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Let's say that the Bayes network consists of node K that represents knowledge. Since knowledge is evaluated using questions, each question is represented by the node related to the knowledge node (Q1, Q2). Regarding the types of questions, let's say there are inverse questions that are supposed to be related and form a cycle.

enter image description here If it is ensured in the testing process that specific inverse question (e.g. Q2) is never asked if the corresponding question (e.g. Q1) is not answered correctly, is it allowed to drop one relation (Q2->Q1) and solve the cycle problem?

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Bayesian networks need to be acyclic. You don't put an arrow on every related node. Because, it is always the case that if $X$ is related to $Y$, $Y$ is related to $X$. Otherwise it'd be an undirected graph, or a directed graph with every relation doubled. Conceptually, causal relations make more sense when building the graph. And, the builder of the graph makes assumptions. If the resulting graph violates the acyclic property, then Bayesian Networks might not be a good framework to solve the problem.

For your case, I think the following sentence describes $P(Q_2|Q_1=0)=0$ which is a causal dependence and maybe encoded as an arrow from $Q_1$ to $Q_2$, and the inverse one can be dropped.

If it is ensured in the testing process that specific inverse question (e.g. Q2) is never asked if the corresponding question (e.g. Q1) is not answered correctly,

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  • $\begingroup$ Thank you for your answer. Because of its acyclic property, Bayes networks are probably not the best approach to solve this problem. However, this is the example of how the cycles can be treated during the design process of the Bayes network - find the assumptions that change the problem environment, but not the problem itself. $\endgroup$
    – hello
    Feb 23, 2020 at 9:58
  • $\begingroup$ Hi, do you want to say that conditional probability P(Q2|Q1=0) impacts the existence of the relation between the nodes? Is it possible that the previous probability is different that zero, but the relation still does not exist between Q2->Q1? If this option is not possible in the model application, these conditional probabilities are never used? $\endgroup$
    – hello
    Feb 25, 2020 at 22:34

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