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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)

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  • $\begingroup$ The equality is an immediate consequence of the fact that $\tau^n \uparrow \tau$ a.s. Note that $X_t \to \infty$ a.s. $\endgroup$ Jun 1 at 22:00
  • $\begingroup$ @DieterKadelka I got the point. Now I write down the claim that I wish to prove (see the post above). Do you think the result is still true? $\endgroup$
    – Neymar
    Jun 1 at 22:38
  • $\begingroup$ Hello @Neymar , you should ask this in a new question. $\endgroup$ Jun 2 at 8:58
  • $\begingroup$ @DieterKadelka Thanks. I post the new question at mathoverflow.net/questions/394332/… $\endgroup$
    – Neymar
    Jun 2 at 9:11
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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 $$ \mathbb P[\tau = \infty] - \mathbb P[\tau^n = \infty] \to 0 . $$

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  • $\begingroup$ Very nice reasoning! My question aims to find a counterexample for some claim that I am unable to prove. Now I have written down my claim. Do you think it is still true? $\endgroup$
    – Neymar
    Jun 1 at 22:34
  • $\begingroup$ This claim has been rewritten in a new post, see mathoverflow.net/questions/394332/… $\endgroup$
    – Neymar
    Jun 2 at 9:11

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