- #1
carllacan
- 274
- 3
Hi!
I just want to check that I'm getting this right.
Suppose we have a DAG for a Bayesian network:
V S
\ /
\ /
C
Each variable is discrete and has two possible values, named v1 and v2 and similar for the others.
We know the priors P(v) and P(s) and also the conditional on C, P(c1,v1,s1), P(c1,v1,s2)...
Then [itex]P(v1|c1)=\frac{P(c1,v1,s1)+P(c1,v1,s2)}{P(c1,v1,s1)+P(c1,v1,s2)+P(c1,v2,s1)+P(c1,v2,s2)}[/itex]
And [itex]P(v1|c1,s1)= P(c1,vs,s1)[/itex]
Is that right?
Also, is the fact that P(v1|c1,s1) and P(v1|c1,s2) are different mean that the conditional probabilities of V are not independent of its non-descendants?
And therefore that is what makes this DAG not satisfy the Markov Assumption?
Thank you for your time.
I just want to check that I'm getting this right.
Suppose we have a DAG for a Bayesian network:
V S
\ /
\ /
C
Each variable is discrete and has two possible values, named v1 and v2 and similar for the others.
We know the priors P(v) and P(s) and also the conditional on C, P(c1,v1,s1), P(c1,v1,s2)...
Then [itex]P(v1|c1)=\frac{P(c1,v1,s1)+P(c1,v1,s2)}{P(c1,v1,s1)+P(c1,v1,s2)+P(c1,v2,s1)+P(c1,v2,s2)}[/itex]
And [itex]P(v1|c1,s1)= P(c1,vs,s1)[/itex]
Is that right?
Also, is the fact that P(v1|c1,s1) and P(v1|c1,s2) are different mean that the conditional probabilities of V are not independent of its non-descendants?
And therefore that is what makes this DAG not satisfy the Markov Assumption?
Thank you for your time.