P(S|~M) vs. P(S)

["Subbarao Kambhampati" <rao@asu.edu>]

P(S|~M) vs. P(S)

      Someone
(If I remember faces right, Brandon Mechtley)  asked whether it is
really easier to assess P(S|~M) compared to P(S). I tried to venture a
weak answer in the class. 

In general, there don't seem to be
good and convincing arguments that it will always be easier to assess
P(S|~M) compared to P(S). (So I oversold my 
case :-(

The
most reasonable explanation as to why we look at P(S|~M) type
probabilities rather than P(S)  that I can offer on further reflection
is that we are interested in computing posterior probability
distribution of a variable (rather than just its maximal probable
value).  If the patient's disease can be one of 7 types. There may be a
prior distribution over these diseases, and after seeing some evidence,
the doctor wants to get the posterior *distribution* on the diseases
(not just the most likely disease but the distribution). If we are
doing, we will anyway be needing 
probabilities of type P(S|disease= di)  (note that P(S|~M) can be seen as P(S|M=false)).


I added a slide to the class notes making this point.

cheers
Rao