26.12.2013 Views

AI - a Guide to Intelligent Systems.pdf - Member of EEPIS

AI - a Guide to Intelligent Systems.pdf - Member of EEPIS

AI - a Guide to Intelligent Systems.pdf - Member of EEPIS

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

BIAS OF THE BAYESIAN METHOD<br />

73<br />

Domain experts do not deal easily with conditional probabilities and quite<br />

<strong>of</strong>ten deny the very existence <strong>of</strong> the hidden implicit probability (0.3 in our<br />

example).<br />

In our case, we would use available statistical information and empirical<br />

studies <strong>to</strong> derive the following two rules:<br />

IF the starter is bad<br />

THEN the symp<strong>to</strong>m is ‘odd noises’ {with probability 0.85}<br />

IF the starter is bad<br />

THEN the symp<strong>to</strong>m is not ‘odd noises’ {with probability 0.15}<br />

To use the Bayesian rule, we still need the prior probability, the probability<br />

that the starter is bad if the car does not start. Here we need an expert judgement.<br />

Suppose, the expert supplies us the value <strong>of</strong> 5 per cent. Now we can apply the<br />

Bayesian rule (3.18) <strong>to</strong> obtain<br />

pðstarter is bad j odd noisesÞ ¼<br />

0:85 0:05<br />

0:85 0:05 þ 0:15 0:95 ¼ 0:23<br />

The number obtained is significantly lower than the expert’s estimate <strong>of</strong> 0.7<br />

given at the beginning <strong>of</strong> this section.<br />

Why this inconsistency? Did the expert make a mistake?<br />

The most obvious reason for the inconsistency is that the expert made different<br />

assumptions when assessing the conditional and prior probabilities. We may<br />

attempt <strong>to</strong> investigate it by working backwards from the posterior probability<br />

pðstarter is bad j odd noisesÞ <strong>to</strong> the prior probability pðstarter is badÞ. In our case,<br />

we can assume that<br />

pðstarter is goodÞ ¼1<br />

pðstarter is badÞ<br />

From Eq. (3.18) we obtain:<br />

pðHÞ ¼<br />

pðHjEÞpðEj:HÞ<br />

pðHjEÞpðEj:HÞþpðEjHÞ½1<br />

pðHjEÞŠ<br />

where:<br />

pðHÞ ¼pðstarter is badÞ;<br />

pðHjEÞ ¼pðstarter is bad j odd noisesÞ;<br />

pðEjHÞ ¼pðodd noises j starter is badÞ;<br />

pðEj:HÞ ¼pðodd noises j starter is goodÞ.<br />

If we now take the value <strong>of</strong> 0.7, pðstarter is badjodd noisesÞ, provided by the<br />

expert as the correct one, the prior probability pðstarter is badÞ would have

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!