Interaction in Choquet inegral model

[ ∑
=
=
K⊆N\A
[
µ(N) +
[ ∑
a∑
p=0,
p even
a∑
a∑
p=2,
p even
∑
L⊆A,
l=a−p
∑
L⊆A,
l=a−p
∑
K⊆N\A p=0, L⊆A,
p even l=a−p−1
[ ∑
≥ µ(N) −
K⊆N\A
]
µ(K ∪ L)
[ ∑
−
µ((N \ A) ∪ L) +
]
µ(K ∪ L)
a∑
p=0,
p even
∑
L⊆A,
l=a−p−1
K⊆N\A
]
µ(K ∪ L)
Therefore, (n − a + 1)!k! × I µ A ≥ µ(N) − [
[ ∑
We still have to prove that µ(N) −
∑
a∑
p=0,
p even
KN\A
K⊆N\A
∑
∑
L⊆A,
l=a−p−1
a∑
p=0,
p even
K⊆N\A
a∑
p=0,
p even
∑
L⊆A,
l=a−p
a∑
p=0,
p even
∑
L⊆A,
l=a−p−1
]
µ(K ∪ L)
]
µ(K ∪ L) -
∑
L⊆A,
l=a−p−1
]
µ(K ∪ L) .
]
µ(K ∪ L) > 0.
Let K ⊆ N \ A, p ∈ {0, . . . , a} even number and L ⊆ A with l = a − p − 1. We have
L A, therefore K ∪ L N, then by hypothesis not(a K∪L T C I∪M a N ). Thus a N ∈ B m
and there exists l K∪L ∈ {0, 1, . . . , m − 1} such that a K∪L ∈ B lK∪L .
Then µ(K ∪ L) = 1 α (2n)l K∪L
or µ(K ∪ L) = 0, so in both cases we have µ(K ∪ L) ≤
1
α (2n)l K∪L
.
∑ a∑ ∑
Therefore,
µ(K ∪ L) ≤
∑ a∑ ∑ 1
α (2n)l K∪L
≤
∑
K⊆N\A
K⊆N\A
a∑
p=0,
p even
∑
p=0,
p even
L⊆A,
l=a−p−1
L⊆A,
l=a−p−1
K⊆N\A
p=0,
p even
1
α (2n)m−1 = 1 α (2)n−1 (2n) m−1 .
L⊆A,
l=a−p−1
Moreover, we have 1 α (2)n−1 (2n) m−1 = 2n−1 (2n) m−1
(2n) n (2n) m = 2 n−1 (2n) m−1
2(2) n−1 (n) n (2n)(2n) m−1
= 1 < 1 = µ(N).
4nn+1 [ ∑
Hence we have µ(N) >
[ ∑
µ(N) −
K⊆N\A
a∑
p=0,
p even
∑
L⊆A,
l=a−p−1
K⊆N\A
a∑
p=0,
p even
∑
L⊆A,
l=a−p−1
]
µ(K ∪ L) > 0.
]
µ(K ∪ L) , i.e.,
We can therefore conclude that (n − a + 1)! × I µ A > 0, i.e., Iµ A > 0.
The following lemma gives a simplified condition of previous condition
not(a S T C I∪M a N ) for all S N.
Lemma 2. The previous condition [not(a S T C I∪M a N ) for all S N] is equivalent at
[
not(aN\{i} T C I∪M a N ) for all i ∈ N ] .
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