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Let $(f_n)_{n\ge 1}$ be a sequence of non-decreasing and continuous functions defined on $\mathbb R_+$ and taking values in $[0,1]$. Further, for each $t\ge 0$, $n\mapsto f_n(t)$ is non-decreasing. Denote by $f$ the point-wise limit of $(f_n)_{n\ge 1}$ and by $W$ a standard Brownian motion. For every $x>0$ (saying $x=1$), can we prove that the following convergence

$$\lim_{n\to\infty}\mathbb P[\inf_{0\le s\le t}(x+W_s-f_n(s))\le 0] = \mathbb P[\inf_{0\le s\le t}(x+W_s-f(s))\le 0]$$

holds for every $t\ge 0$?

PS : The direction $$\lim_{n\to\infty}\mathbb P[\inf_{0\le s\le t}(W_s-f_n(s))\le -x] \le \mathbb P[\inf_{0\le s\le t}(W_s-f(s))\le -x]$$ is trivial. For the other direction, my idea is as follows : Denote by $\tau_n$ (resp. $\tau$) the first hitting time of $W-f_n$ (resp. $W-f$) at $-x$. Then

$$\mathbb P[\inf_{0\le s\le t}(W_s-f(s))\le -x]-\mathbb P[\inf_{0\le s\le t}(W_s-f_n(s))\le -x] = \mathbb P[\tau\le t, \tau_n>t]$$

as $\tau\le \tau_n$. Rewrite

\begin{eqnarray} \mathbb P[\tau\le t, \tau_n>t] &=& \int_{[0,t]}\mathbb P[\tau_n>t|\tau=s]\mathbb P[\tau\in ds]\\ &=&\int_{[0,t)}\mathbb P[\tau_n>t|\tau=s]\mathbb P[\tau\in ds] + \mathbb P[\tau_n>t|\tau=t]\mathbb P[\tau=t]. \end{eqnarray}

How to estimate $\mathbb P[\tau_n>t|\tau=s]$? Is $\mathbb P[\tau=t]=0$?

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  • $\begingroup$ Something seems wrong with the set-up. Already $W_0=0$, and $f_n(0)\geq 0$, $f(0)\geq 0$, so all the probabilities are $1$. $\endgroup$ Jun 28, 2022 at 18:13
  • $\begingroup$ the infimum has continuity properties so you can probably just use usual measure continuity williewong.wordpress.com/2011/11/01/continuity-of-the-infimum. Otherwise, try some version of dominated/monotone convergence. $\endgroup$ Jun 28, 2022 at 18:25
  • $\begingroup$ @JamesMartin Thanks for pointing out this mistake. Yes, this is a trivial situation and I have made the modification $\endgroup$
    – user478657
    Jun 29, 2022 at 8:06
  • $\begingroup$ @ThomasKojar Thanks for the comment. Do you mind to specify a bit more? $\endgroup$
    – user478657
    Jun 29, 2022 at 8:34

1 Answer 1

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Denote by $\tau_n$ (resp. $\tau$) the first hitting time of $W-f_n$ (resp. $W-f$) at $-x$. Then $$\mathbb P[\inf_{0\le s\le t}(W_s-f_n(s))\le -x]=\mathbb P[\tau_n\le t],\quad \mathbb P[\inf_{0\le s\le t}(W_s-f(s))\le -x]=\mathbb P[\tau\le t].$$ One has by assumption $\tau\le \tau_n$ and thus $$ 0\le \mathbb P[\tau\le t]-\mathbb P[\tau_n\le t] =\mathbb P[\tau\le t, \tau_n>t] = \int_{(0,t]}\mathbb P[\tau_n>t|\tau=s]\mathbb P[\tau\in ds], $$ where the last equality holds as $\tau>0$. Restricted on the set $\{\tau=s\}$ for $s\in (0, t]$, one has

$$\{\tau_n>t\}\subset \{W_s-f_n(s)> -x\}\cap \{W_s-f(s)\le -x\}.$$

This gives

$$\mathbb P[\tau_n>t|\tau=s] \le \mathbb P[f(s)-x<W_s \le f_n(s)-x] \to 0 \mbox{ as } n\to\infty,$$

and we may conclude by the dominated convergence theorem.

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