Results 1 to 6 of 6

Math Help - Conditional probability of a joint independent distribution

  1. #1
    Newbie
    Joined
    Jul 2012
    From
    United States
    Posts
    6

    Conditional probability of a joint independent distribution

    Hi

    From the definition of independence I know that if A and B are independent P(A\cap B)=P(A)*P(B).

    But if I have a conditional probability of a joint probability, such as P(A\cap B|R) and I make the same assumption that A and B are independent, can I now say that P(A\cap B|R)=P(A|R)*P(B|R)?

    It seems correct to me, but I'm concerned there may be some weird interaction I'm not thinking of.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Jul 2012
    From
    United States
    Posts
    6

    Re: Conditional probability of a joint independent distribution

    For what it's worth, here's as much of a proof as I can work out

    P(A\cap B|R)=\frac{P(A\cap B\cap R)}{P(R)} by one of the definitions of a conditional probability. And since by the same definition

    P(A|R)*P(B|R)=\frac{P(A\cap R)*P(B\cap R)}{P(R)*P(R)} The only way that

    P(A\cap B|R)=P(A|R)*P(B|R) under the assumption that A and B are independent is if

    P(A\cap B\cap R) \stackrel{?}{=}\frac{P(A\cap R)*P(B\cap R)}{P(R)} under the same assumption. And you can use the same identity in the opposite direction to reduce the right side to

    P(A\cap B\cap R) \stackrel{?}{=} P(A|R)*P(B\cap R) which I'm not sure about
    Last edited by Salain; July 31st 2012 at 02:45 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2012
    From
    United States
    Posts
    6

    Re: Conditional probability of a joint independent distribution

    Ok I believe I've got the rest of it, but please check my work since I'm definitely not an expert and it's important that I get the correct result!

    Where I left off before,
    P(A\cap B\cap R) \stackrel{?}{=} P(A|R)*P(B\cap R) and since P(A\cap R)=P(R)*P(B|R) we can further expand to

    P(A\cap B\cap R) \stackrel{?}{=} P(A|R)*P(B|R)*P(R) and finally from the chain rule of probability we know that

    P(A\cap B\cap R) = P(A|B\cap R)*P(B|R)*P(R) which by our assumption of independence between A and B reduces to

    P(A\cap B\cap R) = P(A|R)*P(B|R)*P(R) , so by the chain rule the two are equivalent and therefore P(A\cap B|R)=P(A|R)*P(B|R) when A and B are independent
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,925
    Thanks
    1764
    Awards
    1

    Re: Conditional probability of a joint independent distribution

    Quote Originally Posted by Salain View Post
    From the definition of independence I know that if A and B are independent P(A\cap B)=P(A)*P(B).
    But if I have a conditional probability of a joint probability, such as P(A\cap B|R) and I make the same assumption that A and B are independent, can I now say that P(A\cap B|R)=P(A|R)*P(B|R)?
    Consider the set of digits: \{0,1,2,3,4,5,6,7,8,9\}.
    Select one digit at random. Let A be the event that the digit is prime, B be the event that the digit is less than five, and R be the event that the digit is even.
    Are A~\&~B independent? Why or why not?

    Now what about (A\cap R)~\&~(B\cap R)~?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2012
    From
    United States
    Posts
    6

    Re: Conditional probability of a joint independent distribution

    So I tried this and ended up getting a slight imbalance in the example which I don't think you intended. There are 3 even numbers and 2 odd numbers below 5, and 3 odd numbers and 2 evens above 5. So, I believe, none of the events A, B or R are independent of any of the others. If I changed the example to be 2..9 and ask <=5, then I get that A (even) and B (<=5) are independent of each other, but not of R (prime). In that case I can replicate the conditions of the original problem and ask if P(A,B|R) is equal to P(A|R)*P(B|R) and by my calculations they're not. The numbers were as close as they could be, but when you've only got 5 numbers which are not prime in a set it's simply not possible to pull any two probabilities from it, multiply, and have the number equal a single probability from it.

    That's good to know, but it doesn't show me where the proof went wrong or what the real answer is. Any insights?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2012
    From
    United States
    Posts
    6

    Re: Conditional probability of a joint independent distribution

    Believe I have it figured out. There are a few equivalent answers

    P(A\cap B|R)=P(R|A\cap B)*P(B|R) which doesn't rely on the assumption of independence between A and B, and

    P(A\cap B|R)=P(R|A\cap B)*\frac{p(A)*p(B)}{p(R)}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Joint distribution, conditional distribution
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: December 3rd 2011, 09:44 AM
  2. Replies: 3
    Last Post: August 13th 2011, 01:10 PM
  3. Conditional and Joint Probability
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: December 5th 2010, 10:41 AM
  4. Help with conditional probability of joint density functions.
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: November 18th 2009, 12:06 AM
  5. Replies: 8
    Last Post: November 15th 2009, 08:59 AM

Search Tags


/mathhelpforum @mathhelpforum