Let $\ell^n: [0,\infty)\to [0,1]$ be right-continuous and increasing functions s.t. $\ell^n(0)=0$. Given $x>0$ and Brownian motion $(B_t)_{t\ge 0}$, can we prove $$\limsup_{n\to\infty}\mathbb P[\exists s\in [0,t]:~ x+B_s\le \ell^n(s)]\le \mathbb P[\exists s\in [0,t]:~ x+B_s\le \limsup_{n\to\infty}\ell^n(s)],\quad \forall t>0?$$

PS : It appears that the pathwise inequality $$\limsup_{n\to\infty} {\bf 1}_{\{\exists s\in [0,t]: x+B_s\le \ell^n(s)\}}\le {\bf 1}_{\{\exists s\in [0,t]: x+B_s\le \limsup_{n\to\infty}\ell^n(s)\}}$$ does not hold.

  • $\begingroup$ The case $\ell^n $ be a series of constants , and looking at $\limsup_{n\to\infty} {\bf 1}_{\{ B_1\le \ell^n\}} $ simple, but perhaps something important is lost. $\endgroup$
    – mike
    Aug 5 at 10:22
  • $\begingroup$ @mike If $\ell^n$ are constant, then the reflection principle yields the desired result by straightforward computation. Even $\ell^n$ are equicontinuous, the inequality above can be derived $\endgroup$
    – Neymar
    Aug 5 at 11:10

$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$This is not quite obvious, and it has hardly anything to do with the reverse Fatou lemma.

Indeed, for all $s\in[0,t]$, let \begin{equation*} l_n(s):=\sup_{m\colon m\ge n}\ell^m(s), \end{equation*} so that \begin{equation*} \ell^n(s)\le l_n(s)\downarrow l(s):=\limsup_n\ell^n(s). \tag{0} \end{equation*} So, \begin{align*} &\limsup_n P(\exists s\in[0,t]\ x+B_s\le\ell^n(s)) \\ \le &\limsup_n P(\exists s\in[0,t]\ x+B_s\le l_n(s)) \\ =&\lim_n P(\exists s\in[0,t]\ x+B_s\le l_n(s))=P(A), \end{align*} where \begin{align*} A&:=\{\forall n\ \exists s\in[0,t]\ x+B_s\le l_n(s)\} \\ & =\{\forall n\ge m\ \exists s\in[0,t]\ x+B_s\le l_n(s)\}; \end{align*} here and in what follows, $m$ is any natural number. So, it is enough to show that \begin{equation*} P(A)\overset{\text{(?)}}\le P(C), \tag{1} \end{equation*} where \begin{equation*} C:=\{\exists s\in[0,t]\ x+B_s\le l(s)\}. \end{equation*} We shall actually show that \begin{equation*} P(A\setminus C)\overset{\text{(?)}}=0, \tag{2} \end{equation*} which will of course imply (1).

Suppose that event $A$ occurs. For all $n$, let
\begin{equation*} s_n:=\inf\{s\in[0,t]\colon x+B_s\le l_n(s)\}; \end{equation*} of course, $s_n$ is a random variable (r.v., with values in $[0,t]$ on $A$), depending on the random path of the Brownian motion $(B_t)$. Moreover, since $l_n(s)\downarrow l(s)$ for all $s\in[0,t]$, we have \begin{equation*} s_n\uparrow s_* \end{equation*} for some r.v. $s_*$, with values in $[0,t]$ on $A$.

Consider first the case when $A$ occurs and $s_n<s_*$ for all $n$. Then for all $n$ there is some $t_n\in[s_n,s_*)$ such that $x+B_{t_n}\le l_n(t_n)$. Also, $l_n(s)$ is nondecreasing in $s\in[0,t]$. So, for all $n$, we have $x+B_{t_n}\le l_n(s_*)$. So, \begin{equation*} x+B_{s_*}=\lim_n(x+B_{t_n})\le\lim_n l_n(s_*)=l(s_*). \end{equation*} Thus, \begin{equation*} A\cap\{\forall n\ s_n<s_*\}\subseteq C. \tag{2.5} \end{equation*}

If $A$ occurs and $s_n=s_*=t$ for some $n$ (and hence for all large enough $n$), then for such $n$ we have $x+B_t\le l_n(t)$ and hence $x+B_t\le l(t)$. Thus, \begin{equation*} A\cap\{\exists n\ s_n=s_*=t\}\subseteq C. \tag{2.75} \end{equation*}

If $A$ occurs and if $s_n=s_*<t$ for some $n$ (and hence for all large enough $n$) and if $s_*$ is a point of discontinuity of the function $l$, then for large enough $n$ we have $x+B_{s_*}=x+B_{s_n}\le l_n(s_*+)$, so that $x+B_{s_*}\le l^+(s_*)$, where $l^+(s):=\lim_n l_n(s+)$. So, \begin{equation*} x+B_d\le l^+(d) \tag{3} \end{equation*} at some point $d\in D$, where $D$ is the set of all points of discontinuity of the nondecreasing function $l$; there are at most countably many such points.
Note that for all $u\in[0,t)$ and all $s\in(u,t]$ we have $l(s)=\lim_n l_n(s)\ge\lim_n l_n(u+)=l^+(u)$. So, $l(s)\ge l^+(u)$ for all $u\in[0,t)$ and all $s\in(u,t]$. Now suppose that $C$ does not occur, so that $x+B_s>l(s)$ for all $s\in[0,t]$ and hence $x+B_s> l^+(d)$ for each $d\in D$ and all $s\in(d,t]$. In view of, say, the (local) law of the iterated logarithm for the Brownian motion, for each $d\in[0,t)$ the event $\{x+B_d\le l^+(d),\ x+B_s>l^+(d)\ \forall s\in(d,t]\}$ has the zero probability. In view of (3) and because the set $D$ is at most countable, we conclude that \begin{equation*} P(A\cap\{\exists n\ s_n=s_*<t,s_*\in D\}\setminus C)=0. \tag{4} \end{equation*}

Suppose finally that $A$ occurs and $s_n=s_*<t$ for some $n$ (and hence for all large enough $n$) and $s_*$ is a point of continuity of the function $l$. Take now any real $\ep>0$. Then there is some real $\de>0$ such that $l(s_*+\de)\le l(s_*)+\ep$. So, for all large enough $n$, \begin{equation*} x+B_{s_*}=x+B_{s_n}\le l_n(s_n+\de)=l_n(s_*+\de)\to l(s_*+\de)\le l(s_*)+\ep. \end{equation*} Letting now $\ep\downarrow0$, we get $x+B_{s_*}\le l(s_*)$. So, \begin{equation*} A\cap\{\exists n\ s_n=s_*<t,s_*\notin D\}\subseteq C. \tag{5} \end{equation*}

Collecting (2.5), (2.75), (4), and (5), we confirm (2), as desired.

  • $\begingroup$ Thank you very kindly for your answer. This is not the first time that I get help from your behalf. Thanks a lot! I think there are two typos in your arguments: $\endgroup$
    – Neymar
    Aug 6 at 12:30
  • $\begingroup$ In the first case, it should be $t_n\in [s_n,s_*)$ instead of $t_n\in (s_n,s_*)$; In the second case, it should be $l_n(t)\to l(t)$ instead of $l_n(t)=l(t)$. Could you have a check such that I accept your solution? $\endgroup$
    – Neymar
    Aug 6 at 12:32
  • $\begingroup$ @Neymar : Thank you for your comments. I have fixed the typos. $\endgroup$ Aug 6 at 13:41
  • $\begingroup$ Thanks again for your help. If you don't mind, may I ask you to take a look at my question arising in previous post mathoverflow.net/questions/398321/… ? $\endgroup$
    – Neymar
    Aug 6 at 13:43
  • $\begingroup$ @Neymar : I saw that question, but don't have good ideas about it. $\endgroup$ Aug 6 at 13:46

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