I don't understand the the last sentence of a proof of the Markov property for Lévy processes given in Jochen Wengenroth's textbook "Wahrscheinlichkeitstheorie" (de Gruyter, 2008). I will appreciate the community's help.
In what follows I will state the theorem and outline the proof. I will then describe the two points that I do not understand. I will then describe my attempts to tackle these difficulties.
I
The following is taken from p. 142 (the first sentence) and from p. 144 (the rest) in Wengenroth's book (translated from German).
Given a filtration $\mathcal{F} = (\mathcal{F}_t)_{t \geq 0}$, the right-continuous filtration $\mathcal{F}^+$ is defined as follows: $\mathcal{F}^+_t \equiv \bigcap_{s > t} \mathcal{F}_s$.
A right-continuous stochastic process $(X_t)_{t \geq 0}$ that is adapted to the filtration $\mathfrak{F}$ and such that, for every $t \geq 0$, $X_t$ takes values in a separable, normed space, is called an $\mathcal{F}$-Lévy process.
Theorem 7.14 (Strong Lévy property) Let $X = (X_t)_{t \geq 0}$ be a Lévy process adapted to the filtration $\mathcal{F}$ and let $\tau$ be a real-valued $\mathcal{F}$ stopping time. Then the following process $Y_t \equiv X_{\tau + t} - X_\tau$ is independent of $\mathcal{F}^+_\tau = (\mathcal{F}^+)_\tau$ and satisfies $Y \overset{d}{=} X$.
The following is an outline of the proof of the theorem, adapted from pp. 144-145 in Wengenroth's book. The full proof (in German) can be found at this link.
Step 1 For every $n \in \{0, 1, 2, \dots\}$ we define the stopping time $\tau_n$ as follows $\tau_n \equiv k/2^n$ whenever $\tau \in [(k-1)/2^n, k/2^n)$. It is shown that the processes $Y^n$ defined by $Y^n_t \equiv X_{\tau_n + t} - X_{\tau_n}$ are independent of $\mathcal{F}_{\tau_n}$, respectively, and satisfy $Y^n \overset{d}{=} X$.
Step 2 Since $\tau \leq \tau_n \rightarrow \tau$, we have that $Y^n_t \rightarrow Y_t$ (pointwise on $\Omega$), whence $Y \overset{d}{=} Y^n \overset{d}{=} X$, and therefore, since $\mathcal{F}^+_\tau \subseteq \mathcal{F}_{\tau_n}$, the independence of $Y$ and $\mathcal{F}^+_\tau$ follows.
QED
II
The two points that I don't understand in the proof appear in step 2.
- Why is it the case that $Y \overset{d}{=} Y^n$?
- Why are $Y$ and $\mathcal{F}^+_\tau$ independent?
III
Following are my (unsuccessful) attempts at tackling the two points that I don't understand.
If we think of $Y^n$ and $Y$ as measurable functions from $\Omega$ to the set of right-continuous functions, then we may use the well-known fact that almost-sure convergence entails convergence in distribution. However, this does not explain why Wengenroth writes $Y \overset{d}{=} Y^n$ rather than $Y^n \overset{d}{\rightarrow} Y$. Another caveat is that the only metric I am familiar with on the set of right-continuous functions (actually on the set of continuous functions) is the metric of uniform convergence, or some variation thereof, whereas the convergence of $Y^n$ to $Y$ is only pointwise.
Since $\sigma(Y^n) \subseteq \bigvee Y^n = \sigma(\bigcup \sigma(Y^n))$, $Y$ is $\bigvee Y^n$-measurable, so it suffices to show that $\bigvee Y^n$ and $\mathcal{F}^+_\tau$ are independent. If $\bigcup \sigma(Y^n)$ was a $\pi$-system, this would follow from that fact that $\sigma(Y^n)$ and $\mathcal{F}^+_\tau$ are independent (see, for instance, lemma 3.6 on p. 50 in Kallenberg's "Foundations of Modern Probability", 2nd edition, Springer 2002). However, is $\bigcup \sigma(Y^n)$ a $\pi$-system?
In what follows I will prove that $Y \overset{d}{=} X$ and that $Y$ and $\mathcal{F}^+_\tau$ are independent, using only theorems and definitions presented in Wengenroth's book.
I
In the following section I will prove that $Y \overset{d}{=} X$ (see theorem 8 below).
Definition: $C_b(\mathcal{X})$, the set of continuous, bounded functions [p. 83] Given a metric space $(\mathcal{X}, d)$ with the induced Borel $\sigma$-algebra $\mathcal{B}$, we denote with $C_b(\mathcal{X})$ the set consisting of all continuous, bounded functions $f: \mathcal{X} \rightarrow \mathbb{R}$ ($\mathbb{R}$ taken to be equipped with the standard Euclidean metric).
Definition: $Q_m \overset{w}{\rightarrow} Q$, Weak convergence [p. 83] A sequence $(Q_m)_{m \in \mathbb{N}_1}$ of distributions on the measurable metric space $(\mathcal{X}, \mathcal{B})$ (cf. definition 1 above) is said to be weakly convergent to a distribution $Q$ over $(\mathcal{X}, \mathcal{B})$, or in symbols $Q_m \overset{w}{\rightarrow} Q$, iff $$ \int f\ dQ_m \rightarrow \int f\ dQ $$ for every $f \in C_b(\mathcal{X})$.
Lemma: Uniqueness of limits-in-distribution Let $(\mathcal{X}, \mathcal{B})$ be a measurable, metric space (cf. definition 1), let $(Q_m)_{m \in \mathbb{N}_1}$ be a sequence of distributions on this space and let $P$ and $Q$ be distributions over this space.
(a) If $Q_m \overset{w}{\rightarrow} P$ and $Q_m \overset{w}{\rightarrow} Q$, then $P = Q$.
(b) Let $Q_m = Q$ for every $m \in \mathbb{N}_1$. Then $Q_m \overset{w}{\rightarrow} P$ iff $P = Q$.
Proof
(a) This result is shown on p. 84 using the Portmanteau theorem.
(b) The "only if" direction follows from the definition of weak convergence. The "if" direction follows from the "only if" direction together with part (a).
QED
Theorem: Uniqueness of measures [Theorem 1.5, p. 8] Let $\mu$ and $\nu$ be two $\sigma$-finite measures on the $\sigma$-algebra $\mathcal{A}$. If $\mu$ and $\nu$ coincide on a $\pi$-system that generates $\mathcal{A}$, then $\mu = \nu$.
Proof A proof is given on p. 8. QED
Definition: Distribution-determining [p. 121] A set $\mathcal{M}$ of measurable functions from a general measurable space $(\mathcal{X}, \mathcal{B})$ (not necessarily a metric space) into $\mathbb{C}$ is called distribution-determining w.r.t. $(\mathcal{X}, \mathcal{B})$, iff every $f \in \mathcal{M}$ is integrable w.r.t. every probability measure on $(\mathcal{X}, \mathcal{B})$ and if, whenever $P$ and $Q$ are probability measures on $(\mathcal{X}, \mathcal{B})$ that satisfy $\int f\ dP = \int f\ dQ$ for all $f \in \mathcal{M}$, we have $P = Q$.
Definition: $\mathcal{P}_0$, finite subsets [p. 129] Let $T$ be a set. We denote by $\mathcal{P}_0$ the set of non-empty, finite subsets of $T$: $\mathcal{P}_0 \equiv \{I \subseteq T \mid: I \text{ is finite }, I \neq \emptyset\}$.
Example: Distribution-determining sets [p. 121] Let $(\mathcal{X}, \mathcal{B})$ be a measurable space.
(a) If $\mathcal{E}$ is a $\pi$-system generator of $\mathcal{B}$, then $\{I_E \mid: E \in \mathcal{E}\}$ is distribution-determining.
(b) If $\mathcal{X}$ is metric space and $\mathcal{B}$ is the induced $\sigma$-algebra, then $C_b(\mathcal{X})$ is distribution-determining.
(c) If $\mathcal{X}$ is a metric space, $\mathcal{B}$ is the induced $\sigma$-algebra and $I$ is a non-empty index set, then the set $\{f\circ \pi_J :\mid J \in \mathcal{P}_0,\ f \in C_b(\mathcal{X}^J)\}$ is distribution-determining for the product space $(\mathcal{X}^I, \mathcal{B}^I)$. ($\pi_J$ is the projection of $\mathcal{X}^I$ on $\mathcal{X}^J$.)
Proof
(a) Since for every $A \in \mathcal{B}$, $E(I_A) = P(A)$, the result follows from the uniqueness of measure theorem.
(b) Follows from the uniqueness of limits-in-distribution lemma.
(c) Let $P$ and $Q$ be probability measures on $(\mathcal{X}^I, \mathcal{B}^I)$, such that for every $J \in \mathcal{P}_0$ and every $f \in C_b(\mathcal{X}^J)$, $\int f\circ \pi_J\ dP = \int f\circ \pi_J\ dQ$. Then, by the previous example, $P$ and $Q$ coincide on the $\sigma$-algebra $\pi_J^{-1}(\mathcal{B}^J)$. Since the collection $\bigcup_{J \in \mathcal{P}_0}\pi_J^{-1}(\mathcal{B}^J)$ is a $\pi$-system that generates $\mathcal{B}^I$, we obtain $P = Q$ from the uniqueness of measures theorem.
QED
Theorem: $Y \overset{d}{=} X$ Using the notation defined in the original Question, $Y \overset{d}{=} X$.
Proof By the last example and by the uniqueness of limits-in-distribution lemma, it suffices to show that for every $J \in \mathcal{P}_0$, $X_J \overset{d}{=} Y_J$ (where, for $J = \{j_1, j_2, \dots, j_n\}$, $X_J \equiv (X_{j_1}, X_{j_2}, \dots, X_{j_n})$, and likewise $Y_J$.) Indeed, on the one hand, since for every $n \in \mathbb{N}_1$, $X \overset{d}{=} Y^n$ (cf. step 1 in the original question), we have for every $n \in \mathbb{N}_1$, $X_J \overset{d}{=} Y^n_J$; while on the other hand, since $Y^n \rightarrow Y$ pointwise on $\Omega$ (cf. step 2 in the original question), we have in particular $Y^n_J \rightarrow Y_J$ almost surely, whence $Y^n_J \overset{d}{\rightarrow} Y_J$ (justification - in the comment below.) Now, by the uniqueness of limits-in-distribution, $X_J \overset{d}{=} Y_J$, QED.
Comment The following claim:
can be justified by reference to Olav Kallenberg's book "Foundations of Modern Probability", 2nd. edition (Springer, 2002), namely using Lemma 4.2 ("subsequence criterion") on p. 63 together with Lemma 4.7 ("Convergence in probability and in distribution") on p. 66.
Wengenroth gives analogous results in theorem 4.6 ("Subsequence criterion") on p. 66 and theorem 5.3 ("Weak convergence and convergence-in-probability") on p. 85, however Wengenroth does not make it as clear as Kallenberg does, that the only assumption on the space where the functions take values is that it is a metric space.
II
In the following section I will prove that $Y$ and $\mathcal{F}^+_\tau$ are independent (see theorem 5 below).
Theorem: Conditional dominated convergence [Theorem 6.10.10, p. 113] Let $X, X_1, X_2, \dots$ be integrable functions defined on the probability space $(\Omega, \mathcal{A}, P)$, and let $Y: (\Omega, \mathcal{A}) \rightarrow (\mathcal{Y}, \mathcal{C})$ be a measurable function from $\Omega$ into the measurable space $(\mathcal{Y}, \mathcal{C})$. If $X_n \rightarrow X$ and $|X_n| \leq Z$ $P$-almost surely for some $Z \in \mathcal{L}_1(\Omega, \mathcal{A}, P)$, then $E(X_n |Y) \rightarrow E(X|Y)$ $P$-almost surely.
Proof A proof is given on p. 116. QED
Corollary: Conditional dominated convergence w.r.t. a $\sigma$-algebra Let $X, X_1, X_2, \dots$ be integrable functions defined on the probability space $(\Omega, \mathcal{A}, P)$, and let $\mathcal{C}$ be some sub-$\sigma$-algebra of $\mathcal{A}$. If $X_n \rightarrow X$ and $|X_n| \leq Z$ $P$-almost surely for some $Z \in \mathcal{L}_1(\Omega, \mathcal{A}, P)$, then $E(X_n |\mathcal{C}) \rightarrow E(X|\mathcal{C})$ $P$-almost surely.
Proof In the conditional dominated convergence theorem, take $Y: (\Omega, \mathcal{A}) \rightarrow (\Omega, \mathcal{C})$ to be the identity: $Y(\omega) := \omega$. QED
Theorem: Independence and conditional expectation [Theorem 6.15, p. 122] Two measurable functions $X$ and $Y$ defined on the measurable space $(\Omega, \mathcal{A})$ with values in the measurable spaces $(\mathcal{X}, \mathcal{B})$ and $(\mathcal{Y}, \mathcal{C})$, respectively, are independent iff $E(f(X) | Y) = E(f(X))$ for all functions $f$ belonging to a distribution determining set w.r.t. $(\mathcal{X}, \mathcal{B})$ (cf. definition I.5).
Proof A proof is given on p. 122. QED
Corollary: Independence and conditional expectation w.r.t. a $\sigma$-algebra Let $X$ be a measurable function defined on the measurable space $(\Omega, \mathcal{A})$ with values in the measurable space $(\mathcal{X}, \mathcal{B})$ and let $\mathcal{C}$ be a sub-$\sigma$-algebra of $\mathcal{A}$. Then $X$ and $\mathcal{C}$ are independent iff $E(f(X) | \mathcal{C}) = E(f(X))$ for all functions $f$ belonging to a distribution-determining set w.r.t. $(\mathcal{X}, \mathcal{B})$ (cf. definition I.5).
Proof In the Independence and conditional expectation theorem, take $Y: (\Omega, \mathcal{A}) \rightarrow (\Omega, \mathcal{C})$ to be the identity: $Y(\omega) := \omega$. QED
Theorem: $Y$ and $\mathcal{F}^+_\tau$ are independent Using the notation introduced in the original question, $Y$ and $\mathcal{F}^+_\tau$ are independent.
Proof Denote by $(\mathcal{X}, \mathcal{B})$ the separable, normed space in which $X$, $Y^n$ and $Y$ take values ($\mathcal{B}$ is the induced $\sigma$-algebra). We will only use the fact that this space is metric. By corollary 4 and by example I.7 (c) it suffices to show, for every $J \in \mathcal{P}_0$ and for every $f \in C_b(\mathcal{X}^J)$, that $$ E(f(Y_J) | \mathcal{F}^+_\tau) = E(f(Y_J)) \tag{*} $$ $P$-almost surely. Let then $J \in \mathcal{P}_0$ and let $f \in C_b(\mathcal{X}^J)$. Since, for every $n \in \mathbb{N}_1$, $Y^n$ is independent of $\mathcal{F}_{\tau_n}$ (cf. step 1 in the original question), and since, for every $n \in \mathbb{N}_1$, $\mathcal{F}^+_\tau \subseteq \mathcal{F}_{\tau_n}$ (cf. step 2 in the original question), we have, for every $n \in \mathbb{N}_1$, that $Y^n$ is independent of $\mathcal{F}^+_\tau$. Therefore, by corollary 4, for all $n \in \mathbb{N}_1$, $E(f(Y^n_J) | \mathcal{F}^+_\tau) = E(f(Y^n_J))$ $P$-almost surely. Hence, since, by theorem I.8, for every $n \in \mathbb{N}_1$, $Y^n \overset{d}{=} Y$, we have, for every $n \in \mathbb{N}_1$, $$ E(f(Y^n_J) | \mathcal{F}^+_\tau) = E(f(Y_J)) \tag{**} $$ $P$-almost surely. Since $Y^n \rightarrow Y$ pointwise on $\Omega$ (cf. step 2 in the original question) and since $f$ is continuous, we have $f(Y^n_J) \rightarrow f(Y_J)$ almost surely. Since $f$ is bounded, we may apply corollary 2 to obtain $$ E(f(Y^n_J) | \mathcal{F}^+_\tau) \rightarrow E(f(Y_J) | \mathcal{F}^+_\tau) \tag{***} $$ $P$-almost surely. Combining (**) and (***) we get (*), QED.