ML approach

Under certain conditions, the J-function is independent of $ H_0$ as long as $ H_0$ remains within the ``region of sufficiency" (ROS) for $ {\bf z}$ (See Section 2.3.3). Then, $ H_0$ can even ``float" with the data as long as it remains in the ROS. The ROS can be spanned by a parametric model $ p({\bf x}; {\bf h})$ as long as (a) $ p({\bf x};H_0) = p({\bf x}; {\bf h}_0)$ for some parameter $ {\bf h}_0$, and (b) $ {\bf z}$ is a sufficient statistic for $ {\bf h}$.

We can easily meet these conditions using the multi-variate TED distribution (6.5) with $ \alpha$$ ={\bf A} {\bf h}$. Condition (a) is met by $ {\bf h}={\bf0}$. Condition (b) is met since (6.5) can be written $ p({\bf x}; {\bf h}) = f({\bf z},{\bf h}),$ for some function $ f$. It therefore follows that

$\displaystyle \frac{p({\bf x}; H_0)}{p({\bf z}; H_0)} = \frac{p({\bf x}; {\bf h})}{p({\bf z}; {\bf h})},$ (16.16)

for any $ {\bf h}$, where $ p({\bf z}; {\bf h})$ is defined as the distribution of $ {\bf z}$ when $ {\bf x}\sim p({\bf x}; {\bf h})$. Since the ratio (16.16) does not depend on $ {\bf h}$, it makes sense to place $ {\bf h}$ at the point where $ p({\bf z}; {\bf h})$ can be easily evaluated, and that is the point where both $ p({\bf x}; {\bf h})$ and $ p({\bf z}; {\bf h})$ have their maximum value, that is to say at the maximum likelihood (ML) point

$\displaystyle \hat{{\bf h}} = \arg \max_{{\bf h}} p({\bf x}; {\bf h}).$

At this point, we can apply the central limit theorem to find $ p({\bf z}; \hat{{\bf h}})$. The mean is given by

$\displaystyle {\cal E}\{{\bf z}\} = {\bf A}^\prime {\cal E}\{{\bf x}\}
= {\bf A}^\prime \lambda( {\bf A} \hat{{\bf h}}),$

where $ {\cal E}\{ \; \}$ is expected value, and $ \lambda(\;)$ is the TED mean (6.3). Note that under $ p({\bf x}; \omega)$,

$\displaystyle {\cal E}\{x_i^2\} =
\frac{2}{\alpha^2}
\left[
\frac{1-\frac{1}{2} \left( \alpha^2 -2\alpha + 2\right) e^\alpha}{1-e^\alpha}
\right],$

where $ \alpha$$ ={\bf A} {\bf h}$ [77,78]. From this, we can solve for the variance of $ x_i$,

$\displaystyle {\rm var}(x_i) = {\cal E}\{x_i^2\}-({\cal E}\{x_i\})^2 = \frac{1}{\alpha^2} - \frac{e^\alpha}{(e^\alpha -1)^2}.$ (16.17)

The covariance of $ {\bf z}$ is therefore

$\displaystyle {\rm cov}({\bf z}) = {\bf A}^\prime \Lambda {\bf A},$

where $ \Lambda$ is the diagonal matrix with elements (16.17). Finally, then, we apply (16.16) at $ {\bf h}=\hat{{\bf h}}$, and $ p({\bf x}; H_0)=1$, to get

$\displaystyle \frac{1}{p({\bf z}; H_0)} = \frac{p({\bf x}; \hat{{\bf h}})} {{\c...
...bf A}^\prime \lambda( {\bf A} \hat{{\bf h}}), {\bf A}^\prime \Lambda {\bf A})},$ (16.18)

where $ {\cal N}({\bf z};$   $ \mu$$ , {\bf R})$ is the Gaussian distribution with mean $ \mu$ and covariance $ {\bf R}$. This approach can then be compared numerically with the reciprocal of (2.12).

Baggenstoss 2017-05-19