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Why a Random Reward in One-step Dynamics MDP?


What is the Q function and what is the V function in reinforcement learning?What is the reward function in the 10 armed test bed?Reward dependent on (state, action) versus (state, action, successor state)Cannot see what the “notation abuse” is, mentioned by author of bookWhat is the difference between “expected return” and “expected reward” in the context of RL?How is that possible that a reward function depends both on the next state and an action from current state?How is Importance-Sampling Used in Off-Policy Monte Carlo Prediction?Time horizon T in policy gradients (actor-critic)Reinforcement learning: Discounting rewards in the REINFORCE algorithmAbout applying time series forecasting to problems better suited for reinforcement learning, like toy example “Jack's car rental”













5












$begingroup$


I am reading the 2018 book by Sutton & Barto on Reinforcement Learning and I am wondering the benefit of defining the one-step dynamics of an MDP as
$$
p(s',r|s,a) = Pr(S_t+1,R_t+1|S_t=s, A_t=a)
$$

where $S_t$ is the state and $A_t$ the action at time $t$. $R_t$ is the reward.



This formulation would be useful if we were to allow different rewards when transitioning from $s$ to $s'$ by taking an action $a$, but this does not make sense. I am used to the definition based on $p(s'|s,a)$ and $r(s,a,s')$, which of course can be derived from the one-step dynamics above.



Clearly, I am missing something. Any enlightenment would be really helpful. Thx!










share|improve this question











$endgroup$











  • $begingroup$
    Could you explain why, to you, that "allow different rewards when transitioning from 𝑠 to 𝑠′ by taking an action 𝑎" does not make sense? It makes sense to me, but I cannot explain it to you, unless you give more details about what is wrong with the idea to you
    $endgroup$
    – Neil Slater
    Mar 16 at 22:39










  • $begingroup$
    My understanding is that given a starting state and a target state, reachable by applying action $a$, there is only a single reward. If we have multiple rewards, then we are allowing the Markov Chain model (thought as a graph) being a multi-graph where we can go from $s$ to $s'$ (with $a$) over an edge with reward $r$ and another with reward $r'$. I thought this is not the right model ... but again ... I might be wrong ...
    $endgroup$
    – RLSelfStudy
    Mar 16 at 22:46















5












$begingroup$


I am reading the 2018 book by Sutton & Barto on Reinforcement Learning and I am wondering the benefit of defining the one-step dynamics of an MDP as
$$
p(s',r|s,a) = Pr(S_t+1,R_t+1|S_t=s, A_t=a)
$$

where $S_t$ is the state and $A_t$ the action at time $t$. $R_t$ is the reward.



This formulation would be useful if we were to allow different rewards when transitioning from $s$ to $s'$ by taking an action $a$, but this does not make sense. I am used to the definition based on $p(s'|s,a)$ and $r(s,a,s')$, which of course can be derived from the one-step dynamics above.



Clearly, I am missing something. Any enlightenment would be really helpful. Thx!










share|improve this question











$endgroup$











  • $begingroup$
    Could you explain why, to you, that "allow different rewards when transitioning from 𝑠 to 𝑠′ by taking an action 𝑎" does not make sense? It makes sense to me, but I cannot explain it to you, unless you give more details about what is wrong with the idea to you
    $endgroup$
    – Neil Slater
    Mar 16 at 22:39










  • $begingroup$
    My understanding is that given a starting state and a target state, reachable by applying action $a$, there is only a single reward. If we have multiple rewards, then we are allowing the Markov Chain model (thought as a graph) being a multi-graph where we can go from $s$ to $s'$ (with $a$) over an edge with reward $r$ and another with reward $r'$. I thought this is not the right model ... but again ... I might be wrong ...
    $endgroup$
    – RLSelfStudy
    Mar 16 at 22:46













5












5








5





$begingroup$


I am reading the 2018 book by Sutton & Barto on Reinforcement Learning and I am wondering the benefit of defining the one-step dynamics of an MDP as
$$
p(s',r|s,a) = Pr(S_t+1,R_t+1|S_t=s, A_t=a)
$$

where $S_t$ is the state and $A_t$ the action at time $t$. $R_t$ is the reward.



This formulation would be useful if we were to allow different rewards when transitioning from $s$ to $s'$ by taking an action $a$, but this does not make sense. I am used to the definition based on $p(s'|s,a)$ and $r(s,a,s')$, which of course can be derived from the one-step dynamics above.



Clearly, I am missing something. Any enlightenment would be really helpful. Thx!










share|improve this question











$endgroup$




I am reading the 2018 book by Sutton & Barto on Reinforcement Learning and I am wondering the benefit of defining the one-step dynamics of an MDP as
$$
p(s',r|s,a) = Pr(S_t+1,R_t+1|S_t=s, A_t=a)
$$

where $S_t$ is the state and $A_t$ the action at time $t$. $R_t$ is the reward.



This formulation would be useful if we were to allow different rewards when transitioning from $s$ to $s'$ by taking an action $a$, but this does not make sense. I am used to the definition based on $p(s'|s,a)$ and $r(s,a,s')$, which of course can be derived from the one-step dynamics above.



Clearly, I am missing something. Any enlightenment would be really helpful. Thx!







machine-learning reinforcement-learning






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 10 mins ago









Esmailian

1,686115




1,686115










asked Mar 16 at 21:59









RLSelfStudyRLSelfStudy

283




283











  • $begingroup$
    Could you explain why, to you, that "allow different rewards when transitioning from 𝑠 to 𝑠′ by taking an action 𝑎" does not make sense? It makes sense to me, but I cannot explain it to you, unless you give more details about what is wrong with the idea to you
    $endgroup$
    – Neil Slater
    Mar 16 at 22:39










  • $begingroup$
    My understanding is that given a starting state and a target state, reachable by applying action $a$, there is only a single reward. If we have multiple rewards, then we are allowing the Markov Chain model (thought as a graph) being a multi-graph where we can go from $s$ to $s'$ (with $a$) over an edge with reward $r$ and another with reward $r'$. I thought this is not the right model ... but again ... I might be wrong ...
    $endgroup$
    – RLSelfStudy
    Mar 16 at 22:46
















  • $begingroup$
    Could you explain why, to you, that "allow different rewards when transitioning from 𝑠 to 𝑠′ by taking an action 𝑎" does not make sense? It makes sense to me, but I cannot explain it to you, unless you give more details about what is wrong with the idea to you
    $endgroup$
    – Neil Slater
    Mar 16 at 22:39










  • $begingroup$
    My understanding is that given a starting state and a target state, reachable by applying action $a$, there is only a single reward. If we have multiple rewards, then we are allowing the Markov Chain model (thought as a graph) being a multi-graph where we can go from $s$ to $s'$ (with $a$) over an edge with reward $r$ and another with reward $r'$. I thought this is not the right model ... but again ... I might be wrong ...
    $endgroup$
    – RLSelfStudy
    Mar 16 at 22:46















$begingroup$
Could you explain why, to you, that "allow different rewards when transitioning from 𝑠 to 𝑠′ by taking an action 𝑎" does not make sense? It makes sense to me, but I cannot explain it to you, unless you give more details about what is wrong with the idea to you
$endgroup$
– Neil Slater
Mar 16 at 22:39




$begingroup$
Could you explain why, to you, that "allow different rewards when transitioning from 𝑠 to 𝑠′ by taking an action 𝑎" does not make sense? It makes sense to me, but I cannot explain it to you, unless you give more details about what is wrong with the idea to you
$endgroup$
– Neil Slater
Mar 16 at 22:39












$begingroup$
My understanding is that given a starting state and a target state, reachable by applying action $a$, there is only a single reward. If we have multiple rewards, then we are allowing the Markov Chain model (thought as a graph) being a multi-graph where we can go from $s$ to $s'$ (with $a$) over an edge with reward $r$ and another with reward $r'$. I thought this is not the right model ... but again ... I might be wrong ...
$endgroup$
– RLSelfStudy
Mar 16 at 22:46




$begingroup$
My understanding is that given a starting state and a target state, reachable by applying action $a$, there is only a single reward. If we have multiple rewards, then we are allowing the Markov Chain model (thought as a graph) being a multi-graph where we can go from $s$ to $s'$ (with $a$) over an edge with reward $r$ and another with reward $r'$. I thought this is not the right model ... but again ... I might be wrong ...
$endgroup$
– RLSelfStudy
Mar 16 at 22:46










2 Answers
2






active

oldest

votes


















3












$begingroup$

In general, $R_t+1$ is is a random variable with conditional probability distribution $Pr(R_t+1=r|S_t=s,A_t=a)$. So it can potentially take on a different value each time action $a$ is taken in state $s$.



Some problems don't require any randomness in their reward function. Using the expected reward $r(s,a,s')$ is simpler in this case, since we don't have to worry about the reward's distribution. However, some problems do require randomness in their reward function. Consider the classic multi-armed bandit problem, for example. The payoff from a machine isn't generally deterministic.



As the basis for RL, we want the MDP to be as general as possible. We model reward in MDPs as a random variable because it gives us that generality. And because it is useful to do so.






share|improve this answer









$endgroup$




















    1












    $begingroup$

    State is just an observation of the environment, in many case, we can't get all the variables to fully describe the environment(or maybe it's too time-consuming or space consuming to cover every thing). Just imagine you are designing an robot, you can't and don't need to define a state covering the direction of wind, the density of the atmosphere etc.



    So, although you are in the same state(the same just means the variables you care about have the same value, but not all dynamics of the environment), you are not totally in the same environment.



    So, we can say that, from one particular state to another particular state, the reward may be different, because the state is not the environment, and the environment can't never be the same, because time is flowing






    share|improve this answer








    New contributor




    苏东远 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$












    • $begingroup$
      Very good explanation!
      $endgroup$
      – Esmailian
      15 mins ago










    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    In general, $R_t+1$ is is a random variable with conditional probability distribution $Pr(R_t+1=r|S_t=s,A_t=a)$. So it can potentially take on a different value each time action $a$ is taken in state $s$.



    Some problems don't require any randomness in their reward function. Using the expected reward $r(s,a,s')$ is simpler in this case, since we don't have to worry about the reward's distribution. However, some problems do require randomness in their reward function. Consider the classic multi-armed bandit problem, for example. The payoff from a machine isn't generally deterministic.



    As the basis for RL, we want the MDP to be as general as possible. We model reward in MDPs as a random variable because it gives us that generality. And because it is useful to do so.






    share|improve this answer









    $endgroup$

















      3












      $begingroup$

      In general, $R_t+1$ is is a random variable with conditional probability distribution $Pr(R_t+1=r|S_t=s,A_t=a)$. So it can potentially take on a different value each time action $a$ is taken in state $s$.



      Some problems don't require any randomness in their reward function. Using the expected reward $r(s,a,s')$ is simpler in this case, since we don't have to worry about the reward's distribution. However, some problems do require randomness in their reward function. Consider the classic multi-armed bandit problem, for example. The payoff from a machine isn't generally deterministic.



      As the basis for RL, we want the MDP to be as general as possible. We model reward in MDPs as a random variable because it gives us that generality. And because it is useful to do so.






      share|improve this answer









      $endgroup$















        3












        3








        3





        $begingroup$

        In general, $R_t+1$ is is a random variable with conditional probability distribution $Pr(R_t+1=r|S_t=s,A_t=a)$. So it can potentially take on a different value each time action $a$ is taken in state $s$.



        Some problems don't require any randomness in their reward function. Using the expected reward $r(s,a,s')$ is simpler in this case, since we don't have to worry about the reward's distribution. However, some problems do require randomness in their reward function. Consider the classic multi-armed bandit problem, for example. The payoff from a machine isn't generally deterministic.



        As the basis for RL, we want the MDP to be as general as possible. We model reward in MDPs as a random variable because it gives us that generality. And because it is useful to do so.






        share|improve this answer









        $endgroup$



        In general, $R_t+1$ is is a random variable with conditional probability distribution $Pr(R_t+1=r|S_t=s,A_t=a)$. So it can potentially take on a different value each time action $a$ is taken in state $s$.



        Some problems don't require any randomness in their reward function. Using the expected reward $r(s,a,s')$ is simpler in this case, since we don't have to worry about the reward's distribution. However, some problems do require randomness in their reward function. Consider the classic multi-armed bandit problem, for example. The payoff from a machine isn't generally deterministic.



        As the basis for RL, we want the MDP to be as general as possible. We model reward in MDPs as a random variable because it gives us that generality. And because it is useful to do so.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 17 at 0:39









        Philip RaeisghasemPhilip Raeisghasem

        2135




        2135





















            1












            $begingroup$

            State is just an observation of the environment, in many case, we can't get all the variables to fully describe the environment(or maybe it's too time-consuming or space consuming to cover every thing). Just imagine you are designing an robot, you can't and don't need to define a state covering the direction of wind, the density of the atmosphere etc.



            So, although you are in the same state(the same just means the variables you care about have the same value, but not all dynamics of the environment), you are not totally in the same environment.



            So, we can say that, from one particular state to another particular state, the reward may be different, because the state is not the environment, and the environment can't never be the same, because time is flowing






            share|improve this answer








            New contributor




            苏东远 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$












            • $begingroup$
              Very good explanation!
              $endgroup$
              – Esmailian
              15 mins ago















            1












            $begingroup$

            State is just an observation of the environment, in many case, we can't get all the variables to fully describe the environment(or maybe it's too time-consuming or space consuming to cover every thing). Just imagine you are designing an robot, you can't and don't need to define a state covering the direction of wind, the density of the atmosphere etc.



            So, although you are in the same state(the same just means the variables you care about have the same value, but not all dynamics of the environment), you are not totally in the same environment.



            So, we can say that, from one particular state to another particular state, the reward may be different, because the state is not the environment, and the environment can't never be the same, because time is flowing






            share|improve this answer








            New contributor




            苏东远 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$












            • $begingroup$
              Very good explanation!
              $endgroup$
              – Esmailian
              15 mins ago













            1












            1








            1





            $begingroup$

            State is just an observation of the environment, in many case, we can't get all the variables to fully describe the environment(or maybe it's too time-consuming or space consuming to cover every thing). Just imagine you are designing an robot, you can't and don't need to define a state covering the direction of wind, the density of the atmosphere etc.



            So, although you are in the same state(the same just means the variables you care about have the same value, but not all dynamics of the environment), you are not totally in the same environment.



            So, we can say that, from one particular state to another particular state, the reward may be different, because the state is not the environment, and the environment can't never be the same, because time is flowing






            share|improve this answer








            New contributor




            苏东远 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$



            State is just an observation of the environment, in many case, we can't get all the variables to fully describe the environment(or maybe it's too time-consuming or space consuming to cover every thing). Just imagine you are designing an robot, you can't and don't need to define a state covering the direction of wind, the density of the atmosphere etc.



            So, although you are in the same state(the same just means the variables you care about have the same value, but not all dynamics of the environment), you are not totally in the same environment.



            So, we can say that, from one particular state to another particular state, the reward may be different, because the state is not the environment, and the environment can't never be the same, because time is flowing







            share|improve this answer








            New contributor




            苏东远 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            share|improve this answer



            share|improve this answer






            New contributor




            苏东远 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            answered 2 hours ago









            苏东远苏东远

            111




            111




            New contributor




            苏东远 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.





            New contributor





            苏东远 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            苏东远 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.











            • $begingroup$
              Very good explanation!
              $endgroup$
              – Esmailian
              15 mins ago
















            • $begingroup$
              Very good explanation!
              $endgroup$
              – Esmailian
              15 mins ago















            $begingroup$
            Very good explanation!
            $endgroup$
            – Esmailian
            15 mins ago




            $begingroup$
            Very good explanation!
            $endgroup$
            – Esmailian
            15 mins ago

















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