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8

A stationary policy is a policy that does not change. Although strictly that is a time-dependent issue, that is not what the distinction refers to in reinforcement learning. It generally means that the policy is not being updated by a learning algorithm. If you are working with a stationary policy in reinforcement learning (RL), typically that is because ...


7

The MDP defines the environment (which corresponds to the task that you need to solve), so it defines e.g. the states of the environment, the actions that you can take in those states, the probabilities of transitioning from one state to the other and the probabilities of getting a reward when you take a certain action in a certain state. The policy ...


6

Is the optimal policy always stochastic (that is, a map from states to a probability distribution over actions) if the environment is also stochastic? No. An optimal policy is generally deterministic unless: Important state information is missing (a POMDP). For example, in a map where the agent is not allowed to know its exact location or remember ...


5

I would say no. For example, consider the multi-armed bandit problem. So, you have $n$ arms which all have a probability of giving you a reward (1 point, for example), $p_i$, $i$ being between 1 and $n$. This is a simple stochastic environment: this is a one state environment, but it is still an environment. But obviously the optimal policy is to choose ...


4

$\pi(s)$ does not mean $q(s,a)$ here. $\pi(s)$ is a policy that represents probability distribution over action space for a specific state. $q(s,a)$ is a state-action pair value function that tells us how much reward do we expect to get by taking action $a$ in state $s$ onwards. For the value iteration on the right side with this update formula: $v(s) \...


4

A deterministic policy is a function of the form $\pi_{\mathbb{d}}: S \rightarrow A$, that is, a function from the set of states of the environment, $S$, to the set of actions, $A$. The subscript $_{\mathbb{d}}$ only indicates that this is a ${\mathbb{d}}$eterministic policy. For example, in a grid world, the set of states of the environment, $S$, is ...


4

for example, the "greedy policy" always chooses the action with the highest expected return, no matter which state we are in The "no matter which state we are in" there is generally not true; in general, the expected return depends on the state we are in and the action we choose, not just the action. In general, I wouldn't say that a policy is a mapping ...


3

In the book, the phrase "generate the data" refers to the data from observations about states, actions, next states and rewards, that then get used to make value estimate updates. In both the SARSA and Q learning pseudocode from the book, there is a behaviour policy that selects the next action to take. Other than the initial start state, this ...


2

I guess I'm having difficulty grasping the concept that the goodness of a state changes depending on how an agent got there It doesn't. The value of a state changes depending on what the agent will do next. That is where the dependency on the policy comes in, not in past behaviour, but expectations of future behaviour. The future behaviour depends on the ...


2

A stationary policy is the one that does not depend on time. Meaning that the agent will take the same decision whenever certain conditions are met. This stationary policy may be probabilistic which implies that the probability of choosing an action remains the same. It may take different decisions but the probability remains the same. A Stationary ...


2

Consider a very simple grid-world, consisting of 4 cells, where an agent starts in the bottom-left corner, has actions to move North/East/South/West, and receives a reward $R = 1$ for reaching the top-right corner, which is a terminal state. We'll name the four cells $NW$, $NE$, $SW$ and $SE$ (for north-west, north-east, south-west and south-east). We'll ...


2

The output layer of the network contains one unit, telling me the Q value of the provided state with the assumption that the action taken in that state will be determined by the policy. Typically in Reinforcement Learning, the symbol $Q$ is used when you calculate an action value, and if you are evaluating for a specific policy, it is noted $q_{\pi}(s,a)$ ...


2

A stationary policy, $\pi_t$, is a policy that does not change over time, that is, $\pi_t = \pi, \forall t \geq 0$, where $\pi$ can either be a deterministic function, $\pi: S \rightarrow A$ (a deterministic policy), or a conditional density, $\pi(A \mid S)$ (a stochastic policy). A non-stationary policy is a policy that is not stationary. More precisely, $\...


2

The good news is that: Your MDP appears valid, with well-defined states, actions. It has state transition and reward functions (which you have implemented as matrices). There is nothing else to add, it's a full MDP. You could use this MDP to evaluate a policy, using a variety of reinforcement learning (RL) methods suitable for finite discrete MDPS. For ...


2

First of all, $Q_\pi(s, a)$ IS DEFINED AS the value (i.e. the expected return) of taking some action $a$ in some state $s$, AND THEN following some given policy $\pi$ (until e.g. the end of the game or your life). In other words, suppose that you take action $a$ in state $s$, AND THEN use the policy $\pi$ to behave in the world until you die, then $Q_\pi(s, ...


2

The $\epsilon$-greedy policy is a policy that chooses the best action (i.e. the action associated with the highest value) with probability $1-\epsilon \in [0, 1]$ and a random action with probability $\epsilon $. The problem with $\epsilon$-greedy is that, when it chooses the random actions (i.e. with probability $\epsilon$), it chooses them uniformly (i.e. ...


2

Why do RL implementations converge on one action? If the optimal policy shouldn't always select the same action in the same state, i.e., if the optimal policy isn't deterministic (e.g., in the case of the rock paper scissors, the optimal policy cannot be deterministic because any intelligent player would easily memorize your deterministic policy, so, after ...


2

So, here's is the question: Is it true that a non-stationary policy must satisfy this condition? $$ \forall i, j \in \mathbb{N}, s \in S, \pi (i, s) = \pi(j, s) $$ With your custom notation (which certainly isn't common, but seems reasonable)... I assume you meant to say that a stationary policy must satisfy that condition, rather than that a non-...


2

It seems to me that you're thinking about the parameters a and b as being characteristic of the agent that's moving in the environment (therefore determining the final policy), but they are actually a characteristic of the environment. Think of a frozen lake. You want to pass the lake but there is a hole five meters in front of you. Let's say you have boots ...


2

Value function: How good it is to be in a state $s$ following policy $\pi$. There are different value functions. There's the state value function, often denoted as $v(s)$ (or $V(s)$), so it's a function of only one variable, i.e. $s$ (a state). There's the state-action value function $q(s, a)$ (or $Q(s, a$)). A value function is a function, so it's not a ...


2

Aside from the points raised in nbro's answer, I'd like to point out that for a single MDP (a single instance of a "problem"), it may be sensible to study it from perspectives that include no policy at all, or multiple different policies. For instance, if I have an MDP, I may be interested in studying it by looking at various inherent properties of the ...


2

Your policy gradient algorithms appear to be working as intended. All standard MDPs have one or more deterministic optimal solutions, and those are the policies that solvers will converge to. Making any of these policies more random will often reduce their effectiveness, making them sub-optimal. So once consistently good actions are discovered, the learning ...


2

You appear to comparing the value table update steps in policy iteration and value iteration, which are both derived from Bellman equations. Policy iteration In policy iteration, a policy lookup table is generated, which can be arbitrary. It usually maps a deterministic policy $\pi(s): \mathcal{S} \rightarrow \mathcal{A}$, but can also be of the form $\pi(a|...


2

Yes, in general any linear combination of probability distributions between optimal policies is also an optimal policy. In fact any combination with each state treated separately will also be an optimal policy. This can be seen using the equation for optimal deterministic policy in terms of optimal value function: $$\pi^*(s) = \text{argmax}_a [\sum_{r,s'}p(r,...


1

I think most of it is correct. Q function(also called state-action value, or just action value): How good it is to be in a state S and perform action A while following policy π. It uses reward to measure the state-action value This is a bit off. Q function basically tells you how good it is to be in state S and perform action A, and follow policy $\pi$ ...


1

The formula in question uses a function N(state, action) that defines a visit count of a state-action pair (introduced on page 3). To describe how it is used, lets first describe the steps of AlphaGo Zero as a whole. There are 4 "phases" to the Monte-Carlo tree search in AlphaGo Zero as depicted in Figure 2. The first 3 expand and update the tree and ...


1

You can simply train a policy from the inputs to predict the actions in your dataset. You can use the cross entropy loss for this, i.e. maximize the the log probability that the policy assigns to the actions in the data set when given the corresponding inputs. This is called behavioral cloning. The result is an approximation of the behavioral policy that ...


1

Both value iteration (VI) and policy iteration (PI) algorithms are guaranteed to converge to the optimal policy, so it is expected that you get similar policies from both algorithms (if they have converged). However, they do this differently. VI can be seen as truncated version of PI. Let me first illustrate the pseudocode of both algorithms (taken from ...


1

If a policy is fixed, it is said that an MDP becomes an MRP. I would change the phrasing slightly here, to: If a policy is fixed, an MDP can be accurately modeled as an MRP. Why is this so? Aren't the transitions and rewards still parameterized by the action and current state? In other words, aren't the transition and reward matrices still cubes? The ...


1

Bias is not necessarily bad, even though the term bias usually has a negative connotation. In fact, in machine learning, inductive bias is quite important and necessary. For example, if you want to learn a function $f(x) = y$, where $x \in \mathcal{X}$ and $y \in \mathcal{Y}$, you often just have a finite dataset $\mathcal{D} = \{ (x_i, y_i)\}_{i=1}^N$, ...


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