Solving for asymptotic mean of MCMC-UMS

We can apply the method of section 5.3.2 approximate the asymptotic mean of MCMC-UMS for doubly-bounded data. In other words, we propose to model the distribution of MCMC-UMS generated data by (6.6). We then maximize the entropy of this distribution over the mean under the constraint that the mean is on the manifold. More precisely, we propose to maximize the entropy of (6.6) over $ \lambda$ subject to (5.12). Unfortunately, the entropy is written in terms of both $ \alpha$ and $ \lambda$. But, it can be shown that for a given $ \lambda$, there is a unique $ \alpha$ [41]. This is true in one or more dimensions. Therefore, $ \alpha$ and $ \lambda$ are alternative parameterizations for the multivariate truncated exponential distribution.

In the same manner as in Sections 4.4 and 5, we use $ {\bf u}$ as the free variable under the constraint (5.12). So, to maximize (6.6), we need the derivatives of (6.6) with respect to the elements of $ {\bf u}$. Using the derivative chain-rule, we can write the first derivative of (6.6) with respect to $ u_i$

$\displaystyle H_{db}^{u_i} =
H_{db}^{\alpha_k} \left( \frac{{\rm d...
...bda_k}{{\rm d} \alpha_k}\right)^{-1}
\frac{\partial \lambda_k}{\partial u_i},

where, from (6.6)

$\displaystyle H_{db}^{\alpha_k} = -\frac{1}{\alpha_k}
+ \alpha_k \frac{e^{\alpha_k}}{ \left(e^{\alpha_k}-1\right)^2}.$

From (6.3) ,

$\displaystyle \frac{{\rm d}\lambda_k}{{\rm d} \alpha_k}
= \frac{1}{\alpha_k^2} - \frac{e^{\alpha_k}}{ \left(e^{\alpha_k}-1\right)^2}.$

And, from (5.10),

$\displaystyle \frac{{\rm d}\lambda_k}{{\rm d} u_i} = B_{k,i}.$

After (a lot of) cancellations, we get

$\displaystyle H_{db}^{u_i} = - \sum_{k=1}^N \; \alpha_k B_{k,i},$ (6.7)

resulting in the condition for maximization of entropy:

$\displaystyle \sum_{k=1}^N \; \alpha_k B_{k,i} = 0, \;\;\; \forall i.$ (6.8)

Baggenstoss 2017-05-19