Generation of Samples from $ G({\bf x};H_0,T,g)$

Indeed, $ G({\bf x};H_0,T,g)$ is a generative model. In this book, we will continually discuss the generative process of creating synthetic samples of $ {\bf x}$ by drawing samples from $ G({\bf x};H_0,T,g)$. Generation of random samples from $ G({\bf x};H_0,T,g)$ has several purposes:
  1. Validation of feature information content. Good re-producibility of the essential character and intelligibility of the input data indicates good feature extraction.
  2. The Use of PDF projection in sampling methods (importance sampling, rejection sampling, MCMC methods).
  3. PDF sculpting (Section 12.3);
Generation of data from $ G({\bf x};H_0,T,g)$ is accomplished using the following process ([5], Section 2.1)
  1. Draw a sample $ {\bf z}^*$ from $ g({\bf z})$,
  2. Determine the manifold $ {\cal M}({\bf z}^*)$, which is the set of all points $ {\bf x}$ that map to $ {\bf z}^*$ through transformation $ T({\bf x})$:

    $\displaystyle {\cal M}({\bf z}^*) = \{ {\bf x}: \; \; T({\bf x})={\bf z}^*, \; \; {\bf x}\in{\cal X} \},$ (2.4)

    where $ {\cal X}$ is the set of valid input data samples $ {\bf x}$. It is common to call $ {\cal M}({\bf z}^*)$ a manifold or level set 2.2.

  3. draw a sample $ {\bf x}$ from $ {\cal M}({\bf z}^*)$ according to a distribution proportional to $ p({\bf x}\vert H_0)$.
Note that drawing a sample $ {\bf x}$ from $ {\cal M}({\bf z}^*)$ according to a distribution proportional to $ p({\bf x}\vert H_0)$ can be regarded as a a posteriori distribution of $ {\bf x}$ given $ {\bf z}$. But, it is not a proper distribution since all its probability mass exists on $ {\cal M}({\bf z}^*)$ which has zero volume, and so must have infinite value. If we restrict our analysis just to the set $ {\cal M}({\bf z}^*)$, we can write down a representative distribution, called the manifold distribution $ \mu({\bf x};H_0,T,{\bf z})$,

$\displaystyle \mu({\bf x};H_0,T,{\bf z})=\frac{p({\bf x}\vert H_0) \; \delta(T(...
...f z}) }{\int_x p({\bf x}\vert H_0) \;\delta(T({\bf x})-{\bf z}){\rm d}{\bf x}},$ (2.5)

where $ \delta(T({\bf x})-{\bf z})=1$ when $ {\bf x}\in {\cal M}({\bf z})$ and is zero otherwise. Clearly

$\displaystyle \int_{{\bf x}\in {\cal M}({\bf z})} \; \mu({\bf x};H_0,T,{\bf z}) = 1.$

Intuitively, the manifold distribution is just a distribution on $ {\cal M}({\bf z}^*)$ that is proportional to $ p({\bf x}\vert H_0)$.

Baggenstoss 2017-05-19