Kernels for $L$

For $E_L$ we have the truncated equation

$$L = {{\bf C}} N.$$ (688)

Normalizing the exponential of the solution gives

$$L(x,y) = \sum_i^n {{\bf C}} (x,y;x_i,y_i) - \ln \int \! dy^\prime \, e^{\sum_i^n {{\bf C}} (x,y^\prime ;x_i,y_i)} ,$$ (689)

or

$$L = {\bf C} N -\ln {\bf I}_X {\bf e}^{{\bf C}N} .$$ (690)

Notice that normalizing $L$ according to Eq. (689) after each iteration the truncated equation (688) is equivalent to a one-step iteration with uniform $P^{(0)}$ = $e^{L^{(0)}}$ according to

$$L^1 = {{\bf C}} N + {\bf C}{\bf P}^{(0)} \Lambda_X,$$ (691)

where only $({\bf I}-{\bf C}{\bf K})L$ is missing from the nontruncated equation (683), because the additional $y$-independent term ${\bf C}{\bf P}^{(0)} \Lambda_X$ becomes inessential if $L$ is normalized afterwards.

Lets us consider as example the choice ${{\bf C}}$ = $-{\bf H}^{-1}(\phi^{(0)})$ for uniform initial $L^{(0)}=c$ corresponding to a normalized $P$ and ${{\bf K}}L^{(0)}$ = $0$ (e.g., a differential operator). Uniform $L^{(0)}$ means uniform $P^{(0)}=1/v_y$, assuming that $v_y = \int \!dy$ exists and, according to Eq. (137), $\Lambda_X$ = $N_X$ for ${{\bf K}}L^{(0)}$ = $0$. Thus, the Hessian (161) at $L^{(0)}$ is found as

$${\bf H}(L^{(0)}) = - \left( {\bf I} - \frac{{\bf I}_X}{v_y}... ...I}_X}{v_y} \right) \frac{{\bf N}_X}{v_y} = -{{\bf C}}^{-1} ,$$ (692)

which can be invertible due to the presence of the second term.

Another possibility is to start with an approximate empirical log-density, defined as

$$L^\epsilon_{\rm emp} = \ln P^\epsilon_{\rm emp} ,$$ (693)

with $P^\epsilon_{\rm emp}$ given in Eq. (239). Analogously to Eq. (686), the empirical log-density may for example also be smoothed and correctly normalized again, resulting in an initial guess,

$$L^{(0)} = {\bf C} L^\epsilon_{\rm emp} - \ln {\bf I}_X {\bf e}^{{\bf C} L^\epsilon_{\rm emp}} .$$ (694)

Similarly, one may let a kernel ${\bf C}$, or its normalized version $\tilde {\bf C}$ defined below in Eq. (698), act on $P_{\rm emp}$ first and then take the logarithm

$$L^{(0)} = \ln (\tilde {\bf C} P^\epsilon_{\rm emp}) .$$ (695)

Because already $\tilde {\bf C} P_{\rm emp}$ is typically nonzero it is most times not necessary to work here with $P^\epsilon_{\rm emp}$. Like in the next section $P_{\rm emp}$ may be also be replaced by $\tilde P_{\rm emp}$ as defined in Eq. (238).