Independent increments and stationary increments, Lévy process Prop:Let $(X_t)$ be a $\mathbb{R}^d$ valued stochastic process with transition probability $P_t(x,dy)$. We assume there exsist probabilities on $\mathbb{R}^d$ $\{m_t\}_{t\geq 0}$ s.t., $P_t (x,B)=m_t(B-x)$, $m_0=\delta _0$(Dirac measure), $m_t*m_s=m_{t+s}$, $m_t(B)=m_t(-B)$, $m_t$ convergent to $\delta _0$ weakly as $t\to 0$.
Then $X_t$ has independent increments and stationary increments.
I know it is Markov process becasue of regularity of Dirichlet form. However, I cannot prove independent and stationary increments. How can I prove it?
 A: Denote by $\mathcal{F}_t := \sigma(X_s; s \leq t)$ the canonical filtration and set $$\mathbb{P}^x(X_t \in B) := P_t(x,B).$$ You have already shown that $(X_t)_{t \geq 0}$ is a Markov process, and therefore we have $$\mathbb{E}^x (f(X_t) \mid \mathcal{F}_s) = \mathbb{E}^{X_s} (f(X_{t-s})) \tag{1}$$ for any $s \leq t$ and any bounded measurable function $f$.  This gives
\begin{align*} \mathbb{E}^x(e^{i \xi (X_t-X_s)} \mid \mathcal{F}_s) &= e^{-i \xi X_s} \mathbb{E}^x(e^{i \xi X_t} \mid \mathcal{F}_s) \\ &\stackrel{(1)}{=} e^{-i \xi X_s} \mathbb{E}^{X_s} e^{i \xi X_{t-s}} \\ &= h(X_{t-s}) \tag{2}\end{align*} for any $\xi \in \mathbb{R}^d$ where $$h(y) := \mathbb{E}^{y}(e^{i \xi (X_{t-s}-y)}).$$ Since \begin{align*} \mathbb{P}^y(X_{r} \in B) = P_r(y,B) = m_r(B-y) &= P_r(0,B-y) \\ &= \mathbb{P}^0(X_r \in B-y) \\ &= \mathbb{P}^0(y+X_r \in B) \end{align*} for any $r \geq 0$ and $y \in \mathbb{R}^d$, it follows that $$\mathbb{E}^y f(X_r) = \mathbb{E}^0 f(y+X_r)$$ for any bounded measurable function $f$. Consequently, we get $$h(y) = \mathbb{E}^0(e^{i \xi X_{t-s}}). \tag{3}$$ Plugging this into $(2)$ shows that $$\mathbb{E}^x (e^{i \xi (X_t-X_s)} \mid \mathcal{F}_s) = \mathbb{E}^0 e^{i \xi X_{t-s}}. \tag{4}$$ This identity allows us to conclude that $(X_t)_{t \geq 0}$ has stationary and independent increments:


*

*stationarity of increments: Taking the expectation of both sides in $(4)$ yields $$\mathbb{E}^xe^{i \xi (X_t-X_s)} = \mathbb{E}^0 e^{i \xi X_{t-s}}. \tag{5}$$ For $s=0$ this gives $$\mathbb{E}^x e^{i \xi (X_t-x)} = \mathbb{E}^0 e^{i \xi X_t} \tag{6}$$ for any $t \geq 0$. Hence, $$\mathbb{E}^x e^{i \xi (X_t-X_s)} \stackrel{(5)}{=} \mathbb{E}^0 e^{i \xi X_{t-s}} \stackrel{(6)}{=} \mathbb{E}^x e^{i \xi (X_{t-s}-x)}, \qquad s \leq t \tag{7}$$  which shows that $X_t-X_s \sim X_{t-s}-X_0$, i.e. $(X_t)_{t \geq 0}$ has stationary increments.

*independence of increments: Fix $s \leq t$. By the tower property, we have $$\mathbb{E}^x e^{i \eta X_s + i \xi (X_t-X_s)} = \mathbb{E}^x \bigg[ \mathbb{E}^x \big( e^{i \eta X_s + i \xi (X_t-X_s)} \mid \mathcal{F}_s \big) \bigg]$$ and so \begin{align*} \mathbb{E}^x e^{i \eta X_s + i \xi (X_t-X_s)} &= \mathbb{E}^x \bigg[ e^{i \eta X_s} \mathbb{E}^x \big( e^{i \xi (X_t-X_s)} \mid \mathcal{F}_s \big) \bigg] \\ &\stackrel{(4)}{=} \mathbb{E}^x( e^{i \eta X_s}) \mathbb{E}^0 (e^{i \xi X_{t-s}}) \\ &\stackrel{(7)}{=} \mathbb{E}^x( e^{i \eta X_s}) \mathbb{E}^x (e^{i \xi (X_t-X_s)})\end{align*} and therefore $X_t-X_s$ and $X_s$ are independent, see e.g. here for details.
Remark: The idea of the proof is taken from the monograph Lévy Matters III by Böttcher, Schilling & Wang (Theorem 2.6).
