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If $r<s$, coupling by extra independent squared normals gives ${\chi }_s^2={\chi }_r^2+W$ with $W\ge 0$, so every fixed upper quantile increases with the degrees of freedom.

For real random variables on one probability space, fix a continuity point $x$ of the cumulative distribution function $F_X$. For every $\epsilon >0$, inclusions of events giveTake the lower and upper limits as $n\to \infty$, using convergence in probability, and then let $\epsilon \downarrow 0$. Continuity at $x$ makes both bounds $F_X(x)$, proving convergence in distribution. The two modes of convergence have different information: convergence in probability concerns the coupling on one probability space, whereas convergence in distribution concerns only the marginal probability distributions.

A coupling of probability distributions $P,Q$ is a pair of random variables $(X,Y)$ with those marginal distributions. On a finite set, and more generally on a standard Borel space, the least achievable mismatch isA maximal coupling attains the minimum; maximal here refers to agreement. An arbitrary coupling can have a much larger mismatch.^[ref]

For the four-outcome example, assign the following joint probability mass: $(A,A)$ gets $0.1$, $(B,B)$ gets $0.1$, $(C,C)$ gets $0.2$, and $(D,D)$ gets $0.3$. The remaining $0.3$ must have $X=A$; place $0.2$ on $(A,B)$ and $0.1$ on $(A,C)$. Row sums give $P$, column sums give $Q$, and the diagonal contains $0.7$. Agreement can reach 70%; mismatch cannot fall below 30%.

To see the lower bound, for any event $E$,The two indicator functions agree whenever $X=Y$. To attain the bound on a finite set, keep mass $\min \{P(x),Q(x)\}$ at $(x,x)$ and couple the remaining excesses. The two residual probability distributions live on disjoint subsets, so their part necessarily mismatches. When the probability distributions are identical, the residual part disappears entirely.

This is a choice of joint construction, not a claim that independently drawn observations are usually equal. Two independent random variables with the same continuous probability distribution can disagree with probability one even though their total variation distance is zero. The construction of a maximal coupling deliberately introduces statistical dependence to reveal how similar the marginal laws are.

For any coupling of random variables, the coupling characterization of total variation distance gives $d_{\mathrm{TV}}(\mathcal{L}(X),\mathcal{L}(Y))\le \mathbb{P}(X\ne Y)$, where $\mathcal{L}$ denotes the law of a random variable. If two coupled Markov chains stay identical after their coalescence time $T$, then $\{X_t\ne Y_t\}\subseteq \{T>t\}$ andThe persistent agreement is essential: a first meeting time does not give this bound if the chains can separate again. A coalescing coupling turns a difficult comparison of probability distributions into a concrete question about how long a joint simulation takes to merge.

If a measure $\rho$ has finite Green energy, thenconverges in $L^2$ for any orthonormal basis $(e_n)$ of $H_0^1(D)$ and independent standard normal coefficients ${\xi }_n$. The limit is centered Gaussian with variance equal to the Green energy of $\rho$ and does not depend on the chosen basis.

For a canonically normalized massless phase scalar field with shift $\pi \mapsto \pi +f\alpha$, the Noether current is $J^{\mu }=f{\partial }^{\mu }\pi$. With invariant one-particle normalization, the displayed matrix element follows from $⟨0|\pi (x)|\pi (p)⟩=e^{-ipx}$. Its divergence is $-f{p}^2{e}^{-ipx}=0$ on the massless shell, despite a nonzero current matrix element. This is the elementary current coupling of a Goldstone boson; conservation constrains its momentum contraction, not every component separately.

With ${\beta }_0=-ϵg_0$, the chain rule gives $\beta (g)={\beta }_{0}dg/d{g}_0$. Inverting the displayed coupling expansion yields $g_0=g-a_1{g}^2/ϵ+(2a_1^2-b_1)g^3/{ϵ}^2$. Substitution gives $\beta =-ϵg-a_1{g}^2+2(a_1^2-b_1)g^3/ϵ+O(g^4)$, so finiteness requires the displayed double-pole condition. An omitted simple-pole term $c_1{g}_0^3/ϵ$ supplies a finite $-2c_1{g}^3$ term; higher-pole consistency does not by itself determine the finite two-loop beta function.

The time reversal of a Brownian motion conditioned to exit at a boundary point can be coupled to radial ${\mathrm{SLE}}_2$ so that its loop-erasure is the radial curve. The coupling requires the Brownian path's first hit of each initial radial segment to occur at its tip. It is an existence theorem for a coupling and does not assert that the Brownian trace equals the simple radial curve, or that an elementary deterministic finite-loop algorithm is defined for all continuum paths.

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.

The noisy voter model updates a chosen vertex either by replacing its spin with fresh random noise or by copying a neighboring spin. A positive refresh probability destroys conserved opinions and permits contractive coupling bounds even on nonregular graphs.

Let $o(\omega )$ be the number of open edges, $k(\omega )$ the number of connected components in the open subgraph, counting isolated vertices, and $m=|E|$. Summing the joint weight over spins requires a constant spin on every open component. There are exactly $q^{k(\omega )}$ choices. Thus the edge marginal isThis is the random-cluster measure, and the joint normalizing constant is $Z_{\mathrm{RC}}$.

Summing instead over edge configurations factors independently over edges. For a fixed spin configuration the resulting weight issince $1-p=e^{-\beta }$. The factor $e^{-\beta m}$ is independent of the spins. Hence the spin marginal is exactly the ferromagnetic Potts measure in the conventionIts partition function satisfies $Z_{\mathrm{RC}}=e^{-\beta m}{Z}_{\mathrm{P}}$.

Conditionally on $\sigma$, edge states are independent: an edge with different endpoint spins is closed with probability one, while one with equal endpoint spins is open with probability $p$. Conditionally on $\omega$, color its $k(\omega )$ open components independently and uniformly from $\{1,\dots ,q\}$. Equivalently, every compatible spin configuration has conditional probability $q^{-k(\omega )}$, and all incompatible configurations have probability zero. These are the two conditional laws of the Edwards-Sokal coupling.

If $x,y$ belong to the same open component, their colors agree surely; otherwise the two independent component colors agree with probability $1/q$. ThereforeRearranging proves the spin-connectivity identity for the Potts model:

Phase transitions refer to infinite-volume limits, not a single finite graph. The identity shows that connectivity is exactly the excess spin agreement above its independent-color baseline. More directly, use a wired random-cluster boundary condition and fix the corresponding boundary spin to color $1$. Its component has color $1$, while every finite component has a uniform color. In the compatible infinite-volume coupling,Consequently normalized spontaneous magnetization isThe onset of a wired infinite cluster and the onset of spontaneous magnetization in the Potts model are the same transition under $p=1-e^{-\beta }$, giving ${\beta }_c(q)=-\log (1-p_c(q))$. This also transfers any jump or continuity of that order parameter between the two descriptions; it makes no assumption about behavior exactly at the critical point.

Fix $\epsilon >0$ and a finite epsilon-net $x_1,\dots ,x_N$ in $X$. Choose a Borel map $r:X\to \{x_1,\dots ,x_N\}$ with $d(x,r(x))\le \epsilon$, for example the nearest point with ties broken by index. Its cells $A_i$ give discrete probability measures with masses $p_i=P(A_i)$ and $q_i=Q(A_i)$. LetThe feasible set is nonempty and compact, so the minimum exists. Pushing any coupling forward by $(r,r)$ changes its cost by at most $2\epsilon$, hence $W_{\epsilon }\le W+2\epsilon$. Conversely, lift a discrete coupling by the measureTerms with zero row or column masses vanish. This lift has marginals $P,Q$, and its cost differs by at most $2\epsilon$. Therefore

Here is the finite linear programming duality proof. Write $d_{ij}=d(x_i,x_j)$ and consider the convex coneIt is closed and finitely generated. One elementary justification of closedness is to reduce each nonnegative representation to linearly independent generators, eliminating a coefficient along any linear dependence; after choosing a subsequence using the same subset of generators, their independent coordinates have bounded, convergent coefficients whenever the represented vectors converge. For $\eta >0$, the point $(p,q,W_{\epsilon }-\eta )$ lies outside $C$. Separation from this closed convex cone gives coefficients $(a,b,\tau )$ withIf $\tau =0$, summing the middle inequalities against any feasible $\pi$ contradicts the last one. Thus $\tau >0$, and $f_i=-a_i/\tau$, $g_j=-b_j/\tau$ obey $f_i+g_j\le d_{ij}$ with dual objective greater than $W_{\epsilon }-\eta$. The weak inequality already proved gives the reverse bound. Letting $\eta \downarrow 0$ proves finite strong duality.

Reduce this discrete dual to a single Lipschitz function. For a feasible pair put ${\varphi }_i={\min }_j(d_{ij}-g_j)$. The triangle inequality gives $|{\varphi }_i-{\varphi }_l|\le d_{il}$, while $f_i\le {\varphi }_i$ and $g_i\le -{\varphi }_i$. ConsequentlyConversely, any $1$-Lipschitz vector $\varphi$ gives a feasible pair $(\varphi ,-\varphi )$. Adding a constant leaves its objective unchanged, so impose ${\varphi }_1=0$. The resulting feasible set is compact, and its maximum is $W_{\epsilon }$.

For a maximizing vector, the McShane extension theorem has the following explicit proof in this situation. DefineThe triangle inequality proves that this extension is $1$-Lipschitz, and the discrete Lipschitz condition makes $\varphi (x_i)={\varphi }_i$. Thus $(\varphi ,-\varphi )$ is a continuous feasible pair. Because $|\varphi (x)-\varphi (r(x))|\le \epsilon$,Letting $\epsilon \downarrow 0$, together with part i, proves $m_d(P,Q)=W(P,Q)$. The infimum defining $W$ is also attained: the set of couplings is a closed subset of the compact space of probability measures on $X\times X$, and integrating the continuous cost is continuous for the weak-star topology.

Use the Lipschitz constant convention $\Vert f{\Vert }_L={\sup }_{x\ne y}|f(x)-f(y)|/d(x,y)$ in the final quantity; it is a seminorm, with constants having value zero. Also interpret the coupling space as $\mathcal{P}(X\times X)$. The printed $\mathcal{P}(X,Y)$ has an undefined $Y$. The third quantity has a duplicated printed label; it is placed in part iii here, with the fourth in part iv.

First prove weak optimal transport duality. For any feasible continuous pair $f,g$ and any coupling of probability distributions $\pi$ with marginals $P,Q$,Taking the supremum over pairs and infimum over couplings gives $m_d(P,Q)\le W(P,Q)$. These quantities are finite: $P\otimes Q$ is a coupling and the continuous cost $d$ is bounded on the compact metric space $X\times X$. The zero pair is feasible, so $0\le m_d\le W\le \mathrm{diam}X$.

The reverse inequality is established in part ii by finite approximation and a proof of the required finite linear programming duality. Parts iii and iv then show that all the dual formulations agree.

Put $D=d_2(\mu ,\nu )$ and $z_t(x,y)=(1-t)x+ty$. The marginal distributions of the optimal transport plan $\Pi$ give ${\mu }_0=\mu$ and ${\mu }_1=\nu$. Alsoso ${\mu }_t$ has a finite second absolute moment. For $0\le s<t\le 1$, the pushforward measure $(z_s,z_t{)}_{\#}\Pi$ is a coupling of ${\mu }_s,{\mu }_t$. ThereforeApply the triangle inequality for the second Wasserstein distance along ${\mu }_0,{\mu }_s,{\mu }_t,{\mu }_1$:All inequalities are equalities, including when $D=0$. Thus this displacement interpolation is a constant speed metric geodesic:

Set $q=y-x$ and $z_t=x+tq$. The definition of pushforward measure gives the confinement energy $\mathcal{V}[{\mu }_t]=\int V(z_t)d\Pi$. The chain rule and differentiation under the integral sign, whenever justified, giveFor example, bounded Hessian matrix is sufficient: it gives $|\nabla V(z)|\le C(1+|z|)$, and the first two differentiated integrands are bounded by $C^{\prime }(1+|x{|}^2+|y{|}^2)$. The dominated convergence theorem applies because $\Pi$ has finite second absolute moments. Compactly supported endpoints also suffice because all interpolating points lie in one compact set.

The printed growth assumption alone does not ensure a classical second derivative. Here is a counterexample within the stated assumptions. On the line take the bounded nonnegative smooth functionand the probability measures $\mu ={\sum }_n{p}_n{\delta }_{n-1/2}$, $\nu ={\sum }_n{p}_n{\delta }_{n+1/2}$. Both have finite second absolute moments. Their translation transport plan is optimal: its cost is one, while every coupling has cost at least the square of its mean displacement, also one. At $t_0=1/2$, $z_{t_0}=n$, so $F(t)=\mathcal{V}[{\mu }_t]$ satisfies $F(t_0)=0$. Each summand is nonnegative, and the Fatou lemma givesThus the requested classical identity requires an additional integrability assumption on the Hessian matrix; quadratic growth controls the energy but not its second derivative.

For a confinement potential $V$ that is a $\lambda$-strongly convex function, the derivative formula, where valid, yieldsThe underlying displacement convexity conclusion is valid under the printed assumptions without that extra differentiation hypothesis. Apply convexity of $V(x)-\lambda |x{|}^2/2$ pointwise and integrate:The same argument on every subinterval makes $t\mapsto \mathcal{V}[{\mu }_t]-\lambda D^2{t}^2/2$ convex, so the second derivative lower bound also holds in the sense of distributions.

SetSymmetry here means $W(-z)=W(z)$, so $\nabla W(0)=0$ and $K(0)=0$. The bounded Hessian matrix and the fact that $J$ is an orthogonal matrix giveThus the particle system is ${\dot{X}}^i=N^{-1}{\sum }_{j}K(X^i-X^j)$. On the configuration space with norm $\Vert X{\Vert }_{\max }={\max }_i|X^i|$, its vector field has Lipschitz constant at most $2L$:The Picard-Lindelöf theorem gives local existence and uniqueness. Linear growth and the Gronwall inequality give $\Vert X(t){\Vert }_{\max }\le e^{2Lt}\Vert X(0){\Vert }_{\max }$, which rules out escape to infinity in finite time and extends the solution to every $t\ge 0$.

The associated empirical measure isHere each ${\delta }_{X^i(t)}$ is a Dirac measure. The finite first absolute moment follows from $\int |x|d{\mu }_t^N=N^{-1}{\sum }_i|X^i(t)|<\infty$. Pairing equal particle labels defines a coupling, soConsequently

A finite first moment suffices for the velocity, but may not suffice for the printed potential. For example, $W(z)=|z{|}^2/2$ has all the stated properties. If $\mu$ is the law of $(S,0)$ with density $2s^{-3}$ on $s\ge 1$, then $\mathbb{E}S=2$ but $\mathbb{E}{S}^2=\infty$, and $(W\ast \mu )(x)=\infty$ for every $x$. Thus the literal definition of $V$ needs a finite second absolute moment, or another hypothesis ensuring integrability of $W$.

For the specified class ${\mathcal{P}}_1$, define the velocity fields directly by $b_{\mu }=K\ast \mu$ and $b_{\hat{\mu }}=K\ast \hat{\mu }$. Equivalently, use the renormalized interaction potentialThe bounded Hessian matrix and $\nabla W(0)=0$ implyThus $V_{\mathrm{ren}}$ is finite, is twice continuously differentiable in space, and has $\nabla V_{\mathrm{ren}}=\nabla W\ast \mu$, by dominated differentiation under the integral sign. Its rotated negative gradient is exactly $b_{\mu }$. Whenever $W\ast \mu$ is finite, the two potentials differ only by an $x$-independent constant and give the same transport equation.

We check the hypotheses for characteristic representation of a measure-valued continuity equation. For every $x,y$,For any coupling $\gamma \in \Gamma ({\mu }_t,{\hat{\mu }}_t)$, the Lipschitz condition on $K$ also givesTaking the infimum over couplings proves the uniform boundApplying the same estimate at two times of one curve proves time continuity of $b_{\mu }$, uniformly in $x$. Also $|m_1({\mu }_t)-m_1({\mu }_s)|\le d_1({\mu }_t,{\mu }_s)$, so $m_1({\mu }_t)$ is bounded on compact time intervals. Spatial partial derivatives of $b_{\mu }$ are continuous and bounded by $L$, using dominated convergence in space and weak convergence of probability measures in time for the bounded continuous kernel $DK$. All corresponding facts hold for $\hat{\mu }$.

The characteristic representation of a measure-valued continuity equation says that a continuous velocity field, continuously differentiable and globally Lipschitz in space and with locally uniform linear growth, has a global characteristic flow map $X_{\mu }(t;s,a)$ solvingFor each $f\in {\mathcal{P}}_1$, its pushforward measure is the unique weak solution in $C([0,\infty ),{\mathcal{P}}_1)$ of the associated continuity equation. The hypotheses have all been verified above, and the fields have zero divergence, so these are also the two required transport equations. Explicitly,For completeness, the Gronwall inequality givesso these pushforward measures have finite first absolute moments. Integrating $|X_{\mu }(t;0,a)-X_{\mu }(s;0,a)|$ against $f$ proves continuity in the Wasserstein distance. The chain rule proves the weak identity. For uniqueness at a fixed terminal time $t$, use the backward test function $\phi (s,x)=\psi (X_{\mu }(t;s,x))$, where $\psi$ is continuously differentiable with compact support. It satisfies ${\partial }_s\phi +b_{\mu }\cdot \nabla \phi =0$ and has compact spatial support uniformly for $0\le s\le t$, by the backward linear-growth bound. The weak identity with temporal cutoffs therefore gives $\int \psi d{\rho }_t=\int \psi (X_{\mu }(t;0,a))df(a)$, identifying every weak solution with the displayed pushforward measure.

Finally compare the two characteristic curves with the same initial point $a$. Put ${\Delta }_t=X_{\mu }(t;0,a)-X_{\hat{\mu }}(t;0,a)$. The spatial Lipschitz condition and the boxed velocity estimate givefor almost every time; the integral form gives the same conclusion at zeros of $\Delta$. The Gronwall inequality impliesThe pushforward measure $(X_{\mu }(t;0,\cdot ),X_{\hat{\mu }}(t;0,\cdot ){)}_{\#}f$ is a coupling of ${\rho }_t,{\hat{\rho }}_t$. Integrating the last bound against the probability measure $f$ provesThe same constant $L$ depends only on the uniform bound for the Hessian matrix of $W$, and is independent of the prescribed curves and their common initial data.

Convergence in distribution to $Y$ means convergence of cumulative distribution functions at every continuity point of the limiting distribution function:Convergence at a jump of $F_Y$ is not required. Only the laws are involved, so this definition does not require a particular coupling of the variables.

The update defines a time-homogeneous Markov process: given its present value, the next value depends only on independent fresh input. Part (b)(i) identifies an invariant law: initializing with the rate-one exponential distribution preserves that distribution at every time. Part (b)(iii) shows convergence of the one-time laws to this invariant law from every deterministic initial value.

For a countable state space, the analogous countable-state Markov chain convergence theorem says that an irreducible Markov chain that is aperiodic and positive recurrent converges to its unique stationary distribution from every initial state. Here the state space is continuous, so that countable-state theorem does not directly supply the proof; the explicit coupling does. The two conclusions have the same equilibrium interpretation.

Let ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. The conjugate momentum fields are ${\pi }_{\varphi }={\dot{\varphi }}^{†}$ and ${\pi }_{{\varphi }^{†}}=\dot{\varphi }$. Inserting the given mode expansion of a free field into $i\int d^{3}x({\dot{\varphi }}^{†}\varphi -{\varphi }^{†}\dot{\varphi })$ produces spatial Dirac delta factors. The $b^{†}{c}^{†}$ and $cb$ terms vanish after integration: they contain ${\omega }_{\mathbf{p}}-{\omega }_{\mathbf{q}}$ on ${\delta }^3(\mathbf{p}+\mathbf{q})$, and ${\omega }_{-\mathbf{p}}={\omega }_{\mathbf{p}}$. The number-conserving terms instead add their two frequency factors, cancelling the mode normalization. The un-subtracted ordered expression isThe last term is an infinite vacuum constant, not a finite physical vacuum charge. Choosing the Fock vacuum to have zero Noether charge, equivalently using normal ordering, givesA state containing $n_c$ quanta created by $c^{†}$ and $n_b$ by $b^{†}$ has Noether charge $n_c-n_b$. Thus the two families describe opposite unit Noether charges; with this source's current the $c^{†}$ quantum has positive Noether charge and the $b^{†}$ quantum negative Noether charge. This is an internal Noether charge; identifying it with electric charge requires a specified electromagnetic coupling. The vacuum subtraction of a complex scalar charge leaves all commutators unchanged.^[ref]

Scope of the expansion: a fixed-frequency expansion with time-independent oscillator coefficients is the free-field or asymptotic-field form. For arbitrary interacting $f$, it can be used as canonical oscillator coordinates at a chosen time, rather than as the exact interacting time dependence. For example, $f(s)=m^{2}s+\lambda s^2$ gives $(\square +m^2)\varphi +2\lambda {\varphi }^{\ast }\varphi \varphi =0$, which a generic free plane-wave superposition does not satisfy. The Noether charge and its generator action remain valid in the interacting theory; the symmetry does not depend on using a free Hamiltonian.