Hitting-Time Bounds for Asynchronous Q-Learning

Practical Tips

Practical tips and how-to guidance will be added by our editorial team.

中文解读

中文解读待补充：本站将优先为睡眠改善、失眠治疗、助眠方法等高价值文章补充中文说明。

Article Summary

We study the last iterate of standard asynchronous Q-learning from a single trajectory of a finite Markov decision process. The trajectory follows a fixed behavior policy whose state--action chain is irreducible and may be periodic. For discount factor $γ$, write $H=(1-γ)^{-1}$, and let $μ_{\min}$ be the minimum stationary state--action probability and $\mathfrak h_\star$ the largest expected hitting time of the reverse state--action chain. We prove that zero-initialized Q-learning with a constant step size achieves sup-norm error at most $\varepsilon$ with high probability using $\widetilde{O}(H^4/(μ_{\min}\varepsilon^2)+\mathfrak h_\star)$ transitions, for $0<\varepsilon\le1$. The statistical term retains the quartic effective-horizon dependence of standard Q-learning, while initialization cost is additive in $\mathfrak h_\star$. For aperiodic behavior chains, a comparison with mixing time yields a transient term $\widetilde{O}(t_{\mathrm{mix}}/μ_{\min})$. An empirical state--action count rule selects a constant step size for a subsequent training run and attains the same rate without prior knowledge of model parameters, target accuracy, or confidence level. The output is the last training iterate, with both stages included in the sample count. The proof uses two deterministic Bellman envelopes to separate statistical error from fixed-policy initialization error and adapts reverse-time hitting-cost compensation from TD learning to state--action propagation. An exploration lower bound holds for arbitrary predictable step sizes. Together with the existing statistical lower bound and its extension to pilot-selected constant steps, it yields matching mean-square rates up to logarithmic factors on the stated large-sample parameter classes.

Sources & References

• Original source
• Full article (PDF)

Comments