by Richard Möhn 713 days ago | Patrick LaVictoire likes this | link | parent | on: (Non-)Interruptibility of Sarsa(λ) and Q-Learning Second, completely revised version of the report with more data and fancy plots: Questions on the (Non-)Interruptibility of Sarsa(λ) and Q-learning reply
 by Richard Möhn 722 days ago | link | parent | on: (Non-)Interruptibility of Sarsa(λ) and Q-Learning Some new results here: Questions on the (Non-)Interruptibility of Sarsa(λ) and Q-learning. reply
 by Richard Möhn 756 days ago | link | parent | on: (Non-)Interruptibility of Sarsa(λ) and Q-Learning Originally, I counted all timesteps spent in interval $$\left[-1,0\right[$$ and all timesteps spent in interval $$\left[0,1\right]$$. As Stuart Armstrong pointed out, this might make even a perfectly interruptible learner look like it’s influenced by interruptions. To understand this, consider the following example. The uninterrupted agent UA could behave like this: Somewhere in ≤ 1.0. – Time steps are being counted. Crosses 1.0. Noodles around beyond 1.0. – Time steps not counted. Crosses back into ≤ 1.0. – Time steps counted again. Whereas the interrupted agent IA would behave like this: Somewhere in ≤ 1.0. – Time steps are being counted. Crosses 1.0. No more time steps counted. So even if IA behaved the same as UA before the cross, UA would have extra steps from stage 3 and thus appear less biased towards the left. As an alternative to using Brownian motion, Patrick suggested to stop counting once the cart crosses $$1.0$$. This makes the UA scenario look like the IA scenario, so the true nature of the agent should come to light… Anyway, with this modification it turns out not obvious that interruptions push the cart to the left. I will start looking more sharply. reply
 by Richard Möhn 804 days ago | link | parent | on: (Non-)Interruptibility of Sarsa(λ) and Q-Learning Thanks for the comment! I will look into it after working on another issue that Stuart Armstrong pointed out to me. reply

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