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For a finite Markov decision process under fixed policy $\pi$, bounded reward vector $r^{\pi }$ and $0\le \gamma <1$, splitting the discounted return at its first step gives $V^{\pi }=r^{\pi }+\gamma P^{\pi }{V}^{\pi }$. A stochastic matrix has supremum-norm operator bound one, so ${\sum }_{k\ge 0}(\gamma P^{\pi }{)}^k$ converges and $V^{\pi }=(I-\gamma P^{\pi }{)}^{-1}{r}^{\pi }$. This is policy evaluation, not optimization over actions.

Temporal-difference learning uses sampled reward-plus-continuation predictions to approach a Bellman expectation equation. Policy evaluation must be distinguished from finding an optimal policy. Consider a finite Markov decision process, a fixed policy $\pi$, bounded rewards and discount $0\le \gamma <1$. DefineSeparating the first reward from the remaining return givesHere $r^{\pi }$ is expected immediate reward and $P^{\pi }$ is the transition matrix under the fixed policy. In matrix form, $V^{\pi }=(I-\gamma P^{\pi }{)}^{-1}{r}^{\pi }$. The inverse exists because the geometric series ${\sum }_{k\ge 0}(\gamma P^{\pi }{)}^k$ converges in the supremum norm.

Define the Bellman expectation operator $T^{\pi }V=r^{\pi }+\gamma P^{\pi }V$. A stochastic matrix averages component differences, soThe Banach fixed-point theorem gives a unique fixed point and convergence of exact repeated backups. This establishes why the equation has a well-defined solution in the stated discounted setting; it does not require the intelligent agent to know its transition matrix.

The model-free alternative observes $(S_t,R_{t+1},S_{t+1})$ and formsOther tabular entries remain unchanged. Under the same policy and Markov assumptions, conditioning on the current value estimates givesThus the sample error has the correct mean direction for correcting the current Bellman mismatch. The defining idea is bootstrapping: a later prediction supplies part of the earlier prediction's target, so learning need not wait for a final outcome.^[ref]

In tabular temporal-difference policy evaluation, convergence requires appropriate sampling and step sizes, not merely writing the update. Standard sufficient conditions include bounded rewards, infinitely many visits to each relevant state and, along visits to each state, ${\sum }_t{\alpha }_t(s)=\infty$ with ${\sum }_t{\alpha }_t(s{)}^2<\infty$. Diminishing steps allow continuing correction while limiting accumulated sampling noise. Constant steps generally track a noisy estimate instead of converging exactly. The target's dependence on the present values also means that this is not ordinary gradient descent on an independent, fixed supervised label. These tabular guarantees should not be transferred automatically to nonlinear function approximation or off-policy training.

For an exact illustrative problem, alternate deterministically between states $A$ and $B$, receive reward $1$ on leaving $A$ and $0$ on leaving $B$, and take $\gamma =1/2$. ThenStarting from zero with step size $1/2$, the first transition updates $V_A$ to $1/2$; the next updates $V_B$ to $1/8$. At the exact solution, both transition errors are zero. These values are a model calculation, not neural measurements.

This solves evaluation of the specified policy. The optimal Bellman equation instead maximizes over possible actions. An intelligent agent can combine policy evaluation with policy iteration or an actor-critic improvement mechanism. Estimating the value of a poor fixed policy does not by itself make that policy optimal. Episodic tasks use terminal-state value zero; undiscounted continuing tasks require additional conditions or a different formulation.

A proposed neural implementation uses a representation of the situation to drive a value-predicting population, with a reward prediction error modulating recently active synapses. An eligibility trace can preserve a record of prior activity so delayed errors influence earlier states. Temporal representation is indispensable: if the population does not distinguish when a reward is expected, it cannot generate the appropriate omission signal at that time.

Primate recordings during acquisition of successive delayed-response tasks found that dopamine-neuron reward responses were more prominent during learning than after performance was established, while significant cues elicited phasic responses. This supports prediction-sensitive signalling but also allows attentional and motivational contributions.^[ref] The temporal-difference interpretation explains how a reward response can diminish as reward becomes predicted and a response can develop at an informative cue; a worse-than-expected omission gives a negative error. The physiological interpretation is a quantitative hypothesis, not an identification of every dopamine signal with a single critic variable.^[ref]

Causal evidence comes from reward-timed stimulation of dopamine neurons in rat blocking and extinction tasks. Artificial activation, intended to supply a positive prediction-error-like event, increased subsequent cue-driven reward seeking. This strengthens the teaching-signal interpretation beyond response correlation. It does not show that all components of the Bellman computation reside in dopamine neurons or that the entire algorithm is unique.^[ref]

A probability distribution on the countable state space $\mathcal{S}$ is precisely a row vector satisfyingA transition matrix is a stochastic matrix, meaningEvery row therefore gives the conditional probability distribution of the next state given the present state. These conditions do not require $\pi$ to be an invariant distribution of $P$.

The transition matrix is $P=(p_{ij}{)}_{i,j\in S}$, where $p_{ij}=P(X_{n+1}=j\mid X_n=i)$. It is row-stochastic: $p_{ij}\ge 0$ and ${\sum }_j{p}_{ij}=1$ for each $i$. Initial states of zero probability can be specified by the chain's transition mechanism, so the matrix describes the chain from any starting state.

With column vectors, entry $P_{ij}$ is the fraction moving from region $j$ to region $i$. The first column is therefore $(1/2,1/4,1/4{)}^{\mathsf{T}}$, the second is $(1/2,1/2,0{)}^{\mathsf{T}}$, and the third is $(1/2,0,1/2{)}^{\mathsf{T}}$. HenceEach column sums to one, as required for this convention for a stochastic matrix. Multiplying by $P$ conserves the total resources.

The transition matrix acts on a column probability vector in the order Central, Headington, Botley. Thus column $j$ records the destinations of buses starting at station $j$. The Central-to-Headington, Headington-to-Botley and Botley-to-Central entries giveEach column sums to one, as required for this column convention for a stochastic matrix.

A probability vector lists the probabilities of the outcomes in a finite sample space. Its entries are nonnegative and sum to one, so a column probability vector is a point of a probability simplex. Row and column conventions describe the same probability distribution, but they determine whether a stochastic matrix acts on the right or its transpose acts on the left.

Every finite Markov chain has a stationary distribution, without requiring Markov chain irreducibility or aperiodicity. For a row-stochastic matrix $P$, put $B=P^T$ and choose a column probability vector $q$. Every $B^{j}q$ lies in the probability simplex. Their arithmetic averages $x_N=N^{-1}{\sum }_{j=0}^{N-1}{B}^{j}q$ therefore lie in this compact set, while telescoping givesA convergent subsequence has a limit $x$ in the probability simplex. Continuity of matrix multiplication gives $Bx=x$, so $\pi =x^T$ is a stationary distribution. The individual iterates need not converge: a deterministic two-state swap alternates its point masses forever.

For a time-homogeneous continuous-time Markov chain, $P_{ij}(t)={\mathbb{P}}_i(X_t=j)$ and the matrices satisfy $P(s+t)=P(s)P(t)$, the Chapman-Kolmogorov equation. On a finite state space with Q-matrix $Q$, the matrix exponential gives $P(t)=e^{tQ}$.

The distinction from a discrete-time transition matrix is the elapsed-time parameter: $P(t)$ describes where the process is after $t$ time units, allowing any number of intervening jumps. Each $P(t)$ is a stochastic matrix, with $P(0)=I$. A Q-matrix instead records instantaneous rates and has row sums zero. Sampling at a fixed spacing $\Delta$ yields a Discrete-time Markov chain with transition matrix $P(\Delta )$. See Hao Wu's MIT lecture 21, transition semigroups and lecture 22, infinitesimal rates.