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 $$ \lim_{n \to \infty} \mathbb P[\tau^n = \infty \iff \tau = \infty] = 1 . $$

  • $\begingroup$ This is rather sketchy. If need be, I'll be able to add further details after the weekend. $\endgroup$ – Mateusz Kwaśnicki Jun 2 at 22:21
  • $\begingroup$ 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 $n\ge 1$. $\endgroup$ – Neymar Jun 3 at 6:02
  • $\begingroup$ 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 $\endgroup$ – Neymar Jun 3 at 6:04
  • $\begingroup$ I look forward to more details of your answer. $\endgroup$ – Neymar Jun 3 at 13:42
  • $\begingroup$ I updated the answer. Let me know if anything remains unclear. $\endgroup$ – Mateusz Kwaśnicki Jun 7 at 9:51

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