In mathematics, the word operator can refer to several distinct but related concepts. An operator can be defined as a function between two vector spaces, it can be defined as a function where the domain and the codomain are the same, or it can be defined as a function from functions (which are vectors) to other functions (for example, the differential operator), that is, a high-order function (if you are familiar with functional programming).

  • What is the Bellman operator in reinforcement learning (RL)?
  • Why do we even need it?
  • How is the Bellman operator related to the Bellman equations in RL?

1 Answer 1


The notation I'll be using is from two different lectures by David Silver and is also informed by these slides.

The expected Bellman equation is $$v_\pi(s) = \sum_{a\in \cal{A}} \pi(a|s) \left(\cal{R}_s^a + \gamma\sum_{s' \in \cal{S}} \cal{P}_{ss'}^a v_\pi(s')\right) \tag 1$$

If we let $$\cal{P}_{ss'}^\pi = \sum\limits_{a \in \cal{A}} \pi(a|s)\cal{P}_{ss'}^a \tag 2$$ and $$\cal{R}_{s}^\pi = \sum\limits_{a \in \cal{A}} \pi(a|s)\cal{R}_{s}^a \tag 3$$ then we can rewrite $(1)$ as

$$v_\pi(s) = \cal{R}_s^\pi + \gamma\sum_{s' \in \cal{S}} \cal{P}_{ss'}^\pi v_\pi(s') \tag 4$$

This can be written in matrix form

$$\left. \begin{bmatrix} v_\pi(1) \\ \vdots \\ v_\pi(n) \end{bmatrix}= \begin{bmatrix} \cal{R}_1^\pi \\ \vdots \\ \cal{R}_n^\pi \end{bmatrix} +\gamma \begin{bmatrix} \cal{P}_{11}^\pi & \dots & \cal{P}_{1n}^\pi\\ \vdots & \ddots & \vdots\\ \cal{P}_{n1}^\pi & \dots & \cal{P}_{nn}^\pi \end{bmatrix} \begin{bmatrix} v_\pi(1) \\ \vdots \\ v_\pi(n) \end{bmatrix} \right. \tag 5$$

Or, more compactly,

$$v_\pi = \cal{R}^\pi + \gamma \cal{P}^\pi v_\pi \tag 6$$

Notice that both sides of $(6)$ are $n$-dimensional vectors. Here $n=|\cal{S}|$ is the size of the state space. We can then define an operator $\cal{T}^\pi:\mathbb{R}^n\to\mathbb{R}^n$ as

$$\cal{T^\pi}(v) = \cal{R}^\pi + \gamma \cal{P}^\pi v \tag 7$$

for any $v\in \mathbb{R}^n$. This is the expected Bellman operator.

Similarly, you can rewrite the Bellman optimality equation

$$v_*(s) = \max_{a\in\cal{A}} \left(\cal{R}_s^a + \gamma\sum_{s' \in \cal{S}} \cal{P}_{ss'}^a v_*(s')\right) \tag 8$$

as the Bellman optimality operator

$$\cal{T^*}(v) = \max_{a\in\cal{A}} \left(\cal{R}^a + \gamma \cal{P}^a v\right) \tag 9$$

The Bellman operators are "operators" in that they are mappings from one point to another within the vector space of state values, $\mathbb{R}^n$.

Rewriting the Bellman equations as operators is useful for proving that certain dynamic programming algorithms (e.g. policy iteration, value iteration) converge to a unique fixed point. This usefulness comes in the form of a body of existing work in operator theory, which allows us to make use of special properties of the Bellman operators.

Specifically, the fact that the Bellman operators are contractions gives the useful results that, for any policy $\pi$ and any initial vector $v$,

$$\lim_{k\to\infty}(\cal{T}^\pi)^k v = v_\pi \tag{10}$$

$$\lim_{k\to\infty}(\cal{T}^*)^k v = v_* \tag{11}$$

where $v_\pi$ is the value of policy $\pi$ and $v_*$ is the value of an optimal policy $\pi^*$. The proof is due to the contraction mapping theorem.

  • $\begingroup$ why the limit given by applying the bellman optimal operator is the value function of the optimal policy? $\endgroup$
    – piero
    Commented Sep 24, 2023 at 3:47
  • $\begingroup$ @piero: good question. The operator T acts on the Value function V(s) which is itself a function over states. The operator T acting on V gives the V of the next state. If T keeps acting on an initial V ad infinitum, then it gives the V of the "final state", which is the end value of the entire policy. $\endgroup$ Commented Dec 2, 2023 at 5:35
  • $\begingroup$ Actually the domains of T is $\mathbb{R}^{S \times A} \rightarrow \mathbb{R}^{S \times A}$ where S = state space, A = action space. $\endgroup$ Commented Dec 2, 2023 at 6:05

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