# Is the local time of the reflected SDE Markovian?

Consider a stochastic differential equation: $$X_0=x\in\mathbb{R}$$, $$$$dX_t=b(X_t)dt+\sigma(X_t)dB^t+dL_t.$$$$ where $$L$$ is nondecreasing and $$B$$ is a Brownian motion. Suppose $$L$$ is the unique process such that $$X_t$$ is reflected at some boundary $$x^*\in\mathbb{R}$$ such that $$X_t\geq x^*$$ a.s. for all $$t\geq 0$$. This is known as the Skorokhod problem and $$L$$ (the local time of $$X$$ spent at $$x^*$$) has the expression $$$$L_t=\sup_{0\leq s \leq t}\left(X_s-L_s\right)^-.$$$$ My question is: Is the process $$L$$, $$X$$-Markovian?

My intuition is no. Because although $$L$$ `knows' to increase only when $$X_t$$ is at $$x^*$$, but as for how much it has to increase to reflect $$X$$, $$L$$ needs to know the dynamic of $$B$$.

$$L$$ is not a Markov process on its own, but the pair $$(L,X)$$ is Markovian. This because of the additivity property $$L_{t+s} = L_t+L_s(\theta_t),$$ in which $$\theta_t$$ stands for the time-shifted path $$(X_{u+t})_{u\ge 0}$$. From this (and the Markov property of $$X$$) it follows that the conditional distribution of $$(L_{s+t},X_{s+t})_{s\ge 0}$$, given the past of $$X$$ up to time $$t$$, depends only on $$L_t$$ and $$X_t$$.
• Thank you! Would you know if $L_t\in\sigma(X_t)$ i.e. $L_t$ is measurable with respect to the sigma-algebra generated by $X_t$ for all $t\geq 0$? Commented Jun 25, 2021 at 17:35
• No. $L_t$ accounts for all of the visiting of $X$ to $x$ up to time $t$, so it depends on more of the past-to-time-$t$ than just $X_t$. Commented Jun 27, 2021 at 0:30