Lévy processのジャンプはいつまでも起き続ける

The following might be nontrivial so I will write down a proof.

Claim
Assume that the filtration $(\mathcal{F}_t)_{t\geq 0}$ satisfies the usual conditions. Let $(X_t)_{t\geq 0}$ be a Lévy process (with càdlàg sample paths) adapted to $(\mathcal{F}_t)$ and $\Lambda$ be a Borel set in $\mathbb{R}$ such that $0\notin \bar{\Lambda}$ . Define the random times
\begin{align*}
T^0_\Lambda &= 0, \\
T^{n}_\Lambda &= \inf\{t>T^{n-1}_\Lambda;\Delta X_t\in\Lambda\}, \quad n=1,2,\dots
\end{align*}
, where $\Delta X_t := X_t-X_{t-}$ . Then the followings hold:
(i) $T^n_\Lambda>T^{n-1}_\Lambda$ a.s. for $n=1,2,\dots$
(ii) Every $T^n_\Lambda$ is an $(\mathcal{F}_t)$ -stopping time.
(iii) $\lim_{n\to\infty} T^n_\Lambda=\infty$ a.s.

Proof.
(i) Since $0\notin \bar{\Lambda}$ , $d:=\inf\{|y|;y\in\Lambda\}>0$ holds. Then $\Delta X_t\in\Lambda$ implies $|\Delta X_t|\geq d$ .
By right continuity of $X$ , there is a $\delta_1>0$ such that
\begin{align*}
\left|X_t-X_{T^{n-1}_\Lambda}\right| \leq \frac{d}{6}
\end{align*}
for all $T^{n-1}_\Lambda < t < T^{n-1}_\Lambda+\delta_1$ .
For any $T^{n-1}_\Lambda < t < T^{n-1}_\Lambda+\delta_1$ , taking a sufficiently small $\delta_2 > 0$ satisfying $T^{n-1}_\Lambda < t-\delta_2 < T^{n-1}_\Lambda + \delta_1$ yields, by càdlàgness of $X$ ,
\begin{align*}
\left| X_{t-{\delta_2}}-X_{T^{n-1}_\Lambda} \right|,|X_{t-{\delta_2}}-X_{t-}| \leq \frac{d}{6}.
\end{align*}
Thus, it holds that
\begin{align*}
\left|\Delta X_t\right|
&\leq |X_t-X_{T^{n-1}_\Lambda}| + |X_{t-{\delta_2}}-X_{T^{n-1}_\Lambda}| + |X_{t-{\delta_2}}-X_{t-}| \\
&\leq \frac{d}{6}+\frac{d}{6}+\frac{d}{6} = \frac{d}{2} < d.
\end{align*}
Then we obtain $T^n_\Lambda\geq T^{n-1}_\Lambda+\delta_1>T^{n-1}_\Lambda$ .

(ii) We use the result from Karatzas-Shreve; Proposition 2.26.
It holds for $t\geq 0$ , since $(\Delta X_t)_{t\geq 0}$ is $(\mathcal{F}_t)$ -adapted,
\begin{align*}
\{ T^n_{\Lambda} \leq t \}
&= \bigcup_{k_1,\dots,k_n\in\mathbb{N}} \bigcap_{j=1}^n \left( \{T_{k_j}\leq t\} \cap \{ \Delta X_{T_{k_j}}\in\Lambda \} \right) \\
&\in \mathcal{F}_t,
\end{align*}
where, by the proposition above, $(T_k)_{k\geq 1}$ is the sequence of stopping times exhausting the jumps of $X$ . Thus we see all $T^n_\Lambda$ are stopping times.

(iii) Assume that there is a $K>0$ such that $T^n_\Lambda \leq K$ for all $n\geq 1$ . Then, $(T^n_\Lambda)$ is increasing and bounded. Thus, $T:=\lim_{n\to\infty}T^n_\Lambda\leq K$ exists.
By càdlàgness of $X$ , it follows $\lim_{n\to\infty}X_{T^n_\Lambda}=X_{T-}$ . Since $t\mapsto X_{t-}$ is left continuous, we get $\lim_{t\to\infty}X_{T^n_\Lambda-}=X_{T-}$ .
Hence, $|\Delta X_{T^n_\Lambda}|=|X_{T^n_\Lambda}-X_{T^n_\Lambda-}|\to 0$ as $n\to\infty$ . However, $\lim_{n\to\infty}|\Delta X_{T^n_\Lambda}|\geq d$ . Those are contradiction. □

やがてめぐになる

さんすうとえいごのおべんきょうブログ。ほかにもかきたいことをかきます。

Lévy processのジャンプはいつまでも起き続ける