MHB Probability calculation with Bayesian Networks

[tmt1]

Given this base data (taken from Graphical Models )$P(C) = 0.5$
$P(\lnot C) = 0.5$

$P(R | C) = 0.8$
$P(R | \lnot C) = 0.2$
$P(\lnot R | C) = 0.2$
$P(\lnot R | \lnot C) = 0.8$

$P(S | C) = 0.1$
$P(S | \lnot C) = 0.5$
$P( \lnot S | \lnot C) = 0.5$
$P( \lnot S | C) = 0.9$

$P(W | \lnot S, \lnot R) = 0.0$
$P(W | S, \lnot R) = 0.9$
$P(W | \lnot S, R) = 0.9$
$P(W | S, R) = 0.99$
$P(\lnot W | \lnot S, \lnot R) = 1.0$
$P(\lnot W | \ S, \lnot R) = 0.1$
$P(\lnot W | \lnot S, R) = 0.1$
$P(\lnot W | S, R) = 0.01$

Now, I need to calculate $P(S | W)$ or $P(S = 1 | W = 1)$ which is equal to

$\frac{P(S = 1, W = 1)}{P(W = 1)}$

or

$\frac{\sum_{c, r}^{} P(C = c, S = 1, R = r, W = 1)}{ P(W = 1)}$

I'm not sure how to begin calculating this, I think I have to use the chain rule though.

I think we need to find all the permutations of of c and r, which is these 4:

$P(R | C) = 0.8$
$P(R | \lnot C) = 0.2$
$P(\lnot R | C) = 0.2$
$P(\lnot R | \lnot C) = 0.8$

So in the first example, $P(R | C) = 0.8$, then $P(C) = 0.5$ and $P(R) = 0.8$ so for the first iteration of the sigma expression it would be

$P(C = 0.5, S = 1, R = 0.8, W = 1)$ and then I need to find the chain rule for this permutation? How would this be calculated?

 

[I like Serena]

Hi tmt,

The article explains, with the assumed independence of R and S, and with the assumed independence of W and C, that:
$$P(C,S,R,W)=P(C)\,P(S|C)\,P(R|C)\,P(W|S,R)$$
And that:
\begin{array}{lcl}
P(S=1,W=1)
&=& \sum_{c,r} P(C=c, S=1, R=r, W=1) \\
&=& P(C=0, S=1, R=0, W=1) + P(C=0, S=1, R=1, W=1) \\
&& + P(C=1, S=1, R=0, W=1) + P(C=1, S=1, R=1, W=1) \\
&=& P(\lnot C, S, \lnot R, W) + ... \\
&=& P(\lnot C)\,P(S|\lnot C)\,P(\lnot R|\lnot C)\,P(W|S,\lnot R) + ... \\
&=& 0.5 \cdot 0.5 \cdot 0.8 \cdot 0.9 + ... \\
&=& 0.2781
\end{array}