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In order to be able to use a basic possibility function as a Body of Evidence in the Dempster-Shafer Theory of Evidence, it is needed to transform the function to its Möbius representation.

There is a transformation for discrete possibility functions which is based on sorting the possibility values. I couldn't find a continuous counterpart to the transformation. Are Möbius functions inherently discrete?


Definition of Möbius transform from "G. Klir, Uncertainty and Information: Foundations of Generalized Information Theory, Hoboken N.J.: Wiley-Interscience, 2006":

Every set function

$\mu: \mathcal{P}(X) \rightarrow R$

where X is a finite set, can be uniquely represented by another set function

$^{\mu}m: \mathcal{P}(X) \rightarrow R $

via the formula

$^{\mu}m(A) = \sum_{B | B \subseteq A} (-1)^{|A-B|} \mu(B)$

for all $A \in \mathcal{P}(X)$. This formula is called Möbius transform and function $^{\mu}m(A)$ is called a Möbius representation of $\mu$.

(Another reference: http://en.wikipedia.org/wiki/Fuzzy_measure_theory)


Edit: A possibility function is a function defined on elements of $X$ and assign a number between 0 and 1 to each member:

$r: X \rightarrow [0,1], \quad \max_{x \in X} r(x) = 1$

To measure possibility of subsets of $X$, the possibility measure is defined on $\mathcal{C}$, a family of subsets of $X$ with a good algebraic structure (e.g. a $\sigma$-algebra):

  • $\operatorname{Pos}: \mathcal{C} \rightarrow [0,1]$
  • $\operatorname{Pos}(\emptyset) = 0,$
  • $\operatorname{Pos}(X) = 1, $
  • $\forall A,B \in \mathcal{C} : \operatorname{Pos}(A \cup B) = \max (\operatorname{Pos}(A), \operatorname{Pos}(B)) $

This is the set function which I want to transform. With this definition, we can define $\operatorname{Pos}$ based on possibility function:

$\forall A \in \mathcal{C}: \operatorname{Pos}(A) = \sup_{x \in A}(r(x))$

In order to transform the possibility measure $\operatorname{Pos}$, $P(X)$ in the definition of $\mu$ must be changed so that: $\mu: \mathcal{C} \rightarrow [0,1]$.

My problem: In case $X=[0,1]$, the subsets in $\mathcal{C}$ are not countable. So, the summation, and $|A-B|$ in the definition of the transformation are meaningless!

In conclusion, I want $X$ to be uncountable (but with finite bounds).

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I don't think functional-analysis is the right tag, but I am not sure what it should be replaced with –  Yemon Choi Jan 4 '12 at 16:09
    
This question is in the context of Possibility and Evidence theories. The set function μ is the general form of the possibility and belief measures in these theories. I couldn't find these theories among tags. But added another two tags which are relevant. –  SadeghD Jan 4 '12 at 18:12
    
(Not set-theory.) –  Andres Caicedo Jan 4 '12 at 22:03

1 Answer 1

up vote 1 down vote accepted

(edit) Okay, so as far as i can see you want to find a replacement for the mobius transform, but for a $\sigma$-algebra. In fact I'm going to guess that your $\sigma$-algebra is the measurable sets in the unit interval, based on what you've said.

The most general setting I know of in which you can define a Möbius function is a locally finite, partially ordered set (see, for example, http://en.wikipedia.org/wiki/Incidence_algebra). So it sounds like you're out of luck. The measurable sets definitely don't form a locally finite poset.

However, I really don't think you've asked the right question yet. You probably would get better answers than mine if you frame your question in terms of measure theory, rather than Möbius inversion. For instance, the wikipedia article seems to imply that I should think of Möbius inversion as "analagous to differentiation", and convolution with the zeta function as "analagous to integration". I don't really find this too helpful, but that's what it says. Maybe you're looking for some kind of derivative? Just a guess.

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I edited my question, and clarified that I want X to be uncountable, but finite. –  SadeghD Jan 7 '12 at 8:48
    
That doesn't make sense I'm afraid... –  Benjamin Young Jan 7 '12 at 11:46
    
Yes, being uncountable contradicts with being finite. I correct myself. X is infinite but because of being uncountable, not because of having infinite bounds. Besides that, the question looks to have an answer. Is Mobius representation of a set function restricted to finite sets? –  SadeghD Jan 7 '12 at 12:56

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