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Answer by Mateusz Kwaśnicki for Strong Markov property, independence, regular...
Let $\tau$ be a Markov time, and define the usual $\sigma$-algebras: $$\mathcal F^{<\tau} = \sigma\{X_t^{-1}(E) \cap \{t < \tau\} : t \geq 0, \, E \text{ — Borel}\}$$ and $$\begin{aligned}...
View ArticleAnswer by RaphaelB4 for Strong Markov property, independence, regular...
Strong Markov property says that under the condition that $\forall i, y(\tau_i)=y_i$, on each $[\tau_i,\tau_{i-1}]$ the process $X_t$ is just a brownian motion starting at $(i,y_i)$ and ending at...
View ArticleStrong Markov property, independence, regular conditional probability
I have a question on an argument appearing in this paper P.SettingLet $S=(1,\infty) \times (-1,1) \subset \mathbb{R}^2$ be a split, and let $X=(\{X_t\},\{P_x\}_{x \in S})$ be a Brownian motion in $S$...
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