Expectation involving local time of Brownian motion

Question

I have a problem related to expectation of stopped local time where it is shown that $$\mathbb{E}_x\left(L^y_{T_a \wedge T_b}\right)=2(x-a)\frac{b-y}{b-a}.$$ for $a \leq x \leq y \leq b$, $T_z = \inf\{t\geq 0:B_t =z\}$, $B_t$ is one dimensional Brownian motion and $L_t^y$ its local time in $z$ until $t$.

Let $T=T_a \wedge T_b$ and $t\geq 0$. Is there a similar expression for $\mathbb{E}_x\left(1_{\{T < t\}}L^y_{T}\right)$? I have tried to modify the proof of the original post but this didn't work. Does anybody have an idea?

Answer 1

You'll have more luck with the complementary object $E_x[1_{\{T>t\}}L^y_T]$. On the event $\{T>t\}$ you have $L^y_T=L^y_t+L^y_T\circ\theta_t$, where $\theta_t$ is the shift operator on Brownian paths. The expectations of these two terms can each be expressed in terms of the transition density of the Brownian motion killed when it first reaches $a$ or $b$, the first term directly and the second making use of the formula you have displayed. (When local time is normalized appropriately, you have $E_x[L_t^y] = p_t(x,y)$, with $p_t(x,y)$ the Brownian transition density; likewise for $E_x[1_{\{T>t\}}L_t^y]$ but using the transition density of the killed motion.)

Answer 2

Let me try to perform the first steps based on the approach posted by John Dawkins. We consider the complementary object $$E_x[1_{\{T>t\}}L^y_T]=E_x[1_{\{T>t\}}(L^y_t+L^y_T\circ \theta_t)]=E_x[1_{\{T>t\}}L^y_t]+E_x[1_{\{T>t\}}L^y_T\circ \theta_t]$$ and use the Markov property $$=E_x[1_{\{T>t\}}L^y_t]+E_x[1_{\{T>t\}}E_t[L_T^y]]=E_x[1_{\{T>t\}}L^y_t]+E_t[L_T^y]P_x(T>t).$$ It remains to calculate the first expectation of the lhs $E_x[1_{\{T>t\}}L^y_t]$.

I am no expert in local times so the fact that $E_x[L_t^y] = p_t(x,y)$ mentioned by John Dawkins was new to me and I couldn't find any reference for this. How do I show this statement and how can I find a similar expression for $E_x[1_{\{T>t\}}L_t^y]$?