Dynkin Hunt formula

Question

I have a question about Dynkin Hunt formula.

Last day, I found a formula in this paper enter link description here. The formula is the equation (2.5) in this paper, which is called Dynkin Hunt formula. I know Dynkin formula. Dynkin formula states \begin{align*} R_{\alpha}f(x)=R_{\alpha}^{U}f(x)+E_{x}[e^{-\alpha \tau_{U}}R_{\alpha}f(X_{\tau_{U}})], \end{align*} where $\{X_t,P_x\}$ is a Markov(Hunt) process defined on a topological space $E$ and $R_{\alpha}$ is its resolvent. $R_{\alpha}^{U}f(x)$ is defined by \begin{equation*} R_{\alpha}^{U}f(x)=E_{x}[\int_{0}^{\tau_{U}}e^{-\alpha t}f(X_t)\,dt], \end{equation*} where $\tau_{U}$ is the first exit time from an open set $U$. In other words, $R_{\alpha}^{U}$ is the resolvent of subprocess $(X_{t}^{U},P_x)$ of $(X_t,P_x)$ on $U$.

You can see this formula in Dirichlet Forms and Symmetric Markov Processes by M. Fukushima, Y. Oshima, and M. Takeda. The equation (4.1.6) in this book is Dynkin formula.

Question

Assume $\{X_t\}$ has transition probability density function $p(t,x,y)$ and transition density function of $\{X_{t}^{U}\}$ is denoted by $p^{U}(t,x,y)$.

Can we show the Dynkin-Hunt formula \begin{equation*} p(t,x,y)=p^{U}(t,x,y)+E_{x}[1_{\{\tau_{U} \le t\}}p(t-\tau_{U},x,y)] \end{equation*} for every $t>0$, $x,y \in E$?

My attempt

Assume resolvent density $R_{\alpha}(x,y)$ is continuous in $y$. Then, from Dynkin formula, we see \begin{equation*} R_{\alpha}(x,y)=R_{\alpha}^{U}(x,y)+E_{x}[e^{-\alpha \tau_{U}}r_{\alpha}(X_{\tau_{Y}},y)] \end{equation*} for every $\alpha>0,x,y \in E$. Therefore, \begin{align*} \int_{0}^{\infty}e^{-\alpha t}p(t,x,y)\,dt &=\int_{0}^{\infty}e^{-\alpha t}p^{U}(t,x,y)\,dt \\ &+\int_{0}^{\infty}e^{-\alpha t}E_{x}[1_{\{\tau_{U} \le t\}}p(t-\tau_{U},X_{\tau_{U}},y)]\,dt \end{align*} Therefore, for every $x,y \in E$, \begin{equation*} p(t,x,y)=p^{U}(t,x,y)+E_{x}[1_{\{\tau_{U} \le t\}}p(t-\tau_{U},x,y)] \end{equation*} holds for a.e. $t$. Can we refine this equation? That is, can we show \begin{equation*} p(t,x,y)=p^{U}(t,x,y)+E_{x}[1_{\{\tau_{U} \le t\}}p(t-\tau_{U},x,y)] \end{equation*} for every $t>0$, $x,y \in E$?

If you know related works about Dynkin-Hunt formula, please let me know.

Answer 1

The authors of the paper you cite mis-state the matter slightly. Dynkin's formula is indeed a direct consequence of the strong Markov property. Your Laplace inversion argument (with no assumption of continuity) shows that for each fixed $x$, $$ p(t,x,y)=p^{U}(t,x,y)+E_{x}[1_{\{\tau_{U} \le t\}}p(t-\tau_{U},x,y)] $$ for $m\otimes\lambda$-a.e. $(y,t)$, where $m$ is the symmetry measure of $X$ and $\lambda$ is Lebesgue measure on $(0,\infty)$. The key observation now is that the two sides of this identity, as functions of $(y,t)$, are finely continuous with respect to the space-time process $(X_t, r-t)_{0\le t<r}$, for which $m\otimes\lambda$ is a reference measure. As such, the two sides of the identity agree for all $(y,t)$. A nice discussion of such methods can be found in the paper "Excursions of dual processes" by R.K. Getoor and M.J. Sharpe [Advances in Mathematics 45 (1982) 259–309]. A detailed discussion of the "Dynkin-Hunt" identity, in the context of Brownian motion, appears in the book From Brownian motion to Schrödinger's equation of K.L. Chung and Z.X. Zhao. In the paper you cite, the proces $X$ is symmetric, so arguments based on quasi-continuity come into play; perhaps this is what the authors had in mind.