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Math Help - Bayesian network and distributions

  1. #1
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    Bayesian network and distributions

    Hi,

    I'm new when it comes to Bayesion networks and would be glad if someone could give me some hints on a problem.

    I have the following probability distributions:

    P(A)
    P(B)
    P(X|C)
    P(Y|C)
    P(C|A,B)

    I would like to find P(A|X=true, Y= false). The domain of all variables are: true, false.


    I have no idea how to start on this? Any tips?
    Last edited by PeterPan2009; September 24th 2012 at 09:09 AM.
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  2. #2
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    Re: Bayesian network and distributions

    Hey PeterPan2009.

    Just to get things starting, can you show us what you've tried?
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  3. #3
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    Re: Bayesian network and distributions

    Quote Originally Posted by chiro View Post
    Hey PeterPan2009.
    Just to get things starting, can you show us what you've tried?
    Hi chiro!

    Well, to be honest, I'm not sure how to start. That is, I'm not sure about the strategy to use to simplify the problem - in terms of the probabilities and distributions I know. The first step may be to use Bayes' rule like this (I'm not sure this is valid though):


    P(A|X=true, Y=false) = \frac{P(X=true,Y=false|A)\times P(A)}{P(X=true, Y=false)}

    And then marginalise over some variable?
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  4. #4
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    Re: Bayesian network and distributions

    This is what I have so far (I don't know if this is correct):

    P(A|X=true, Y=false)

    Using Bayes' rule:
    = \frac{P(X=true, Y=false|A) P(A)}{P(X=true, Y=false)}

    = \alpha P(X=true, Y=false|A) P(A)


    Using marginalisation on C:
    = \alpha \bigg[ \sum_{c \in Domain(C)} P(X=true, Y=false, C=c|A) \bigg] P(A)

    = \alpha \bigg[ \sum_{c \in Domain(C)} P(X=true, Y=false |C=c, A) P(C=c|A) \bigg] P(A)


    Using P(X,Y|C,A)P(C|A)=P(X,Y|C)P(C|A):
    = \alpha \bigg[ \sum_{c \in Domain(C)} P(X=true, Y=false, C=c) P(C=c|A) \bigg] P(A)


    Using marginalisation on B:
    = \alpha \bigg[ \sum_{c \in Domain(C)} P(X=true, Y=false, C=c) \bigg[ \sum_{b \in Domain(B)} P(C=c, B=b|A)\bigg] \bigg] P(A)

    = \alpha \bigg[ \sum_{c \in Domain(C)} P(X=true, Y=false, C=c) \bigg[ \sum_{b \in Domain(B)} P(C=c|B=b, A) P(B=b|A))\bigg] \bigg] P(A)

    Now I don't know what to do?
    Last edited by PeterPan2009; September 25th 2012 at 07:37 AM.
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  5. #5
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    Re: Bayesian network and distributions

    Can you sum any of the variables out (since summing out all probabilities of a particular random variable sums out the variable entirely)?
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  6. #6
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    Re: Bayesian network and distributions

    Updated:

    P(A|X=true, Y=false)

    Using Bayes' rule:
    = \frac{P(X=true, Y=false|A) P(A)}{P(X=true, Y=false)}

    = \alpha P(X=true, Y=false|A) P(A)


    Using marginalisation on C:
    = \alpha \bigg[ \sum_{c \in Domain(C)} P(X=true, Y=false, C=c|A) \bigg] P(A)

    = \alpha \bigg[ \sum_{c \in Domain(C)} P(X=true, Y=false |C=c, A) P(C=c|A) \bigg] P(A)


    Using P(X,Y|C,A)P(C|A)=P(X,Y|C)P(C|A):
    = \alpha \bigg[ \sum_{c \in Domain(C)} P(X=true, Y=false | C=c) P(C=c|A) \bigg] P(A)


    Using marginalisation on B:
    = \alpha \bigg[ \sum_{c \in Domain(C)} P(X=true, Y=false | C=c) \bigg[ \sum_{b \in Domain(B)} P(C=c, B=b|A)\bigg] \bigg] P(A)

    = \alpha \bigg[ \sum_{c \in Domain(C)} P(X=true, Y=false | C=c) \bigg[ \sum_{b \in Domain(B)} P(C=c|B=b, A) P(B=b|A))\bigg] \bigg] P(A)

    Hm, how do I sum out any variables. I have the distribution for P(C=c|B=b, A), but I dont have e.g. P(X=true, Y=false|C=c) only P(X=true|C=c) and P(y=false|C=c)?
    Last edited by PeterPan2009; September 25th 2012 at 11:42 PM.
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  7. #7
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    Re: Bayesian network and distributions

    I think I figured this out. The important thing is to be aware of is that nodes in a network can be conditionally independent.
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