# Convergence of the probabilities that drifted Brownian motion with jump never hits zero

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)

• The equality is an immediate consequence of the fact that $\tau^n \uparrow \tau$ a.s. Note that $X_t \to \infty$ a.s. Jun 1, 2021 at 22:00
• @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?
– user128095
Jun 1, 2021 at 22:38
• Hello @Neymar , you should ask this in a new question. Jun 2, 2021 at 8:58
• @DieterKadelka Thanks. I post the new question at mathoverflow.net/questions/394332/…
– user128095
Jun 2, 2021 at 9:11

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 $$\mathbb P[\tau = \infty] - \mathbb P[\tau^n = \infty] \to 0 .$$