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An MPS parent Hamiltonian is a sum of local projectors onto the orthogonal complement of the tensor's local support. The MPS is a frustration-free ground state.

An MPS tensor is injective after blocking when products of its physical matrices span the full virtual matrix algebra.

Caliper matching rejects candidate pairs whose distance exceeds a tolerance.

Covariate balance means treatment groups have similar pretreatment covariate distributions.

If a $k$-regular graph on $n$ vertices has a maximum matching of size $m$, its unmatched vertices are independent. Counting their incident edges gives $k(n-2m)\le 2(k-1)m$, and hence $m\ge kn/(4k-2)$.

For a finite graph $G$, the matching number isIt is the cardinality of a maximum cardinality matching. The vertex capacity constraint gives $0\le \nu (G)\le \lfloor |V(G)|/2\rfloor$. The worked path graph examples have $\nu (P_4)=2$ and $\nu (P_6)=3$. A perfect matching reaches $|V(G)|/2$; the empty graph has matching number zero.

A maximal matching is a matching in a graph to which no extra edge can be added while retaining all current selected edges. Equivalently, no edge joins two unmatched vertices in a matching; those vertices form an independent set. Every maximum cardinality matching is a maximal matching, but the converse fails.

Take the path graph $P_4$ with vertices $1,2,3,4$ and edges $(1,2),(2,3),(3,4)$. A greedy algorithm that first selects the middle edge produces $M=\{(2,3)\}$. Both remaining edges touch a matched vertex, so this size-one matching in a graph is maximal. Yet $N=\{(1,2),(3,4)\}$ has two edges. No matching in a graph on four vertices has more than two edges, so $N$ is a maximum cardinality matching. The two remaining free vertices of $M$, namely $1$ and $4$, are not adjacent; that does not prevent a longer augmenting path in a matching from joining them.

Maximality is useful as a quick stopping rule and a modest guarantee. For any maximal matching $M$ and maximum cardinality matching $N$ in a finite simple graph, every edge of $N$ touches an endpoint of an edge of $M$. Assign it to one such edge of $M$. Since $N$ has disjoint endpoints, each edge of $M$ receives at most two assignments, proving $|N|\le 2|M|$. The P₄ example reaches equality. Thus maximality guarantees at least half the largest cardinality, but says nothing comparable about arbitrary edge benefits. Its selected endpoints also form a vertex cover, as explained under vertex cover.

A perfect matching is a matching in a graph that covers every vertex exactly once. Its edges have no shared endpoints, and every vertex is an endpoint of one selected edge. It is useful when a collection must be divided into disjoint pairs with no element left over.

In a bipartite graph with equally many vertices on its two sides, a perfect matching can assign every person one allowed job and every job one person. The example under permanent of a matrix has three people and three jobs, with exactly three valid assignments. The blue edges below show those three perfect matchings; a crossing in the drawing is not an extra vertex.

Every selected edge covers two vertices, so a graph with an odd integer number of vertices has no perfect matching. An even integer number is necessary but does not guarantee existence: an isolated vertex cannot be covered. A perfect matching also need not be unique. For example, the complete bipartite graph $K_{2,2}$ has two perfect matchings, corresponding to the two ways of assigning two people to two jobs.

Counting all perfect matchings is a different question from finding one. For a zero-one biadjacency matrix of a balanced bipartite graph, the count is its permanent of a matrix. The nonzero determinant criterion for a perfect matching supplies an existence test, but cancellation means that a zero determinant does not rule one out.

The finite-horizon Snell envelope of rewards $Z_n$ is defined backward by $V_N=Z_N$ and $V_n=\max \{Z_n,\mathbb{E}[V_{n+1}\mid {\mathcal{F}}_n]\}$. It is the least supermartingale dominating $Z$.

In discrete time, a process $A_n$ is predictable or previsible when $A_n$ is ${\mathcal{F}}_{n-1}$-measurable for every $n\ge 1$. In continuous time, predictability means measurability with respect to the predictable sigma-algebra.

A random time $\tau$ is a stopping time when $\{\tau \le n\}\in {\mathcal{F}}_n$ for every $n$.

An integrable adapted process $(V_n)$ is a supermartingale when $\mathbb{E}[V_{n+1}\mid {\mathcal{F}}_n]\le V_n$.

If $(M_n)$ is a martingale and $U_N[a,b]$ counts its completed upcrossings of $[a,b]$ by time $N$, thenEquivalent conventions may add an initial endpoint term.

If every increment ${\xi }_n=M_n-M_{n-1}$ of a discrete martingale takes values in $\{-1,1\}$, thenIterated conditioning shows that the increments are independent symmetric signs, so $M_n-M_0$ is a simple symmetric random walk.

For a square-integrable martingale starting at zero,For a continuous local martingale stopped so that its quadratic variation is integrable, the Itô isometry makes the right-hand side $4\mathbb{E}[M{]}_t$.

This is the $p=2$ specialization of Doob's martingale inequality.

An adapted integrable process $(M_t{)}_{t\ge 0}$ is a continuous-time martingale when $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$ for every $s\le t$.

A stationary distribution is a probability distribution $\pi$ preserved by a Markov chain's evolution. For a Discrete-time Markov chain in the row convention,Thus $\pi$ is a normalized nonnegative left eigenvector of the transition matrix for eigenvalue one. Starting with $X_0$ distributed as $\pi$ makes every $X_n$ have that same probability distribution. For a time-homogeneous Markov chain, the path probabilities are also unchanged by shifting all observation times together: the process has strict stationarity. This does not mean its individual paths are constant. See Hao Wu's MIT lecture 2, stationary distributions.

For the sunny/rainy transition matrix, write $\pi =(a,1-a)$. The sunny component of $\pi P=\pi$ gives $a=0.8a+0.4(1-a)$, so $0.6a=0.4$ andMoreover,These equal stationary flows verify detailed balance and show that the Markov chain is reversible. The diagonal detailed balance equations hold automatically. Only $1/3$ of the stationary days are rainy, but $0.6$ of the days following a rainy day are rainy: stationarity permits dependence.

A finite Markov chain that is an irreducible Markov chain has a unique stationary distribution. If it is also an aperiodic Markov chain, ${\mu }_0{P}^n\to \pi$ for every initial probability distribution. The sunny/rainy example satisfies both conditions: both directions are possible and each state has a positive holding probability of a Markov chain. See Hao Wu's MIT lecture 4, convergence theorem. For an infinite state space, the additional positive recurrence hypothesis in the countable-state Markov chain convergence theorem matters.

Existence of a stationary distribution alone does not give this convergence. Consider the deterministic two-cycleHere $\pi C=\pi$, but starting in the first state gives alternating probability distributions $(1,0),(0,1),(1,0),\dots$. Its period of a state in a Markov chain is two. The running fraction of visits to either state nevertheless tends to $1/2$, illustrating the Markov chain ergodic theorem. Starting with $\pi$ instead keeps the one-time probability distributions stationary even though every path alternates.

In a finite, time-homogeneous Discrete-time Markov chain with row-stochastic transition matrix $P$, write $i\to j$ when $j$ is accessible from $i$, meaning $(P^n{)}_{ij}>0$ for some $n\ge 0$. States communicate, written $i\leftrightarrow j$, when both $i\to j$ and $j\to i$. Communication is an equivalence relation, and a communicating class is an entire equivalence classThus communicating classes form a partition of the state space; a smaller mutually communicating subset is not itself a class. See Maria Cameron's Markov Chains notes, section 1.2.

The allowance of zero steps matters: $P^0=I$, the identity matrix, makes $i\to i$ true even for a state with no possible positive-time return. Symmetry is built into communication. For transitivity, concatenate a positive-probability path from $i$ to $j$ with one from $j$ to $k$, and do the same in reverse. These observations prove the equivalence relation directly. One-way accessibility alone is not symmetric and does not define the classes; see Accessible state of a Markov chain.

A closed communicating class satisfies $P_{ij}=0$ for every $i\in C$ and $j\notin C$. Once entered, it cannot be left. A singleton class is an absorbing state exactly when it is closed. An open communicating class has an outgoing transition; after a path leaves it, it cannot return, since that would make the visited outside states communicate with it. The internal motion of a closed class can nevertheless continue forever.

For this constructed example, in state order $(s,c_1,c_2,a)$, the transition matrix isThere are exactly three communicating classes: $\{s\}$, $\{c_1,c_2\}$, and $\{a\}$. Only the first is open. The second is closed but neither of its states is an absorbing state; they alternate deterministically. Although $s\to c_2$ via $c_1$, the reverse accessibility fails. A direct matrix power check gives $(P^2{)}_{s,c_2}=1/4$ and $(P^2{)}_{c_2,s}=0$.

Starting at $s$, let $\tau$ be the first hitting time of either closed class. Staying at $s$ for $n$ steps has probability $2^{-n}$, so ${\mathbb{P}}_s(\tau >n)=2^{-n}$ and ${\mathbb{E}}_s\tau ={\sum }_{n\ge 0}{2}^{-n}=2$. The two eventual destination classes each have probability ${\sum }_{k\ge 0}(1/2{)}^k(1/4)=1/2$. Hence the hitting probability of the absorbing state $a$ is only $1/2$, and its unconditional expected hitting time from $s$ is infinite: on half the paths it is never hit. In particular, entering a closed class with certainty does not imply entering an absorbing state with certainty.

Classifying the state space separates two questions: which closed region will eventually be entered, and how does the Markov chain move inside that region? In this example the restriction to $\{c_1,c_2\}$ is an irreducible Markov chain with period two; the restriction to $\{a\}$ is a one-state irreducible Markov chain with period one. The whole Markov chain is reducible. The absorbing state example illustrates a different possibility: every closed class is a singleton, so eventual class selection is also terminal outcome selection.

A continuous-time Markov chain waits an exponential time of rate $q_i$ in state $i$ and then jumps according to probabilities $p_{ij}=q_{ij}/q_i$.