Bayesian network and distributions

• Sep 24th 2012, 08:26 AM
PeterPan2009
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? :)
• Sep 25th 2012, 12:01 AM
chiro
Re: Bayesian network and distributions
Hey PeterPan2009.

Just to get things starting, can you show us what you've tried?
• Sep 25th 2012, 02:40 AM
PeterPan2009
Re: Bayesian network and distributions
Quote:

Originally Posted by chiro
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):

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

And then marginalise over some variable?
• Sep 25th 2012, 07:30 AM
PeterPan2009
Re: Bayesian network and distributions
This is what I have so far (I don't know if this is correct):

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

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

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

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

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

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

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

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

Now I don't know what to do?
• Sep 25th 2012, 08:12 PM
chiro
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)?
• Sep 25th 2012, 11:40 PM
PeterPan2009
Re: Bayesian network and distributions
Updated:

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

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

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

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

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

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

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

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

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)?
• Sep 27th 2012, 12:06 AM
PeterPan2009
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.