Understanding policy gradient theorem - What does it mean to take gradients of reward wrt policy parameters? The Next CEO of Stack Overflow2019 Community Moderator ElectionSemi-gradient TD(0) Choosing an ActionDoes employment of engineered immediate rewards in RL introduce a non-linear problem to an agent?How does action get selected in a Policy Gradient Method?Potential-based reward shaping in DQN reinforcement learningPolicy Gradient Methods - ScoreFunction & Log(policy)Dueling DQN - why should we decompose and then combine them back into?Policy Gradients - gradient Log probabilities favor less likely actions?RL's policy gradient (REINFORCE) pipeline clarificationReinforcement learning for continuous state and action space
If the heap is initialized for security, then why is the stack uninitialized?
Can a single photon have an energy density?
Anatomically Correct Strange Women In Ponds Distributing Swords
How can I quit an app using Terminal?
WOW air has ceased operation, can I get my tickets refunded?
How can I open an app using Terminal?
Robert Sheckley short story about vacation spots being overwhelmed
How to start emacs in "nothing" mode (`fundamental-mode`)
How do I solve this limit?
Customer Requests (Sometimes) Drive Me Bonkers!
Why were Madagascar and New Zealand discovered so late?
How to write the block matrix in LaTex?
Go Pregnant or Go Home
Unreliable Magic - Is it worth it?
Anatomically Correct Mesopelagic Aves
How to make a software documentation "officially" citable?
How should I support this large drywall patch?
What is the point of a new vote on May's deal when the indicative votes suggest she will not win?
How do scammers retract money, while you can’t?
How to use tikz in fbox?
How did people program for Consoles with multiple CPUs?
Should I tutor a student who I know has cheated on their homework?
Why did we only see the N-1 starfighters in one film?
Which organization defines CJK Unified Ideographs?
Understanding policy gradient theorem - What does it mean to take gradients of reward wrt policy parameters?
The Next CEO of Stack Overflow2019 Community Moderator ElectionSemi-gradient TD(0) Choosing an ActionDoes employment of engineered immediate rewards in RL introduce a non-linear problem to an agent?How does action get selected in a Policy Gradient Method?Potential-based reward shaping in DQN reinforcement learningPolicy Gradient Methods - ScoreFunction & Log(policy)Dueling DQN - why should we decompose and then combine them back into?Policy Gradients - gradient Log probabilities favor less likely actions?RL's policy gradient (REINFORCE) pipeline clarificationReinforcement learning for continuous state and action space
$begingroup$
I am looking for a little clarity on what the policy gradient theorem means. My confusion lies in the fact that the reward $R$ in reinforcement learning is non-differentiable in the policy parameters. As that is the case how does the central objective of policy gradients, finding the gradients of Reward $R$ wrt the parameters of policy function even make sense?
machine-learning reinforcement-learning policy-gradients
$endgroup$
bumped to the homepage by Community♦ 6 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
I am looking for a little clarity on what the policy gradient theorem means. My confusion lies in the fact that the reward $R$ in reinforcement learning is non-differentiable in the policy parameters. As that is the case how does the central objective of policy gradients, finding the gradients of Reward $R$ wrt the parameters of policy function even make sense?
machine-learning reinforcement-learning policy-gradients
$endgroup$
bumped to the homepage by Community♦ 6 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
I am looking for a little clarity on what the policy gradient theorem means. My confusion lies in the fact that the reward $R$ in reinforcement learning is non-differentiable in the policy parameters. As that is the case how does the central objective of policy gradients, finding the gradients of Reward $R$ wrt the parameters of policy function even make sense?
machine-learning reinforcement-learning policy-gradients
$endgroup$
I am looking for a little clarity on what the policy gradient theorem means. My confusion lies in the fact that the reward $R$ in reinforcement learning is non-differentiable in the policy parameters. As that is the case how does the central objective of policy gradients, finding the gradients of Reward $R$ wrt the parameters of policy function even make sense?
machine-learning reinforcement-learning policy-gradients
machine-learning reinforcement-learning policy-gradients
asked Feb 26 at 14:52
MiloMinderbinderMiloMinderbinder
280110
280110
bumped to the homepage by Community♦ 6 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♦ 6 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We want to find the gradient of policy "return" $V$ wrt. parameters of the policy $theta$. Where the return $V$ could be written as "how good is an action $Q$ $times$ probability of taking that action $pi$".
Consider the policy gradient, $nabla_theta V = sum_a Q nabla_theta pi + pi nabla_theta Q$
The first term tells us to adjust the action probability proportionally to how good it is. To me it could read "if an action yields good, take more". That is to move the peak of $pi$ to match the peak of $Q$. This is a reasonable thing to do. But of course since $Q$ cannot directly guide us toward its peak, it is up to our $pi$ to luckily stumble upon the high peak of $Q$. This emphasizes the importance of exploratory nature of $pi$.
The second term is vice versa. That is to move the peak of $Q$ to match the peak of $pi$. This is much harder a task because $Q$ is a function of both action and policy, $Q_pi_theta(s, a)$. We clearly don't have this in a differentiable form i.e. we don't have a universal $Q$ function over the space of all possible $pi$.
We now have a partial gradient from the first term but we have yet to estimate the second term.
Turns out, the second term can be recursively written solely in the form of the first term but with subsequent actions and states.
$$
nabla_theta V_0 = sum Q_0 nabla_theta pi_0 + sum Q_1 nabla_theta pi_1 + sum Q_2 nabla_theta pi_2 + dots
$$
That is to get good policy i.e. policy gradient we only need to move the peaks of $pi$ to match the peaks of $Q$ not only the first (state, action) but also for all subsequent (state, action)'s. This yields the same result as if we differentiate through the $Q$.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "557"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdatascience.stackexchange.com%2fquestions%2f46266%2funderstanding-policy-gradient-theorem-what-does-it-mean-to-take-gradients-of-r%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We want to find the gradient of policy "return" $V$ wrt. parameters of the policy $theta$. Where the return $V$ could be written as "how good is an action $Q$ $times$ probability of taking that action $pi$".
Consider the policy gradient, $nabla_theta V = sum_a Q nabla_theta pi + pi nabla_theta Q$
The first term tells us to adjust the action probability proportionally to how good it is. To me it could read "if an action yields good, take more". That is to move the peak of $pi$ to match the peak of $Q$. This is a reasonable thing to do. But of course since $Q$ cannot directly guide us toward its peak, it is up to our $pi$ to luckily stumble upon the high peak of $Q$. This emphasizes the importance of exploratory nature of $pi$.
The second term is vice versa. That is to move the peak of $Q$ to match the peak of $pi$. This is much harder a task because $Q$ is a function of both action and policy, $Q_pi_theta(s, a)$. We clearly don't have this in a differentiable form i.e. we don't have a universal $Q$ function over the space of all possible $pi$.
We now have a partial gradient from the first term but we have yet to estimate the second term.
Turns out, the second term can be recursively written solely in the form of the first term but with subsequent actions and states.
$$
nabla_theta V_0 = sum Q_0 nabla_theta pi_0 + sum Q_1 nabla_theta pi_1 + sum Q_2 nabla_theta pi_2 + dots
$$
That is to get good policy i.e. policy gradient we only need to move the peaks of $pi$ to match the peaks of $Q$ not only the first (state, action) but also for all subsequent (state, action)'s. This yields the same result as if we differentiate through the $Q$.
$endgroup$
add a comment |
$begingroup$
We want to find the gradient of policy "return" $V$ wrt. parameters of the policy $theta$. Where the return $V$ could be written as "how good is an action $Q$ $times$ probability of taking that action $pi$".
Consider the policy gradient, $nabla_theta V = sum_a Q nabla_theta pi + pi nabla_theta Q$
The first term tells us to adjust the action probability proportionally to how good it is. To me it could read "if an action yields good, take more". That is to move the peak of $pi$ to match the peak of $Q$. This is a reasonable thing to do. But of course since $Q$ cannot directly guide us toward its peak, it is up to our $pi$ to luckily stumble upon the high peak of $Q$. This emphasizes the importance of exploratory nature of $pi$.
The second term is vice versa. That is to move the peak of $Q$ to match the peak of $pi$. This is much harder a task because $Q$ is a function of both action and policy, $Q_pi_theta(s, a)$. We clearly don't have this in a differentiable form i.e. we don't have a universal $Q$ function over the space of all possible $pi$.
We now have a partial gradient from the first term but we have yet to estimate the second term.
Turns out, the second term can be recursively written solely in the form of the first term but with subsequent actions and states.
$$
nabla_theta V_0 = sum Q_0 nabla_theta pi_0 + sum Q_1 nabla_theta pi_1 + sum Q_2 nabla_theta pi_2 + dots
$$
That is to get good policy i.e. policy gradient we only need to move the peaks of $pi$ to match the peaks of $Q$ not only the first (state, action) but also for all subsequent (state, action)'s. This yields the same result as if we differentiate through the $Q$.
$endgroup$
add a comment |
$begingroup$
We want to find the gradient of policy "return" $V$ wrt. parameters of the policy $theta$. Where the return $V$ could be written as "how good is an action $Q$ $times$ probability of taking that action $pi$".
Consider the policy gradient, $nabla_theta V = sum_a Q nabla_theta pi + pi nabla_theta Q$
The first term tells us to adjust the action probability proportionally to how good it is. To me it could read "if an action yields good, take more". That is to move the peak of $pi$ to match the peak of $Q$. This is a reasonable thing to do. But of course since $Q$ cannot directly guide us toward its peak, it is up to our $pi$ to luckily stumble upon the high peak of $Q$. This emphasizes the importance of exploratory nature of $pi$.
The second term is vice versa. That is to move the peak of $Q$ to match the peak of $pi$. This is much harder a task because $Q$ is a function of both action and policy, $Q_pi_theta(s, a)$. We clearly don't have this in a differentiable form i.e. we don't have a universal $Q$ function over the space of all possible $pi$.
We now have a partial gradient from the first term but we have yet to estimate the second term.
Turns out, the second term can be recursively written solely in the form of the first term but with subsequent actions and states.
$$
nabla_theta V_0 = sum Q_0 nabla_theta pi_0 + sum Q_1 nabla_theta pi_1 + sum Q_2 nabla_theta pi_2 + dots
$$
That is to get good policy i.e. policy gradient we only need to move the peaks of $pi$ to match the peaks of $Q$ not only the first (state, action) but also for all subsequent (state, action)'s. This yields the same result as if we differentiate through the $Q$.
$endgroup$
We want to find the gradient of policy "return" $V$ wrt. parameters of the policy $theta$. Where the return $V$ could be written as "how good is an action $Q$ $times$ probability of taking that action $pi$".
Consider the policy gradient, $nabla_theta V = sum_a Q nabla_theta pi + pi nabla_theta Q$
The first term tells us to adjust the action probability proportionally to how good it is. To me it could read "if an action yields good, take more". That is to move the peak of $pi$ to match the peak of $Q$. This is a reasonable thing to do. But of course since $Q$ cannot directly guide us toward its peak, it is up to our $pi$ to luckily stumble upon the high peak of $Q$. This emphasizes the importance of exploratory nature of $pi$.
The second term is vice versa. That is to move the peak of $Q$ to match the peak of $pi$. This is much harder a task because $Q$ is a function of both action and policy, $Q_pi_theta(s, a)$. We clearly don't have this in a differentiable form i.e. we don't have a universal $Q$ function over the space of all possible $pi$.
We now have a partial gradient from the first term but we have yet to estimate the second term.
Turns out, the second term can be recursively written solely in the form of the first term but with subsequent actions and states.
$$
nabla_theta V_0 = sum Q_0 nabla_theta pi_0 + sum Q_1 nabla_theta pi_1 + sum Q_2 nabla_theta pi_2 + dots
$$
That is to get good policy i.e. policy gradient we only need to move the peaks of $pi$ to match the peaks of $Q$ not only the first (state, action) but also for all subsequent (state, action)'s. This yields the same result as if we differentiate through the $Q$.
answered Feb 27 at 3:59
PhizazPhizaz
62
62
add a comment |
add a comment |
Thanks for contributing an answer to Data Science Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdatascience.stackexchange.com%2fquestions%2f46266%2funderstanding-policy-gradient-theorem-what-does-it-mean-to-take-gradients-of-r%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown