"The first time a continuous local martingale grows in absolute value beyond $n$" is a localizing sequence

Question

How can it be shown that, for a continuous local martingale $X$ defined w.r.t. the filtered probability space $(\Omega, \mathcal{A}, P; \mathcal{F})$, the stopping times $\tau_n := \inf \{t \geq 0 \mid: |X_t| \geq n\}$ ($n \in \{0, 1, 2, \dots\}$) are localising, i.e. for every $n$, the stopped process $X^{\tau_n}$ is a martingale?

Background: This claim is stated without proof in a textbook on stochastic processes ("Wahrscheinlichkeitstheorie" by Jochen Wengenroth, de Gruyter 2008, in the definition of Itô integral on p. 179 and again in the proof of theorem 9.5.4 on p. 180).

Attempt at a solution: Letting $\sigma_n$ be a localizing sequence for $X$, $\sigma_n\wedge\tau_n$ is localising for $X$.

Answer 1

Take $\tau_n$ as in your definition and let's look at the process $X^{\tau_n}$.

For all $n$ this is not only a local martingale (easy to this point) but a bounded one. The fact is that a bounded local martingale is a martingale follows from the fact that a local martingale is a martingale if and only it is of class (DL) (see theorem 1).

Being a local martingale of class (DL) means that for any family of stopping times $\mathcal{T}$ bounded by a positive real number say $a$ we have that $(X^{\tau_n}_\rho)_{\rho\in\mathcal{T}}$ is uniformly integrable.

Now let's take such a family $\mathcal{T}$ and observe that $X^{\tau_n}_\rho$ is uniformly (i.e. for any $\rho$) bounded by $n$, as a uniformly bounded family of random variables is uniformly integrable (*) we are done proving that $X^{\tau_n}$ is of class (DL).

We have shown that the stopped processes $X^{\tau_n}$ form a sequence of martingales and as $\tau_n$ increasingly tends to $ \infty$ almost surely, $\tau_n$ is an announcing sequence for $X$ which the claim we were looking for.

(*)a quick proof : see the definition of U.I. on the planetmath webpage, we have for any $\epsilon>0$ and $\rho\in \mathcal{T}$ :

$E[X^{\tau_n}_\rho.1[X_\rho>n]]=0<\epsilon$ meaning exactly that the family $(X_\rho)_{\rho\in \mathcal{T}}$ is U.I.

Answer 2

The following answer is based entirely on ideas suggested by TheBridge in his answer. The result is not proved in all the generality stipulated in the original post, but the restrictions are sufficient for my purposes.

Lemma 1 A bounded local martingale is a martingale.

Proof Let $X = (X_t)_{t \in [0,\infty)}$ be a bounded local martingale w.r.t. the filtered probability space $(\Omega, \mathcal{A}, P; \mathcal{F})$. Let $\sigma_1, \sigma_2, \dots$ be a sequence of real-valued $\mathcal{F}$-stopping times that weakly increases to infinity, such that for each $n \in \{1, 2, \dots\}$ the stopped process $X^{\sigma_n}$ is a martingale w.r.t. said probability space. Then for every $t \in [0, \infty)$ $X_t = \lim_{n \rightarrow \infty} X^{\sigma_n}_t$ and $(X_t^{\sigma_n})_{n \in \{1, 2, \dots\}}$ are uniformly bounded. Hence, by the conditional dominated convergence theorem, given $s, t \in [0, \infty)$ with $s < t$, $$ E(X_t | \mathcal{F}_s) = E(\lim_{n \rightarrow \infty} X^{\sigma_n}_t | \mathcal{F}_s) = \lim_{n \rightarrow \infty} E(X^{\sigma_n}_t | \mathcal{F}_s) = \lim_{n \rightarrow \infty} X^{\sigma_n}_s = X_s $$ Q.E.D.

Lemma 2 If $X$ is a continuous local martingale w.r.t. the filtered probability space $(\Omega, \mathcal{A}, P; \mathcal{F})$, and if $\sigma$ is a real valued $\mathcal{F}$-stopping time, then $X^\sigma$ is a local martingale w.r.t. said probability space.

Proof Let $\rho_1, \rho_2, \dots$ be a sequence of real-valued $\mathcal{F}$-stopping times that weakly increases to infinity and such that for each $n \in \{1, 2, \dots\}$ $X^{\rho_n}$ is a martingale w.r.t. the given probability space. Then, by the optional stopping theorem, $(X^{\rho_n})^\sigma$ is a martingale, for every $n \in \{1, 2, \dots\}$. But $(X^{\rho_n})^\sigma = X^{\rho_n \wedge \sigma} = (X^\sigma)^{\rho_n}$, and therefore $\rho_n$ is a localizing sequence for $X^\sigma$, Q.E.D.

Theorem Letting $X$ and $\tau_1, \tau_2, \dots$ be as in the original question, and assuming, furthermore, that $X_0=0\mathbb{1}_\Omega$ and that the $\tau_n$'s are real-valued, then $(\tau_n)_{n \in \{1, 2, \dots\}}$ is a localizing sequence for $X$, i.e. it is weakly increasing to infinity and for every $n \in \{1, 2, \dots\}$ the stopped process $X^{\tau_n}$ is a martingale w.r.t. the given probability space.

Proof The fact that the sequence $(\tau_n)_{n \in \{1, 2, \dots\}}$ weakly increases to infinity follows readily from its definition. The fact that this is a sequence of $\mathcal{F}$-stopping times results from the fact that $X$ is continuous, $I := (-\infty, n] \cup [n, \infty)$ is closed, and $\{|X_t| \geq n\} = \{X_t \in I\}$ (cf. the third example for a stopping time here, in German). Now let $n \in \{1, 2, \dots\}$. Then, by lemma 2, $X^{\tau_n}$ is a local martingale. But, by $\tau_n$'s definition, $X^{\tau_n}$ is bounded (specifically, since $X$ is continuous, $|X^{\tau_n}_t| \leq n$ for every $t \in [0, \infty)$). Hence, by lemma 1, $X^{\tau_n}$ is a martingale, Q.E.D.