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:
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.
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.
