The PDF $ p({\bf z}\vert H_0)$ can be derived if we first orthogonalize the features. Let $ \rho = {\bf x}^\prime \left\{ {\bf I}-{\bf A} \left({\bf A}^\prime{\bf A}
\right)^{-1} {\bf A}^\prime\right\}
{\bf x},$ which can also be written $ \rho = t_2({\bf x}) - {\bf x}^\prime {\bf A} \left({\bf A}^\prime{\bf A}
\right)^{-1} {\bf A}^\prime {\bf x}.$ This is the energy in $ {\bf x}$ orthogonal to the columns of $ {\bf A}$. Accordingly, $ \rho$ is statistically independent of $ {\bf z}_A$ under $ H_0$. Thus, $ p(\rho, {\bf z}_A\vert H_0) = p(\rho\vert H_0) \; p({\bf z}_A \vert H_0).$ Also, $ p(\rho\vert H_0)$ is chi-squared with $ N-D$ degrees of freedom,

$\displaystyle p(\rho\vert H_0)=\frac{\rho^{(\kappa/2-1)} \; e^{-\rho/2}}{2^{\kappa/2} \; \Gamma(\kappa/2)},$

where $ \kappa=N-D$ and $ p({\bf z}_A \vert H_0)$ is the Gaussian distribution with mean $ {\bf0}$ and co-variance $ {\bf A}^\prime{\bf A}$:

$\displaystyle p({\bf z}_A\vert H_0) = \left(2\pi\right)^{-\frac{D}{2}}
\; \vert...
...frac{1}{2}{\bf z}_A^\prime \left({\bf A}^\prime{\bf A}
\right)^{-1} {\bf z}_A}

And, since $ (\rho,{\bf z}_A)$ can be obtained from $ {\bf z}$ using a linear transformation with Jacobian of determinant 1, we can write $ p({\bf z}\vert H_0) = p(\rho\vert H_0) \; p({\bf z}_A \vert H_0).$ The J-function is just $ p({\bf x}\vert H_0)/p({\bf z}\vert H_0).$ For more information see the function software/module_lin_gauss.m.

Baggenstoss 2017-05-19