The maximal possibility degree is affected to the maximal utility

The second polynomial case of possibilistic Choquet integrals (denoted by Max-Class) con- cerns possibilistic lotteries where the maximal possibility degree namely 1 is affected to the maximal utility in the lottery. This class is defined as follows:

Definition 4.3 Let U = {u1, . . . , un} be the set of possible utilities where umax is the maximal utility in a possibilistic lottery L such that umax ≤ un. In the case of Max-Class, each lottery L ∈ L is as follows: L = hλ1/u1, . . . , 1/umaxi.

Proposition 4.9 DT-OPT-ChΠ (resp. DT-OPT-ChN with α = 1) is polynomial in the case of Max-Class.

Proof. [Proof of Proposition 4.9]

In what follows, we present the proof in numerical setting (the same principle is valid for the ordinal setting).

Necessity-based Choquet integrals

We will consider the case where α = 1

• Case 1: Let us consider three lotteries L, L0 _{and L” having the same maximal utility} un:

L = hλ1/u1, . . . , 1/uni, L0 = hλ01/u1, . . . , 1/uni and L” = hλ”1/u1, . . . , 1/uni. Then,

ChN(L) = u1+ · · · + (un− un−1)(1 − max(λ1, . . . , λn−1)) ChN(L0) = u1+ · · · + (un− un−1)(1 − max(λ01], . . . , λ0n−1))

L1 = hα/L, β/L”i and L2 = hα/L0, β/L” are two compound lotteries, we have: L1 = hmax(λ1, βλ”1)/u1, . . . , 1/uni

L2 = hmax(λ01, βλ”1)/u1, . . . , 1/uni ChN(L1) = u1+ . . .

+ (un− un−1)(1 − max(max(λ1, βλ”1), . . . , max(λn−1, βλ”n−1))) ChN(L2) = u1+ . . .

+ (un− un−1)(1 − max(max(λ01, βλ”1), . . . , max(λ0n−1, βλ”n−1))) ⇒ If ChN(L) > ChN(L0) then ChN(L1) ≥ ChN(L2)

• Case 2: L and L0 have the same maximal utility denoted by umax and L00 has as maximal utility ui.

– If umax> ui

L = hλ1/u1, . . . , λi/ui, . . . , 1/umaxi L0 = hλ0₁/u1, . . . , λ0i/ui, . . . , 1/umaxi and L” = hλ”1/u1, . . . , λ”i/ui, . . . , 1/umaxi. Then,

ChN(L) = u1 + · · · + (ui − ui−1)(1 − max(λ1, . . . , λi−1)) + · · · + (umax − umax−1)(1 − max(λ1, . . . , λmax−1))

ChN(L0) = u1 + · · · + (ui − ui−1)(1 − max(λ01, . . . , λ0i−1)) + · · · + (umax − umax−1)(1 − max(λ01, . . . , λ0max−1))

L” = hλ”1/u1, . . . , 1/uii.

L1 = hα/L, β/L”i and L2 = hα/L0, β/L” are two compound lotteries, we have: L1 = hmax(λ1, βλ”1)/u1, . . . , 1/ui, . . . , 1/umaxi

L2 = hmax(λ01, βλ”1)/u1, . . . , 1/ui, . . . , 1/umaxi ChN(L1) = u1+ . . .

+ (ui− ui−1)(1 − max(max(λ1, βλ”1), . . . , max(L[ui−1], βL00[ui−1])))

ChN(L2) = u1+ . . .

+ (un− un−1)(1 − max(max(λ01, βλ”1), . . . , max(λ0i−1, βλ”i−1))) ⇒ If ChN(L) > ChN(L0) then ChN(L1) ≥ ChN(L2). – If umax< ui

L = hλ1/u1, . . . , 1/umaxi, L0 = hλ0₁/u1, . . . , 1/umaxi and

L00= hλ”1/u1, . . . , λ”max/umax, . . . , 1/uii. Then,

ChN(L) = u1+ · · · + (umax− umax−1)(1 − max(λ1, . . . , λmax−1)) ChN(L0) = u1+ · · · + (umax− umax−1)(1 − max(λ01, . . . , λ0max−1))

L1 = hα/L, β/L”i and L2 = hα/L0, β/L” are two compound lotteries, we have: L1 = hmax(λ1, βλ”1)/u1, . . . , 1/umax, . . . , 1/uii

L2 = hmax(λ01, βλ”1)/u1, . . . , 1/umax, . . . , 1/uii

ChN(L1) = u1+ · · · + (umax− umax−1)(1 − max(max(λ1, βλ”1), . . . , max(λmax−1, βλ”max−1)))

ChN(L2) = u1+ · · · + (umax− umax−1)(1 − max(max(λ01, βλ”1), . . . , max(λ0_max−1], βλ”max−1)))

⇒ If Ch_N(L) > ChN(L0) then ChN(L1) ≥ ChN(L2)

• Case 3: L and L” have the same maximal utility denoted by u_max and L0 has as maximal utility ui.

– If umax> ui

L = hλ1/u1, . . . λi/ui, . . . , 1/umaxi, L0 = hλ0₁/u1, . . . , 1/ui> and

L” = hλ”1/u1, . . . , λ”i/ui, . . . , 1/umaxi. Then,

ChN(L) = u1 + · · · + (ui − ui−1)(1 − max(λ1, . . . , λi−1)) + · · · + (umax − umax−1)(1 − max(λ1, . . . , λmax−1))

ChN(L0) = u1+ · · · + (ui− ui−1)(1 − max(λ01, . . . , λ0i−1)).

L1 = hα/L, β/L”i and L2 = hα/L0, β/L” are two compound lotteries, we have: L1 = hmax(λ1, βλ”1)/u1, . . . , max(λi, βλ”i)/ui, . . . , 1/umaxi

L2 = hmax(λ01, βλ”1)/u1, . . . , 1/uii ChN(L1) = u1+ . . .

+ (ui− ui−1)(1 − max(max(λ1, βλ”1), . . . , max(λi−1, βλi−1))) + · · · + (umax− umax−1)(1 − max(max(λ1, βλ”1),

. . . , max(λmax−1, βλ”max−1))) ChN(L2) = u1+ . . .

+ (ui− ui−1)(1 − max(max(λ01, βλ”1), . . . , max(λ”i−1, βλ”i−1))) ⇒ If Ch_N(L) > ChN(L0) then ChN(L1) ≥ ChN(L2).

– If umax< ui

L = hλ1/u1, . . . , 1/umaxi,

L0 = hλ0₁/u1, . . . , λ0max/umax, . . . , 1/uii and L” = hλ”1/u1, . . . , 1/umaxi. Then,

ChN(L) = u1+ · · · + (umax− umax−1)(1 − max(λ1, . . . , λmax−1))

ChN(L0) = u1+ · · · + (umax− umax−1)(1 − max(λ01, . . . , λ0max−1)) + · · · + (ui− ui−1)(1 − max(λ01, . . . , λ0i−1))

L1 = hα/L, β/L”i and L2 = hα/L0, β/L” are two compound lotteries, we have: L1 = hmax(λ1, βλ”1)/u1, . . . , 1/umaxi

L2 = hmax(λ01, βλ”1)/u1, . . . , 1/umax, . . . , 1/uii

ChN(L1) = u1+ · · · + (umax− umax−1)(1 − max(max(λ1, βλ”1), . . . , max(λmax−1, βλ”max−1)))

ChN(L2) = u1+ · · · + (umax− umax−1)(1 − max(max(λ01, βλ”1), . . . , max(λ0_max1, βλ”max−1)))

• Case 4: L has a maximal utility denoted by umax and L0 and L00 have the same maximal utility ui. – If umax> ui L = hλ1/u1, . . . , λi/ui, . . . , 1/umaxi, L0 = hλ0₁/u1, . . . , 1/uiiand L”= hλ”1/u1, . . . , 1/uii. Then,

ChN(L) = u1 + · · · + (ui − ui−1)(1 − max(λ1, . . . , λi−1)) + · · · + (umax − umax−1)(1 − max(λ1, . . . , λmax−1))

ChN(L0) = u1+ · · · + (ui−ui−1)(1 − max(λ01, . . . , λ0i−1)) ⇒ ChN(L) > ChN(L0)

L1 = hα/L, β/L”i and L2 = hα/L0, β/L” are two compound lotteries, we have: L1 = hmax(λ1, βλ”1)/u1, . . . , max(λi, β)/ui, . . . , 1/umaxi

L2 = hmax(λ01, βλ”1)/u1, . . . , 1/ui > ChN(L1) = u1+ . . .

+ (ui− ui−1)(1 − max(max(λ1, βλ”1), . . . , max(λi−1, β) + (umax− umax−1)(1 − max(max(λ1, βλ”1), . . . , max(λmax−1, βλ”max−1)))

ChN(L2) = u1+ . . .

+ (ui− ui−1)(1 − max(max(λ01, βλ”1), . . . , max(λ0_i−1, βλ”i−1))) ⇒ Ch_N(L1) > ChN(L2).

– If umax< ui

L = hλ1/u1, . . . , 1/umaxi,

L0 = hλ0₁/u1, . . . , λ0max/umax, . . . , 1/uii and L00= hλ”1/u1, . . . , λ”max/umax, . . . , 1/uii. Then,

ChN(L) = u1+ · · · + (umax− umax−1)(1 − max(λ1, . . . , λmax−1))

ChN(L0) = u1+ · · · + (umax− umax−1)(1 − max(λ01, . . . , λ0max−1)) + · · · + (ui− ui−1)(1 − max(λ01, . . . , λ0i−1))

ChN(L0) > ChN(L)

L1 = hα/L, β/L”i and L2 = hα/L0, β/L” are two compound lotteries, we have: L1 = hmax(λ1, βλ”1)/u1, . . . , 1/umaxi

L2 = hmax(λ01, βλ”1)/u1, . . . , max(λ0max, βλ”max), . . . , 1/uii ChN(L1) = u1+ · · · + (umax− umax−1)(1 − max(max(λ1, βλ”1), . . . , max(λmax−1, βλ”max−1)))

ChN(L2) = u1+ · · · + (umax− umax−1)(1 − max(max(λ01, βλ”1),

. . . , max(λ0_max−1, βλ”max−1) + · · · + (ui − ui−1)(1 − max(max(λ0₁, βλ”1), . . . , max(λ0i−1, βλ”i−1)))

⇒ ChN(L2) > ChN(L1).

Possibility-based Choquet integrals

Let three possibilistic lotteries L = hλ1/u1, . . . , 1/uni, L0 = hλ0₁/u1, . . . , 1/uni and L” = hλ”1/u1, . . . , 1/uni. Then,

ChΠ(L) = u1+ (u2− u1) ∗ 1 + · · · + (un− un−1) ∗ 1 and ChΠ(L0) = u1+ (u2− u1) ∗ 1 + · · · + (un− un−1) ∗ 1. ⇒ Ch_Π(L) = ChΠ(L0). There are two cases:

• Case 1: If α = 1

L1 = h1/L, β/L”i and L2= h1/L0, β/L” are two compound lotteries, we have: L1 = hmax(λ1, βλ”1)/u1, . . . , max(1, β)/uni, and

L2 = hmax(λ01, βλ”1)/u1, . . . , max(1, β)/uni. ⇒ ChΠ(L1) = ChΠ(L2).

• Case 2: If β = 1: similar to the first case.