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let $(\Omega, \mathcal{A}, P)$ be a probability space and $(N_t)_{t \geq 0}$ a right-continuous counting process with jumps of size 1, $N_0 = 0$ and canonical filtration $\mathcal{F}_t := \sigma( N_u \ | \ u \in [0,t] )$. Consider the first jump $T_1 := \inf \left\{ t > 0 \ | \ N_t = 1 \right\}$ which is a stopping time with respect to $\mathcal{F}_t$. Let $\mathcal{F}_{T_1} = \left\{ A \in \mathcal{A} \ | \ A \cap \left\{ T_1 \leq t \right\} \in \mathcal{F}_t \ \forall t \geq 0 \right\}$ denote its stopped $\sigma$-algebra. Since $T_1$ is $\mathcal{F}_{T_1}$-measurable it follows that $\sigma(T_1) \subseteq \mathcal{F}_{T_1}$.

Is it true that $\mathcal{F}_{T_1} = \sigma(T_1)$?

Intuitively, the information gained up to time $T_1$ is just given by $T_1$ itself, since $T_2$ is activated only when the first arrival happens.

This statement seems to be so natural, that I assumed it to find in in the literature or by googling around. But I haven't found any proof for that and I also can't create a proof on my own.

Take for example $A = \left\{ T_2 \leq s \right\}$ and try to contradict that $A \in \mathcal{F}_{T_1}$. Then $A \in \mathcal{F}_s$ and consider the set $S = \left\{ t > 0 \ | \ A \not\in \mathcal{F}_t \right\}$. Assume that $0 \in S$ and take $t_0 := \sup S$ such that $t_0 < s$. If $\left\{ T_2 \leq s \right\} \in \mathcal{F}_{T_1}$ then $\left\{ T_2 \leq s \right\} \cap \left\{ T_1 \leq t \right\} \in \mathcal{F}_t$ for all $t < t_0$ but $\left\{ T_2 \leq s \right\} \not\in \mathcal{F}_t$, which seems a little bit strange but does not leed to a contradiction so far. Translated verbally, this means that at all times $t < t_0$ I can observe whether $T_1 \leq t$ and $T_2 \leq s$ simultaneously, but I can't observe $T_2 \leq s$. If quantifying with probability measures then the probability of $\left\{ T_2 \leq s \right\} \cap \left\{ T_1 \leq t \right\}$ given $\mathcal{F}_t$ can be less then 1.

Maybe it is necessary to consider $\Omega$ as the space of right-continuous functions $D([0, \infty), \mathbb{R})$, since the claim is a statement on the level of $\sigma$-algebras and not a statement of an "almost surely" kind.

yada
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1 Answers1

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Quite generally, if the canonical filtration $(\mathcal F_t)$ is generated by the process $N=(N_t)$, then, for every stopping time $T$, the sigma-algebra $\mathcal F_T$ is generated by the process $N^T=(N^T_t)_t$ obtained by stopping $N$ at time $T$. That is, $N^T_t=N_{\min\{T,t\}}$ for every $t\geqslant0$. In your case $N^T$ is $\sigma(T)$-measurable since $N^T_t=\mathbf 1_{T\leqslant t}$ for every $t\geqslant0$, hence $\mathcal F_T\subseteq\sigma(T)$. The other inclusion being valid in general, this indeed proves that $\mathcal F_T=\sigma(T)$.

Did
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  • Thank you for the hint. The fact that $\mathcal{F}_T$ is generated by $N^T$ is precisely the point with which I am struggling. It seems that the authors in this post also have this problem. Do you have a reference for this fact? I believe the correctness for this statement if $N$ is progressively measurable (e.g. right-continuous). But for general $N$ this statement might be wrong. – yada Sep 15 '14 at 07:53
  • There seems to be a specific reference to a proof in the page you link to. – Did Sep 15 '14 at 07:57
  • The reference there is to the book of Bain and Crisan. It seems like here the authors also have problems with the proof that is shown in that book. Maybe you know some reference which you can trust? And also, it seems like there must be an assumption that $N$ is progressively measurable in order to be able to deduce such a statement. – yada Sep 15 '14 at 09:35
  • The specific reference on the page is to IV.100 of Probabilities and Potential (Dellacherie and Meyer). – Did Sep 15 '14 at 09:41
  • Dellacherie and Meyer stick to a special sample space $\Omega$, namely to cadlag functions $\omega : [0, \infty) \to \overline{E}$. For this special choice of $\Omega$ and the canonical process $X_t(\omega) = \omega(t)$ together with the result IV.100 applied in the proof of Bain and Crisan the statement is true. But for a general $\Omega$ with probability measure $P$, I still think that the statement only holds in an "almost surely" kind measured by $P$, since the law of $P$ can be given by the canonical process. – yada Sep 15 '14 at 11:58
  • I had a look at the proof of Bain and Crisan. They take $A \in \mathcal{F}_T$ (i.e. $T_A$ is a stopping time) which is equivalent by Dellacherie and Meyer that ($\omega \in A, T(\omega) = T(\omega'), X_s(\omega) = X_s(\omega') \forall s \leq T(\omega) => \omega' \in A$). Now they say that in order to show that $A \in \sigma(X^T_t | t \geq 0)$ it is necessary to show that for $\omega, \omega' \in \Omega$ with $X_s(\omega) = X_s(\omega')$ for all $s \leq T(\omega)$ it holds that if $\omega \in A$ then $\omega' \in A$ which is just a rephrasement of $A$ being in $\mathcal{F}_T$! – yada Sep 19 '14 at 14:39