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Consider a stochastic differential equation: $X_0=x\in\mathbb{R}$, \begin{equation} dX_t=b(X_t)dt+\sigma(X_t)dB^t+dL_t. \end{equation} 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 \begin{equation} L_t=\sup_{0\leq s \leq t}\left(X_s-L_s\right)^-. \end{equation} 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 $B$.

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$L$ 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 $X_t$.

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  • $\begingroup$ 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 $t\geq 0$? $\endgroup$
    – Amira
    Commented Jun 25, 2021 at 17:35
  • $\begingroup$ 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 $X_t$. $\endgroup$ Commented Jun 27, 2021 at 0:30

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