Strong Markov Property for Lévy processes

Question

I do not understand one step in the proof of Theorem 32 in Chapter 1.

How can we conclude that $Y_t = X_{T + t} - X_{T}$ is independent from $\mathcal{F}_T$ using the identity $$\mathbb{E}\left(\textbf{1}_{A}\exp\left(i\sum_{j = 1}^{n}u_j(X_{T + t_j} - X_{T + t_{j-1}})\right)\right) = \mathbb{P}(A)\prod_{j = 1}^{n}f_{t_j - t_{j-1}}(u_j).$$ I understand that we can choose the $u_j$ such that $$\mathbb{E}\left(\textbf{1}_{A}\exp\left(i\sum_{j = 1}^{n}u_j(X_{T + t_j} - X_{T})\right)\right) = \mathbb{E}\left(\exp\left(i\sum_{j = 1}^{n}u_j(X_{T + t_j} - X_{T})\right)\right)\mathbb{P}(A).$$ Which implies that $$\mathbb{E}\left(\left.\exp\left(i\sum_{j = 1}^{n}u_j(X_{T + t_j} - X_{T})\right)\right| \mathcal{F}_{T}\right) = \mathbb{E}\left(\exp\left(i\sum_{j = 1}^{n}u_j(X_{T + t_j} - X_{T})\right)\right).$$ But how can we then prove that $Y_t$ is independent of $\mathcal{F}_T$?

Thank you for your time!

Answer 1

Recall Kac' theorem:

Let $U:\Omega \to \mathbb{R}^d$ and $V:\Omega \to \mathbb{R}^k$ be two random vectors. Then $U$ and $V$ are independent if, and only if, $$\forall \xi \in \mathbb{R}^d, \eta \in \mathbb{R}^k\::\: \mathbb{E}\exp(i \xi U + i \eta V) = \mathbb{E}\exp(i \xi U) \mathbb{E}\exp(i \eta V).$$

Now fix $0<t_1<\ldots<t_n$ and $A \in \mathcal{F}_T$, and define $U:=1_A$ and $V:=(Y_{t_1},\ldots,Y_{t_n})$. If $\xi \in \mathbb{R}$ and $\eta \in \mathbb{R}^n$, then

\begin{align*} \mathbb{E}\exp(i \xi U + i \eta V) &= \mathbb{E}\left[ e^{i \xi 1_A} \exp \left( i \sum_{j=1}^n \eta_j (X_{t_j+T}-X_{t_j}) \right) \right] \\ &= e^{i \xi} \mathbb{E}\left[ 1_A \exp \left( i \sum_{j=1}^n \eta_j (X_{t_j+T}-X_{t_j}) \right) \right] \\ &\quad +\mathbb{E}\left[ 1_{A^c} \exp \left( i \sum_{j=1}^n \eta_j (X_{t_j+T}-X_{t_j}) \right) \right].\end{align*}

(The last identity follows from the fact that $e^{i\xi 1_A}=e^{i\xi} 1_A+1_{A^c}$ - this is elementary, just consider separately the cases $\omega \in A$ and $\omega \notin A$.) Having this, we can apply the identities from the proof in the book to get

\begin{align*} \mathbb{E}\exp(i \xi U + i \eta V) &= e^{i\xi} \mathbb{P}(A)\mathbb{E}\left[ \exp \left( i \sum_{j=1}^n \eta_j (X_{t_j+T}-X_{t_j}) \right) \right] \\ &\quad + \mathbb{P}(A^c) \mathbb{E}\left[ \exp \left( i \sum_{j=1}^n \eta_j (X_{t_j+T}-X_{t_j}) \right) \right] \\ &= \mathbb{E}e^{i \xi 1_A} \mathbb{E}\left[ \exp \left( i \sum_{j=1}^n \eta_j (X_{t_j+T}-X_{t_j}) \right) \right] \\ &= \mathbb{E}\exp(i \xi U) \mathbb{E}\exp(i \eta V).\end{align*}

By Kac' theorem, $1_A$ and $(Y_{t_1},\ldots,Y_{t_n})$ are independent. Since $A \in \mathcal{F}_T$ and $0<t_1<\ldots<t_n$ are arbitrary, this implies that $(Y_t)_{t \geq 0}$ and $\mathcal{F}_T$ are independent.