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I found this exercise online:

enter image description here

I am stuggling with the last part of the second exercise, that is I am not able to show that $\tau = \sup_i \tau_i$. Obviously we have that $\tau \ge \sup_i \tau_i$, but I struggle to show the other inequality. Any tips?

Also, I have heard that the second case also holds under the restriction from the first case, that is right continuity, or maybe we need to assume cadlag-trajectories. Do you see how we can show the second case assuming this?

My idea for proving this was this: I was able to show it easily if $\tau=0,\infty$. So I can exclude those cases, and look at any $\epsilon>0$. Then I must have that in the interval $[0,\tau-\epsilon]$ atleast one $G_i$ is not hit.

Can you please help me?

saz
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user119615
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2 Answers2

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Since $G_i \supseteq B$, we have $\tau_i \leq t$ and therefore, as you already noted,

$$\sup_{i \in \mathbb{N}} \tau_i \leq \tau.$$

It remains to show that

$$\tau \leq \sup_{i \in \mathbb{N}} \tau_i.$$

Fix $\omega \in \Omega$. Without loss of generality, we may assume that the right-hand side is finite (otherwise there is nothing to prove), i.e.

$$T(\omega):= \sup_{i \in \mathbb{N}} \tau_i(\omega) <\infty.$$

Since $\tau_1 \leq \tau_2 \leq \ldots$, it follows from the very definition of "sup" that $\lim_{i \to \infty} \tau_i(\omega) = T(\omega)$. Hence, as $(X_t)_{t \geq 0}$ has continuous sample paths,

$$X_{T(\omega)}(\omega) = \lim_{i \to \infty} X_{\tau_i(\omega)}(\omega).$$

Since

$$|X_{T(\omega)}-b| \leq |X_{T(\omega)}-X_{\tau_i(\omega)}(\omega)| + |X_{\tau_i(\omega)}(\omega)-b|$$

for any $b \in B$, we find

$$d(X_{T(\omega)}(\omega),B) \leq |X_{T(\omega)}-X_{\tau_i(\omega)}(\omega)| + \underbrace{d(X_{\tau_i(\omega)}(\omega),B)}_{\leq i^{-1}} \xrightarrow[]{i \to \infty} 0,$$

i.e.

$$d(X_{T(\omega)}(\omega),B)=0.$$

As $B$ is closed, this entails $X_{T(\omega)}(\omega) \in B$. By the very definition of $\tau$, this means that $\tau(\omega) \leq T(\omega)$.

As the proof shows we just need left-continuity of the sample paths and not necessarily continuity.

saz
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  • Thank you very much, do you know if it is easy to change it to right-continuous? – user119615 Mar 30 '16 at 06:26
  • @user119615 No, I don't think so. It is easy to change it to left-continuity, but, as far as I can see, not to right-continuity. However, the statement does hold true for right-continuous processes (this is a particular case of the so-called Début theorem). Have a look at this question: http://math.stackexchange.com/q/1514488/ – saz Mar 30 '16 at 08:07
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    Just to be clear: right-continuity is still needed for the conclusion at that bullet point, just not for the $\tau=\sup \tau_i$ property – Bananach Jan 30 '20 at 23:29
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Think about where $\lim_i X_{\tau_i}$ has to be; begin by observing that $X_{\tau_i}\in\overline G_i$ on $\{\tau_i<\infty\}$.

John Dawkins
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