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Unconditional probabilities do not exist? Discuss!

  1. May 28, 2012 #1
    I am somewhat annoyed by the term "unconditional probability", in that all probabilities are indeed conditional on filtration (an information set if you would like), without which, a probability is inadequately defined as though plucked out of thin air based on no logical information.

    I make the above conjecture based on the definition of a probability space (Ω,[itex]\scriptstyle \mathcal{F}[/itex],P), where [itex]\scriptstyle \mathcal{F}[/itex] is the filtration.

    What I'm noticing however is that in explaining the conditional probability P(A|B), most undergraduate texts refer to P(A) or P(B) as the unconditional probability, where the right terminology should be [in the case of P(A)] the probability of A unconditional on B. This dumbing down of terminology serves to help non-statistics students in understanding probability theory, but it gives them an incorrect philosophy of what a probability is.

    I'm opened to being convinced that an unconditional probability exists though.

    What are your thoughts?
     
    Last edited: May 28, 2012
  2. jcsd
  3. May 28, 2012 #2
    Imagine a set with 1 woman and 9 men, I say that P("being a man")=9/10 and I say it is conditioned to nothing.

    If you say otherwise, you have to tell us what is that probability conditioned to.
     
  4. May 28, 2012 #3
    In that example, the filtration is the information that there are 1 woman and 9 men.

    So in actual fact, P("being a man" | [itex]\scriptstyle \mathcal{F}[/itex] = {"1 woman and 9 men"}) = 9/10, or under a simplified notation, P("being a man") = 9/10.

    If the filtration is changed to "2 women and 8 men", then the above P("being a man") ≠ 9/10. Hence P("being a man") is conditional on its filtration [itex]\scriptstyle \mathcal{F}[/itex] = {"1 woman and 9 men"}, as it changes when the filtration changes.

    The point I wish to make is that all probabilities are conditional on its filtration. Lazy notation has caused us to ignore this omnipresent conditioning in probability, as such it is incorrect to call a probability an unconditional probability, it is only unconditional on a specific event (but remains conditional to others, such as its filtration).
     
    Last edited: May 28, 2012
  5. May 28, 2012 #4
    I think this is going to be a bit about semantics but, anyway, when I understand that something is conditioned to something else, I expect that something else to be a random variable.

    You're treating the filtration like if it was a random variable itself, but I see it as a property of the set; I do not expect one man to turn into a woman spontaneously in the set. But again, if you want to say that, I guess is just about semantics, but to me is a tautological exercise; like saying number 5 is not unconditioned because it is conditioned by its fiveness.
     
    Last edited: May 28, 2012
  6. May 28, 2012 #5
    But isn't it reasonable to suggest that the death of one man in the set would then change the P("being a man")? Or perhaps if we're measuring the probability between islands of populations and find a variation in the P("being a man")?

    In this sense, the filtration is an uncertain condition and should be treated as a random variable.
     
  7. May 28, 2012 #6
    Yeah, but you're just changing the sample space Ω and, by doing so, changing the example I gave you. If you want I can add that the men and women in my sample are immortal, or simply consider a universe where you only have 1 black ball and 9 white balls... and they are eternal too.

    My point is that talking about unconditioned probabilities makes sense, I don't think this "dumbs down" the subject for students or that we are just lazy as you claim.
     
  8. May 28, 2012 #7

    chiro

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    Just use B=U for P(A|B) and you get the normal value and valid interpretation of P(A), where U is the universal state-space, and P(U) = 1.
     
  9. May 28, 2012 #8
    And what is U conditioned to?
     
  10. May 28, 2012 #9

    chiro

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    You can condition it to anything you want: it won't change a thing.

    Since P(U|B) = P(U AND B)/P(B) = P(B)/P(B) = 1 since B is a subset of U.
     
  11. May 28, 2012 #10
    So basically:

    P(B) = P(B|U) = P(B|U^B) = P(B|B)

    So B is conditioned by itself like 5 is conditioned by its "fiveness"

    I still think unconditioned probabilities makes a lot of sense.
     
  12. May 28, 2012 #11

    chiro

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    But all a conditioned probability is, is simply conditioning something with respect a specific set. Seeing in this way helps people make sense what conditioning actually means.

    Conditioning something on U means the interpretation that is no constrain on the conditioning, but conditioning on some subset means that it is in the context of that particular set.
     
  13. May 28, 2012 #12

    Hurkyl

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    Filtration? [itex]\scriptstyle \mathcal{F}[/itex] is a σ-algebra on which the measure P is defined, and the only reason we have such a thing is because for many spaces, it is logically contradictory for P to be defined on all subsets of Ω and still satisfy the measure axioms.




    It may be the case that you have a reasonable idea in your head, but if you do, you're trying to shoehorn it in someplace where it doesn't belong.
     
  14. Jun 1, 2012 #13

    Jano L.

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    I would say that in everyday use, all probabilities are conditional, we just do not state all the conditions explicitly. They are assumed. When asked, we can express the statement carefully with conditioning, to make it compatible with the probability based on sets.

    Example:

    All members of a group of 10 persons draw sticks.

    The probability that the person who draws the shortest stick is a man, /provided we know in the group there are 9 men and 1 woman/, is 9/10.

    The discussed probability is conditioned by the information in /.../.
     
  15. Jun 1, 2012 #14
    Aren't all problems in probability given in the context of a problem? Doesn't this imply that all probability has to be conditional (the condition being the context of the problem)? Would noting that condition for every problem be, quite frankly, a pain in the ***? Then again I am not a mathematician so perhaps you are speaking about something more specific to the field.
     
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