We consider a continuous-time multi-arm bandit problem (CTMAB), where the learner can sample arms any number of times in a given interval and obtain a random reward from each sample, however, increasing the frequency of sampling incurs an additive penalty/cost. Thus, there is a tradeoff between obtaining large reward and incurring sampling cost as a function of the sampling frequency. The goal is to design a learning algorithm that minimizes regret, that is defined as the difference of the payoff of the oracle policy and that of the learning algorithm. CTMAB is fundamentally different than the usual multi-arm bandit problem (MAB), e.g., even the single-arm case is non-trivial in CTMAB, since the optimal sampling frequency depends on the mean of the arm, which needs to be estimated. We first establish lower bounds on the regret achievable with any algorithm and then propose algorithms that achieve the lower bound up to logarithmic factors. For the single-arm case, we show that the lower bound on the regret is $\Omega((\log T)^2/\mu)$, where $\mu$ is the mean of the arm, and $T$ is the time horizon. For the multiple arms case, we show that the lower bound on the regret is $\Omega((\log T)^2 \mu/\Delta^2)$, where $\mu$ now represents the mean of the best arm, and $\Delta$ is the difference of the mean of the best and the second-best arm. We then propose an algorithm that achieves the bound up to constant terms.
翻译:我们考虑的是连续时间的多武器强盗问题(CTMAB),在这个问题中,学习者可以在给定间隔内对武器做多少次取样,并从每个抽样中获得随机的奖励,但是,增加取样的频率会产生累加惩罚/成本。因此,在获得大量奖励和产生取样成本之间有一个权衡,这是取样频率的一个函数。我们首先在任何算法所能实现的遗憾上设定较低的界限,然后提出达到低于逻辑系数的算法。在单武器案中,我们表明,遗憾的下限是美元(MAB),例如,即使是单武器案件在CTMAB中是非三重的,因为最佳取样频率取决于手臂的平均值。我们首先在任何算法中设定较低的遗憾界限,然后提出达到低于逻辑系数的下限值的算法。对于单武器案,我们表示的下限是美元(MONG(MGT) = 美元 2/\\ mu) 美元,而武器的中位值值值是美元 和 武器的下限值是平均值。