# Strong Markov property, independence, regular conditional probability

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

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...

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.

• 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. – Mateusz Kwaśnicki Sep 24 '19 at 8:17
• @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 $\{1\} \times (-1,1)$. – sharpe Sep 24 '19 at 8:52
• Thanks for clarification! (By the way, I misread the formula with conditioning, please ignore the last point of my previous comment.) – 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 $$\mathcal G \vee \mathcal G_1 \vee \mathcal G_2$$.

• This was written somewhat hastily, sorry. Please do not hesitate to ask for clarification. – Mateusz Kwaśnicki Sep 26 '19 at 9:26
• 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? – sharpe Sep 26 '19 at 16:47
• I added some clarification. If someone has a better reference, feel free to edit it into the answer. – Mateusz Kwaśnicki Sep 26 '19 at 17:16
• Thank you for your reply. I was not educated enough. – sharpe Sep 26 '19 at 17:25
• I learned a lot from your answer. Thank you for teaching me carefully. – 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 $$j\neq i$$).

• 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 ($j \neq i$)" – sharpe Sep 24 '19 at 8:18
• I have at least an intuitive reason. Thank you very much. – sharpe Sep 24 '19 at 9:09
• 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! – Mateusz Kwaśnicki Sep 24 '19 at 9:17
• "$dX_t$ on $[\tau_j,\tau_{j-1}]$ are independent." This is a good expression. – sharpe Sep 24 '19 at 9:24
• I do not know how to formulate this expression, though. – sharpe Sep 24 '19 at 9:25