Klenke's construction of Brownian motion

Question

Why does Klenke's concise construction of Brownian motion via probability transition kernels satisfy the motion's characterizing properties, equations (14.17) and (14.18)?

(results referenced in the construction)

Answer 1

Let $0=t_0<t_1<\cdots <t_n \in \mathbb R$ ($0<n$) and define $J_0:=\left\{t_0, t_1, \dots, t_n\right\}$, $J:=\left\{t_1, \dots, t_n\right\}$.

Define $\underline{X}_0 := \left(X_{t_0}, X_{t_1}, X_{t_2}, \dots , X_{t_n}\right)$, $\underline X := \left(X_{t_1}, X_{t_2}, \dots , X_{t_n}\right)$, $\Delta\underline{X}_0 := \left(X_{t_1}-X_{t_0}, X_{t_2}-X_{t_1}, \dots, X_{t_n}-X_{t_{n-1}}\right)$. Note that, since $\mathbb P_{X_0}=\mathcal N_{0,0}$ (i.e. $X_0=0\space\space a.s.$), $\mathbb P_{\Delta\underline{X}_0}=\mathbb P_{\left(X_{t_1}, X_{t_2}-X_{t_1}, \dots, X_{t_n}-X_{t_{n-1}}\right)}$.

Let $\underline Y := \left(Y_1, Y_2, \dots, Y_n\right)$ be independent and such that $\mathbb{P}_{Y_i} = \mathcal{N}_{0, t_{i-1}-t_i}, \space i=1,\dots,n$. I'll show that $\mathbb{P}_{\Delta\underline{X}_0}=\mathbb{P}_\underline{Y}$.

From Corollary 14.44, $\mathbb{P}_{\underline{X}_0}=\delta_0\otimes\bigotimes_{i=1}^n\kappa_{t_i-t_{i-1}}$. To each rectangle $Q_0=B_{t_0}\times \underbrace{B_{t_1}\times\cdots\times B_{t_n}}_{=:Q}\in \times_{i=0}^n \mathcal B$, we get $\mathbb{P}_{\underline{X}_0}\left(Q_0\right)=\mathbb 1_{B_{t_0}}(0)\cdot\bigotimes_{i=1}^n\kappa_{t_i-t_{i-1}}\left(0, Q\right)$. So, based on Theorem 14.28, $\mathbb{P}_{\underline{X}_0}\left(Q_0\right)=\mathbb 1_{B_{t_0}}\left(0\right)\cdot\mathbb{P}_\underline{S}\left(Q\right)$, where $\underline{S}=\left(S_1,S_2,\dots,S_n\right), \space S_m:=\sum_{i=1}^m Y_i \space\space \left(m=1,\dots,n\right)$. Hence $\mathbb P_{\underline{X}}\left(Q\right)=P_{\underline{X}_0}\left(\mathbb R \times Q\right)=P_{\underline{S}}\left(Q\right)$. As the rectangles $Q$ comprise a $\pi$-system that generates $\mathcal B_J$, $\mathbb P_{\underline{X}}=P_{\underline{S}}$ (cf. Lemma 1.42, "Uniqueness by an $\cap$-closed generator", p. 20).

So $\mathbb{P}_{\Delta \underline{X}}=\mathbb{P}_{\left(S_1, S_2-S_1, \dots, S_n-S_{n-1}\right)}=\mathbb{P}_\underline{Y}$.