Dempster-Shafer Belief

1A frame of discernment

A frame of discernment \Theta is the finite set of mutually exclusive, exhaustive answers the case is allowed to have. The campus lab alarm uses \Theta = \{\mathrm{Fire},\, \mathrm{Fault}\}: either there is a fire, or the sensor is faulting, and those are the only two stories we will entertain. A subset of \Theta is a claim that is still compatible with more than one story: \{\mathrm{Fire}\} is a committed claim, \Theta itself is 'I am not saying which'.

The frame is a modelling decision, not a fact about the lab. If a third story — a drill — is possible and you left it out, every number this lesson computes is a number about the wrong set of answers. Bayes needs the same honesty about its sample space; Dempster-Shafer does not relax that, it changes what you are allowed to do with subsets.

2Mass, belief, plausibility

A basic probability assignment — a mass function m — spreads one unit of credit over the non-empty subsets of \Theta. Sensor A reports m_1(\{\mathrm{Fire}\}) = 0.6 and m_1(\Theta) = 0.4: it puts 0.6 on fire and leaves 0.4 uncommitted. It does not put 0.4 on Fault. That leftover is ignorance, not a vote for the rival.

Belief \mathrm{Bel}(A) is the mass already committed to A and its subsets — credit that cannot escape A. Plausibility \mathrm{Pl}(A) is one minus the belief in the complement — credit that could still move into A. For Sensor A, \mathrm{Bel}(\{\mathrm{Fire}\}) = 0.6 and \mathrm{Pl}(\{\mathrm{Fire}\}) = 1 - \mathrm{Bel}(\{\mathrm{Fault}\}) = 1 - 0 = 1. The interval [0.6,\, 1] is the claim: at least 0.6, and up to everything, could still be fire. Bayes would have forced the 0.4 onto Fault; this calculus refuses that force.

Figure. Belief is the sage segment (0.6). Plausibility is the full bar (1.0). The warm gap is the 0.4 sitting on Θ — ignorance, not Fault. Segment values are the masses from the ledger.

3Combining two agreeing sensors

Dempster's rule combines two mass functions. For each pair of subsets, one from each source, put the product of their masses on the intersection. Sum every pair that lands on the same intersection. If some pairs land on the empty set, their total K is conflict; divide the remaining masses by 1 - K so they add to one again. Pairs that land on \Theta stay uncommitted.

Sensor B reports m_2(\{\mathrm{Fire}\}) = 0.7, m_2(\Theta) = 0.3. Every non-empty intersection of a pair from m_1 and m_2 is either \{\mathrm{Fire}\} or \Theta — nothing lands on empty, so K = 0. The combined mass on fire is 0.6\times 0.7 + 0.6\times 0.3 + 0.4\times 0.7 = 0.88, and the leftover on \Theta is 0.4\times 0.3 = 0.12. Two sensors that both refuse to vote for Fault, at different strengths, leave less ignorance than either did alone.

Figure. Mass on {Fire} before and after the combination. The two sources (0.6, 0.7) become 0.88, the ledger total. The leftover 0.12 on Θ is not drawn as a fourth bar — it is 1 − 0.88.

4Conflict, and when not to combine

Now let Sensor B vote the other way: m_2(\{\mathrm{Fault}\}) = 0.7, m_2(\Theta) = 0.3. The pair \{\mathrm{Fire}\} \cap \{\mathrm{Fault}\} is empty and carries mass 0.6 \times 0.7 = 0.42. That 0.42 is K. The surviving masses are renormalised by 1 - K = 0.58: fire gets 0.18 / 0.58 = 9/29, fault gets 14/29, and \Theta keeps 6/29. The combination did not average 0.6 with 0.7; it threw away the contradictory 0.42 and re-scaled what was left.

When K is close to 1, two sources barely agree on any non-empty intersection and the renormaliser is a large number multiplying a small remainder. That is a warning, not a sharper answer. Do not combine a smoke detector with a sensor that only ever reports Fault and call the result a probability. Reach for Bayes when you do have likelihoods and a prior on singletons — the earlier uncertainty lessons. Reach for this calculus when the honest report is a committed subset plus leftover ignorance, and drop the combination when K says the sources do not share a frame.

Figure. m1({Fire})=0.6 meets m2({Fault})=0.7. The empty intersection carries K = 0.6 × 0.7 = 0.42 and is discarded. Survivors renormalised by 0.58: Fire 9/29, Fault 14/29, Θ 6/29. That 0.42 is a warning, not a sharper answer.