Consider a change of measure on $\mathcal{F}_{t}$ defined by the restriction of two probability measures of the form \begin{align} \frac{dQ_{t}(\theta)}{dP_{t}}=\exp^{ \theta A_{t}-\kappa(\theta) S_{t}} \end{align} for all $t>0$ So we can think of $\kappa(\theta) S_{t}$ being the norming factor. $S_{t}$ is a continuous non-decreasing function with $S_{t}\rightarrow \infty$ $Q-a.s.$ $S_{0}=0$ and $A_{t}$ a cadlag function. For a random-change of time $\sigma_{u}:=\inf\{t:S_{t}>u\}$ we know since $S_{t}$ is a continuous increasing function, it has inverse $S_{\sigma_{u}}=u$. If $S$ strictly increasing it holds that $\sigma_{S_{t}}=t$ and thus $A_{t}=A_{\sigma_{S_{t}}}$. Now lets assume we doesn't have the strictly increasing setup, so we have problems at $[\sigma_{u-},\sigma_{u}]$.

However in the Paper "Exponential families of stochastic processes and Lévy processes" by Kücher, Sorensen

the authorsright below equation (5.12) (on page 14 in the pdf file / resp. page 224 in the Journal) that if the partial derivative (wrt to $\theta$) of log-likelihood given by $\dot{l}_{t}:=A_{t}-\dot{\kappa}(\theta)S_{t}$ is a square integrable Martingle with mean zero and quadratic characteristic (predictable quadratic characteristic) $\ddot{\kappa}(\theta)S_{t}$. Thus if $S_{t}$ is constant, $\dot{l}_{t}$ is constant and thus $A_{t}$ is constant. Thus $A_{t}=A_{\sigma_{S_{t}}}$ even on $[\sigma_{u-},\sigma_{u}]$. I actually don't get the statement, since i only found literature according for continuous local martingales $M_{t}$, the intervals of constancy are the same as the ones of $<M>_{t}=[M]_{t}$. But we dont have this setup.

This is a more precisely reformulation of the post: link

@Admins: Should i delete the previous post then?