Download A Methodology for Uncertainty in Knowledge-Based Systems by Kurt Weichselberger PDF

By Kurt Weichselberger

In this ebook the ensuing use of likelihood idea is proposed for dealing with uncertainty in professional platforms. it truly is proven that tools violating this advice could have risky outcomes (e.g., the Dempster-Shafer rule and the tactic utilized in MYCIN). the need of a few standards for an accurate combining of doubtful info in professional structures is proven and compatible ideas are supplied. the chance is considered that period estimates are given rather than targeted information regarding percentages. For combining details containing period estimates ideas are supplied that are invaluable in lots of cases.

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They are attributed to subsets not only to singletons. 21) to basic probability numbers. 21) generated? Using, because of "independence", the product rule, one obtains at first: P(E ~1)) =PI(Eil) " - , - " Pl(Eil) , whereby E(1) denotes an "event" which may - in analogy to the random experiment - be described as follows: The first source leads to Ell, the second to Ei2 and so on. Our knowledge that all sources contain information about the same topic, excludes all E(11, with the exception of those for which Ell .

2 4 ; P(E3) = L 3 = 0 . 2 3 ; P(E4) = L 4 = 0 . 25 :t P(E2) = U 2 = 0 . 2 4 ; P(E3) = U 3 = 0 . 3 6 ; P(E4) = L 4 = 0 . 2 8 ~ P ( E 1 ) = 0 . 1 2 : s5 P(E2) = U 2 = 0 . 2 4 ; = 0 . 1 2 : s2 P(E3) = L 3 = 0 . 2 3 ; P(E4) = U 4 = 0 . 4 1 ~ P ( E t ) P(E2) = L 2 = 0 . 1 5 ; P(E3) = U 3 = 0 . 3 6 ; P(E4) = U 4 = 0 . 08 : t So the two equations, each causing degeneracy, reduce the number of corners by two, but the same process reduces the number of failing candidates, which was 12 in the case of non-degeneracy, to eight.

1 2 + 0 . 2 4 + 0 . 41 = 1 W e shall d e s i g n a t e c o n s t e l l a t i o n s which p r o d u c e t h e s a m e corner using t h e s a m e i n d e x of s. P(E,) = L , = 0 . 1 2 ; P(E2) =L2 = 0 . 1 5 P(E1) = L , = 0 . 1 2 ; P(Eu) =152 = 0 . 2 4 ; P(E3) = ; P(E3) =Us = 0 . 3 6 ~ L3 = 0 . 2 3 ~ P(E,) : U, = 0 . 1 7 ; r ( E 2 ) : L2 = 0 . 1 5 ; P ( E a ) = L 3 = 0 . 2 3 ~ P(E1) =UI = 0 . 1 7 P(E1) = U a = 0 . 1 7 ; P(E,) = L , = 0 . 1 2 P(E4) = 0 . 41 : s2 P(E4) = 0 . 4 5 : t ; P(e2) =U2 = 0 .

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