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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?

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2 Answers 2

up vote 8 down vote accepted

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

Well, not very easily I am afraid, since $\alpha$ is not always a stopping time for the filtration $(\mathcal F_t)_{t\geqslant0}$ defined by $\mathcal F_t=\sigma(X_s;0\leqslant s\leqslant t)$.

For a counterexample, consider some Bernoulli random variable $U$ such that $P(U=1)=\frac12$ and $P(U=0)=\frac12$ and define the process $(X_t)_{t\geqslant0}$ as follows:

  • If $U=1$ then $X_t=t$ for every $t\geqslant0$.
  • If $U=0$ then $X_t=t$ for every $0\leqslant t\lt1$ and $X_t=2-t$ for every $t\geqslant1$.

Then $\alpha=\inf\{t;|X_t|\gt1\}$ is $\alpha=1$ if $U=1$ and $\alpha=3$ if $U=0$ hence $\{\alpha\leqslant1\}=\{U=1\}$. For every $s\leqslant1$, the random variable $X_s=s$ is deterministic, hence $\mathcal F_1=\{\varnothing,\Omega\}$ is the trivial sigma-algebra. The event $\{U=1\}$ is not in $\mathcal F_1$ hence $\alpha$ is not a stopping time for the filtration $(\mathcal F_t)_{t\geqslant0}$.

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Thanks Did, indeed $\alpha$ is not stopping time in this example. –  Ron Jul 24 at 17:58
    
@Ilya, thanks I forgot this. –  Ron Jul 24 at 18:24
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(+1). It might be worth mentioning that $\alpha$ is a stopping time with respect to the filtration $(\mathcal{F}_{t+})_{t \geq 0}$. –  saz Aug 22 at 20:11
    
@saz Quite so. Thanks. –  Did Aug 22 at 23:13

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.

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Thanks a lot for the proof. –  Ron Apr 5 '13 at 11:54
    
@Shaik: you're welcome –  Ilya Apr 5 '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 at 15:58
    
@Did: you certainly have a fan :) I'll read your answer now. –  Ilya Jul 24 at 16:04

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