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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) \...


3

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 ...


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-...


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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 ...


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

A policy can be stochastic or deterministic. A deterministic policy is a function of the form $\pi_{\text{deterministic}}: S \rightarrow A$, that is, a function from the set of states to the set of actions. A stochastic policy is a map of the form $\pi_{\text{stochastic}} : S \rightarrow P(A)$, where $P(A)$ is a set of probability distributions ($P(A) = \{ ...


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

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 ...


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 $\mathcal{D} = \{ (x_i, y_i)\}_{i=1}^N$, ...


1

Having low variance is important in general as it reduces the number of samples needed to obtain accurate estimates. This is the case for all statistical machine learning, not just reinforcement learning. In general, if you are estimating a mean or expected quantity by taking many samples, the variation in the error is proportional to $\frac{\sigma}{\sqrt{N}...


1

If your game agent performs any kind of advance learning from self play or database of moves, that will generate parameters for some kind of model (e.g. a table of expected values, or neural network weights to select a preferred action). This is unavoidable, and if you want to re-use the results of that machine learning, you absolutely have to store the ...


1

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 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, $\pi_i$ may not ...


1

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 ...


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Why are they comparing state value function to action value function? It is because $v_{\pi}(s)$ and $q_{\pi}(s,a)$ measure the same quantity at different stages of the trajectory. By comparing the values at the same $s$ and modifying how $a$ is selected, the proof makes assertions about how that choice impacts the value. It is important to recall that $v_{...


1

I think you're overthinking it. I've never seen a formalisation of the concept of "stationary policy" (apart from yours). However, in general, "stationary" means that something does not change (over time). In the context of reinforcement learning, you can interpret it in such a way that it is consistent with the context where you find this expression, unless ...


1

You are right: a stationary policy is independent of time. It is basically a mapping from states to actions. Despite the point in time in which the agent observes the state s it will select an action a.


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