From theory to practice

I found that squaring the semblance values ${\rm semb}(\gamma)$ produced a better result than the semblance values themselves. My resulting cdf $a_{\rm cdf}(\gamma_i)$ function was then  

$$a_{\rm cdf}(\gamma_i)= \frac{ \sum_{\gamma_0}^{\gamma^i} {\... ...^2(\gamma)} {\sum_{\gamma_0}^{\gamma^n} {\rm semb}^2(\gamma)} .$$ (11)

In addition, I need to account for the fact that I am not scanning over all possible $\gamma$ values. I introduced a constant parameter g_max that scaled my $\bf c$ values so that that one standard deviation of $\bf c$might correspond to 2-5 standard deviations $\bf a$.

Finally, we are going to a precondition the model Fomel et al. (1997) with the inverse of $\bf A$ and solve the problems in terms of the variable $\bf p= \bf A\bf m$,

$$\begin{eqnarray} \bf n &\approx&\bf A^{-1}\bf p\nonumber \\ \bf 0&\approx&\epsilon \bf p .\end{eqnarray}$$ (12)