By substituting the optimal policy $\pi_{\star}$ into the Bellman equation, we get the Bellman equation for $v_{\pi_{\star}}(s)=v_{\star}(s)$:

$$ v_{\star}(s) = \sum\limits_a \pi_{\star}(a|s) \sum\limits_{s'} \sum_r p(s', r | s, a)[r + \gamma v_{\star}(s')]$$

From the above equation, how can we obtain the this one?

$$ v_{\star}(s) = \max\limits_a \sum\limits_{s'} \sum\limits_r p(s', r|s,a) [r + \gamma v_{\star}(s')]$$


1 Answer 1


Your first equation is the definition of any state value function, so it must also be definition of the optimal state value function $v_*$.

The second equation is the definition of $v_*$ in terms of the state-action value function $\color{green}{q_*}$.

In reality, the first equation is also the definition of $v_*$ in terms of $\color{green}{q_*}$, which is what you want to see :)

First, note that

\begin{align} v_\pi(s) &= \sum_a \pi(a \mid s) \color{blue}{\sum_{s'} \sum_r p(s', r|s, a) [r + \gamma v_\pi(s')]} \\ &= \sum_a \pi(a \mid s) \color{blue}{q_\pi(s, a)} \tag{1}\label{1}. \end{align} Now, we denote the optimal state value function as $v_{\pi_*} = v_*$. If we plug $v_*$, $\pi_*$ and $\color{green}{q_*}$ in the equation above, we get your first equation, but we write it as

\begin{align} v_*(s) &= \sum_a \pi_*(a \mid s) \color{green}{q_*(s, a)} \tag{2}\label{2}. \end{align} For finite MDPs, the optimal policy is deterministic, i.e. it chooses one action (the optimal one) with probability $1$, so that means that \ref{2} can be written as

\begin{align} v_*(s) &= 0 \color{green}{q_*(s, a_1)} + \cdots + 1 \color{green}{q_*(s, a_*)} + 0 \color{green}{q_*(s, a_N)} \\ &= \color{green}{q_*(s, a_*)} \\ &= \max_a \color{green}{q_*(s, a)} \\ &=\max_a \color{green}{ \sum_{s'} \sum\limits_r p(s', r|s,a) [r + \gamma v_{*}(s')]} \tag{3}\label{3}, \end{align} where $a_* = \text{argmax}_a\pi_*(a \mid s) = \text{argmax}_a \color{green}{q_*(s, a)}$ is the optimal action. By definition, the optimal action in state $s$ is the one that leads to the highest expected return. See also this answer.

Finally, note that optimal value functions are unique for finite MPDs, so $\color{green}{q_*}$ and $v_*$ are unique.

  • $\begingroup$ still, there are some typos and unclear points: (1) $a_*$ is an action value and $\pi_{*}(s)$ is a probability value and hence they cannot be equal as you wrote in the bottom line. (2) why do we have $q_{*}(s,a_*)= \max_{a}q_{*}(s,a)?$ $\endgroup$ Jan 24, 2023 at 3:45
  • $\begingroup$ @DSPinfinity I used my notation inconsistently. If $\pi_*(s)$ is a probability distribution, then $\pi_*(s)$ would be a probability. However, there, I used $\pi_*$ to refer to a function (or decision rule) that is derived from the stochastic policy. $a_* = \text{argmax}_a \pi_*(s, a)$. $a_*$ is an action, not an action value, $q_*(s, a_*)$ would be the (action) value of $a_*$ in $s$. You have $q_*(s, a_*) = \max_a q_*(s, a)$ because the highest action value in state $s$ of the optimal value function is associated with the optimal action, $a_*$ That's the definition of the optimal action. $\endgroup$
    – nbro
    Jan 24, 2023 at 8:49
  • $\begingroup$ @DSPinfinity Maybe this can also help. $\endgroup$
    – nbro
    Jan 24, 2023 at 8:51

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