Showing natural filtration of counting process is the same as filtration at $n-1$'st jump $T_{n-1}$ given the event $\{T_{n-1}\le t < T_n\}$

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Consider a simple counting process $N(t)$ and $(\mathcal{F}_t)_{t\ge 0}$ the natural filtration and let $T_1,T_2,...,$ be the associated jumps, which are all $\mathcal{F}_t$-stopping times. I want to show that the two trace $\sigma$-algebras coincide: $$\mathcal{F}_t|_{\{T_{n-1}\le t < T_n\}}=\mathcal{F}_{T_{n-1}}|_{\{T_{n-1}\le t < T_n\}}$$

Where $\mathcal{F}_{T_n-1}=\{A\in \bigcup_{s\ge 0}\mathcal{F}_s : A\cap\{T_{n-1}\le s\}\in\mathcal{F}_s \forall s\ge0 \}$
I.e that the information at time $t$ is the same as the information at the $n-1$'st jump given that no jumps appear between $T_{n-1}$ and $t$.

This seems very intuitive, but I've really been struggling with how to show it rigorously. I really don't know where to begin, the sets in the above $\sigma $-algebras seem too complicated to work with. I would greatly appreciate some help or hints.