How do we use strong Markov property to get the formula

Question

Suppose $X(t)$ is a homogeneous Markov process and satisfies the strong Markov property. Assume that $U\subset\mathbb{R}^{n}$ is an open and bounded set, denote the boundary of $U$ by $\Gamma$, $\tau_{U}$ is a stopping time defined by the time that $X(t)$ first leaves $U$. Let $A\in\mathcal{B}(\mathbb{R}^{n})$. The textbook claims that

$\mathbb{P}^{x}(X(t)\in A,\tau_{U}\leq t)=\int_{\Gamma}\int_{u=0}^{t}\mathbb{P}^{x}(\tau_U\in\mathrm{d}u,X(\tau_u)\in\mathrm{d}z)\mathbb{P}^{z}(X(t-u)\in A)$

Here $\mathbb{P}^{x}(X(t)\in A,\tau_{U}\leq t):=\mathbb{P}(X(t)\in A,\tau_{U}\leq t|X(0)=x)$ . I guess the formula means that $x$ takes time $\tau_U$ to some point z, which is on the boundary of $U$(that is $\Gamma$), then $z$ takes time $t-\tau_U$ to $A$.

My idea is $\mathbb{P}^{x}(X(t)\in A,\tau_{U}\leq t)=\mathbb{E}^{x}[\mathbb{E}^{x}[(1_{A}(X(t)))1_{\tau_U\leq t}|X(t)]]=\mathbb{E}^{x}[1_{\tau_U\leq t}\mathbb{E}^{x}[(1_{A}(X(t)))|X(t)]]$. It doesn't seem right. Can you give me some hints? Thanks!

Answer 1

We need to condition on the $\sigma$-algebra generated by the stopping time $\tau_U$, instead of $X(t)$.

Preliminaries

Recall the definition of the $\sigma$-algebra of events prior to $\tau_U$: $$ \mathcal{F}_{\tau_U}:=\{A\in\mathcal{F}\mid A\cap\{\tau_U\le t\}\in\mathcal{F}_t\}. $$ Note $\{\tau_U\le t\}\in\mathcal{F}_{\tau_U}$.$\renewcommand{\P}{\,\mathbb{P}}\newcommand{\E}{\,\mathbb{E}} \newcommand{\R}{\mathbb{R}}\newcommand{\F}{\mathcal{F}} \newcommand{\abs}[1]{\lvert#1\rvert}\newcommand{\Abs}[1]{\left|#1\right|}\newcommand{\ABs}[1]{\biggl|#1\biggr|}\newcommand{\norm}[1]{\|#1\|}\newcommand{\Norm}[1]{\left\|#1\right\|}\newcommand{\NOrm}[1]{\biggl\|#1\biggr\|}\newcommand{\Brace}[1]{\left\{#1\right\}}\newcommand{\BRace}[1]{\biggl\{#1\biggr\}}\newcommand{\paren}[1]{\left(#1\right)}\newcommand{\Paren}[1]{\biggr(#1\biggl)}\newcommand{\bracket}[1]{\langle#1\rangle}\newcommand{\brac}[1]{\langle#1\rangle}\newcommand{\Bracket}[1]{\left\langle#1\right\rangle}\newcommand{\Brac}[1]{\left\langle#1\right\rangle}\newcommand{\bra}[1]{\left\langle#1\right|}\newcommand{\ket}[1]{\left|#1\right\rangle}\newcommand{\Square}[1]{\left[#1\right]}\newcommand{\SQuare}[1]{\biggl[#1\biggr]}$

The strong Markov property can be expressed as: $$ \E^x[1_{\Brace{\tau_U<\infty}}f(X_{\tau_U+t})|\F_{\tau_U}]=1_{\Brace{\tau_U<\infty}}\E^{X_{\tau_U}}[f(X_t)], $$ where $f:\R^n\to\R$ is any bounded measurable function. Here, we assume that we don't need to bother with $1_{\Brace{\tau_U<\infty}}$ since $\tau_U$ is finite. To ensure this finiteness, we need additional condition on $\{X_t\}$ in addition to what you described, but no worries.

Answer

First conditioning on $\F_{\tau_U}$, and then using the strong Markov property, we conclude:

\begin{align*} \P^x[X_t\in A,\tau_U\le t]&=\E^x\SQuare{1_{\Brace{\tau_U\le t}}\E^x[1_{\Brace{X_t\in A}}|\F_{\tau_U}]}\\ &=\E^x\SQuare{1_{\Brace{\tau_U\le t}}\P^{X_{\tau_U}}[X_{t-\tau_U}\in A]}. \end{align*}

The last expectation can be readily expressed as the following integral. Note that we have two random variables, $\tau_U$ and $X_{\tau_U}$, in the expectation. \begin{align*} \int^t_0\int_{U^\complement}\P^z[X_{t-u}\in A]\P^x[\tau_U\in du,X_u\in dz]. \end{align*} Under additional assumptions, such as $\{X_t\}$ a.s. having continuous sample paths, the inner integral can be taken on $\Gamma$, instead of the entire $\R^n\setminus U$.