tag:blogger.com,1999:blog-3863347052747918849.post3467794326030554431..comments2023-06-25T03:02:40.664-07:00Comments on Blog for Fall 2007 ASU CSE 471/598 Introduction to AI: [Thinking cap questions on bayes networks]Subbarao Kambhampatihttp://www.blogger.com/profile/08449853328445416609noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-3863347052747918849.post-47708175292396861052007-10-17T09:27:00.000-07:002007-10-17T09:27:00.000-07:000.1 We would need 2^5 entries for the Joint distri...0.1 We would need 2^5 entries for the Joint distribution. But as we can have any value between 0 and 1 for any of them there could be infinite entries satisfying it? <BR/>0.2 X||Y|Z where X, Y, Z are subsets of variables gives us exponential different conditional Independence Assertions. Not sure of the exact value though as if we start with 2^5 * 2^5 * 2^5 ; it makes little sense to say that the same set is independent of itself given itself? Thus maybe we need to consider smaller sets for Y and Z as 2^5 * 2^4 * 2^3?<BR/>1. Yes, we can check if any of Independence property of the network does not hold for example:<BR/>P(MaryCalls|Alarm ) != P(MaryCalls| Alarm, Earthquake, Burglary) these probabilities can be calculated from the joint distribution and verified.<BR/>2. If the conditional properties for a particular case say Alarm add up to more than one you would have inconsistent joint probability distribution. I.e. if he says 0.5 for all his answers and you end up trying to fill the table for Alarm with all four entries as 0.5 then it cannot be allowed as the sum would have to be < 1.<BR/>4. In choosing B1 where CIA(B1) is superset, we have more conditional Independence i.e. the graph connections are sparse than the actual case. We can use that to simplify the problem.However B2 where CIA(B2) is a subset, has more restrictions, but we can't be sure that the added edges/connections benefit us to draw any other conclusions. <BR/>-InaInahttps://www.blogger.com/profile/11976076041015667969noreply@blogger.com