Let $k , h: \mathbb R_+\to [0,1]$ be non-decreasing and right continuous s.t. $k(t)\le h(t)$ for all $t\ge 0$. Define $\tau_{k}$ (resp. $\tau_h$) by

$$\tau_k : = \inf\{t\ge 0:2+\beta t+ W_t \le k(t)\}\quad \left(\mbox{resp. } \tau_h : = \inf\{t\ge 0:2+\beta t+ W_t \le h(t)\}\right),$$

where $\beta>0$ and $(W_t)_{t\ge 0}$ is a standard Brownian motion. Then $\tau_k\ge \tau_h$ holds by definition. If there exist $t_2>t_1\ge 0$ s.t. $k(t)<h(t)$ for all $t\in [t_1,t_2]$, can we prove

$$\mathbb P[\tau_{k}=\infty]>\mathbb P[\tau_{h}=\infty]?$$

I strongly believe the strict inequality holds, but I'm unable to show it rigorously.

  • $\begingroup$ You also have $k(t)\le a<b\le h(t)$ on a suitable time interval, and the probability of hitting $(a,b)$ during that interval is non-zero. $\endgroup$ Oct 8, 2021 at 17:12
  • $\begingroup$ @ChristianRemling Thanks for the hint. I think I know how to prove the strict inequality $\endgroup$
    – GJC20
    Oct 9, 2021 at 16:44
  • $\begingroup$ @YuvalPeres I think $\lim_{t\to\infty}X^k_t=\infty$ with $X^k_t:=2+\beta t+W_t-k(t)$, while $\mathbb P[\tau_k=\infty]=\mathbb P[X^k_t>0, \forall t\ge 0]>1$ $\endgroup$
    – GJC20
    Oct 9, 2021 at 16:49
  • $\begingroup$ I deleted my comment. $\endgroup$ Oct 9, 2021 at 16:50

1 Answer 1


$\newcommand{\R}{\mathbb{R}}\newcommand{\be}{\beta}\newcommand{\al}{\alpha}\newcommand{\ga}{\gamma}$Let $k_1:=h$, $k_2:=k$, \begin{equation*} g_i(t):=2+\be t-k_i(t), \tag{1} \end{equation*} $B_t:=-W_t$, \begin{equation*} \tau_i:=\inf\{t\ge0\colon B_t\ge g_i(t)\}, \tag{2} \end{equation*} $i\in\{1,2\}$.

It is assumed that $g_1$ and $g_2$ are right-continuous (r.c.), $g_1\le g_2$, and \begin{equation*} g_1(u)<g_2(u) \tag{3} \end{equation*} for some real $u\ge0$. It is also assumed that $\be>0$ and \begin{equation} 0\le k_1,k_2\le1. \tag{3.5} \end{equation}

We want to show that then $P(\tau_1=\infty)<P(\tau_2=\infty)$, that is, \begin{equation*} P(\tau_1<\infty=\tau_2)\overset{\text{(?)}}>0. \tag{4} \end{equation*}

Since $g_1$ and $g_2$ are r.c., (3) implies that for some $w\in(u,\infty)$ and some real $a,b$ we have \begin{equation} g_1<a<b<g_2\text{ on the interval }[u,w]. \tag{5} \end{equation} Take any \begin{equation} v\in(u,w). \tag{6} \end{equation} Consider the events \begin{equation} \begin{aligned} A&:=\{|B_t|<1\ \forall t\in[0,u]\}, \\ B&:=A\cap\{\tau_1\le v\}, \\ C&:=B\cap\{B_t<h_2(t)\ \forall t\in[\tau_1,w]\}, \\ D&:=C\cap\{B_t<1+\be t\ \forall t\ge w\}, \end{aligned} \tag{7} \end{equation} where \begin{equation} h_2(t):=b-c(t-\tau_1),\quad c:=\frac b{w-v}. \tag{8} \end{equation}

For $\al,\ga,T$ in $[0,\infty]$ and $L\in\R$, consider the probabilities \begin{equation} \begin{aligned} p_{\al,\ga,L,T}&:=P\big(-\al+Lt<B_t<\ga+Lt\ \forall t\in(0,T)\big), \\ q_{\ga,L,T}&:=p_{\infty,\ga,L,T}=P\big(B_t<\ga+Lt\ \forall t\in(0,T)\big), \\ q_{\ga,L}&:=q_{\ga,L,\infty}=P\big(B_t<\ga+Lt\ \forall t\in(0,\infty)\big). \end{aligned} \tag{9} \end{equation} Then \begin{equation} \begin{aligned} p_{\al,\ga,L,T}>0&\text{ if }\al>0,\ga>0,T<\infty, \\ q_{\ga,L,T}\in(0,1)&\text{ if }\ga>0,T<\infty, \\ q_{\ga,L}>0&\text{ if }\ga>0,L>0. \end{aligned} \tag{10} \end{equation}

So, \begin{equation} P(A)=p_{1,1,0,u}>0. \tag{11} \end{equation}

By (1) and (3.5), \begin{equation*} g_i(t)\ge1+\be t\ge1 \tag{12} \end{equation*} for $t\ge0$. So, by (5), \begin{equation} b>1>0, \tag{13} \end{equation} and on the event $A$ we have $\tau_1>u$ and hence,
\begin{equation} P(B)=P(A)P(B|A)\ge P(A)(1-q_{1+b,0,v-u})>0, \tag{14} \end{equation} by (5), (10), and (11).

Next, by (8), (13), and (6), $c>0$ and hence $h_2(t)\le b<g_2(t)$ on the event $B$ for all $t\in[\tau_1,w]$, by (5). So, \begin{equation} \tau_2>w\text{ on the event $C$.} \tag{15} \end{equation} Moreover, \begin{equation} P(C)=P(B)P(C|B)\ge P(B)q_{b-a,-c,w-u}>0, \tag{16} \end{equation} by (5), (10), and (14).

Further, on the event $C$ we have $B_w<h_2(w)\le b-c(w-v)=0$, by (8), (7), and the condition $c>0$. Therefore, \begin{equation} P(D)=P(C)P(D|C)\ge P(C)q_{1,\be}>0, \tag{17} \end{equation} by (10) and (16). Also, by (15) and (12), we have $\tau_1<\infty=\tau_2$ on the event $D$.

Thus, \begin{equation*} P(\tau_1<\infty=\tau_2)\ge P(D)>0, \end{equation*} which proves (4). $\quad\Box$

(The condition for $k$ and $h$ to be nondecreasing was not needed or used in this proof.)

The picture below shows my crude rendering of a path (gray) of $(B_t)$ in the event $D$; plus graphs of $g_1$ (dotted, red) and $g_2$ (dotted, blue) corresponding to $\be=0.2$, $h_1(t)=\min(1,\max(0,t-1))$, and $h_2(t)=h_1(t)/2$; horizontal segments at levels $a=1.65$ (red) and $b=1.92$ (blue) over the interval $[u,w]=[1.7,1.9]$; plus two horizontal segments (dotted, black) at levels $-1$ and $1$ over the interval $[0,u]=[0,1.7]$; and the graph of $h_2$ (dotted, black) with slope $-c=-19.2$.

enter image description here

  • $\begingroup$ Nice proof. It appears more elegant than the mine. Thanks a lot $\endgroup$
    – GJC20
    Oct 11, 2021 at 17:45

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