Tag: independence

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

Setting

Let $ S=(1,\infty) \times (-1,1) \subset \mathbb{R}^2$ and let $ X=(\{X_t\},\{P_x\}_{x \in S})$ be a diffusion process on $ S$ . Imagine something like the Brownian motion on $ 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$ .

The author consider random variables of the form \begin{align*} R=\int_{0}^{\tau_1}\frac{1}{r(s)^2}\,ds,\quad R_k=\int_{\tau_{k}}^{\tau_{k-1}}\frac{1}{r(s)^2}\,ds,\quad k \ge 2. \end{align*}

My question

Let $ (n,y) \in (2,\infty) \times (-1,1)$ and let $ \{y_k\}_{k=1}^{n-1} \in (-1,1)^{n-1}$ .

• The author claims that the random variables $ \{R_k\}_{k=2}^{n}$ are independent under $ P_{n,y}(\cdot \mid y(\tau_k)=y_k,\ k=1,\cdots,n-1)$ . Here, $ P_{n,y}(\cdot \mid y(\tau_k)=y_k,\ k=1,\cdots,n-1)$ is defined as follows: for any events $ A$ and Borel subset $ B \subset (-1,1)^{n-1}$

\begin{align*} &P_{n,y}(A \cap \{(y(\tau_1),\cdots, y(\tau_{n-1})) \in B\} )\ &=\int_{B}P_{n,y}(A \mid y(\tau_k)=y_k,\ k=1,\cdots,n-1)\,d\nu(y_1,\cdots, y_n),\ &\nu=\text{ the distribution of } (y(\tau_1),\cdots, y(\tau_{n-1})). \end{align*}

• The author seems to use the strong Markov property for \begin{align*} E_{n,y}\left[\int_{\tau_k}^{\tau_{k-1}}\frac{1}{r(s)^2}\,ds \mid y(\tau_k)=y_k,\ k=1,\cdots,n-1 \right]. \end{align*} to obtain the following: \begin{align*} &E_{n,y}\left[\sum_{k=2}^{n}\int_{\tau_{k}}^{\tau_{k-1}}\frac{1}{r(s)^2}\,ds \mid y(\tau_k)=y_k,\ k=1,\cdots,n-1 \right]\ &=\sum_{k=2}^{n}E_{k,y_{k}}\left[\int_{\tau_k}^{\tau_{k-1}}\frac{1}{r(s)^2}\,ds \mid y(\tau_{k-1})=y_{k-1} \right]. \end{align*}

However, I do not know how to use the strong Markov property. The author seems to consider conditioning on future events. Is this possible?