$x$, where $\Phi$ and $\phi$ are the standard normal distribution and density functions, respectively, and $P$ is an even, quadratic polynomial.

For example, if ${\hat{\theta}}$ denotes the mean of a sample of size $n$ from a population with mean ${\theta}$, and if data from the population have finite third moment and a nonlattice distribution, then (\[eq1.1\]) holds with $P(x)={\frac{1}{6}}{\beta}(1-x^2)$, where ${\beta}$ is the standardised skewness of the population. On the other hand, still in the case of the mean of a population with finite third moment, if the population is lattice then an extra, discontinuous term has to be added to (\[eq1.1\]).

This extra term reflects the discrete continuity correction that statisticians are often obliged to introduce when approximating a lattice distribution, for example the binomial distribution or the distribution of a sum of Poisson variates, using the smooth normal distribution: $$\label{eq1.2} P(T\leq x)=\Phi(x)+n^{-1/2}P(x) \phi(x)
+n^{-1/2}d_n(x) \phi(x)+o \bigl(n^{-1/2} \bigr),$$ where $d_n(x)=(e_0/{\sigma}) \psi_n(x)$ denotes the discontinuous term in the Edgeworth expansion, $e_0$ is the maximal span of the lattice, ${\sigma}^2$ is the population variance, $$\psi_n(x)=\psi \bigl\{(x-\xi_n) {\sigma}n^{1/2}
/e_0 \bigr\}, \qquad \xi_n= \bigl(e_0 /{\sigma}n^{1/2} \bigr) \bigl\{{\tfrac{1}{2}}-\psi(n x_0/e_0)
\bigr\},$$ $\psi(x)=\lfloor x\rfloor-x+{\frac{1}{2}}$, $\lfloor x\rfloor$ is the largest integer not strictly exceeding $x$, and it 