We study learning algorithms for the classical Markovian bandit problem with discount. We explain how to adapt PSRL [24] and UCRL2 [2] to exploit the problem structure. These variants are called MB-PSRL and MB-UCRL2. While the regret bound and runtime of vanilla implementations of PSRL and UCRL2 are exponential in the number of bandits, we show that the episodic regret of MB-PSRL and MB-UCRL2 is $\tilde{O}(S\sqrt{nK})$ where $K$ is the number of episodes, $n$ is the number of bandits and $S$ is the number of states of each bandit (the exact bound in S, n and K is given in the paper). Up to a factor $\sqrt S$, this matches the lower bound of $\Omega(\sqrt{SnK})$ that we also derive in the paper. MB-PSRL is also computationally efficient: its runtime is linear in the number of bandits. We further show that this linear runtime cannot be achieved by adapting classical non-Bayesian algorithms such as UCRL2 or UCBVI to Markovian bandit problems. Finally, we perform numerical experiments that confirm that MB-PSRL outperforms other existing algorithms in practice, both in terms of regret and of computation time.
翻译:我们用折扣来研究古典Markovian盗匪问题的算法。 我们解释如何调整PSRL[ 24]和UCRL2[2], 以利用问题结构。 这些变式称为MB- PSRL和MB-UCRL2。 虽然PSRL和UCRL2的香草执行的遗憾约束和运行时间在盗匪数量上成倍上升,但我们显示,MB- PSRL和MB-UCRL2的同比差是$\Omega(sqrt{SnK})和MB-UCRL2的同比差为$(S\qrt{O}(S\qrt{nK})(S\qrt{nK})的同比值较低。 MB-PSRL的同比值也是计算效率: 美元运行时间线性是麻匪数目中的线性时间, 美元, 美元是每个土匪数目。我们进一步显示, AS-L 的轨迹性演算法无法通过我们BRBSBR的不计数。