Time horizon T in policy gradients (actor-critic) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) 2019 Moderator Election Q&A - Questionnaire 2019 Community Moderator Election ResultsWhat is “experience replay” and what are its benefits?What is the Q function and what is the V function in reinforcement learning?Why do we normalize the discounted rewards when doing policy gradient reinforcement learning?Reinforcement Learning - What's the formula for the value functionWhat is the difference between “expected return” and “expected reward” in the context of RL?How is Importance-Sampling Used in Off-Policy Monte Carlo Prediction?Deep RL: Proximal policy optimization gradient calculationPolicy Gradients - gradient Log probabilities favor less likely actions?RL's policy gradient (REINFORCE) pipeline clarificationNatural actor-critic with Q function approximation

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Time horizon T in policy gradients (actor-critic)



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
2019 Moderator Election Q&A - Questionnaire
2019 Community Moderator Election ResultsWhat is “experience replay” and what are its benefits?What is the Q function and what is the V function in reinforcement learning?Why do we normalize the discounted rewards when doing policy gradient reinforcement learning?Reinforcement Learning - What's the formula for the value functionWhat is the difference between “expected return” and “expected reward” in the context of RL?How is Importance-Sampling Used in Off-Policy Monte Carlo Prediction?Deep RL: Proximal policy optimization gradient calculationPolicy Gradients - gradient Log probabilities favor less likely actions?RL's policy gradient (REINFORCE) pipeline clarificationNatural actor-critic with Q function approximation










2












$begingroup$


I am currently going through the Berkeley lectures on Reinforcement Learning. Specifically, I am at slide 5 of this lecture.



At the bottom of that slide, the gradient of the expected sum of rewards function is given by
$$
nabla J(theta) = frac1N sum_i=1^N sum_t=1^T nabla_theta logpi_theta(a_i,t vert s_i,t) (Q(s_i,t,a_i,t) - V(s_i,t))
$$
The q-value function is defined as
$$Q(s_t,a_t) = sum_t'=t^T mathbbE_pi_theta[r(s_t',a_t')vert s_t,a_t]$$
At first glance, this makes sense, because I compare the value of taking the chosen action $a_i,t$ to the average value in time step $t$ and can evaluate how good my action was.



My question is: a specific state $s_spec$ can occur in any timestep, for example, $s_1 = s_spec = s_10$. But isn't there a difference in value depending on whether I hit $s_spec$ at timestep 1 or 10 when $T$ is fixed? Does this mean that for every state there is a different q value for each possible $t in 0,ldots,T$? I somehow doubt that this is the case, but I don't quite understand how the time horizon $T$ fits in.



Or is $T$ not fixed (perhaps it's defined as the time step in which the trajectory ends in a terminal state - but that'd mean that during trajectory sampling, each simulation would take a different number of timesteps)?










share|improve this question











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bumped to the homepage by Community 27 mins ago


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  • $begingroup$
    I think I have found something helpful - John Schulman notes in his thesis: !screenshot So it sounds like either I sample trajectories with variable length (arrival in a terminal state) or I encode the time step into the state and the state value function will also take the time step into consideration?
    $endgroup$
    – Dummie Variable
    Sep 15 '18 at 13:49
















2












$begingroup$


I am currently going through the Berkeley lectures on Reinforcement Learning. Specifically, I am at slide 5 of this lecture.



At the bottom of that slide, the gradient of the expected sum of rewards function is given by
$$
nabla J(theta) = frac1N sum_i=1^N sum_t=1^T nabla_theta logpi_theta(a_i,t vert s_i,t) (Q(s_i,t,a_i,t) - V(s_i,t))
$$
The q-value function is defined as
$$Q(s_t,a_t) = sum_t'=t^T mathbbE_pi_theta[r(s_t',a_t')vert s_t,a_t]$$
At first glance, this makes sense, because I compare the value of taking the chosen action $a_i,t$ to the average value in time step $t$ and can evaluate how good my action was.



My question is: a specific state $s_spec$ can occur in any timestep, for example, $s_1 = s_spec = s_10$. But isn't there a difference in value depending on whether I hit $s_spec$ at timestep 1 or 10 when $T$ is fixed? Does this mean that for every state there is a different q value for each possible $t in 0,ldots,T$? I somehow doubt that this is the case, but I don't quite understand how the time horizon $T$ fits in.



Or is $T$ not fixed (perhaps it's defined as the time step in which the trajectory ends in a terminal state - but that'd mean that during trajectory sampling, each simulation would take a different number of timesteps)?










share|improve this question











$endgroup$




bumped to the homepage by Community 27 mins ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.














  • $begingroup$
    I think I have found something helpful - John Schulman notes in his thesis: !screenshot So it sounds like either I sample trajectories with variable length (arrival in a terminal state) or I encode the time step into the state and the state value function will also take the time step into consideration?
    $endgroup$
    – Dummie Variable
    Sep 15 '18 at 13:49














2












2








2


1



$begingroup$


I am currently going through the Berkeley lectures on Reinforcement Learning. Specifically, I am at slide 5 of this lecture.



At the bottom of that slide, the gradient of the expected sum of rewards function is given by
$$
nabla J(theta) = frac1N sum_i=1^N sum_t=1^T nabla_theta logpi_theta(a_i,t vert s_i,t) (Q(s_i,t,a_i,t) - V(s_i,t))
$$
The q-value function is defined as
$$Q(s_t,a_t) = sum_t'=t^T mathbbE_pi_theta[r(s_t',a_t')vert s_t,a_t]$$
At first glance, this makes sense, because I compare the value of taking the chosen action $a_i,t$ to the average value in time step $t$ and can evaluate how good my action was.



My question is: a specific state $s_spec$ can occur in any timestep, for example, $s_1 = s_spec = s_10$. But isn't there a difference in value depending on whether I hit $s_spec$ at timestep 1 or 10 when $T$ is fixed? Does this mean that for every state there is a different q value for each possible $t in 0,ldots,T$? I somehow doubt that this is the case, but I don't quite understand how the time horizon $T$ fits in.



Or is $T$ not fixed (perhaps it's defined as the time step in which the trajectory ends in a terminal state - but that'd mean that during trajectory sampling, each simulation would take a different number of timesteps)?










share|improve this question











$endgroup$




I am currently going through the Berkeley lectures on Reinforcement Learning. Specifically, I am at slide 5 of this lecture.



At the bottom of that slide, the gradient of the expected sum of rewards function is given by
$$
nabla J(theta) = frac1N sum_i=1^N sum_t=1^T nabla_theta logpi_theta(a_i,t vert s_i,t) (Q(s_i,t,a_i,t) - V(s_i,t))
$$
The q-value function is defined as
$$Q(s_t,a_t) = sum_t'=t^T mathbbE_pi_theta[r(s_t',a_t')vert s_t,a_t]$$
At first glance, this makes sense, because I compare the value of taking the chosen action $a_i,t$ to the average value in time step $t$ and can evaluate how good my action was.



My question is: a specific state $s_spec$ can occur in any timestep, for example, $s_1 = s_spec = s_10$. But isn't there a difference in value depending on whether I hit $s_spec$ at timestep 1 or 10 when $T$ is fixed? Does this mean that for every state there is a different q value for each possible $t in 0,ldots,T$? I somehow doubt that this is the case, but I don't quite understand how the time horizon $T$ fits in.



Or is $T$ not fixed (perhaps it's defined as the time step in which the trajectory ends in a terminal state - but that'd mean that during trajectory sampling, each simulation would take a different number of timesteps)?







machine-learning deep-learning reinforcement-learning policy-gradients actor-critic






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Aug 29 '18 at 8:16







Dummie Variable

















asked Aug 28 '18 at 14:56









Dummie VariableDummie Variable

362




362





bumped to the homepage by Community 27 mins ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.







bumped to the homepage by Community 27 mins ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.













  • $begingroup$
    I think I have found something helpful - John Schulman notes in his thesis: !screenshot So it sounds like either I sample trajectories with variable length (arrival in a terminal state) or I encode the time step into the state and the state value function will also take the time step into consideration?
    $endgroup$
    – Dummie Variable
    Sep 15 '18 at 13:49

















  • $begingroup$
    I think I have found something helpful - John Schulman notes in his thesis: !screenshot So it sounds like either I sample trajectories with variable length (arrival in a terminal state) or I encode the time step into the state and the state value function will also take the time step into consideration?
    $endgroup$
    – Dummie Variable
    Sep 15 '18 at 13:49
















$begingroup$
I think I have found something helpful - John Schulman notes in his thesis: !screenshot So it sounds like either I sample trajectories with variable length (arrival in a terminal state) or I encode the time step into the state and the state value function will also take the time step into consideration?
$endgroup$
– Dummie Variable
Sep 15 '18 at 13:49





$begingroup$
I think I have found something helpful - John Schulman notes in his thesis: !screenshot So it sounds like either I sample trajectories with variable length (arrival in a terminal state) or I encode the time step into the state and the state value function will also take the time step into consideration?
$endgroup$
– Dummie Variable
Sep 15 '18 at 13:49











1 Answer
1






active

oldest

votes


















0












$begingroup$

In this case, I think it doesn't matter when you reach $s_spec$, but how the q-value gets updated because of taking an action at that state.
Therefore there shouldn't be different q-values for each possible $tin 0, ..., T$, only q-values for each possible actions.
I'm sure it does make a difference for being at a state at a specific timestep, but it's the agents job to learn this by using the RL algorithms (like policy gradient method in the lecture).



In regards to $T$ being fixed or not, horizon $T$ can be infinite or fixed to a finite number.
For example, if $T$ is fixed to $10$, the agent should learn a policy that maximizes the total discounted rewards in the finite amount of time, but it may not be the most optimal policy. When $T$ is infinite, there is more time to explore and figure out the most optimal policy.



The closest method I know that takes notice to when the state-action pair was encountered is Experience Replay that is used in DQN.



I'm also learning Reinforcement Learning right now! I recommend Deep RL Bootcamp since they give you labs in Python which are really intuitive.






share|improve this answer











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    1 Answer
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    0












    $begingroup$

    In this case, I think it doesn't matter when you reach $s_spec$, but how the q-value gets updated because of taking an action at that state.
    Therefore there shouldn't be different q-values for each possible $tin 0, ..., T$, only q-values for each possible actions.
    I'm sure it does make a difference for being at a state at a specific timestep, but it's the agents job to learn this by using the RL algorithms (like policy gradient method in the lecture).



    In regards to $T$ being fixed or not, horizon $T$ can be infinite or fixed to a finite number.
    For example, if $T$ is fixed to $10$, the agent should learn a policy that maximizes the total discounted rewards in the finite amount of time, but it may not be the most optimal policy. When $T$ is infinite, there is more time to explore and figure out the most optimal policy.



    The closest method I know that takes notice to when the state-action pair was encountered is Experience Replay that is used in DQN.



    I'm also learning Reinforcement Learning right now! I recommend Deep RL Bootcamp since they give you labs in Python which are really intuitive.






    share|improve this answer











    $endgroup$

















      0












      $begingroup$

      In this case, I think it doesn't matter when you reach $s_spec$, but how the q-value gets updated because of taking an action at that state.
      Therefore there shouldn't be different q-values for each possible $tin 0, ..., T$, only q-values for each possible actions.
      I'm sure it does make a difference for being at a state at a specific timestep, but it's the agents job to learn this by using the RL algorithms (like policy gradient method in the lecture).



      In regards to $T$ being fixed or not, horizon $T$ can be infinite or fixed to a finite number.
      For example, if $T$ is fixed to $10$, the agent should learn a policy that maximizes the total discounted rewards in the finite amount of time, but it may not be the most optimal policy. When $T$ is infinite, there is more time to explore and figure out the most optimal policy.



      The closest method I know that takes notice to when the state-action pair was encountered is Experience Replay that is used in DQN.



      I'm also learning Reinforcement Learning right now! I recommend Deep RL Bootcamp since they give you labs in Python which are really intuitive.






      share|improve this answer











      $endgroup$















        0












        0








        0





        $begingroup$

        In this case, I think it doesn't matter when you reach $s_spec$, but how the q-value gets updated because of taking an action at that state.
        Therefore there shouldn't be different q-values for each possible $tin 0, ..., T$, only q-values for each possible actions.
        I'm sure it does make a difference for being at a state at a specific timestep, but it's the agents job to learn this by using the RL algorithms (like policy gradient method in the lecture).



        In regards to $T$ being fixed or not, horizon $T$ can be infinite or fixed to a finite number.
        For example, if $T$ is fixed to $10$, the agent should learn a policy that maximizes the total discounted rewards in the finite amount of time, but it may not be the most optimal policy. When $T$ is infinite, there is more time to explore and figure out the most optimal policy.



        The closest method I know that takes notice to when the state-action pair was encountered is Experience Replay that is used in DQN.



        I'm also learning Reinforcement Learning right now! I recommend Deep RL Bootcamp since they give you labs in Python which are really intuitive.






        share|improve this answer











        $endgroup$



        In this case, I think it doesn't matter when you reach $s_spec$, but how the q-value gets updated because of taking an action at that state.
        Therefore there shouldn't be different q-values for each possible $tin 0, ..., T$, only q-values for each possible actions.
        I'm sure it does make a difference for being at a state at a specific timestep, but it's the agents job to learn this by using the RL algorithms (like policy gradient method in the lecture).



        In regards to $T$ being fixed or not, horizon $T$ can be infinite or fixed to a finite number.
        For example, if $T$ is fixed to $10$, the agent should learn a policy that maximizes the total discounted rewards in the finite amount of time, but it may not be the most optimal policy. When $T$ is infinite, there is more time to explore and figure out the most optimal policy.



        The closest method I know that takes notice to when the state-action pair was encountered is Experience Replay that is used in DQN.



        I'm also learning Reinforcement Learning right now! I recommend Deep RL Bootcamp since they give you labs in Python which are really intuitive.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Sep 18 '18 at 15:10

























        answered Sep 18 '18 at 15:00









        haruishiharuishi

        85




        85



























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