Balanced vs. totally balanced cooperative games

Question

I would like to understand what here on the page 89 is the difference of totally balanced game vs. balanced one. More precisely, I would like to see a game which is balanced but not totally balanced. Where $U$ ranges through for totally balanced game? Also, is this guess correct: game is balanced iff it has a nonempty core?

Answer 1

The definition and notation of the above reference is not quite understandable on a quick glance, so I restate here the definitions and major statements of these concepts.

We say that a game $v \in \mathcal{G}^{n}$ is balanced if for every balanced collection $\mathcal{B}$ with weights $\{w_{S}\}_{S \in \mathcal{B}}$, we obtain \begin{equation} \label{eq:bal_propA} \sum_{S \in \mathcal{B}}\,w_{S}\,v(S) \le v(N). \end{equation} A game $v \in \mathcal{G}^{n}$ is called to be totally balanced if all subgames $\langle\,S,v_{S}\,\rangle, \emptyset \neq S \subseteq N$ are balanced. Notice that a subgame $\langle\,S,v_{S}\,\rangle$ is specified by $v_{S}(T)=v(T)$ for all $T \subseteq S$. Bondareva (1963) and Shapley (1967) proved independently from one another that a game $v \in \mathcal{G}^{n}$ is balanced iff the core of the game is non-empty, that is, $\mathcal{C}(N,v) \neq \emptyset$. Implying that a game is totally balanced iff any subgame has a non-empty core.

The class of market and flow games are totally balanced games. Moreover, the class of wedge games are totally balanced too, which are defined by

$v=\bigwedge\{{\boldsymbol\lambda}^{\,1},\ldots,{\boldsymbol\lambda}^{\,r}\}$,

whereas the ${\boldsymbol\lambda}^{\,k}$'s are assignment vectors. Hence, games that do not belong to these classes of games and have a non-empty core cannot be expected to be totally balanced.