An Example

This section gives an example illustrating the ideas discussed in the previous two sections. Assume $m = 3$, ${\bf a} = (-\infty, -\infty, -\infty)$, ${\bf b} = (1, 4, 2)$ and

$$\Sigma = \left( \begin{array}{ccc} 1&\frac{3}{5}&\frac{1}{... ...4}{5}&\frac{2}{3} \\ 0&0&\frac{2}{3} \end{array} \right) .$$

Then, after the first transformation,

$$F({\bf a, b}) = \frac{1}{\sqrt{(2\pi)^3}} \int_{-\infty}^{1... ...}} \int_{-\infty}^{3-y_1/2-y_2} e^{-\frac{y_3^2}{2}} d{\bf y}.$$

After the second transformation,

$$F({\bf a, b}) = \int_{0}^{\Phi(1)} \int_{0}^{\Phi(5-3\Phi^... ...} \int_{0}^{\Phi(3-\Phi^{-1}(z_1)/2-\Phi^{-1}(z_2))} d{\bf z}.$$

Finally,

$$\begin{eqnarray*} F({\bf a, b}) & = & \Phi(1) \int_{0}^{1} \Phi(5-3\Phi^{-1}(w_... ...-1}(w_2\Phi(5-3\Phi^{-1}(w_1\Phi(1))/4))) \int_{0}^{1}d{\bf w}. \end{eqnarray*}$$

If variables $\theta_2$ and $\theta_3$ are interchanged to sort ${\bf b}$, then

$$\Sigma = \left( \begin{array}{ccc} 1&\frac{1}{3}&\frac{3}{... ...rt{2}}{5} \\ 0&0&\frac{2\sqrt{2}}{5} \end{array} \right) .$$

After the first transformation,

$$F({\bf a, b}) = \frac{1}{\sqrt{(2\pi)^3}} \int_{-\infty}^{1... ...-3y_1/2-\sqrt{2}y_2)/\sqrt{2}} e^{-\frac{y_3^2}{2}} d{\bf y}.$$

And after the final transformation,

$$\begin{eqnarray*} F({\bf a, b}) & = & \Phi(1) \int_{0}^{1} \Phi(\frac{3-\Phi^{-... ...(w_1\Phi(1))/2}) {\sqrt{2}})}{\sqrt{2}}) \int_{0}^{1}d{\bf w}. \end{eqnarray*}$$

If the Monte-Carlo algorithm is used with this example (where $F({\bf a, b}) \doteq 0.82798$), the computed sample variance for the original problem is approximately 0.0016, while the computed sample variance for the reordered variable version of the problem is approximately 0.000064. These results imply that in order to compute $F({\bf a, b})$ to some prescribed accuracy level, the original form of the problem would require approximately 25 times more computation than the reordered variable version of the problem.