Why a Lattice Needs a Correction
A continuous Edgeworth approximation treats probability as if it were spread smoothly along the real line. A lattice-valued random variable instead places its probability at separated points. When a CDF is formed, the last included point contributes a boundary effect: integrating a smooth density only up to that point misses about half of the probability cell centered there. The correction is of the same order as the first skewness correction, so it cannot be ignored in a first-order approximation.
We derive the effect for \(S_n\sim\operatorname{Binomial}(n,p)\), with fixed \(0<p<1\), and write \(q=1-p\). Its mean and standard deviation are \(np\) and \(\sigma=\sqrt{npq}\). The lattice spacing after standardization is \(h=1/\sigma\). The binomial model gives a convenient setting in which both the local expansion and the CDF correction can be made explicit.
Local Edgeworth Approximation for the Binomial
Set \(x_j=(j-np)/\sigma\), and let \(\phi(x)=(2\pi)^{-1/2}e^{-x^2/2}\) be the standard normal density. The standardized third cumulant of the sum is \(\lambda_3=(q-p)/\sigma\). With \(H_3(x)=x^3-3x\), the first-order local expansion is
The factor \(h\) matters: this is a probability at one lattice point, not a density value. Here is a form of the approximation strong enough to sum into a CDF without relying on an uncontrolled accumulation of errors.
To see why this bound has the stated form, apply Stirling's formula with remainder to \(n!,j!\), and \((n-j)!\) in \(P(S_n=j)=\binom njp^jq^{n-j}\). On the range \(|x_j|\le n^{1/12}\), both \(j\) and \(n-j\) are bounded below by a positive constant times \(n\), for large \(n\). Thus the Stirling remainders are \(O(n^{-1})\). Taylor expansion of the logarithms about \(j=np\) gives the normal log-density term \(-x_j^2/2\), the standardized skewness term \(\lambda_3H_3(x_j)/6\), and a remaining log error bounded by \(C n^{-1}(1+|x_j|^4)\). The skewness term is bounded in absolute value by \(C n^{-1/2}(1+|x_j|^3)\), which tends uniformly to zero on this range. Exponentiating, its quadratic contribution is bounded by \(C n^{-1}(1+|x_j|^6)\), giving the displayed probability bound.
Worked Example: A Binomial Point Probability
Take \(S_{20}\sim\operatorname{Binomial}(20,0.4)\) and approximate \(P(S_{20}=10)\). Here \(q=0.6\), \(\mu=np=8\), \(\sigma=\sqrt{4.8}\approx2.1909\), and \(h=1/\sigma\approx0.4564\). Thus
The Hermite polynomial value is \(H_3(x_{10})=x_{10}^3-3x_{10}\approx-1.978\), and \(\phi(x_{10})\approx0.2630\). Substitution into the local expansion gives
For comparison, the exact probability is \(\binom{20}{10}(0.4)^{10}(0.6)^{10}\approx0.1171\). This is a small-\(n\) illustration of the calculation, not a claim that the asymptotic remainder bound is numerically sharp at \(n=20\).
From Lattice Sums to the Continuity-Corrected CDF
For an integer \(k\), put \(z_k=(k+\tfrac12-np)/\sigma=x_k+h/2\). The half-step is the upper edge of the cell centered at \(x_k\). The following result gives the CDF approximation, including an error bound that accounts for the full sum rather than just a bounded standardized region.
where the error is bounded in absolute value by \(C/n\), uniformly over integer \(k\). Here \(\Phi\) is the standard normal CDF.
Proof. First restrict attention to lattice points satisfying \(|x_j|\le n^{1/12}\). The local expansion's remainders summed over any subset of these points are bounded by
The sum is bounded uniformly in \(n\): a Gaussian times a fixed polynomial is integrable, and its Riemann sums on a grid of mesh \(h\to0\) are uniformly bounded. The points outside this range do not cause a hidden CDF error. Hoeffding's inequality gives
The corresponding normal and polynomial-times-normal tails are also exponentially small in \(n^{1/6}\). Consequently, truncating the probability sum and the approximating sums to the stated range costs \(o(n^{-1})\), uniformly in the cutoff \(k\).
Now sum the two leading local terms up to \(j=k\). The points \(x_j\) are cell centers with spacing \(h\), and the union of cells through \(x_k\) ends at \(z_k=x_k+h/2\). For a twice differentiable function \(g\) with integrable second derivative, Taylor's theorem on each cell shows that the difference between its midpoint value times \(h\) and its integral over that cell is bounded by a constant times \(h^2\int_{\text{cell}}|g''(u)|\,du\). Summing gives an \(O(h^2)\) bound, uniformly over partial sums. Apply this to \(g=\phi\), for which \(\int_{\mathbb R}|\phi''(u)|\,du<\infty\). Hence
Apply the same midpoint argument to \(g=H_3\phi\), whose second derivative is also integrable. Its sum is its integral through \(z_k\), up to \(O(h^2)\). Since it is multiplied by \(\lambda_3/6=O(n^{-1/2})\), this quadrature error is \(O(n^{-3/2})\). Direct differentiation gives
so \(\int_{-\infty}^{z_k}H_3(u)\phi(u)\,du=(1-z_k^2)\phi(z_k)\). Combining the summed local remainders, the tail bounds, and these two integral approximations proves the theorem. \(\square\)
Worked Example: A Continuity-Corrected Binomial CDF
Let \(S_{100}\sim\operatorname{Binomial}(100,0.3)\), and approximate \(P(S_{100}\le35)\). The mean is \(30\), \(\sigma=\sqrt{21}\approx4.5826\), and \(\lambda_3=0.4/\sqrt{21}\approx0.0873\). The continuity-corrected standardized cutoff is
Using \(\Phi(1.2002)\approx0.8850\) and \(\phi(1.2002)\approx0.1942\), the skewness term is
Therefore \(P(S_{100}\le35)\approx0.8837\), using unrounded values. The corresponding upper-tail approximation is \(P(S_{100}\ge36)=1-P(S_{100}\le35)\approx0.1163\). The sign of the skewness adjustment is negative here because the cutoff lies to the right of one standard deviation and \(1-z_{35}^2<0\).
The Periodic Term and Other Cutoffs
The half-step is the simplest expression of a periodic lattice correction. To see the periodicity, consider an integer-valued variable and a real, not necessarily integer, cutoff \(t\). Write \(m=\lfloor t\rfloor\), so the event \(S_n\le t\) is exactly \(S_n\le m\). Its continuity-corrected boundary is \(m+\tfrac12\), not \(t\). Relative to integrating to \(t\), the shift is \(m+\tfrac12-t\), which repeats with period one as \(t\) moves across successive lattice cells. A first-order Taylor expansion of the normal integral makes the boundary contribution approximately \((m+\tfrac12-t)\phi((t-np)/\sigma)/\sigma\), with the standardized location understood to the accuracy of this first-order expansion. This is the periodic, sawtooth-shaped correction; at an integer cutoff it reduces to the familiar half-step.
For a lattice with maximal span \(d\), the same geometry uses cells of width \(d\): a sum through a lattice point is matched to an integral ending halfway between that point and the next point of the supporting lattice. The standardized cell width is \(d/\sigma\) when \(\sigma\) is the standard deviation of the variable being approximated. The precise periodic term depends on where the requested cutoff falls within a lattice cell.
Worked Example: What the Half-Step Changes
For the binomial model in the previous example, compare the boundary at \(k=35\) with and without the half-step. Without it, the standardized cutoff would be \((35-30)/\sqrt{21}\approx1.0911\); with it, the cutoff is approximately \(1.2002\). Since \(\Phi(1.0911)\approx0.8624\), omitting the half-step changes the leading normal CDF term by about \(0.02259\). This is much larger than the skewness adjustment, about \(0.00125\), and illustrates why a continuous CDF approximation can be poorly aligned with a discrete cutoff even when the normal shape is otherwise reasonable.
The half-step does not alter the underlying binomial event: \(S_{100}\le35\) still includes exactly the integers \(0,\ldots,35\). It changes only the endpoint of the continuous approximation so that the integral covers the same collection of lattice cells.
Common Mistake and Practical Use
A common but invalid argument is to derive a local expansion only for bounded \(x_j\), sum it from the bottom of the support through \(k\), and claim an \(O(n^{-1})\) CDF remainder. A pointwise error bound on bounded standardized ranges says nothing about points outside those ranges, and even a small local error can accumulate when many terms are added. The correction is to use a local bound on a growing range, sum its weighted errors, and separately control the omitted tails. The binomial proof above does all three; the stated CDF error therefore follows from more than a formal integration of the local expansion.
The continuity-corrected Edgeworth formula is useful when a binomial CDF or tail probability is needed and a normal approximation alone is not adequate. Its main components have distinct roles: \(\Phi(z_k)\) supplies the continuous bulk approximation, the half-step aligns the boundary with the lattice, and the skewness term adjusts for binomial asymmetry. The \(O(n^{-1})\) statement assumes fixed \(p\) strictly between zero and one; it does not assert reliable relative accuracy for extremely small tail probabilities.
Check Your Understanding
Use the definitions and approximations in this tutorial to answer the following.
- For \(S_n\sim\operatorname{Binomial}(n,p)\), write the standardized lattice spacing and the continuity-corrected standardized cutoff for \(P(S_n\le k)\).
- Why is it not valid to sum a pointwise local expansion proved only for bounded standardized values over the entire binomial support?
- For which values of \(z\) is the first-order binomial skewness correction to the CDF positive, and for which values is it negative, assuming \(p<q\)?
- Explain how a periodic boundary term arises when the cutoff is real rather than an integer.
- What two kinds of error control are needed to turn the local expansion into a CDF approximation with a justified remainder?