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Consider the stochastic processes $$X_t^n=1+t+B_t-\Lambda_t^n \quad \mbox{and} \quad X_t=1+t+B_t-\lambda(t), \quad \forall t\ge 0,$$

where $(B_t)_{t\ge 0}$ is a standard Brownian motion, $(\Lambda^n_t)_{t\ge 0}$ is a right-continuous and increasing stochastic process taking values in $[0,1]$, $\lambda:[0,\infty)\to [0,1]$ is a right-continuous and increasing function. If $\lim_{n\to\infty}\rho(\Lambda^n, \lambda)=0$ almost surely, where

$$\rho(\Lambda^n, \lambda) := \inf\big\{\epsilon>0:~ \lambda(t+\epsilon)+\epsilon \ge \Lambda^n_t \ge \lambda(t-\epsilon)-\epsilon,\quad \forall t\in\mathbb R\big\},$$

and $\Lambda^n_t:= 0=: \lambda(t)$ for all $t<0$. Could we show

$$\lim_{n\to\infty}\mathbb P(X^n_t>0, \forall t\ge 0) = \mathbb P(X_t>0, \forall t\ge 0)?$$

Any solutions, remarks or references are highly appreciated!

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  • $\begingroup$ A very natural idea is to consider $X^{\pm \epsilon}_t=1+t+B_t-\big(\lambda(t\pm\epsilon)\pm\epsilon\big)$. While I do not find any results/reference to compare $X^{\pm\epsilon}$ and $X$. $\endgroup$ – Neymar Nov 15 '20 at 18:44

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