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Let

  • $(\Omega,\mathcal A)$ and $(E,\mathcal E)$ be measurable spaces
  • $(Y_n)_{n\in\mathbb N_0}$ be a $(E,\mathcal E)$-valued stochastic process on $(\Omega,\mathcal A)$
  • $(N_t)_{t\ge0}$ be a $\mathbb N_0$-valued stochastic process on $(\Omega,\mathcal A)$

Don't know if it matters, but in my application, $Y$ is a Markov chain and $N$ is a Poisson process, independent from $Y$.

Moreover, let $$\mathcal F^N_t:=\sigma(N_s:0\le s\le t)\;\;\;\text{for }t\ge0$$ and $$\mathcal F^Y_n:=\sigma(Y_m:m\in\left\{0,\ldots,n\right\})\;\;\;\text{for }n\in\mathbb N_0.$$

Now, let $$X_t:=Y_{N_t}\;\;\;\text{for }t\ge0$$ and $$\mathcal F^X_t:=\sigma(X_s:0\le s\le t)\;\;\;\text{for }t\ge0$$

How can we show that $$\sigma\left(\left\{A\cap B\cap\left\{N_t=n\right\}:A\in\mathcal F^N_t,n\in\mathbb N_0\text{ and }B\in\mathcal F^Y_n\right\}\right)=\mathcal F_t:=\sigma(\mathcal F_t^N\cup\mathcal F_t^X)$$ for all $t\ge0$?

Let $t\ge0$. I even fail to show that if $A\in\mathcal F^N_t$, $n\in\mathbb N_0$ and $B\in\mathcal F^Y_n$, then $A\cap B\cap\left\{N_t=n\right\}\in\mathcal F_t$. Though, it would be clear, if $B\cap\left\{N_t=n\right\}\in\mathcal F^X_t$. I guess we somehow need to use that $X_t$ coincindes with $Y_n$ on $\left\{N_t=n\right\}$.

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  • $\begingroup$ One condition I guess should be that $Y_m\neq Y_{m+1}$ a.s. otherwise there is no way one can see the time when $N_{t}=m+1$ occures. $\endgroup$
    – RaphaelB4
    Commented Nov 7, 2020 at 10:57
  • $\begingroup$ @RaphaelB4 It should hold in general. Please take a look at p. 164 of Ethier/Kurtz. They write that the claim holds by the Dynkin class theorem. (BTW, it would be helpful if I could at least show that $N_{s+t}-N_s$ is independent of $\mathcal F^X_s$, which seems intuitively plausible. Do you have any idea regarding that?) $\endgroup$
    – 0xbadf00d
    Commented Nov 7, 2020 at 11:12

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