Independence assumptions and Bayesian updating
\textit{E. P. D. Pednault}, \textit{S. W. Zucker} and \textit{L. V. Muresan} [ibid. 16, 213-222 (1981; Zbl 0465.62005)] proved the following theorem: If the hypotheses \(H_ 1,H_ 2,...,H_ N\) with \(N>2\) are complete and mutually exclusive, i.e. \(\sum^{N}_{i=1}P(H_ i)=1\), and if the assumptions \(P(E_ 1,...,E_ m| H_ i)=\prod^{n}_{j=1}P(F_ j| H_ i)\) and \(P(E_ 1,...,E_ m| H_ i)=\prod^{m}_{j=1}P(E_ j| H_ i)\) are satisfied then for any sequence of evidence sentences \(E_ 1,...,E_ m\) and for every subsequence \(E_ i,...,E_ k\) and every \(H_ i\) we have \[ P(H_ i| E_ i,...,E_ k)/P(\bar H_ i| E_ i,...,E_ k)=P(H_ i)/P(\bar H_ i) \] (no updating occurs). The author proves by using a simple counterexample that the above theorem is false. The error in the proof of Pendault et al. lies in their use of an incorrect result by \textit{A. B. S. Hussain} [IEEE Trans. Comput. C-21, 318-320 (1972; Zbl 0242.68072)]. It is not clear whether the result claimed by Pendault et al. holds if in addition for all i, \(P(H_ i| E_ 1,...,E_ k)\neq 0\).
scientific article; zbMATH DE number 3921702