# First jump time of Poisson process (and general right-continuous processes).

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

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

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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