An Example

Figure 1: Example Graphs
\begin{figure}\centering\scalebox{.75}{\includegraphics{test64c.ps}}\end{figure}

We now illustrate some of the methods described in this section with an example. Consider the three-dimensional problem where ${\bf a}= (-3,-2,-1)^t$, ${\bf b}= (2,2,2)^t$, $\nu = 5$ and

\begin{displaymath}
\mbox{\boldmath $\Sigma$}= \left[
\begin{array}{ccc}1&12/13&...
...1& 0& 0 12/13&5/13& 0 -3/5&-16/25&12/15\end{array}\right].
\end{displaymath}

The SV integration region is determined by ${\bf a}\leq C{\bf y}\leq {\bf b}$:

\begin{displaymath}
-3 \leq y_1 \leq 2,  \
\frac{-2-12y_1/13}{5/13} \leq y_2\l...
...5+16y_2/25}{12/25}\leq y_3\leq\frac{2+3y_1/5+16y_2/25}{12/25},
\end{displaymath}

with $y_1=u_1$, $y_2 = u_2\sqrt{\frac{5+u_1^2}{6}}$, $y_3 = u_3\sqrt{\frac{5+u_1^2}{6}\frac{6+u_2^2}{7}}$, so

\begin{eqnarray*}
{\bf T}({\bf a}, {\bf b}, \mbox{\boldmath$\Sigma$}, 5)
&=&
\...
...}
\frac{K_7^{(1)}}{(1+\frac{u_1^2}{7})^{\frac{8}{2}}}
d{\bf u},
\end{eqnarray*}



with ${\tilde a}_2(u_1)=\sqrt{\frac{6}{5+u_1^2}}(-\frac{2}{5}-\frac{12}{5}u_1)$, ${\tilde b}_2(u_1)=\sqrt{\frac{6}{5+u_1^2}}(\frac{26}{5}-\frac{12}{5}u_1)$,
${\tilde a}_3(u_1,u_2)=
\sqrt{\frac{6}{5+u_1^2}\frac{7}{6+u_2^2}}
(-\frac{25}{12}+\frac{5}{4}u_1+\frac{4}{3}u_2\sqrt{\frac{6}{5+u_1^2}})$, and ${\tilde b}_3(u_1,u_2)=
\sqrt{\frac{6}{5+u_1^2}\frac{7}{6+u_2^2}}
( \frac{25}{6}+\frac{5}{4}u_1+\frac{4}{3}u_2\sqrt{\frac{6}{5+u_1^2}})$. Then we use $d_1=t_5(-3)$, $e_1=t_5(2)$, $z_1 = d_1+(e_1-d_1)w_1$, $u_1 = t_5^{-1}(z_1)$, $d_2=t_6({\tilde a}_2(u_1))$,
$e_2=t_6({\tilde b}_2(u_1)$, $z_2 = d_2+(e_2-d_2)w_2$, $u_2 = t_6^{-1}(z_2)$, $d_3=t_7({\tilde a}_3(u_1,u_2))$, $e_3=t_7({\tilde b}_3(u_1,u_2))$, so

\begin{displaymath}
{\bf T}({\bf a},{\bf b},\mbox{\boldmath $\Sigma$},5)= (e_1-d...
...tyle\int _0^1(e_2-d_2)\displaystyle\int _0^1(e_3-d_3)dw_2dw_1.
\end{displaymath}

In Figure 1 we provide four graphs. The top left graph is a contour graph for the original density function for this example, with the last variable integrated out. The top right graph is a contour graph for the SR integrand $F$ (see equation 9). The bottom left graph is a contour graph for the SV integrand $(e_1-d_1)(e_2(w_1)-d_2(w_1))(e_3(w_1,w_2)-d_3(w_1,w_2))$. The bottom right graph also shows the SV integrand for the example in this section but variables 1 and 3 have been interchanged. These graphs show the integrands that would be sampled by Monte-Carlo methods that we have described for the transformed problems. The ``RelStD'' quantities given for each graph are relative standard errors (standard error divided by the mean) for the data used to plot each graph.




2004-12-02