Randomly restarted Lévy process is again a Lévy process

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space;
• $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A,\operatorname P)$;
• $E$ be a $\mathbb R$-Banach space;
• $(L_t)_{t\ge0}$ be a $E$-valued process on $(\Omega,\mathcal A,\operatorname P)$.

Remember that $L$ is called $\mathcal F$-Lévy if

1. $L$ is $\mathcal F$-adapted;
2. $L_0=0$;
3. $L_{s+t}-L_s$ and $\mathcal F_s$ are independent for all $t\ge s\ge0$;
4. $L_{s+t}-L_s\sim L_s$ for all $t\ge s\ge0$.

Assume $L$ is $\mathcal F$-Lévy. Let $\tau$ be a $\mathcal F$-stopping time, $\tilde\Omega:=\{\tau<\infty\}$, $\tilde{\mathcal A}:=\left.\mathcal A\right|_{\tilde\Omega}$, $\tilde{\operatorname P}:=\left.\operatorname P\right|_{\tilde\Omega}$, $$\mathcal G_t:=\mathcal F_{\tau+t}\;\;\;\text{for }t\ge0$$ and $$X_t(\omega):=L_{\tau+t}(\omega)-L_\tau(\omega)\;\;\;\text{for }(\omega,t)\in\tilde\Omega\times[0,\infty).$$

How can we show that $X$ is a $\mathcal G$-Lévy process on $(\tilde\Omega,\tilde{\mathcal A},\tilde{\operatorname P})$?

(1.) and (2.) are clearly trivial.

I think the easiest way to show (3.) and (4.) is to approximate $\tau$ in a suitable way. So, let's first assume that $\tau$ is finite and $k:=\left|\tau(\Omega)\right|\in\mathbb N$. Then, $$\tau(\Omega)=\{t_1,\ldots,t_k\}\tag1$$ for some $0\le t_1<\cdots<t_k$. Since $\{\tau=t_i\}\in\mathcal F_{t_i}\subseteq\mathcal F_{t_i+s}$, we easily obtain \begin{equation}\begin{split}\operatorname P\left[X_{s+t}-X_s\in B\right]&=\sum_{i=1}^k\operatorname P\left[\tau=t_i,L_{t_i+s+t}-L_{t_i+s}\in B\right]\\&\sum_{i=1}^k\operatorname P\left[\tau=t_i\right]\operatorname P\left[L_{t_i+s+t}-L_{t_i+s}\in B\right]\\&\operatorname P\left[L_t\in B\right]\sum_{i=1}^k\operatorname P\left[\tau=t_i\right]=\operatorname P\left[L_t\in B\right]\end{split}\tag2\end{equation} for all $B\in\mathcal B(E)$ and $s,t\ge0$; which is (4.).

Analogously¹, since $L_{t_i+s+t}-L_{t_i+s}$ and $\mathcal F_{t_i+s}$ are independent and $\{\tau=t_i\}\in\mathcal F_{t_i}\subseteq\mathcal F_{t_i+s}$ for all $i\in\{1,\ldots,k\}$, \begin{equation}\begin{split}\operatorname P\left[X_{s+t}-X_s\in B\mid\mathcal G_s\right]&=\sum_{i=1}^k\operatorname P\left[\tau=t_i,L_{t_i+s+t}-L_{t_i+s}\in B\mid\mathcal F_{\tau+s}\right]\\&=\sum_{i=1}^k1_{\left\{\:\tau\:=\:t_i\:\right\}}\operatorname P\left[L_{t_i+s+t}-L_{t_i+s}\in B\mid\mathcal F_{t_i+s}\right]\\&=\sum_{i=1}^k1_{\left\{\:\tau\:=\:t_i\:\right\}}\operatorname P\left[L_{t_i+s+t}-L_{t_i+s}\in B\right]\\&=\sum_{i=1}^k\operatorname P\left[\tau=t_i,L_{t_i+s+t}-L_{t_i+s}\in B\right]\\&=\operatorname P\left[X_{s+t}-X_s\in B\right]\end{split}\tag3\end{equation} almost surely for all $B\in\mathcal B(E)$; which is (3.).

Can we derive the general case by approximating $\tau$ with $\mathcal F$-stopping times of the formerly considered form?

EDIT 1: Both, $(2)$ and $(3)$, should hold line by line when $\tau$ is finite and $\tau(\Omega)$ is countable. We simply need to use the sums $\sum_{r\in\tau(\Omega)}\operatorname P\left[\tau=r,L_{r+s+t}-L_{r+s}\in B\right]$ and $\sum_{r\in\tau(\Omega)}\operatorname P\left[\tau=r,L_{r+s+t}-L_{r+s}\in B\mid\mathcal F_{\tau+s}\right]$ instead.

EDIT 2: Now assume $\tau$ is only finite. Let $\tau_n$ be a $\mathcal F$-stopping time² on $(\Omega,\mathcal A,\operatorname P)$ for $n\in\mathbb N$, such that $\tau_n(\Omega)$ is countable and $$\tau_n\ge\tau_{n+1}\tag4$$ for all $n\in\mathbb N$ and $$\tau_n\xrightarrow{n\to\infty}\tau\tag5.$$

Let $X^{(n)}_t:=L_{\tau_n+t}-L_{\tau}$ and $\mathcal G^{(n)}_t:=\mathcal F_{\tau_n+t}$ for $t\ge0$. By $(4)$, $$\mathcal G^{(n)}_t\supseteq\mathcal G^{(n+1)}_t\;\;\;\text{for all }t\ge0\tag6$$ for all $n\in\mathbb N$. Now assume $L$ is right-continuous. Then, by $(4)$ and $(5)$, $$X^{(n)}_t\xrightarrow{n\to\infty}X_t\;\;\;\text{for all }t\ge0\tag7.$$

Let $B\in\mathcal B(E)$ and $s,t\ge0$. By what we've already shown, $$\operatorname P\left[X^{(n)}_{s+t}-X^{(n)}_s\in B\mid\mathcal G^{(n)}_s\right]=\operatorname P\left[X^{(n)}_{s+t}-X^{(n)}_s\in B\right]\tag8$$

Using $(7)$ and the dominated convergence theorem, the right-hand side of $(8)$ should converge to $\operatorname P\left[X_{s+t}-X_s\in B\right]$.

What can we do with the left-hand side? Maybe $(6)$ is the crucial ingredient which allows us to obtain convergence to $\operatorname P\left[X_{s+t}-X_s\in B\mid\mathcal G_s\right]$ as desired ...

Now, in order to conclude for general finite $\tau$, I guess we need to assume right-continuity, but how do we then need to argue exactly?

Remark: I'm also unsure whether we need to impose further assumptions on $(\mathcal F_t)_{t\ge0}$ like completeness or right-continuity.

---

¹ If $Y\in\mathcal L^1(\operatorname P;E)$, then $$\operatorname E\left[1_{\left\{\:\tau\:=\:t\:\right\}}Y\mid\mathcal F_\tau\right]=1_{\left\{\:\tau\:=\:t\:\right\}}\operatorname E\left[X\mid\mathcal F_t\right]\;\;\;\text{almost surely}.$$

² We could, for example, take

Answer 1

First of all, note that, say, $Y_n \to Y$ almost surely does not imply $\mathbb{P}(Y_n \in B) \to \mathbb{P}(Y \in B)$. The latter holds only if $B$ is such that $\mathbb{P}(Y \in \partial B)=0$. You can easily see this issue if you consider for instance $Y_n := \frac{1}{n}$ and $B=\{0\}$. As a consequence, we cannot take, in general, the limit on the right-hand side of (8).

---

Let $f$ be a bounded continuous function. Since $X_t^{(n)} \to X_t$, we have, by the dominated convergence theorem,

$$\mathbb{E}(f(X_{s+t}-X_s) \mid \mathcal{G}_s) = \lim_{n \to \infty} \mathbb{E}(f(X_{s+t}^{(n)}-X_s^{(n)}) \mid \mathcal{G}_s). \tag{1}$$

From $\tau_n \geq \tau$ we see that $\mathcal{G}_s^{(n)} = \mathcal{F}_{\tau_n+s} \supseteq \mathcal{F}_{\tau+s}=\mathcal{G}_s$. Consequently, by the tower property of conditional expectation,

$$ \mathbb{E}(f(X_{s+t}^{(n)}-X_s^{(n)}) \mid \mathcal{G}_s) = \mathbb{E} \bigg [ \mathbb{E}(f(X_{s+t}^{(n)}-X_s^{(n)}) \mid \mathcal{G}_s^{(n)}) \mid \mathcal{G}_s \bigg].$$

By your earlier considerations for stopping times taking only finitely many values, we can compute the right-hand side:

\begin{align*} \mathbb{E}(f(X_{s+t}^{(n)}-X_s^{(n)}) \mid \mathcal{G}_s) &= \mathbb{E} \bigg[ \mathbb{E}(f(X_{s+t}^{(n)}-X_s^{(n)})) \mid \mathcal{G}_s\bigg] \\ &= \mathbb{E}(f(X_{s+t}^{(n)}-X_s^{(n)})). \end{align*}

Plugging this into $(1)$ and using once more the dominated convergence theorem and the right-continuity of the sample paths, we arrive at

$$\mathbb{E}(f(X_{s+t}-X_s) \mid \mathcal{G}_s) = \mathbb{E}(f(X_{t+s}-X_s)),$$

which should be all you need. (At least for the case $\mathbb{P}(\tau<\infty)=1$, which you were considering.)

---

Proving the assertion for stopping times which may take the value $+\infty$ requires some more work. Define $\tau_n := \min\{\tau \wedge n\}$ and denote by $X^{(n)}$ the corresponding restarted Lévy process with filtration $\mathcal{G}^{(n)}$. Then, by the previous step of the proof,

$$\mathbb{E}(f(X_{t+s}^{(n)}-f(X_s^{(n)}) \mid \mathcal{G}_s^{(n)}) = \mathbb{E}(f(X_{t+s}^{(n)}-X_s^{(n)}))= \mathbb{E}(f(L_t)).$$

Since $\{\tau \leq n\} \in \mathcal{F}_{\tau} \cap \mathcal{F}_n = \mathcal{F}_{\tau \wedge n} \subseteq \mathcal{G}_s^{(n)}$, we can multiply both sides by $1_{\{\tau \leq n\}}$ to obtain that

$$\mathbb{E}(1_{\{\tau \leq n\}} f(X_{t+s}^{(n)}-f(X_s^{(n)}) \mid \mathcal{G}_s^{(n)})=1_{\{\tau \leq n\}} \mathbb{E}(f(L_t)). \tag{2}$$

We would like to let $n \to \infty$. To this end, we first show that

$$\mathbb{E}(1_{\{\tau \leq n\}} f(X_{t+s}^{(n)}-f(X_s^{(n)}) \mid \mathcal{G}_s^{(n)}) \xrightarrow[]{L^1} \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s). \tag{3}$$

By the triangle inequality,

\begin{align*} &\mathbb{E}\bigg|\mathbb{E}(1_{\{\tau \leq n\}} f(X_{t+s}^{(n)}-f(X_s^{(n)}) \mid \mathcal{G}_s^{(n)})- \mathbb{E}(1_{\{\tau <\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s)\bigg| \\&\leq \mathbb{E}\bigg|\mathbb{E}(1_{\{\tau \leq n\}} f(X_{t+s}^{(n)}-f(X_s^{(n)}) \mid \mathcal{G}_s^{(n)})- \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s^{(n)})\bigg|\\ &\quad +\mathbb{E}\bigg| \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s^{(n)})- \mathbb{E}(1_{\{\tau<\infty\}}f(X_{t+s}-X_s) \mid \mathcal{G}_s)\bigg| \\ &=: \Delta_1+\Delta_2. \end{align*}

For the first term we see, using the tower property,

$$\Delta_1 \leq \mathbb{E}(|1_{\{\tau \leq n\}} f(X_{t+s}^{(n)}-X_s^{(n)})-1_{\{\tau<\infty\}} f(X_{t+s}-X_s)|).$$

If $\omega \in \{\tau<\infty\}$, then $X_{r}^{(n)}(\omega)=X_r(\omega)$ for $n=n(\omega)$ sufficiently large, and so the dominated convergence theorem yields $I_1 \to 0$ as $n \to \infty$. On the other hand, $\mathcal{G}_s = \sigma(\bigcup_n \mathcal{G}_s^{(n)})$, see this question, and so Lévy's upwards theorem yields $I_2 \to 0$ as $n \to \infty$. This then proves $(3)$. Because of $(3)$, we can choose an almost surely convergent subsequence

$$\mathbb{E}(1_{\{\tau \leq n_k\}} f(X_{t+s}^{(n_k)}-f(X_s^{(n_k)}) \mid \mathcal{G}_s^{(n_k)}) \xrightarrow[]{\text{a.s.}} \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s). \tag{4}$$

Letting $n \to \infty$ in (2) now gives

$$ \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s) = 1_{\{\tau<\infty\}} \mathbb{E}(f(L_t)). \tag{5}$$

Taking expectation on both sides yields

$$ \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s)) = \mathbb{P}(\tau<\infty) \mathbb{E}(f(L_t)),$$

i.e.

$$\mathbb{E}(f(L_t)) = \frac{1}{\mathbb{P}(\tau<\infty)} \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s)).$$

Plugging this into $(5)$ shows that

$$ \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s) \mid \mathcal{G}_s) = 1_{\{\tau<\infty\}} \frac{1}{\mathbb{P}(\tau<\infty)} \mathbb{E}(1_{\{\tau<\infty\}} f(X_{t+s}-X_s)).$$

If we define a probability measure $\tilde{P}(A) := \frac{\mathbb{P}(A \cap \{\tau<\infty\})}{\mathbb{P}(\tau<\infty)}$ on $\tilde{\Omega} := \{\tau<\infty\}$, then this is equivalent to

$$\mathbb{E}_{\tilde{\mathbb{P}}}(f(X_{t+s}-X_s) \mid \mathcal{G}_s) = \mathbb{E}_{\tilde{\mathbb{P}}}(f(X_{t+s}-X_s)).$$