# How to construct a simple Bayes network?

Suppose I'm interested in the probability that event $$A$$ will occur. I'm uncertain about $$P(A)$$, but I believe that it has a uniform distribution on $$[0.1,.9]$$. Moreover, I know that event $$B$$ and $$C$$ make event $$A$$ more likely, specifically, I know the following:

$$P(B|A)=0.55\hspace{0.2cm} (\text{odds of}\hspace{0.1cm}1.222), P(B|\neg A)=0.5\hspace{0.2cm} (\text{odds of}\hspace{0.1cm}1)$$ $$P(C|A)=0.75\hspace{0.2cm} (\text{odds of}\hspace{0.1cm}3), P(C|\neg A)=0.6 \hspace{0.2cm}(\text{odds of}\hspace{0.1cm}1.5)$$

Moreover, I cannot observe events $$B$$ and $$C$$, but believe that $$P(B)=0.75$$ and $$P(C)=0.9$$. How should I update my credence regarding $$P(A)$$? I believe Bayes networks are relevant to this type of question, and I think I have an intuitive sense for how this work, but I'm not too certain.

My attempt, using the the products of Bayes-factors, is as follows. My Bayes factor for the likelihood of $$A$$ occurring vs not occurring given that $$B$$ occurs is (in odds) $$B_f(B)=\frac{1.222}{1}=1.222$$

My Bayes factor for the likelihood of $$A$$ occurring vs not occurring given that $$C$$ occurs is (in odds) $$B_f(C)=\frac{3}{1.5}=2$$

But I don't know for certain that events $$B$$ and $$C$$ occur, so I weight these bayes factors according to their probabilities, giving the following total Bayes Factor:

$$0.75*1.222+(1-0.75)+0.9*2+(1-0.9)= 3.0665$$

I add $$1-P(A)$$ and $$1-P(B)$$ because not observing $$A$$ and $$B$$ should not update my credence (I think). Since my initial distribution about $$P(A)$$ was the uniform distribution, which in odds was [0.11111111111, 9], updating this by multiplying by the Bayes factor gives [0.3018,27.59]. In probabilities this is roughly

$$[0.23,0.97].$$

• Any useful reading material would also be appreciated! – pafnuti Oct 20 '18 at 10:02