# Question concerning an inequality on probabilities of hitting times in a paper

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.

• 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.
– mike
Aug 5 at 10:22
• @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 Aug 5 at 11:10


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 $$s_n for all $$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

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 $$s_n=s_* for some $$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_*

Suppose finally that $$A$$ occurs and $$s_n=s_* for some $$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_*

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

• 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: Aug 6 at 12:30
• 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? Aug 6 at 12:32
• @Neymar : Thank you for your comments. I have fixed the typos. Aug 6 at 13:41
• 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/… ? Aug 6 at 13:43
• @Neymar : I saw that question, but don't have good ideas about it. Aug 6 at 13:46