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Multiple agents finitely repeated inspection game with dismissals

机译:多个特工有限重复检查与解雇游戏

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This paper deals with an inspection game between a single inspector and several independent (potential) violators over a finite-time horizon. In each period, the inspector gets a renewable inspection resource, which cannot be saved and used in future periods. The inspector allocates it to inspect the (potential) violators. Each violator decides in each period whether to violate or not, and in what probability. A violation may be detected by the inspector with a known and positive probability. When a violation is detected, the responsible violator is "dismissed" from the game. The game terminates when all the violators are detected or when there are no more remaining periods. An efficient method to compute a Nash equilibrium for this game is developed, for any possible value of the (nominal) detection probability. The solution of the game shows that the violators always maintain their detection probability below 0.5.
机译:本文讨论了在有限时间范围内单个检查员与几个独立的(潜在)违规者之间的检查游戏。在每个期间,检查员都会获得可更新的检查资源,该资源无法在以后的期间中保存和使用。检查员将其分配给检查(潜在)违规者。每个违反者都会在每个时期内决定是否违反,以及以什么概率。检查员可以已知和肯定的概率检测到违规情况。当检测到违规时,负责的违规者被“解雇”出游戏。当检测到所有违规者或没有其他剩余时间时,游戏终止。对于(标称)检测概率的任何可能值,开发了一种有效的方法来计算该游戏的纳什均衡。游戏的解决方案表明,违规者始终将其检测概率保持在0.5以下。

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