research42 min studyLesson 770 in Mathematics

Sample complexity and transport statistics

Connect empirical Wasserstein error to moments and dimension without treating every transport estimate as a root-n statistic.

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What you’ll understand

  • State an almost-sure empirical convergence result with the required finite moment.
  • Compute exact two-point Wasserstein errors and expectation bounds.
  • Derive a geometric dimension-dependent lower bound and separate statistical and computational errors.

An empirical distribution is a measure, so its transport distance from the population measures sampling error directly. Consistency is reassuring but does not specify how quickly that distance shrinks. The rate depends on the cost exponent, dimension, tails, and distributional structure. A two-point example exposes the exponent dependence exactly, while a simple volume argument shows why a finite cloud cannot represent high-dimensional continuous mass with arbitrarily small error. These calculations also clarify what a distance between two observed samples can actually establish.

Consistency needs the moment matching the distance

Let X₁, X₂, and so on be independent identically distributed samples from μ on real d-space, and let μₙ put equal mass on the first n observations. If μ has finite p-th absolute moment for p ≥ 1, empirical weak convergence and the strong law applied to |X|ᵖ combine with the Wasserstein convergence criterion to give Wₚ(μₙ, μ) tending to zero almost surely. The p-th moment is not optional in this statement.

Almost-sure convergence is asymptotic and does not give a universal finite-sample confidence radius. Quantitative results depend on dimension and on stronger moment or tail assumptions. A light-tailed law on a line and a high-dimensional heavy-tailed law need not share a rate. Even for compactly supported distributions, geometric approximation costs can dominate simple scalar averaging rates, especially when the dimension is large.

μ^n=1n∑i=1nδXi,Xi∼iidμ∈Pp(Rd)⟹Wp(μ^n,μ)→0 almost surely\widehat\mu_n=\frac1n\sum_{i=1}^n\delta_{X_i},\qquad X_i\overset{\mathrm{iid}}\sim\mu\in\mathcal P_p(\mathbb R^d)\quad\Longrightarrow\quad W_p(\widehat\mu_n,\mu)\to0\ \text{almost surely}
The conclusion uses empirical weak convergence and the strong law for the finite p-th moment. It does not specify a dimension-independent convergence rate.↔ Scroll wide equations horizontally.

Two points reveal how taking the root changes a rate

Suppose μ has success mass q on location one and mass 1 − q on zero. If K successes appear in n draws, the empirical success mass is K/n. Exactly the mass mismatch must move distance one, giving Wₚ to the power p equal to |K/n − q| for every p ≥ 1. The empirical mean is a root-n statistic, but taking the p-th root turns it into a different transport scaling.

Cauchy–Schwarz bounds the expected absolute mismatch by the square root of q(1 − q)/n. Concavity of the square root then bounds expected W₂ by the fourth root of q(1 − q)/n. These are expectation bounds, not deterministic limits on an observed error. A particular sample can exceed them without contradiction. Nor is this two-point calculation a universal rate theorem for arbitrary continuous distributions.

Wp(μ^n,μ)p=∣K/n−q∣,EW1≤q(1−q)n,EW2≤(q(1−q)n)1/4W_p(\widehat\mu_n,\mu)^p=|K/n-q|,\qquad \mathbb E W_1\le\sqrt{\frac{q(1-q)}n},\qquad \mathbb E W_2\le\left(\frac{q(1-q)}n\right)^{1/4}
The ground support is {0, 1}, K is binomial with parameters n and q, and both bounds concern expectations over repeated independent samples.↔ Scroll wide equations horizontally.

A volume argument makes dimension unavoidable

Let μ be uniform on the unit cube in dimension d and let ν have at most N support points, with any weights. A union of N Euclidean balls of radius r has volume at most Nκ_d rᵈ, where κ_d is the volume of the unit ball. Choose that expression equal to one-half. At least half the cube’s volume then lies at distance at least r from every support point. Every coupling must move that mass by at least r, giving W₁ at least r/2.

This is a deterministic lower bound for every N-point approximation, not an upper bound for empirical sampling, and it needs no randomness or restriction on the atom weights. It shows an N to the minus one-over-d obstruction as dimension grows. In statistical use, two independent empirical clouds from the same population can still have positive transport distance. Distinguish that sampling variation from support discretization, entropic regularization bias, and incomplete numerical solution before attributing an observed distance to a population change.

W1(μ,ν)≥12(2κdN)−1/d(μ=Unif⁡([0,1]d), ∣supp⁡ν∣≤N)W_1(\mu,\nu)\ge\frac12(2\kappa_dN)^{-1/d}\qquad\bigl(\mu=\operatorname{Unif}([0,1]^d),\ |\operatorname{supp}\nu|\le N\bigr)
The ball-union volume bound leaves at least half the source mass at distance r = (2κ_dN)^(−1/d) from the target support. This supplies a lower bound for arbitrary target weights.↔ Scroll wide equations horizontally.

An observed error versus an expectation bound

Take a true success probability q = 1/2 and n = 16 observations on {0, 1}. Suppose K = 12 successes are observed. Compute W₁ and W₂ to the population and compare their expectation bounds.

  1. The empirical success mass is 12/16 = 3/4, so its excess over the true mass one-half is one-quarter. Any coupling must move exactly that mass between the two locations at unit distance.

  2. Thus W₁ = 1/4, W₂ squared = 1/4, and W₂ = 1/2 for this observed sample. The mismatch is a mass amount, while the rooted distance depends on the chosen exponent.

  3. Here q(1 − q)/n = 1/64, giving E W₁ ≤ 1/8 and E W₂ ≤ (1/64)^(1/4) = 1/(2√2), approximately 0.354. The observed values exceed these expectation upper bounds, which is allowed: an average bound is not a sample-by-sample bound.

The result

The realized errors are W₁ = 0.25 and W₂ = 0.5. Expectation estimates describe repetition of the experiment rather than forbidding this outcome.

Explore the model

Interactive lab

Interactive lab · probability

Sampling noise in a transport mass mismatch

Pause and Reset, then set both success probability p = 0.5 and trials n = 16. Predict mean success count 8 and variance 4. Interpret the displayed K = 12 bar as an empirical transport error W₁ = 0.25 and W₂ = 0.5 on support {0, 1}; compare it with the central count K = 8, which gives zero mismatch. The chart shows exact binomial probabilities, not fresh random samples, confidence intervals, or a Wasserstein computation.

Sampling noise in a transport mass mismatchThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

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 forA nonzero empirical transport distance can arise entirely from sampling even when the population distribution is fixed.

Check your understanding

What does the cube volume argument prove about an N-atom approximation?

Try it yourself

Let μ be uniform on the unit square and let ν have at most 100 atoms anywhere in the plane. Use the volume argument to bound W₁ below. Explain why this does not give a confidence radius for 100 independent samples.

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

Give me a hint

In two dimensions the unit-ball volume is π. Choose 100πr² = 1/2 and charge distance r to at least half the source mass.

Compare with the solution

Choose r = 1/√(200π), approximately 0.039894. The union of 100 radius-r disks covers at most half a unit of area, even before intersecting with the square, so at least half the uniform source mass is at distance at least r from every target atom. Any coupling costs at least r/2 = 1/(2√(200π)), approximately 0.019947. Arbitrary atom weights do not alter this support obstruction. The result is a deterministic lower bound for every 100-atom target, including empirical measures; a confidence radius would require a probabilistic upper-tail statement under a specified sampling law, which this argument does not provide.

Take it with you

  • Empirical Wₚ consistency follows from weak convergence and the matching finite-moment strong law.
  • Transport roots, distributional geometry, and dimension alter sampling rates.
  • Statistical variation must be distinguished from discretization, regularization bias, and numerical residual.

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