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I am confused about the definition of the optimal value ($V^*$) and optimal action-value (Q*) in reinforcement learning, so I need some clarification, because some blogs I read on Medium and GitHub are inconsistent with the literature.

Originally, I thought the optimal action value, $Q^*$, represents you performing the action that maximizes your current reward, and then acting optimally thereafter.

And the optimal value, $V^*$, being the average $Q$ values in that state. Meaning that if you're in this state, the average "goodness" is this.

For example: If I am in a toy store and I can buy a pencil, yo-yo, or Lego.

Q(toy store, pencil) = -10
Q(toy store, yo-yo) = 5
Q(toy store, Lego) = 50

And therefore my $Q^* = 50$

But my $V^*$ in this case is:

V* = -10 + 5 + 50 / 3 = 15

Representing no matter what action I take, the average future projected reward is $15$.

And for the advantage of learning, my baseline would be $15$. So anything less than $0$ is worse than average and anything above $0$ is better than average.

However, now I am reading about how $V^*$ actually assumes the optimal action in a given state, meaning $V^*$ would be 50 in the above case.

I am wondering which definition is correct.

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I am wondering which definition is correct.

The asterisk * in both the definitions stands for "optimal" in the sense of "value when following the optimal policy"

So this one is correct:

$V^*$ actually assumes the optimal action in a given state, meaning $V^*$ would be $50$ in the above case

However, you have got the definition of Q slightly wrong.

I think this is because you are omitting the parameters.

The state value function uses the state as a parameter, $V_{\pi}(s)$, it returns the value of being in state $s$ and following a fixed policy $\pi$. The * is used to denote following an optimal policy.

The action value function has two parameters - a state and an action that is possible in that state, $Q_{\pi}(s, a)$, it returns the value of being in state $s$, taking action $a$ (regardless of whether it is the best action or not) and following the policy $\pi$ after that point.

Your assertion in the question:

And therefore my $Q^* = 50$

is wrong, or rather not meaningful, as you have not stated the parameters. You already list all the possible values of Q with the parameters. You could say $\text{max}_a Q(\text{toy store}, a) = 50$, or to choose the best action $\pi(\text{toy store}) = \text{argmax}_a Q(\text{toy store}, a) = \text{Lego}$

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    $\begingroup$ Ok I see. So V* assumes the agent takes the best possible action in the state, and then follows the optimal policy thereafter. Does that also mean the baseline at each state should be V* and that A(s, a) = 0 for the optimal case (Q(s,a) - V*), and always negative for all other cases? $\endgroup$
    – Rui Nian
    Aug 24, 2018 at 20:40
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    $\begingroup$ @RuiNian: Yes, the maximum Advantage when you already have an optimal policy is zero. $\endgroup$ Aug 24, 2018 at 20:48

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