I want to find a cooperative, characteristic function TU game $v$ (of at best of 3 or 4 players;2 players seem impossible to me) for which the prenucleolus is different from the nucleolus. I do not know if this is a research question or if it is more trivial as stated by the rules of this site. The prenucleolus is as usual defined by lexicographic minimization over ordered preimputations of excesses of all coalitions while the nucleolus is defined by lexicographic minimization over imputations. It seems that then the prenucleolus will be lexicographically strictly less than nucleolus. I just need an easy example of (pre)-nuclei which are not the same.
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The set of imputations may be empty, while the set of preimputations is never empty. Hence the nucleolus may be empty, while the prenucleolus is never empty. This happens, for example, in the two-player game v(1)=v(2)=0 and v(1,2)=-1. If you look for a game where both nucleolus and prenucleolus are nonempty and yet they differ, Exercise 20.15 in "Game Theory" by Maschler-Solan-Zamir asks to compute the nucleolus and the prenucleolus of the three-player game where v(1,2)=1 and v(S)=0 for every other coalition S. The nucleolus is (0,0,0), since this is the only imputation. I do not recall what is the prenucleolus, but my guess is that it is (1/4,1/4,-1/2), where the maximal excess is 1/2.
The sequel is an edit that includes my other clarification:
In the prenucleolus, symmetric players obtain the same payoff, hence the prenucleolus has the form $(x,x,-2x)$ for some real number $x$.
The prenucleolus is the preimputation that minimizes the maximal excesses (in lexicographic order). Let us start by finding the $x$ that minimizes the maximal excess $max_S E(S,x)$.
For each coalition $S$ draw the graph $x \mapsto E(S,x) = v(S) - x(S)$. This graph appears below. The maximal excess for a given $x$ is the upper contour $\max\{ 1-2x,2x\}$, whose minimum is attained at $x=1/4$. Hence the prenucleolus is $(\frac{1}{4},\frac{1}{4},-\frac{1}{2})$.
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$\begingroup$ How did you arrive at the maximal excess $-\frac{1}{2}$ ? $\endgroup$ Commented Nov 10, 2019 at 16:29
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1$\begingroup$ In the prenucleolus, symmetric players obtain the same payoff, hence the prenucleolus has the form (x,x,-2x) for some real number x. One can draw (on the same graph) the excess function E(S,x) = v(S) - x(S), for every nonempty coalition S. Do it. Find the point x where the maximum of these 7 functions is maximal. You will find that the maximum is attained at x=1/4, and the maximum is 1/2. Therefore the nucleolus is (1/4,1/4,-1/2). $\endgroup$– EilonCommented Nov 11, 2019 at 20:48
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$\begingroup$ My guess was that $S=\{1,2\}$. But then the maximum is at $x=0$. Is there a better $S$ ? $\endgroup$ Commented Nov 12, 2019 at 19:54
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$\begingroup$ I do not follow what if we take $S=\{1\}$ and $x_1=-1000 000 \ , x_2=x_3=500 000$. Then $x(\{1,2,3\})=0$, BUT $E(S,x)$ is virtually unbounded. What is wrong with this objection here? $\endgroup$ Commented Nov 13, 2019 at 19:18
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