7

In general the different reward functions $R(s)$, $R(s, a)$ and $R(s, a, s')$ are not equivalent mathematically, so you will not find any formal proof. It is possible for the functions to resolve to the same value in a specific MDP, if, for instance, you use $R(s, a, s')$ and the value returned only depends on $s$, then $R(s, a, s') = R(s)$. This is not true ...


6

Rather than the survey by Liu et al. recommended above, I'd suggest you read the following survey paper for an overview of MORL (disclaimer - I was a co-author on this, but I genuinely think it is a much more useful introduction to this area) Roijers, D. M., Vamplew, P., Whiteson, S., & Dazeley, R. (2013). A survey of multi-objective sequential decision-...


5

1) Is there any way to set the initial Q-values for the actions? You can generally do this, but you cannot specify specific weights for specific actions in specific states. Not through the network weights directly, at least. That would defeat the purpose of using backpropagation to optimize the weights and find the optimal parameters and Q-values. 2) Is ...


5

If I understood correctly you're looking at a Multi-Objective Reinforcement Learning (MORL). Keep in mind however that many scientist will often follow the reward hypothesis (Sutton and Barto) which says that All of what we mean by goals and purposes can be well thought of as the maximization of the expected value of the cumulative sum of a received scalar ...


5

Markov decision problems are usually defined with a reward function $r:\mathcal{S}\times\mathcal{A}\rightarrow\mathbb{R}$, and in these cases the rewards are expected to be scalar real values. This makes reinforcement learning (RL) easier, for example when defining a policy $\pi(s,a)=\arg\max_a Q(s,a)$, it is clear what is the maximum of the Q-factors in ...


4

If you have multiple types of rewards (say, R1 and R2), then it is no longer clear what would be the optimal way to act: it can happen that one way of acting would maximize R1 and another way would maximize R2. Therefore, optimal policies, value functions, etc., would all be undefined. Of course, you could say that you want to maximize, for example, R1+R2, ...


4

If your objective is for the agent to attain some goal (say, reaching a target), then a valid reward function is to assign a reward of 1 when the goal is attained and 0 otherwise. The problem with this reward function is that it's too sparse, meaning the agent has little guidance on how to modify their behavior to become better at attaining said goal, ...


4

Designing reward functions Designing a reward function is sometimes straightforward, if you have knowledge of the problem. For example, consider the game of chess. You know that you have three outcomes: win (good), loss (bad), or draw (neutral). So, you could reward the agent with $+1$ if it wins the game, $-1$ if it loses, and $0$ if it draws (or for any ...


4

There seem to be two different ideas in this question here: What's the impact / importance of our choice for reward values? What's the impact / importance of our choice for initial value estimates (how do we initialise our table of $Q(s, a)$ values in the case of a simple, tabular RL algorithm like Sarsa or $Q$-learning)? The reward values are typically ...


4

My question is, would $r_1 =r_2$? That's usually up to you as the designer of the system. Usually when you declare that you have "a deterministic environment", you imply that both $s'$ and $r$ are fixed values depending on $(s,a)$. So in your examples, you would expect your observations to also have $r_1 = r_2$ However, it is possible to define a MDP ...


4

Let $R(s)$ denote a probability distribution over rewards that our agent may get in some MDP as a reward for entering a state $s$. The easiest case is to demonstrate that we can also choose to write this as $R(s, a)$ or $R(s, a, s')$: simply take $\forall a: R(s, a) = R(s)$, or $\forall a \forall s': R(s, a, s') = R(s)$, as also described in Neil's answer. ...


3

What if a scalar reward is insufficient, or its unclear on how to collapse a multi-dimensional reward to a single dimension. Example, for someone eating a burger, both taste and cost are important. Agents may prioritize taste and cost differently, so its not clear on how to aggregate the two. It is also not clear on how a subjective categorical taste value ...


3

Reinforcement learning (RL) control maximises the expected sum of rewards. If you change the reward metric, it will change what counts as optimal. Your reward functions are not the same, so will in some cases change the priority of solutions. As a simple example, consider a choice between trajectories with costs A(0,4,4,4) and B(1,1,1,1). In the original ...


3

I think I may be in position to answer my own question. The Bellman equation (for the optimal policy) for a MDP with $r(s,a,s')$ rewards would look like this: $$V(s) = \max_a \left\{ \sum_{s'} p(s'|s,a)(r(s,a,s') + \gamma V(s')) \right\} $$ $$V(s) = \max_a \left\{ \sum_{s'} p(s'|s,a) \cdot r(s,a,s') + \gamma \sum_{s'} p(s'|a,s) \cdot V(s') \right\} $$ Now, ...


3

No, the substitution you suggest based on Equation (3.4) is not correct because you forgot about the $\sum_{r \in \mathcal{R}}$ in the right-hand side Equation (3.4). Equation (3.4) says (leaving out the middle part): $$p(s' \vert s, a) \doteq \sum_{r \in \mathcal{R}} p(s', r \vert s, a).$$ If you plug this into Equation (3.6) to substitute the ...


3

In Reinforcement Learning (RL), a reward function is part of the problem definition and should: Be based primarily on the goals of the agent. Take into account any combination of starting state $s$, action taken $a$, resulting state $s'$ and/or a random amount (a constant amount is just a random amount with a fixed value having probability 1). You should ...


3

Generally speaking, is it better for rewards to be a scalar, or is using matrices okay? Rewards need to be scalar, real values to match to standard theory of Markov decision processes (MDPs) and reinforcement learning (RL) methods. Although it is possible to accumulate matrices in various ways, by e.g. simple matrix addition, and come up with an analog for ...


3

Andrew Y. Ng (yes, that famous guy!) et al. proved, in the seminal paper Policy invariance under reward transformations: Theory and application to reward shaping (ICML, 1999), which was then part of his PhD thesis, that potential-based reward shaping (PBRS) is the way to shape the natural/correct sparse reward function (RF) without changing the optimal ...


3

Question 1 The taylor expansion of $\frac{1}{1-\gamma}$ at $\gamma= 0$ is as follows $$\frac{1}{1-\gamma} = 1 + \gamma + \gamma^2 + \dots$$ When you multiply by $1-\gamma$ you get $$ 1 = (1-\gamma)(1 + \gamma + \gamma^2 + \dots)$$ Which can be equivalently written as $$1 = (1-\gamma)\sum_\limits{i=0}^{\infty}\gamma^i$$ Hence we can see that by multiplying ...


2

Yes, you are right. It is somehow an arbitrary choice, although you should consider the reasonable numerical ranges of your activation functions if you decide to go beyond the values +/- 1. You can also have a think about whether you want to add a small reward for the agent reaching states that are near the goal, if you have an environment where such states ...


2

In this case, the word "system" refers to a Markov decision process (MDP), which is the mathematical model used to represent the reinforcement learning (RL) problem or, in general, a decision making problem. Recall that, in RL, the problem consists in finding an (optimal) policy, which is a policy that allows the agent to collect the highest amount of reward ...


2

The same $\gamma = 0.9$ that you use in the definition $F \doteq \gamma \Phi(s') - \Phi(s)$ should also be used as the discount factor in computing returns for multi-step trajectories. So, rather than simply adding up all the rewards for your different time-steps for the different trajectories, you should discount them by $\gamma$ for every time step that ...


2

In a toy environment, this is a choice you can make relatively freely, depending on what you want to achieve with the learning challenge. It may help if you think through what the actual consequences for making the "wrong" move are in your environment. There are a few self-consistent options: The move simply cannot be made and count as playing the ...


2

In reinforcement learning (RL), an immediate reward value must be returned after each action, along with the next state. This value can be zero though, which will have no direct impact on optimality or setting goals. Unless you are modifying the reward scheme to try and make an environment easier to learn (called reward shaping), then you should be aiming ...


2

I agree with Tomasz that the approach you are describing falls within the field of MORL. For a solid introduction MORL I would recommend the survey by Roijers, D. M., Vamplew, P., Whiteson, S., & Dazeley, R. (2013). A survey of multi-objective sequential decision-making. Journal of Artificial Intelligence Research, 48, 67-113. https://www.jair.org/index....


2

Having only a non-zero reward at the very end is not uncommon. When rewards are sparse, it becomes a bit harder to learn compared to having lots of different rewards along the way, but for your problem, the goal state is always reached, so that should not be a problem. (The real problem with sparse rewards is that, if an agent can do a lot of exploration ...


2

What are the pros and cons of sparse and dense rewards in reinforcement learning? It is unusual to refer to this difference as "pros and cons" because that term is often used to make comparisons between difference choices. Assuming you have a specific problem to solve, then whether or not the rewards are naturally sparse or dense is not a choice. ...


2

The Bellman optimality equation is given by $$q_*(s,a) = \sum_{s' \in \mathcal{S}, r \in \mathcal{R}}p(s',r \mid s,a)(r + \gamma \max_{a'\in\mathcal{A}(s')}q_*(s',a')) \tag{1}\label{1}.$$ If the reward is multiplied by a constant $c > 0 \in \mathbb{R}$, then the new optimal action-value function is given by $cq_*(s, a)$. To prove this, we just need to ...


1

If by, I can compute the reward given $(a_1, a_2, \dots, a_n)$ you simply mean that your game is deterministic, this is absolutely fine. I feel another answer had assumed you were implying your terminal reward is a matter of some sequence. RL does, however, struggle more greatly in games with indeterminable reward until the terminal state, however, it is ...


1

You have some freedom to re-define reward schemes, whilst still describing the same goals for an agent. How this works depends to some degree on whether you are dealing with an episodic or continuing problem. Episodic problems An episodic problem ends, and once an agent reaches the terminal state, it is guaranteed zero rewards from that point on. The optimal ...


Only top voted, non community-wiki answers of a minimum length are eligible