This question is now concerning stochastic processes. Let $(X_t)_{t\geq0}$ be defined on the probability space $(\Omega,\mathcal{F},P)$ with $\mathcal{F}_t=\sigma(X_s:s\leq t)$. Assume that for the restircted measures it holds $P_{|\mathcal{F}_t}\ll Q_{|\mathcal{F}_t}$.

Then $$f(\mathcal{F}_t)=\frac{dP_{|\mathcal{F_t}}}{dQ_{|\mathcal{F}_t}} \tag1$$ is $\mathcal{F}_t$ measurable by radon-nikodym and thus is a.s. a function of $(X_s)_{s\leq t}$, say $g((X_s)_{s\leq t})$.

Can we conclude that the following holds, considering the push forward measures $P^{(/X_s)(\omega))_{s\leq t}}$?

$$f(\mathcal{F}_t)(\omega)=\frac{dP_{|\mathcal{F_t}}}{dQ_{|\mathcal{F}_t}}(\omega)=\frac{dP^{(X_s(\omega))_{s\leq t}}}{dQ^{(X_s(\omega))_{s\leq t}}}=g((X_s(\omega)_{s\leq t}) \tag2$$

What can we say about this situation? What does this mean?

Are the pushforward measures induced by the sample path $(X_s)_{s\leq t}$? And can we think of the Radon-Nikodym-Density of $P\ll Q$ on $\mathcal{F}_t$ as the Likelihood-function of the joint-density of $(X_s)_{s\leq t}$?

Or do we say that (1) can be expressed almost surely as a function of the sample path $(X_s)_{s\leq t}$?

What I know from more elementary probability theory

Let $X$ be a random variable mapping to $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ and $\mathcal{F}^{X}=X^{-1}(\mathcal{B}(\mathbb{R}))$. Assume there is a measure $Q$, such that $P_{|\mathcal{F}^{X}}\ll Q_{|\mathcal{F}^{X}}$. We define the Radon-Nikodym density as


If this is case for the pushforward measures it holds $P^{X}\ll Q^{X}$, where $Q^{X}(B)=Q(A)$ and $A:=X^{-1}(B)\in \mathcal{F}^{X}$. We can define a Radon-Nikodym density as


Now the transformation theorem states, that

$$f(\mathcal{F}^{X})(\omega)=g(X(\omega))\quad Q-a.s.$$

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    $\begingroup$ As far as i can see the first and the second statement are the same. In the first statement you consider the random variable: $Y : \Omega \to \mathbb{R}^{[0,t]},$ with $Y(\omega) = (X_s (\omega))_{0 \le s \le t}$ $\endgroup$ – Kore-N Jun 20 '16 at 8:05
  • $\begingroup$ @Cornelis yeah i thought of the same. But is (2) fullfilled? Lets say $P\ll Q$ on $\mathcal{F}_t$ then the result would say, that the push forward measures of $P$ and $Q$ are induced by the whole path process $(X_s)_{s\leq t}$ generating $\mathcal{F}_t$, right? $\endgroup$ – ziT Jun 20 '16 at 8:53
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    $\begingroup$ (2) should be fullfilled because your "elementary" approach fives you a g such that: $f(\mathcal{F}^{Y})(\omega)=g(Y(\omega))\quad Q-a.s.$ Now you need to prove that $f(\mathcal{F}^{Y}) = f(\mathcal{F}_{t}).$ But this follows from the definition of product sigma-field. $\endgroup$ – Kore-N Jun 20 '16 at 9:02
  • $\begingroup$ @Cornelis ok thanks. If one thinks a little bit further, does this mean, that $P\ll Q$ on $\mathcal{F}_t$ corresponds to the likelihood-function of observering the process $(X_s)_{s\leq t}$. Since in (2) on $(\mathbb{R}^{[0,t]},\mathcal{B})$ the push forward measure is induced by the path. Which we can see as the joint distribution of all finite dimensional distribution ( which in analogy to discrete time, defines the likelihood observing the process.) $\endgroup$ – ziT Jun 20 '16 at 9:21

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