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Interval-type and affine arithmetic-type techniques for handling uncertainty in expert systems

机译:间隔类型和仿射算术类型技术,用于处理专家系统中的不确定性

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Expert knowledge consists of statements S-j (facts and rules). The facts and rules are often only true with some probability. For example, if we are interested in oil, we should look at seismic data. If in 90% of the cases, the seismic data were indeed helpful in locating oil, then we can say that if we are interested in oil, then with probability 90% it is helpful to look at the seismic data. In more formal terms, we can say that the implication "if oil then seismic" holds with probability 90%. Another example: a bank A trusts a client B, so if we trust the bank A, we should trust B too; if statistically this trust was justified in 99% of the cases, we can conclude that the corresponding implication holds with probability 99%. If a query Q is deducible from facts and rules, what is the resulting probability p(Q) in Q? We can describe the truth of Q as a propositional formula F in terms of S-j, i.e., as a combination of statements S-j linked by operators like &, v, and -; computing p(Q) exactly is NP-hard, so heuristics are needed. Traditionally, expert systems use technique similar to straightforward interval computations: we parse F and replace each computation step with corresponding probability operation. Problem: at each step, we ignore the dependence between the intermediate results F-j; hence intervals are too wide. Example: the estimate for P(A v - A) is not 1. Solution: similar to affine arithmetic, besides P(F-j), we also compute P(F-j&F-i) (or P(F-ji &•center dot center dot& F-jd)), and on each step, use all combinations of 1 such probabilities to get new estimates. Results: e.g., P(A - A) is estimated as 1. (c) 2005 Elsevier B.V. All rights reserved.
机译:专家知识由陈述S-j(事实和规则)组成。事实和规则通常只有几率才适用。例如,如果我们对石油感兴趣,则应查看地震数据。如果在90%的情况下,地震数据确实有助于寻找石油,那么可以说,如果我们对石油感兴趣,那么概率为90%的地震数据是有帮助的。用更正式的术语来说,我们可以说“如果先石油,然后地震”的含义以90%的概率成立。另一个例子:银行A信任客户B,因此,如果我们信任银行A,我们也应该信任B。如果从统计学上讲,这种信任在99%的案例中是合理的,那么我们可以得出结论,相应的蕴涵以99%的概率成立。如果查询Q可从事实和规则推论得出,则Q中的结果概率p(Q)是多少?我们可以根据S-j将Q的真值描述为命题公式F,即由&,v和-;等运算符链接的语句S-j的组合。精确地计算p(Q)是NP难的,因此需要启发式。传统上,专家系统使用类似于简单区间计算的技术:我们解析F并将每个计算步骤替换为相应的概率运算。问题:在每一步,我们都忽略中间结果F-j之间的依赖关系;因此间隔太宽。示例:P(A v-A)的估计值不是1。解决方案:与仿射算法相似,除了P(Fj),我们还计算P(F-j&F-i)(或P(F-ji && BULL; center点中心点&F-jd)),并在每个步骤上使用1这样的概率的所有组合来获得新的估计值。结果:例如,P(A-A)估计为1。(c)2005 Elsevier B.V.保留所有权利。

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