I have a question on an argument appearing in this paper [P](http://www.ams.org/journals/tran/2000-352-06/S0002-9947-00-02594-0/S0002-9947-00-02594-0.pdf).


**Setting**


Let $S=(1,\infty) \times (-1,1) \subset \mathbb{R}^2$ be a split, and let $X=(\{X\_t\},\{P\_x\}\_{x \in S})$ be a Brownian motion in $S$ conditioned to hit $\{1 \} \times (-1,1)$.


We denote by $r(t)$, $y(t)$ the first coordinate process of $X$ and the second coordinate process of $X$, respectively. Let $\tau\_r=\inf\{t>0 \mid r(t)=r\}$, $r \ge 1$.


**Question**


We consider random variables:
\begin{align\*}
R\_{k}=\int\_{\tau\_{k}}^{\tau\_{k-1}}\frac{1}{r(s)^2}\,ds,\quad 2 \le k \le n. 
\end{align\*}


I would like to ask why $\{R\_k\}\_{k=2}^{n}$ are independent under $P\_{n,y}(\cdot \mid y(\tau\_j)=y\_j,\quad j=1,\cdots,n-1)$ for each $n \ge 2$ and $y \in (-1,1)$.


The author of the above paper claims that strong Markov property yields the independence. However, I couldn't know how to use it. The event $\{y(\tau\_j) \in B\_j,\quad j=1,\cdots, n-1\}$, $B\_j \in \mathcal{B}((-1,1))$ seems like future events... 


**ADD**


As RaphaelB4 said, under $P\_{n,y}(\cdot \mid y(\tau\_j)=y\_j,\quad j=1,\cdots,n-1)$, $X\_t$ on $[\tau\_i,\tau\_{i-1}]$ and $X\_{t}$ on $[\tau\_{j},\tau\_{j-1}]$ $(i \neq j)$ would be independent. But, I don't know how to prove this statement.



Strong Markov property, independence, regular conditional probability
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 $$\begin{aligned} \mathcal F\_{\geqslant\tau} & = \sigma\{X\_{\tau + t}^{-1}(E) : t \geq 0, \, E \text{ — Borel}\} \\ & = \sigma\{X\_t^{-1}(E) \cap \{t \geqslant \tau\} : t \geq 0, \, E \text{ — Borel}\} . \end{aligned} $$ (There are some regularity issues involved in the last equality, but let us ignore them.) Strong Markov property asserts that $\mathcal F^{<\tau}$ and $\mathcal F\_{\geqslant \tau}$ are conditionally independent, given $\sigma\{X\_\tau\}$.


Let $0 = \tau\_0 < \tau\_1 < \tau\_2 < \ldots < \tau\_n < \tau\_{n+1} = \infty$ (compared to the original question, the order of $\tau\_j$ is reversed here). Define in a similar way
$$ \mathcal F\_j := \mathcal F\_{\geqslant \tau\_{j-1}}^{<\tau\_j} = \sigma\{X\_t^{-1}(E) \cap \{\tau\_{j-1} \leqslant t < \tau\_j\} : t \geq 0, \, E \text{ — Borel}\}$$
By the strong Markov property, $\bigvee\_{j \leqslant k} \mathcal F\_j$ and $\bigvee\_{j > k} \mathcal F\_j$ are conditionally independent, given $\sigma\{X\_{\tau\_k}\}$ — and therefore also given a larger $\sigma$-algebra $\mathcal G := \sigma\{X\_{\tau\_j} : j = 1, 2, \ldots, n\}$ (see Lemma below; note that each $X\_{\tau\_i}$ is measurable with respect to either $\bigvee\_{j \leqslant k} \mathcal F\_j$ (if $i \leqslant k$) or $\bigvee\_{j > k} \mathcal F\_j$ (if $i \geqslant k$)).


The above property implies that the family of $\sigma$-algebras $\mathcal F\_1, \mathcal F\_2, \ldots, \mathcal F\_n, \mathcal F\_{n+1}$ is conditionally independent, given $\mathcal G$ (a rigorous argument is somewhat tiresome, though). It remains to note that the random variable $R\_j = \int\_{\tau\_{j-1}}^{\tau\_j} \phi(X\_s) ds$ for an arbitrary numerical function $\phi$ is measurable with respect to $\mathcal F\_j$ (these random variables correspond to the random variables $R\_j$ in the question, in a reversed order).




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Notation: $\mathcal F < \mathcal G$ means that $\mathcal F$ is a sub-$\sigma$-algebra of $\mathcal G$; $\mathcal F \vee \mathcal G$ is the smallest $\sigma$-algebra which contains $\mathcal F$ and $\mathcal G$.


**Lemma** *(see Theorem 1.19 in [lecture notes](http://web.math.ku.dk/~erhansen/Markov_04/doku/noter/condindep_1.ps) by Ernst Hansen)*: If $\mathcal F\_1$ and $\mathcal F\_2$ are conditionally independent given $\mathcal G$, and $\mathcal G\_1 < \mathcal F\_1$, $\mathcal G\_2 < \mathcal F\_2$, then $\mathcal F\_1$ and $\mathcal F\_2$ are conditionally independent given $\mathcal G \vee \mathcal G\_1 \vee \mathcal G\_2$.


