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The Coase Theorem, the Nonempty Core, and the Legal Neutrality Principle

  • Bertrand Crettez EMAIL logo
Veröffentlicht/Copyright: 17. Oktober 2019
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Abstract

The Coase theorem states that where there are externalities and no transaction costs resource allocation is Pareto-optimal and independent of the stakeholders’ legal position. This result has been challenged many times. In the cooperative game approach to resource allocation, the refutation is made by constructing a three-person game which has an empty core under one set of liability rules—which implies that optimal allocations are coalitionally unstable–and a nonempty core under another set. In this example, however, the probability that the core is non-empty is rather high (5/6). Yet, even if coalitionally stable Pareto-optimal arrangements are likely, to establish the plain validity of the Coase theorem it must be shown that the legal neutrality statement also holds. We show that for the three-person cooperative game example mentioned above, the probability that the two assertions of the Coase theorem hold can be as low as 3/8.

JEL Classification: C71; C78; D23; D62; K0

Acknowledgements

I am grateful to Régis Deloche and Gabrielle Smart for very helpful comments on a previous version of this paper. I also thank a referee for stimulating comments and suggestions.

Appendix

A Proof of Proposition 2

Proof

[13]

To establish the result we shall first assume that θ1v(AB)/v(ABC) is first drawn from a uniform distribution. Let θ4 denote vl(C)/v(ABC). For each realization θ1 the probability that vl(C)/v(ABC) be lower than 1v(AB)/v(ABC), namely θ41θ1 is 1θ1. Now we must check that inequation (12) holds. But we must take care of the fact that vl(C)v(ABC)v(AC)v(ABC)θ2 and vl(C)v(ABC)v(BC)v(ABC)θ3 (the superadditivity property holds by assumption).

Therefore, the probability that the legal neutrality property holds when θ1 is given and θ41θ1 is the probability that: θ4θ2, θ4θ3andθ2+θ32θ1. This probability is equal to the shaded area in Figure 1. We have

(28)Pθ4θ2,θ4θ3,θ2+θ32θ1θ41θ1=11θ112θ1θ21dθ3dθ20θ40θ4dθ2dθ3.

We can check that:

(29)Pθ4θ2,θ4θ3,θ2+θ31θ1θ41θ1=1θ122θ42.

Now the probability that the legal neutrality property holds, θ1 being given, is

(30)01θ1Pθ4θ2,θ4θ3,θ2+θ32θ1θ41θ1dθ4,=01θ11θ122θ42dθ4,=1θ1θ122+θ132(1θ1)33.

Finally, the probability that the legal neutrality property holds is equal to

(31)P[θ4θ2θ4θ2,θ42θ1,θ2+θ31θ1]=011θ1θ122+θ132(1θ1)33dθ1=18.

   □

Figure 1: Probability that the legal neutrality property holds (θ1 and θ4 being given).
Figure 1:

Probability that the legal neutrality property holds (θ1 and θ4 being given).

B Proof of Proposition 3

To compute the probability that the legal neutrality principle holds let us first assume that θ1, θ2 and θ3 are defined as in the preceding part of the appendix. Let us also define θ2=vl(AC)v(ABC) and θ3=vl(BC)v(ABC). The probability that the legal neutrality principle holds is equal to the probability that θ2+θ32θ1, θ2θ21, θ3θ31 and θ2+θ32θ1. To compute this probability we shall first compute the probability that θ2+θ32θ1 given that θ2+θ32θ1, θ1, θ2 and θ3 being fixed. To proceed there are four zones (see Figure 2). We shall compute the aforementioned probability in each of these zones.

Figure 2: Probability that θ2+θ3≤2−θ1\theta_2 + \theta_3 \leq 2- \theta_1.
Figure 2:

Probability that θ2+θ32θ1.

Denote by Piθ2θ21,θ3θ31,θ1+θ2+θ32θ1+θ2+θ32 the probability referring to zone i (i = I,II,III,IV).

  1. Zone I

    We can check that

    (32)PI=θ22θ1θ3θ32θ1θ2dθ3dθ2,
    (33)=2θ1θ3222θ1θ3θ2+θ222.
  2. Zone II

    (34)PII=1θ12θ1θ3θ32θ1θ2dθ3dθ2+θ21θ1θ31dθ3dθ2,
    (35)=2θ1θ3221θ122(1θ3)θ2.
  3. Zone III

    (36)PIII=θ211θ12θ1θ2dθ3dθ2+θ21θ31θ1dθ3dθ2,
    (37)=12θ2+θ222+1θ1θ3(1θ2).
  4. Zone IV

    (38)PIV=θ21θ1θ31dθ3dθ2+1θ11θ31θ1dθ3dθ2+1θ111θ12θ1θ2dθ3dθ2,
    (39)=(1θ3)1θ1θ2+1θ1θ3θ112+θ1+1θ122.

Now we have

(40)P[θ2θ2,θ3θ3,θ1+θ2+θ3θ1+θ2+θ32]=01P(θ1)dθ1,

where

(41)01P(θ1)dθ1=1θ111θ12θ1θ2PI(θ1,θ2,θ3)dθ3dθ2+01θ11θ11PII(θ1,θ2,θ3)dθ3dθ2+1θ1101θ1PIII(θ1,θ2,θ3)dθ3dθ2+01θ101θ1PIV(θ1,θ2,θ3)dθ3dθ2,

where we have stressed that the Pi are all functions of θ1, θ2 and θ3. We next compute each integral in the above sum.

To compute the first integral we first get after tedious computations

(42)1θ12θ1θ2PI(θ1,θ2,θ3)dθ3=1θ236θ2(2θ1)(1θ2)+θ22(2θ1θ2)2(1θ1)2+θ222(1θ2)

Then we obtain

(43)1θ111θ12θ1θ2PI(θ1,θ2,θ3)dθ3dθ2=124(1θ1)6+(1θ1)241θ136+(1θ1)424.

We also have

(44)01θ11θ11PII(θ1,θ2,θ3)dθ3dθ2=1θ16+(1θ1)412(1θ1)24,
(45)1θ1101PIII(θ1,θ2,θ3)dθ3dθ2=1θ16(1θ1)24+(1θ1)412,

and

(46)1θ1101PIV(θ1,θ2,θ3)dθ3dθ2=θ1(1θ1)22+θ1(1θ1)32+(1θ1)44.

Using these expressions, we finally get

(47)01P(θ1)dθ1=19120

Finally, the probability that the legal neutrality principle holds (and the Coase theorem) is:

(48)P[θ1+θ2+θ32]×01P(θ1)dθ1=56×19120=19144.

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Supplementary Material

The online version of this article offers supplementary material (DOI:https://doi.org/10.1515/rle-2018-0027).


Published Online: 2019-10-17

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