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Negative probability measure

  1. Oct 10, 2007 #1


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    Has anybody ever heard of some mathematical application of negative probabilities ? What problems arise from allowing negative probabilities ? Of course I know it is counterintuitive, but is there any chance for a reinterpretation of probability (maybe resulting in something very different) that allows negative values ?
    Last edited: Oct 10, 2007
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  3. Oct 10, 2007 #2


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  4. Oct 10, 2007 #3

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    Negative probabilities violate one of the three axioms of probability theory as developed by Kolmogorov. A negative probability is akin to a negative magnitude. It doesn't make sense.
  5. Oct 10, 2007 #4


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    In what respect do you say it doesn't make sense ? One can always drop axioms without getting contradictions. One just gets more general realizations.
  6. Oct 12, 2007 #5
    How would negative probabilities make sense, when the theory is connected to the
    real world?
    Or if they don't, how can you make sure, that they only appear as intermediate results and not as final results of a calculation?
  7. Oct 12, 2007 #6
    They don't necessarily need to make sense when connected to any generic probability problem in the real world though. In general, when you relax axioms, the theories that you get do not apply to the same problems anymore. Occasionally you will have that the new theories will apply to some parts of the problems that the old theory could be used with (for instance, some finite fields can be used to determine properties of integer arithmetic, but not just any finite field can be used for this and you cannot talk about all of integer arithmetic with finite fields)

    As for the original question:
    Probability theory is just the theory of taking measurements in some set so that (with m being the measurement function)
    1: measurements all lie in [0,1]
    2: if [tex]A \subset B[/tex], then [tex]m(A) \leq m(B)[/tex] (Is this one necessary or can you show it from the other two? I can't remember)
    3: If A and B are disjoint sets, then [tex]m(A \cup B) = m(A) + m(B)[/tex]
    4: the measure of the entire space is 1

    I'm assuming that you just mean to throw out the 1st axiom since if you get rid of the last one as well, you could be measuring anything (E.g. the amount of money each person/group of people has with negative values meaning that the person is in debt)

    If you keep the last one and just throw out the requirement that measurements be non-negative, you still get a meaningful theory of measurements (Though you may need to add the requirement that [tex]m(\varnothing) = 0[/tex] then). The only example I can think of off the top of my head is to consider the wealth of a population and measure the fraction of the total wealth that each person/group has (with negative values being debt). This example however doesn't immediately relate nicely back to probabilities. I can't promise that any example will, though I suspect that there are some that do.
  8. Oct 15, 2007 #7
    [tex]A \subset B[/tex], then [tex]m(A) \leq m(B)[/tex] Look at B-A. Then A and B-A are disjoint.
  9. Oct 15, 2007 #8
    Thanks. I was a little tired when I wrote my last post, so I didn't see that.

    Of course I should have realized that it's also false if negative measures and measures over 1 are allowed

    Since if m(S) = 1 and there exists a subset N of S with m(N) < 0, then we must have that m(S\N) > 1. However, S\N is a subset of S
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