Transferable belief model

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Consider the following classical problem of information fusion. A patient has an illness that can be caused by three different factors A, B and C. Doctor 1 says that the patient's illness is very likely to be caused by A (very likely, meaning probability p = 0.95), but B is also possible but not likely (p = 0.05). Doctor 2 says that the cause is very likely C (p = 0.95), but B is also possible but not likely (p = 0.05). How is one to make one's own opinion from this ?

Bayesian updating the first opinion with the second (or the other way round) implies certainty that the cause is B. Dempster's rule of combination lead to the same result. This can be seen as paradoxical, since although the two doctors point at different causes A and C, they both agree that B is not likely. (For this reason the standard Bayesian approach is to adopt Cromwell's rule and avoid the use of 0 or 1 as prior probabilities.)

The transferable belief model (TBM) is an elaboration on the Dempster-Shafer theory of evidence developed by the late Dr. Philippe Smets, based on the intuition that in the situation above, the result should allocate most of the belief weight to the empty set (i.e. neither A, B, nor C). Technically, this would be done by using the TBM conjunction rule for non-interactive sources of information, which is the same as Dempster's rule of combination without renormalization.

While most other theories adhere to the axiom the probability (or belief mass) of the empty set is always zero, there is another intuitive reason to drop this axiom: the open-world assumption. It applies when the frame of reference is not exhaustive, so there are reason to believe that an event not described in this frame of reference will occur. For example, when tossing a coin one usually assumes that Head or Tail will occur, so that $\scriptstyle \Pr(\mathrm{Head})+\Pr(\mathrm{Tail})=1$ . The open-world assumption is that the coin can be stolen in mid-air, disappear, break apart or otherwise fall sideway so that neither Head nor Tail occurs, so that $\scriptstyle \Pr(\varnothing)+\Pr(\mathrm{Head})+\Pr(\mathrm{Tail})=1$ .