Coupling

A coupling of probability distributions $\mu ,\nu$ is a joint probability distribution of random variables $(X,Y)$ whose marginal distributions are $\mu ,\nu$. It lets us choose the statistical dependence between two random constructions while preserving the individual laws. This choice can make a comparison much easier: giving two simulations the same random input may keep their paths close, even though separately generating both paths would hide that relationship.^[ref]

For example, two fair coin tosses can be generated independently, with mismatch probability $1/2$, or by tossing once and setting $Y=X$, with mismatch probability zero. Both constructions have the same two Bernoulli distributions as marginals. Their joint laws differ. A maximal coupling makes the values agree as often as possible; a coalescing coupling arranges permanent agreement of two processes after a random time. Minimizing expected geometric separation instead leads to Wasserstein distance.

Suppose a finite-alphabet transition kernel is uniformly positive and histories agreeing in their most recent $r$ symbols have total variation distance at most $b_r$. Choose decreasing bounds with $b_r<1$ and summable tail. A maximal coupling can be dominated by a match-length chain that increments with probability $1-b_r$ and resets with probability $b_r$. Its chance of never resetting is ${\prod }_{r\ge 0}(1-b_r)>0$. Successive failed attempts therefore end almost surely, leaving a final infinite run of matches. The dominating chain does not depend on the histories, so future-block laws forget the remote past uniformly. This proves uniqueness of a two-sided law with the prescribed transition kernel without assuming stationarity first.

Two simple symmetric random walks on the planar integer lattice starting at sites of the same parity can be coupled to agree after a finite time almost surely. Run them independently until they meet, then give them identical increments. Their difference walk has return probability $p_{2n}(0,0)$ at time $n$; the divergent return sum proves Markov-chain recurrence and makes the meeting time finite.

On a finite partially ordered state space, $\mu ⪯\nu$ means $\int fd\mu \le \int fd\nu$ for every real-valued order-preserving function $f$. Equivalently, every increasing subset has no larger probability under $\mu$ than under $\nu$. A coupling supported on pairs $(X,Y)$ with $X\le Y$ proves this relation. This concerns order of probability laws, rather than stochastic asymptotic order notation.

On a finite distributive configuration lattice, probability weights satisfying ${\mu }_1(\omega \vee \eta ){\mu }_2(\omega \wedge \eta )\ge {\mu }_1(\omega ){\mu }_2(\eta )$ for all pairs give stochastic domination ${\mu }_1\ge {\mu }_2$. The result holds for nonnegative weights as well: apply the four functions theorem to ${\mu }_1{1}_{A^c},{\mu }_2{1}_A,{\mu }_1{1}_A,{\mu }_2{1}_{A^c}$ for an increasing event $A$, obtaining ${\mu }_1(A^c){\mu }_2(A)\le {\mu }_1(A){\mu }_2(A^c)$.

The cross-lattice sufficient condition for stochastic comparison of two measures:The join and meet are coordinatewise maximum and minimum on a Boolean configuration cube. Normalization factors cancel, so the condition can be checked using unnormalized positive weights.

A synchronous coupling drives two solutions of a stochastic differential equation by the same Brownian motion. Their difference has noise coefficient $\sigma (X)-\sigma (Y)$. A dissipativity inequality for the drift and this coefficient gives the mean-square contraction of synchronously coupled diffusions.

If $2(x-y)(b(x)-b(y))+(\sigma (x)-\sigma (y){)}^2\le -k(x-y{)}^2$, the Itô formula for the squared difference and the Gronwall inequality give $\mathbb{E}|X_t-Y_t{|}^2\le e^{-kt}\mathbb{E}|X_0-Y_0{|}^2$. Localization of $e^{kt}|X_t-Y_t{|}^2$ as a nonnegative local supermartingale gives the same estimate under suitable existence hypotheses.

If $|b|\le B$, then $2(x-y)(b(x)-b(y))+(\sigma (x)-\sigma (y){)}^2\ge -4B|x-y|$. It cannot also be at most $-k|x-y{|}^2$ for every pair on the real line when $k>0$. Global strict dissipativity thus requires relaxing global boundedness of the drift; linear confining drifts give natural compatible examples.

If two coupled processes approach each other in L2 and one family of marginals is uniformly tight, then their expectations of any bounded continuous test function approach each other. On a large compact set use uniform continuity; outside it use boundedness; control excessive coupling separation by the second-moment Markov bound. A constant expectation for one process then identifies the other limit. No global derivative or Lipschitz bound is necessary.

A coalescing coupling of probability distributions of two Markov chains keeps them identical after a meeting time $T$. The coupling inequality for total variation gives $\Vert P^t(x,\cdot )-P^t(y,\cdot ){\Vert }_{\mathrm{TV}}\le \mathbb{P}(T>t)$. A uniform bound on $\mathbb{E}T$, combined with Markov inequality, bounds the mixing time of a Markov chain.

Suppose an involution $R$ of a Markov chain satisfies $P(Rx,Ry)=P(x,y)$. Start one chain at $x$, set the other equal to its image under $R$ until the first hitting time of the fixed-point set of $R$, and subsequently use identical transitions. This is a coalescing coupling, because the symmetry preserves each marginal transition matrix and both chains coincide at that hitting time.

For $r,s\in [0,1]$ and a random variable $U$ with the uniform distribution on $[0,1]$, set $B_r=1_{\{U\le r\}}$ and $B_s=1_{\{U\le s\}}$. Each has the corresponding Bernoulli distribution, and $\mathbb{P}(B_r\ne B_s)=|r-s|$, because disagreement occurs precisely between the two thresholds. This is a maximal coupling of the two laws.

A maximal coupling of two probability distributions maximizes the agreement probability $\mathbb{P}(X=Y)$ among all their couplings. Equivalently, it minimizes $\mathbb{P}(X\ne Y)$. The coupling characterization of total variation distance identifies the minimum mismatch with the total variation distance of the two marginal laws. The term maximal concerns agreement, rather than the size of a correlation coefficient or closeness in a geometric metric.

On a finite set, retain all common probability mass at equal pairs and match the excesses elsewhere. The explicit four-outcome example in coupling characterization of total variation distance realizes this construction and checks its marginal distributions. A maximal coupling need not be unique: different joint arrangements of the residual mass can have the same optimal mismatch.

For a metric $\rho$, the first Wasserstein distance is

In Euclidean space, push an optimal transport plan $\pi$ forward by $(x,y)\mapsto (1-t)x+ty$. The resulting probability measures form a constant speed curve for the p-Wasserstein distance:If the plan is induced by a transport map $T$, this becomes the displayed title formula.

On a Polish space $(X,\rho )$, probability measures with finite first moments satisfyThe supremum runs over functions with Lipschitz constant at most one. Fixing $f(x_0)=0$ ensures integrability from the moment assumption.

For continuous functions $f,g$ on a compact metric space satisfying $f(x)+g(y)\le d(x,y)$, set $h(x)={\min }_y(d(x,y)-g(y))$. The triangle inequality makes $h$ $1$-Lipschitz, while the constraint gives $f\le h$ and evaluation at $y=x$ gives $g\le -h$. Thus the feasible pair $(h,-h)$ has at least as large an objective against any two probability measures. This explains why metric optimal transport duality reduces to a single $1$-Lipschitz function.

For $1\le p<\infty$ and probability measures with finite $p$th absolute moments,where $\rho$ is the ground metric and $\pi$ ranges over transport plans. For $p=2$ this is the second Wasserstein distance.

For probability measures with finite second moments on a metric space,

For probability measures ${\mu }_n,\mu$ on the real line with finite second moments, $W_2({\mu }_n,\mu )\to 0$ exactly when ${\mu }_n$ converges weakly to $\mu$ and their second moments converge. The common-quantile coupling is optimal and satisfies

A Markov kernel is Wasserstein contractive with rate $\alpha >0$ when $W_{\rho }(\mu P,\nu P)\le e^{-\alpha }{W}_{\rho }(\mu ,\nu )$ for all probability distributions $\mu ,\nu$.