I have a question on an argument appearing in this paper P.


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 $\tau_r=\inf\{t>0 \mid r(t)=r\}$, $r \ge 1$.


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...


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.

  • $\begingroup$ What happens to $X$ at the boundary, that is, at $\{-1,1\} \times (1,\infty)$? If it is reflected, everything works nicely; however, in this case one would rather speak about BM in $[-1,1] \times (1,\infty)$. If it is absorbed (or killed), the claimed result does not seem to be true. $\endgroup$ – Mateusz Kwaśnicki Sep 24 '19 at 8:17
  • $\begingroup$ @MateuszKwaśnicki Thank you for your comment. $X$ is a absorbing BM on $S$ conditioned to hit $\{1\} \times (-1,1)$. So, $X$ does not hit $\{-1,1\} \times (1,\infty)$ before arriving at $\{1\} \times (-1,1)$. $\endgroup$ – sharpe Sep 24 '19 at 8:52
  • $\begingroup$ Thanks for clarification! (By the way, I misread the formula with conditioning, please ignore the last point of my previous comment.) $\endgroup$ – Mateusz Kwaśnicki Sep 24 '19 at 9:13

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).

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 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 $\mathcal G \vee \mathcal G_1 \vee \mathcal G_2$.

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  • $\begingroup$ This was written somewhat hastily, sorry. Please do not hesitate to ask for clarification. $\endgroup$ – Mateusz Kwaśnicki Sep 26 '19 at 9:26
  • $\begingroup$ Thank you for your kind reply. You wrote " — and therefore also given a larger $\sigma$-algebra $\mathcal G := \sigma\{X_{\tau_j} : j = 1, 2, \ldots, n\}$." Why this holds? I might have misunderstood? $\endgroup$ – sharpe Sep 26 '19 at 16:47
  • $\begingroup$ I added some clarification. If someone has a better reference, feel free to edit it into the answer. $\endgroup$ – Mateusz Kwaśnicki Sep 26 '19 at 17:16
  • $\begingroup$ Thank you for your reply. I was not educated enough. $\endgroup$ – sharpe Sep 26 '19 at 17:25
  • $\begingroup$ I learned a lot from your answer. Thank you for teaching me carefully. $\endgroup$ – sharpe Sep 26 '19 at 17:29

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 $(i-1,y_{i-1})$. This evolution only depend on the starting point and the ending point and not at all what append before $\tau_i$ or after $\tau_{i-1}$. Then $X_t$ on $[\tau_i,\tau_{i-1}]$ and $[\tau_j,\tau_{j-1}]$ are independant (for $j\neq i$).

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  • $\begingroup$ Thank you for your comment, Ho do we formulate "$X_t$ on $[\tau_i,\tau_{I-1}]$ and $[\tau_j,\tau_{j-1}]$ are independent for ($j \neq i$)" $\endgroup$ – sharpe Sep 24 '19 at 8:18
  • $\begingroup$ I have at least an intuitive reason. Thank you very much. $\endgroup$ – sharpe Sep 24 '19 at 9:09
  • $\begingroup$ One could possibly write "$dX_t$ on $[\tau_j,\tau_{j-1}]$ are independent", but it is somewhat sloppy to say that "$X_t$ on $[\tau_j,\tau_{j-1}]$ are independent": in particular, $X_{\tau_j}$ and $X_{\tau_i}$ are dependent! $\endgroup$ – Mateusz Kwaśnicki Sep 24 '19 at 9:17
  • $\begingroup$ "$dX_t$ on $[\tau_j,\tau_{j-1}]$ are independent." This is a good expression. $\endgroup$ – sharpe Sep 24 '19 at 9:24
  • $\begingroup$ I do not know how to formulate this expression, though. $\endgroup$ – sharpe Sep 24 '19 at 9:25

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