# Brownian Motion with a Singular Drift

Dante DeBlassie
Adina Oprisan
Robert G. Smits
Department of Mathematical Sciences
New Mexico State University
P. O. Box 30001
Department 3MB
Las Cruces, NM 88003-8001
<EMAIL_ADDRESS>
<EMAIL_ADDRESS>
<EMAIL_ADDRESS>
> We study the effect of a power law drift on Brownian motion in the positive
> half-line, where the order of the drift at $0$ and $\infty$ is different.

_2000 Mathematical Subject Classification_. Primary 60J60, 60J65, 60J55.
Secondary 60F10.

_Key Words and Phrases_. Brownian motion, Brownian motion with drift, half-
line, exit time, stochastic comparison, large deviations, $h$-transform.

_Running Title:_ Brownian motion with a singular drift.

##  1 Introduction

For a real-valued process $X_{t}$, let $\tau_{R}(X)$ be the first time the
process hits $R$. Specialize $X_{t}$ to be a Brownian motion with a power law
drift: the process satisfying the stochastic differential equation

$\left\\{\begin{array}[]{l}dX_{t}=dB_{t}-\beta X_{t}^{-p}\,dt,\quad
t<\tau_{0}(X)\\\ \vskip 3.0pt plus 1.0pt minus 1.0pt\cr
X_{0}=x>0,\end{array}\right.$

where $\beta\not=0$ and $p>0$.

Note that when $\beta=0$, $X_{t}$ is merely Brownian motion and it is well-
known (Feller 1971) that

$P_{x}(\tau_{0}(X)>t)=\frac{2}{\sqrt{2\pi}}\int_{0}^{x/\sqrt{t}}e^{-u^{2}/2}\,dt.$

When $p=1$, it is known that for $\beta>-\frac{1}{2}$,

$P_{x}(\tau_{0}(X)>t)=\frac{2^{1/2-\beta}}{\Gamma(1/2+\beta)}\int_{0}^{x/\sqrt{t}}u^{2\beta}e^{-u^{2}/2}\,dt,$

and for $\beta\leq-\frac{1}{2}$,

$P_{x}(\tau_{0}(X)=\infty)=1$

(Göing-Jaeschke and Yor (2003)). Using Theorem 1.1 of Chapter 5 in Pinsky
(1995), it is not hard to show if $\beta<0$, then

$\left\\{\begin{array}[]{ll}P_{x}(\tau_{0}(X)=\infty)=1,&\quad\text{for
$p>1$}\\\ &\\\ 0<P_{x}(\tau_{0}(X)=\infty)<1,&\quad\text{for
$p<1$.}\end{array}\right.$

DeBlassie and Smits (2009) proved the following results for $\beta>0$.

  * •

For $p>1$,

$\left\\{\begin{array}[]{ll}E_{x}[\tau_{0}(X)^{q}]<\infty,&\quad\text{if
$q<1/2$,}\\\ &\\\ E_{x}[\tau_{0}(X)^{q}]=\infty,&\quad\text{if
$q>1/2$}.\end{array}\right.$

  * •

For $p<1$,

$\lim_{t\to\infty}t^{-(1-p)/(1+p)}P_{x}(\tau_{0}(X)>t)=-\gamma(p,\beta),$

where

$\gamma(p,\beta)=\tfrac{1}{2}p^{-2p/(1+p)}\beta^{2/(1+p)}\left[B\left(\tfrac{1}{2},\tfrac{1-p}{2p}\right)+B\left(\tfrac{3}{2},\tfrac{1-p}{2p}\right)\right]B\left(\tfrac{1}{2},\tfrac{1-p}{2p}\right)^{-(1-p)/(1+p)}.$

In particular, there is a phase transition of sorts as the power $p$ passes
through the value $1$: roughly power law behavior (with the same power) of the
tail versus subexponential behavior (with a different power).

For the last case of $\beta>0$ and $0<p<1$, it is not hard to show (we
indicate the argument in the next section) that so long as the form of the
drift is $\beta x^{-p}$ for large values of $x$, one can change the drift to
be of the form $\alpha x^{-q}$, $0<q<1,$ for small values of $x$, keeping it
bounded in between, and still have the process hit $0$ almost surely. This
holds even if the multiplier $\alpha$ is negative. A natural question
suggested by this is to determine how much effect this change has on the
asymptotic behavior of the time to hit $0$. We were surprised to learn that
the behavior near the origin does not influence the time to hit $0$, at least
on the logarithmic level.

We now state our results precisely. Consider the diffusion $X_{t}$ given by

(1.1) $\left\\{\begin{array}[]{l}dX_{t}=dB_{t}+b(X_{t})\,dt\\\ \vskip 3.0pt
plus 1.0pt minus 1.0pt\cr X_{0}=x>0,\end{array}\right.$

where $B_{t}$ is one-dimensional Brownian motion and for some $0<M_{1}<M_{2}$,
$\alpha\in\mathbb{R}$, $\beta>0$ and $0<p,q<1$,

(1.2) $b(x)=\left\\{\begin{array}[]{ll}-\alpha x^{-q},&\qquad 0<x\leq M_{1}\\\
\text{bounded measurable},&\qquad M_{1}<x<M_{2}\\\ -\beta x^{-p},&\qquad
M_{2}\leq x.\end{array}\right.$

By standard facts, the law of $X_{t}$ on the space of continuous paths in
$(0,\infty)$ exists uniquely up to an explosion time. Given $a\in[0,\infty]$,
let

$\tau_{a}(X)=\inf\\{t>0:X_{t}=a\\}.$

Below, we will show the explosion time is finite:

(1.3) $P_{x}(\tau_{0}(X)<\infty)=1.$

Here is our main theorem.

######  Theorem 1.1.

Under the condition (1.2), the solution $X_{t}$ of (1.1) satisfies

$\lim_{t\to\infty}t^{-(1-p)/(1+p)}\log
P_{x}(\tau_{0}(X)>t)=-\,\gamma(p,\beta),$

where

(1.4)
$\gamma(p,\beta)=\tfrac{1}{2}p^{-2p/(1+p)}\beta^{2/(1+p)}\left[B\left(\tfrac{1}{2},\tfrac{1-p}{2p}\right)+B\left(\tfrac{3}{2},\tfrac{1-p}{2p}\right)\right]B\left(\tfrac{1}{2},\tfrac{1-p}{2p}\right)^{-(1-p)/(1+p)}.$

Our result extends to drifts of the form $\alpha(x)x^{q}\ell_{1}(x)$ near $0$
and $\beta(x)x^{p}\ell_{2}(x)$ near $\infty$, where $\ell_{1}$ and $\ell_{2}$
are slowly varying at $0$ and $\infty$, respectively, and the the coefficients
$\alpha(x)$ and $\beta(x)$ have limiting values $\alpha_{0}$ and
$\beta_{0}>0$. In the context of regular variation, one would say
$x^{q}\ell_{1}(x)$ is regularly varying at $0$ from the right, with index $q$,
and $x^{p}\ell_{2}(x)$ is regularly varying at infinity, with index $p$.

To be precise, a measurable function $\ell>0$ is slowly varying at infinity if
for each $\lambda>0$,

$\ell(\lambda x)/\ell(x)\to 1,\text{ as $x\to\infty$.}$

Similarly, measurable function $\ell>0$ is slowly varying at zero from the
right if for each $\lambda>0$,

$\ell(\lambda x)/\ell(x)\to 1,\text{ as $x\to 0^{+}$.}$

######  Theorem 1.2.

Suppose $\alpha:(0,\infty)\to\mathbb{R}$ and $\beta:(0,\infty)\to(0,\infty)$
are continuous and $p,q\in(0,1)$. Assume:

  * •

$\alpha(x)$ is bounded in a neighborhood of $0$ and
$\lim_{x\to\infty}\beta(x)=\beta_{0}>0$ exists;

  * •

$\ell_{1}$ is continuous and slowly varying at zero from the right;

  * •

$\ell_{2}$ is continuous and slowly varying at infinity.

Suppose for some $0<M_{1}<M_{2}$,

(1.5) $b(x)=\left\\{\begin{array}[]{ll}-\alpha(x)x^{q}\ell_{1}(x),&\qquad
0<x\leq M_{1}\\\ \text{\rm bounded measurable},&\qquad M_{1}<x<M_{2}\\\
-\beta(x)x^{p}\ell_{2}(x),&\qquad M_{2}\leq x.\end{array}\right.$

Then the solution $X_{t}$ of (1.1) satisfies

$\lim_{t\to\infty}t^{-(1-p)/(1+p)}\log
P_{x}(\tau_{0}(X)>t)=-\,\gamma(p,\beta_{0}),$

where $\gamma(p,\beta_{0})$ is from (1.4).

In the sequel, for a stochastic process $Z_{t}$ in $(0,\infty)$, for
$0<\varepsilon<M$, we will write

$\tau_{\varepsilon,M}(Z)=\tau_{\varepsilon}(Z)\wedge\tau_{M}(Z).$

##  2 Preliminaries

First we prove zero is hit almost surely.

######  Lemma 2.1.

For $X_{t}$ as in (1.1) with $b$ as in (1.2),

$P_{x}(\tau_{0}(X)<\infty)=1\text{ for all $x>0$.}$

This will be a consequence of the following special case.

######  Lemma 2.2.

For $X_{t}$ from (1.1), where $b$ from (1.2) is continuous,

$P_{x}(\tau_{0}(X)<\infty)=1\text{ for all $x>0$.}$

###### Proof.

This is a consequence of the proof of a very general result (part (ii) in
Theorem 1.1 of Chapter 5) in Pinsky (1995). It is a bit cumbersome to verify
the hypotheses there, so we will present the proof in our very special case.
Let

$L=\frac{1}{2}\frac{d^{2}}{dx^{2}}+b(x)\frac{d}{dx}$

be the differential operator corresponding to $X_{t}$. For any
$0<r_{1}<r_{2}$, write

$\tau_{r_{1},r_{2}}=\tau_{r_{1},r_{2}}(X).$

By Theorem 1.1 in Chapter 2 of Pinsky, $\tau_{r_{1},r_{2}}<\infty$ a.s.
$P_{x}$ for any $x\in(r_{1},r_{2})$. Define

$f(x)=\int_{0}^{x}\exp\left(-2\int_{0}^{z}b(y)\,dy\right)\,dz,\quad x>0$

(note the inner integral is finite because $0<q<1$). Then $f\in
C^{2}(0,\infty)$ and $Lf=0$. By Itô’s formula,

$E_{x}[f(X(\tau_{r_{1},r_{2}}))]=f(x).$

This implies

(2.1)
$P_{x}(\tau_{r_{1}}<\tau_{r_{2}})=\frac{f(x)-f(r_{2})}{f(r_{1})-f(r_{2})}.$

The key fact is that for large $r_{2}$,

$\displaystyle f(r_{2})$
$\displaystyle\geq\int_{M_{2}}^{r_{2}}\exp\left(-2\int_{0}^{z}b(y)\,dy\right)\,dz$
$\displaystyle=\int_{M_{2}}^{r_{2}}\exp\left(-2\int_{0}^{M_{2}}b(y)\,dy-2\int_{M_{2}}^{z}b(y)\,dy\right)\,dz$
$\displaystyle=\exp\left(-2\int_{0}^{M_{2}}b(y)\,dy\right)\int_{M_{2}}^{r_{2}}\exp\left(-2\int_{M_{2}}^{z}b(y)\,dy\right)\,dz$
$\displaystyle=C\int_{M_{2}}^{r_{2}}\exp\left(2\beta\int_{M_{2}}^{z}y^{-p}\,dy\right)\,dz$
(where $C<\infty$, since $b$ is integrable on any interval $(0,M)$)
$\displaystyle=C\int_{M_{2}}^{r_{2}}\exp\left(\tfrac{2\beta}{1-p}\left(z^{1-p}-M_{2}^{1-p}\right)\right)\,dz.$

Since $1-p$ and $\tfrac{2\beta}{1-p}$ are both positive, we see
$f(r_{2})\to\infty$ as $r_{2}\to\infty$. Using this in (2.1) gives
$P_{x}(\tau_{r_{1}}<\infty)=1$, and then letting $r_{1}\downarrow 0$ gives
$P_{x}(\tau_{0}<\infty)=1$, as claimed. ∎

Proof of Lemma 2.1. Given $b$ as in (1.2), since $b$ is bounded on
$[M_{1},M_{2}]$, we can choose $\widetilde{M_{1}}<M_{1}$ and
$\widetilde{M_{2}}>M_{2}$, along with a corresponding continuous $\tilde{b}$
satisfying (1.2) for $\widetilde{M_{1}}$, $\widetilde{M_{2}}$, such that
$b\leq\tilde{b}$ on $(0,\infty)$. Then for
$d\widetilde{X}_{t}=dB_{t}+\tilde{b}(\widetilde{X}_{t})\,dt$,
$\widetilde{X}_{0}=x$, by the Comparision Theorem for SDEs (Ikeda and Watanabe
(1989), Theorem 1.1 in Chapter VI), $X_{t}\leq\widetilde{X}_{t}$ almost surely
and so

$1=P_{x}(\tau_{0}(\widetilde{X})<\infty)\leq P_{x}(\tau_{0}(X)<\infty),$

and so $P_{x}(\tau_{0}(X)<\infty)=1$, as claimed. $\square$

The next results pertain to the case when $b\in C^{3}(0,\infty)$.

######  Lemma 2.3.

Suppose $b$ from (1.2) is in $C^{1}$. Then for

(2.2) $h(x)=\exp\left(\int_{0}^{x}b(y)\,dy\right),\,x>0,$

(2.3) $h(x)$ is bounded on $[M_{1},M_{2}]$;

(2.4) $h(x)$ is bounded below away from $0$ on $[M_{1},M_{2}]$;

(2.5) $h(x)=\exp\left(-\tfrac{\alpha}{1-q}x^{1-q}\right)\text{on
$(0,M_{1}]$;}$

for some positive constant $C$,

(2.6) $h(x)=C\exp\left(-\tfrac{\beta}{1-p}x^{1-p}\right),\,x>M_{2}.$

###### Proof.

For some constants $C_{1}$ and $C_{2}$,

$\displaystyle\int_{0}^{x}b(y)\,dy$
$\displaystyle=-\tfrac{\alpha}{1-q}x^{1-q}I_{(0,M_{1}]}(x)+\left[C_{1}+\int_{M_{1}}^{x}b(y)\,dy\right]I_{(M_{1},M_{2})}(x)$
(2.7)
$\displaystyle+\left[C_{2}-\tfrac{\beta}{1-p}x^{1-p}\right]I_{[M_{2},\infty)}(x).$

The assertions of the Lemma follow immediately from this. ∎

######  Lemma 2.4.

Suppose $b$ from (1.2) is in $C^{1}$ and $\alpha\geq 0$. Then for

(2.8) $V=-\tfrac{1}{2}(b^{2}+b^{\prime}),$

we have $V\leq VI_{(M_{1},\infty)}$.

###### Proof.

Since $\alpha\geq 0$ and $0<q<1$, on the interval $(0,M_{1}]$ we have

$V(x)=-\frac{\alpha^{2}}{2x^{2q}}-\frac{\alpha q}{2x^{q+1}}\leq 0.$

The conclusion is immediate. ∎

The next result we will need follows from the formula (1.1) in the proof of
Theorem 1.1 from Section 4.1 in the book by Pinsky (1995). To satisfy the
hypotheses he requires, we take the drift to be in $C^{3}$.

######  Lemma 2.5.

Suppose $X_{t}$ is from (1.1) in the introduction, where $X_{0}=x>0$ and $b$
given by (1.2) is in $C^{3}$. Then for $h$ from (2.2) and $V$ from (2.8),

$P_{x}(\tau_{0}(X)>t)=E_{x}\left[\exp\left(\int_{0}^{t}V(B_{s})\,ds\right)h(B_{t})I_{\tau_{0}(B)>t}\right].$

###### Proof.

Let $L=\tfrac{1}{2}\tfrac{d^{2}}{dx^{2}}+V$. Note $h$ is positive and
$L$-harmonic on $(0,\infty)$. The $h$-transform $L^{h}$ of $L$ is defined by

$L^{h}f=\frac{1}{h}L(hf).$

It is easy to show that

$L^{h}=\frac{1}{2}\frac{d^{2}}{dx^{2}}+b\frac{d}{dx}.$

Let $0<\varepsilon<M<\infty$ be such that $x\in(\varepsilon,M)$. Since $V$ is
bounded on $[\varepsilon,M]$, the formula (1.1) in the proof of Theorem 1.1 in
Section 4.1 of Pinsky (1995) holds: for each continuous $f$ with compact
support in $(\varepsilon,M)$,

$\displaystyle\frac{1}{h(x)}E_{x}$
$\displaystyle\left[\exp\left(\int_{0}^{t}V(B_{s})\,ds\right)(hf)(B_{t})I(\tau_{\varepsilon,M}(B)>t)\right]$
$\displaystyle=E_{x}\left[\exp\left(\int_{0}^{t}\frac{Lh}{h}(X_{s})\,ds\right)f(X_{t})I(\tau_{\varepsilon,M}(X)>t)\right]$
$\displaystyle=E_{x}\left[f(X_{t})I(\tau_{\varepsilon,M}(X)>t)\right],$

since $Lh=0$. From this we get

$\frac{1}{h(x)}E_{x}\left[\exp\left(\int_{0}^{t}V(B_{s})\,ds\right)h(B_{t})I(\tau_{\varepsilon,M}(B)>t)\right]=P_{x}(\tau_{\varepsilon,M}(X)>t).$

By Monotone Convergence and Lemma 2.1, we can let $\varepsilon\downarrow 0$
and $M\uparrow\infty$ to get

$\frac{1}{h(x)}E_{x}\left[\exp\left(\int_{0}^{t}V(B_{s})\,ds\right)h(B_{t})I_{\tau_{0}(B)>t}\right]=P_{x}(\tau_{0}(X)>t),$

as claimed. ∎

##  3 Lower Bound

The main result of this section is the following lower bound.

######  Theorem 3.1.

Let $X_{t}$ be as in (1.1), where $b$ is from (1.2). Then

$\liminf_{t\to\infty}t^{-(1-p)/(1+p)}\log
P_{x}(\tau_{0}(X)>t)\geq-\,\gamma(p,\beta),$

where $\gamma(p,\beta)$ is from (1.4).

We first prove a special case.

######  Theorem 3.2.

Let $X_{t}$ be as in (1.1), where $b$ from (1.2) satisfies $\alpha\geq 0$ and
$b\in C^{3}$. Then

$\liminf_{t\to\infty}t^{-(1-p)/(1+p)}\log
P_{x}(\tau_{0}(X)>t)\geq-\,\gamma(p,\beta),$

where $\gamma(p,\beta)$ is from (1.4).

The proof of the theorem is long, so we break it up into pieces. By Lemma 2.5

(3.1)
$P_{x}(\tau_{0}(X)>t)=E_{x}\left[\exp\left(\int_{0}^{t}V(B_{s})\,ds\right)h(B_{t})I_{\tau_{0}(B)>t}\right].$

Define

$K_{0}=\\{\omega\in C_{0}:\int_{0}^{1}|\dot{\omega}_{u}|^{2}\,du<\infty\\}$

and

(3.2)
$F(\omega)=\frac{\beta^{2}}{2}\int_{0}^{1}|\omega_{u}|^{-2p}\,du+\frac{\beta}{1-p}|\omega_{1}|^{1-p}+\frac{1}{2}\int_{0}^{1}\dot{\omega}_{u}^{2}\,du,\quad\omega\in
K_{0}.$

Let $g\in K_{0}$ with $g\geq 0$. Let $\delta>0$ be given and set
$\tilde{g}=g+\delta$,

(3.3) $\varepsilon=t^{-(1-p)/(1+p)},$

and

(3.4) $Z_{u}=B_{u}-\tilde{g}(u)/\sqrt{\varepsilon}.$

For $Z_{0}>0$, we have

(3.5) $\tau_{0}(Z)>1\Longrightarrow\tau_{0}(B)>1$ (3.6)
$\tau_{0}(Z)>1\Longrightarrow
B_{u}\geq\tilde{g}(u)/\sqrt{\varepsilon},\,u\in[0,1].$

Now scale and use (3.5):

$\displaystyle P_{x}(\tau_{0}(X)>t)$
$\displaystyle=E_{x}\left[\exp\left(\vphantom{\int_{0}^{t}}\right.\right.\left.\left.\int_{0}^{t}V(B_{s})\,ds\right)h(B_{t})I_{\tau_{0}(B)>t}\right]$
$\displaystyle=E_{x/\sqrt{t}}\left[\exp\left(t\int_{0}^{1}V(\sqrt{t}B_{u})\,du\right)h(\sqrt{t}B_{1})I_{\tau_{0}(B)>1}\right]$
(3.7) $\displaystyle\geq
E_{x/\sqrt{t}}\left[\exp\left(t\int_{0}^{1}V(\sqrt{t}B_{u})\,du\right)h(\sqrt{t}B_{1})I_{\tau_{0}(Z)>1}\right].$

######  Lemma 3.3.

For $\tau_{0}(Z)>1$ and $\varepsilon$ from (3.3),

$t\int_{0}^{1}V(\sqrt{t}B_{u})\left[I_{\sqrt{t}B_{u}<M_{1}}+I_{M_{2}<\sqrt{t}B_{u}}\right]\,du\geq
G_{1}(\tilde{g},\varepsilon),$

where

$\lim_{\varepsilon\to
0}\varepsilon\,G_{1}(\tilde{g},\varepsilon)=-\frac{\beta^{2}}{2}\int_{0}^{1}\tilde{g}(u)^{-2p}\,du.$

###### Proof.

By (2.8),

$V(x)\left[I_{0<x<M_{1}}+I_{M_{2}<x}\right]=-\left(\frac{\alpha^{2}}{2x^{2q}}+\frac{q\alpha}{2x^{q+1}}\right)I_{0<x<M_{1}}-\left(\frac{\beta^{2}}{2x^{2p}}+\frac{p\beta}{2x^{p+1}}\right)I_{M_{2}<x}.$

This is nonpositive and increasing, since $\alpha\geq 0$ and $\beta>0$. By
(3.6), for $u\in[0,1]$, $B_{u}\geq\tfrac{\tilde{g}(u)}{\sqrt{\varepsilon}}$,
and so $V\left(\sqrt{t}B_{u}\right)\geq
V\left(\sqrt{\tfrac{t}{\varepsilon}}\,\tilde{g}(u)\right)$. Thus for
$\tau_{0}(Z)>1$,

$\displaystyle t\int_{0}^{1}V(\sqrt{t}B_{u})$
$\displaystyle\left[I_{\sqrt{t}B_{u}<M_{1}}+I_{M_{2}<\sqrt{t}B_{u}}\right]\,du\geq
t\int_{0}^{1}V\left(\sqrt{\tfrac{t}{\varepsilon}}\,\tilde{g}(u)\right)\left[I_{\sqrt{t}B_{u}<M_{1}}+I_{M_{2}<\sqrt{t}B_{u}}\right]\,du$
$\displaystyle=-t\int_{0}^{1}\left[\frac{\alpha^{2}}{2}\left(\sqrt{\tfrac{t}{\varepsilon}}\,\tilde{g}(u)\right)^{-2q}+\frac{q\alpha}{2}\left(\sqrt{\tfrac{t}{\varepsilon}}\,\tilde{g}(u)\right)^{-q-1}\right]I_{\sqrt{t}B_{u}<M_{1}}\,du$
$\displaystyle\qquad-t\int_{0}^{1}\left[\frac{\beta^{2}}{2}\left(\sqrt{\tfrac{t}{\varepsilon}}\,\tilde{g}(u)\right)^{-2p}+\frac{p\beta}{2}\left(\sqrt{\tfrac{t}{\varepsilon}}\,\tilde{g}(u)\right)^{-p-1}\right]I_{M_{2}<\sqrt{t}B_{u}}\,du$
$\displaystyle\geq-t\int_{0}^{1}\left[\frac{\alpha^{2}}{2}\left(\sqrt{\tfrac{t}{\varepsilon}}\,\tilde{g}(u)\right)^{-2q}+\frac{q\alpha}{2}\left(\sqrt{\tfrac{t}{\varepsilon}}\,\tilde{g}(u)\right)^{-q-1}\right]I_{\sqrt{t/\varepsilon}\tilde{g}(u)<M_{1}}\,du$
$\displaystyle\qquad-t\int_{0}^{1}\left[\frac{\beta^{2}}{2}\left(\sqrt{\tfrac{t}{\varepsilon}}\,\tilde{g}(u)\right)^{-2p}+\frac{p\beta}{2}\left(\sqrt{\tfrac{t}{\varepsilon}}\,\tilde{g}(u)\right)^{-p-1}\right]\,du$
(since $I_{M_{2}<\sqrt{t}B_{u}}\leq 1$ and since $\tau_{0}(Z)>1$ implies
$I_{\sqrt{t}B_{u}<M_{1}}=I_{\sqrt{t}(Z_{u}+\tilde{g}(u)/\sqrt{\varepsilon})<M_{1}}\leq
I_{\sqrt{t/\varepsilon}\tilde{g}(u)<M_{1}}$ for $u\in[0,1]$)
$\displaystyle=-\frac{1}{2}\int_{0}^{1}\left[\alpha^{2}\varepsilon^{(2q-p-1)/(1-p)}\tilde{g}(u)^{-2q}+q\alpha\varepsilon^{(q-p)/(1-p)}\tilde{g}(u)^{-q-1}\right]I_{\tilde{g}(u)<M_{1}\varepsilon^{1/(1-p)}}\,du$
(3.8)
$\displaystyle\qquad-\frac{1}{2}\int_{0}^{1}\left[\beta^{2}\varepsilon^{-1}\tilde{g}(u)^{-2p}+p\beta\tilde{g}(u)^{-p-1}\right]\,du,$

where we have used (3.3) to replace the variable $t$ in terms of the variable
$\varepsilon$. Define

$G_{1}(\tilde{g},\varepsilon)=\text{RHS\eqref{eq3.8}.}$

Then

$\displaystyle\varepsilon\,G_{1}(\tilde{g},\varepsilon)$
$\displaystyle=-\frac{1}{2}\int_{0}^{1}\left[\alpha^{2}\varepsilon^{2(q-p)/(1-p)}\tilde{g}(u)^{-2q}+q\alpha\varepsilon^{(1+q-2p)/(1-p)}\tilde{g}(u)^{-q-1}\right]I_{\tilde{g}(u)<M_{1}\varepsilon^{1/(1-p)}}\,du$
(3.9)
$\displaystyle\qquad-\frac{1}{2}\int_{0}^{1}\left[\beta^{2}\tilde{g}(u)^{-2p}+p\beta\varepsilon\tilde{g}(u)^{-p-1}\right]\,du.$

Now if $\varepsilon<(\delta/M_{1})^{1-p}$, then for any $u\in[0,1]$,
$\tilde{g}(u)\geq\delta>M_{1}\varepsilon^{1/(1-p)}$. Thus the first integral
in (3) is $0$. The integrand in the second integral is bounded because
$\tilde{g}\geq\delta$ and $p>0$. It follows that

$\lim_{\varepsilon\to
0}\varepsilon\,G_{1}(\tilde{g},\varepsilon)=-\frac{1}{2}\int_{0}^{1}\beta^{2}\tilde{g}(u)^{-2p}\,du,$

as claimed. ∎

Having studied

$t\int_{0}^{1}V(\sqrt{t}B_{u})\left[I_{\sqrt{t}B_{u}<M_{1}}+I_{M_{2}<\sqrt{t}B_{u}}\right]\,du,$

we now turn attention to

$t\int_{0}^{1}V(\sqrt{t}B_{u})\left[I_{M_{1}\leq\sqrt{t}B_{u}\leq
M_{2}}\right]\,du.$

######  Lemma 3.4.

For any positive $C_{1}$ and $\gamma$, as $\varepsilon\to 0$,

$\displaystyle P_{x/\sqrt{t}}$
$\displaystyle\left(C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\,\tilde{g}(u)<M_{2}}\,du+\frac{\beta}{1-p}\left(\sqrt{t}B_{1}\right)^{1-p}+\frac{1}{\sqrt{\varepsilon}}\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}>\frac{\gamma}{\varepsilon},\,\tau_{0}(B)>1\right)$
$\displaystyle\qquad=o\left(P_{x/\sqrt{t}}\left(\tau_{0}(B)>1\right)\right).$

###### Proof.

Since $\tau_{0}(B)>1$ implies $B_{u}>0$ for $u\in[0,1]$,

$\displaystyle P_{x/\sqrt{t}}$
$\displaystyle\left(C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\,\tilde{g}(u)<M_{2}}\,du>\frac{\gamma}{3\varepsilon},\,\tau_{0}(B)>1\right)$
$\displaystyle\leq
e^{-\gamma/3\varepsilon}E_{x/\sqrt{t}}\left[\exp\left(C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\,\tilde{g}(u)<M_{2}}\,du\right)I_{\tau_{0}(B)>1}\right]$
$\displaystyle\leq
e^{-\gamma/3\varepsilon}E_{x/\sqrt{t}}\left[\exp\left(C_{1}t\int_{0}^{1}I_{\sqrt{t/\varepsilon}\,\tilde{g}(u)<M_{2}}\,du\right)I_{\tau_{0}(B)>1}\right]$
(3.10) $\displaystyle\leq
e^{-\gamma/3\varepsilon}\exp\left(C_{1}t\int_{0}^{1}I_{\sqrt{t/\varepsilon}\,\delta<M_{2}}\,du\right)P_{x/\sqrt{t}}(\tau_{0}(B)>1),$

(since $\tilde{g}\geq\delta$). But by (3.3),

$\dfrac{t}{\varepsilon}=\varepsilon^{-(1+p)/(1-p)-1}\to\infty\text{ as
}\varepsilon\to 0,$

so for small $\varepsilon$, $I_{\sqrt{t/\varepsilon}\,\delta<M_{2}}=0$ and we
get

(3.11)
$P_{x/\sqrt{t}}\left(C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\,\tilde{g}(u)<M_{2}}\,du>\frac{\gamma}{3\varepsilon},\,\tau_{0}(B)>1\right)\leq
e^{-\gamma/3\varepsilon}.$

Next, for some constant $C$ (whose exact value might change from line to
line), independent of $\varepsilon$,

$\displaystyle P_{x/\sqrt{t}}$
$\displaystyle\left(\frac{\beta}{1-p}\left(\sqrt{t}B_{1}\right)^{1-p}>\frac{\gamma}{3\varepsilon}\right)=P_{x/\sqrt{t}}\left(B_{1}>C\left(\frac{\gamma}{\varepsilon}\right)^{1/(1-p)}t^{-1/2}\right)$
$\displaystyle\leq\exp\left(-C\varepsilon^{-1/(1-p)}t^{-1/2}\right)E_{x/\sqrt{t}}\left[e^{B_{1}}\right]$
(3.12)
$\displaystyle=\exp\left(-C\varepsilon^{-1/2}+x\varepsilon^{(1+p)/2(1-p)}+1/2\right),$

using (3.3) to write $t$ in terms of $\varepsilon$.

Finally,

$\displaystyle P_{x/\sqrt{t}}$
$\displaystyle\left(\frac{1}{\sqrt{\varepsilon}}\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}>\frac{\gamma}{3\varepsilon}\right)\leq
e^{-\gamma/3\sqrt{\varepsilon}}\,E_{x/\sqrt{t}}\left[\exp\left(\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}\right)\right]$
(3.13)
$\displaystyle=e^{-\gamma/3\sqrt{\varepsilon}}\,\exp\left(\frac{1}{2}\int_{0}^{1}\tilde{g}^{\prime}(u)^{2}\,du\right),$

since under $P_{x/\sqrt{t}}$, $\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}$ is
normal with mean $0$ and variance $\int_{0}^{1}\tilde{g}^{\prime}(u)^{2}\,du$.
Combining (3.11)–(3), we get

$\displaystyle P_{x/\sqrt{t}}$
$\displaystyle\left(C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\,\tilde{g}(u)<M_{2}}\,du+\frac{\beta}{1-p}\left(\sqrt{t}B_{1}\right)^{1-p}+\frac{1}{\sqrt{\varepsilon}}\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}>\frac{\gamma}{\varepsilon},\,\tau_{0}(B)>1\right)$
$\displaystyle\leq
P_{x/\sqrt{t}}\left(C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\,\tilde{g}(u)<M_{2}}\,du>\frac{\gamma}{3\varepsilon},\,\tau_{0}(B)>1\right)$
$\displaystyle\qquad+P_{x/\sqrt{t}}\left(\frac{\beta}{1-p}\left(\sqrt{t}B_{1}\right)^{1-p}>\frac{\gamma}{3\varepsilon},\,\tau_{0}(B)>1\right)$
$\displaystyle\qquad\qquad+P_{x/\sqrt{t}}\left(\frac{1}{\sqrt{\varepsilon}}\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}>\frac{\gamma}{3\varepsilon},\,\tau_{0}(B)>1\right)$
(3.14) $\displaystyle\leq
e^{-\gamma/3\varepsilon}+C_{2}\exp\left(-C\varepsilon^{-1/2}+x\varepsilon^{(1+p)/2(1-p)}\right)+C_{3}e^{-\gamma/3\sqrt{\varepsilon}},$

where the constants are independent of $\varepsilon$. From Feller (1971), as
$t\to\infty$ (or equivalently, as $\varepsilon\to 0$)

$\displaystyle P_{x/\sqrt{t}}$
$\displaystyle\left(\tau_{0}(B)>1\right)=\frac{2}{\sqrt{2\pi}}\,\int_{0}^{x/\sqrt{t}}e^{-u^{2}/2}\,du\sim\frac{2}{\sqrt{2\pi}}\frac{x}{\sqrt{t}}$
(3.15) $\displaystyle=C_{4}\,\varepsilon^{(1+p)/2(1-p)},$

where $C_{4}$ is independent of $\varepsilon$ and we have used (3.3) to write
$t$ in terms of $\varepsilon$. Combining (3)–(3) yields the conclusion of the
Lemma. ∎

######  Lemma 3.5.

For any $C_{1}>0$ and $\gamma>0$,

$\displaystyle\lim_{\varepsilon\to 0}\varepsilon\,\log P_{x/\sqrt{t}}$
$\displaystyle\left(C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\tilde{g}(u)<M_{2}}\,du+\frac{\beta}{1-p}\left(\sqrt{t}B_{1}\right)^{1-p}\right.$
$\displaystyle\left.\qquad+\frac{1}{\sqrt{\varepsilon}}\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}\leq\frac{\gamma}{\varepsilon},\tau_{0}(B)>1\right)=0.$

###### Proof.

Write

$M=C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\tilde{g}(u)<M_{2}}\,du+\frac{\beta}{1-p}\left(\sqrt{t}B_{1}\right)^{1-p}+\frac{1}{\sqrt{\varepsilon}}\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}.$

Then we want to show

(3.16) $\lim_{\varepsilon\to 0}\varepsilon\,\log
P_{x/\sqrt{t}}\left(M\leq\frac{\gamma}{\varepsilon},\,\tau_{0}(B)>1\right)=0.$

Notice

$\displaystyle P_{x/\sqrt{t}}\left(M\leq\frac{\gamma}{\varepsilon},\right.$
$\displaystyle\left.\,\tau_{0}(B)>1\vphantom{\frac{\gamma}{\varepsilon}}\right)=P_{x/\sqrt{t}}\left(\tau_{0}(B)>1\right)-P_{x/\sqrt{t}}\left(M>\frac{\gamma}{\varepsilon},\,\tau_{0}(B)>1\right)$
$\displaystyle=P_{x/\sqrt{t}}\left(\tau_{0}(B)>1\right)\left[1-P_{x/\sqrt{t}}\left(M>\frac{\gamma}{\varepsilon},\,\tau_{0}(B)>1\right)\bigg{/}P_{x/\sqrt{t}}\left(\tau_{0}(B)>1\right)\right].$

It follows that

$\displaystyle\varepsilon\,\log P_{x/\sqrt{t}}$
$\displaystyle\left(M\leq\frac{\gamma}{\varepsilon},\,\tau_{0}(B)>1\right)=\varepsilon\,\log
P_{x/\sqrt{t}}\left(\tau_{0}(B)>1\right)$
$\displaystyle+\varepsilon\,\log\left[1-P_{x/\sqrt{t}}\left(M>\frac{\gamma}{\varepsilon},\,\tau_{0}(B)>1\right)\bigg{/}P_{x/\sqrt{t}}\left(\tau_{0}(B)>1\right)\right].$

Equation (3.16) follows from this, using Lemma 3.4 and (3). ∎

######  Lemma 3.6.

For any $C_{1}>0$,

$\displaystyle\liminf_{\varepsilon\to 0}\varepsilon\,$ $\displaystyle\log
E_{x/\sqrt{t}}\left[\exp\left(-C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\tilde{g}(u)<M_{2}}\,du-\frac{\beta}{1-p}\left(\sqrt{t}B_{1}\right)^{1-p}\right.\right.$
$\displaystyle\left.\left.-\frac{1}{\sqrt{\varepsilon}}\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}\right)I_{\tau_{0}(B)>1}\right]\geq
0.$

###### Proof.

For any $\gamma>0$,

$\displaystyle\liminf_{\varepsilon\to 0}\varepsilon\,$ $\displaystyle\log
E_{x/\sqrt{t}}\left[\exp\left(-C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\tilde{g}(u)<M_{2}}\,du-\frac{\beta}{1-p}\left(\sqrt{t}B_{1}\right)^{1-p}\right.\right.$
$\displaystyle\left.\left.\qquad-\frac{1}{\sqrt{\varepsilon}}\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}\right)I_{\tau_{0}(B)>1}\right]$
$\displaystyle\geq\liminf_{\varepsilon\to 0}\varepsilon\,\log
E_{x/\sqrt{t}}\left[e^{-\gamma/\varepsilon}I\left(C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\tilde{g}(u)<M_{2}}\,du+\frac{\beta}{1-p}\left(\sqrt{t}B_{1}\right)^{1-p}\right.\right.$
$\displaystyle\left.\left.\qquad+\frac{1}{\sqrt{\varepsilon}}\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}\leq\frac{\gamma}{\varepsilon}\right)I_{\tau_{0}(B)>1}\right]$
$\displaystyle=\liminf_{\varepsilon\to 0}\left[-\gamma+\varepsilon\,\log
P_{x/\sqrt{t}}\left(C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\tilde{g}(u)<M_{2}}\,du+\frac{\beta}{1-p}\left(\sqrt{t}B_{1}\right)^{1-p}\right.\right.$
$\displaystyle\left.\left.\qquad+\frac{1}{\sqrt{\varepsilon}}\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}\leq\frac{\gamma}{\varepsilon},\tau_{0}(B)>1\right)\right]$
$\displaystyle\geq-\,\gamma+0,$

by Lemma 3.5. Let $\gamma\to 0$ to finish. ∎

######  Lemma 3.7.

We have

$\displaystyle\liminf_{\varepsilon\to 0}\varepsilon\,\log E_{x/\sqrt{t}}$
$\displaystyle\left[\exp\left(t\int_{0}^{1}V(\sqrt{t}B_{u})I_{M_{1}<\sqrt{t}B_{u}<M_{2}}\,du\right)h(\sqrt{t}B_{1})I_{\tau_{0}(Z)>1}\right]$
$\displaystyle\geq-\frac{\beta}{1-p}\,\tilde{g}(1)^{1-p}-\frac{1}{2}\int_{0}^{1}\tilde{g}^{\prime}(u)^{2}\,du.$

###### Proof.

Since $V$ is bounded on $[M_{1},M_{2}]$, there is $C_{1}>0$ such that $|V|\leq
C_{1}$ on that interval. Then

$\displaystyle E_{x/\sqrt{t}}$
$\displaystyle\left[\exp\left(t\int_{0}^{1}V(\sqrt{t}B_{u})I_{M_{1}<\sqrt{t}B_{u}<M_{2}}\,du\right)h(\sqrt{t}B_{1})I_{\tau_{0}(Z)>1}\right]$
$\displaystyle\geq
E_{x/\sqrt{t}}\left[\exp\left(t\int_{0}^{1}V(\sqrt{t}B_{u})I_{M_{1}<\sqrt{t}B_{u}<M_{2}}\,du\right)h(\sqrt{t}B_{1})I_{M_{2}<\sqrt{t}B_{1}}I_{\tau_{0}(Z)>1}\right]$
$\displaystyle\geq
CE_{x/\sqrt{t}}\left[\exp\left(-C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}<M_{2}}\,du-\frac{\beta}{1-p}(\sqrt{t}B_{1})^{1-p}\right)I_{M_{2}<\sqrt{t}B_{1}}I_{\tau_{0}(Z)>1}\right]$
(where $C$ is from (2.6) and using that $I_{M_{1}<\sqrt{t}B_{u}<M_{2}}\leq
I_{\sqrt{t}B_{u}<M_{2}}$) $\displaystyle\geq
CE_{x/\sqrt{t}}\left[\exp\left(-C_{1}t\int_{0}^{1}I_{\sqrt{t}Z_{u}+\sqrt{t/\varepsilon}\tilde{g}(u)<M_{2}}\,du-\frac{\beta}{1-p}\left(\sqrt{t}Z_{1}+\sqrt{t/\varepsilon}\tilde{g}(1)\right)^{1-p}\right)\cdot\right.$
$\displaystyle\qquad\left.\vphantom{\int_{0}^{1}}\cdot
I_{M_{2}<\sqrt{t/\varepsilon}\tilde{g}(1)}I_{\tau_{0}(Z)>1}\right]$ (using
(3.4) and that
$I_{M_{2}<\sqrt{t}B_{1}}=I_{M_{2}<\sqrt{t}Z_{1}+\sqrt{t/\varepsilon}\tilde{g}(1)}\geq
I_{M_{2}<\sqrt{t/\varepsilon}\tilde{g}(1)}$ for $\tau_{0}(Z)>1$)
$\displaystyle\geq
CE_{x/\sqrt{t}}\left[\exp\left(-C_{1}t\int_{0}^{1}I_{\sqrt{t}Z_{u}+\sqrt{t/\varepsilon}\tilde{g}(u)<M_{2}}\,du-\frac{\beta}{1-p}\left(\sqrt{t}Z_{1}\right)^{1-p}\right.\right.$
$\displaystyle\qquad\left.\left.-\frac{\beta}{1-p}\left(\sqrt{t/\varepsilon}\,\tilde{g}(1)\right)^{1-p}\right)I_{M_{2}<\sqrt{t/\varepsilon}\tilde{g}(1)}I_{\tau_{0}(Z)>1}\right]$
(using that $(a+b)^{1-p}\leq a^{1-p}+b^{1-p}$ since $0<p<1$)
$\displaystyle=C\exp\left(-\frac{\beta}{1-p}\left(\sqrt{t/\varepsilon}\,\tilde{g}(1)\right)^{1-p}\right)I_{M_{2}<\sqrt{t/\varepsilon}\tilde{g}(1)}\cdot$
$\displaystyle\qquad\cdot
E_{x/\sqrt{t}}\left[\exp\left(-C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\tilde{g}(u)<M_{2}}\,du-\frac{\beta}{1-p}\left(\sqrt{t}B_{1}\right)^{1-p}\right.\right.$
$\displaystyle\qquad\qquad\left.\left.-\frac{1}{\sqrt{\varepsilon}}\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}-\frac{1}{2\varepsilon}\int_{0}^{1}\tilde{g}^{\prime}(u)^{2}\,du\right)I_{\tau_{0}(B)>1}\right]$
(by the Cameron-Martin-Girsanov Theorem)
$\displaystyle=C\exp\left(-\frac{\beta}{1-p}\left(\sqrt{t/\varepsilon}\,\tilde{g}(1)\right)^{1-p}-\frac{1}{2\varepsilon}\int_{0}^{1}\tilde{g}^{\prime}(u)^{2}\,du\right)I_{M_{2}<\sqrt{t/\varepsilon}\tilde{g}(1)}\cdot$
$\displaystyle\qquad\cdot
E_{x/\sqrt{t}}\left[\exp\left(-C_{1}t\int_{0}^{1}I_{\sqrt{t}B_{u}+\sqrt{t/\varepsilon}\tilde{g}(u)<M_{2}}\,du-\frac{\beta}{1-p}\left(\sqrt{t}B_{1}\right)^{1-p}\right.\right.$
(3.17)
$\displaystyle\qquad\qquad\left.\left.-\frac{1}{\sqrt{\varepsilon}}\int_{0}^{1}\tilde{g}^{\prime}(u)\,dB_{u}\right)I_{\tau_{0}(B)>1}\right].$

From this and Lemma 3.6, we get

$\displaystyle\liminf_{\varepsilon\to 0}\varepsilon\,$ $\displaystyle\log
E_{x/\sqrt{t}}\left[\exp\left(t\int_{0}^{1}V(\sqrt{t}B_{u})I_{M_{1}<\sqrt{t}B_{u}<M_{2}}\,du\right)h(\sqrt{t}B_{1})I_{\tau_{0}(Z)>1}\right]$
$\displaystyle\geq\liminf_{\varepsilon\to 0}\varepsilon\,\left(\log
C-\frac{\beta}{1-p}\left(\sqrt{t/\varepsilon}\,\tilde{g}(1)\right)^{1-p}-\frac{1}{2\varepsilon}\int_{0}^{1}\tilde{g}^{\prime}(u)^{2}\,du+\log
I_{M_{2}<\sqrt{t/\varepsilon}\tilde{g}(1)}\right)+0$
$\displaystyle=\liminf_{\varepsilon\to 0}\left(\varepsilon\,\log
C-\frac{\beta}{1-p}\tilde{g}(1)^{1-p}-\frac{1}{2}\int_{0}^{1}\tilde{g}^{\prime}(u)^{2}\,du+\varepsilon\,\log
I_{M_{2}<\varepsilon^{-1/(1+p)}\tilde{g}(1)}\right)$ (using (3.3) to
substitute for $t$ in terms of $\varepsilon$)
$\displaystyle=-\frac{\beta}{1-p}\,\tilde{g}(1)^{1-p}-\frac{1}{2}\int_{0}^{1}\tilde{g}^{\prime}(u)^{2}\,du,$

as desired, where we have used that

$\lim_{\varepsilon\to 0}\,I_{M_{2}<\varepsilon^{-1/(1+p)}\tilde{g}(1)}=1,$

since $\tilde{g}(1)=g(1)+\delta>0$. ∎

Proof of Theorem 3.2. It was shown in DeBlassie and Smits (2007) that

$\gamma(p,\beta)=\inf_{\omega\in K_{0}}F(\omega)=\mathop{\inf_{\omega\in
K_{0}}}_{\omega\geq 0}F(\omega),$

where $F$ is from (3.2) (see the very end of Section 5 and Theorem 6.1 in that
article). Hence to prove Theorem 3.2 we need to show

$\liminf_{t\to\infty}t^{-(1-p)/(1+p)}\log
P_{x}(\tau_{0}(X)>t)\geq-\mathop{\inf_{\omega\in K_{0}}}_{\omega\geq
0}F(\omega).$

Now by (3.3) and (3),

$\displaystyle\liminf_{t\to\infty}\,$ $\displaystyle t^{-(1-p)/(1+p)}\log
P_{x}(\tau_{0}(X)>t)$ $\displaystyle\geq\liminf_{\varepsilon\to
0}\varepsilon\,\log
E_{x/\sqrt{t}}\left[\exp\left(t\int_{0}^{1}V(\sqrt{t}B_{u})\,du\right)h(\sqrt{t}B_{1})I_{\tau_{0}(Z)>1}\right]$
$\displaystyle\geq\liminf_{\varepsilon\to 0}\varepsilon\,\log
E_{x/\sqrt{t}}\left[\exp\left(G_{1}(\tilde{g},\varepsilon)+t\int_{0}^{1}V(\sqrt{t}B_{u})I_{M_{1}<\sqrt{t}B_{u}<M_{2}}\,du\right)h(\sqrt{t}B_{1})I_{\tau_{0}(Z)>1}\right]$
(by Lemma 3.3) $\displaystyle=\liminf_{\varepsilon\to
0}\left[\varepsilon\,G_{1}(\tilde{g},\varepsilon)+\varepsilon\,\log
E_{x/\sqrt{t}}\left[\exp\right.\left(t\int_{0}^{1}V(\sqrt{t}B_{u})I_{M_{1}<\sqrt{t}B_{u}<M_{2}}\,du\right)h(\sqrt{t}B_{1})I_{\tau_{0}(Z)>1}\right]$
$\displaystyle\geq-\frac{\beta^{2}}{2}\int_{0}^{1}\tilde{g}(u)^{-2p}\,du-\frac{\beta}{1-p}\,\tilde{g}(1)^{1-p}-\frac{1}{2}\int_{0}^{1}\tilde{g}^{\prime}(u)^{2}\,du.$
(by Lemmas 3.2 and 3.7).

Recalling $\tilde{g}=g+\delta$, where $\delta>0$ was arbitrary, we can let
$\delta\to 0$ to end up with

$\displaystyle\liminf_{t\to\infty}\,t^{-(1-p)/(1+p)}$ $\displaystyle\log
P_{x}(\tau_{0}(X)>t)\geq-\frac{\beta^{2}}{2}\int_{0}^{1}g(u)^{-2p}\,du-\frac{\beta}{1-p}\,g(1)^{1-p}-\frac{1}{2}\int_{0}^{1}g^{\prime}(u)^{2}\,du.$
$\displaystyle=-F(g)$

Since $g\geq 0$ in $K_{0}$ was arbitrary, the desired lower bound in Theorem
3.2 holds. $\square$

Proof of Theorem 3.1. Since $b$ is bounded below on $[M_{1},M_{2}]$, we can
choose $C<0$ such that $C\leq b$ on $[M_{1},M_{2}]$. Then for some
$x_{1}<M_{1}$ and $x_{2}>M_{2}$, we can choose $\tilde{b}\in C^{3}(0,\infty)$
such that

$\tilde{b}(x)=\left\\{\begin{array}[]{ll}-|\alpha|x^{-q},&\quad
x\in(0,x_{1}]\\\ -\beta x^{-p},&\quad x\in[x_{2},\infty),\end{array}\right.$

and $b\geq\tilde{b}$ on $(0,\infty)$. By the Comparison Theorem, if
$dY_{t}=dB_{t}+\tilde{b}(Y_{t}),\,Y_{0}=x$, then

$P_{x}(\tau_{0}(X)>t)\geq P_{x}(\tau_{0}(Y)>t).$

The desired lower bound follows from Theorem 3.2 applied to $Y$. $\square$

##  4 Upper Bound: A Special $C^{3}$ Case

The main theorem of this section is the following special case. It will be
crucial in proving the general case.

######  Theorem 4.1.

Let $X_{t}$ be from (1.1) where $b$ from (1.2) satisfies $\alpha\geq 0$, $b\in
C^{3}$ and $b^{\prime}\geq 0$ on $(0,\infty)$. Then

$\limsup_{t\to\infty}\log P_{x}(\tau_{0}(X)>t)\leq-\,\gamma(p,\beta),$

where $\gamma(p,\beta)$ is from (1.4).

###### Proof.

Let $h$ be from (2.2) and $V$ from (2.8). By Lemma 2.4 and our hypotheses that
$b^{\prime}\geq 0$ and $\alpha\geq 0$,

$\displaystyle V(x)$ $\displaystyle\leq V(x)I_{(M_{1},\infty)}(x)$
$\displaystyle=V(x)I_{[M_{2},\infty)}(x)-\tfrac{1}{2}\left(b^{2}(x)+b^{\prime}(x)\right)I_{(M_{1},M_{2})}(x)$
(4.1) $\displaystyle\leq V(x)I_{[M_{2},\infty)}(x).$

By Lemma 2.3, since $\beta>0$,

(4.2) $h(x)\leq C\exp\left(-\tfrac{\beta}{1-p}x^{1-p}\right),\quad x>0.$

By Lemma 2.5, (4), scaling and translation,

$\displaystyle P_{x}(\tau_{0}(X)>t)$
$\displaystyle=E_{x}\left[\exp\left(\int_{0}^{t}V(B_{s})\,ds\right)h(B_{t})I_{\tau_{0}(B)>t}\right]$
$\displaystyle\leq
E_{x}\left[\exp\left(\int_{0}^{t}(VI_{[M_{2},\infty)}(B_{s})\,ds\right)h(B_{t})I_{\tau_{0}(B)>t}\right]$
$\displaystyle=E_{0}\left[\exp\left(t\int_{0}^{1}(VI_{[M_{2},\infty)}(\sqrt{t}(B_{u}+x/\sqrt{t}))\,du\right)\cdot\right.$
(4.3) $\displaystyle\left.\vphantom{\int_{0}^{t}}\hskip 54.2025pt\cdot
h(\sqrt{t}(B_{1}+x/\sqrt{t}))I(\tau_{0}(B+x/\sqrt{t})>1)\right].$

Writing $\varepsilon=t^{-(1-p)/(1+p)}$ and
$x_{\varepsilon}=x\varepsilon^{1/(1-p)}$, we have

$\sqrt{t}(B_{u}+x/\sqrt{t})=\sqrt{t}B_{u}+x=\varepsilon^{-1/(1-p)}(\varepsilon^{1/2}B_{u}+x_{\varepsilon}).$

By (2.8) and (1.2),

$\displaystyle VI_{[M_{2},\infty)}(ax)$
$\displaystyle=-\tfrac{1}{2}\left(\beta^{2}a^{-2p}x^{-2p}+p\beta
a^{-p-1}x^{-p-1}\right)I_{[a^{-1}M_{2},\infty)}(x)$
$\displaystyle\leq-\tfrac{1}{2}\beta^{2}a^{-2p}x^{-2p}\,I_{[a^{-1}M_{2},\infty)}(x).$

Thus

$\displaystyle(VI_{[M_{2},\infty)})$
$\displaystyle(\sqrt{t}B_{u}+x)=(VI_{[M_{2},\infty)})(\varepsilon^{-1/(1-p)}(\varepsilon^{1/2}B_{u}+x_{\varepsilon}))$
(4.4)
$\displaystyle\leq-\tfrac{1}{2}\beta^{2}\varepsilon^{2p/(1-p)}(\varepsilon^{1/2}B_{u}+x_{\varepsilon})^{-2p}\,I_{\left[\varepsilon^{1/(1-p)}M_{2},\infty\right)}(\varepsilon^{1/2}B_{u}+x_{\varepsilon}).$

By (4.2),

$\displaystyle h(\sqrt{t}B_{1}+x)$ $\displaystyle\leq
C\exp\left(-\tfrac{\beta}{1-p}(\sqrt{t}B_{1}+x)^{1-p}\right)$ (4.5)
$\displaystyle=C\exp\left(-\tfrac{\beta}{1-p}\varepsilon^{-1}(\varepsilon^{1/2}B_{1}+x_{\varepsilon})^{1-p}\right).$

Substituting $t=\varepsilon^{-(1+p)/(1-p)}$ into (4) and using (4)–(3.5) gives

$\displaystyle P_{x}(\tau_{0}(X)>t)$ $\displaystyle\leq
CE_{0}\left[\exp\left(-\tfrac{\beta^{2}}{2}\varepsilon^{-1}\int_{0}^{1}\left(\varepsilon^{1/2}B_{u}+x_{\varepsilon}\right)^{-2p}I_{\left[\varepsilon^{1/(1-p)}M_{2},\infty\right)}(\varepsilon^{1/2}B_{u}+x_{\varepsilon})\,du\right.\right.$
$\displaystyle\qquad\left.\left.-\tfrac{\beta}{1-p}\varepsilon^{-1}\left(\varepsilon^{1/2}B_{1}+x_{\varepsilon}\right)^{1-p}\right)I(\tau_{0}(\varepsilon^{1/2}B+x_{\varepsilon})>1)\right]$
$\displaystyle\leq
CE_{0}\left[\exp\left(-\tfrac{\beta^{2}}{2}\varepsilon^{-1}\int_{0}^{1}\left(\varepsilon^{1/2}B_{u}+x_{\varepsilon}\right)^{-2p}I_{\left[\varepsilon^{1/(1-p)}M_{2},\infty\right)}(\varepsilon^{1/2}B_{u}+x_{\varepsilon})\,du\right.\right.$
$\displaystyle\qquad\left.\left.-\tfrac{\beta}{1-p}\varepsilon^{-1}\left|\varepsilon^{1/2}B_{1}+x_{\varepsilon}\right|^{1-p}\right)\right].$

Writing $Q_{\varepsilon}$ for the law on $C([0,\infty),\mathbb{R})$ of
$\sqrt{\varepsilon}B$ under $P_{0}$, this becomes

$\displaystyle P_{x}(\tau_{0}(X)>t)$ $\displaystyle\leq
CE^{Q_{\varepsilon}}\left[\exp\left(-\tfrac{\beta^{2}}{2}\varepsilon^{-1}\int_{0}^{1}\left(\omega_{u}+x_{\varepsilon}\right)^{-2p}I_{\left[\varepsilon^{1/(1-p)}M_{2},\infty\right)}(\omega_{u}+x_{\varepsilon})\,du\right.\right.$
$\displaystyle\qquad\left.\left.-\tfrac{\beta}{1-p}\varepsilon^{-1}\left|\omega_{1}+x_{\varepsilon}\right|^{1-p}\vphantom{\int_{0}^{1}}\right)\right]$
$\displaystyle=CE^{Q_{\varepsilon}}\left[\exp\left(-\tfrac{1}{\varepsilon}J_{\varepsilon}(\omega)\right)\right],$

where

$\displaystyle J_{\varepsilon}(\omega)$
$\displaystyle=\tfrac{\beta^{2}}{2}\int_{0}^{1}\left(\omega_{u}+x_{\varepsilon}\right)^{-2p}I_{\left[\varepsilon^{1/(1-p)}M_{2},\infty\right)}(\omega_{u}+x_{\varepsilon})\,du+\tfrac{\beta}{1-p}\left|\omega_{1}+x_{\varepsilon}\right|^{1-p}.$

Set

$J(\omega)=\tfrac{\beta^{2}}{2}\int_{0}^{1}|\omega_{u}|^{-2p}\,du+\tfrac{\beta}{1-p}|\omega_{1}|^{1-p}$

if the integral is finite, otherwise set $J(\omega)=\infty$. Then $J$ is lower
semicontinuous on

(4.6) $C_{0}=\\{\omega:[0,1]\to\mathbb{R}\,|\,\omega\text{ is continuous and
$\omega(0)=0$}\\}$

and if $\omega_{n}\to\omega$ in $C_{0}$ as $n\to\infty$, then

$\mathop{\liminf_{n\to\infty}}_{\varepsilon\to
0^{+}}J_{\varepsilon}(\omega_{n})\geq J(\omega).$

If $\omega\in C_{0}$ is absolutely continuous, denote its derivative by
$\dot{\omega}_{u}$. Then for

$K_{0}=\\{\omega\in C_{0}:\int_{0}^{1}|\dot{\omega}_{u}|^{2}\,du<\infty\\},$

by Varadhan’s Theorem (1984, Theorem 2.3),

$\displaystyle\limsup_{\varepsilon\to 0^{+}}\varepsilon\log
E^{Q_{\varepsilon}}\left[\exp\left(-\tfrac{1}{\varepsilon}J_{\varepsilon}(\omega)\right)\right]$
$\displaystyle\leq-\inf_{\omega\in
C_{0}}\left[J(\omega)+\tfrac{1}{2}\int_{0}^{1}(\dot{\omega}_{u})^{2}\,du\right]$
$\displaystyle=-\inf_{\omega\in
K_{0}}\left[J(\omega)+\tfrac{1}{2}\int_{0}^{1}|\dot{\omega}_{u}|^{2}\,du\right].$

By Theorem 6.1 in DeBlassie and Smits (2007), the infimum is
$\gamma(p,\beta)$. This completes the proof.

∎

##  5 Transformation of Drift

In this section we set the stage to extend Theorem 4.1 to a drift with
nonnegative $\alpha$ and $b$ that can take on positive values on
$[M_{1},M_{2}]$. There will be no increasing or $C^{3}$ conditions imposed. We
will also set things up for the case of negative $\alpha$.

First we prove a variant of the formula in Lemma 2.5 that is applicable to
discontinuous drifts.

######  Lemma 5.1.

Suppose $b_{X}$ and $b_{Y}$ satisfy (1.2) with the same $M_{1}$ and $M_{2}$.
If $b_{X}$ and $b_{Y}$ are continuous on $(0,\infty)\backslash\\{M_{2}\\}$ and
each restricted to $(M_{1},M_{2})$ has a $C^{1}$ extension to $[M_{1},M_{2}]$,
then the function

(5.1) $g(y)=\int_{0}^{y}\left(b_{X}-b_{Y}\right)(z)\,dz$

is a linear combination of convex functions with generalized second derivative
$\mu$ given by

(5.2)
$\mu(A)=(b_{X}-b_{Y})(M_{2}-)\,\delta_{M_{2}}(A)-\int_{A}\left(\left(b_{X}^{\prime}-b_{Y}^{\prime}\right)I_{(M_{1},M_{2})}\right)(a)\,da$

for any Borel set $A\subseteq(0,\infty)$. Here $\delta_{M_{2}}$ is the unit
point mass at $M_{2}$.

###### Proof.

By our assumptions on $b_{X}$ and $b_{Y}$, $b_{X}-b_{Y}$ is of bounded
variation on any bounded open subset $U$ of $(0,\infty)$, hence $g$ is a
linear combination of convex functions on $U$ (Roberts and Varberg (1973),
page 23). We now show its generalized second derivative is given by (5.2).

Since $g^{\prime}=b_{X}-b_{Y}$ a.e., it suffices to show for each $H\in
C_{0}^{\infty}(0,\infty)$,

(5.3) $\int\left(H^{\prime}(b_{X}-b_{Y})\right)(a)\,da=\int H(a)\,d\mu(a).$

To this end, let $H$ be so given. Then denoting the $C^{1}$ extensions of
$b_{X}$, $b_{Y}$ on $(M_{1},M_{2})$ to $[M_{1},M_{2}]$ by $\tilde{b}_{X}$,
$\tilde{b}_{Y}$, respectively,

$\displaystyle\int H^{\prime}(b_{X}-b_{Y})$ $\displaystyle=\int
H^{\prime}(\tilde{b}_{X}-\tilde{b}_{Y})I_{(M_{1},M_{2})}$ (since $b_{X}=b_{Y}$
on $(0,M_{1}]\cup[M_{2},\infty)$) $\displaystyle=\lim_{\varepsilon\to
0^{+}}\int
H^{\prime}(\tilde{b}_{X}-\tilde{b}_{Y})I_{(M_{1}+\varepsilon,M_{2}-\varepsilon)}$
$\displaystyle=\lim_{\varepsilon\to
0^{+}}\left[(H(\tilde{b}_{X}-\tilde{b}_{Y}))(M_{2}-\varepsilon)-(H(\tilde{b}_{X}-\tilde{b}_{Y}))(M_{1}+\varepsilon)-\int_{M_{1}+\varepsilon}^{M_{2}-\varepsilon}H(\tilde{b}_{X}^{\prime}-\tilde{b}_{Y}^{\prime})\right]$
$\displaystyle=\lim_{\varepsilon\to
0^{+}}\left[(H(b_{X}-b_{Y}))(M_{2}-\varepsilon)-\left(H(b_{X}-b_{Y})\right)(M_{1}+\varepsilon)-\int_{M_{1}+\varepsilon}^{M_{2}-\varepsilon}H(b_{X}^{\prime}-b_{Y}^{\prime})\right]$
$\displaystyle=\left(H(b_{X}-b_{Y})\right)(M_{2}-)-\int_{M_{1}}^{M_{2}}H(b_{X}^{\prime}-b_{Y}^{\prime})$
(since $b_{X}$, $b_{Y}$ are continuous at $M_{1}$ and coincide there, and
$\tilde{b}_{X}^{\prime}-\tilde{b}_{Y}^{\prime}=b_{X}^{\prime}-b_{Y}^{\prime}$
is bounded on $(M_{1},M_{2})$)
$\displaystyle=H(M_{2})(b_{X}-b_{Y})(M_{2}-)-\int_{M_{1}}^{M_{2}}H(b_{X}^{\prime}-b_{Y}^{\prime})$

(since $H$ is continuous). This gives (5.2). ∎

######  Lemma 5.2.

Let $x>0$ and let

(5.4) $\left\\{\begin{array}[]{ll}dX_{t}=dB_{t}+b_{X}(X_{t})\,dt,&\quad
X_{0}=x\\\ \\\ dY_{t}=dB_{t}+b_{Y}(Y_{t})\,dt,&\quad
Y_{0}=x\end{array}\right.$

be such that $b_{X}$ and $b_{Y}$ satisfy (1.2) with the same $M_{1}$ and
$M_{2}$. If $b_{X}$ and $b_{Y}$ are continuous on
$(0,\infty)\backslash\\{M_{2}\\}$ and each restricted to $(M_{1},M_{2})$ has a
$C^{1}$ extension to $[M_{1},M_{2}]$, then for $g$ as in (5.1) and

$H=\left(b_{X}^{2}-b_{Y}^{2}\right)-\left(b_{X}^{\prime}-b_{Y}^{\prime}\right)I_{(M_{1},M_{2})}$

we have

$\displaystyle
P_{x}(\tau_{0}(X)>t)=E_{x}\left[\exp\left(g(Y_{t})-g(x)-\tfrac{1}{2}\int_{0}^{t}H(Y_{s})\,ds\right.\right.$
(5.5)
$\displaystyle\left.\left.-\tfrac{1}{2}(b_{X}-b_{Y})(M_{2}-)\,\ell_{t}^{M_{2}}(Y)\vphantom{\int_{0}^{t}}\right)I(\tau_{0}(Y)>t)\right],$

where $\ell_{t}^{M_{2}}(Y)$ is the local time of $Y$ at $M_{2}$.

###### Proof.

The idea is similar to the one used by Pinsky (1995) to get the formula cited
in Lemma 2.5: use the Girsanov Theorem and eliminate the stochastic integral
in the martingale measure. Our reference for the particular form of the
Girsanov Theorem that we use is Theorem 8.6.6 in Oksendal (2007).

Let $0<\varepsilon<M_{1}\leq M_{2}<M$ such that $x\in(\varepsilon,M)$. Then
since $b_{X}-b_{Y}$ is bounded on $(\varepsilon,M)$, the “Local Novikov
Condition” holds:

$E_{x}\left[\exp\left(\tfrac{1}{2}\int_{0}^{t\wedge\tau_{\varepsilon,M}(Y)}(b_{X}-b_{Y})^{2}(Y_{s})\,ds\right)\right]<\infty.$

It follows that

$M_{t}=\exp\left(\int_{0}^{t\wedge\tau_{\varepsilon,M}(Y)}(b_{X}-b_{Y})(Y_{s})\,dB_{s}-\tfrac{1}{2}\int_{0}^{t\wedge\tau_{\varepsilon,M}(Y)}(b_{X}-b_{Y})^{2}(Y_{s})\,ds\right)$

is a martingale. By Girsanov’s Theorem,

$\displaystyle P_{x}(\tau_{\varepsilon,M}(X)>t)$
$\displaystyle=E_{x}\left[M_{t}I(\tau_{\varepsilon,M}(Y)>t)\right]$
$\displaystyle=E_{x}\left[\exp\left(\int_{0}^{t}(b_{X}-b_{Y})(Y_{s})\,dB_{s}\right.\right.$
(5.6)
$\displaystyle\qquad\left.\left.-\tfrac{1}{2}\int_{0}^{t}(b_{X}-b_{Y})^{2}(Y_{s})\,ds\right)I(\tau_{\varepsilon,M}(Y)>t)\right].$

Now we eliminate the stochastic integral in (5). By Lemma 5.1, we can apply
the Itô-Tanaka formula (Revuz and Yor (1991)) to get for
$\tau_{\varepsilon,M}(Y)>t$,

$g(Y_{t})=g(x)+\int_{0}^{t}g_{-}^{\prime}(Y_{s})\left[dB_{s}+b_{Y}(Y_{s})\,ds\right]+\tfrac{1}{2}\int\ell^{a}_{t}(Y)\mu(da),$

where $g_{-}^{\prime}$ is the left derivative and $\mu(da)$ is the generalized
second derivative of $g$ from (5.2) in Lemma 5.1. By our hypotheses on $b_{X}$
and $b_{Y}$, $g_{-}^{\prime}=b_{X}-b_{Y}$ a.e. on $(0,\infty)$. Using this,
the occupation times formula (Revuz and Yor (1991)) and Lemma 5.1 gives, for
$\tau_{\varepsilon,M}(Y)>t$,

$\displaystyle g(Y_{t})$
$\displaystyle=g(x)+\int_{0}^{t}(b_{X}-b_{Y})(Y_{s})\left[dB_{s}+b_{Y}(Y_{s})\,ds\right]$
$\displaystyle\qquad+\tfrac{1}{2}\left[(b_{X}-b_{Y})(M_{2}-)\,\ell^{M_{2}}_{t}(Y)-\int_{0}^{t}((b_{X}^{\prime}-b_{Y}^{\prime})I_{(M_{1},M_{2})})(Y_{s})\,ds\right]$

Solving for the stochastic integral, we get, for $\tau_{\varepsilon,M}(Y)>t$,

$\displaystyle\int_{0}^{t}(b_{X}-b_{Y})(Y_{s})\,dB_{s}$
$\displaystyle=g(Y_{t})-g(x)-\int_{0}^{t}((b_{X}-b_{Y})b_{Y})(Y_{s})\,ds$
$\displaystyle\qquad+\tfrac{1}{2}\int_{0}^{t}\left((b^{\prime}_{X}-b^{\prime}_{Y})I_{(M_{1},M_{2})}\right)(Y_{s})\,ds-\tfrac{1}{2}(b_{X}-b_{Y})(M_{2}-)\,\ell_{t}^{M_{2}}(Y).$

Substituting this into the exponential in (5) gives

$\displaystyle P_{x}(\tau_{\varepsilon,M(X)}>t)$
$\displaystyle=E_{x}\left[\exp\left(g(Y_{t})-g(x)-\int_{0}^{t}((b_{X}-b_{Y})b_{Y})(Y_{s})\,ds\right.\right.\hskip
72.26999pt$
$\displaystyle\qquad+\tfrac{1}{2}\int_{0}^{t}\left((b^{\prime}_{X}-b^{\prime}_{Y})I_{(M_{1},M_{2})}\right)(Y_{s})\,ds-\tfrac{1}{2}(b_{X}-b_{Y})(M_{2}-)\,\ell_{t}^{M_{2}}(Y)$
$\displaystyle\qquad\qquad\left.\left.-\tfrac{1}{2}\int_{0}^{t}(b_{X}-b_{Y})^{2}(Y_{s})\,ds\right)I(\tau_{\varepsilon,M}(Y)>t)\right]$
$\displaystyle=E_{x}\left[\exp\left(g(Y_{t})-g(x)-\tfrac{1}{2}\int_{0}^{t}H(Y_{s})\,ds\right.\right.$
$\displaystyle\qquad\left.\left.-\tfrac{1}{2}(b_{X}-b_{Y})(M_{2}-)\,\ell_{t}^{M_{2}}(Y)\vphantom{\int_{0}^{t}}\right)I(\tau_{\varepsilon,M}(Y)>t)\right].$

By Monotone Convergence and Lemma 2.1, upon letting $\varepsilon\downarrow 0$
and $M\uparrow\infty$, we get the desired conclusion. ∎

Next, we prove a variant of the previous results applicable to the case
$\alpha<0$.

######  Lemma 5.3.

Let $b_{1}$ and $b_{2}$ satisfy (1.2) with the same $\beta>0$, $M_{1}$ and
$M_{2}$, but the corresponding $\alpha$’s—call them $\alpha_{1}$ and
$\alpha_{2}$, respectively—are different. Suppose $b_{1}$ and $b_{2}$ are
continuous on $(0,\infty)$ and for some $M\in(M_{1},M_{2})$, $b_{1}-b_{2}$
restricted to $(M_{1},M)$ and $(M,M_{2})$ has $C^{1}$ extensions to
$[M_{1},M]$ and $[M,M_{2}]$, respectively. Then the function

(5.7) $g(y)=\int_{0}^{y}\left(b_{1}-b_{2}\right)(z)\,dz$

is a linear combination of convex functions with generalized second derivative
$\mu$ given by

(5.8)
$\mu(A)=-\int_{A}\left(\left(b_{1}^{\prime}-b_{2}^{\prime}\right)I_{(M_{1},M_{2})}\right)(a)\,da$

for any Borel set $A\subseteq(0,\infty)$.

###### Proof.

This is similar to the proof of Lemma 5.1, except because of the continuity of
$b_{1}$ and $b_{2}$ on $(0,\infty)$, there will be no point mass term in the
generalized second derivative. ∎

######  Lemma 5.4.

Let $x>0$ and let

(5.9) $\left\\{\begin{array}[]{ll}dX_{t}=dB_{t}+b_{X}(X_{t})\,dt,&\quad
X_{0}=x\\\ \\\ dY_{t}=dB_{t}+b_{Y}(Y_{t})\,dt,&\quad
Y_{0}=x\end{array}\right.$

be such that $b_{1}$ and $b_{2}$ satisfy (1.2) with the same $\beta>0$,
$M_{1}$ and $M_{2}$, but the corresponding $\alpha$’s—call them $\alpha_{1}$
and $\alpha_{2}$, respectively—are different. Suppose $b_{1}$ and $b_{2}$ are
continuous on $(0,\infty)$ and for some $M\in(M_{1},M_{2})$, $b_{1}-b_{2}$
restricted to $(M_{1},M)$ and $(M,M_{2})$ has $C^{1}$ extensions to
$[M_{1},M]$ and $[M,M_{2}]$. Then for $g$ as in (5.7) and

$H=\left(b_{1}^{2}-b_{2}^{2}\right)-\left(b_{1}^{\prime}-b_{2}^{\prime}\right)I_{(M_{1},M_{2})\backslash\\{M\\}}$

we have

(5.10)
$P_{x}(\tau_{0}(X)>t)=E_{x}\left[\exp\left(g(Y_{t})-g(x)-\tfrac{1}{2}\int_{0}^{t}H(Y_{s})\,ds\right)I(\tau_{0}(Y)>t)\right].$

###### Proof.

This is a simple modification of the proof of Lemma 5.1, where again, because
of the continuity of $b_{1}$ and $b_{2}$ on $(0,\infty)$, there will be no
local time term. ∎

##  6 Upper Bound

The main result of this section is the following theorem, which combined with
Theorem 3.1 will prove Theorem 1.1.

######  Theorem 6.1.

Let $X_{t}$ be as in (1.1), where $b$ is from (1.2). Then

$\limsup_{t\to\infty}t^{-(1-p)/(1+p)}\log
P_{x}(\tau_{0}(X)>t)\leq-\,\gamma(p,\beta),$

where $\gamma(p,\beta)$ is from (1.4).

We break up the proof into the cases when $\alpha>0$, $\alpha<0$ and
$\alpha=0$.

_First case: $\alpha>0$._

We remove the assumptions that $b\in C^{3}$ and $b^{\prime}\geq 0$ in Theorem
4.1.

######  Lemma 6.2.

Under the condition (1.2), if $\alpha>0$, then the solution $X_{t}$ of (1.1)
satisfies

$\limsup_{t\to\infty}\log P_{x}(\tau_{0}(X)>t)\leq-\,\gamma(p,\beta),$

where $\gamma(p,\beta)$ is from (1.4).

The proof uses several lemmas. Here is the basic setup. Let $0<x_{1}<x_{2}$,
$y_{1}<y_{2}<0$ and $\gamma>0$ be given. Consider the lines

(6.1) $\tilde{b}_{X}(x)=\frac{\gamma-y_{1}}{x_{2}-x_{1}}(x-x_{1})+y_{1},\quad
x>0$

and

(6.2) $\tilde{b}_{Y}(x)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})+y_{1},\quad
x>0.$

Then $\tilde{b}_{X}$ passes through the points $(x_{1},y_{1})$ and
$(x_{2},\gamma)$ and $\tilde{b}_{Y}$ passes through $(x_{1},y_{1})$ and
$(x_{2},y_{2})$.

######  Lemma 6.3.

For $\gamma$ sufficiently large,

$\tilde{b}_{X}^{2}-\tilde{b}_{Y}^{2}-(\tilde{b}_{X}^{\prime}-\tilde{b}_{Y}^{\prime})\geq
0\text{ on $[x_{1},x_{2}]$.}$

###### Proof.

We have

$\tilde{b}_{X}^{2}-\tilde{b}_{Y}^{2}=\left(\frac{x-x_{1}}{x_{2}-x_{1}}\right)^{2}\left[(\gamma-
y_{1})^{2}-(y_{2}-y_{1})^{2}\right]+\frac{2y_{1}(\gamma-
y_{2})}{x_{2}-x_{1}}(x-x_{1})$

and $\tilde{b}_{X}^{2}-\tilde{b}_{Y}^{2}$ takes on its minimum value
$-\frac{y_{1}^{2}(\gamma-y_{2})^{2}}{(\gamma-y_{1})^{2}-(y_{2}-y_{1})^{2}}$ at
the point $x=x_{1}-\frac{y_{1}(\gamma-y_{2})(x_{2}-x_{1})}{(\gamma-
y_{1})^{2}-(y_{2}-y_{1})^{2}}$ (note that $\gamma-y_{1}>y_{2}-y_{1}>0$). In
particular, on $[x_{1},x_{2}]$,

$\displaystyle(\tilde{b}_{X}^{2}-\tilde{b}_{Y}^{2})(x)$
$\displaystyle+(\tilde{b}_{X}^{\prime}-\tilde{b}_{Y}^{\prime})(x)\geq-\frac{y_{1}^{2}(\gamma-
y_{2})^{2}}{(\gamma-y_{1})^{2}-(y_{2}-y_{1})^{2}}+\frac{\gamma-
y_{1}}{x_{2}-x_{1}}-\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$\displaystyle=-\frac{y_{1}^{2}(\gamma-y_{2})^{2}}{(\gamma-
y_{1})^{2}-(y_{2}-y_{1})^{2}}+\frac{\gamma-y_{2}}{x_{2}-x_{1}}$
$\displaystyle=(\gamma-y_{2})\left[-\frac{y_{1}^{2}(\gamma-y_{2})}{(\gamma-
y_{1})^{2}-(y_{2}-y_{1})^{2}}+\frac{1}{x_{2}-x_{1}}\right]$ $\displaystyle\geq
0$

for large $\gamma$. ∎

######  Lemma 6.4.

Let $dZ_{t}=dB_{t}+b_{Z}(Z_{t})\,dt$, where $b_{Z}$ satisfies (1.2) with
$b_{Z}=C>0$ on $(M_{1},M_{2})$. Then

$\limsup_{t\to\infty}\log P_{x}(\tau_{0}(Z)>t)\leq-\,\gamma(p,\beta),$

where $\gamma(p,\beta)$ is from (1.4).

###### Proof.

By our hypotheses,

(6.3) $b_{Z}(x)=\left\\{\begin{array}[]{ll}-\alpha x^{-q},&\quad
x\in(0,M_{1}]\\\ C,&\quad x\in(M_{1},M_{2})\\\ -\beta x^{-p},&\quad
x\in[M_{2},\infty).\end{array}\right.$

In Lemma 5.2, choose $(x_{2},y_{2})=(M_{2},-\beta M_{2}^{-p})$ and
$(x_{1},y_{1})=\left(\tfrac{M_{1}}{K},-\alpha\left(\tfrac{M_{1}}{K}\right)^{-q}\right)$
where $K>1$ is so large that $y_{1}<y_{2}$. Notice this tells us
$y_{1}<-\alpha M_{1}^{-q}$. Then we can choose $\gamma>0$ so large that the
conclusion of Lemma 5.2 holds for $\tilde{b}_{X}$ and $\tilde{b}_{Y}$ given by
(6.1) and (6.2):

(6.4)
$\tilde{b}_{X}^{2}-\tilde{b}_{Y}^{2}-(\tilde{b}_{X}^{\prime}-\tilde{b}_{Y}^{\prime})\geq
0\text{ on $[x_{1},x_{2}]$.}$

Define

(6.5) $b_{X}(x)=\left\\{\begin{array}[]{ll}-\alpha x^{-q},&\quad
x\in(0,x_{1}]\\\ \tilde{b}_{X}(x),&\quad x\in(x_{1},x_{2})=(x_{1},M_{2})\\\
-\beta x^{-p},&\quad x\in[x_{2},\infty)=[M_{2},\infty)\end{array}\right.$

and

(6.6) $b_{Y}(x)=\left\\{\begin{array}[]{ll}-\alpha x^{-q},&\quad
x\in(0,x_{1}]\\\ \tilde{b}_{Y}(x),&\quad x\in(x_{1},x_{2})=(x_{1},M_{2})\\\
-\beta x^{-p},&\quad x\in[x_{2},\infty)=[M_{2},\infty).\end{array}\right.$

Then by definition of $b_{X}$, $b_{Y}$ and (6.4),

(6.7)
$b_{X}^{2}-b_{Y}^{2}-(b_{X}^{\prime}-b_{Y}^{\prime})I_{(x_{1},x_{2})}\geq
0\text{ on $(0,\infty)$.}$

Notice this continues to hold if $\gamma$ is made larger. Since
$x_{1}=\frac{M_{1}}{K}<M_{1}$, we can make $\gamma$ larger, if necessary, so
that the line segment $\\{(x,C):M_{1}<z<M_{2}\\}$ (recall $C$ is from (6.3))
lies to the right of the line through the points $(x_{1},y_{1})$ and
$(x_{2},\gamma)$. Since $x_{1}<M_{1}$, we have $b_{Z}\leq b_{X}$ on
$(0,\infty)$. By the Comparison Theorem,

(6.8) $P_{x}(\tau_{0}(Z)>t)\leq P_{x}(\tau_{0}(X)>t).$

We now apply Lemma 5.2 to $b_{X}$, $b_{Y}$ given by (6.5), (6.6),
respectively, but with $M_{1},M_{2}$ in the lemma taken to be $x_{1},x_{2}$.
First check the hypotheses of the Lemma:

  * •

$b_{X}$ and $b_{Y}$ are continuous on $(0,\infty)\backslash\\{x_{2}\\}$ since

$\lim_{x\to x_{1}^{-}}b_{X}(x)=y_{1}=\lim_{x\to x_{1}^{+}}b_{X}(x)\text{ and
}\lim_{x\to x_{1}^{-}}b_{Y}(x)=y_{1}=\lim_{x\to x_{1}^{+}}b_{Y}(x).$

  * •

Since each is linear on $(x_{1},x_{2})$, each restricted to $(x_{1},x_{2})$
has a $C^{1}$ extension to $[x_{1},x_{2}]$.

Thus Lemma 5.2 applies and so for

$g(z)=\int_{0}^{x}(b_{X}-b_{Y})(y)\,dy$

and

$H=(b_{X}^{2}-b_{Y}^{2})-(b_{X}^{\prime}-b_{Y}^{\prime})I_{(x_{1},x_{2})},$

we have

$\displaystyle
P_{x}(\tau_{0}(X)>t)=E_{x}\left[\exp\left(g(Y_{t})-g(x)-\tfrac{1}{2}\int_{0}^{t}H(Y_{s})\,ds\right.\right.$
(6.9)
$\displaystyle\left.\left.-\tfrac{1}{2}\left(b_{X}-b_{Y}\right)(M_{2}-)\,\ell_{t}^{M_{2}}(Y)\vphantom{\int_{0}^{t}}\right)I(\tau_{0}(Y)>t)\right].$

Now by (6.7), $H\geq 0$, and by definition of $b_{X}$ and $b_{Y}$,

$\displaystyle(b_{X}-b_{Y})(M_{2}-)$ $\displaystyle=\lim_{x\to
x_{2}^{-}}\left(\tilde{b}_{X}(x)-\tilde{b}_{Y}(x)\right)\text{ (by
\eqref{eq6.5}--\eqref{eq6.6})}$
$\displaystyle=\tilde{b}_{X}(x_{2})-\tilde{b}_{Y}(x_{2})\text{
($\tilde{b}_{X}$ and $\tilde{b}_{Y}$ are continuous)}$ $\displaystyle=\gamma-
y_{2}\text{ (by \eqref{eq6.1}--\eqref{eq6.2})}$ $\displaystyle>0.$

Then (6) becomes

(6.10) $P_{x}(\tau_{0}(X)>t)\leq
E_{x}\left[\exp\left(g(Y_{t})-g(x)\right)I(\tau_{0}(Y)>t)\right].$

Since $b_{X}=b_{Y}$ on $(x_{1},x_{2})^{c}$ and since
$b_{X}-b_{Y}=\tilde{b}_{X}-\tilde{b}_{Y}$ is bounded on $(x_{1},x_{2})$, we
see

$g(z)=\int_{0}^{z}(b_{X}-b_{Y})(y)\,dy\leq\sup_{(x_{1},x_{2})}(\tilde{b}_{X}-\tilde{b}_{Y})(x_{2}-x_{1}),\quad
z>0.$

Hence for some positive constant $C_{1}$ independent of $t$, (6.10) becomes

(6.11) $P_{x}(\tau_{0}(X)>t)\leq C_{1}P_{x}(\tau_{0}(Y)>t).$

The crucial point is that

$\sup_{(0,\infty)}b_{Y}<0.$

This holds because:

  * •

on $(0,x_{1}]$, $b_{Y}(x)=-\alpha x^{-q}\leq-\alpha x_{1}^{-p}$;

  * •

on $(x_{1},x_{2})$, by (6.2) and that $y_{1}<y_{2}<0$,
$b_{Y}=\tilde{b}_{Y}\leq y_{2}<0$;

  * •

on $(x_{2},\infty)$, $b_{Y}(x)=-\beta x^{-p}\leq-\beta x_{2}^{-p}$.

Thus Theorem 4.1 applies to the process $Y$, and we conclude

$\limsup_{t\to\infty}\log P_{x}(\tau_{0}(Y)>t)\leq-\,\gamma(p,\beta),$

where $\gamma(p,\beta)$ is from (1.4). Combined with (6.8) and (6.11), this
gives us

$\limsup_{t\to\infty}\log P_{x}(\tau_{0}(Z)>t)\leq-\,\gamma(p,\beta),$

as desired. ∎

Proof of Lemma 6.2. Let $X_{t}$ be from (1.1), where $b$ satisfies (1.2) and
$\alpha\geq 0$. Let $C=\sup_{[M_{1},M_{2}]}|b|$. Suppose
$dZ_{t}=dB_{t}+b_{Z}(Z_{t})\,dt$, where $b_{Z}$ satisfies (1.2) with $b_{Z}=C$
on $(M_{1},M_{2})$. Then $b_{X}\leq b_{Z}$ and so by the Comparison Theorem,
$P_{x}(\tau_{0}(X)>t)\leq P_{x}(\tau_{0}(Z)>t)$. The desired upper bound
follows upon applying Lemma 6.4 to $Z$. $\square$

_Second case: $\alpha<0$._

######  Lemma 6.5.

Under the condition (1.2), if $\alpha<0$, then the solution $X_{t}$ of (1.1)
satisfies

$\limsup_{t\to\infty}\log P_{x}(\tau_{0}(X)>t)\leq-\,\gamma(p,\beta),$

where $\gamma(p,\beta)$ is from (1.4).

For the proof, we need the following special case.

######  Lemma 6.6.

Let $\alpha<0$. Assume $b\in C^{3}$ satisfies condition (1.2). Suppose that
for some $x_{1}\in(0,M_{2})$, $b(x_{1})=0$ and $b$ is strictly decreasing on
$(0,x_{1})$. Then the solution $X_{t}$ of (1.1) satisfies

$\limsup_{t\to\infty}\log P_{x}(\tau_{0}(X)>t)\leq-\,\gamma(p,\beta),$

where $\gamma(p,\beta)$ is from (1.4).

###### Proof.

Define

(6.12) $b_{Y}(x)=\left\\{\begin{array}[]{ll}-b(x),&\quad x\in(0,x_{1}]\\\
b(x),&\quad x\in(x_{1},\infty),\end{array}\right.$

and let $Y_{t}$ solve

$dY_{t}=dB_{t}+b_{Y}(Y_{t})\,dt,\quad Y_{0}=x.$

Now $b$ and $b_{Y}$ are continuous on $(0,\infty)$ and $b-b_{Y}$ restricted to
$(M_{1},x_{1})$ and $(x_{1},M_{2})$ has $C^{1}$ extensions to $[M_{1},x_{1}]$
and $[x_{1},M_{2}]$, respectively. Then by Lemma 5.4, taking $M$ there to be
$x_{1}$, for

$g(z)=\int_{0}^{z}\left(b-b_{Y}\right)(y)\,dy$

and

$H=\left(b^{2}-b_{Y}^{2}\right)-\left(b^{\prime}-b_{Y}^{\prime}\right)I_{(M_{1},M_{2})\backslash\\{x_{1}\\}},$

we have

(6.13)
$P_{x}(\tau_{0}(X)>t)=E_{x}\left[\exp\left(g(Y_{t})-g(x)-\tfrac{1}{2}\int_{0}^{t}H(Y_{s})\,ds\right)I(\tau_{0}(Y)>t)\right].$

On the other hand, $b^{2}=b_{Y}^{2}$ on $(0,\infty)$ and

$\displaystyle(b^{\prime}-b_{Y}^{\prime})(x)$
$\displaystyle=\left\\{\begin{array}[]{ll}2b^{\prime}(x),&\quad
x\in(M_{1},x_{1})\\\ 0,&\quad x\in(x_{1},\infty)\end{array}\right.$
$\displaystyle\leq 0.$

In particular, $H\geq 0$ and (6.13) becomes

(6.14) $P_{x}(\tau_{0}(X)>t)\leq
E_{x}\left[\exp\left(g(Y_{t})-g(x)\right)I(\tau_{0}(Y)>t)\right].$

Since $b=b_{Y}$ on $(x_{1},\infty)$ and $b-b_{Y}=2b$ is nonnegative and
integrable on $(0,x_{1})$, we see $g$ is bounded. Hence for some positive
constant $C_{1}$ independent of $t$, (6.14) becomes

(6.15) $P_{x}(\tau_{0}(X)>t)\leq C_{1}P_{x}(\tau_{0}(Y)>t).$

By Lemma 6.2 applied to $Y$,

$\limsup_{t\to\infty}\log P_{x}(\tau_{0}(X)>t)\leq\limsup_{t\to\infty}\log
P_{x}(\tau_{0}(Y)>t)\leq-\,\gamma(p,\beta),$

where $\gamma(p,\beta)$ is from (1.4). ∎

Proof of Lemma 6.5. Let $X_{t}$ be from (1.1), where $b$ satisfies (1.2) and
$\alpha<0$. Let $C>\sup_{[M_{1},M_{2}]}|b|$ and choose $\alpha_{1}<\alpha$
such that $-\alpha_{1}>CM_{2}^{q}$. Then choose $b_{Y}\in C^{3}$ having the
following properties: for some $M_{3},M_{4}\in(M_{2},\infty)$ with
$M_{3}<M_{4}$,

  * •

$b_{Y}(x)=-\alpha_{1}x^{-q}$ on $(0,M_{2}]$;

  * •

$b_{Y}$ is strictly decreasing on $[M_{2},M_{3}]$;

  * •

$b_{Y}(M_{3})=0$;

  * •

$b_{Y}=b$ on $[M_{4},\infty)$.

Suppose $dY_{t}=dB_{t}+b_{Y}(Y_{t})\,dt$. Then $b\leq b_{Y}$ and so by the
Comparison Theorem, $P_{x}(\tau_{0}(X)>t)\leq P_{x}(\tau_{0}(Y)>t)$. The
process $Y$ satisfies the conditions of Lemma 6.6 and the desired upper bound
follows by applying that lemma to $Y$. $\square$

_Third case: $\alpha=0$._

Define

$b_{Y}(x)=\left\\{\begin{array}[]{ll}x^{-q},&\quad x\in(0,M_{1}]\\\
b(x),&\quad x\in(M_{1},\infty),\end{array}\right.$

and let $Y_{t}$ solve

$dY_{t}=dB_{t}+b_{Y}(Y_{t})\,dt,\quad Y_{0}=x.$

Then by the Comparison Theorem,

$P_{x}(\tau_{0}(X)>t)\leq P_{x}(\tau_{0}(Y)>t).$

Since the situation of $\alpha<0$ holds for $b_{Y}$, that case implies

$\limsup_{t\to\infty}\log P_{x}(\tau_{0}(X)>t)\leq\limsup_{t\to\infty}\log
P_{x}(\tau_{0}(Y)>t)\leq-\,\gamma(p,\beta),$

once again. $\square$

##  7 Proof of Theorem 1.2

We need the following bounds due to Potter. See Theorem 1.5.6 in Bingham,
Goldie and Teugels (1989).

######  Theorem 7.1.

(a) If $\ell$ is slowly varying at infinity, then for any choice of $A>1$ and
$\delta>0$, there is $M>0$ such that

$\ell(y)/\ell(x)\leq A\max\left((y/x)^{\delta},(y/x)^{-\delta}\right),\quad
x,y\geq M.$

(b) If $\ell$ is slowly varying at zero from the right, then for any choice of
$A>1$ and $\delta>0$, there is $M>0$ such that

$\ell(y)/\ell(x)\leq A\max\left((y/x)^{\delta},(y/x)^{-\delta}\right),\quad
x,y\leq M.$

The following is a consequence of the Representation Theorem for slowly
varying functions (Bingham, Goldie and Teugels (1989), Theorem 1.3.1).

######  Theorem 7.2.

(a) If $\ell$ is slowly varying at infinity, then for each $\delta>0$

$\lim_{x\to\infty}x^{-\delta}\ell(x)=0\text{ and
}\lim_{x\to\infty}x^{\delta}\ell(x)=\infty.$

(b) If $\ell$ is slowly varying from the right at $0$, then for each
$\delta>0$

$\lim_{x\to 0^{+}}x^{\delta}\ell(x)=0\text{ and }\lim_{x\to
0^{+}}x^{-\delta}\ell(x)=\infty.$

We will also need the following fact to prove Theorem 1.2.

######  Lemma 7.3.

Let $\delta>0$ be so small that $0<p-\delta<p+\delta<1$ and
$0<q-\delta<q+\delta<1$. By decreasing $M_{1}$ and increasing $M_{2}$ if
necessary, we have

$x^{-\delta}\leq\ell_{2}(x)\leq x^{\delta},\quad x\geq M_{2}$

and

$x^{\delta}\leq\ell_{1}(x)\leq x^{-\delta},\quad x\leq M_{1}.$

###### Proof.

By making $M_{2}$ bigger if necessary, by Theorem 7.2(a),

(7.1) $M_{2}^{-\delta}\ell_{2}(M_{2})\leq 1/2<2\leq
M_{2}^{\delta}\ell_{2}(M_{2}).$

Then by Theorem 7.1(a), again making $M_{2}$ bigger if necessary, we have for
$x\geq M_{2}$,

$\displaystyle\ell_{2}(x)$ $\displaystyle\leq
2\max\left((x/M_{2})^{\delta},(x/M_{2})^{-\delta}\right)\ell_{2}(M_{2})$
$\displaystyle=2(x/M_{2})^{\delta}\ell_{2}(M_{2})$
$\displaystyle=2x^{\delta}M_{2}^{-\delta}\ell_{2}(M_{2})$ $\displaystyle\leq
2x^{\delta}(1/2)$ $\displaystyle=x^{\delta}$

and

$\displaystyle\ell_{2}(M_{2})$ $\displaystyle\leq
2\max\left((M_{2}/x)^{\delta},(M_{2}/x)^{-\delta}\right)\ell_{2}(x)$
$\displaystyle=2(x/M_{2})^{\delta}\ell_{2}(x)$
$\displaystyle\leq(M_{2}^{\delta}\ell_{2}(M_{2}))(x/M_{2})^{\delta}\ell_{2}(x)$
$\displaystyle=x^{\delta}\ell_{2}(M_{2})\ell_{2}(x).$

Rearranging the latter, we have

$x^{-\delta}\leq\ell_{2}(x).$

Combined, we have

(7.2) $x\geq M_{2}\implies x^{-\delta}\leq\ell_{2}(x)\leq x^{\delta}.$

Similarly, by making $M_{1}$ smaller if necessary, by Theorems 7.2(b) and
7.1(b) we have

(7.3) $x\leq M_{1}\implies x^{\delta}\leq\ell_{1}(x)\leq x^{-\delta}.$

∎

_Proof of Theorem 1.2._ Let $\varepsilon>0$ be small so that
$0<\beta_{0}-\varepsilon<\beta_{0}+\varepsilon<1$, and if necessary, decrease
$M_{1}$ and increase $M_{2}$ so that Lemma 7.3 continues to apply and we have

$\alpha_{0}=\sup_{(0,M_{1}]}|\alpha|<\infty$

and

$x\geq M_{2}\implies
0<\beta_{0}-\varepsilon<\beta(x)<\beta_{0}+\varepsilon<1.$

Define

$\widetilde{b}(x)=\left\\{\begin{array}[]{ll}\alpha_{0}x^{q-\delta},&\qquad
0<x\leq M_{1}\\\ b(x),&\qquad M_{1}<x<M_{2}\\\
-(\beta_{0}+\varepsilon)x^{p+\delta},&\qquad M_{2}\leq x,\end{array}\right.$

and

$\overline{b}(x)=\left\\{\begin{array}[]{ll}-\alpha_{0}x^{q-\delta},&\qquad
0<x\leq M_{1}\\\ b(x),&\qquad M_{1}<x<M_{2}\\\
-(\beta_{0}-\varepsilon)x^{p-\delta},&\qquad M_{2}\leq x.\end{array}\right.$

Let $\widetilde{X}_{t}$ be the solution of

$d\widetilde{X}_{t}=dB_{t}+\widetilde{b}(\widetilde{X}_{t})\,dt,\quad\widetilde{X}_{0}=x$

and let $\overline{X}_{t}$ be the solution of

$d\overline{X}_{t}=dB_{t}+\overline{b}(\overline{X}_{t})\,dt,\quad\overline{X}_{0}=x.$

Since $\overline{b}\leq b\leq\widetilde{b}$, by the Comparison Theorem we have

$P_{x}(\tau_{0}(\overline{X})>t)\leq P_{x}(\tau_{0}(X)>t)\leq
P_{x}(\tau_{0}(\widetilde{X})>t).$

But Theorem 1.1 applies to $\widetilde{X}_{t}$ and $\overline{X}_{t}$ and so
we get

$\displaystyle-\gamma(p-\delta,\beta_{0}-\varepsilon)$
$\displaystyle\leq\liminf_{t\to\infty}t^{-(1-p)/(1+p)}\log
P_{x}(\tau_{0}(X)>t)$
$\displaystyle\leq\limsup_{t\to\infty}t^{-(1-p)/(1+p)}\log
P_{x}(\tau_{0}(X)>t)$ $\displaystyle-\gamma(p+\delta,\beta_{0}+\varepsilon).$

Let $\varepsilon$ and $\delta$ go to $0$ to get the desired conclusion.
$\square$

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