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Conditional Probability Proof

  1. Jul 1, 2014 #1
    I'd like some help understanding a proof, from http://www.statlect.com/cndprb1.htm. Properties are introduced, which a conditional probability ought to have:
    1) Must satisfy properties of probability measures:
    a) for any event E, 0≤P(E)≤1;
    b) P(Ω)=1;
    c) Sigma-additivity: Let {E1, E2, ... En, ....} be a sequence of events, where i≠j implies Ei and Ej are mutually exclusive, then P([itex]\bigcup_{n=1}^∞ E_n[/itex]) = [itex]\sum_{n=1}^∞ P(E_n)[/itex].​
    2)P(I|I)=1
    3) If [itex]E \subseteq I[/itex] and [itex]F \subseteq I[/itex], and P(I) is greater than 0, then [itex]\frac {P(E|I)}{P(F|I)} = \frac {P(E)}{P(F)}[/itex].​

    Then a proof is given for the proposition: Whenever P(I) is positive, P(E|I) satisfies the four properties above if and only if P(E|I) = [itex]\frac {P(E \cap I)}{P(I)}[/itex].

    I'm having a hard time following the proof of the "only if" part. That is, if P(E|I) satisfies the four properties above, then P(E|I) = [itex]\frac {P(E \cap I)}{P(I)}[/itex].

    Here's a quote:
    Now we prove the 'only if' part. We prove it by contradiction. Suppose there exists another conditional probability [itex]\bar{P}[/itex] that satifies the four properties. Then There Exists an even E such that:
    [itex]\bar{P}(E|I) ≠ P(E|I)[/itex]

    It can not be noted that [itex]E \subseteq I[/itex], otherwise we would have:
    [itex]\frac {\bar{P}(E|I)}{\bar{P}(I|I)} = \frac {\bar{P}(E|I)}1 ≠ \frac {P(E|I)}1 = \frac {P(E \cap I)}{P(I)} = \frac {P(E)}{P(I)} [/itex]

    *which would be a contradiction, since if [itex]\bar{P}[/itex] was a conditional probability, it would satisfy:
    [itex]\frac {\bar{P}(E|I)}{\bar{P}(I|I)} = \frac {P(E)}{P(I)}[/itex]​

    The proof by contradiction, seems more like a proof of the uniqueness of a conditional probability.

    Anyways, I'm not really seeing the statement, *. How is it that [itex]\frac {\bar{P}(E|I)}{\bar{P}(I|I)} = \frac {P(E)}{P(I)}[/itex]?
     
  2. jcsd
  3. Jul 1, 2014 #2

    mfb

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    For "only if", you assume that ##\bar P(E|I)## satisfies the four properties. In particular, it satisfies the third one with F=I.
     
  4. Jul 1, 2014 #3
    OH....

    So Property 3 asserts that: If [itex]E \subseteq I[/itex] and [itex]F \subseteq I[/itex], where P(F) is positive, then for any probability A, [itex]\frac {A(E|I)}{A(F|I)} = \frac {P(E)}{P(F)}[/itex]?

    BTW, the other property was this: If [itex]E \subseteq I^C[/itex] then P(E|I)=0.
     
  5. Jul 2, 2014 #4

    mfb

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    Oh, I thought the RHS would have bars as well, they are hard to see. Maybe a typo there and it should be ##\bar P##.
     
  6. Jul 3, 2014 #5
    Yes, I agree, but...

    If [itex] \frac {\bar{P}(E|I)}{\bar{P}(I|I)} = \frac {\bar{P}(E)}{\bar{P}(I)} [/itex], then there is no contradiction.

    So I'm looking for a justification for the statement *.

    I'm fairly certain that I quoted the author correctly, but if there is doubt, you can always visit statlect.com. The notes on Conditional probability are in the Fundamentals of probability theory section.
     
    Last edited: Jul 3, 2014
  7. Jul 9, 2014 #6

    I think the notation for probabilities and conditional probabilities is not the same and therefore there is no typo.
     
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