Mixing
Does the Prenucleolus of $[N,nu]$ for at least one Player $iin L^*$ increase if $nu^prime(L^*)>nu(L^*)$?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Consider any cooperative TU game $[N,nu]$ in characteristic function form, where $N$ is a set of Players $iin N$ and $nu:2^NtomathbbR$ is a characteristic function assigning some real worth to every coalition $Lsubseteq N$. Then, let's define some concepts (all definitions are from Maschler, Solan and Zamir, Chapter 20)
EXCESS
For every vector $mathbfxinmathbbR^N$ and every coalition $Lsubseteq N$.
begingather*
e(L,x)=nu(L)-sum_iin Lx_i
endgather*
is called the excess of coalition $L$ at $mathbfx$.
Given a vector $mathbfxinmathbbR^N$, we compute the excess of all the coalitions at $mathbfx$, and we write them in decreasing order from left to right,
$theta(x)=(e(L_1,mathbfx),e(L_2,mathbfx),...,e(L_2^N,mathbfx))$,
where $L_1,L_2,...,L_2^N$ are all the coalitions, indexed such that
$e(L_1,mathbfx)geqslant e(L_2,mathbfx)geqslant ...geqslant e(L_2^N,mathbfx)$.
SET OF PREIMPUTATIONS
Given any game $[N,nu]$ the set of vectors $mathbfxinmathbbR^N$ that satisfy $sum_iin Nx_i=nu(N)$ is the set of preimputations of $[N,nu]$, denoted $X^0(N,nu)$.
LEXICOGRAPHIC RELATION
Let $a=(a_1,a_2,...,a_d)$ and $b=(b_1,b_2,...,b_d)$ be two vectors in $R^d$. Then, $aô°— geqslant_L b$ if either $a=b$, or there exists an integer $k$, $1leqslant kleqslant d$,such that $a_k>b_k$, and $a_i=b_i$ for every $1leqslant i<k$. This order relation is termed the lexicographic order.
PRENUCLEOLUS
The Prenucleolus of the game $[N,nu]$, denoted $mathcalP(N,nu,X^0(N,nu))$ is the set
begingather*
mathcalP(N,nu,X^0(N,nu))= xin X^0(N,nu): theta(x) leqslant_L theta(y), forall yin X^0(N,nu)
endgather*
For more details about these concepts, see Maschler, Solan and Zamir, Chapter 20. Now, take any coalition $L^*subsetneq N$ of some game $[N,nu]$ and define a new game $[N,nu^prime]$ in which $nu^prime(L)=nu(L)$ for all $Lsubseteq Nbackslash L^*$ and $nu^prime(L)=nu(L)+epsilon$ for some $epsilon>0$. Then, my question is the following:
Does it necessarily exist at least one Player $iin L^*$ for whom $mathcalP_i(N,nu^prime,X^0(N,nu^prime))>mathcalP_i(N,nu,X^0(N,nu))$?
My intuition says yes, but I can't prove it myself. If the answer is negative, a counter-example will be enough. If it is affirmative, an intuition of why that is the case will also be enough.
game-theory
asked Jul 20 at 9:55
Héctor
247210
add a comment |Â
up vote
1
down vote
favorite
Consider any cooperative TU game $[N,nu]$ in characteristic function form, where $N$ is a set of Players $iin N$ and $nu:2^NtomathbbR$ is a characteristic function assigning some real worth to every coalition $Lsubseteq N$. Then, let's define some concepts (all definitions are from Maschler, Solan and Zamir, Chapter 20)
EXCESS
For every vector $mathbfxinmathbbR^N$ and every coalition $Lsubseteq N$.
begingather*
e(L,x)=nu(L)-sum_iin Lx_i
endgather*
is called the excess of coalition $L$ at $mathbfx$.
Given a vector $mathbfxinmathbbR^N$, we compute the excess of all the coalitions at $mathbfx$, and we write them in decreasing order from left to right,
$theta(x)=(e(L_1,mathbfx),e(L_2,mathbfx),...,e(L_2^N,mathbfx))$,
where $L_1,L_2,...,L_2^N$ are all the coalitions, indexed such that
$e(L_1,mathbfx)geqslant e(L_2,mathbfx)geqslant ...geqslant e(L_2^N,mathbfx)$.
SET OF PREIMPUTATIONS
Given any game $[N,nu]$ the set of vectors $mathbfxinmathbbR^N$ that satisfy $sum_iin Nx_i=nu(N)$ is the set of preimputations of $[N,nu]$, denoted $X^0(N,nu)$.
LEXICOGRAPHIC RELATION
Let $a=(a_1,a_2,...,a_d)$ and $b=(b_1,b_2,...,b_d)$ be two vectors in $R^d$. Then, $aô°— geqslant_L b$ if either $a=b$, or there exists an integer $k$, $1leqslant kleqslant d$,such that $a_k>b_k$, and $a_i=b_i$ for every $1leqslant i<k$. This order relation is termed the lexicographic order.
PRENUCLEOLUS
The Prenucleolus of the game $[N,nu]$, denoted $mathcalP(N,nu,X^0(N,nu))$ is the set
begingather*
mathcalP(N,nu,X^0(N,nu))= xin X^0(N,nu): theta(x) leqslant_L theta(y), forall yin X^0(N,nu)
endgather*
For more details about these concepts, see Maschler, Solan and Zamir, Chapter 20. Now, take any coalition $L^*subsetneq N$ of some game $[N,nu]$ and define a new game $[N,nu^prime]$ in which $nu^prime(L)=nu(L)$ for all $Lsubseteq Nbackslash L^*$ and $nu^prime(L)=nu(L)+epsilon$ for some $epsilon>0$. Then, my question is the following:
Does it necessarily exist at least one Player $iin L^*$ for whom $mathcalP_i(N,nu^prime,X^0(N,nu^prime))>mathcalP_i(N,nu,X^0(N,nu))$?
My intuition says yes, but I can't prove it myself. If the answer is negative, a counter-example will be enough. If it is affirmative, an intuition of why that is the case will also be enough.
game-theory
asked Jul 20 at 9:55
Héctor
247210
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider any cooperative TU game $[N,nu]$ in characteristic function form, where $N$ is a set of Players $iin N$ and $nu:2^NtomathbbR$ is a characteristic function assigning some real worth to every coalition $Lsubseteq N$. Then, let's define some concepts (all definitions are from Maschler, Solan and Zamir, Chapter 20)
EXCESS
For every vector $mathbfxinmathbbR^N$ and every coalition $Lsubseteq N$.
begingather*
e(L,x)=nu(L)-sum_iin Lx_i
endgather*
is called the excess of coalition $L$ at $mathbfx$.
Given a vector $mathbfxinmathbbR^N$, we compute the excess of all the coalitions at $mathbfx$, and we write them in decreasing order from left to right,
$theta(x)=(e(L_1,mathbfx),e(L_2,mathbfx),...,e(L_2^N,mathbfx))$,
where $L_1,L_2,...,L_2^N$ are all the coalitions, indexed such that
$e(L_1,mathbfx)geqslant e(L_2,mathbfx)geqslant ...geqslant e(L_2^N,mathbfx)$.
SET OF PREIMPUTATIONS
Given any game $[N,nu]$ the set of vectors $mathbfxinmathbbR^N$ that satisfy $sum_iin Nx_i=nu(N)$ is the set of preimputations of $[N,nu]$, denoted $X^0(N,nu)$.
LEXICOGRAPHIC RELATION
Let $a=(a_1,a_2,...,a_d)$ and $b=(b_1,b_2,...,b_d)$ be two vectors in $R^d$. Then, $aô°— geqslant_L b$ if either $a=b$, or there exists an integer $k$, $1leqslant kleqslant d$,such that $a_k>b_k$, and $a_i=b_i$ for every $1leqslant i<k$. This order relation is termed the lexicographic order.
PRENUCLEOLUS
The Prenucleolus of the game $[N,nu]$, denoted $mathcalP(N,nu,X^0(N,nu))$ is the set
begingather*
mathcalP(N,nu,X^0(N,nu))= xin X^0(N,nu): theta(x) leqslant_L theta(y), forall yin X^0(N,nu)
endgather*
For more details about these concepts, see Maschler, Solan and Zamir, Chapter 20. Now, take any coalition $L^*subsetneq N$ of some game $[N,nu]$ and define a new game $[N,nu^prime]$ in which $nu^prime(L)=nu(L)$ for all $Lsubseteq Nbackslash L^*$ and $nu^prime(L)=nu(L)+epsilon$ for some $epsilon>0$. Then, my question is the following:
Does it necessarily exist at least one Player $iin L^*$ for whom $mathcalP_i(N,nu^prime,X^0(N,nu^prime))>mathcalP_i(N,nu,X^0(N,nu))$?
My intuition says yes, but I can't prove it myself. If the answer is negative, a counter-example will be enough. If it is affirmative, an intuition of why that is the case will also be enough.
game-theory
asked Jul 20 at 9:55
Héctor
247210
Consider any cooperative TU game $[N,nu]$ in characteristic function form, where $N$ is a set of Players $iin N$ and $nu:2^NtomathbbR$ is a characteristic function assigning some real worth to every coalition $Lsubseteq N$. Then, let's define some concepts (all definitions are from Maschler, Solan and Zamir, Chapter 20)
EXCESS
For every vector $mathbfxinmathbbR^N$ and every coalition $Lsubseteq N$.
begingather*
e(L,x)=nu(L)-sum_iin Lx_i
endgather*
is called the excess of coalition $L$ at $mathbfx$.
Given a vector $mathbfxinmathbbR^N$, we compute the excess of all the coalitions at $mathbfx$, and we write them in decreasing order from left to right,
$theta(x)=(e(L_1,mathbfx),e(L_2,mathbfx),...,e(L_2^N,mathbfx))$,
where $L_1,L_2,...,L_2^N$ are all the coalitions, indexed such that
$e(L_1,mathbfx)geqslant e(L_2,mathbfx)geqslant ...geqslant e(L_2^N,mathbfx)$.
SET OF PREIMPUTATIONS
Given any game $[N,nu]$ the set of vectors $mathbfxinmathbbR^N$ that satisfy $sum_iin Nx_i=nu(N)$ is the set of preimputations of $[N,nu]$, denoted $X^0(N,nu)$.
LEXICOGRAPHIC RELATION
Let $a=(a_1,a_2,...,a_d)$ and $b=(b_1,b_2,...,b_d)$ be two vectors in $R^d$. Then, $aô°— geqslant_L b$ if either $a=b$, or there exists an integer $k$, $1leqslant kleqslant d$,such that $a_k>b_k$, and $a_i=b_i$ for every $1leqslant i<k$. This order relation is termed the lexicographic order.
PRENUCLEOLUS
The Prenucleolus of the game $[N,nu]$, denoted $mathcalP(N,nu,X^0(N,nu))$ is the set
begingather*
mathcalP(N,nu,X^0(N,nu))= xin X^0(N,nu): theta(x) leqslant_L theta(y), forall yin X^0(N,nu)
endgather*
For more details about these concepts, see Maschler, Solan and Zamir, Chapter 20. Now, take any coalition $L^*subsetneq N$ of some game $[N,nu]$ and define a new game $[N,nu^prime]$ in which $nu^prime(L)=nu(L)$ for all $Lsubseteq Nbackslash L^*$ and $nu^prime(L)=nu(L)+epsilon$ for some $epsilon>0$. Then, my question is the following:
Does it necessarily exist at least one Player $iin L^*$ for whom $mathcalP_i(N,nu^prime,X^0(N,nu^prime))>mathcalP_i(N,nu,X^0(N,nu))$?
My intuition says yes, but I can't prove it myself. If the answer is negative, a counter-example will be enough. If it is affirmative, an intuition of why that is the case will also be enough.
game-theory
asked Jul 20 at 9:55
Héctor
247210
asked Jul 20 at 9:55
Héctor
247210
asked Jul 20 at 9:55
Héctor
247210
asked Jul 20 at 9:55
Héctor
247210
247210
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your intuition is not in general true. Due to the work of
H. Reijnierse and J. Potters. The B-Nucleolus of TU-Games. Games and Economic Behaviour, 24:77–96, 1998,
it is known that for zero-normalized games with v(N) > 0, a collection of at most $2 (n - 1)$ coalitions admits the determination of the (pre-)nucleolus. Thus, changing the worth of some of these coalitions will change the outcome of the (pre-)nucleolus. However, those coalitions which do not determine the outcome of the (pre-)nucleolus can be changed without harm. Or to put it differently, at most $2 (n-1)$ coalitions are needed to replicate the (pre-)nucleolus, the others can be discarded.
Of course, the hardest work is to determine these coalition in advance. This is ongoing research.
More details to this topic can be found in an extract of my new book project that can be downloaded from the following URL:
Chapter: Replication of the Pre-Nucleolus
Hope this will help.
answered Jul 20 at 11:08
Holger I. Meinhardt
728147
Thank you for your answer and for pointing me to these useful sources. As I understand it, changing the worth of a coalition may or may not affect the Prenucleolus vector, thus making my intuition false. However, suppose that changing the worth of this coalition affects the Prenucleolus. Then, is my intuition correct or still may be false?
– Héctor
Jul 20 at 11:42
1
Notice that from a pure logical point of view your conjecture must hold for all coalitions to become a true statement. Thus, selecting an arbitrary coalition should change the outcome. However, as I have pointed out, this is not in general the case. Hence, your conjecture is incorrect. Nevertheless, reformulating your conjecture w.r.t. to the determining coalitions makes it correct.
– Holger I. Meinhardt
Jul 20 at 12:18
Thank you very much: this has been incredibly helpful.
– Héctor
Jul 20 at 12:32
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your intuition is not in general true. Due to the work of
H. Reijnierse and J. Potters. The B-Nucleolus of TU-Games. Games and Economic Behaviour, 24:77–96, 1998,
it is known that for zero-normalized games with v(N) > 0, a collection of at most $2 (n - 1)$ coalitions admits the determination of the (pre-)nucleolus. Thus, changing the worth of some of these coalitions will change the outcome of the (pre-)nucleolus. However, those coalitions which do not determine the outcome of the (pre-)nucleolus can be changed without harm. Or to put it differently, at most $2 (n-1)$ coalitions are needed to replicate the (pre-)nucleolus, the others can be discarded.
Of course, the hardest work is to determine these coalition in advance. This is ongoing research.
More details to this topic can be found in an extract of my new book project that can be downloaded from the following URL:
Chapter: Replication of the Pre-Nucleolus
Hope this will help.
answered Jul 20 at 11:08
Holger I. Meinhardt
728147
Thank you for your answer and for pointing me to these useful sources. As I understand it, changing the worth of a coalition may or may not affect the Prenucleolus vector, thus making my intuition false. However, suppose that changing the worth of this coalition affects the Prenucleolus. Then, is my intuition correct or still may be false?
– Héctor
Jul 20 at 11:42
1
Notice that from a pure logical point of view your conjecture must hold for all coalitions to become a true statement. Thus, selecting an arbitrary coalition should change the outcome. However, as I have pointed out, this is not in general the case. Hence, your conjecture is incorrect. Nevertheless, reformulating your conjecture w.r.t. to the determining coalitions makes it correct.
– Holger I. Meinhardt
Jul 20 at 12:18
Thank you very much: this has been incredibly helpful.
– Héctor
Jul 20 at 12:32
add a comment |Â
up vote
1
down vote
accepted
Your intuition is not in general true. Due to the work of
H. Reijnierse and J. Potters. The B-Nucleolus of TU-Games. Games and Economic Behaviour, 24:77–96, 1998,
it is known that for zero-normalized games with v(N) > 0, a collection of at most $2 (n - 1)$ coalitions admits the determination of the (pre-)nucleolus. Thus, changing the worth of some of these coalitions will change the outcome of the (pre-)nucleolus. However, those coalitions which do not determine the outcome of the (pre-)nucleolus can be changed without harm. Or to put it differently, at most $2 (n-1)$ coalitions are needed to replicate the (pre-)nucleolus, the others can be discarded.
Of course, the hardest work is to determine these coalition in advance. This is ongoing research.
More details to this topic can be found in an extract of my new book project that can be downloaded from the following URL:
Chapter: Replication of the Pre-Nucleolus
Hope this will help.
answered Jul 20 at 11:08
Holger I. Meinhardt
728147
Thank you for your answer and for pointing me to these useful sources. As I understand it, changing the worth of a coalition may or may not affect the Prenucleolus vector, thus making my intuition false. However, suppose that changing the worth of this coalition affects the Prenucleolus. Then, is my intuition correct or still may be false?
– Héctor
Jul 20 at 11:42
1
Notice that from a pure logical point of view your conjecture must hold for all coalitions to become a true statement. Thus, selecting an arbitrary coalition should change the outcome. However, as I have pointed out, this is not in general the case. Hence, your conjecture is incorrect. Nevertheless, reformulating your conjecture w.r.t. to the determining coalitions makes it correct.
– Holger I. Meinhardt
Jul 20 at 12:18
Thank you very much: this has been incredibly helpful.
– Héctor
Jul 20 at 12:32
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your intuition is not in general true. Due to the work of
H. Reijnierse and J. Potters. The B-Nucleolus of TU-Games. Games and Economic Behaviour, 24:77–96, 1998,
it is known that for zero-normalized games with v(N) > 0, a collection of at most $2 (n - 1)$ coalitions admits the determination of the (pre-)nucleolus. Thus, changing the worth of some of these coalitions will change the outcome of the (pre-)nucleolus. However, those coalitions which do not determine the outcome of the (pre-)nucleolus can be changed without harm. Or to put it differently, at most $2 (n-1)$ coalitions are needed to replicate the (pre-)nucleolus, the others can be discarded.
Of course, the hardest work is to determine these coalition in advance. This is ongoing research.
More details to this topic can be found in an extract of my new book project that can be downloaded from the following URL:
Chapter: Replication of the Pre-Nucleolus
Hope this will help.
answered Jul 20 at 11:08
Holger I. Meinhardt
728147
Your intuition is not in general true. Due to the work of
H. Reijnierse and J. Potters. The B-Nucleolus of TU-Games. Games and Economic Behaviour, 24:77–96, 1998,
it is known that for zero-normalized games with v(N) > 0, a collection of at most $2 (n - 1)$ coalitions admits the determination of the (pre-)nucleolus. Thus, changing the worth of some of these coalitions will change the outcome of the (pre-)nucleolus. However, those coalitions which do not determine the outcome of the (pre-)nucleolus can be changed without harm. Or to put it differently, at most $2 (n-1)$ coalitions are needed to replicate the (pre-)nucleolus, the others can be discarded.
Of course, the hardest work is to determine these coalition in advance. This is ongoing research.
More details to this topic can be found in an extract of my new book project that can be downloaded from the following URL:
Chapter: Replication of the Pre-Nucleolus
Hope this will help.
answered Jul 20 at 11:08
Holger I. Meinhardt
728147
answered Jul 20 at 11:08
Holger I. Meinhardt
728147
answered Jul 20 at 11:08
Holger I. Meinhardt
728147
answered Jul 20 at 11:08
Holger I. Meinhardt
728147
728147
Thank you for your answer and for pointing me to these useful sources. As I understand it, changing the worth of a coalition may or may not affect the Prenucleolus vector, thus making my intuition false. However, suppose that changing the worth of this coalition affects the Prenucleolus. Then, is my intuition correct or still may be false?
– Héctor
Jul 20 at 11:42
1
Notice that from a pure logical point of view your conjecture must hold for all coalitions to become a true statement. Thus, selecting an arbitrary coalition should change the outcome. However, as I have pointed out, this is not in general the case. Hence, your conjecture is incorrect. Nevertheless, reformulating your conjecture w.r.t. to the determining coalitions makes it correct.
– Holger I. Meinhardt
Jul 20 at 12:18
Thank you very much: this has been incredibly helpful.
– Héctor
Jul 20 at 12:32
add a comment |Â
Thank you for your answer and for pointing me to these useful sources. As I understand it, changing the worth of a coalition may or may not affect the Prenucleolus vector, thus making my intuition false. However, suppose that changing the worth of this coalition affects the Prenucleolus. Then, is my intuition correct or still may be false?
– Héctor
Jul 20 at 11:42
1
Notice that from a pure logical point of view your conjecture must hold for all coalitions to become a true statement. Thus, selecting an arbitrary coalition should change the outcome. However, as I have pointed out, this is not in general the case. Hence, your conjecture is incorrect. Nevertheless, reformulating your conjecture w.r.t. to the determining coalitions makes it correct.
– Holger I. Meinhardt
Jul 20 at 12:18
Thank you very much: this has been incredibly helpful.
– Héctor
Jul 20 at 12:32
Thank you for your answer and for pointing me to these useful sources. As I understand it, changing the worth of a coalition may or may not affect the Prenucleolus vector, thus making my intuition false. However, suppose that changing the worth of this coalition affects the Prenucleolus. Then, is my intuition correct or still may be false?
– Héctor
Jul 20 at 11:42
Thank you for your answer and for pointing me to these useful sources. As I understand it, changing the worth of a coalition may or may not affect the Prenucleolus vector, thus making my intuition false. However, suppose that changing the worth of this coalition affects the Prenucleolus. Then, is my intuition correct or still may be false?
– Héctor
Jul 20 at 11:42
1
1
Notice that from a pure logical point of view your conjecture must hold for all coalitions to become a true statement. Thus, selecting an arbitrary coalition should change the outcome. However, as I have pointed out, this is not in general the case. Hence, your conjecture is incorrect. Nevertheless, reformulating your conjecture w.r.t. to the determining coalitions makes it correct.
– Holger I. Meinhardt
Jul 20 at 12:18
Notice that from a pure logical point of view your conjecture must hold for all coalitions to become a true statement. Thus, selecting an arbitrary coalition should change the outcome. However, as I have pointed out, this is not in general the case. Hence, your conjecture is incorrect. Nevertheless, reformulating your conjecture w.r.t. to the determining coalitions makes it correct.
– Holger I. Meinhardt
Jul 20 at 12:18
Thank you very much: this has been incredibly helpful.
– Héctor
Jul 20 at 12:32
Thank you very much: this has been incredibly helpful.
– Héctor
Jul 20 at 12:32
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857481%2fdoes-the-prenucleolus-of-n-nu-for-at-least-one-player-i-in-l-increase%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password