Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conditional probability with joint condition

  1. Oct 11, 2014 #1
    So say I have Pr(Z|X&Y)

    I'm guessing that it follows the standard Pr(A|B)=[Pr(B|A)Pr(A)]/Pr(B)

    So Pr(Z|X&Y)=[Pr(X&Y|Z)Pr(Z)]/Pr(X&Y)?


    Also, if X&Y are independent, then would I get Pr(X&Y|Z)=Pr(X|Z)Pr(Y|Z)?
     
    Last edited: Oct 11, 2014
  2. jcsd
  3. Oct 11, 2014 #2

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    What? If A=B, this would say that Pr(A)=1.
    No. A and B may be independent in general but not independent given Z. Example: toss two coins. A=coin 1 is heads. B=coin 2 is heads. Z=coins 1 and 2 match. Obviously Z forces a dependence between A and B.
     
    Last edited: Oct 11, 2014
  4. Oct 11, 2014 #3
    Thanks, that example about independence makes things make more sense now.

    Oh sorry about that. I forgot to divide by the conditional. But does Bayes Rule still work for the case I presented? Given two joint events, X&Y?
     
  5. Oct 12, 2014 #4

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    By "divide by the conditional", I assume you mean "divide the right side by P(A)". Yes, Bayes' Rule always works. The only exception to the form of Bayes' Rule that you are using is if P(A) = 0. Then both P(B|A) and the division of the right side by 0 are problems. (Some statements of Bayes' Rule talk about a partitioning of the space. If you use ever that form, make sure that condition is met.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Conditional probability with joint condition
Loading...