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The Positive solution

Figure 2.15: An implementation of the Positive solution, defined for a distribution $
G\in {\cal U} \cup \widehat{\cal M}
\cup \left\{F \Big\vert r^F \neq \frac{3}{2} \mbox{ and } m_3^F < 2m_2^F - 1\right\}.
$
\begin{figure}\begin{center}
\begin{tabular}[hbt]{\vert l\vert}\hline
($n$, $p$,...
...\mu_1^X$, $\mu_2^X$, $\mu_3^X$)\\
\hline
\end{tabular}
\end{center}\end{figure}

The Positive solution is defined for almost all the input distributions in ${\cal PH}_3$. Specifically, it is defined for all the distributions in

\begin{displaymath}
{\cal U} \cup \widehat{\cal M}
\cup \left\{F \Big\vert r^F \neq \frac{3}{2} \mbox{ and } m_3^F < 2m_2^F - 1\right\}.
\end{displaymath}

Figure 2.15 shows an implementation of the Positive solution. Below, we elaborate on the Positive solution, and prove an upper bound on the number of phases used in the Positive solution.

When the input distribution $G$ is in ${\cal U} \cup \Bar{\cal M}$, the EC distribution produced by the Minimal solution does not have a mass probability at zero. When $G$ is in $\left\{F \mid m_3^F<2m_2^F-1 \mbox{ and } r^F>\frac{3}{2}\right\}$, $G$ can be well-represented by a two-phase Coxian$^+$ PH distribution, whose parameters are given by (2.4)-(2.5). Below, we focus on input distributions $G\in\left\{F \Big\vert m_3^F < 2m_2^F-1 \mbox{ and } r^F<\frac{3}{2} \right\}$.

We first consider the first approach of using a mixture of an EC distribution (with no mass probability at zero), $X$, and an exponential distribution, $W$, (i.e. $\lambda_Z=\infty$). Let

\begin{displaymath}
\widehat{\cal L}_N = \left\{ F \Big\vert \frac{N+3}{N+2}m_2^...
... \frac{N+2}{N+1}m_2^F \mbox{ and } m_3^F < 2m_2^F -1 \right\}.
\end{displaymath}

Given a distribution $G\in \widehat{\cal L}_N$, we seek $m_2^X$, $m_3^X$, $0<p<1$, and $w>0$ such that
$\displaystyle \frac{N+2}{N+1}$ $\textstyle \leq$ $\displaystyle m_2^X < \frac{N+1}{N}$ (2.9)
$\displaystyle m_3^X$ $\textstyle =$ $\displaystyle 2m_2^X - 1$ (2.10)
$\displaystyle m_2^G$ $\textstyle =$ $\displaystyle \frac{pm_2^X+2(1-p)w^2}{(p+(1-p)w)^2}$ (2.11)
$\displaystyle m_3^G$ $\textstyle =$ $\displaystyle \frac{pm_2^Xm_3^X+6(1-p)w^3}{(p+(1-p)w)(pm_2^X+2(1-p)w^2)}$ (2.12)

Note that (2.9)-(2.10) guarantee that we can find, via the Complete solution, an ($N+1$)-phase EC distribution, $X$, such that $X$ has no mass probability at zero and has normalized moments $m_2^X$ and $m_3^X$. Let $W$ be the exponential distribution with $\mu_1^W = w / \mu_1^X$. (2.11)-(2.12) guarantee that, by choosing $\mu_1^X$ appropriately, $G$ is well-represented by a distribution $Y$, where

\begin{displaymath}
Y(\cdot) = pX(\cdot)+(1-p)W(\cdot).
\end{displaymath}

The following lemma provides conditions on the input distribution for which the first approach is defined.

Lemma 5   Suppose

\begin{displaymath}
G\in \widehat{\cal L}_N\quad\mbox{and}\quad
\frac{(N+1)m_2^G+(N+4)}{2(N+2)} \leq r^G
\end{displaymath}

for $N\geq 1$ (see Figure 2.16(a)). Let

\begin{eqnarray*}
w & = & \frac{2-m_2^G}{4\left(\frac{3}{2}-r^G\right)}\\
p & =...
... 4(2m_2^G-1-m_3^G)}\\
m_2^X & = & 2w\\
m_3^X & = & 2m_2^X - 1.
\end{eqnarray*}

Then, $w>0$, $0<p<1$, and conditions (2.9)-(2.12) are satisfied.

A proof of Lemma 5 is postponed to Appendix A.

Figure 2.16: Two regions in $\cal{L}_N$ that input $G$ can fall into under the Positive solution: (a) $G$ is well-represented by a mixture of an EC distribution $X$ and an exponential distribution $W$; (b) $G$ is well-represented by the convolution of an EC distribution $X$ and an exponential distribution $Z$.
\includegraphics[width=0.7\linewidth]{fig/PositiveIdea1.eps}
(a)
\includegraphics[width=0.7\linewidth]{fig/PositiveIdea2.eps}
(b)

The key idea behind Lemma 5 is to fix some of the parameters so that the set of equations becomes simpler and yet there exists a unique solution. The difficulty in finding closed form solutions is that we are given a system of nonlinear equations with high degree (2.10)-(2.12), and the solutions are not unique. By fixing some of the parameters, the system of equations can be reduced to have a unique solution. We find that $w$ given by Lemma 5 has nice characteristics. First, $m_2^X$ leads to a very simple expression: $m_2^X = 2w$. Second, with this expression of $m_2^X$, the expression involving $r^G$ is significantly simplified:

\begin{displaymath}
r^G = \frac{2pr^X+3(1-p)w}{2(p+(1-p)w)}.
\end{displaymath}

Now, solving (2.10)-(2.12) for $p$ and $w$ is a relatively easy task, and $p$ and $w$ immediately gives $m_2^X$ and $m_3^X$. Although Lemma 5 allows us to find a simple closed form solution, the set of input distributions defined for Lemma 5 is rather small. This necessitates the second approach of using the convolution of an EC distribution and an exponential distribution. Note that the second approach alone does not suffice, either. Applying the first approach to a small set of input distributions and applying the second approach to the rest of the input distribution, in fact, lead to simpler closed form solutions by both approaches.

Next, we consider the second approach of using the convolution of an EC distribution (with mass probability at zero) and an exponential distribution (i.e. $\lambda_W=\infty$). Given a distribution $G\in \widehat{\cal L}_N$ (we assume $r^G \neq \frac{3}{2}$), we seek $m_2^X$, $m_3^X$, and $z>0$ such that

$\displaystyle m_2^X$ $\textstyle \geq$ $\displaystyle \frac{N+2}{N+1}$ (2.13)
$\displaystyle m_3^X$ $\textstyle =$ $\displaystyle \frac{N+3}{N+2}m_2^X$ (2.14)
$\displaystyle m_2^G$ $\textstyle =$ $\displaystyle \frac{m_2^X+2z+2z^2}{(1+z)^2}$ (2.15)
$\displaystyle m_3^G$ $\textstyle =$ $\displaystyle \frac{m_2^Xm_3^X+3m_2^Xz+6z^2+6z^3}{(m_2^X+2z+2z^2)(1+z)}$ (2.16)

Note that (2.13)-(2.14) guarantees that we can find, via the Complete solution, an ($N+1$)-phase EC distribution, $X$, such that $X$ has no mass probability at zero and has normalized moments $m_2^X$ and $m_3^X$. Let $Z$ be the exponential distribution whose first moment is $\mu_1^Z = z \mu_1^X$, where $\mu_1^X$ is the first moment of $X$. (2.15)-(2.16) guarantee that, by choosing $\mu_1^X$ appropriately, $G$ is well-represented by the convolution of $X$ and $Z$.

The following lemma provides conditions on the input distribution for which the second approach is defined.

Lemma 6   Suppose

\begin{displaymath}
G\in \widehat{\cal L}_N\quad\mbox{and}\quad
r^G < \frac{(N+1)m_2^G+(N+4)}{2(N+2)}
\end{displaymath}

for $N\geq 1$ (see Figure 2.16(b)). If $m_2^G = 2$, we choose

\begin{eqnarray*}
z & = & \frac{m_3^G-2\frac{N+3}{N+2}}{3-m_3^G}\\
m_2^X & = & 2(1+z)\\
m_3^X & = & \frac{N+3}{N+2}m_2^X.
\end{eqnarray*}

If $m_2^G \neq 2$, we choose

\begin{eqnarray*}
z & = & \frac{m_2^G\left((m_3^G-3)-2\frac{N+3}{N+2}(m_2^G-2)\r...
...)\left(m_2^G(1+z)-2z\right)\\
m_3^X & = & \frac{N+3}{N+2}m_2^X.
\end{eqnarray*}

Then, $z>0$ and conditions (2.13)-(2.16) are satisfied.

A proof of Lemma 6 is postponed to Appendix A.


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