# Bayesian/causal networks: just a fast check

1. May 26, 2014

### carllacan

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 $P(v1|c1)=\frac{P(c1,v1,s1)+P(c1,v1,s2)}{P(c1,v1,s1)+P(c1,v1,s2)+P(c1,v2,s1)+P(c1,v2,s2)}$
And $P(v1|c1,s1)= P(c1,vs,s1)$

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?

2. May 27, 2014

### abitslow

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.
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?

3. May 28, 2014