Joint MGF of $ {\bf y}$.

We start with the joint MGF of $ {\bf y}$, which is (see [46], p. 68, eq. B-18)

\begin{displaymath}\begin{array}{l} g_0(\alpha_1, \alpha_2 \ldots \alpha_N) = \\...
..._1 +\alpha_2 + \cdots + \alpha_N \over N} \right) } \end{array}\end{displaymath} (7.8)

for $ {\rm Re}(\alpha_1)<1$, $ {\rm Re}(\alpha_1+\alpha_2)<2$, ... , $ {\rm Re}(\alpha_1+\alpha_2+\cdots + \alpha_N)<N$. We can rewrite (7.8) as

$\displaystyle g_0(\alpha_1, \alpha_2 \ldots \alpha_N) =
{1 \over \prod_{n=1}^N \phi_n( \alpha_1,\alpha_2 \ldots \alpha_N)},
$

where

$\displaystyle \phi_n( \alpha_1,\alpha_2 \ldots \alpha_N)
= 1-\frac{1}{n} \sum_{p=1}^n \alpha_p, \;\;\; 1\leq n \leq N.
$

Alternatively,

$\displaystyle \phi_n( \alpha_1,\alpha_2 \ldots \alpha_N)
= 1-\sum_{p=1}^N \; q_{np}\;\alpha_p, \;\;\; 1\leq n \leq N,
$

where

$\displaystyle q_{np} = \left\{ \begin{array}{lcl} \frac{1}{n}&\mbox{for}&1\leq p\leq n  \\
0&\mbox{for}&n< p\leq N \end{array}\right\}, n=1,2 \ldots N.
$

Define the $ N$-by-$ N$ matrix $ {\bf Q}=[q_{np}]$ and $ \alpha$$ =[\alpha_1\;\alpha_2 \cdots \alpha_N]^\prime$. Then,

   $\displaystyle \mbox{\boldmath$\phi$}$$\displaystyle ($$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle ) \stackrel{\mbox{\tiny$\Delta$}}{=}
\left[ \begin{array}{c}
\p...
...
\end{array}\right] =
{\underline{\bf 1}}- {\bf Q} \mbox{\boldmath$\alpha$},
$

where $ {\underline{\bf 1}}$ is an $ N$-by-1 column vector of ones. Thus, $ g_0($$ \alpha$$ )$ is the reciprocal of the product of the elements of $ \phi$$ ($$ \alpha$$ )$, denoted by

$\displaystyle g_0($$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle ) = {1 \over {\rm prod}({\underline{\bf 1}}- {\bf Q} \mbox{\boldmath$\alpha$})},$ (7.9)

where $ {\rm prod}(\;)$ is the product of the elements of the argument.

Baggenstoss 2017-05-19