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can anyone help me on this?

I would like to show that the vector reward:

$(v(1,2)+v(1,3)-v(1,2,3), v(1,2,3)-v(1,3), v(1,2,3)-v(1,2))$ is an imputation of the core

Knowing that:

$v(1,2)+v(1,3)+v(2,3)\le2(1,2,3)$

and

$v(1,2)+v(1,3)-v(1,2,3)\ge v(1)$

I know that it needs to respect the group and individual rationality but I do not know how to start. Can anyone give an hint?

Thanks

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An imputation $\mathbf{x}$ is in the core of a three person game, if it satisfies the following set of inequalities

$$v(12) + v(13) +v (23) \le 2 v(123) \qquad v(1) + v(23) \le v(123) \qquad v(2) + v(13) \le v(123) \qquad v(3) + v(12) \le v(123).$$

Now, the following imputation is given

$$\mathbf{x}=(x_{1},x_{2},x_{3})=(v(12)+v(13)-v(123),v(123)-v(13),v(123)-v(12)). $$

We have to check that these inequalities are satisfied by $\mathbf{x}$ to conclude that this imputation belongs to the core

Under the mentioned assumptions and that the game is super-additive, we obtain now

$$ v(2) \le x_{2} = v(123) - v(13) \qquad v(3) \le x_{3} = v(123) - v(12). $$

Moreover, it holds that

$$ \sum_{i \in N} x_{i} = v(12) + v(13) - v(123) + v(123) -v(13)+v(123)-v(12) = v(123).$$

Similar, we get

$$x(12) = x_{1}+x_{2} = v(12) \qquad x(13) = x_{1}+x_{3} = v(13) \qquad x(23) = x_{2}+x_{3} = v(23). $$

This shows that the imputation $\mathbf{x}$ is in the core.

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  • $\begingroup$ Makes sense, I though that for the individual rationality I needed to show for each $x_i$ and not for pairs. But now it makes more sense. thank you very much for your help. Additionally, you showed that both properties ( group and individual rationality hold) Should I also show that the vector is in the core? $\endgroup$ – user290335 Dec 12 '17 at 14:04
  • $\begingroup$ I think you showed that in the first part, Am I correct? $\endgroup$ – user290335 Dec 12 '17 at 14:06
  • $\begingroup$ This was already done. You take $\mathbf{x}$ and substitute everything in the inequalities. If you don't get a contradiction, then it is in the core. If an inequality is violated. then you have established that an imputation is not in core. It seems to me that you are confused by the fact that the imputation $\mathbf{x}$ is defined by the coalitional values. $\endgroup$ – Holger I. Meinhardt Dec 12 '17 at 14:11

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