AA {ClaimsProblems}

Average of awards rule

Description

This function returns the awards vector assigned by the average of awards rule (AA) to a claims problem.

Usage

AA(E, d, name = FALSE)

Arguments

E

The endowment.

d

The vector of claims.

name

A logical value.

Details

Let E\ge 0 be the endowment to be divided and d\in \mathcal{R}^n the vector of claims with d\ge 0 and such that \sum_{i=1}^{n} d_i\ge E,\; the sum of claims exceeds the endowment.

A vector x=(x_1,\dots,x_n) is an awards vector for the claims problem (E,d) if 0\le x \le d and satisfies the balance requirement, that is, \sum_{i=1}^{n}x_i=E the sum of its coordinates is equal to E. Let X(E,d) be the set of awards vectors for (E,d).

The average of awards rule assigns to each claims problem (E,d) the expectation of the uniform distribution defined over the set of awards vectors, that is, the centroid of X(E,d).

Let \mu be the (n-1)-dimensional Lebesgue measure and V(E,d)=\mu (X(E,d)) the measure (volume) of the set of awards X(E,d). The average of awards rule assigns to each problem (E,d) the awards vector given by:

AA(E,d)=\frac{1}{V(E,d)}\int_{X(E,d)} x d\mu

The average of awards rule corresponds to the core-center of the associated coalitional (pessimistic) game.

Value

The awards vector selected by the AA rule. If name = TRUE, the name of the function (AA) as a character string.

References

Gonzalez-Díaz, J. and Sánchez-Rodríguez, E. (2007). A natural selection from the core of a TU game: the core-center. International Journal of Game Theory, 36(1), 27-46.

Mirás Calvo, M.Á., Quinteiro Sandomingo, C., and Sánchez-Rodríguez, E. (2022). The average-of-awards rule for claims problems. Soc Choice Welf. doi: 10.1007/s00355-022-01414-6

Mirás Calvo, M.Á., Núñez Lugilde, I., Quinteiro Sandomingo, C., and Sánchez-Rodríguez, E. (2020). An algorithm to compute the core-center rule of a claims problem with an application to the allocation of CO2 emissions. Working paper.

See Also

allrules, CD, setofawards, coalitionalgame

Examples

E=10
d=c(2,4,7,8)
AA(E,d)
#The average of awards rule is self-dual: AA(E,d)=d-AA(D-E,d)
D=sum(d)
d-AA(D-E,d)