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.


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.


1 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.

  • $\begingroup$ Thank you for telling me the reference! Corollary (3.14) in this paper may be what I was looking for. $\endgroup$
    – sharpe
    Commented Jul 9, 2017 at 9:19

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