A no-free-lunch theorem for multitask learning
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多任务学习的无免费午餐定理
- 研究问题
- 研究多个任务共享最优分类器时，增加任务数量能否改善目标分布上的学习性能，以及缺乏分布信息会如何限制自适应聚合。
- 方法与思路
- 作者比较具有分布信息时的极小极大速率与未知分布信息时的可实现速率，并提出利用任务到目标分布距离排序的聚合方法。
- 材料陈述的贡献
- 摘要称，在所考察情形中，固定每任务样本量时不存在保证随任务数改善速率的自适应算法；额外排序信息可支持近最优聚合。
- 阅读关注点
- 应核对任务相关性和距离的定义、不可能性结论覆盖的分布类，以及排序信息的要求、聚合规则和近最优中的误差因子。
- 阅读边界
- 仅据稿件版本未注明的 OpenAlex 摘要；不可能性结论有其特定情形和信息限制，不能解释为所有多任务学习均无收益或推断刊本变化。
原文摘要
Multitask learning and related areas such as multisource domain adaptation address modern settings where data sets from N related distributions {Pt} are to be combined toward improving performance on any single such distribution D. A perplexing fact remains in the evolving theory on the subject: while we would hope for performance bounds that account for the contribution from multiple tasks, the vast majority of analyses result in bounds that improve at best in the number n of samples per task, but most often do not improve in N. As such, it might seem at first that the distributional settings or aggregation procedures considered in such analyses might be somehow unfavorable; however, as we show, the picture happens to be more nuanced, with interestingly hard regimes that might appear otherwise favorable. In particular, we consider a seemingly favorable classification scenario where all tasks Pt share a common optimal classifier h∗, and which can be shown to admit a broad range of regimes with improved oracle rates in terms of N and n. Some of our main results are: ∙ We show that, even though such regimes admit minimax rates accounting for both n and N, no adaptive algorithm exists, that is, without access to distributional information, no algorithm can guarantee rates that improve with large N for n fixed. ∙ With a bit of additional information, namely, a ranking of tasks {Pt} according to their distance to a target D, a simple rank-based procedure can achieve near optimal aggregations of tasks’ data sets, despite a search space exponential in N. Interestingly, the optimal aggregation might exclude certain tasks, even though they all share the same h∗.