Given a probability space $(\Omega,\mathcal{F},P)$. Let $(X_t)_{t\geq0}$ be a stochastic process defined on it with cadlag paths, lets say on $(\mathcal{X},\mathcal{B}(X))$. Let be $\mathcal{F}_{t}$ be the natural filtration of $(X_t)$. Assume that $Q$ is a $\sigma$-finite measure on $\mathcal{F}_t$ for each $t\geq 0$. Denote $P^t$ the restriction of $P$ to $\mathcal{F}_t$. Then by Radon-Nikodym on $\mathcal{F}_t$ is defined by $$ L(\mathcal{F}_t)(\omega)=\frac{dP^{t}}{dQ^{t}}(\omega) \tag 1 $$ on $(\Omega,\mathcal{F})$ and this expression is $\mathcal{F}_t$ measurable. Note this is a martingale so for $s<t$ we have $\mathcal{F}_s\subseteq \mathcal{F}_t$ and thus $$ L(\mathcal{F}_s)=E_{Q}[L(\mathcal{F}_t)|\mathcal{F}_s] $$ Let $X:=(X_s)_{s\in[0,t]}$ be the trajectory on $[0,t]$. Then we have with the pushforward measure/induced measure $P^{X}$ on $(\mathcal{X},\mathcal{B}(X))$ the equality $$ L(\mathcal{F}_t)(\omega)=\frac{dP^{t}}{dQ^{t}}(\omega)=\frac{dP^{X}}{dQ^{X}} (X(\omega))\quad Q-a.s. $$ by theorem of measure transformation.
How is this setup change, when we take the right continuous filtration $\tilde{\mathcal{F}}_t=\bigcap_{s>t}\sigma(X_u:u\leq s)$. This is often done in statistical inference for stochastic processes. Lets define the radon nikodym derivative like above with respect to this filtration. Can we get a representation of $L(\tilde{\mathcal{F}}_t)$ analogous to the one above? Is $$ L(\tilde{\mathcal{F}}_t)(\omega)=\frac{dP^{X(\omega)}}{dQ^{X(\omega)}} \tag2 $$ still valid? Note $X_t$ is cadlag. Or is this defined otherwise?
In many applications (change of measure theory of stochastic processes) the likelihood-functions i've seen coincide.
Best regards