Editing Dempster–Shafer theory (section)

== Dempster–Shafer as a generalisation of Bayesian theory  ==
As in Dempster–Shafer theory, a Bayesian belief function  <math>\operatorname{bel}: 2^X \rightarrow [0,1] \,\!</math> has the properties <math>\operatorname{bel}(\emptyset) = 0</math> and <math>\operatorname{bel}(X) = 1</math>. The third condition, however, is subsumed by, but relaxed in DS theory:<ref name="SH76"/>{{rp|p. 19}}
:<math>\text{If } A \cap B = \emptyset, \text{ then} \operatorname{bel}(A \cup B) = \operatorname{bel}(A) + \operatorname{bel} (B).</math>

Either of the following conditions implies the Bayesian special case of the DS theory:<ref name="SH76"/>{{rp|p. 37,45}}
* <math>\operatorname{bel}(A) + \operatorname{bel}(\bar{A}) = 1 \text{ for all } A \subseteq X.</math>
* For finite ''X'', all focal elements of the belief function are singletons.

As an example of how the two approaches differ, a Bayesian could model the color of a car as a probability distribution over (red, green, blue), assigning one number to each color. Dempster–Shafer would assign numbers to each of (red, green, blue, (red or green), (red or blue), (green or blue), (red or green or blue)). These numbers do not have to be coherent; for example, Bel(red)+Bel(green) does not have to equal Bel(red or green).

Thus, Bayes' conditional probability can be considered as a special case of Dempster's rule of combination.<ref name="SH76"/>{{rp|p. 19f.}} However, it lacks many (if not most) of the properties that make Bayes' rule intuitively desirable, leading some to argue that it cannot be considered a generalization in any meaningful sense.<ref>Dezert J., Tchamova A., Han D., Tacnet J.-M., [https://www.researchgate.net/profile/Florentin_Smarandache/publication/273062785_Advances_and_Applications_of_DSmT_for_Information_Fusion_Collected_Works_Volume_4/links/54f9b7940cf29a9fbd7c508f.pdf#page=196 Why Dempster's fusion rule is not a generalization of Bayes fusion rule], Proc. Of Fusion 2013 Int. Conference on Information Fusion, Istanbul, Turkey, July 9–12, 2013</ref> For example, DS theory violates the requirements for [[Cox's theorem]], which implies that it cannot be considered a coherent (contradiction-free) generalization of [[classical logic]]—specifically, DS theory violates the requirement that a statement be either true or false (but not both). As a result, DS theory is subject to the [[Dutch Book]] argument, implying that any agent using DS theory would agree to a series of bets that result in a guaranteed loss. (Note: Some of the criticism was later found to be erroneous and inappropriate.<ref name="xu2025"/>).