Let $(\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0},\mathbb P)$ a filtered probability space satisfying the usual conditions and $\mathbb Q$ a further probability measure with $\mathbb P\ll\mathbb Q$. By the Radon-Nikodym theorem we know of the existence of a $\mathcal F$-measureable function $$\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$$
so that $\mathbb P(A)=\int_A \frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}d\mathbb Q$ for $A\in\mathcal F$. My question is, if we now restict the measures for each $t$ to the sub-$\sigma$-algebras $\mathcal F_t$ of the filtration, what can we say about the Radon Nikodym derivatives
$$S_t:=\frac{\mathrm d\mathbb P|_{\mathcal F_t}}{\mathrm d\mathbb Q|_{\mathcal F_t}}.
$$
My guess is, that the Radon Nikodym derivative $S_t$ has to be 'very similar' to the 'original one' $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$, since I know from this post that if $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$ is already $\mathcal F_t$-measureable, it wouldn't change by the restiction. But what can we say in genereal? Are there some interesting results regrading this? Also I would realy like to have a picturesque description of how the restriction effects on the Radon Nikodym derivative. And can we conclude from the 'usual conditions' that the function $t\mapsto S_t$ is càdlag?
EDIT: Tanks to the comment of Kavi Rama Murthy, I know that $S_t$ is the conditional expectation of $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$ given $\mathcal F_t$ – to be exactly the conditional $\mathbb Q$-expectation. Here is my proof:
For all $A\in\mathcal F_t$ holds
$$\int_A \frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}\mathrm d\mathbb Q=\mathbb P(A)=\mathbb P|_{\mathcal F_t}(A)=\int_A \frac{\mathrm d\mathbb P|_{\mathcal F_t}}{\mathrm d\mathbb Q|_{\mathcal F_t}}\mathrm d\mathbb Q|_{\mathcal F_t}=\int_A \frac{\mathrm d\mathbb P|_{\mathcal F_t}}{\mathrm d\mathbb Q|_{\mathcal F_t}}\mathrm d\mathbb Q.
$$
Thus we know that $S$ is a uniformly integrable $\mathbb Q$-Martingale, i.e.
$$\forall 0\leq s\leq t:\mathbb E_{\mathbb Q}\left(\left.S_t\right|\mathcal F_s\right)=S_s\qquad\text{and}\qquad \mathbb E_{\mathbb Q}\left(\left.\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}\right|\mathcal F_t\right)=\frac{\mathrm d\mathbb P|_{\mathcal F_t}}{\mathrm d\mathbb Q|_{\mathcal F_t}}.
$$
This answers pretty good the question in which kind $S$ and $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$ are similar. Also by this we can assue $S$ to be càdlàg. But this brings me up to a new question: Is $S$ also a uniformly integrable $\mathbb P$-Martingale?
Here are my thoughts about it: If we had the additional assumption $\mathbb Q\ll\mathbb P$ (i.e. the measures are equivalent) and $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$ is $\mathbb P$-integrable (or equivalent $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}\in L^2(\mathbb Q)$), the statement would hold since $\mathbb E_{\mathbb Q}\left(\left.\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}\right|\mathcal F_t\right)=\mathbb E_{\mathbb P}\left(\left.\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}\right|\mathcal F_t\right)$ by using the Baye's rule for conditional expectations.
It is clear that the condition $\frac{\mathrm d\mathbb P}{\mathrm d\mathbb Q}$ is $\mathbb P$-integrable always is requiered, as it would be implied by the martingale property anyway. But I think this requirement is not sattisfied in general, is it? The other requirement $\mathbb Q\ll\mathbb P$ is very restrictive. So is it possible to proof anything without this?
Best Answer
$S_t$ is the conditional expectation of $\frac {d\mathbb P} {d\mathbb Q}$ given $\mathcal F_t$ (w.r.t the probability space $(\Omega, \mathcal F, \mathbb Q)$). This is an easy consequence of the definition of conditional expectation.