The PIC is nothing but the dual of the normalized Shannon entropy and thus is actually unit less. **P IC**(P) takes its values in [0, 1]. **P IC**(P) is maximum, i.e. P ICmax = 1 with any deterministic probability and it is minimum, i.e. P ICmin = 0, with the uniform probability over the frame Θ. The simple relationships between H(P) and **P IC**(P) are **P IC**(P) =
1 − (H(P)/Hmax) and H(P) = Hmax · (1 − **P IC**(P)).

VII. EXAMPLES AND COMPARISONS ON A 2D FRAME
Due to the space limitation constraint, all details of derivations are voluntarily omitted here but they will appear in [8].

In this section, we work with the 2D frame Θ = {**A, B**}.

A. Example 1 (Shafer's model and a general source)
Since one assumes Shafer's model, GΘ = 2Θ =
{∅**, A, B, A** ∪ B}. The non-Bayesian quantitative belief mass is given in Table I. Table II presents the results of the different mappings and their PIC sorted by increasing order.

One sees that **DSmP**ǫ→0 provides same result as **P rBel** and P IC(**DSmP**ǫ→0) is greater than the PIC values obtained with P rNP L, BetP, CuzzP, **P rPl** and **P raPl**.

![0_image_0.png](0_image_0.png)

TABLE I
QUANTITATIVE INPUTS FOR EXAMPLE 1

| A              | B      | P IC(.)   |        |
|----------------|--------|-----------|--------|
| P rNP l(.)     | 0.5625 | 0.4375    | 0.0113 |
| BetP(.)        | 0.6000 | 0.4000    | 0.0291 |
| CuzzP(.)       | 0.6000 | 0.4000    | 0.0291 |
| P rP l(.)      | 0.6375 | 0.3625    | 0.0553 |
| P raP l(.)     | 0.6375 | 0.3625    | 0.0553 |
| P rHyb(.)      | 0.6825 | 0.3175    | 0.0984 |
| DSmPǫ=0.001(.) | 0.7492 | 0.2508    | 0.1875 |
| P rBel(.)      | 0.7500 | 0.2500    | 0.1887 |
| DSmPǫ=0(.)     | 0.7500 | 0.2500    | 0.1887 |
| TABLE II       |        |           |        |

B. Example 2 (Shafer's model and the totally ignorant source)
Let's assume Shafer's model and the vacuous bba characterizing the totally ignorant source, i.e. m(A ∪ B) = 1. It can be verified that all mappings coincide with the uniform probability measure over singletons of Θ, except **P rBel** which is mathematically not defined in that case. This result can be easily proved for any size of the frame Θ with |Θ| > 2.

C. Example 3 (Shafer's model and a probabilistic source)
Let's assume Shafer's model and let's see what happens when applying all the transformations on a probabilistic source10 which commits a belief mass only to singletons of 2 Θ, i.e. a Bayesian mass [4]. It is intuitively expected that all transformations are idempotent when dealing with probabilistic sources, since actually there is no reason/need 10This has obviously no practical interest since the source already provides a probability measure, nevertheless this is very interesting to see the theoretical behavior of the transformations in such case.

to modify m(.) (the input mass) to obtain a new subjective probability measure since Bel(.) associated with m(.) is already a probability measure. So if we consider for example the uniform Bayesian mass defined by mu(A) = mu(B) =
1/2, it is very easy to verify in this case, that almost all transformations coincide with the (probabilistic) input mass as expected, so that the idempotency property is satisfied. Only Cuzzolin's transformation fails to satisfy this property because in **CuzzP**(.) formula (9) one gets 0/0 indeterminacy since all ∆(.) = 0 in (9). This remark is valid whatever the dimension of the frame Θ is, and for any Bayesian mass (not only for uniform belief mass).

## D. Example 4 (Shafer'S Model And Non-Bayesian Mass)

Let's assume Shafer's model and the non-Bayesian mass
(more precisely the simple support mass) given in Table III.

We summarize in Table IV, the results obtained with all transformations. One sees that P IC(**DSmP**ǫ→0) is maximum among all PIC values. **P rBel**(.) does not work correctly since it can not have a division by zero. We use NaN acronym here standing for *Not a Number*11; even overcoming it12, P rBel does not do a fair redistribution of the ignorance m(A ∪ B) = 0.6 because B does not receive anything from the mass 0.6, although B is involved in the ignorance A ∪ B. All m(A ∪ B) = 0.6 was unfairly redistributed to A only.

![0_image_1.png](0_image_1.png)

TABLE III
QUANTITATIVE INPUTS FOR EXAMPLE 4
A B **P IC**(.)
P rBel(.) 1 NaN NaN P rNP l(.) 0.6250 0.3750 0.0455 BetP(.) 0.7000 0.3000 0.1187 CuzzP(.) 0.7000 0.3000 0.1187 P rP l(.) 0.7750 0.2250 0.2308 P raP l(.) 0.7750 0.2250 0.2308 P rHyb(.) 0.8650 0.1350 0.4291 DSmPǫ=0.001(.) 0.9985 0.0015 0.9838 DSmPǫ=0(.) 1 0 1 TABLE IV
RESULTS FOR EXAMPLE 4.

The best result is an *adequate probability*, not the biggest PIC in this case. This is because P(B) deserves to receive some mass from m(A ∪ B), so the most correct result is done by DSmPǫ=0.001 in Table IV (of course we can choose any other very small positive value for ǫ if we want). Always when a singleton whose mass is zero, but it is involved in an ignorance whose mass is not zero, then ǫ (in **DSmP** formula
(11)) should be different from zero.

11we could also use the standard "N/A" standing for "does not apply".

12since the *direct* derivation of **P rBel**(B) cannot be done from the formula (5) because of the undefined form 0/0, we could however force it to **P rBel**(B) = 0 since **P rBel**(B) = 1 − **P rBel**(A) = 1 − 1 = 0, and consequently we indirectly take P IC(**P rBel**) = 1.