Let $X\_t=2+t+W\_t$ for $t\ge 0$, where $(W\_t)\_{t\ge 0}$ is a standard Brownian motion. For every $n\ge 1$, set $X^n\_t:=X\_t-{\bf 1}\_{t\ge n}$. Denote respectively


$$\tau:=\inf\{t\ge 0:~ X\_t\le 0\}\quad \mbox{and} \quad \tau^n:=\inf\{t\ge 0:~ X^n\_t\le 0\}.$$


Could we prove or disprove $\lim\_{n\to\infty}\mathbb P[\tau^n=\infty]=\mathbb P[\tau=\infty]$?


**Further Question** : I wish to prove the similar convergence result. This question can be found at [Convergence of the probabilities that drifted Brownian motion with jump never hits zero (continuation)](https://mathoverflow.net/questions/394332/convergence-of-the-probabilities-that-drifted-brownian-motion-with-jump-never-hi)



Convergence of the probabilities that drifted Brownian motion with jump never hits zero
Just use the Markov property at time $n$.


Write $f(x)$ for the probability that $x+t+W\_t$ never hits $(-\infty, 0]$. Then
$$ \mathbb P[\tau = \infty] = \mathbb E[\mathbb 1\_{\{\tau > n\}} f(X\_n)] $$
and
$$ \mathbb P[\tau^n = \infty] = \mathbb E[\mathbb 1\_{\{\tau > n\}} f(X\_n - 1)] , $$
so the difference of the two is
$$ \mathbb E[\mathbb 1\_{\{\tau > n\}} (f(X\_n) - f(X\_n - 1))] . $$
Now $X\_n \to \infty$ and $\mathbb 1\_{\{\tau > n\}} \to \mathbb 1\_{\{\tau = \infty\}}$ with probability one as $n \to \infty$, and $f(x) - f(x - 1) \to 0$ as $x \to \infty$. Thus, the integrand converges to zero with the probability one, and the dominated convergence theorem implies that indeed
$$ \mathbb P[\tau = \infty] - \mathbb P[\tau^n = \infty] \to 0 . $$


