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

This question can be seen as a continuation of my question at Convergence of the probabilities that drifted Brownian motion with jump never hits zero

Let $$(W_t)_{t\ge 0}$$ be a standard Brownian motion and define processes

$$X^n_t:=2+t+W_t-\ell^n(t) \quad \mbox{and} \quad X_t:=2+t+W_t-\ell(t),\quad \mbox{for all } t\ge 0,$$

where $$(\ell^n)_{n\ge 1}$$ and $$\ell$$ are right-continuous and non-decreasing functions s.t. $$\ell^n(0)=\ell(0)=0$$ and $$0\le \ell^n(t), \ell(t)\le 1$$ for all $$t\ge 0$$. If $$\lim_{n\to\infty}\ell^n(t)=\ell^n(t)$$ holds for all the points of continuity of $$\ell$$, can we prove

$$\lim_{n\to\infty}\mathbb P[\tau^n=\infty]=\mathbb P[\tau=\infty]?$$

Here $$\tau^n:=\inf\{t\ge 0:~ X^n_t\le 0\}$$ and $$\tau:=\inf\{t\ge 0:~ X_t\le 0\}$$.

Personal thoughts : My idea is the following. Take a sequence $$(T_m)_{m\ge 1}$$ diverging to $$\infty$$ s.t. $$\ell$$ is continuous at every $$T_m.$$ Then

$$\big|\mathbb P[\tau^n=\infty]-\mathbb P[\tau=\infty]\big|\le \big|\mathbb P[\tau^n>T_m]-\mathbb P[\tau^n=\infty]\big|+\big|\mathbb P[\tau^n>T_m]-\mathbb P[\tau>T_m]\big|+\big|\mathbb P[\tau>T_m]-\mathbb P[\tau=\infty]\big|.$$

If we are able to show the first and third terms can be uniformly small as $$m\to\infty$$, i.e. for any $$\epsilon>0$$, there exists $$m_{\epsilon}$$ s.t.

$$\big|\mathbb P[\tau^n>T_m]-\mathbb P[\tau^n=\infty]\big|+\big|\mathbb P[\tau>T_m]-\mathbb P[\tau=\infty]\big|\le \epsilon,\quad \mbox{for all } m\ge m_{\epsilon}.\quad \quad (\ast)$$

Then it suffices to show for fixed $$m_{\epsilon}$$, one has

$$\lim_{n\to\infty}\big|\mathbb P[\tau^n>T_{m_{\epsilon}}]-\mathbb P[\tau>T_{m_{\epsilon}}]\big|=0.$$

But I don't know how to prove $$(\ast)$$.

This is essentially the same as in the previous question, but requires strong Markov property rather than the usual one, and a somewhat different auxiliary function. Let $$\sigma_\delta = \inf\{t > \delta : t + W_t < -\delta\}$$ be the hitting time of $$(\delta, \infty) \times (-\infty, -\delta)$$ by the bi-variate process $$(t, t + W_t)$$ started at $$0$$. All that we need is that $$\sigma_\delta$$ goes to zero as $$\delta \to 0^+$$ with probability one. This is due to oscillatory behaviour of $$t + W_t$$ at small times: given any $$t > 0$$, with probability one there is $$s \in (0, t)$$ such that $$s + W_s < -s$$ (by the law of the iterated logarithm, for example), and it follows that $$\sigma_\delta \leqslant s$$ when $$\delta \in (0, s)$$.

Define the auxiliary function $$f(\delta) = \mathbb P[\sigma_\delta < \infty] = \mathbb P[\delta + t + W_t < 0 \text{ for some } t \in [\delta, \infty)] .$$ Let $$T > 0$$ and let $$\delta_n$$ denote the Kolmogorov distance between $$\ell^n$$ and $$\ell$$ over $$[0, R]$$. By assumption, $$\delta_n$$ goes to zero. Furthermore, it is rather easy to see that if $$\tau_n < T$$, then $$\tau \le \tau^n + \sigma_{\delta_n} \circ \theta_{\tau^n}$$ (where $$\theta_t$$ is the usual shift operator). Indeed, have a look at the picture:

The purple region lies entirely below the blue line. Thus, before hitting the purple region at time $$t = \tau^n + \sigma_{\delta_n} \circ \theta_{\tau^n}$$ the process $$2 + t + W_t$$ necessarily crosses the blue line $$x = \ell(t)$$ at time $$t = \tau$$ — and this is the desired inequality.

Thus, $$\mathbb P[\tau^n < T, \tau = \infty] \leqslant \mathbb P[\tau^n < T, \, \sigma_{\delta_n} \circ \theta_{\tau^n} = \infty] \leqslant \mathbb P[\sigma_{\delta_n} = \infty] ,$$ and the right-hand side goes to zero. It follows that $$\lim_{n \to \infty} \mathbb P[\tau^n < T, \tau = \infty] = 0 ,$$ A very similar argument shows that $$\mathbb P[\tau^n = \infty, \tau < T]$$ goes to zero.

Now we employ the fact that $$t + W_t$$ goes to infinity as $$t \to \infty$$, and $$\ell^n$$ are uniformly bounded. By choosing $$T$$ large enough, we can make the probability that $$2 + t + W_t < 1$$ for some $$t \geqslant T$$ less than any given $$\epsilon > 0$$ (again by the law of the iterated logarithm). Thus, $$\mathbb P[T \leqslant \tau^n < \infty, \tau = \infty] \leqslant \mathbb P[2 + t + W_t < 1 \text{ for some } t \geqslant T] < \epsilon$$ and similarly $$\mathbb P[\tau^n = \infty, T \leqslant \tau < \infty] < \epsilon .$$ We conclude that $$\limsup_{n \to \infty} \mathbb P[\tau^n < \infty, \tau = \infty] \leqslant \epsilon$$ and $$\limsup_{n \to \infty} \mathbb P[\tau^n = \infty, \tau < \infty] \leqslant \epsilon .$$ Since $$\epsilon > 0$$ is arbitrary, we get the desired conclusion $$\lim_{n \to \infty} \mathbb P[\tau^n = \infty \iff \tau = \infty] = 1 .$$

• This is rather sketchy. If need be, I'll be able to add further details after the weekend. – Mateusz Kwaśnicki Jun 2 at 22:21
• Many thanks for the solution. While I think there is a small mistake here : $\lim_{n\to\infty}\delta_n=0$ may not hold. We only know that $\lim_{n\to\infty}\ell^n(t)=\ell(t)$ for all the continuity points of $\ell$, and this does not imply $\lim_{n\to\infty}\delta_n=0$. Taking the example in my previous post, with $\ell^n(t)={\bf 1}_{\{t\ge n\}}$ and $\ell(t)=0$, one has $\delta_n\ge 1$ for all $n\ge 1$. – Neymar Jun 3 at 6:02
• However, inspired by your argument in my previous post, I succeed in proving $(\ast)$ for sufficiently large $m$, uniformly in $n$. This allows me finally to show the desired result – Neymar Jun 3 at 6:04
• I look forward to more details of your answer. – Neymar Jun 3 at 13:42
• I updated the answer. Let me know if anything remains unclear. – Mateusz Kwaśnicki Jun 7 at 9:51