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Do parallel environments improve the agent's ability to learn or does it not really make a difference? Yes they can make a difference. There are two ways improvement is seen: Collecting data from multiple trajectories at once reduces correlation in the dataset. This improves convergence for online learning systems like neural networks, which work best with ...


4

You are correct in the question that in RL terms chess a game of chess where the agent is one player, and the other player has an unknown state is a partially observable environment. Chess played like this is not a fully observable environment. I did not use the term "fully observable game" or "fully observable system" above , because ...


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Personally, I would choose the following two as the most important: epsilon: When using an epsilon-greedy policy, epsilon determines how often the agent should explore and how often it should exploit. Balancing exploration and exploitation is crucial for the success of the learning agent. Too little exploration might not teach anything to the agent and too ...


3

First, note that the current state does not determine the next state. What determines the next state are the dynamics of the environment, which, in the context of reinforcement learning and, in particular, MDPs, are encoded in the probability distribution $p(s', r \mid s, a)$. So, if the agent is in a certain state $s$, it could end up in another state $s'$, ...


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Reinforcement Learning is really fun because the agent will find any bug in your implementation and will exploit it. >>> take_left(0) 0 >>> take_left(1) -4 The agent figured out your bug with negative values and exploits negative indexing to get to the target faster.


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One renowned example for the specified case is SeqGAN Modeling the data generator as a stochastic policy in reinforcement learning (RL), SeqGAN bypasses the generator differentiation problem by directly performing gradient policy update. The RL reward signal comes from the GAN discriminator judged on a complete sequence, and is passed back to the ...


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Terminal state is always the same in the sense that it represents the same thing, that the episode is over. They don’t need to be the exact same state; for instance you could have an $n$ by $n$ grid world where the top right and bottom left states are terminal as when you reach these your agent dies. These are both terminal but not the same state. For chess ...


2

The basic algorithms (MC control, Q-learning, or Dyna-Q) seemed to all be based on solving whichever specific maze the agent was trained on. All RL algorithms are based on creating solutions to a defined state and action space. If you limit your state space representation and training to a single maze, then that is what will be learned. This is no different ...


2

Q-learning can learn about the greedy policy (the policy that we define as $\pi(s) = \arg\max_a Q(s, a)$) whilst following some arbitrary exploratory policy because Q-learning is an off-policy algorithm. In Q-learning, we are updating our values of $Q(s, a)$ using a bootstrapped value from one time step in the future. This means that we don't need to worry ...


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You should read this study https://arxiv.org/abs/2006.05990 which does some empirical study on this question, specifically for on-policy, continuous action space DRL. It suggests that discount factor and learning rate are the two most important parameters to tune, followed by the width of the policy/value functions. That study also reports that it's very ...


2

I am not sure if it is standard notation, but Sutton & Barto use a convention that a function of a random variable is a new random variable that maps between values of the old variable to values of the new one using the function, and without affecting probability distribution (other than the function could be one-way hence probabilties may effectively ...


2

Policy and value iteration both require you to, for each possible transition and each corresponding possible reward at each state, compute a statistic of $r + \gamma V(s')$. In order for this to be tractable, you need for there to be at most finitely many states, actions, possible rewards, and possible transitions at each state. You also need to know the ...


2

Starting with rewards, states don't have rewards in general. A reward is a number returned at a certain step of the MDP. If you arrange things in sequence over a whole time step $s, a, r, s'$ for state, action, reward, next state, then the reward $r$ is allowed to depend on all three of $s, a, s'$, and it can also be from a random distribution of real ...


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What is the difference between a reward and a value for a given state? Let us say that an agent took an action from state A and reached state B and got a score R. This instantaneous score the agent received on reaching state B is called the reward. Now let me introduce you to the concept of Returns, Assume that an agent followed a particular trajectory: ...


1

Here is how I managed to construct a reward function in one of my projects, where I trained an RL model for a self-driving robot that has only a single camera to navigate through a tunnel: $$ R = \left\{ \begin{array}{ll} d_m - 3 - \left| d_l - d_r \right| & \text{if not terminal state} \\ -100 & \text{otherwise} \end{array} \right. $$ where $d_m$ ...


1

The bandit problem is an MDP. You can make the same argument about needing data to learn in the stateful MDP setting. The thing is, the data you need (the past rewards in this case) was drawn iid (conditioned on the arm) and is not actually a trajectory. For instance, once you learn an optimal policy, you no longer need to gather data and the sequence of ...


1

Yes you can map the output onto [0,1] as you indicate. You should treat this as a modification to the environment. I.e. imagine that the environment takes actions in [-1, 1] instead of [0,1]. No you don't need to change any equations.


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Since my question arose from my incomprehension of $v(S_{t + 1})$ and since I got clarifications on it by Neil Slater, I thought I'd go back to this question and try to answer it again. So I'm assuming that $v(S_{t + 1})$ is a random variable made by the composition of the state-value function $v$ and the random variable $S_{t + 1}$. Since $v(s) = \mathbb{E}[...


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Actually, distributional RL is a well-studied field in deep RL. Generally speaking, distributional RL needs more computing sources (1.1X), because of the quantile head. We can also find new distributional RL literature in NeurIPS2020 :).


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There are problems with both the approach and the expressions that you have. I don't want to just give the correct solution, though, that's an exercise for you to go through and learn from your own experience trying to accomplish it. Instead, let me illustrate that your expressions for $v(s_i)$ are wrong. To do that we'll just do a Monte-Carlo estimate for a ...


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I read section 2.2 of Sutton and Barto, and I understand your confusion: the $\epsilon$-greedy algorithm is not defined precisely on page 27-28. Selecting an action randomly "every once in awhile" with probability $\epsilon$ means selecting an action randomly with probability $\epsilon$ at each timestep and selecting an action greedily with ...


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