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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)

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Perhaps you could be more specific about where you get stuck trying to verify these statements? – Nate Eldredge Aug 7 '12 at 18:11
@NateEldredge: Take for instance the claim that $\left(X_{t_i}-X_{t_{i-1}}\right)_{i=1,\dots,n}$ are independent (14.17). I'd try to prove it by induction. Now, since $t_0=0$ and $X_0\equiv 0$ by definition, $X_{t_1}-X_{t_0} = X_{t_1}$ . So showing that $X_{t_1}-X_{t_0}$ and $X_{t_2}-X_{t_1}$ are independent is tantamount to showing that $X_{t_1}$ and $X_{t_2}-X_{t_1}$ are independent. Great! Now what? – Evan Aad Aug 7 '12 at 18:58
@NateEldredge: A small progress: I've managec to show that $\forall t \in \left[0,\infty\right)\bullet X_t\sim \mathcal N_{0,t}$. – Evan Aad Aug 7 '12 at 20:16
@NateEldredge: OK, i think i've figured out the answer. Thanks! – Evan Aad Aug 7 '12 at 20:32
Glad to hear it! You might like to post your own answer here and then accept it. – Nate Eldredge Aug 7 '12 at 20:41
up vote 1 down vote accepted

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

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Neither of your expressions for $\mathbb{P}_\underline{X}$ is quite right (and they are not the same!). You want $\mathbb{P}_\underline{X}=\delta_0\kappa_{t_1}\otimes\bigotimes_{i=2}^n\kappa_{t‌​_i-t_{i-1}}$. – Byron Schmuland Aug 8 '12 at 12:37
@ByronSchmuland: I think i got your point. I (hopefully) corrected my proof accordingly. – Evan Aad Aug 8 '12 at 15:58

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