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

  • Thread starter Thread starter prashantgolu
  • Start date Start date
  • Tags Tags
    Bayesian Networks
prashantgolu
Messages
50
Reaction score
0
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...
 
Physics news on Phys.org
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: P(A|B,C,D)= \frac{A\cap (B\cup C\cup D)}{B\cup C\cup D}
 
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:
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.

Similar threads

Replies
72
Views
4K
Replies
7
Views
285
Replies
5
Views
1K
Replies
4
Views
1K
Replies
8
Views
2K
Replies
12
Views
6K
Replies
2
Views
2K
Back
Top