Bayesian Networks: P(a,b,c,d) Calculation

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    Bayesian Networks
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Discussion Overview

The discussion revolves around the calculation of joint probabilities in Bayesian networks, specifically focusing on the expression P(a,b,c,d) given certain dependencies among the variables. Participants explore the implications of conditional independence and the correct formulation of probabilities in the context of two Bayesian networks.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that P(a,b,c,d) can be expressed as P(a)*P(b|a)*P(c|a,b)*P(d|a,b,c) and seeks confirmation of this formulation.
  • Another participant questions the validity of the proposed simplification to P(a)*P(b|a)*P(c)*P(d|c), noting that the left term is not a conditional probability.
  • A clarification is made regarding the notation, where a participant states that ',' refers to intersection and provides a basic probability relationship P(x,y)=P(x|y)*P(y).
  • A subsequent reply acknowledges the correctness of the first line of the proposed equation but expresses uncertainty about the assumptions of conditional independence that would justify the simplification in the second line.
  • The same participant questions the assumption that P(c|a,b)=P(c) and seeks clarification on how the dependency of d on c influences this relationship.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the simplifications proposed. There is a clear disagreement regarding the assumptions of conditional independence and the implications for the joint probability expression.

Contextual Notes

Participants express uncertainty about the assumptions required for the simplifications, particularly regarding the independence of the two Bayesian networks and the relationships among the variables involved.

prashantgolu
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suppose i have 2 bayesian networks...
a->b and c->d ie b depends on a and d depends on b
now P(a,b,c,d)=P(a)*P(b|a)*P(c|a,b)*P(d|a,b,c)
=P(a)*P(b|a)*P(c)*P(d|c)

Am i right...?
please reply..i need it as quickly as possible...
 
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prashantgolu said:
suppose i have 2 bayesian networks...
a->b and c->d ie b depends on a and d depends on b
now P(a,b,c,d)=P(a)*P(b|a)*P(c|a,b)*P(d|a,b,c)
=P(a)*P(b|a)*P(c)*P(d|c)

Am i right...?
please reply..i need it as quickly as possible...

It's not clear what you are trying to say since the term on the left is not a conditional probability.

Note: [tex]P(A|B,C,D)= \frac{A\cap (B\cup C\cup D)}{B\cup C\cup D}[/tex]
 
Last edited:
By ',' i mean intersection
P(x,y)=p(x|y)*p(y)
like this...
 
prashantgolu said:
suppose i have 2 bayesian networks...
a->b and c->d ie b depends on a and d depends on b
now P(a,b,c,d)=P(a)*P(b|a)*P(c|a,b)*P(d|a,b,c)
=P(a)*P(b|a)*P(c)*P(d|c)

Am i right...?
please reply..i need it as quickly as possible...

OK, then the first line looks correct. However I don't know how the assumptions you're making regarding conditional independence allow you to get your simplification on the second line. If the two networks (a,b), (c,d) are conditionally independent, then I believe your simplifications are correct if they are taken individually. However, I'm not sure why you would combine them since they have no common variables.

EDIT: In other words, am I supposed to know that P(c|a,b)=P(c)? How does the fact that d depends on c tell me that? (I assume that you made a mistake when you stated d depends on b since your formula indicates d depends on c.)
 
Last edited:

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