = 0  (16)

Summing (10) y (12) and using λ_{3}^{i} = 0,

c′_{i}(s_{i}^{∗})+βπ_{i}f′(0)+λ_{1}^{i} [c′′_{i}(s_{i}^{∗}) + π_{i}f′′(0)]−λ_{2}^{i}−λ_{4}−βπ_{i}f′(0)−λ_{1}^{i}π_{i}f′′(0)+λ_{2}^{i} = 0

c′_{i}(s_{i}^{∗}) + λ_{1}^{i} [c′′_{i}(s_{i}^{∗})] − λ_{4} = 0

From (11)

λ_{1}^{i} = −\frac{μ}{f′(0)}

Then

c′_{i}(s_{i}^{∗}) − \frac{μ}{f′(0)} [c′′_{i}(s_{i}^{∗})] − λ_{4}^{i} = 0

Since π_{i}^{∗} = π_{i} (s_{i}^{∗}) = \frac{−c′_{i}(s_{i}^{∗})}{f′(0)},

\frac{dπ_{i} (s_{i}^{∗})}{ds_{i}} = \frac{−c′′_{i}(s_{i}^{∗})}{f′(0)}

Then,

c′_{i}(s_{i}^{∗}) + μ\frac{dπ_{i} (s_{i}^{∗})}{ds_{i}} − λ_{4} = 0

QED Note: If the optimal policy induces compliance the optimal policy does

not imply an allocation of emissions such that marginal abatement costs are equal, unless c′′_{i}(s_{i}^{∗}) are equal for all i.

Proposition 3: If the optimal policy (s_{i}^{∗}, π_{i}^{∗}) induces non-compliance, it is given by

c′_{i}(e_{i}^{∗}) + βπ_{i}f′(e_{i}^{∗} − s_{i}^{∗}) + \frac{μ + βf(e_{i}^{∗} − s_{i}^{∗})}{∂e_{i}/∂π_{i}} − λ_{4}^{i} = 0  (17)
c′_{i}(e_{i}^{∗}) + π_{i}f′(e_{i}^{∗} − s_{i}^{∗}) = 0  (18)
[μ + βf(e_{i}^{∗} − s_{i}^{∗})] \frac{f′′(e_{i}^{∗} − s_{i}^{∗})}{f′(e_{i}^{∗} − s_{i}^{∗})} − βf′(e_{i}^{∗} − s_{i}^{∗}) ≥ 0  (19)
[[μ + βf(e_{i}^{∗} − s_{i}^{∗})] \frac{f′′(e_{i}^{∗} − s_{i}^{∗})}{f′(e_{i}^{∗} − s_{i}^{∗})} − βf′(e_{i}^{∗} − s_{i}^{∗})] s_{i} = 0  (20)

Proof of Proposition 3: Under non-compliance, the FOC of the regulator’s problem are

c′_{i}(e_{i}^{∗}) + βπ_{i}f′(e_{i}^{∗} − s_{i}^{∗}) + λ_{1}^{i} [c′′_{i}(e_{i}) + π_{i}f′′(e_{i} − s_{i})] − λ_{2}^{i} − λ_{4} = 0 (21)

μ + βf(e_{i}^{∗} − s_{i}^{∗}) + λ_{1}^{i}f′(e_{i}^{∗} − s_{i}^{∗}) = 0  (22)
−βπ_{i}f′(e_{i}^{∗} − s_{i}^{∗}) − λ_{1}^{i}π_{i}f′′(e_{i}^{∗} − s_{i}^{∗}) + λ_{2}^{i} − λ_{3}^{i} = 0  (23)

c′_{i}(e_{i}^{∗}) + π_{i}f′(e_{i}^{∗} − s_{i}^{∗}) = 0  (24)
−e_{i} + s_{i} < 0; λ_{2}^{i} = 0  (25)
−s_{i} ≤ 0; λ_{3}^{i} ≥ 0; λ_{3}^{i}s_{i} = 0  (26)
E − ∑_{i} e_{i} = 0  (27)

From (22)

λ_{1}^{i} = −\frac{μ + βf(e_{i}^{∗} − s_{i}^{∗})}{f′(e_{i}^{∗} − s_{i}^{∗})}

Substituting into (23) and using λ_{2}^{i} = 0,

[μ + βf(e_{i}^{∗} − s_{i}^{∗})] \frac{π_{i}f′′(e_{i}^{∗} − s_{i}^{∗})}{f′(e_{i}^{∗} − s_{i}^{∗})} − βπ_{i}f′(e_{i}^{∗} − s_{i}^{∗}) = λ_{3}^{i} ≥ 0

which is the third equation in proposition 3. Then, using 21

c′_{i}(e_{i}^{∗}) + βπ_{i}f′(e_{i}^{∗} − s_{i}^{∗}) + λ_{1}^{i} [c′′_{i}(e_{i}) + π_{i}f′′(e_{i} − s_{i})] − λ_{2}^{i} − λ_{4}^{i} = 0

and

λ_{1}^{i} = −\frac{μ + βf(e_{i}^{∗} − s_{i}^{∗})}{f′(e_{i}^{∗} − s_{i}^{∗})}

we obtain

c′_{i}(e_{i}^{∗}) + βπ_{i}f′(e_{i}^{∗} − s_{i}^{∗}) − \frac{μ + βf(e_{i}^{∗} − s_{i}^{∗})}{f′(e_{i}^{∗} − s_{i}^{∗})} [c′′_{i}(e_{i}) + π_{i}f′′(e_{i} − s_{i})] − λ_{4}^{i} = 0

Remembering that ∂e_{i}/∂π_{i} = −f′/ (c′′_{i} + π_{i}f′′) , we obtain

c′_{i}(e_{i}^{∗}) + βπ_{i}f′(e_{i}^{∗} − s_{i}^{∗}) + \frac{μ + βf(e_{i}^{∗} − s_{i}^{∗})}{∂e_{i}/∂π_{i}} − λ_{4} = 0

which is the first equation of proposition 3. Finally, combining(23) with (26), we obtain

[−βπ_{i}f′(e_{i}^{∗} − s_{i}^{∗}) − λ_{1}^{i}π_{i}f′′(e_{i}^{∗} − s_{i}^{∗})] s_{i} = 0

QED

Note that assuming s_{i}^{∗} > 0, the optimal policy (s_{i}^{∗}, π_{i}^{∗}) that induces noncompliance is given by

c′_{i}(e_{i}^{∗}) + βπ_{i}^{∗}f′(e_{i}^{∗} − s_{i}^{∗}) + \frac{μ + βf(e_{i}^{∗} − s_{i}^{∗})}{∂e_{i}^{∗}/∂π_{i}} − λ_{4} = 0  (28)
c′_{i}(e_{i}^{∗}) + π_{i}^{∗}f′(e_{i}^{∗} − s_{i}^{∗}) = 0  (29)
[μ + βf(e_{i}^{∗} − s_{i}^{∗})] \frac{f′′(e_{i}^{∗} − s_{i}^{∗})}{f′(e_{i}^{∗} − s_{i}^{∗})} − βf′(e_{i}^{∗} − s_{i}^{∗}) = 0  (30)

Here we can see again that under lnear penalties condition (30) never holds. It is never optimal to induce non-compliance.

3 The Choice of the Appropriate Penalties

"First we derive the most appropriate shape of the penalties under the two possible scenarios: complaince and non-compliance. Next, we select the socially preferred scenario."

The result for the first Objective is Proposition 4:

Proposition 4: If the optimal pociy induces compliance, the best shape of the gravity fine is such that the linear component is set as high as possible and the progressive component is zero. Conversley, if the optimal policy induces non-compliance, then the best shape of the gravity fine is such that the linear component is zero and the progressive component is set as high as possible. ^{1}

A MI NO ME DA ESTO. A MI ME DA: IF THE OPRIMAL POL-ICY INDUCES COMPLIANCE, THE LINEAR COMPONENT IS SET AS HIGH AS POSSIBLE AND THE PRGRESSIVE COMPONENT IS IR-RELEVANT (IN EQUILIBRIUM). IF THE OPTIMAL POLICY INDUCES NON-COMPLIANCE, IT DEPENDS ON A CONDITION BELOW.

Proof of Proposition 4:

1"In any case, the optimal non-gravity component (F_{0}) is zero" is the last part of the proposition in Arguedas (2007).

The fine f(e − s) (a second order degree polinomial function) can be represented by the Maclaurin series (expansion around e − s = 0) by

f(e − s) = f′(0)(e − s) + \frac{f′′(0)}{2}(e − s)^{2}

f′(0) is the linear gravity component.

f′′(0) is the progressive gravity component

"we restrict ourselves to linear quadratic penalty functions"

If the optimal policy induces compliance, sanctioning costs are zero. By Proposition 2

π_{i}^{∗} = π_{i} (s_{i}^{∗}) = \frac{−c′_{i}(s_{i}^{∗})}{f′(0)}

"the larger the linear gravity component the lower the minimum probability to achieve compliance and therefore the social costs. Therefore, the optimal fine is one on which f′(0) is as high as possible and f′′(0) is as low as possible, since only the first component affects the probability." (En realidad es más correcto, o correcto decir que f′′(0) puede ser cualquier cosa, en equilibrio, y por ende mi comentario arriba a la Proposition 4).

If the optimal policy induces non-compliance,

f(e_{i}^{∗} − s_{i}^{∗}) = f′(0)(e_{i}^{∗} − s_{i}^{∗}) + \frac{f′′(0)}{2}(e_{i}^{∗} − s_{i}^{∗})^{2}

In order to know how the shape of the penalty function affects social costs, we differentiate the Lagrangean with respect to f′(0) and f′′(0)

dL = \frac{dL}{df′(0)}df′(0) + \frac{dL}{df′′(0)}df′′(0)

L = ∑_{i} c_{i}(e_{i}) + μ ∑_{i} π_{i} + β ∑_{i} π_{i} [f′(0)(e_{i} − s_{i}) + \frac{f′′(0)}{2}(e_{i} − s_{i})^{2}]

+ ∑_{i} λ_{1}^{i} [c′_{i}(e_{i}) + π_{i} [f′(0) + f′′(0)(e_{i} − s_{i})]] + ∑_{i} λ_{2}^{i} [−e_{i} + s_{i}] − ∑_{i} λ_{3}^{i}s_{i} + λ_{4} [E − ∑_{i} e_{i}]

dL = [β ∑_{i} π_{i}(e_{i} − s_{i}) + λ_{1}^{i}π_{i}] df′(0)+[β ∑_{i} π_{i} \frac{(e_{i} − s_{i})^{2}}{2} + λ_{1}^{i}π_{i}(e_{i} − s_{i})] df′′(0)

\frac{dL}{df′(0)} = [β ∑_{i} π_{i}(e_{i} − s_{i}) + ∑_{i} λ_{1}^{i}π_{i}]+[β ∑_{i} π_{i} \frac{(e_{i} − s_{i})^{2}}{2} + ∑_{i} λ_{1}^{i}π_{i}(e_{i} − s_{i})] \frac{df′′(0)}{df′(0)}

In order to keep total fines collected, and therefore the sanctioning costs, constant for every firm, the regulator must choose choose df′(0) and df′′(0) such that

df = (e_{i} − s_{i})df′(0) + \frac{(e_{i} − s_{i})^{2}}{2} df′′(0) = 0

Or

df′(0) + \frac{(e − s)}{2}df′′(0) = 0

\frac{(e − s)}{2} = −\frac{df′(0)}{df′′(0)}

Then,

\frac{dL}{df′(0)} = [β ∑_{i} π_{i}(e_{i} − s_{i}) + λ_{1}^{i}π_{i}]+[β ∑_{i} π_{i} \frac{(e_{i} − s_{i})^{2}}{2} + λ_{1}^{i}π_{i}(e_{i} − s_{i})] \frac{df′′(0)}{df′(0)}

\frac{dL}{df′(0)} = [β ∑_{i} π_{i}(e_{i} − s_{i}) + λ_{1}^{i}π_{i}]+[−β ∑_{i} π_{i}(e_{i} − s_{i})\frac{df′(0)}{df′′(0)} + λ_{1}^{i}π_{i}(e_{i} − s_{i})] \frac{df′′(0)}{df′(0)}

\frac{dL}{df′(0)} = [β ∑_{i} π_{i}(e_{i} − s_{i}) + λ_{1}^{i}π_{i}]+[−n\frac{df′(0)}{df′′(0)}β ∑_{i} π_{i}(e_{i} − s_{i}) − 2λ_{1}^{i}π_{i} \frac{df′(0)}{df′′(0)}] \frac{df′′(0)}{df′(0)}

\frac{dL}{df′(0)} = β ∑_{i} π_{i}(e_{i} − s_{i}) + λ_{1}^{i}π_{i} − nβ ∑_{i} π_{i}(e_{i} − s_{i}) − 2λ_{1}^{i}π_{i}

\frac{dL}{df′(0)} = (1 − n) β ∑_{i} π_{i}(e_{i} − s_{i}) − λ_{1}^{i}π_{i}

Assuming (1 < n) the first term (1 − n) βΣ_{i}π_{i}(e_{i} − s_{i}) is ≤ 0. On the other hand, λ_{1}^{i} = − (μ + βf(e_{i}^{∗} − s_{i}^{∗})) /f′(e_{i}^{∗} − s_{i}^{∗}) ≤ 0, therefore the second term −λ_{1}^{i}π_{i} is ≥ 0. Therefore, unlike the case of one firm (Arguedas, 2007) this expression cannot be signed.

INTUITION?????????????????

If true, it is no longer valid that if optimal policy induces non-compliance, f′(0) must be as low as possible and f′′(0) be as high as possible.

Proposition 5: the optimal policy (s_{i}^{∗}, π_{i}^{∗}, f^{∗}) induces compliance and it is characterized by:

c′_{i}(s_{i}^{∗}) − λ_{4}^{i} − μ\frac{c′′_{i}(s_{i}^{∗})}{\overline{a}} = 0

π_{i}^{∗} = −\frac{c′′_{i}(s_{i}^{∗})}{\overline{a}}

F (e_{i}^{∗} − s_{i}^{∗}) = \overline{a}(e_{i}^{∗} − s_{i}^{∗})

where \overline{a} > 0 is the largest. In other words, the optimal polciy induces compliance and have linear

fines (idem Stranlund, 2007).

Proof of Proposition 5 ("adapted from Stranlund (2007)"). "First, we prove that the optimal policy induces compliance": Assume to the contrary that the optimal policy does not induce compli-

ance, and call it (s_{i}^{n}, π_{i}^{n}, f^{n}). Arguedas, following her proposition (4), assumes that the form of the f^{n} = \frac{\overline{b}}{2}(e_{i}^{n}−s_{i}^{n})^{2}. Given that proposition (4) is not longer always valid with n > 1 (if i am correct), i assume f^{n} = g(e − s) + \frac{\overline{b}}{2}(e − s)^{2}, where as in Arguedas, \overline{b} is the (largest possible) f′′(0) given by the legislation.

If the policy induces non-compliance, we know from Proposition (3) that

c′_{i}(e_{i}^{n}) + π_{i}^{n} [g + \overline{b}(e_{i}^{n} − s_{i}^{n})] = 0

Now consider an alternative policy (s_{i}^{c}, π_{i}^{c}, f^{c}), such that s_{i}^{n} = e_{i}^{n} and π_{i}^{c} = π_{i}^{n} and

f^{c′}(0) = f^{n′}(e_{i}^{n} − s_{i}^{n}) = g + \overline{b}(e_{i}^{n} − s_{i}^{n})