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Are proofs needed for definitions? Conditional Probability

  1. Sep 9, 2012 #1
    My probability class has me wondering about pure math questions now. We started with the axioms and are slowly building up the theory. Everything was fine but then a definition of Conditional Probability [tex] P[A|B] = \frac{P[AB]}{P} [/tex] appeared and it's just not sitting right with me. I know that formula works because in simple problems I can usually see the answer. I'm just not seeing how it works or why a proof isn't needed..
     
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  3. Sep 9, 2012 #2

    Stephen Tashi

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    Axiomatic development and having intuition are two different topics. If you take an "Platonic" view of mathematics, things like "conditional probability" or "integers" exist in some sort of reality and the formal definitions are merely attempts to summarize their properties. However, this Platonic outlook is not the way modern mathematics is done. From the modern point of view the things defined ARE what the definitions say they are and nothing more can be asserted about their properties unless it is proven.

    You seem to have a Platonic idea about what conditional probability is and you are apparently asking for some sort of proof that the formal definition agrees with this Platonic view. Informal explanations can be given about why the formal definition does (or does not) agree with the commonly intuitive idea of a concept, but it isn't meaningful to ask for "proof" of a definition.

    If you have one precisely formulated definition for a concept, you can ask whether the definition is logically equivalent to another precisely formulated definition. But I think what your asking for is an intuitive explanation of why the formal definition agrees with your ideas about what conditional probability should be. Can you state what ideas you have that the formal definition seems to contradict?
     
  4. Sep 10, 2012 #3
    I haven't found any issues with the formula other than I don't understand it intuitively yet. In the book we're using in class, there has been a proof for every theorem but definitions don't have proofs. I'm just confused why definitions don't need a proof.
     
  5. Sep 10, 2012 #4

    Stephen Tashi

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    From the point of view of proof and logic, definitions are arbitrary. They are simply conventions. So there is nothing to prove about a definition. For example, if a book has a passage that says "Let X be a random variable" there is no reason the book needs to prove that X is a random variable. It is simply adopting the convention that "X" denotes a random variable.

    Definitions can be right or wrong from a sociological and cultural point of view. Someone might criticize a definition by saying "That's not how most peopel deifne a ...". However, that type of controversy is subjective so you can't use a proof to settle such matters.

    Occasionally books make definitions that only make sense if certain claims are already proven. For example if we say "The number 0 is defined to be the number such that for any real number x, x + 0 = x", this use of the word "the" could be taken to mean that there is one and only one number 0. However that fact that there is only one number with the properties of 0 does need a proof. In rigorous mathematical books, a definition of "a" zero is defined and then a proof is given that there is only one zero, which justifies speaking of "the" number zero.

    Likewise, defining the conditiona probability [itex] P(A|B) [/itex] to be [itex] \frac{ P(A \cap B)}{P(B)} [/itex] describes the thing defined as unique ("the") and a "probability", but the definition itself is not a proof that the quantity [itex] \frac{P(A \cap B)}{P(B)} [/itex] is unique or that it is a number in the interval [0,1] as a probability ought to be. So that aspect of the definition does need some proof. However, I think those things are easy to establish.
     
  6. Sep 10, 2012 #5

    chiro

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    The easiest way to think about this is that we are looking at one probability relative to another.

    Looking at one relative to another means we are constraining our space to B which means that we look at our B is our new "universal space" for a conditional probability.

    So when you find a non-conditional probability you are doing this with respect to some pre-defined universal set with all the properties but when you do it with respect to some arbitrary set (which is a subset of the universal set), then it means that the intersection is not P(A and U) = P(A) but instead P(A and B) = whatever that is.

    That's all you are doing: you are looking at a probability with a new "universe" where your universe now becomes an arbitrary subset of the true universe as opposed to just the universe.

    The universal set plugged in gives you an unconditional probability since:

    P(A|U) = P(A and U)/P(U) = P(A)/1 [Since A is a subset of U) = P(A).
     
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