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I've read that the first jump time of the Poisson process is totally inaccessible (definition at the bottom for anyone interested). This made me wonder if the first jump time is a stopping time. I think the answer is yes. (If the answer was no, the first jump time would automatically be totally inaccessible and people probably wouldn't bother mentioning totally inaccessibility.) More generally, I think the answer is yes for any right-continuous process.

EDIT: Lost1 has pointed out that that the answer is easily yes for a Poisson process. That still leaves general right-continuous processes.

Let $(\mathcal{F}_t)_{t \geq 0}$ be a filtration of a probability space. A random variable $T$ is called an stopping time if $\left\{T \leq t \right\} \in \mathcal{F}_t$ for all $t \geq 0$. For us, $\mathcal{F}_t = \sigma(X_{s} | s \leq t)$.

Here's my attempt to show the first jump time for a right-continuous process is a stopping time. The first jump time is defined to be $T = \inf\{t>0 | X_t \neq X_0 \}$. We need $\{ T \leq t \} \in \mathcal{F}_{t}$. I don't know how to prove this, but I do know that $$ \{ T < t \} = \{ X_s \neq X_0 \text{ for some $s < t$} \} = \bigcup_{r < t, \, r \in \mathbb{Q}} \{ X_r \neq X_0 \} \in \mathcal{F}_{t}, $$ where the last equality is by right-continuity. It's not clear to me how to write $\{ T \leq t \}$ in a way that shows it belongs to $\mathcal{F}_{t}$.

By the way, here is the definition of a totally inaccessible stopping time. If $T_n$ are stopping times and $T_n \uparrow T$, then $T$ is an stopping time and we call it predictable. If $P(S = T < \infty) = 0$ for all predictable stopping times $S$, we say $T$ is totally inaccessible.

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Am I insane or isn't $\{T\leq t \} = \{ X_t>X_0 \}$ which looks like a totally measurable with respect to $\mathcal{F}_t$? –  Lost1 Apr 12 '13 at 11:22
    
You only need this intersection stuff if you have a process which actually can jump up then down like a brownian motion. Poisson process only jump upward... –  Lost1 Apr 12 '13 at 11:23
    
@Lost1 I'd like to consider processes that jump down too. You are right that $\{ T \leq t\} = \{ X_t > X_0 \}$ for a Poisson process. –  Richard Polt Apr 12 '13 at 12:11
    
then at t, it is either $X_t\neq X_0$ or it has jumped up at least once and jumped down at least once already. Well the latter can only occur if a jump happened before t because 2 jumps cannot happen at once... If it jumped up once strictly before t, then it is just the expression you wrote down already –  Lost1 Apr 12 '13 at 12:18
    
Poisson processes only jump upward, maybe you should rephrase your question... –  Lost1 Apr 12 '13 at 12:18

2 Answers 2

First, I cannot understand what you are trying to prove? Are you trying to prove $T$ is a stopping time? Why is the notion of totally inaccessible here?

Poisson first jump times are not predictable. If $H_t$ is predictable, then it $H_t\in\mathcal{F}_{t-}$. This is clearly not the case for jump processes.

The intuition is that, if a stopping time is predictable, you will know that you are getting close to the stopping time before it occurs (for example for a Brownian motion), whereas Poisson jumps do not happen this way. If it hasn't happened for 5 minutes, it is not any more likely to happen than 5 minutes ago

Also why are you trying to show $\{T<t\}\in\mathcal{F_t}$ when you have given $\{T\leq t\}\in\mathcal{F_t}$ as your definiton as a stopping time. The former is not equivalent to $T$ being a stopping time, without extra knowledge about the filtration etc.

Am I insane or isn't $\{T\leq t \} = \{ X_t>X_0 \}$ which looks like a totally measurable with respect to $\mathcal{F}_t$?

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You are not insane (that is, according to the very specific criterion you introduced...). –  Did Apr 12 '13 at 13:30

Here's some comments which hopefully will yield the answers to your questions.

First, as regards measurability of the first jump time, I don't know about any one-line proofs, but at least there is a relatively short elementary proof in the paper "An elementary proof that the first hitting time of an open set by a jump process is a stopping time" (on ArXiV, 2012). In here, it is shown that for any càdlàg process $X$, right-continuous filtration $(\mathcal{F})_{t\ge0}$ and open set $U$, the variable

$T = \inf\{t\ge0\mid \Delta X_t \in U\}$

is a stopping time. As $\mathbb{R}\setminus\{0\}$ is open, the result follows. As was pointed out in the comments, the case of a Poisson process is simpler because of the structure of the sample paths.

Second, as regards total inaccessibility of the first jump time of a Poisson process: This follows from a general theorem by Meyer, stating that all Feller processes only jump at totally inaccessible times. The statement of the theorem can be found in the books by Protter or Rogers & Williams, though no proofs are given. To give a proof for the case of a Poisson process, we first prove a lemma.

$\mathbf{Lemma.}$ Let $M$ be a càdlàg local martingale. Assume that $M$ only has nonnegative jumps. Then $\Delta M_T=0$ almost surely for all predictable stopping times $T$.

$\mathbf{Proof.}$ Consider the case where $M$ is uniformly integrable (the general case follows by localization). Consider a predictable stopping time $T$. Then $E(\Delta M_T\mid\mathcal{F}_{T-})=0$. As $\Delta M\ge0$ by assumption, this implies $\Delta M_T=0$ almost surely. I don't have a precise reference for the claim about the conditional expectation, but it can be found as part of the PFA theorem in chapter six of the book by Rogers & Williams, or otherwise proven by taking an announcing sequence and applying the discrete-time martingale convergence theorem.

Now consider a Poisson process $N$. Let $M_t = N_t - t$. Then $M$ is a local martingale with nonnegative jumps. Let $T$ be the first jump time of $N$, then $\Delta M_T = \Delta N_T = 1$. Let $T = T_F \land T_{F^c}$ be the decomposition of $T$ into its predictable and totally inaccessible parts $T_F$ and $T_{F^c}$. By the lemma, we must have $T_F=\infty$ almost surely on $F$, as otherwise $\Delta N_T=0$ by the lemma. Thus, $T=T_{F^c}$ almost surely, so $T$ is totally inaccessible.

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