# Comparison of probabilities that drifted Brownian motion never hits barriers

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. $$k(t) for all $$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.

• 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. Oct 8, 2021 at 17:12
• @ChristianRemling Thanks for the hint. I think I know how to prove the strict inequality Oct 9, 2021 at 16:44
• @YuvalPeres I think $\lim_{t\to\infty}X^k_t=\infty$ with $X^k_t:=2+\beta t+W_t-k(t)$, while $\mathbb P[\tau_k=\infty]=\mathbb P[X^k_t>0, \forall t\ge 0]>1$ Oct 9, 2021 at 16:49
• I deleted my comment. Oct 9, 2021 at 16:50


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

We want to show that then $$P(\tau_1=\infty), that is, $$\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 $$$$g_1 Take any $$$$v\in(u,w). \tag{6}$$$$ Consider the events \begin{aligned} A&:=\{|B_t|<1\ \forall t\in[0,u]\}, \\ B&:=A\cap\{\tau_1\le v\}, \\ C&:=B\cap\{B_t where $$$$h_2(t):=b-c(t-\tau_1),\quad c:=\frac b{w-v}. \tag{8}$$$$

For $$\al,\ga,T$$ in $$[0,\infty]$$ and $$L\in\R$$, consider the probabilities \begin{aligned} p_{\al,\ga,L,T}&:=P\big(-\al+Lt Then \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}

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

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

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

Further, on the event $$C$$ we have $$B_w, by (8), (7), and the condition $$c>0$$. Therefore, $$$$P(D)=P(C)P(D|C)\ge P(C)q_{1,\be}>0, \tag{17}$$$$ by (10) and (16). Also, by (15) and (12), we have $$\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 $$-c=-19.2$$.

• Nice proof. It appears more elegant than the mine. Thanks a lot Oct 11, 2021 at 17:45