Continuity Corrections for Lattice Sums

Why Discrete Boundaries Need an Adjustment

A normal distribution assigns probability to intervals, while an integer-valued sum assigns probability to individual points. To approximate a probability such as \(P(S=50)\), it is therefore natural to associate the integer 50 with the interval of real numbers closer to 50 than to either neighboring integer: \([49.5,50.5]\). This half-unit adjustment is the continuity correction.

The adjustment also matters for tails and intervals. The event \(S\leq k\) includes the entire probability mass at \(k\), so its normal counterpart should extend to \(k+0.5\), not stop at \(k\). For an interval \(a\leq S\leq b\), the corrected boundaries are \(a-0.5\) and \(b+0.5\). These choices preserve the integer values included by the original event.

Suppose \(S\) has mean \(\mu\) and positive variance \(\sigma^2\), and let \(G\sim N(\mu,\sigma^2)\). For a unit-lattice variable, the usual continuity-corrected approximations are:

Writing these in terms of the standard normal distribution function \(\Phi\) gives, for example,

For a lattice with span \(h\), replace \(0.5\) by \(h/2\), and use the lattice points belonging to that lattice. If the lattice has offset \(c\), its points have the form \(c+jh\); the cell centered at such a point has boundaries \(c+jh-h/2\) and \(c+jh+h/2\).

Why the Half-Unit Adjustment Works

The cell interpretation can be made exact by adding an independent uniform random variable. Let \(S\) be integer-valued and let \(U\sim\operatorname{Uniform}(-0.5,0.5)\) be independent of \(S\). The added jitter spreads the probability at each integer evenly over that integer's cell.

Proof. If \(S\leq k\), then \(S+U\leq k+0.5\), since \(U\leq0.5\). If \(S\geq k+1\), then \(S+U\geq k+0.5\). Equality at this boundary requires \(S=k+1\) and \(U=-0.5\), an event of probability zero because a uniform random variable has no point masses. Thus the two events agree except possibly on a probability-zero event, proving the first identity.

For the second identity, when \(S=k\), \(S+U\) lies in \([k-0.5,k+0.5]\) with probability one. When \(S\leq k-1\) or \(S\geq k+1\), \(S+U\) can enter that interval only at an endpoint, which requires \(U=0.5\) or \(U=-0.5\). Those endpoint events have probability zero. Therefore the events also agree up to probability zero, proving the second identity. \(\square\)

This exact jitter argument explains the boundaries, but it does not itself make \(S+U\) normally distributed. The normal approximation enters when we approximate the distribution of the jittered sum—or, in the usual calculation, use a normal curve with the same mean and variance as \(S\). The distinction matters: the identities are exact; replacing the jittered distribution by a normal one is an approximation.

Point Probabilities as Normal Cell Areas

For a continuous normal variable, the probability of a cell is the difference between two normal distribution function values. There is also a useful density interpretation: for a narrow cell centered at \(k\), its probability is approximately the cell width times the density at \(k\).

Proof. Taylor's theorem about \(k\) gives, for each \(t\in[-h/2,h/2]\),

Integrate over the symmetric interval. The linear term integrates to zero because \(\int_{-h/2}^{h/2}t\,dt=0\). The remainder is bounded by

This proves the result. \(\square\)

For a normal density, the corrected approximation to \(P(S=k)\) is the normal probability over the cell, not merely the normal density evaluated at \(k\). The result above explains why the cell probability is close to \(h\) times that density when the density changes little across the cell. For \(h=1\), the standard formula is

Worked Examples

When the Correction Helps—and What It Cannot Guarantee

A continuity correction is useful when a normal approximation is reasonable and the event has discrete boundaries. It often improves finite-sample calculations by aligning the continuous region with the integer values included in the event. For a sum with lattice span \(h\), use \(h/2\), not \(1/2\); for example, if a sum can take only even integers, its natural lattice span is 2 and the cell half-width is 1.

A central limit theorem alone does not guarantee that a normal approximation to each point probability is accurate. Convergence in distribution controls probabilities of fixed intervals whose standardized endpoints remain separated; a single lattice point corresponds to an interval whose standardized width shrinks as the variance grows. Point-probability approximations generally need stronger local information, as supplied by a suitable local limit theorem, or must be treated as practical finite-sample approximations rather than consequences of the CLT alone.

The normal approximation can also be poor when the distribution is strongly skewed, the variance is small, or the event lies far into a tail. The correction fixes a boundary mismatch; it does not repair a poor normal model. In practice, check the model's mean and variance, use the correct lattice span, and consider an exact calculation when it is readily available.

Common Mistake

A frequent error is to write \(P(S\leq k)\approx\Phi((k-\mu)/\sigma)\) because \(k\) is the largest allowed integer. That uses the center of the last included mass as the normal cutoff, rather than the boundary after that mass's cell. For a unit-lattice sum, the corrected cutoff is \(k+0.5\). Conversely, for \(P(S<k)\), the last included integer is \(k-1\), so the corrected cutoff is \(k-0.5\). First identify exactly which integers the event includes; then place the normal boundary halfway between the last included and first excluded lattice points.