Conventional imitation learning assumes access to the actions of demonstrators, but these motor signals are often non-observable in naturalistic settings.

We provide a probabilistic interpretation of attention and show that the standard dotproduct attention in transformers is a special case of Maximum APosteriori (MAP) inference. The proposed approach suggests the use of Expectation Maximization algorithms for online adaptation of key and value model parameters. This approach is useful for cases in which external agents, e.g., annotators, provide inference-time information about the correct values of some tokens, e.g., the semantic category of some pixels, and we need for this new information to propagate to other tokens in a principled manner. We illustrate the approach on an interactive semantic segmentation task in which annotators and models collaborate online to improve annotation efficiency. Using standard benchmarks, we observe that key adaptation boosts model performance ( 10% mIoU) in the low feedback regime and value propagation improves model responsiveness in the high feedback regime.

This appendix contains all proofs of the results mentioned in the main body of the paper, plus further results which have been omitted there due to space limits. We recall the following lemma which upper bounds the probability measure of the ball around a point x X that contains its kth nearest neighbors. The proof immediately follows from the multiplicative Chernoff bound (see, e.g., Lemma 3.2 in [28]). When combined with Assumption 5.1 we obtain the following corollary. Corollary A.2. Suppose that the measure-smoothness assumption (Assumption 5.1) holds with parameters λ, Cλ, k k.

When the expected rewards of all arms are the same, we know that the arm with the lowest index will be chosen and thus the first K pulls will be π1 = 1,...,πK = K. We will complete the proof through induction. Suppose that the greedy pull sequence is periodic with π1 = 1,...,πK = K and πt+K = πt until time h>K. We will show that πh+1 = 1 if πh = K and πh+1 = πh + 1 otherwise. When k0 = 0 (i.e., πh = K), all arms have been pulled exactly ntimes as of time h. Therefore, by (3), at time h+ 1, arm 1 has the highest expected reward and will be chosen.