Average enough independent random variables, rescale, and a Gaussian emerges. The central limit theorem describes the limit; a more delicate question asks how quickly we reach it. Even two symmetric, bounded, unit-variance distributions can approach the Gaussian at different powers of the sample size.
Compare a fair coin taking values \(\pm1\) with a continuous uniform variable on \([-\sqrt3,\sqrt3]\). Both have mean zero, variance one, and zero skewness. Yet their normalized sums satisfy
\[ \begin{aligned} W_1(\operatorname{Law}(S_n^{\mathrm B}),\gamma) &\sim\frac{1}{2\sqrt n},\\ W_1(\operatorname{Law}(S_n^{\mathrm U}),\gamma) &\sim\frac{C_{\mathrm U}}n, \end{aligned} \]where \(\gamma=\mathcal N(0,1)\) and
\[ C_{\mathrm U} =\frac{1+4e^{-3/2}}{10\sqrt{2\pi}} \approx0.0755. \]The difference comes from the small staircases in the coin-sum CDF. Their width is still of order \(n^{-1/2}\), regardless of symmetry. This is the phenomenon behind Propositions 10.3–10.4 of Optimal Transport for Machine Learners [1].
Wasserstein distance is an area between CDFs
For probability laws \(\mu,\nu\) on the line with finite first moments,
\[ \begin{aligned} W_1(\mu,\nu) &=\inf_{(X,Y):\,X\sim\mu,\,Y\sim\nu}\mathbb E|X-Y|\\ &=\int_{\mathbb R}|F_\mu(x)-F_\nu(x)|\,dx. \end{aligned} \]The optimal coupling matches equal quantiles. The area between quantile graphs equals the area between the CDF graphs, which gives the second formula [2]. The absolute value matters: positive and negative CDF errors cannot cancel.
Let \(X_1,X_2,\ldots\) be independent copies of a centered, unit-variance variable \(X\), and put \(S_n=n^{-1/2}\sum_{i=1}^nX_i\). Write \(F_n\) for its CDF and \(\Phi,\phi\) for the standard Gaussian CDF and density. We need to understand the shape of \(F_n-\Phi\), not just bound its maximum.
A lattice leaves a sawtooth
A lattice of span \(h\) is a set \(a+h\mathbb Z\). The maximal span is the largest spacing of a lattice containing the support. For a fair \(\pm1\) variable, it is \(2\), not \(1\): the normalized sum has adjacent values separated by \(2/\sqrt n\).
Theorem 1 (the symmetric lattice rate). Suppose \(X\) is symmetric, centered, has variance one, and has finite support contained in a lattice of maximal span \(h>0\). Then
\[ W_1(\operatorname{Law}(S_n),\gamma) =\frac{h}{4\sqrt n}+o(n^{-1/2}). \]Thus the fair-coin constant is \(1/2\).
Proof: average the absolute continuity correction
The lattice Edgeworth expansion gives
\[ \begin{aligned} F_n(x)-\Phi(x) &=\frac h{\sqrt n}\, \psi\!\left(\frac{\sqrt n\,x-na}{h}\right)\phi(x)+r_n(x),\\ \psi(u)&=\tfrac12-\{u\},\qquad \|r_n\|_{L^1}=o(n^{-1/2}), \end{aligned} \]where \(\{u\}\) is the fractional part. Values at the jump points have no effect on the integral. Lattice local-limit and Edgeworth theory give this half-cell correction with \(\|r_n\|_\infty=O(n^{-1})\) [3], [4]. A uniform fourth-moment bound then yields the stated \(L^1\) remainder by the tail argument below. Symmetry removes the smooth skewness term at order \(n^{-1/2}\).
The inequality \(\bigl||u+v|-|u|\bigr|\le|v|\) shows that the remainder changes the Wasserstein distance by \(o(n^{-1/2})\). The absolute sawtooth oscillates increasingly rapidly and has mean
\[ \int_0^1\left|\tfrac12-u\right|\,du=\frac14. \]Consequently its integral against \(\phi\) tends to \(1/4\). To see this without a probabilistic argument, first use a compactly supported step function as weight: each interval contains an integer number of full periods plus two negligible partial periods. Then approximate \(\phi\) in \(L^1\), using the uniform bound \(|\psi|\le1/2\). Multiplying by \(h/\sqrt n\) proves the result. \(\square\)
Geometrically, one transports a smooth Gaussian mass within cells of width \(h/\sqrt n\) to lattice sites near their centers. The average distance to the center of an interval of width \(\delta\) is \(\delta/4\). The theorem makes this picture precise and shows why the constant is universal within this symmetric lattice class.
A density exposes the fourth cumulant
For a symmetric law with a density, the staircase disappears. The first possible smooth correction comes from the fourth cumulant
\[ \kappa_4=\mathbb E[X^4]-3. \]The value \(3\) is the fourth moment of a standard Gaussian, so \(\kappa_4\) measures a departure from Gaussian shape after variance and skewness have been matched.
Theorem 2 (the symmetric density rate). Suppose \(X\) is symmetric, centered, has variance one, has a density, and has finite moments of every order. Then
\[ W_1(\operatorname{Law}(S_n),\gamma) =\frac{|\kappa_4|}{24n} \int_{\mathbb R}|x^3-3x|\phi(x)\,dx+o(n^{-1}). \]The moment assumptions are convenient sufficient conditions, rather than a minimal list. If \(\kappa_4=0\), the conclusion is \(o(n^{-1})\); it does not assert a nonzero equivalent.
Proof: the first nonzero smooth Edgeworth term
For fixed frequency \(t\), the characteristic function has the expansion
\[ \mathbb E e^{itS_n} =e^{-t^2/2} \left(1+\frac{\kappa_4t^4}{24n}+o(n^{-1})\right). \]Symmetry removes the cubic term. Fourier inversion suggests a fourth derivative of the Gaussian density; integrating once produces the third Hermite polynomial \(H_3(x)=x^3-3x\).
The Edgeworth theorem, together with a tail estimate, gives the integrated expansion
\[ F_n-\Phi =-\frac{\kappa_4}{24n}H_3\phi+r_n, \qquad \|r_n\|_{L^1}=o(n^{-1}). \]The analytic input is the usual uniform Edgeworth expansion [3], [4]. It applies because a density makes the characteristic function \(\chi(t)=\mathbb E e^{itX}\) tend to zero at infinity, so Cramér's condition \(\limsup_{|t|\to\infty}|\chi(t)|<1\) holds. The fixed-frequency calculation alone would not justify integration over the entire line.
Why the remainders in both theorems can be integrated
Use the uniform Edgeworth expansion through order \(n^{-2}\). Symmetry removes the terms of orders \(n^{-1/2}\) and \(n^{-3/2}\); the remaining correction after the displayed \(n^{-1}\) term is a Gaussian times a polynomial, of order \(n^{-2}\). Thus \(\|r_n\|_\infty=O(n^{-2})\).
Also \(\sup_n\mathbb E|S_n|^8<\infty\). Indeed, expanding the eighth power, independence and centering eliminate every term with a singly occurring index. At most four distinct indices remain, giving \(O(n^4)\) terms before normalization by \(n^4\).
Put \(R_n=n^{1/4}\). Integrating the uniform bound on \([-R_n,R_n]\) gives \(O(n^{-7/4})\). Outside this interval, with \(Z\sim\gamma\),
\[ \begin{aligned} \int_{|x|>R_n}|F_n(x)-\Phi(x)|\,dx &\le\mathbb E(|S_n|-R_n)_+ +\mathbb E(|Z|-R_n)_+\\ &\le\frac{\mathbb E|S_n|^8+\mathbb E|Z|^8}{R_n^7} =O(n^{-7/4}). \end{aligned} \]The tail of \(n^{-1}H_3\phi\) is exponentially small. Combining the central interval and the tails proves \(\|r_n\|_1=o(n^{-1})\). This is the step which a pointwise expansion cannot replace.
For Theorem 1, the same argument uses \(\|r_n\|_\infty=O(n^{-1})\) and uniformly bounded fourth moments. With the same \(R_n=n^{1/4}\), the central integral is \(O(n^{-3/4})\), and the tail is \(O(R_n^{-3})=O(n^{-3/4})\); the sawtooth times \(\phi\) has a Gaussian tail. Thus \(\|r_n\|_1=o(n^{-1/2})\), as required there.
Apply \(\bigl||u+v|-|u|\bigr|\le|v|\) and integrate to obtain the formula. \(\square\)
The continuous uniform constant, explicitly
For \(X\) uniform on \([-\sqrt3,\sqrt3]\), \(\mathbb E[X^4]=9/5\), so \(\kappa_4=-6/5\). The remaining Gaussian integral can be evaluated without numerical quadrature.
Calculation: split at the roots of the Hermite polynomial
The roots of \(H_3(x)=x(x^2-3)\) are \(0,\pm\sqrt3\), and
\[ \frac{d}{dx}\bigl[(x^2-1)\phi(x)\bigr] =-(x^3-3x)\phi(x). \]By symmetry and integration over the two positive intervals,
\[ \int_{\mathbb R}|x^3-3x|\phi(x)\,dx =2\phi(0)+8\phi(\sqrt3). \]Multiplication by \(|\kappa_4|/24=1/20\) gives
\[ C_{\mathrm U} =\frac{\phi(0)+4\phi(\sqrt3)}{10} =\frac{1+4e^{-3/2}}{10\sqrt{2\pi}}. \]There are two distinct mechanisms: a rapidly oscillating lattice error of order \(n^{-1/2}\), and a smooth kurtosis error of order \(n^{-1}\). Both are visible before taking any absolute-value integral.
Numerical experimentation
Choose the number of summands to inspect the CDF error, then compare the convergence curves and the rescaled quantities \(\sqrt n\,W_1\) and \(nW_1\). The first approaches \(1/2\) for Bernoulli input, and the second approaches \(C_{\mathrm U}\) for continuous uniform input.
This distinction matters numerically. If we first replace a uniform variable by a fixed finite grid, we have changed it into a lattice law. At sufficiently large \(n\), the lattice term dominates and the asymptotic rate changes. Increasing the quadrature resolution for the continuous CDF is legitimate; changing the input into discrete mass is a different experiment.
Likewise, the coin characteristic function \(\cos t\) returns to modulus one at arbitrarily large frequencies. It fails Cramér's condition. More generally, merely being nonlattice is weaker than satisfying that condition. Symmetry and arbitrarily many finite moments cannot remove the coin's lattice obstruction.
References
- Gabriel Peyré. Optimal Transport for Machine Learners, Proposition 10.3 and Proposition 10.4. Book manuscript.
- Gabriel Peyré and Marco Cuturi. Computational Optimal Transport. Foundations and Trends in Machine Learning 11(5–6), 355–607, 2019.
- Valentin V. Petrov. Sums of Independent Random Variables. Springer, 1975. Chapters VI–VII.
- Rabi N. Bhattacharya and R. Ranga Rao. Normal Approximation and Asymptotic Expansions. SIAM, 2010; original edition 1976. Chapters 4–5.