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Let $\{X_t, t \ge 0\}$ be a continuous stochastic process and adapted to the filtration $\{\mathcal{F}_t,t\ge 0 \}$ and consider

$$ \alpha = \inf\{t, |X_t|>1\}, $$ the first time the the process $X_t$ leaves the interval $[-1,1]$. Then can you help me to show that $\alpha$ is in fact stopping time?

Did
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Ron
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2 Answers2

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Then can you help me to show that $\alpha$ is in fact stopping time?

Well, this would be difficult, since $\alpha$ is not always a stopping time for the natural filtration $(\mathcal F^X_t)_{t\geqslant0}$ of the process $(X_t)_{t\geqslant0}$, defined by $\mathcal F^X_t=\sigma(X_s\,;\,0\leqslant s\leqslant t)$.

For a counterexample, consider some event $A$ such that $0<P(A)<1$ and define the process $(X_t)_{t\geqslant0}$ as follows:

  • On $A$, $X_t=t$ for every $t\geqslant0$.
  • On $\Omega\setminus A$, $X_t=t$ for every $0\leqslant t\leqslant1$ and $X_t=2-t$ for every $t\geqslant1$.

Then $\alpha=\inf\{t\,;\,|X_t|\gt1\}$ equals $\alpha=1$ on $A$ and $\alpha=3$ on $\Omega\setminus A$.

Thus, $\{\alpha\leqslant1\}=A$. For every $t\leqslant1$, the random variable $X_t$ is deterministic, hence $\mathcal F^X_1=\{\varnothing,\Omega\}$ is the trivial sigma-algebra, but the event $A$ is not in $\mathcal F^X_1$ hence $\{\alpha\leqslant1\}\notin\mathcal F^X_1$, which proves that $\alpha$ is not a stopping time for the filtration $(\mathcal F^X_t)_{t\geqslant0}$.

Did
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The proof below is not correct. See an example by Did. The flaw in the proof: in fact we have $$ \{\alpha>t\} = \bigcup_n\{\alpha \geq t+\frac1n\} = \bigcup_{n}\bigcap_{s\in [0,t+\frac1n]}\{|X_s|\leq 1\} $$ and there appear events $\{|X_s|\leq 1\}$ for $s>t$.

By the definition, a random variable $\tau:\Omega\to [0,\infty]$ is a stopping time if and only if $$ \{\tau \leq t\}\in \mathscr F_t $$ for any $t\in [0,\infty)$. We have in your case $$ \{\alpha \leq t\} = \Omega\setminus \{\alpha > t\} = \Omega\setminus \bigcap_{s\in [0,t]}\{|X_s|\leq 1\} = \Omega\setminus\bigcap_{s\in \Bbb Q\cap [0,t]}\left\{|X_s|\leq 1\right\} \in \mathscr F_t $$ where we pass to the intersection over rational numbers only since $X$ has continuous trajectories.

SBF
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  • Thanks a lot for the proof. – Ron Apr 05 '13 at 11:54
  • @Shaik: you're welcome – SBF Apr 05 '13 at 11:55
  • Is it always true that we pass intersection over rational numbers in your proof? If ${X_t, t \ge 0}$ is right continuous, then even in that case do we pass intersection over rational numbers? – Ron Apr 17 '13 at 12:46
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    Sorry but the proof does not work since ${\alpha > t}$ is not always $$\bigcap_{s\in [0,t]}{|X_s|\leq 1}.$$ – Did Jul 24 '14 at 15:58
  • @Did: you certainly have a fan :) I'll read your answer now. – SBF Jul 24 '14 at 16:04
  • @Did: is that correct then that this even is ${\alpha\geq t}$? – SBF Jul 24 '14 at 16:11
  • Neither, the most one can say is that the intersection is between ${\alpha\gt t}$ and ${\alpha\geqslant t}$. – Did Jul 24 '14 at 16:14
  • @Did: can you elaborate on this, please? If $|X_s|\leq 1$ for all $s\leq t$, then $\alpha$ is at least $t$, so the intersection is a subset of ${\alpha\geq t}$. Now, if $\alpha\geq t$ suppose there exists $s\leq t$ such that $|X_s|>1$, then by continuity there exists $s'<s\leq t$ such that $|X_{s'}|>1$ and hence $\alpha \leq s' <t$. So, if $\alpha \geq t$ then $|X_s|\leq 1$ for all $s\leq t$. Is there a mistake in this reasoning? – SBF Jul 24 '14 at 16:25
  • With continuity, your reasoning is allright. Without continuity, no. – Did Jul 24 '14 at 16:36
  • @Did: I see, we have continuity in the OP, but I'd have to modify $$ {\alpha>t} = \bigcup_n{\alpha \geq t+\frac1n} = \bigcup_{n}\bigcap_{s\in [0,t+\frac1n]}{|X_s|\leq 1} $$ and there appear events ${|X_s|\leq 1}$ for $s>t$. – SBF Jul 24 '14 at 16:43
  • Am I confused or someone just upvoted this answer? – Did Jul 24 '14 at 17:58
  • It is me who upvoted. The explantion of $\alpha$ not being stopping time by Ilya seems logical to me. – Ron Jul 24 '14 at 18:05
  • @Did I'm not sure what was the purpose of your last question – SBF Jul 24 '14 at 18:25
  • I thought the OP already upvoted when they accepted this answer. Note that the explanation added at the beginning of the post points the flaw in the alleged proof that $\alpha$ is a stopping time, not why $\alpha$ is not a stopping time. – Did Jul 24 '14 at 20:29