Wasserstein metrics and weak convergence
Track the moment condition that upgrades weak convergence and distinguish a transport metric from its powered objective.
On this page
The question for today
What you’ll understand
- Define Wₚ on probability measures with finite p-th moment.
- Explain the role of gluing and Minkowski in the triangle inequality.
- Construct a weakly convergent sequence whose transport moments fail to converge.
Weak convergence checks averages of bounded continuous functions. A tiny amount of probability can therefore move very far away while becoming invisible to those tests. Wasserstein distance also charges the displacement needed to move that tail, so its convergence requires moment control. The exponent p determines how strongly distant mass is penalized. This distinction matters whenever distributional convergence is used to justify transport approximations, and it also explains why the squared transport objective is not itself the same metric as its square root.
The root turns a powered cost into a distance
For p at least one, let Pₚ be probability measures on a complete separable metric space with finite integral of distance from a reference point raised to p. The choice of reference point does not change this class. Define Wₚ by minimizing the expected p-th power of distance over couplings and taking the p-th root. The finite-moment condition makes this value finite between every pair of measures in the class.
Symmetry and nonnegativity follow from the ground metric. A zero value forces identical marginals, using existence or an approximating-coupling argument. To prove the triangle inequality, glue two couplings along their shared intermediate marginal and apply the ground triangle inequality followed by Minkowski’s inequality. Gluing creates a joint law of three positions; it is not the same as assuming the two original couplings were independent. We use the resulting metric theorem rather than reproving the disintegration step here.
Weak convergence needs the matching moment convergence
In Euclidean space, convergence in Wₚ is equivalent to weak convergence together with convergence of the p-th moments about a fixed reference point. The moment requirement rules out transport cost hidden in an escaping tail. Uniform boundedness of those moments is not sufficient; their actual convergence to the limit moment, or an appropriate uniform-integrability condition, is required. On a common compact support, moments are bounded continuous tests, so weak convergence already supplies the needed condition.
The direct implication can be understood by comparing each measure with a reference atom: Wₚ(μ, δ₀) raised to p equals its p-th moment. The reverse implication uses truncation to a large ball, controls the tail costs by uniform integrability, and applies compact-support transport convergence inside the ball. This proof pattern explains why a numerical domain truncation needs a tail estimate rather than only weak convergence of its renormalized histogram.
Different exponents and different normalizations matter
A transport problem with cost |x − y|² reports W₂ squared, whereas W₂ itself is its square root. The square is not a metric: among atoms at zero, one, and two, the direct squared cost is four while the sum of the two adjacent squared costs is only two. This does not contradict the metric theorem, whose triangle inequality concerns the rooted quantity. A factor one-half in the cost further rescales the powered value but leaves its optimizing plans unchanged.
Increasing p places greater emphasis on rare long displacements. For fixed measures with enough moments, W₁ is at most W₂ by the usual comparison of L¹ and L² norms under a coupling. Thus W₂ convergence implies W₁ convergence, while the reverse can fail because a sequence retains second-moment tail mass. These comparisons describe the same ground geometry with different integrability demands, not arbitrary scores that can be interchanged without checking assumptions.
Work it through
A vanishing tail with persistent quadratic cost
For integer n ≥ 1, let μₙ = (1 − 1/n)δ₀ + (1/n)δ√n. Compare weak convergence, W₁ distance, and W₂ distance to δ₀.
For every bounded continuous test f, the difference of integrals is (f(√n) − f(0))/n, whose magnitude is at most 2||f||∞/n. Thus μₙ converges weakly to the atom at zero.
There is only one coupling with target δ₀, so W₁ is the first absolute moment: √n/n = 1/√n, which tends to zero.
The second moment is (1/n)(√n)² = 1. Therefore W₂² = 1 and W₂ = 1 for every n, so no W₂ convergence occurs despite weak convergence and a uniformly bounded second moment.
Weak and W₁ convergence hold, but W₂ convergence fails because the second moments remain one instead of converging to the limit moment zero.
The living notebook
Explore the model
Interactive lab · probability
A bounded-support transport distance from a marginal
For each comparison Pause and Reset, then set both controls. Choose p = 0.25 and n = 1; predict W₁ to δ₀ equals 0.25 and W₂ equals 0.5. In a fresh run use p = 0.01 and n = 1, giving W₁ = 0.01 and W₂ = 0.1. Compute these from the displayed masses on {0, 1}; the lab does not itself evaluate Wasserstein distances or model an escaping tail.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. 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 forChanging the transport exponent changes how probability mass contributes, even on a fixed two-point support.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let νₙ = (1 − 1/n²)δ₀ + (1/n²)δₙ. Determine weak convergence and the W₁ and W₂ distances to δ₀. State which moment causes the difference.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
A target atom fixes the coupling, so each powered distance is the corresponding absolute moment of νₙ.
Compare with the solution
Bounded-test differences are at most 2||f||∞/n², so νₙ converges weakly to δ₀. Its first absolute moment is n/n² = 1/n, hence W₁ tends to zero. Its second moment is n²/n² = 1, so W₂² = 1 and W₂ remains one. Every individual νₙ has finite moments of every order, but their second moments do not converge to the limiting zero moment. Finite moments term by term therefore do not guarantee transport convergence of the sequence.
Keep the idea
Take it with you
- Wₚ is the p-th root of the minimized p-power ground-distance cost.
- Weak convergence must be combined with matching moment convergence on unbounded spaces.
- Uniform moment bounds and individual finiteness are weaker than the tail control required for Wₚ convergence.
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