By Joachim Rosenmüller (auth.)

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**Example text**

And thus the crucial sets of m € po. t. fov are those whi ch are thrown into ex" by m, i. }) = Qov . 1, .. ('t:',d€H,:K€Hc)l (H~{l, ... / 0H by m , i . e. the sets H~{1, ... ,t1 from ~ ~{l, QW If ... 1 m i s determined by its "crucial sets" we cannot vary m in order to generate partitions 2 v = -1 (v 1 + v) of v and hence v is extreme. 2. 1. At this stage already the computation of extremes of rl is simplified. For, if we attempt to compute the extremes of ~1 by standard procedures, we are forced to deal with inequalities and equations in 2 n variables, why testing nondegeneracy requires systems of n variables.

There is (at least) one term in the representation (7) that is maximizing. Accordingly, we may partition the sequence ~ such that ~ ... 0.. '----v------' ~ \ (q) ~ •.. ~ € {q that is k(l) { k (accordingly, we assume k(l+I) - 1 ~ 1:' 0 , ~1:' ) (m = o 0 Now, while coalitions 5 are increasingly moving through the marginal value assigned to player i € 5K - 5k _1 is obviously 1:' v (\) - v (5 k -1) = m 1 (51') - - m'&"1 (5 \(-1 ) - ()(, 1:"1 = m~l 1 ex. 't'l ~2r------------/~ / that is, at this stage of the / / coalition forming process, m~l is the governing marginal "1:"1 measure and everybody joining is served according to m The only problem occurs when the process is switching from g~l to Q~l+l , the player jOining at this instant will receive some intermediate value.

I] v = 1 1. d. (X. 2. m hom t ( r-. ,IJ oy 0(... d. ) = (p,~). d. d. representation. 1. m(S) (3) Consider a set C(v) = C(m) = C(y). Obviously we have ~cx i f and only if S 6 Q~ y(S) ~ ~ (p). We may assume S s: (S 6 e) . C(y) = C(v). Now, m(S) ~ cx follows from (3). , then, by homogeneity , there is T 6 S T ~ C(m) = C(v) such that m(T) = oe. ) we have e' 1 1" y(T) > y(S) = ~ which, however, is a contradiction to (3). Thus m(S) and ~ m(S) ex. d. p{O) = q. e. d. ~ ! it follows that ! ~m{Q) m= ~. we conclude that Using 0(, =\-' • m=J • Note: It is well known that a homogeneous weighted majori ty zero sum v = 1 0 m is uniquely represented by (m.