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Let $X=(X_t)_{t\geq 0}$ be a stochastic process on a stochastic basis $(\Omega,\mathcal{F},\mathcal{P},\mathbb{F}=(\mathcal{F}_t)_{t\geq 0})$. Let $\tilde{\mathbb{F}}=(\tilde{\mathcal{F}}_t)_{t\geq 0})$ be a subfiltration.

While it is (probably) not true in general that $$\int_0^tE[f(X_s)\mid \tilde{\mathcal{F}}_s]ds=E[\int_0^tf(X_s)ds\mid\tilde{\mathcal{F}}_t]$$ holds, my question is if $$\int_0^tE[(\mathcal{L}^Xf)(X_s)\mid\tilde{\mathcal{F}}_s]ds=E[\int_0^t(\mathcal{L}^Xf)(X_s)ds\mid\tilde{\mathcal{F}}_t]$$ is true where $X$ is $\mathbb{F}$-adapted and a $(\mathbb{F},\mathcal{P})$-markov process and $\mathcal{L}^X$ the generator of $X$ and $f$ an element of the domain of $\mathcal{L}^X$.

peer
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  • Is $X$ adapted? – Did Aug 01 '17 at 07:47
  • @Did I did some edits to the question. $X$ is $\mathbb{F}$-adapted, but not adapted to the coarser filtration $\tilde{\mathbb{F}}$. Your first comment (which you deleted?) seems to be important if we want to apply this result: https://math.stackexchange.com/questions/305707/exchange-integral-and-conditional-expectation . – peer Aug 01 '17 at 09:02
  • @Did Or is it true if $X$ is adapted to $\tilde{\mathbb{F}}$? – peer Aug 01 '17 at 09:05
  • If $X$ is not $\tilde{\mathbb F}$-adapted, I see no way to expect that $$E(f(X_s)\mid\tilde{\mathcal F}_s)=E(f(X_s)\mid\tilde{\mathcal F}_t)$$ for every $s<t$, likewise with $\mathcal L^Xf$, and then why should we expect the identity you are after, to hold? – Did Aug 01 '17 at 10:04
  • @Did $\tilde{\mathbb{F}}$ is generated by a point process $N$ defined by $N_t=\sum_{i=1}^l1_{{T_i\leq t}}$ where $l\in\mathbb{N}$, $P(T_i>t\mid \mathcal{G}^X_\infty)=\exp(-\int_0^t \lambda(s,X_s)ds)$. But that doesn't imply (to my knowledge) that $X$ is adapted to $\tilde{\mathbb{F}}=\mathbb{F}^N$. $X$ is $\mathbb{F}$-adapted. – peer Aug 01 '17 at 13:53

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