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We consider the following SDE: $$dX_t = 1(X_t = 0) \, dt + 1(X_t >0) \, dB_t, \quad X_0= x > 0,$$ where $(B_t, \, t \ge 0)$ is linear Brownian motion. Let $\tau: = \inf\{t >0: X_t = 0\}$ be the first time at which $X$ hits $0$. It is obvious that $(X_t, \, 0 \le t \le \tau)$ is Brownian motion up to the first hitting to $0$.

Question: Can we say something for $X$ after $\tau$? Is it well-defined?

It is clear that $X$ is not reflected Brownian motion but is also supported on $[0,\infty)$. It seems that there is accumulation of zeros after $\tau$ which would lead to local times.

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  • $\begingroup$ There can never be a "first" zero of Brownian motion. If $B_{t_{0}}=a$, the $B_{t}$ takes the value $a$ an infinite number of times on every interval containing $t_{0}$. $\endgroup$
    – Buzz
    Commented Feb 18, 2022 at 0:45
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    $\begingroup$ @Buzz This true if $t_0$ is deterministic but false in general. Anyway, every continuous function defined on $\mathbb{R}_+$ has a first zero (unless it doesn't have any). $\endgroup$
    – Martin Hairer
    Commented Feb 18, 2022 at 9:23

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This is sticky reflecting Brownian motion, see for example this relatively recent paper. You can alternatively construct it by taking a reflected Brownian motion and then "stretching out" the local time accumulated at the origin, turning it into real time. Different constants in front of the $dt$ term yield different stretching factors.

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