# 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?

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.