research36 min studyLesson 761 in Mathematics

Monge and Kantorovich: maps versus couplings

Formulate mass rearrangement with deterministic maps or joint distributions and understand why allowing mass splitting changes feasibility.

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What you’ll understand

  • Express a transport map through its pushforward constraint.
  • Define a coupling and identify its two fixed marginals.
  • Construct a feasible mass-splitting plan when no deterministic map exists.

Comparing two distributions requires deciding which source locations supply which target locations. The cost depends on those pairings, not just on the separate means or histograms. Monge assigns one destination to every source point. Kantorovich allows a joint distribution of origins and destinations, so one atom’s mass may be split. This enlargement is not merely a numerical convenience: it can turn an impossible matching request into a feasible optimization problem while keeping both marginal distributions exactly fixed.

A map must preserve the prescribed mass

Let μ and ν be probability measures on source and target spaces, and let c(x, y) be a nonnegative measurable cost. A measurable map T is feasible when its pushforward T#μ equals ν: the target mass of every measurable set is the source mass of its preimage. The Monge objective integrates c(x, T(x)) against μ. A source point has one image, so an atomic source cannot divide its mass among several destinations.

The map constraint is nonlinear in the unknown map and may admit no candidate. For example a single source atom always pushes forward to a single target atom, regardless of how its location is chosen. It cannot produce a target containing two atoms with positive masses. This is a structural obstruction, not an optimizer’s failure to search hard enough. If the two distributions have unequal total mass, even the balanced coupling problem below is infeasible; unbalanced transport requires a different formulation.

inf⁡T#μ=ν∫c(x,T(x)) dμ(x),(T#μ)(B)=μ(T−1(B))\inf_{T_\#\mu=\nu}\int c(x,T(x))\,d\mu(x),\qquad (T_\#\mu)(B)=\mu(T^{-1}(B))
The infimum is over measurable deterministic maps satisfying the exact pushforward constraint. If there is no such map, the feasible set is empty.↔ Scroll wide equations horizontally.

A coupling records origins and destinations jointly

A coupling π is a probability measure on the product space whose first marginal is μ and whose second marginal is ν. It records how much source mass is paired with each target region. The independent product μ ⊗ ν is always a coupling, so feasibility is automatic for two probability measures. A transport map induces the special coupling (id, T)#μ, supported on its graph; general couplings need not have this graph form.

In finite spaces, couplings are nonnegative matrices with prescribed row and column sums. The row sums supply source masses and the column sums supply target demands. The objective is linear in the matrix entries. Keeping the marginals fixed is essential: without those constraints, minimizing a nonnegative transport cost would permit sending no mass at all. A coupling’s randomness describes allocation conditional on the origin, not random movement added after solving an unrelated optimization.

Π(μ,ν)={π:(pr⁡1)#π=μ, (pr⁡2)#π=ν},inf⁡π∈Π(μ,ν)∫c(x,y) dπ(x,y)\Pi(\mu,\nu)=\{\pi:(\operatorname{pr}_1)_\#\pi=\mu,\ (\operatorname{pr}_2)_\#\pi=\nu\},\qquad \inf_{\pi\in\Pi(\mu,\nu)}\int c(x,y)\,d\pi(x,y)
The projections select origin and destination coordinates. Every feasible map gives a feasible coupling with the same cost, so the coupling infimum cannot be larger.↔ Scroll wide equations horizontally.

A relaxation preserves the question but enlarges the choices

Kantorovich transport minimizes over a convex set: averaging two couplings preserves their marginals. This convexity enables linear programming, dual certificates, and compactness arguments. It does not imply that every optimal coupling is deterministic. Conditions ensuring a map, such as Brenier’s theorem for quadratic cost with an absolutely continuous source, are additional results. An optimizer’s matrix should not be silently converted into a map by choosing the largest entry of each row, because that can destroy target masses.

The cost function supplies the geometry. Squared Euclidean distance discourages long moves more strongly than ordinary distance, and multiplying all costs by a positive constant changes the reported value without changing minimizers. Neither formulation by itself guarantees smooth paths between endpoints; it specifies only the origin–destination allocation. Later lessons will turn optimal couplings into interpolating distributions and distinguish their static cost from the kinetic action of a time-dependent motion.

A source atom that must split

Let μ be a unit atom at zero and ν put half its mass at −1 and half at 1. Use cost c(x, y) = |x − y|². Compare Monge feasibility with the Kantorovich plan and value.

  1. A deterministic map sends the sole source point zero to one destination T(0), so its pushforward is a single unit atom. It cannot equal the stated two-atom target; the Monge problem has no feasible map.

  2. The marginal constraints force a coupling with mass one-half at (0, −1) and one-half at (0, 1). Its first marginal is the atom at zero and its second marginal is exactly ν.

  3. Each pair has squared distance one, so the plan cost is one-half plus one-half, equal to one. The coupling is unique here because the source is a single atom, making this also the optimal Kantorovich value.

The result

Monge is infeasible, while Kantorovich has a unique feasible plan with cost 1. Splitting changes feasibility without changing either prescribed marginal.

Explore the model

Interactive lab

Interactive lab · markov

A split marginal from one starting state

Pause and Reset, then set transition p = 0.5 and transition q = 0.5. Starting from state zero, predict probabilities (0.5, 0.5) after one Step. This is a concrete conditional allocation of one starting atom into two target atoms, analogous to a coupling row. The lab computes a fixed two-state Markov chain; it neither chooses an optimal cost nor solves a general transport problem.

A split marginal from one starting stateExact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Current readings: P(state 1) at step 0: 0%; Stationary P(state 1): 60%; Chain behavior: Monotone convergence. All parameters are adjustable with the labeled controls below.Two-state Markov distribution after 0 steps: state zero probability 1, state one probability 0.Initial distribution (1, 0) · step 0p = 0.3q = 0.201100%0%1−p1−qP(0)P(1)Monotone convergence
P(state 0)P(state 1)Stationary distribution
P(state 1) at step 0
0%
Stationary P(state 1)
60%
Chain behavior
Monotone convergence
t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forOne origin can supply several destinations in a plan, even when a deterministic map cannot split that origin.

Check your understanding

Why can a unit source atom not be mapped deterministically to two positive target atoms?

Try it yourself

On the locations {0, 1}, let source masses be (1/2, 1/2) and target masses (1/4, 3/4), with squared-distance cost. Find an optimal coupling and explain whether any deterministic source map can achieve the target.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use diagonal transport whenever possible and identify the remaining target deficit. A deterministic image of each half-mass atom has mass in multiples of one-half.

Compare with the solution

Send mass 1/4 from zero to zero, 1/4 from zero to one, and 1/2 from one to one. The matrix is [[1/4, 1/4], [0, 1/2]] and its cost is 1/4. Every coupling must move at least the extra 1/4 demanded at location one, so this value is optimal. A deterministic map from two atoms of mass one-half can create target masses only 0, 1/2, or 1 at a location, never 1/4. Thus there is no feasible Monge map even though the coupling problem has the stated optimizer.

Take it with you

  • A pushforward specifies exactly how a map preserves the target distribution.
  • Couplings have fixed marginals and can split atomic source mass.
  • Relaxing maps to plans creates a convex problem without automatically making the optimizer deterministic.

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