Starting Interval Selection

Let $A_j(t) = \{x_j: x_j \leq t \}$ for the one-sided cases, or $A_j = \{ x_j: \vert x_j\vert \leq t \}$ for the two-sided cases. We let

\begin{displaymath}S_1(t) = \sum_{j=1}^m Prob(A_j^c(t)), \end{displaymath}

where $A_j^c(t)$ is the compliment of the set $A_j(t)$. The Bonferroni bound for $P(t)$ (see Hsu, 1996) is

\begin{displaymath}L'(t) = 1-S_1(t) \leq P(t).\end{displaymath}

A simple upper bound for $P(t)$ is

\begin{displaymath}
P(t) \leq 1- \min_j Prob(A_j^c(t)) = U'(t).
\end{displaymath}

Both of these bounds require only 1-dimensional distribution values. If $t'_a$ and $t'_b$ are determined by solving $U'(t) = 1-\alpha$ and $L'(t) = 1-\alpha$, respectively, then $t_\alpha \in [t'_a,t'_b]$. This bounding interval for $t_\alpha$ can be found directly using the appropriate 1-dimensional inverse distribution function. For example, with the two-sided case,

\begin{displaymath}[t'_a,t'_b]= [t_\nu^{-1}(1-\frac{\alpha}{2}),t_\nu^{-1}(1-\frac{\alpha}{2m})],
\end{displaymath}

where $t_\nu(u) =
\frac{\Gamma(\frac{\nu+1}{2})}{\Gamma(\frac{\nu}{2})\sqrt{\nu\pi}}
\int_{-\infty}^u (1+\frac{s^2}{\nu})^{-\frac{\nu+1}{2}}ds$. Shorter intervals can be found using bivariate distribution values (Dunnett and Sobel, 1954) if a modified Bonferroni bound (Dawson and Sankoff, 1967) is combined with the Hunter-Worsley bound. These bounds are described in the book by Hsu (1966, Appendix A). If we define $S_2(t)$ by

\begin{displaymath}
S_2(t) = \sum_{j<i} Prob(A_j^c(t)\cap A_i^c(t)),
\end{displaymath}

then the modified Bonferroni bounds and Hunter-Worsley guarantee that

\begin{eqnarray*}
L(t) &=& 1 - S_1(t) + \sum_{(i,j)\in T^*} Prob(A_j^c(t)\cap A_i^c(t))\\
\leq& P(t)& \leq 1-2\frac{S_1(t)-S_2(t)/k}{k+1} = U(t).
\end{eqnarray*}



where $k = 1 + \lfloor 2S_2(t)/S_1(t)\rfloor$ and $T^*$ is maximal spanning tree for complete graph of order $m$ with edge weights $Prob(A_j^c(t)\cap A_i^c(t))$. If $U(t_a) = 1-\alpha$ and $L(t_b) = 1-\alpha$ then $t'_a \leq t_a \leq t_\alpha \leq t_b \leq t'_b$. Starting with $[t'_a, t'_b]$, we can use numerical optimization, applied to $L(t)$, to determine $t_b$, then use numerical optimization, applied to $U(t)$ starting with $[t'_a, t_b]$, to determine $t_a$.




2003-02-17