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336 14 Bayesian Networks

Event A Pr(A)

PF and not(Flu) and not(Cold) (0.9999)(0.99)(0.01) = 0.0099

PF and not(Flu) and (Cold) (0.9999)(0.01)(0.10) = 0.0010

(PF and Flu) and not(Cold) (0.0001)(0.99)(0.90) = 0.0001

(PF and Flu) and (Cold) (0.0001)(0.01)(0.95) = 0.0000

not(PF) and not(Flu) and not(Cold) (0.9999)(0.99)(0.99) = 0.9800

not(PF) and not(Flu) and (Cold) (0.9999)(0.01)(0.90) = 0.0090

(not(PF) and Flu) and not(Cold) (0.0001)(0.99)(0.10) = 0.0000

(not(PF) and Flu) and (Cold) (0.0001)(0.01)(0.05) = 0.0000

Table 14.1 A joint probability distribution as the result of a stochastic inference.

Consider first the case of BN inference starting with no evidence at all.

In this case, the PS of the query nodes is computed by summing the terms

of the JPD over all of the random variables that are not in the query set.

For continuous random variables, one must integrate over the probability

density. Consider the diagnostic BN in figure 14.1. Suppose one would like

to know the probability that a patient reports a fever. Integrating over the

temperature node produces this JPD (rounding all results to four decimal

places):

Next, by summing the columns, one obtains the distribution of the PF node

(except for some roundoff error):Pr(PF)=0. 011 ,Pr(not(PF)) = 0. 989.

The process of summing over random variables iscomputing the marginal dis-

tribution.

Now suppose that some evidence is available, such as that the patient is

complaining of a fever, and that the practitioner would like to know whether

the patient has influenza. This is shown in figure 14.2. The evidence pre-

sented to the BN is the fact that the random variable PF is true. The query

is the value of the Flu random variable. The evidence is asserted by con-

ditioning on the evidence. This is where Bayes’ law finally appears and is

the reason why BNs are named after Bayes. To compute this distribution,

one first selects the terms of the JPD that satisfy the evidence, compute the

marginal distribution, and finally normalize to get a PD. This last step is

equivalent to dividing by the probability of the evidence.

To see why this is equivalent to Bayes’ law, consider the case of two Boolean

random variables A and B joined by an edge from A to B. The probability

distribution of a Boolean random variable is determined by just one prob-