I am trying to find out how to decompose conditional loglikelihood of a function to a conditional loglikelihood of an argument plus some reminder terms.
I.e. shortly: $$\mathcal{L}_E(\hat{E};\sigma) = \mathcal{L}_A(F^{-1}(\hat{E});\sigma) + \ldots$$ where $E=F({A})$, i.e. $F$ - my function, $A$ - my argument.
Real decomposition looks like this: $$ \mathcal{L}_E(\hat{E_1},\ldots,\hat{E_N};\sigma) = \mathcal{L}_A(F^{-1}(\hat{E_1}),\dots,F^{-1}(\hat{E_N});\sigma) - \sum^{N}_{i=1}\log \left[\frac{\partial F}{\partial A}\right] ((F^{-1}(E,\sigma)) $$
But for now, I can see how author obtained it. It reminds me kind of Taylor expansion: $$E = F(A) + \int F(A)'dA + \ldots$$ where $F(A)$ should be $ \mathcal{L}_A(F^{-1}(\hat{E_1}),\dots,F^{-1}(\hat{E_N});\sigma) $, $ F(A)' \approx \left[\frac{\partial F}{\partial A}\right] $ and $ A \approx ((F^{-1}(E,\sigma))$.
But this is of course very vaguely written, as: not $F(A)$ but $A$ should be $ \mathcal{L}_A(F^{-1}(\hat{E_1}),\dots,F^{-1}(\hat{E_N});\sigma)$ and I do not write the integral in second case. Moreover we see equal expression, not infinite sum... Could somebody give a hint how to decompose it?
For simplification, I have been looking at case of just one conditional density function, but could not represent it so far.
