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Bayesian/causal networks: just a fast check

by carllacan
Tags: bayesian or causal, check, networks
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carllacan
#1
May26-14, 03:45 PM
P: 172
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.
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abitslow
#2
May27-14, 01:01 PM
P: 140
Sorry, I am not knowledgeable in this area. Its been a while since you posted this, so I thought I'd respond even though I quite possibly won't be very helpful.
Your "diagram"
V S
\ /
\ /
C
is unclear -- did you mean V→C←S ? (you can't use spaces to format a diagram in many forums)
I also don't know what P(c1,vs,s1) is. What does vs mean??
Finally, when you say that P(v1|c1, s1) is different from P(v1|c1,s2) do you mean logically (in the general case) or that you know P(s1) ≠ P(s2) [≠ 0.5 ] ? (yes, generally its different, but its actual value is indeterminate here.)
The other problem that I have is that I don't understand your diagram (again, its probably my own ignorance).
Lets say I am correct in assuming your diagram is V→C←S... (I hope we agree that a Markov Assumption is about future states.) Does this mean that V(t=i) influences C(t=i+1)? or that V(t=i) influences C(t=i) ? (where t is sequential time intervals) (I assume here that V and S are independent variables with some unknown T/F probability). Obviously, only if the past state(s) don't influence the future state(s) do we have a Markov Process.
I hope I haven't totally wasted your time. If your diagram is timeless, then it says nothing about Markov Processes. If it is describing flow of states, then by definition any X→Y means Y is dependent on what X was, and hence isn't Markovian... I think?
carllacan
#3
May28-14, 04:29 PM
P: 172
Thank you for your answer.

Yes, I meant V→C←S, I thought it was going to look as intended.

With P(c1,vs,s1) I meant P(c1,v1,s1), its a typo (I can't edit it). I meant the joint probability of C = c1, V = v1 and S = s1.

My diagram doesn't involve time its a causal network. The Markov Condition requires, according to my textbook (Neapolitan's Learning Bayesian Networks)
Each variable is conditionally independent of the set of all its nondescendants given the set of all its parents.
In my example V depends on S, which is a non-descendant, and I just want to check if this is why the Markov Condition is not true here.


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