Whitened MCMC-UMS

We may improve mixing even further by ``whitening" the problem. Since we have verified that $ \hat{\mbox{\boldmath $\lambda$}}({\bf z}^*)\simeq\overline{{\cal M}({\bf z}^*)}$, we can now predict the mean of the MCMC-UMS procedure and can transform the problem. Let $ \tilde{{\bf x}}$ be the ``whitened" raw input spectrum,

$\displaystyle \tilde{x}_i = x_i/\hat{\lambda}_i, \;\; 1\leq i \leq N,$

and $ \tilde{\bf A}$ the compensated matrix $ \tilde{\bf A}_{i,j} = {\bf A}_{i,j} \hat{\lambda}_i, \;\; 1\leq i \leq N,
\; 1\leq j \leq D,$ which obtains the same feature

$\displaystyle {\bf z}= {\bf A}^\prime {\bf x}= \tilde{\bf A} ^\prime \tilde{{\bf x}}.$

The new linear constraint ``manifold" is:

$\displaystyle \tilde{{\bf x}} : \; \tilde{\bf A}^\prime \tilde{{\bf x}} = {\bf z}^*.$ (5.25)

As a starting vector for $ \tilde{{\bf x}}$, we can use the vector of ones, denoted by $ {\bf 1}$, since clearly $ \tilde{\bf A}^\prime {\bf 1} = \tilde{\bf A} ^\prime \hat{\mbox{\boldmath $\lambda$}}({\bf z}^*) = {\bf z}^*.$ The MCMC-UMS procedure, then should produce random spectral vectors that have approximate mean $ {\bf 1}$ and meet (5.25). These random vectors can be transformed to produce solutions to the original problem

$\displaystyle x_i = \tilde{x}_i \hat{\lambda}_i, \;\; 1\leq i \leq N.$

The vectors produced this way are actually random samples for the original problem because they are in $ {\cal M}({\bf z}^*)$ and the linear whitening operation preserves the uniform distribution. The result of whitening can be seen in Figure 5.12 and the result is dramatic. The change in ESS (seen by drawing a horizontal line) and is more than an order of magnitude for both systematic and random directions.

Baggenstoss 2017-05-19