percentile bootstrap methods applied to lattice distributions are frustrated by the effects of discontinuities; see, e.g., Singh [@Sin81] and Hall [@Hal89].

For brevity, when establishing this property we treat only the context of Theorem \[the2\]. We begin by stating an analogue of (\[eq2.14\]) there, valid when $k=2$. The arguments used to prove part (i) of Theorem \[the2\] can be employed to show that $$\begin{aligned}
\label{eq2.37} &&P \biggl[\frac{S^*-E(S^*\mid{{\mathcal X}})}{\{{\operatorname{var}}(S\mid{{\mathcal X}})\}^{1/2}}\leq x \Bigm|{{\mathcal X}}\biggr] =
\Phi(x)+n^{-1/2}{\frac{1}{6}} {\hat{\beta}} \bigl(1-x^2
\bigr) \phi (x)
\nonumber
\\
&&\hspace*{170pt} {}+(n_1n_2)^{-1/2}{\widehat
K}_n(x)+n^{-1}{\Delta}_1(x),\end{aligned}$$ where, analogously to the definitions in Section \[sec2.1\], $S^*=\sum_j{\bar X}_j^*$; ${\bar X}_j^*=n_j^{-1}\sum_iX_{ji}^*$ and $X_{j1}^*,\ldots
,X_{jn_j}^*$ are drawn by sampling randomly, with replacement, from ${{\mathcal X}}_j=(X_{j1},\ldots,X_{jn_j})$; ${{\mathcal X}}=({{\mathcal X}}_1,\ldots,{{\mathcal X}}_k)$; $${\hat{\beta}}=\frac{n^{1/2}E[\{S^*-E(S^*\mid{{\mathcal X}})\}^3\mid{{\mathcal X}}]}{\{
{\operatorname{var}}(S^*
\mid{{\mathcal X}})\}^{3/2}} ;$$ using (\[eq2.10\]) or (\[eq2.20\]), respectively, as the model for $K_n(x)$, $$\begin{aligned}
{\widehat K}_n(x)&=&{\hat{\gamma}}\sum_{r=0}^{r_0}
\sum_{-\infty<\ell
<\infty} \frac{{\hat\phi}_r ({\bar\nu}_\ell /n_1^{1/2},x )}{
r! n_1^{r/2}} \sum
_{\nu\in{{\mathcal N}}_\ell} \nu_\ell^r \psi \biggl\{{\hat
\xi}_n(x)-\frac{e_2 n_1}{ e_1 n_2} \nu \biggr\} ,
\\
{\widehat K}_n(x)&=&{\hat{\gamma}}\sum_\nu
\phi \biggl(\frac{x}{{\hat c}_1} -\frac{e_2 n_1^{1/2}}{{\hat{\sigma}}_1 n_2} \nu \biggr) \phi \bigl
\{e_2 \bigl({\hat{\sigma}}_2 n_2^{1/2}
\bigr)^{-1}\nu \bigr\} \psi \biggl\{{\hat\xi}_n(x)-
\frac{e_2 n_1}{ e_1 n_2} \nu \biggr\},\end{aligned}$$ where ${\hat{\gamma}}=\prod_{j=1,2}(e_j/{\hat{\sigma}}_j)$, ${\hat\phi}_r(u,x)=({\partial}/{\partial}u)^r {\hat\phi}(u,x)$, ${\hat\phi}(u,x)=\phi\{(x/{\hat c}_1)-{\hat c}_3 u\}\times\phi
({\hat c}_4 u)$, $${\hat c}_j= \biggl(\frac{n_j^{-1}{\hat{\sigma}}_j^2}{ n_1^{-1}{\hat{\sigma}}_1^2+n_2^{-1}{\hat{\sigma}}_2^2} \biggr)^{ 1/2}$$ for $j=1$ and 