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The Complete solution

Consider an arbitrary distribution $G\in {\cal
PH}_3$. Our approach consists of two steps, the first of which involves constructing a baseline EC distribution, and the second of which involves reducing the number of phases in this baseline solution. If $G\in {\cal PH}_3^-$, then the baseline solution used is simply given by the Simple solution. Also, if ${G}\notin {\cal PH}_3^-$ but $G\in \widehat{\cal L}\cup\widehat{\cal M}$, then it turns out that the Simple solution could be defined for this $G$, and this gives the baseline solution. If ${G}\notin {\cal PH}_3^-$ but $G\in \widehat{\cal U}$, then to obtain the baseline EC distribution we first find a distribution ${W}\in {\cal PH}_3^-$ such that $r^W = r^G$ and $m_2^{W} < m_2^{G}$ and then set $p$ such that ${G}$ is well-represented by distribution $Z$, where $Z(\cdot)=pW(\cdot)+(1-p)O(\cdot)$ (see Figure 2.14(a)). The parameters of the EC distribution that well-represents ${W}$ are then obtained by the Simple solution.

Figure 2.14: Ideas in the Complete solution. (a) A distribution $G$ not in ${\cal PH}_3^-$ is well-represented by an EC distribution $W$ mixed with $O$. (b) The set of two-phase Coxian$^+$ PH distributions used is extended.
\includegraphics[width=0.6\linewidth]{fig/MinimalIdea1.eps}
(a)
\includegraphics[width=0.9\linewidth]{fig/MinimalIdea2.eps}
(b)

To reduce the number of phases used in the baseline EC distribution, we exploit the subset of two-phase Coxian$^+$ PH distributions that are not used in the Simple solution. The Simple solution is based on the fact that a distribution ${X}$ is well-represented by a two-phase Coxian$^+$ PH distribution when ${X}\in {\cal U}_0\cup {\cal M}_0$. In fact, a wider range of distributions are well-represented by the set of two-phase Coxian$^+$ PH distributions. In particular, if $X$ is in set $\cal{S} = \left\{ F \Big\vert \frac{3}{2} \leq m_2^{X} \leq 2\mbox{ and }m_3^{X} = 2m_2^{X}-1\right\}$, then ${X}$ is well-represented by a two-phase Coxian$^+$ PH distribution (see Figure 2.14(a)). By exploiting two-phase Coxian$^+$ PH distributions in $\cal{S} \cup {\cal U}_0 \cup {\cal M}_0$, the Complete solution reduces the number of phases used. The above ideas lead to the following solution:

(i) If $G\in \widehat{{\cal U}}\cap {\cal PH^-}$, then the Simple solution provides the parameters ($n$, $p$, $\mu_{X1}$, $\mu_{X2}$, $p_X$).

(ii) If $G\in \widehat{{\cal U}}\cap{\cal (PH^-)}^c$ (see Figure 2.14(a)), where ${\cal (PH^-)}^c$ denotes the complement of ${\cal PH^-}$, then let

\begin{displaymath}
n = \frac{2m_2^G-1}{m_2^G-1},
\end{displaymath} (2.7)

and

\begin{displaymath}
m_2^W = \frac{1}{2}\left(\frac{n-1}{n-2}+\frac{n}{n-1}\right...
..._2^G}m_2^W,\quad\mbox{and}\quad
\mu_1^W = \frac{\mu_1^G}{p_W},
\end{displaymath}

where $p_W = m_2^W / m_2^G$. $G$ is then well-represented by $Z$, where $Z(\cdot) = p_W W(\cdot) + (1-p_W)O(\cdot)$, where $W$ is an EC distribution with normalized moments $m_2^W$ and $m_3^W$ and mean $\mu_1^W$. The parameters ($n$, $\mu_{X1}$, $\mu_{X2}$, $p_X$) of $W$ are provided by the Simple solution. Also, set $p=p_W$, since $W$ has no mass probability at zero.

(iii) If $G\in \widehat{\cal M}\cup \widehat{\cal L}$, then the Simple solution provides the parameters ($n$, $p$, $\mu_{X1}$, $\mu_{X2}$, $p_X$), except that (2.6) is replaced by

\begin{displaymath}
n = \left\{\begin{array}{ll}
\left\lceil \frac{m_2^{G}}{m_2^...
...{if } m_2^G\leq 2,\\
2 & \mbox{otherwise}.
\end{array}\right.
\end{displaymath} (2.8)

The next theorem guarantees that parameters obtained with the reduced $n$ are still feasible.

Theorem 5   If $X$ is in set $\left\{ F \mid \frac{3}{2} \leq m_2^F \leq 2 \mbox{ and } m_3^F = 2m_2^F - 1\right\}$, then $Z=\phi^N(X)$ is in set
$\left\{ F \mid \frac{N+1}{N} \leq m_2^F \leq \frac{N}{N-1} \mbox{ and } m_3^F = 2m_2^F - 1\right\}$.


Proof:By Theorem 3, $m_2^Z$ is a continuous and monotonically increasing function of $m_2^X$, Thus,

\begin{displaymath}
\frac{N+1}{N} \leq m_2^Z \leq \frac{N}{N-1}
\end{displaymath}

follows by simply evaluating $m_2^Z$ at the lower and upper bound of $m_2^X$. $m_3^Z = 2m_2^Z - 1$ follows from Lemma 4.     width 1ex height 1ex depth 0pt


next up previous contents
Next: Analyzing the number of Up: Complete closed form solution Previous: Complete closed form solution   Contents
Takayuki Osogami 2005-07-19