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Bayesian Conditionalization

  1. Jan 22, 2012 #1

    I'm stuck at the following problem on Baysesian conditionalization:

    Prove that conditionalizing once on (A or B) is equal to conditionalizing first on A and then on B. As a hint, I was asked to define H(x) = P(X given A).

    Any help would be appreciated.
  2. jcsd
  3. Jan 22, 2012 #2

    Stephen Tashi

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    Did your text materials say that or is "conditionalizing" some terminology that you invented? Is the problem to prove that P(C | A and B) = P( (C|A)|B ) ? (i.e. "and" instead of "or")
  4. Jan 22, 2012 #3
    Sorry. I did mean "and' not "or". Also, conditionalizing is what is written on the homework.
  5. Jan 23, 2012 #4

    Stephen Tashi

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    Since the usual term is "conditioning" rather than "conditionalizing", it's possible that your instructor or text has a specific technique in mind and has invented this term for it. In that case I'm not enough of a mind reader to know what the problem wants!

    Can you give an example from your class notes where conditionalizing was used?
    Perhaps the problem expects you do some kind of mechanical manipulations that avoid detailed thought.


    Some other thoughts (which may not be what the problem has in mind):

    The problem of showing P( (C|A)|B) = P(C|A and B) is actually hard to reason about in a logical fashion. (In an advanced course that uses measure theory, it is very hard ! Let's assume you are in an introductory course.) The notation P(C) is usually used to denote the probability of an event, which is a set in some "space of outcomes". However when we write P(C|A) , the "C|A" has no convenient interpretation as an event in the "space of outcomes"where the events C and A take place.

    P(C|A) is usually defined to be P(C and A)/ P(A), so its a P-of-something that isn't an event. at least it is not an event in the original space of outcomes.

    If you want to think about C|A as an event, you must think about a different space of outcomes. The possible space of outcomes for C|A is the set A, not the original space. Within the space A, "C|A" is the set consisting of the events in C that are also in A. As a set, C|A is the set (A and C), but we can't say P(C|A) = P(C and A) since, as events, C|A and (C and A) denote events in different spaces of outcomes.

    We still face the task of interpreting notation like (C|A)|B. One line of reasoning is that the C|A changes the space of outcomes to A. Then the "| B" changes the space of outcomes that part of B which is in space of outcomes A, i.e. the space of outcomes is (A and B). So (C|A)|B denotes the same thing as C|(A and B). Hence P( (C|A)|B) = P(C|(A andB)).

    The hint to let H(X) = P(X gven A) may be some way to detour around having to interpret notation like P( (X|A) | B). I'm not sure why this is necessary.
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