5 ms·

Don't you need some prior to conclude that? Or is this like a frequentist thing?

gizmo686 10y ago
Assume both doctors are rational and make optimal predictions given the information available to them. Let X represent the information that the first doctor, and Y represent the information that doctor second knows. We know that there are only P(C | X) = 0, meaning that you cannot have disease C because the first doctor is positive (and therefore has incontrovertible evidence) that it is not C. Similarly, P(A | Y) = 0, meaning that you cannot have disease A. As a result, the only possibility with a non-zero probability is that you have disease B.
- espeed 10y ago
  Ah, yes. It's the Dempster-Shafer theory example in the case of high conflict. The complete/original example goes as follows: "Suppose that one has two equi-reliable doctors and one doctor believes a patient has either a brain tumor, with a probability (i.e. a basic belief assignment—bba's, or mass of belief) of 0.99; or meningitis, with a probability of only 0.01. A second doctor believes the patient has a concussion, with a probability of 0.99, and believes the patient suffers from meningitis, with a probability of only 0.01. Applying Dempster’s rule to combine these two sets of masses of belief, one gets finally m(meningitis)=1 (the meningitis is diagnosed with 100 percent of confidence)." https://en.wikipedia.org/wiki/Dempster–Shafer_theory#Example_producing_counter-intuitive_results_in_case_of_high_conflict https://en.wikipedia.org/wiki/Dempster–Shafer_theory#Example...
  - defen 10y ago
    I understand the math that is going on there, but I think in the real world I would question the reliability of both of the doctors and begin to wonder if I didn't actually have Lupus. In other words, what mechanism is there for questioning the underlying sample space?
    - gizmo686 10y ago
      That is certainly a possibility. But the situation is much more reasonable if you imagine that the doctors ran different tests. For example, suppose we have a prior probability of: Tumor - .49, Concussion - .49, meningitis - .02. The first doctor performs an MRI that conclusively proves that there is no Tumor. He now conludes that there is a 96% chance of concussion, and 4% chance of meningitis. The second doctor performs a cognitive test that conclusively proves there is no concussion, but a 96% chance of tumor, and 4% chance of menigitis. If we combine these two tests, we have disproven the two post likely causes, so the less likely cause becomes reasonable. This doesn't work out quite so nicely in reality because of false negatives and (as you point out) the non 0% chance of an unspecified disease. Also, ideally you would tell the second doctor about the first test.
- fjdlwlv 10y ago
  Well sure if you add information not in the problem (P(C|X)=0) you can get the answer you want.
  - gizmo686 10y ago
    It is given that P(A | X) = .99 and P(B | X) = .01, because (by assumption) the first doctor gives you perfect predictions given the information X. Clearly P(A or B | X) = 1, so P(C | X) = 0 for any C not in A or B.

Assume both doctors are rational and make optimal predictions given the information available to them. Let X represent the information that the first doctor, and Y represent the information that doctor second knows. We know that there are only P(C | X) = 0, meaning that you cannot have disease C because the first doctor is positive (and therefore has incontrovertible evidence) that it is not C. Similarly, P(A | Y) = 0, meaning that you cannot have disease A. As a result, the only possibility with a non-zero probability is that you have disease B.

espeed 10y ago
Ah, yes. It's the Dempster-Shafer theory example in the case of high conflict. The complete/original example goes as follows: "Suppose that one has two equi-reliable doctors and one doctor believes a patient has either a brain tumor, with a probability (i.e. a basic belief assignment—bba's, or mass of belief) of 0.99; or meningitis, with a probability of only 0.01. A second doctor believes the patient has a concussion, with a probability of 0.99, and believes the patient suffers from meningitis, with a probability of only 0.01. Applying Dempster’s rule to combine these two sets of masses of belief, one gets finally m(meningitis)=1 (the meningitis is diagnosed with 100 percent of confidence)." https://en.wikipedia.org/wiki/Dempster–Shafer_theory#Example_producing_counter-intuitive_results_in_case_of_high_conflict https://en.wikipedia.org/wiki/Dempster–Shafer_theory#Example...
- defen 10y ago
  I understand the math that is going on there, but I think in the real world I would question the reliability of both of the doctors and begin to wonder if I didn't actually have Lupus. In other words, what mechanism is there for questioning the underlying sample space?
  - gizmo686 10y ago
    That is certainly a possibility. But the situation is much more reasonable if you imagine that the doctors ran different tests. For example, suppose we have a prior probability of: Tumor - .49, Concussion - .49, meningitis - .02. The first doctor performs an MRI that conclusively proves that there is no Tumor. He now conludes that there is a 96% chance of concussion, and 4% chance of meningitis. The second doctor performs a cognitive test that conclusively proves there is no concussion, but a 96% chance of tumor, and 4% chance of menigitis. If we combine these two tests, we have disproven the two post likely causes, so the less likely cause becomes reasonable. This doesn't work out quite so nicely in reality because of false negatives and (as you point out) the non 0% chance of an unspecified disease. Also, ideally you would tell the second doctor about the first test.
fjdlwlv 10y ago
Well sure if you add information not in the problem (P(C|X)=0) you can get the answer you want.
- gizmo686 10y ago
  It is given that P(A | X) = .99 and P(B | X) = .01, because (by assumption) the first doctor gives you perfect predictions given the information X. Clearly P(A or B | X) = 1, so P(C | X) = 0 for any C not in A or B.