# 2 bayesian networks

1. Jul 10, 2011

### prashantgolu

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...?

2. Jul 10, 2011

### SW VandeCarr

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: Jul 10, 2011
3. Jul 10, 2011

### prashantgolu

By ',' i mean intersection
P(x,y)=p(x|y)*p(y)
like this...

4. Jul 10, 2011

### SW VandeCarr

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: Jul 11, 2011