The central limit theorem gives a qualitative weak limit. Berry-Esseen theory asks a finite-sample question: how far is the distribution function of a standardized sum from the Gaussian distribution function? The answer reveals both the role of the third absolute moment and the obstruction created by lattice jumps.
Distances between laws
Definition. For probability laws \(P,Q\) on \(\mathbb R\), with distribution functions \(F_P,F_Q\), the Kolmogorov distance is \[ d_K(P,Q)=\sup_{x\in\mathbb R}|F_P(x)-F_Q(x)|. \] The Wasserstein-1 distance is \[ d_W(P,Q)=\sup_{\|f\|_{\mathrm{Lip}}\le1} \left|\int f\,\mathrm{d}P-\int f\,\mathrm{d}Q\right|. \] Kolmogorov distance tests discontinuous half-line indicators and is therefore more sensitive to atoms than \(d_W\).
Fourier smoothing
Lemma (Esseen smoothing inequality). Let \(F\) have characteristic function \(f\), and let \(G\) have bounded density \(G'\) and characteristic function \(g\). There is a universal constant \(C\) such that, for every \(T>0\), \[ \sup_x|F(x)-G(x)| \le \frac1\pi\int_{-T}^{T} \left|\frac{f(t)-g(t)}{t}\right|\,\mathrm{d}t \frac{C\|G'\|_\infty}{T}. \]
Proof outline. Convolve \(F-G\) with a probability kernel whose Fourier transform is supported on \([-T,T]\). Fourier inversion controls the smoothed difference by the integral in the display. The original distribution functions differ from their smoothed versions only inside intervals of width \(O(T^{-1})\); boundedness of \(G'\), together with a comparison argument at points where \(F-G\) is nearly maximal, bounds this loss by \(C\|G'\|_\infty/T\). The factor \(1/t\) is the Fourier transform of a half-line indicator.
The Berry-Esseen bound
Theorem (independent Berry-Esseen bound). Let \(X_1,\ldots,X_n\) be independent, centered random variables with \[ \sigma_k^2=\mathbb EX_k^2,\qquad \rho_k=\mathbb E|X_k|^3<\infty,\qquad s_n^2=\sum_{k=1}^n\sigma_k^2>0. \] Then, for a universal constant \(C\), \[ \sup_x\left| P\!\left(\frac{\sum_{k=1}^nX_k}{s_n}\le x\right)-\Phi(x) \right| \le C\,\frac{\sum_{k=1}^n\rho_k}{s_n^3}. \]
Proof outline. Normalize \(s_n=1\) and write \(\varphi_k(t)=\mathbb Ee^{itX_k}\). Taylor's inequality \[ \left|e^{iu}-1-iu+\frac{u^2}{2}\right| \le\frac{|u|^3}{6} \] gives \[ \varphi_k(t)=1-\frac{\sigma_k^2t^2}{2}+r_k(t), \qquad |r_k(t)|\le\frac{\rho_k|t|^3}{6}. \] A telescoping product comparison, combined with Gaussian decay of the partial products, yields for \(|t|\le c/\max_k(\rho_k/\sigma_k^2)\), \[ \left|\prod_k\varphi_k(t)-e^{-t^2/2}\right| \le C\left(\sum_k\rho_k\right)|t|^3e^{-ct^2}. \] Insert this estimate into Esseen's smoothing inequality. The Fourier integral is bounded by \(C\sum_k\rho_k\), while choosing the truncation frequency of order \((\sum_k\rho_k)^{-1}\) makes the smoothing remainder of the same order. Rescaling restores the factor \(s_n^{-3}\).
Corollary (i.i.d. form). If \(X_i\) are i.i.d. with \(\mathbb EX_i=0\), \(\mathbb EX_i^2=\sigma^2>0\), and \(\rho=\mathbb E|X_i|^3<\infty\), then \[ d_K\!\left( \mathcal L\!\left(\frac{S_n}{\sigma\sqrt n}\right), N(0,1)\right) \le C\frac{\rho}{\sigma^3\sqrt n}. \]
Why the third absolute moment appears
Proposition. If \(X\) is centered and \(\mathbb E|X|^3<\infty\), then \[ \varphi_X(t) =1-\frac{\sigma^2t^2}{2} -i\frac{\mathbb EX^3}{6}t^3+o(|t|^3). \] Without assuming the signed third moment explicitly, one still has \[ \left|\varphi_X(t)-1+\frac{\sigma^2t^2}{2}\right| \le\frac{|t|^3}{6}\mathbb E|X|^3. \]
Proof. Taylor's formula with integral remainder gives \[ e^{iu}=1+iu-\frac{u^2}{2}-i\frac{u^3}{6}+o(|u|^3). \] The quotient of the remainder by \(|u|^3\) is bounded near zero, while for all \(u\) the third-order remainder is bounded by a constant times \(|u|^3\). Substitute \(u=tX\) and apply dominated convergence using \(|X|^3\).
Sharpness and lattice effects
Example (symmetric Bernoulli law). Let \(P(X_i=1)=P(X_i=-1)=1/2\). For even \(n\), the standardized sum has an atom at zero of size \[ P(S_n=0)=2^{-n}\binom{n}{n/2}\sim\sqrt{\frac{2}{\pi n}}. \] Since \(\Phi\) is continuous, at least one of the two one-sided discrepancies adjacent to zero is at least half this jump. Therefore \[ d_K(\mathcal L(S_n/\sqrt n),N(0,1))\ge \frac{c}{\sqrt n}. \] The order \(n^{-1/2}\) cannot be improved uniformly under only a third-moment assumption.
For lattice laws, a continuity correction moves the Gaussian threshold by half a lattice spacing and often removes the largest visible discretization error. It changes the approximation, not the limit theorem.
First Edgeworth correction
Theorem (first-order Edgeworth expansion). Suppose the standardized i.i.d. variables have \(\mathbb EX=0\), \(\mathbb EX^2=1\), third cumulant \(\kappa_3=\mathbb EX^3\), sufficiently many moments, and a nonlattice regularity condition such as Cramer's condition \(\limsup_{|t|\to\infty}|\varphi_X(t)|<1\). Then \[ P(S_n/\sqrt n\le x) = \Phi(x)+\frac{\kappa_3}{6\sqrt n}(1-x^2)\phi(x) +o(n^{-1/2}) \] uniformly in \(x\).
Proof outline. Expand the cumulant generating exponent: \[ n\log\varphi_X(t/\sqrt n) =-\frac{t^2}{2} -i\frac{\kappa_3t^3}{6\sqrt n} +O\!\left(\frac{t^4}{n}\right). \] Exponentiation gives the Gaussian characteristic function plus a cubic correction. Fourier inversion converts multiplication by \((it)^3\) into the third derivative of \(\phi\); integration produces \((1-x^2)\phi(x)\). Cramer's condition controls the high-frequency part uniformly.
Remark. Berry-Esseen is a robust inequality with an explicit error order. An Edgeworth expansion is a signed asymptotic approximation and requires stronger smoothness or nonlattice assumptions. It can be more accurate, but it need not define a monotone distribution function at finite \(n\).