Normalization by parameterization: Error functional $E_z$

Again, normalization can also be ensured by parameterization of $P$ and solving for unnormalized probabilities $z$, i.e.,

$$P(x,y) = \frac{z(x,y)}{\int \!dy\, z(x,y)}, \quad P = \frac{z}{Z_X} .$$ (174)

The corresponding functional reads

$$E_z = - \left(N , \ln \frac{z}{Z_X}\right) + \frac{1}{2}\left( \frac{z}{Z_X}, {{\bf K}}\,\frac{z}{Z_X}\right).$$ (175)

We have

$$\frac{\delta z}{\delta z} = {\bf I}, \quad \frac{\delta Z_X}... ...rac{\delta \ln Z_X}{\delta z} = {\bf Z}_X^{-1} \, {\bf I}_X ,$$ (176)

with diagonal matrix ${\bf z}$ built analogous to ${\bf P}$ and ${\bf Z}_X$, and

$$\frac{\delta P }{\delta z} =\frac{\delta (z/Z_X) }{\delta z... ... z} = {\bf Z}_X^{-1} \left( {\bf P}^{-1} - {\bf I}_X \right),$$ (177)

$$\frac{\delta Z_X^{-1} }{\delta z} = - {\bf Z}_X^{-2} {\bf I... ... \, {\bf Z}_X^{-1} \left( {\bf I} - {\bf P} {\bf I}_X \right).$$ (178)

The diagonal matrices $[{\bf Z}_X , {\bf P} ] = 0$ commute, as well as $[{\bf Z}_X , {\bf I}_X] = 0$, but $[ {\bf P} , {\bf I}_X] \ne 0$. Setting the gradient to zero and using

$$\left( {\bf I} - {\bf P} {\bf I}_X \right)^T = \left( {\bf I} - {\bf I}_X {\bf P} \right) ,$$ (179)

we find

$$0 = -\left( \frac{\delta P}{\delta z} \right)^T \frac{\delta E_z}{\delta P}$$

$$={\bf Z}_X^{-1} \left[ \left( {\bf P}^{-1} - {\bf I}_X \righ... ... \left({\bf I} - {\bf I}_X {\bf P} \right) {{\bf K}} P \right]$$

$$= {\bf Z}_X^{-1} \left( {\bf I} - {\bf I}_X {\bf P} \right) \left( {\bf P}^{-1} N - {{\bf K}} P \right)$$

$$= {\bf Z}_X^{-1} \left( {\bf I} - {\bf I}_X {\bf P} \right)... ...ight) = \left( {\bf G}_P - {\bf\Lambda}_X \right) {Z}_X^{-1} ,$$ (180)

with $P$-gradient $G_P = {\bf P}^{-1} N - {{\bf K}} P$ = $-\delta E_z/\delta P$ of $-E_z$ and ${\bf G}_P$ the corresponding diagonal matrix. Multiplied by ${\bf Z}_X$ this gives the stationarity equation (172).