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Several places in "Optimal Stopping and Free-Boundary Problems" Peskir and Shiryaev make the assumption that a (Markov) process $X = (X_t)_{t\geq 0}$ has sample paths which are right continuous and left continuous over stopping times.

What is the reason for making the distinction between continuity with respect to the time $t$ versus continuity with respect to a stopping time $\tau$? If a process is right and left continuous with respect to $t$ isn't it automatically right and left continuous with respect to a stopping time $\tau$?

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A process which is right continuous can still be left continuous at a fixed time almost surely. For example, a Poisson process $N=(N_t)_{t \ge 0}$ is right continuous, and for any (deterministic) time $t$ it holds that $\lim_{s \uparrow t}N_s=N_t$ a.s., simply because $\mathbb{P}(N \text{ jumps at time } t)=0$. It is in general a stronger statement to say the same is true for stopping times, and this property is known as quasi-left-continuity. As usual, George Lowther's blog has a nice post about this, including the proof that Feller processes are quasi-left-continuous.

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  • $\begingroup$ Thanks that's illuminating. The Poisson process is a Feller process which implies that it is also quasi-left continuous. Is there a process you know of the top of your head which is left-continuous in time but not quasi-left continuous? $\endgroup$
    – deepblue
    Jun 18 '18 at 20:46

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