I have difficulty understanding the following paragraph in the below excerpts from page 4 to page 5 from the paper Dueling Network Architectures for Deep Reinforcement Learning.

The author said "we can force the advantage function estimator to have zero advantage at the chosen action."

For the equation $(8)$ below, is it correct that $A - \max A$ is at most zero?

... lack of identifiability is mirrored by poor practical performance when this equation is used directly.

To address this issue of identifiability, we can force the advantage function estimator to have zero advantage at the chosen action. That is, we let the last module of the network implement the forward mapping

$$Q(s, a; \theta, \alpha, \beta) = V(s; \theta, \beta) + \left( A(s, a; \theta, \alpha) - \max_{a' \in | \mathcal{A} |} A(s, a'; \theta, \alpha) \right). \tag{8}$$

Now, for $a^∗ = \text{arg max}_{a' \in \mathcal{A}} Q(s, a'; \theta, \alpha, \beta) = \text{arg max}_{a' \in \mathcal{A}} A(s, a'; \theta, \alpha)$, we obtain $Q(s, a^∗; \theta, \alpha, \beta) = V (s; \theta, \beta)$. Hence, the stream $V(s; \theta, \beta)$ provides an estimate of the value function, while the other stream produces an estimate of the advantage function.

I would like to request further explanation on Equation 9, when the author wrote what is bracketed between the red parentheses below.

An alternative module replaces the max operator with an average:

$$Q(s, a; \theta, \alpha, \beta) = V (s; \theta, \beta) + \left( A(s, a; \theta, \alpha) − \frac {1} {|A|} \sum_{a' \in \mathcal{A}} A(s, a'; \theta, \alpha) \right). \tag{9}$$

On the one hand this loses the original semantics of $V$ and $A$ because they are now off-target by a constant, but on the other hand it increases the stability of the optimization: with (9) the advantages only need to change as fast as the mean, instead of having to compensate any change to the optimal action’s advantage in (8).

In the paper, to address the identifiability issue, there are two equations used. My understanding is both equations are trying to fix the advantage part - the last module.

For equation $(8)$, are we trying to make $V(s) = Q^*(s)$, as the last module is zero?

For equation $(9)$, the resulting $V(s)$ = true $V(s)$ + mean$(A)$? As the author said "On the one hand this loses the original semantics of $V$ and $A$ because they are now off-target by a constant". And the constant refers to mean$(A)$? Is my understanding correct?

  • $\begingroup$ Please, next time, ask only one question per post. Even if your questions are related, you should ask each in its separate post, so that people can focus on 1 question at time and future users/readers can find the answer to their specific question more quickly. $\endgroup$
    – nbro
    Commented Jan 25, 2023 at 22:08

2 Answers 2


Yes, you're correct, if Equation 8 is used it will only be possible to get estimates $\leq 0$ out of the term

$$\left( A(s, a; \theta, \alpha) - \max_{a' \in \vert \mathcal{A} \vert} A(s, a'; \theta, \alpha) \right).$$

This matches the meaning that we intuitively assign to the $Q(s, a)$, $V(s)$, and $A(s, a)$ estimators (I'm leaving the parameters $\theta$, $\alpha$, and $\beta$ out of those parentheses for the sake of notational brevity). Intuitively, we want:

  • $Q(s, a)$ to estimate the value of being in state $s$ and executing action $a$ for the policy that we are learning about.
  • $V(s)$ to estimate the value of being in state $s$ for the policy that we are learning about.
  • $A(s, a)$ to estimate the advantage of executing action $a$ in state $s$ for the policy that we are learning about.

In the above three points, "the policy that we are learning about" is the greedy policy, the "optimal" policy given what we have learned so far (ideally this would be truly the optimal policy after a long period of training).

In the last point of the three points above, advantage can intuitively be understood as the gain in estimated value if we choose action $a$ over whatever the expected value would be if we were following the policy that we are learning about.

Since we are trying to learn about the greedy policy, we'll ideally (according to our intuition) want the maximum advantage $A(s, a)$ to be equal to $0$; intuitively, the best action is precisely the one we want to execute in our greedy policy, so that best action should not have any relative "advantage". Similarly, all non-optimal actions should have a negative advantage, because they are estimated to be worse than what we estimate to be the optimal action(s).

This intuition is mathematically enforced by using Equation 8 from the paper for training:

$$Q(s, a; \theta, \alpha, \beta) = V(s; \theta, \beta) + \left( A(s, a; \theta, \alpha) - \max_{a' \in \vert \mathcal{A} \vert} A(s, a'; \theta, \alpha) \right).$$

We can consider two cases to explain what this is doing:

  1. Suppose that action $a$ is the best action we could have selected in state $s$ according to our current estimates, i.e. $a = \arg \max_{a' \in \vert \mathcal{A} \vert} A(s, a'; \theta, \alpha)$. Then, the two terms in the large brackets are equal to each other, so the subtraction yields $0$, and the state-action value estimate $Q(s, a)$ equals the state value estimate $V(s)$. This is exactly what we want because we are trying to learn about the greedy policy.

  2. Suppose that action $a$ is worse than the best action we could have selected in state $s$ according to our current estimates, i.e. $A(s, a; \theta, \alpha) < \max_{a' \in \vert \mathcal{A} \vert} A(s, a'; \theta, \alpha)$. Clearly, I've just stated here that the first term in our subtraction is less than the second term in our subtraction... so the subtraction yields a negative number. This means that the state-action value estimate $Q(s, a)$ becomes less than the estimated state value $V(s)$. This is also what we want intuitively, because we started with the assumption that action $a$ was a suboptimal action. Clearly, if we assume that the action $a$ is suboptimal, that should lead to a reduction in the estimated value.

Note that afterwards, when they start explaining Equation 9, they actually intentionally deviate from these standard, intuitive understandings that we have of what the three estimators should represent.

Concerning the additional question about Equation 9:

A major problem in the stability of training processes for Deep Reinforcement Learning algorithms (such as these DQN-based algorithms) is that the update targets contain components that are predictions made by the NN that is being trained. For example, the Dueling DQN architecture in this paper generates $V(s)$ and $A(s, a)$ predictions, which are combined into $Q(s, a)$ predictions, and those $Q(s, a)$ predictions of the network itself are also used (combined with some non-prediction reward observations $r$) in the loss function defined to train the Neural Network.

In other words, the Neural Network's own predictions are a part of its training signal. When these are used to update the Network, this will likely change its future predictions in similar situations, which means that its update target will also actually change when it reaches a similar situation again; this is a moving target problem. We do not have a consistent set of update targets as we would in a traditional supervised learning setting for example (where we have a dataset collected offline with fixed labels as prediction targets). Our targets are moving around during the training process, and this can destabilize learning.

Now, in that explanation following Equation 9, they essentially argue that this "moving target" problem is less bad with Equation 9 than it is with Equation 8, which can result in more stable training. I'm not sure if there is a formal proof of this, but intuitively it does make sense that this would happen in practice.

Suppose that you update your network once based on Equation 8. If your learning step changes the prediction of the advantage $A(s, a)$ of the best action $a$ by a magnitude of $1$ (kind of informal here, hopefully it makes sense what I'm trying to say), this will in turn move future targets for updates also roughly by a magnitude of $1$ (again, quite informal here).

Now, suppose that you update your network once based on Equation 9. It is unlikely that all of the different actions $a$ have their advantage $A(s, a)$ move by the same magnitude and in the same direction as a result of this update. It is more likely that some will move up, some will move down, etc. And even if they all move in the same direction, some will likely move by a smaller magnitude than others. In some sense, Equation 9 "averages out" the movements triggered by the learning update in all of these different advantage estimates, which causes the network's prediction targets overall to simply move more slowly, reducing the moving target problem. At least, that's the intuitive idea. Again, I don't think there is a formal proof that this happens, but it does turn out to often help in practice.

  • $\begingroup$ Really appreciate your detailed explanation. I I have updated my question with further comments. Could you help me out further? Thanks. $\endgroup$
    – Cheng
    Commented Sep 27, 2018 at 1:03
  • $\begingroup$ @Cheng See edit in answer for explanation of Equation 9. As for vector or matrix notation of Equation 8, yeah, that's fine. When talking about the Equations, I find it easier to consider them in scalar representation rather than vector representation, but they can be interpreted as vectors. $\endgroup$
    – Dennis Soemers
    Commented Sep 27, 2018 at 8:38
  • $\begingroup$ @Cheng Does my edit help to make things more clear? or is something unclear still? $\endgroup$
    – Dennis Soemers
    Commented Sep 30, 2018 at 11:40
  • 3
    $\begingroup$ @Cheng SE format is not a discussion board. You should not keep updating the question, and thus mutating the topic. If you have another question, just post a follow-up as a separate Q&A. $\endgroup$
    – BartoszKP
    Commented Oct 2, 2018 at 15:13
  • 2
    $\begingroup$ @Cheng Yes, the purpose of the site is to create a knowledge base for everyone. It should however be a knowledge base consisting of specific questions and specific answers, not chat history between you and the one who answers. Please see: meta.stackexchange.com/questions/43478/… and ai.stackexchange.com/help/how-to-ask . Your question with all the alterations and responses is really hard to follow. The question post should contain only the question. If you want a clarification use comments, if you want to chat - use chat. $\endgroup$
    – BartoszKP
    Commented Oct 3, 2018 at 8:13

I believe that is explained on the prior page:

"Intuitively, the value function $V$ measures the how good it is to be in a particular state $s$. The $Q$ function, however, measures the the value of choosing a particular action when in this state. The advantage function subtracts the value of the state from the $Q$ function to obtain a relative measure of the importance of each action."

Then two paragraphs above were you started your quote:

"However, we need to keep in mind that $Q(s, a; \theta, \alpha, \beta)$ is only a parameterized estimate of the true $Q$-function. Moreover, it would be wrong to conclude that $V (s; \theta, \beta)$ is a good estimator of the state-value function, or likewise that $A(s, a; \theta, \alpha)$ provides a reasonable estimate of the advantage function.

Equation (7) is unidentifiable in the sense that given $Q$ we cannot recover $V$ and $A$ uniquely. To see this, add a constant to $V (s; \theta, \beta)$ and subtract the same constant from $A(s, a; \theta, \alpha)$. This constant cancels out resulting in the same $Q$ value. This lack of identifiability is mirrored by poor practical performance when this equation is used directly."

Another way of looking at it would be:

  • You receive answers to your question

  • Answers receive votes

  • Answerers have reputation

In a perfect world people could vote based on reputation, with a weighing based upon the correctness of the answer.

You could simply look at which answer received the most votes and choose it as correct.

In the real world things don't work that way, things are correct or incorrect whether they are measured or not (think quantum mechanics) and measurement doesn't always reveal the true answer.

See: Parameter Estimation.

The estimate of the advantage is only so good, sometimes it's useful to consider it and in other instances it's useful to reject it - intelligently doing both maximizes it's usefulness.

  • $\begingroup$ Thanks @Rob. I commented in the updated question. If you understand the paper, please do help me out. Thanks. $\endgroup$
    – Cheng
    Commented Sep 25, 2018 at 6:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .