I am following the book "Reinforcement Learning: An Introduction" by Richard Sutton and Andrew Barto, and they give an example of a problem for which the value function can be computed explicitly by solving a system of $\lvert S \rvert $ equations that have $\lvert S \rvert $ unknowns. Each of these $\lvert S \rvert$ equations is given by:

$$v_{\pi}(s) = \sum_{a} \pi(a\rvert s) \sum_{s^{\prime}}\sum_{r} p(s^{\prime}, r \rvert s,a)[r + \gamma v_{\pi}(s^{\prime})] $$

I am having a hard time understanding how one could solve this system of equations. It seems to me as if each equation consists of a summation of an infinite amount of terms and therefore one would not be able to analytically solve them. Could anyone offer any intuition as to how this system of equations could be explicitly solved?


2 Answers 2


First of all, we assume that we have a finite MDP, i.e. the set of states $\mathcal{S}$, the set of actions $\mathcal{A}$ and the set of rewards $\mathcal{R}$ all have a finite number of elements (I didn't think about how the explanations below would extend to other cases, but I suspect you will need differential equations).

For simplicity, let's only consider the value function $v$ (as opposed to the state-action value function $q(s, a)$, but this also applies to $q$). The value function $v$ is defined for all states, i.e. it's a function of the form $v : \mathcal{S} \rightarrow \mathbb{R}$ or, with an alternative notation, $v(s), \forall s \in \mathcal{S}$. So, we can define this function as a vector $\mathbf{v}$ of dimension $|\mathcal{S}| = n$, i.e. $\mathbf{v} \in \mathbb{R}^{|\mathcal{S}|}$, where the $i$th element contains the value of the $i$th state (so we need a function that maps states to indices of this vector, but this is trivial).

The fact that you can represent the value function, in this finite MDP, as a vector should already suggest that you can find this value function by solving a linear system of equations.

However, let me show that by starting with the definition of the value function you also provided

\begin{align} v_{\pi}(s) &= \sum_{a} \pi(a\rvert s) \sum_{s^{\prime}}\sum_{r} p(s^{\prime}, r \rvert s,a)[r + \gamma v_{\pi}(s^{\prime})] \label{1}\tag{1}, \; \forall s \in \mathcal{S} \end{align} which can be expanded as follows \begin{align} v_{\pi}(s) &= \sum_{a} \pi(a\rvert s) \sum_{s^{\prime}}\sum_{r} \left[p(s^{\prime}, r \rvert s,a)r + \gamma p(s^{\prime}, r \rvert s,a) v_{\pi}(s^{\prime}) \right] \\ &= \sum_{a} \pi(a\rvert s) \left[ \sum_{r} \underbrace{\sum_{s^{\prime}} p(s^{\prime}, r \rvert s,a)}_{\text{Marginalization of }p \text{ over } s'}r + \gamma \sum_{s^{\prime}} \underbrace{\sum_{r} p(s^{\prime}, r \rvert s,a) }_{\text{Marginalization of }p \text{ over } r} v_{\pi}(s^{\prime}) \right]\\ &= \sum_{a} \pi(a\rvert s) \left[ \sum_{r} p(r \rvert s, a)r + \gamma \sum_{s^{\prime}}p(s^{\prime} \rvert s,a) v_{\pi}(s^{\prime}) \right] \\ &= \sum_{a} \pi(a\rvert s) \left[ r(s, a) + \gamma \sum_{s^{\prime}}p(s^{\prime} \rvert s,a) v_{\pi}(s^{\prime}) \right] \label{2}\tag{2}, \; \forall s \in \mathcal{S} \end{align} where

  • $\sum_{r} p(r \rvert s, a)r = r(s, a)$ (see this).
  • $\sum_{s^{\prime}}p(s^{\prime}, r \rvert s,a) = p(r \rvert s, a)$ (marginalization)
  • $\sum_{r} p(s^{\prime}, r \rvert s,a) =p(s^{\prime} \rvert s,a) $ (marginalization)

In this form, as in equation \ref{2}, the value function can also be written in a different notation

\begin{align} v_{\pi}(s) &= \sum_{a} \pi(a\rvert s) \left[ R_{s}^a + \gamma \sum_{s^{\prime}}P_{ss'}^a v_{\pi}(s^{\prime}) \right] \label{3}\tag{3}, \; \forall s \in \mathcal{S} \end{align} where

  • $R_{s}^a = r(s, a)$
  • $P_{ss'}^a = p(s^{\prime} \rvert s,a)$

We can still write equation \ref{3} in a "simpler" form as follows \begin{align} v_{\pi}(s) &= \sum_{a} \pi(a\rvert s) R_{s}^a + \gamma \sum_{s^{\prime}} \sum_{a} \pi(a\rvert s) P_{ss'}^a v_{\pi}(s^{\prime}) \\ &= R_{s}^\pi + \gamma \sum_{s^{\prime}} P_{ss'}^\pi v_{\pi}(s^{\prime}) \label{4}\tag{4}, \; \forall s \in \mathcal{S} \end{align} where

  • $\sum_{a} \pi(a\rvert s) R_{s}^a = R_{s}^\pi$
  • $\sum_{a} \pi(a\rvert s) P_{ss'}^a = P_{ss'}^\pi$

We can write the definition of the value function in \ref{4} in matrix form for all states $s \in \mathcal{S}$ as follows

\begin{align} \begin{bmatrix} v_\pi(1) \\ \vdots \\ v_\pi(n) \end{bmatrix}= \begin{bmatrix} {R}_1^\pi \\ \vdots \\ {R}_n^\pi \end{bmatrix} +\gamma \begin{bmatrix} {P}_{11}^\pi & \dots & {P}_{1n}^\pi\\ \vdots & \ddots & \vdots\\ {P}_{n1}^\pi & \dots & {P}_{nn}^\pi \end{bmatrix} \begin{bmatrix} v_\pi(1) \\ \vdots \\ v_\pi(n) \end{bmatrix} \tag{5}\label{5}, \end{align} which can be written in a more compact form as follows \begin{align} \mathbf{v} = \mathbf{r} + \gamma \mathbf{P}\mathbf{v} \tag{6}\label{6}, \end{align} which is a very compact form of the Bellman equation (which is a recursive equation: as you can notice, the $\mathbf{v}$ appears on the left and right of the equals sign) that represents the value function (i.e. the value function can be defined as a recursive equation).

In equation \ref{5}, the unknowns are the $|\mathcal{S}| = n$ values of the value function $v$ and there are $n$ equations, so it should now be clear why we can solve this problem by solving a system of equations. Note that here it's assumed that $\pi$, $r(s, a)$ and $p$ are given and known, which, generally, is not the case, that's why we use algorithms like Q-learning.


Provided you have a finite number of states and actions, then there will not be an infinite number of terms. Therefore the state and action spaces need to be discrete and finite before the quote from the book applies.

I am having a hard time understanding how one could solve this system of equations.

There are a few techniques for solving simulteneous equations.

However, what I would probably do is number all the state values from $v_1 = v_\pi(s_1)$ to $v_{N = |\mathcal{S}|} = v_\pi(s_N)$, and write out each line in order:

$$v_1 = w_{1,1} v_1 + w_{1,2} v_2 + w_{1,3} v_3 + ... w_{1,N} v_N + r_1$$

Where $r_1$ is a constant - it is the expected immediate reward when starting from state $1$, but that is not important. It is the constant offset value you get from resolving the sum that is not multiplied by any $v_i$ unknown variable.

You can discover the values of $w_{i,j}$ by expanding the sum in the Bellman equation for each state in turn.

At that point you can build a matrix of the weights, and solve the linear equations by taking the inverse of the matrix.

[from comments] But if the game has no end then theoretically the sum of expected future rewards should be infinite.

The time series definition of $v_{\pi}(s)$:

$$v_{\pi}(s) = \mathbb{E}_{\pi}[\sum_{k=0}^{\infty} \gamma^k R_{t+k+1} | S_t = s]$$

does not appear in the Bellman equation used to establish the linear equations. This is the main benefit of the Bellman equation, it changes the infinite series view of returns into a set of relations that must hold between the value functions of each state.

  • $\begingroup$ But if the game has no end then theoretically the sum of expected future rewards should be infinite. I understand the we use a discount factor gamma to make this tractable, but in essence this is essentially an infinite sum? Or am I misunderstanding something? $\endgroup$
    – phil
    Commented Jul 7, 2021 at 12:55
  • $\begingroup$ @phil: The Bellman equation does not have any time series in it. That is the point of it, it relates values over single steps. I will try to add that to the answer later $\endgroup$ Commented Jul 7, 2021 at 13:05

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