3.1. Introduction

Let $L_t$ be defined by (4), for $f(x)=k|x|$ and $\gamma=1$. To simplify, we make a time change and consider $X_t \,{:}\,{\raise-1.5pt{=}}\, L_{e^t}$, and subsequently we obtain

\begin{align*}X_{t+{\rm d}t} - X_t &= L_{e^t+e^t {\rm d}t} -L_{e^t}\\&= k|L_{e^t}| {\rm d}t + e^{-t} (B_{t + e^t {\rm d}t} - B_{e^t}) \\&= k|X_t| {\rm d}t + e^{-\frac{t}{2}} {\rm d}B_t,\end{align*}

which is the model we will study. We begin with some definitions: first,

(9) \begin{equation} {\rm d}X_t = k |X_t| {\rm d}t + e^{-\frac{t}{2}}{\rm d} B_t.\end{equation}

We introduce another SDE closely related to the previous one, which will be useful:

(10) \begin{equation}{\rm d}K_t = k K_t {\rm d}t + e^{-\frac{t}{2}}{\rm d} B_t.\end{equation}

It is easy to see that both of these SDEs admit unique strong solutions; for a reference see Theorem 11.2 in Chapter 6 in [Reference Rogers and WilliamsRWW87]. Therefore we can construct $X_t,K_t$ in the classical Wiener space $(\Omega, \mathcal{F}, \mathbb{P})$. The solution for the SDE (10) is given by $ K_t = e^{ k t} (e^{ -t_0k } K_{t_0} + \int_{t_0}^t e^{-s (k +\frac{1}{2} ) } {\rm d} B_s )$. Indeed, substituting in (10) and using Itô’s formula, we get

\begin{align*} {\rm d} K_t &= a'(t) (k_0 + \int_{t_0}^{t} b(s) {\rm d} B_s ) + a(t)b(t){\rm d} B_t \\&= \frac{a'(t)}{a(t)} K_t+ a(t)b(t) {\rm d}B_t ,\end{align*}

where $ a(t)= e^{ k (t-t_0)} $ and $ b(t) = e^{-t(\frac{1}{2}+k) +kt_0 }$, so that $\frac{a'(t)}{a(t)} = k$ and $ a(t)b(t)= e^{-\frac{t}{2}} $.

Proof. This is a direct application of Proposition 2.1.

3.2. Analysis of $\textbf{\textit{X}}_{\textbf{\textit{t}}}$ when $k>1/2$

We start by stating the main result of this section, which we will prove at the end of the subsection.

Now we will show that $(X_t)_{t\geq 1}$ cannot stay negative for all times. This will be accomplished by a direct computation after solving the SDE.

Proof. First, note that the solution of the SDE (9), run from time s with initial condition $X_{s}<0$, coincides with the solution of the SDE $ {\rm d}X_t = -k X_t {\rm d}t + e^{-\frac{t}{2}}{\rm d} B_t $ before $X_t$ hits 0. Formally, we define $\tau_0 = \inf \{ t| t \geq s,\,X_t=0 \}$. Using the same method as when solving SDE (10), we obtain $X_t =e^{-kt} (e^{ks} X_{s} + \int_{s}^{t} e^{u(k-\frac{1}{2}) } {\rm d}B_u ) $ on $\{t <\tau_0 \}$. Set $G_t= \int_{s}^{t} e^{u(k-\frac{1}{2}) } {\rm d}B_u $, and calculate the quadratic variation of $G_t$, namely $\left \langle G_t \right \rangle = ( e^{2t(k-\frac{1}{2}) } -e^{2s(k-\frac{1}{2}) } )/(2k-1) $. Next, we compute the probability of never returning to zero:

\begin{align*}\mathbb{P}(\tau =\infty) & = \mathbb{P}\left(\sup_{s \lt u \lt \infty} X_u \leq 0 \right)\\[6pt] & = \mathbb{P}\left(\sup_{s \lt u \lt \infty} G_u \leq-e^{ks} X_{s} \right)\\[6pt] &= 1- \mathbb{P}\left(\sup_{s \lt u \lt \infty} G_u \gt -e^{ks} X_{s} \right) \\[6pt] &=1- \lim_{t\rightarrow \infty}\mathbb{P}\left(\sup_{s \lt u \lt t} G_u \gt -e^{ks} X_{s} \right) \\[6pt] &= 1- \lim_{t\rightarrow \infty}2 \mathbb{P}\left( G_t \gt -e^{ks} X_{s}\right), \textrm{from\,the\,reflection\,principle} \\[6pt] &= 1- \lim_{t\rightarrow \infty}2 \mathbb{P}\left(N\left( 0, \frac{e^{2t(k-\frac{1}{2})}-e^{2s(k-\frac{1}{2})}}{2k -1} \right) \gt -e^{ks} X_{s} \right) \\[4pt] &=0,\end{align*}

since

\begin{align*}\frac{ e^{2t(k-\frac{1}{2}) } -e^{2s(k-\frac{1}{2}) } }{2k -1 }\to \infty.\\ \end{align*}

We will now prove two important lemmas that are true for solutions of (9) for any $k>0.$

Proof. Assume we start the SDE at time $t_i$ with initial condition $X_{t_i}\geq 0$. Then we see that

\begin{equation*}X_t \geq \int_{t_i}^t k|X_u| {\rm d }u + \int_{t_i}^t e^{-\frac{u}{2}} {\rm d}B_u \geq \int_{t_i}^t e^{-\frac{u}{2}} {\rm d}B_u.\end{equation*}

Set $G_t =\int_{t_i}^t e^{-\frac{u}{2}} {\rm d}B_u$. The quadratic variation of $G_t$ is $\langle G_t\rangle = e^{-t_1} - e^{-t}$. Fix $0<c<1$. Now, observe that we can always choose t big enough so that $\langle G_t \rangle \geq c e^{-t_1}$ for any $t_1$.

Then

\begin{align*}\mathbb{P} \Big( \sup _{t_i<u<t }X_u >e^{-t_1/2 } \Big) &\geq \mathbb{P} \Big( \sup _{t_i<u<t }G_u >e^{-t_1/2 } \Big) \\[4pt] &= 2\mathbb{P} \big( G_t \gt e^{-t_1/2 } \big) \\[4pt] & \geq 2\mathbb{P} \big( N(0,c e^{-t_1}) \gt e^{-t_1/2 } \big) \\[4pt] & = 2\mathbb{P} \big( N(0, c) \gt 1 \big) \gt \gamma \gt 0.\end{align*}

Let $ g(x) = \inf \{ y| e^{-x} - e^{-y} \geq c e^{-x} \}$. Now we can formally define the sequence of the stopping times. The first stopping time is $\tau_1 = \inf \{t |X_{t} \geq 0 \} $; then we define recursively $\tau_{i+1} = \inf \{ t | t> \tau_i, t> g(\tau_i), \, X_t \geq 0 \}$. We also define the associated filtration $ \mathcal{F}_n = \mathcal{F}_{ \tau_n}$, for $n\geq 1$ and $\mathcal{F}_0 =\{ 0 , \Omega \}$. Now let $ A_n =\{ \exists t,\, \tau_{n-1} <t< \tau_{n} \, , \text{ s.t.\ } X_t \geq c e^{- t/2} \}$. By definition $A_n \in \mathcal{F}_n$. We find a lower bound for $ \mathbb{P} (A_n | \mathcal{F}_{n-1} )$:

\begin{align*}\mathbb{P} (A_n | \mathcal{F}_{n-1}) & \geq \mathbb{P} \Big( \sup_{\tau_{n-1} \lt u \lt \tau_n} X_u \gt ce^{-t_{n-1} /2} | \mathcal{F}_{n-1} \Big) \\ &\geq \mathbb{P} \Big( \sup_{\tau_{n-1} \lt u \lt g(\tau_{ n-1})} X_u \gt ce^{-\tau_{n-1} /2} | \mathcal{F}_{n-1}\Big) \\ & \gt \gamma .\end{align*}

On $\{X_t\geq 0 \text{\ i.o.}\}$ the sum $\sum_{n\geq 1} \mathbb{P}(A_n | \mathcal{F}_{n-1} ) $ has infinitely many nonzero terms bigger than $\gamma$; hence $\sum_{n\geq 1} \mathbb{P}(A_n | \mathcal{F}_{n-1} ) = \infty $ a.s. Finally, by Lemma 2.5 (Borel–Cantelli) we conclude.

The next lemma uses the previous lemma to establish that on $\{X_t \geq 0 \text{\ i.o.} \}$ we have $\liminf_{t \rightarrow \infty} X_t >0 $.

Proof. Indeed, if we start the process at time s with initial condition $X_{s} \geq ce^{- \frac{s}{2}}$, then the solution of (9), before hitting 0, is given by

\begin{equation*}X_t= e^{kt} \left(e^{-ks} X_{s} + \int_{s}^{t} e^{-u(k+\frac{1}{2}) } {\rm d}B_u \right)\geq e^{kt} \left(ce^{-s(k +\frac{1}{2} ) } + \int_{s}^{t} e^{-u(k+\frac{1}{2}) } {\rm d}B_u \right) .\end{equation*}

Define $G_t=\int_{s}^{t} e^{-s(k+\frac{1}{2}) } {\rm d}B_s $. We calculate its quadratic variation:

\begin{equation*}\langle G_t \rangle = \dfrac{ e^{-2tk-t } }{ -2k-1 } +\dfrac{ e^{-2sk-s } }{ 2k+1 } .\end{equation*}

Taking $ t \rightarrow \infty $ shows $\langle G_{\infty }\rangle= \dfrac{ e^{-2sk-s} }{2k+1 }$. Therefore,

(11) \begin{align}\mathbb{P} \left(\inf_{s \leq u <\infty } X_u >\frac{c}{2}e^{ -\frac{s}{2} } \right)&=\mathbb{P} \left(\inf_{s \leq u <\infty } e^{ku}\left(ce^{-s\left(k +\frac{1}{2}\right)} + G_u \right) > \frac{c}{2}e^{ -\frac{s}{2} } \right)\nonumber\\[4pt] &\geq \mathbb{P} \left(\inf_{s \leq u <\infty } e^{ks}\left(ce^{-s\left(k +\frac{1}{2}\right)} + G_u \right) > \frac{c}{2}e^{ -\frac{s}{2} } \right)\nonumber \\[4pt] &=\mathbb{P} \left(\inf_{s \leq u <\infty } ce^{-s\left(k +\frac{1}{2}\right)} + G_u > \frac{c}{2}e^{ -s\left(k +\frac{1}{2}\right) } \right)\nonumber \\[4pt] &=\mathbb{P} \left(\inf_{s \leq u <\infty } G_u > -\frac{c}{2}e^{ -s\left(k +\frac{1}{2} \right)} \right)\\[4pt] &= 1- \mathbb{P} \left(\sup_{s \leq u <\infty } G_u >\frac{c}{2}e^{ - s\left(k +\frac{1}{2} \right) } \right)\nonumber \\[4pt] &= 1-2\lim_{t\rightarrow \infty} \mathbb{P} \left(G_t>-\frac{c}{2}e^{ -s\left(k +\frac{1}{2}\right) } \right),\text{ by\,the\,reflection\,principle} \nonumber\\[4pt] &=1-2\mathbb{P} \left(N\left(0, \dfrac{ e^{-s(2k+1)} }{2k+1 }\right) > \frac{c}{2}e^{ -s(k +\frac{1}{2} ) } \right)\nonumber \\[4pt] &=1-2\mathbb{P}\left(N\left(0,\frac{1}{k+1}\right) > \frac{c}{2} \right)>\delta>0. \nonumber\end{align}

We know that on $\{X_t\geq 0 \text{\ i.o.}\}$ the event $\{X_t\geq c e^{-\frac{t}{2}} \text{\ i.o.} \}$ holds a.s. Therefore, on $\{X_t\geq 0 \text{\ i.o.}\}$, if we define $\tau_0=0$ and $\tau_{n+1} = \{t>\tau_n+1| X_t \geq c e^{-\frac{t}{2}} \}$, we see that $\tau_n<\infty$ a.s., and $\tau_n \rightarrow \infty$ a.s. Also, we define the corresponding filtration, namely $\mathcal{F}_n = \sigma( \tau_n)$.

To show that on the event $\{X_t\geq c e^{-\frac{t}{2}} \text{\ i.o.}\}$ the event $A=\{\liminf_{\rightarrow \infty} X_t\leq 0\}$ has probability zero, it suffices to argue that there is a $\delta$ such that $ \mathbb{P}( A|\mathcal{F}_n )<1-\delta$ a.s. for all $n\geq 1$. This is immediate from the previous calculation. Indeed,

\begin{align*}\mathbb{P}( A|\mathcal{F}_n ) & \leq 1-\mathbb{P} \left(\inf_{ \tau_{n}\leq u <\infty } X_u >\frac{c}{2}e^{ -\frac{\tau_n}{2} } |\mathcal{F}_n \right)\\&<1-\delta.\end{align*}

Now we can prove Theorem 3.1.

Proof of Theorem 3.1. From Proposition 3.2 we know that $\{X_t\geq 0 \text{\ i.o.} \}$ has probability 1. Therefore from Lemma 3.2 we deduce $\liminf_{t \rightarrow \infty} X_t >0$ a.s. Consequently, $ \int_{0}^\infty |X_u| { \rm d} u \rightarrow \infty$ a.s. At the same time $ \limsup_{t \rightarrow \infty} \int_{0}^{t} e^{-\frac{u}{2}} {\rm d } B_u < \infty$ a.s.; hence $X_t\rightarrow \infty$ a.s.

3.3. Analysis of $X_t$ when $k<1/2$

As before, $(X_t)_{t\geq 1}$ is the solution of the stochastic differential equation $dX_t= k |X_t|\rm{d}t +e^{-\frac{t}{2}} \text{d}B_t $.

The behavior of $X_t$ when $k<1/2$ is different. The process in this regime can converge to 0 with positive probability. More specifically, we have the following theorem.

Before proving the theorem, we first need to prove a proposition. We will show that the process, starting from a negative value, will never cross 0 with positive probability.

Proof. Define the stopping time $\tau_1 =\inf\{ t\geq s| X_t =0 \}$. As in Proposition 3.2, the solution for $X_t$ started at time s up to time $\tau_1$ is given by $X_t =e^{-kt} (e^{ks} X_{s} + \int_{s}^{t} e^{u(k-\frac{1}{2}) } {\rm d}B_u ) $. We have

\begin{align*}\mathbb{P}( \tau =\infty )&= \mathbb{P}\Big( \sup_{ s <u <\infty } X_u \leq 0 \Big)\\ &= 1-\lim_{t\rightarrow \infty }2 \mathbb{P}\Bigg( N\Bigg( 0, \frac{ e^{2t(k-\frac{1}{2}) } -e^{2t(k-\frac{1}{2}) } }{2k -1 } \Bigg) >-e^{ks} X_{s}\Bigg),\text{ as\,in\,Proposition\,} \linkref{pro3.2}{3.2} \\ &= 1- 2 \mathbb{P}\big( N\big( 0, -e^{2s(k-\frac{1}{2}) } /(2k-1) \big) >-e^{ks} X_{s}\big)\\&=1-\alpha.\end{align*}

Therefore $0<\alpha<1$.

Proof of Theorem 3.2.

1. 1. Define the events $N=\{ \exists s \text{s.t. } X_t <0 \forall t\geq s \}$ and $P=\{X_t \geq 0 \text{\ i.o.} \}$. Of course N and P are disjoint and $\mathbb{P}(P \cup N )=1$. To prove Part $\textit{1}$, we will show that $N\subset \{ X_t \rightarrow 0 \}$ up to a null set and $P= \{ X_t \rightarrow \infty \} $. From Lemma 2.2 we know that $\liminf_{t\rightarrow \infty} X_{t}\geq 0$ a.s.; therefore $N\subset \{X_t\rightarrow 0 \}$ up to a null set.

   To show that $P= \{ X_t \rightarrow \infty \} $, note that Lemma 3.2 shows that on $\{ X_t \geq 0 \text{\ i.o.} \} $, $\liminf_{t \rightarrow \infty }X_t >0$ a.s. Consequently, on $\{X_t \geq 0 \text{\ i.o.}\}$ we have $X_t\to \infty $, as $ \int_{0}^\infty |X_u| { \rm d} u \rightarrow \infty$ and $ \limsup_{t \rightarrow \infty} \int_{0}^{t} e^{-\frac{u}{2}} {\rm d } B_u < \infty$ a.s. Therefore, $P= \{ X_t \rightarrow \infty \} $, which concludes Part $\textit{1}$.

2. 2. The fact that $\mathbb{P}(A)>0$ follows immediately from Proposition 3.3. Now, we will prove that $\mathbb{P}(B)>0$. Define the stopping time $\tau_0=\inf\{t |X_t=0 \}$. Also, define $Y(t,\omega)=1$ if $X_s\geq 0$ for all $s\geq t+1$. Observe that $\{ Y_{\tau_0}=1, \tau_0 < \infty \} \subset P $. Hence, using the strong Markov property,

   \begin{align*} \mathbb{P}(Y_{\tau}=1, \tau < \infty ) &= \int_{0}^{\infty} \mathbb{ P} ( \tau=u ) \mathbb{P}_0 ( X_t\geq 0, \forall t\geq 1 ){\rm d} u\\ &\geq \int_{0}^{\infty} \mathbb{ P} ( \tau=u ) \mathbb{P}_0 ( K_t\geq 0, \forall t\geq 1 ){\rm d} u \quad \text{\,since\,}\,X_t\geq K_t\\ &=\alpha \mathbb{P}_0 ( K_t\geq 0, \forall t\geq 1 ) \\ &>0. \end{align*}

3. 3. This follows immediately from the proof of $\textit{1}$.

Lastly, we prove a proposition that will be used in Section 5.

Proof. Solving the SDE before it hits zero, we find $X_t =e^{-kt} (e^{ks} X_{s} + \int_{s}^{t} e^{u(k-\frac{1}{2}) } {\rm d}B_u ) $ and $Y_t =e^{-k_1t} (e^{k_1s} Y_{s} + \int_{s}^{t} e^{u(k_1-\frac{1}{2}) } {\rm d}B_u ) $. Let $\epsilon >0$. Since the process $G_{t}=\int_{s}^{t} e^{u(k-\frac{1}{2}) } {\rm d}B_u $ has finite quadratic variation, the event $A=\{G_{t}\in(-\epsilon,\epsilon) \forall t>s \}$ has positive probability. Set $\tilde{G}_{t}=\int_{s}^{t} e^{u(k_1-\frac{1}{2}) } {\rm d}B_u $, and define $N_t= G_te^{t(k_1-k)}$. Using Itô’s formula, we find $\text{d}N_t=e^{t(k-\frac{1}{2})} e^{(k_1-k) t} \text{d}B_t+ (k_1-k)e^{(k_1-k) t}G_t\text{d}t$. Therefore,

\begin{equation*}G_te^{(k_1-k) t}= \tilde{G}_t+ \int_{s}^t (k_1-k)e^{(k_1-k) u} G_u\text{d}u .\end{equation*}

So

\begin{equation*}G_te^{(k_1-k) t}-\int_{s}^t (k_1-k)e^{(k_1-k) u} G_u\text{d}u= \tilde{G}_t .\end{equation*}

To bound $|\tilde{G}_t|$ observe that

\begin{align*}-\int_{s}^t (k_1-k)e^{(k_1-k) u} G_u\text{d}u &\leq -\epsilon \int_{s}^t (k_1-k)e^{(k_1-k) u} \text{d}u\\[4pt] &=-\epsilon \left(e^{(k_1-k) t}- e^{(k_1-k) s}\right).\end{align*}

Similarly we obtain $-\int_{s}^t (k_1-k)e^{(k_1-k) u} G_u\text{d}u \geq \epsilon( e^{(k_1-k) t}- e^{(k_1-k) s})$. Thus on A, we obtain the following inequalities:

\begin{equation*}-\epsilon e^{(k_1-k) t}+\epsilon \left(e^{(k_1-k) t}- e^{(k_1-k) s}\right)\leq \tilde{G}_t \leq \epsilon e^{(k_1-k) t}-\epsilon\left(e^{(k_1-k) t}- e^{(k_1-k) s}\right) . \end{equation*}

Simplifying, we obtain $ | \tilde{G}_t | \leq \epsilon e^{(k_1-k) s}\leq \epsilon$. Now we will estimate $X_t$ on A. Using that $\epsilon<|e^{ks}X_s| $ we obtain the upper bound

\begin{align*}X_t &= e^{-kt} \left(e^{ks} X_{s} + \int_{s}^{t} e^{u(k-\frac{1}{2}) } {\rm d}B_u \right ) \\[4pt] &\leq e^{-kt} (e^{ks} X_{s} +\epsilon )\\[4pt] &<0\end{align*}

and the lower bound

\begin{align*}X_t &= e^{-kt} \left(e^{ks} X_{s} + \int_{s}^{t} e^{u(k-\frac{1}{2}) } {\rm d}B_u \right) \\[4pt] &\geq e^{-kt} (-2e^{ks} \epsilon -\epsilon )\\[4pt] &\geq -3\epsilon.\end{align*}

Doing similarly for $Y_t$, we conclude.

4.1. Introduction

As in the previous section, to simplify matters, we will work with reparametrizing $L_t$. Set $\theta (t) = t^{ \frac{1}{1-\gamma} }$, and let $ X_t= L_{ \theta (t)}$. To obtain the SDE that $X_t$ obeys, notice that $ {\rm d} B_{ \theta (t) } = \sqrt{\theta ' (t)} {\rm d} B_t $. Therefore

\begin{align*}{ \rm d }X_t &= \frac{|X_t|^k}{\theta(t)^{\gamma}} \theta'(t){ \rm d}t + \frac{1}{\theta(t)^\gamma} \sqrt{\theta ' (t)}{\rm d} B_t \\&= c_1|X_t|^k { \rm d}t + c_2t^{-\frac{\gamma}{1-\gamma}} \sqrt{\theta ' (t)}{\rm d} B_t \\&=c_1|X_t|^k { \rm d}t + c_2t^{-\frac{\gamma}{2(1-\gamma)}} {\rm d} B_t,\end{align*}

where $c_2^2=c_1= 1/(1-\gamma)$. By abusing the notation we set $ X_t = X_t/c_2$, which satisfies an SDE of the form

(12) \begin{equation}{ \rm d }X_t =c|X_t|^k { \rm d}t + t^{-\frac{\gamma}{2(1-\gamma)}} {\rm d} B_t,\end{equation}

where $k>1$, $\gamma\in(1/2,1)$, and $c\in(0,\infty)$. By a time scaling, we may assume that $X_t$ solves

(13) \begin{equation}{ \rm d }X_t =|X_t|^k { \rm d}t + t^{-\frac{\gamma}{2(1-\gamma)}} {\rm d} B_t,\end{equation}

where $k>1$ and $\gamma\in(1/2,1)$. Notice that the noise is scaled differently. However, it will be evident that only the order of the noise is relevant. The SDE (13) will be the primary focus of the next subsection, and the results will apply to solutions of (12) as well.

We define another process that will be fundamental for our analysis, namely $\smash{Z_t=-\frac{X_t}{h(t)}}$, where $h(t)= -t^{\frac{1}{1-k}} $. Next, we find the SDE that $Z_t$ satisfies.

Proposition 4.1. Suppose that $(X_t)_{t\geq 1}$ solve (12), and set $C(c)=\frac{1}{c(k-1)}$, $h(t)= -t^{\frac{1}{1-k}} $. Then the process $Z_t= -\frac{X_t}{h(t)}$ satisfies

(14) \begin{equation} Z_t - Z_s= \int_{ s }^{t}c\dfrac{X_u}{h(u) } \left(C\frac{|h(u)|^k }{h(u)} -\frac{|X_u| ^k }{X_u}\right) { \rm d } u + \int_{ s }^{t} -\frac{1}{h(u)}u^{-\frac{\gamma}{2(1-\gamma)}} { \rm d }B_u. \end{equation}

Also, before $X_t$ hits zero we get a solution purely in terms of $Z_t$:

(15) \begin{equation} Z_t - Z_s= \int_{ s }^{t}c|h(u)|^{k-1} Z_u \left(C- (-Z_u) ^ {k-1}\right) { \rm d } u + \int_{ s }^{t} -\frac{1}{h(u)}u^{-\frac{\gamma}{2(1-\gamma)}} { \rm d }B_u . \end{equation}

Proof. Recall that since h(t) is a continuous function, the covariance $ \langle h(t) , Z_t \rangle$ is 0. Using Itô’s formula we obtain

\begin{equation*} {\rm d} Z_t = -\frac{1}{h(t)} {\rm d} X_t + X_t { \rm d } \left(-\frac{1}{h(t) } \right) .\end{equation*}

Thus,

\begin{align*}Z_t - Z_s &= \int_{ s }^{t} -\frac{1}{h(u)} c|X_u| ^ k { \rm d } u + \int_{ s }^{t} -\frac{1}{h(u)}u^{-\frac{\gamma}{2(1-\gamma)}} { \rm d }B_u + \int_{ s }^{t} X_u\frac{h'(u)}{ h(u) ^2} {\rm d }u \\[8pt] &= \int_{ s }^{t} X_u\frac{h'(u)}{ h(u) ^2} -\frac{1}{h(u)} c|X_u| ^ k { \rm d } u + \int_{ s }^{t} -\frac{1}{h(u)}u^{-\frac{\gamma}{2(1-\gamma)}} { \rm d }B_u \\[8pt] &=\int_{ s }^{t}c\dfrac{X_u}{h(u) } \left(\frac{h'(u)}{c h(u) } - \frac{|X_u| ^k }{X_u}\right) { \rm d } u + \int_{ s }^{t} -\frac{1}{h(u)}u^{-\frac{\gamma}{2(1-\gamma)}} { \rm d }B_u \\[8pt] &=\int_{ s }^{t}c\dfrac{X_u}{h(u) } \left( \frac{1}{c(k-1)}\frac{|h(u)|^k }{h(u)} -\frac{|X_u| ^k }{X_u}\right) { \rm d } u + \int_{ s }^{t} -\frac{1}{h(u)}u^{-\frac{\gamma}{2(1-\gamma)}} { \rm d }B_u.\end{align*}

The SDE (15) is an immediate consequence of the last line of the above calculation.

In the next proposition, we describe some properties of the noise for the process $Z_t$, and we give a very important inequality for Subsection 4.2, which relates the order of the deterministic system converging to zero and the order of the remaining noise for $X_t$, i.e. the order of $\left \langle \int_s^\infty u^{-\frac{\gamma}{2(1-\gamma)}} {\rm d} B_u\right \rangle$.

Proof. We calculate its quadratic variation at time t, namely,

\begin{equation*}\langle{G'}_{s,t}\rangle =\int_{ s }^{t} \frac{1}{h(u)^2}u^{-\frac{\gamma}{1-\gamma}} { \rm d }u .\end{equation*}

Notice that by the definition of h(t), we have $h(t)^{-1} = \Theta \left(t^{\frac{1}{k-1}}\right)$; therefore

\begin{equation*}\frac{1}{h(u)^2}u^{-\frac{\gamma}{1-\gamma}} = \Theta \left(u^{\frac{2}{k-1}-\frac{\gamma}{1-\gamma } }\right). \end{equation*}

Consequently, $ \langle {G'}_{s,\infty}\rangle=\infty$ when

\begin{equation*}\frac{2}{k-1}-\frac{\gamma}{1-\gamma } \geq-1 \iff \frac{2}{k-1}+\frac{1}{\gamma-1 } \geq -2.\end{equation*}

In the first regime we have

(16) \begin{align}\frac{1}{2} +\frac{1}{2k} \geq \gamma &\iff \frac{k-1}{2k} \leq 1-\gamma \iff \frac{2k}{k-1} \geq \frac{1}{1-\gamma} \nonumber\\&\iff \frac{2}{k-1}+2 \geq \frac{1}{1-\gamma} \iff\frac{2}{k-1}+\frac{1}{1-\gamma} \geq -2. \end{align}

So, indeed, when $\frac{1}{2} +\frac{1}{2k}\geq \gamma$, $\langle{G'}_{s,\infty}\rangle =\infty.$ Also, from the previous calculation we see that when $\frac{1}{2} +\frac{1}{2k}<\gamma$,

(17) \begin{equation}\frac{2}{k-1}-\frac{\gamma}{1-\gamma } <-1; \end{equation}

therefore when $\frac{1}{2} +\frac{1}{2k}<\gamma$, $\langle{G'}_{s,t}\rangle <\infty$. Finally, rearranging the first inequality of (16), we obtain

\begin{equation*}\frac{1}{k-1}\geq \frac{2\gamma -1}{2(1-\gamma )} .\end{equation*}

The solution of the SDE (13), when $X_t$ is positive, explodes in finite time. However, since we are interested in the behavior of $X_t$ when $X_t<M$ for a positive constant M, we may change the drift when $X_t$ surpasses the value M, which in turn would imply that the SDE (13) admits strong solutions. One way to do this is by studying the SDE whose drift term is equal to $|x|^k$ when $x<M$ and $M^k$ when $x>M$. This SDE can be seen to admit strong solutions for infinite time. The reason is that this process is a.s. bounded from below, and the drift is positive. Also, $X_t$ cannot explode to plus infinity in finite time since the drift is bounded from above when $X_t$ is positive. However, for simplicity, we will use the form shown in (13).

4.2. Analysis of ${\textit{\textbf{X}}}_{\textit{\textbf{t}}}$ when $\frac12 + \frac{1}{2{\textit{\textbf{k}}}} \geq \gamma$, ${\textit{\textbf{k}}} > 1$, and $\gamma \in (\frac12,1)$

The main result of this section is the following theorem.

We will see its proof at the end of this subsection. Now we will prove an important proposition which shows that $X_t$ cannot stay far away from the left of the origin.

Proof. Set

\begin{equation*}{G'}_t =\int_{ s }^{t} -\frac{1}{h(u)}u^{-\frac{\gamma}{2(1-\gamma)}} { \rm d }B_u,\end{equation*}

which corresponds to the noise term of (15). First we will prove that $ \{ X_t \geq C' h(t) \text{\ i.o.} \}$ a.s., where $C'> C^{\frac{1}{k-1}}$ and $C=C(1)=\frac{1}{k-1}$. To do so, we will argue by contradiction. Assume that $A=\{ \exists\,s, X_t<C'\cdot h(t) \forall t>s \}$ has positive measure. Take $\omega \in A$, and find $s(\omega)$ such that $ X_t<C'\cdot h(t)$ for all $t>s$. Notice that this implies that $Z_t <-C'$ for $t>s$. Take $u>s$; since $ \frac{|x|^k}{x}$ is increasing we see that

\begin{equation*}\frac{ |X_u|^k }{X_u}< {C'}^{k-1} \frac{ |h(u)|^k }{h(u )} < C \frac{ |h(u)|^k }{h(u )} .\end{equation*}

This in turn gives

\begin{equation*} C \frac{ |h(u)|^k }{h(u )} -\frac{ |X_u|^k }{X_u}>0.\end{equation*}

Therefore

\begin{equation*}\int_{ s }^{t}\tfrac{X_u}{h(u) } \left(C\frac{ |h(u)|^k }{h(u )} -\frac{ |X_u|^k }{X_u}\right) { \rm d } u >0\end{equation*}

for all $t>0$. However, since the process

\begin{equation*}G'_{w,t} \,{:}\,{\raise-1.5pt{=}}\, \int_w^s -\frac{1}{h(u)}u^{-\frac{\gamma}{2(1-\gamma)}}{ \rm d } B_u \end{equation*}

for any fixed w has infinite quadratic variation, we may find $G'_{s,t} >-Z_s$. Now, from (14) we get

\begin{align*}Z_t &= \int_{ s }^{t}\dfrac{X_u}{h(u) } \left(C\frac{ |h(u)|^k }{h(u )} -\frac{ |X_u|^k }{X_u}\right) { \rm d } u + Z_s +G'_{s,t}\\&>0.\end{align*}

This contradicts the fact that $Z_t<-C'$. Therefore $ \{X_t> C'h(t) \text{\ i.o.} \}$ a.s.

Finally, in Proposition 4.2 Part 3 we have shown that $\frac{1}{k-1}\geq \frac{2\gamma -1}{2(1-\gamma )} $; therefore

\begin{equation*} - t^{ \frac{1}{1-k} }\geq - t^{ \frac{1-2\gamma }{2(1-\gamma)} }.\end{equation*}

So we conclude that there exists a constant $\beta<0$ such that $\{ X_t \geq \beta t^{ \frac{1-2\gamma}{ 2(1-\gamma)}}\ \text{\,i.o.} \} $ holds a.s.

Proof. Set $G_{s,t}=\int_s^t u^{-\frac{\gamma}{2(1-\gamma)}} {\rm d} B_u $, and note that $ \langle G_{s,\infty}\rangle = \Theta (s^{ \frac{1-2\gamma}{ (1-\gamma)} } )$. Fix $\gamma>0$; since $ \langle G_{s,\infty}\rangle = \Theta (s^{ \frac{1-2\gamma}{ (1-\gamma)} } )$ for any $u>0$, it is possible to find $W(u)>u>0$ such that

(18) \begin{equation}{}\mathbb{P}\left(\sup_{u<t<W(u)}G_{u,t} > \gamma u^{ \frac{1-2\gamma}{ 2(1-\gamma)} } \right)> \delta,\end{equation}

for $\delta$ independent of u. Take $\gamma>-\beta$, where $\beta$ is such that $\{X_t\geq \beta t^{\frac{1-2\gamma}{2(1-\gamma)}} \text{\ i.o.} \}$ (as in Proposition 4.3). Now, using the lower bound $ X_t-X_s \geq G_{s,t}$, we obtain

(19) \begin{equation} \mathbb{P}\left(\sup_{s<t<W(s)}X_t-X_s > \gamma s^{ \frac{1-2\gamma}{ 2(1-\gamma)} } \right)\geq \mathbb{P}\left(\sup_{s<t<W(s)}G_{s,t} > \gamma s^{ \frac{1-2\gamma}{ 2(1-\gamma)} } \right)> \delta. \end{equation}

When $X_s\geq \beta s^{\frac{1-2\gamma}{2(1-\gamma)}}$, observe that on the event

\begin{equation*}\Bigg\{ \sup_{s<t<W(s)}X_t-X_s> \gamma s^{ \frac{1-2\gamma}{ 2(1-\gamma)} } \Bigg\}\end{equation*}

there is $\tau_s$ such that

\begin{equation*}X_{\tau_s}\geq X_s+\beta s^{ \frac{1-2\gamma}{ 2(1-\gamma)} } =(\gamma+\beta)s^{ \frac{1-2\gamma}{ 2(1-\gamma)} } \geq 0 .\end{equation*}

Hence, if we choose a sequence of stopping times such that $X_{\tau_n}\geq \beta {\tau_n}^{\frac{1-2\gamma}{2(1-\gamma)}}$ and $\tau_{ n+1}> W(\tau_n)$, we have

\begin{equation*}\mathbb{P}\bigg(\sup_{\tau_n<t<\tau_{ n+1}}G_{\tau_n,t} > \gamma {\tau_n}^{ \frac{1-2\gamma}{ 2(1-\gamma)} }|\mathcal{F}_{\tau_n} \bigg)> \delta.\end{equation*}

So, by Borel–Cantelli (Lemma 2.5), on the events

\begin{equation*}\Bigg\{\sup_{\tau_n<t<\tau_{ n+1}}X_t-X_{\tau_n} > \gamma \tau_n^{ \frac{1-2\gamma}{ 2(1-\gamma)} } \Bigg\}\end{equation*}

we may conclude that $\{X_t\geq 0 \text{\ i.o.}\}$ has probability 1. Define $\tau_n$ as before, except that instead of $X_{\tau_n}\geq \beta {\tau_n}^{\frac{1-2\gamma}{2(1-\gamma)}}$ we set $X_{\tau_n}\geq 0$; by Borel–Cantelli we obtain that $\{X_t\geq \gamma t^{\frac{1-2\gamma}{2(1-\gamma)}} \text{\ i.o.}\}$ has probability 1.

Since $G_{s,t}$ is symmetric and $\langle G_{s,\infty}\rangle = \Theta \left(s^{ \frac{1-2\gamma}{ 1-\gamma} } \right)$, we have

\begin{align*}\mathbb{P}\left(\inf_{s<t<\infty }X_t-X_s > -\frac{\gamma}{2} s^{ \frac{1-2\gamma}{ 2(1-\gamma)} } \right)&\geq \mathbb{P}\left(\inf_{s<t<\infty}G_{s,t} > -\frac{\gamma}{2} s^{ \frac{1-2\gamma}{ 2(1-\gamma)} } \right)\\[5pt] &=1-\mathbb{P}\left(\sup_{s<t<\infty}G_{s,t} > \frac{\gamma}{2} s^{ \frac{1-2\gamma}{ 2(1-\gamma)} } \right)\\&>\delta'>0.\end{align*}

for some $\delta'$ independent of s.

Define $\tau_n$ such that $X_{\tau_n}\geq \gamma {\tau_n}^{\frac{1-2\gamma}{2(1-\gamma)}}$, and set $\mathcal{F}_{\tau_n}=\mathcal{F}_n$. To show that $A=\{\liminf_{\rightarrow \infty} X_t\leq 0\}$ has probability zero, it suffices to argue that there is a $\delta$ such that $ \mathbb{P}( A|\mathcal{F}_n )<1-\delta $ a.s. for all $n\geq 1$. This is immediate from the previous calculation. Indeed,

\begin{align*}\mathbb{P}( A|\mathcal{F}_n ) & \leq 1-\mathbb{P} \left(\inf_{ \tau_{n}\leq u <\infty } X_u -X_{\tau_n}>-\frac{\gamma}{2} {\tau_n}^{ \frac{1-2\gamma}{ 2(1-\gamma)} } |\mathcal{F}_n \right)\\&<1-\delta'.\end{align*}

Proof of Theorem 4.1. Since $X_t$ is a solution of (13), we have $X_t-X_1=\int_{1}^t |X_u|^k \text{d}u+G_{1,t} $. From Corollary 4.1 we know that $\liminf_{t \rightarrow \infty}X_t>0$ a.s.; therefore $\int_{1}^t |X_u|^k \text{d}u \rightarrow \infty$ a.s. However, since $\langle G_{1,\infty}\rangle<\infty$, we have that $\limsup_{t \rightarrow \infty }|G_{1,t}|<\infty$ a.s. Therefore, $X_t \rightarrow \infty$ a.s.

4.3 Analysis of ${\textit{\textbf{X}}}_{\textit{\textbf{t}}}$ when $\frac{1}{2} + \frac{1}{2{\textit{\textbf{k}}}} < \gamma $ and ${\textit{\textbf{k}}} > 1$

We now state the main theorem of this section, which we will prove at the end.

Lemma 4.1. Let $(Z_t)_{t\geq s}$ solve (14), and set

\begin{equation*}G'_{t}=\int_{ s }^{t} -\frac{1}{h(u)}u^{-\frac{\gamma}{2(1-\gamma)}} { \rm d }B_u .\end{equation*}

Suppose that $Z_s>-\left(\frac{C}{k} \right)^{\frac{1}{k-1}}$. Define

\begin{equation*}A= \Big\{G'_t \in (-\epsilon, \epsilon )\forall t \in (s, s+\delta), \text{and } G'_t \in \bigg(-2\epsilon ,-\frac{9}{10}\epsilon \bigg) \forall t \in (s+\delta,\infty)\Big\}.\end{equation*}

Then the following hold:

1. 1. $\mathbb{P}( A) >0$ for every $\epsilon,\delta >0$.

2. 2. For all $\epsilon>0$ small enough, there is $\delta>0$ such that $Z_t<-\dfrac{5\epsilon}{3}$ for every $t \in (s, s+\delta)$ on A.

3. 3. Define $\tau_C = \inf \Big\{t>s| Z_t= - 2 \left(\frac{C}{k} \right) ^{\frac{1}{k-1}} \Big\}$. Then $\tau_C>s+\delta$ a.s., where $\delta$ is the same as in Part 2.

Proof.

1. 1. This is immediate, since in Proposition 4.2 Part 2 we have shown that $\langle G'_{\infty} \rangle <\infty$.

2. 2. The first restriction on $\epsilon$ is such that $Z_s<-3\epsilon$. Next, we begin by defining $f_1$ and $f_2$ on $(s, s+\delta)$ satisfying

   (20) \begin{equation} f'(x) = c|h(x)|^{k-1} f(x) (C-(-f(x)^{k-1}) ), \end{equation}
   where c and C are the same as the parameters of the SDE (14), with initial conditions satisfying
   \begin{equation*} -\left(\frac{C}{k} \right)^{\frac{1}{k-1}} < Z_s +\epsilon< f_1(s ) <-\dfrac{5\epsilon}{3} \end{equation*}
   and
   \begin{equation*} -\left(\frac{C}{k} \right)^{\frac{1}{k-1}}<f_2(s ) < Z_s-\epsilon .\qquad\quad\end{equation*}
   Also, we define the function $ q(x) = x(C- (-x)^{k-1} ) $, whose derivative is $ q'(x) = C-k(-x)^{k-1}$, which implies that q(x) is increasing on $ \big( \big(-\frac{C}{k} \big)^{\frac{1}{k-1}} ,0 \big) $. This function will be important later. We should also note that f is decreasing in intervals where $f(x) \in \big( - \big(\frac{C}{k} \big)^{\frac{1}{k-1}} ,0 \big)$, since there $f'(x)<0$.

   We can pick the $\delta>0$ so that $f_2(t)>- \big(\frac{C}{k} \big)^{\frac{1}{k-1}}$ for every $t \in ( s, s+\delta ) $. We will show that $ Z_t>f_2(t)$ on $( s, s+\delta ) $ by contradiction. Using the SDE (15) for $Z_t$, we get that

   (21) \begin{equation} Z_t- Z_s = \int_{ s }^{t}c|h(u)|^{k-1} Z_u \left( C- (-Z_u )^ {k-1}\right) { \rm d } u + g(t), \end{equation}
   where g(t) is a continuous function such that $ \sup_{t\in ( s, s+\delta ) } |g(t)| \leq \epsilon. $ Assume that $f_2$ and Z become equal at some point, and choose t to be the first time. Using the integral form of (20), and subtracting it from (15), we get
   \begin{align*} 0&=Z_t- f_2(t) \\[2pt]&= \int_{ s }^{t} c|h(u)|^{k-1} Z_u \big(C- (-Z_u )^ {k-1}\big) -c|h(u)|^{k-1} f_2(u) \big(C- (-f_2(u)) ^ {k-1}\big) { \rm d } u \\[2pt]&\quad +Z_s- f_2(s)+ g(t) \\[2pt] &= (t-s)\big(c|h(\xi)|^{k-1} Z_{\xi} \big(C- ( -Z_{\xi}) ^ {k-1}\big) -c|h(\xi)|^{k-1} f_2(\xi) \big(C- (-f_2(\xi)) ^ {k-1}\big) \big) \\[2pt] &\quad +Z_s+g(t) - f_2(s)\\[2pt] &>(t-s)\big(c|h(\xi)|^{k-1} Z_{\xi} \big(C- (-Z_{\xi}) ^ {k-1}\big)-c|h(\xi)|^{k-1} f_2(\xi) \big(C- (-f_2(\xi)) ^ {k-1}\big)\big), \end{align*}
   where in the last line we used that $Z_s+g(t) - f_2(s)> 0$. Since $\xi <t$, we have that $ Z_{\xi}>f_2(\xi) > -\left(\frac{C}{k} \right)^{\frac{1}{k-1}} $, and consequently $ q(Z_{\xi} ) > q( f_2(\xi) ) $, so
   \begin{equation*} |h(\xi)|^{k-1} q(Z_{\xi} ) > |h(\xi)|^{k-1} q( f_2(\xi) ) .\end{equation*}
   Therefore,
   \begin{equation*}0 < c|h(\xi)|^{k-1} Z_{\xi} \left(C- (-Z_{\xi}) ^ {k-1}\right) -c|h(\xi)|^{k-1} f_2(\xi) \left(C- (-f_2(\xi)) ^ {k-1}\right), \end{equation*}
   which gives a contradiction.

   Arguing similarly we can show that $ f_1(t)> Z_t$ on $ (s, s+\delta)$, which completes Part 2.

3. 3. Finally, for Part 3 we observe that $Z_t>f_2(t)$ for $t\in (s,s+\delta)$; hence $\tau_C >s+\delta$ a.s.

Before proving the theorem we will need the following proposition.

Proof. The condition $1/2 + 1/2k <\gamma$, as has already been shown in Section 4.2, implies that $ \langle G'_{\infty} \rangle <\infty$. On the event A as defined in Lemma 2.3, using (14), we get the following upper and lower bounds for all $t\geq s+\delta$:

(22) \begin{equation} -\frac{X_t}{h(t)} \leq -\frac{X_s}{h(s)} +\int_{ s }^{t}c\dfrac{X_u}{h(u) } \left(C\frac{|h(u)|^k }{h(u)} -\frac{|X_u| ^k }{X_u}\right) { \rm d } u -\frac{9}{10}\epsilon,\end{equation}

(23) \begin{equation} -\frac{X_t}{h(t)} \geq -\frac{X_s}{h(s)} +\int_{ s }^{t}c\dfrac{X_u}{h(u) } \left(C\frac{|h(u)|^k }{h(u)} -\frac{|X_u| ^k }{X_u}\right) { \rm d } u -2\epsilon.\end{equation}

Claim: On the event A, $X_t < 0$ for all $t>s$.

Proof: We will argue by contradiction. Assume that $\mathbb{P}(\{ \tau_0 <\infty \}\cap A) >0$. We choose $\epsilon$ such that $ \frac{3\epsilon}{2}<C^{\frac{1}{k-1} } $. Now, define $\tau_{l} = \sup \{ t\leq \tau_0 |-\tfrac{X_t}{h(t) } = -\frac{3\epsilon}{2} \} $ and notice that Lemma 4.1 implies that $ \tau_{ l \epsilon }>s+\delta$, since $Z_t< -\frac{5 \epsilon}{3}$ on $(s,s+\delta)$. Also, on $\{ \tau_0 <\infty \}\cap A$ we have $\tau_{ l}<\infty$. Then from (23) we see that

\begin{equation*} \int_{ s }^{ \tau_{ l}}c\dfrac{X_u}{h(u) } \left(C \frac{|h(u)|^k }{h(u)} -\frac{|X_u| ^k }{X_u}\right) { \rm d } u \leq \frac{X_s}{h(s)} +\frac{\epsilon}{2} .\end{equation*}

Therefore,

(24) \begin{equation} -\frac{X_s}{h(s)} +\int_{ s }^{ \tau_{ l}}c\dfrac{X_u}{h(u) } \left(C\frac{|h(u)|^k }{h(u)} -\frac{|X_u| ^k }{X_u}\right) { \rm d } u -\frac{9}{10}\epsilon \leq -\frac{2\epsilon}{5} .\end{equation}

Now, notice that $ X_t > \frac{3}{2} \epsilon h(t)$ for every $t \in ( \tau_{l},\tau_0 )$, so if $w \in ( \tau_{l},\tau_0 ),$ we get

\begin{equation*} C\frac{|h(w)|^k }{h(w)} -\frac{|X_w| ^k }{X_w} < C\frac{|h(w)|^k }{h(w)}- C\frac{|h(w)|^k }{h(w)} =0, \end{equation*}

and of course $ \frac{X_w}{ h(w)} >0$. So we conclude that

(25) \begin{equation} \int_{ \tau_l }^{\tau_0}c\dfrac{X_u}{h(u) } \left(C \frac{|h(u)|^k }{h(u)} -\frac{|X_u| ^k }{X_u}\right) { \rm d } u <0.\end{equation}

Combining (24) and (25), we get that

\begin{align*}0 & =-\frac{X_{\tau_0}}{h(\tau_0)} \\[6pt] & \leq-\frac{X_s}{h(s)} +\int_{ s }^{\tau_0}c\dfrac{X_u}{h(u) } \left(C \frac{|h(u)|^k }{h(u)} -\frac{|X_u| ^k }{X_u}\right) { \rm d } u -\frac{9}{10}\epsilon \\[6pt] & = -\frac{X_s}{h(s)} +\int_{ s }^{\tau_l}c\dfrac{X_u}{h(u) } \left(C\frac{|h(u)|^k }{h(u)} -\frac{|X_u| ^k }{X_u}\right) { \rm d } u -\frac{9}{10}\epsilon\\[6pt] &\quad + \int_{ \tau_l }^{\tau_0}c\dfrac{X_u}{h(u) } \left(C\frac{|h(u)|^k }{h(u)} -\frac{|X_u| ^k }{X_u}\right) { \rm d } u \\[6pt] & \leq -\frac{2\epsilon}{5},\end{align*}

a contradiction.

We have developed all the tools necessary to prove the theorem.

Proof of Theorem 4.2. Define a stopping time $\sigma=\inf \{ t | Z_t>-\left(\frac{C}{k} \right)^{\frac{1}{k-1}} \}$. If the event $\{\sigma<\infty\}$ has positive probability, then Proposition 4.4 implies that $X_t$ converges to zero with positive probability. Indeed, recall from Lemma 1 that $\liminf_{t\rightarrow \infty}X_t \geq 0$ a.s. Therefore, since on the event A as in Lemma 4.1 we have $\limsup_{t \rightarrow \infty }X_t \leq 0$, we deduce that $\lim_{t\rightarrow \infty }X_t=0$. To finish the proof, it suffices to show that when $\{\sigma<\infty\}$ has zero probability, $X_t\rightarrow 0$ with positive probability. This is easy to see since $\mathbb{P}(\{\sigma<\infty\})=0$ implies that $Z_t$ never hits zero; therefore $\limsup_{t \rightarrow \infty }X_t \leq 0$ on $\{\sigma<\infty \}$.

We now prove a proposition that will be used in the next section.

Proof. Let $\tau_C = \inf \left\{t>s| Z_t= - 2 \left(\frac{C}{k} \right) ^{\frac{1}{k-1}} \right\}$, and define

\begin{equation*}\sigma= \sup \left\{\tau_C >t>s|Z_t= - \left(\frac{C}{k} \right) ^{\frac{1}{k-1}} \right\}.\end{equation*}

We will show that $\tau_C=\infty$ a.s. We assume otherwise, and reach a contradiction. From Lemma 4.1 Part 3, we know that $\tau_C>s+\delta$. Therefore,

\begin{align*}Z_{ \tau_C} &\geq Z_s+ \int_{ s }^{t}c|h(u)|^{k-1} Z_u \left(C- (-Z_u) ^ {k-1}\right) { \rm d } u -2\epsilon \\&\geq Z_{\sigma} -Z_{\sigma}+Z_s+ \int_{ s }^{t}c|h(u)|^{k-1} Z_u \left(C- (-Z_u) ^ {k-1}\right) { \rm d } u -2\epsilon \\&\geq Z_{\sigma} +\frac{9\epsilon}{10} -2\epsilon >-2\left(\frac{C}{k} \right) ^{\frac{1}{k-1}},\end{align*}

the desired contradiction.

6.1. Analysis of $X_t$ when $\frac{1}{2} + \frac{1}{2k} > \gamma $, $k>1$, and $\gamma\in(1/2,1)$

Before proving Theorem 1.4, as described in Section 1.2, we assume that $X_n$ satisfies

(29) \begin{equation} X_{n+1} - X_n\geq\frac{|X_n|^k}{n^{\gamma}} + \frac{Y_{n+1}}{n^{\gamma}}, \qquad k>1 \text{and} \gamma \in (1/2,1), \end{equation}

where the $Y_n$ are a.s. bounded and $\mathbb{E}(Y_{n+1}|\mathcal{F}_n)=0$. In this section we additionally require $Y_n$ to satisfy $ \mathbb{E}(Y_{n+1}^2|\mathcal{F}_n )\geq l>0 $.

Proof. From the restrictions on k we obtain

\begin{align*} \frac{1}{2} +\frac{1}{2k} \geq \gamma &\iff \frac{k-1}{2k}\leq 1-\gamma . \end{align*}

Set $h(t) = -t^{ \frac{1-\gamma}{1-k} }$, and define $Z_n = -\frac{X_n}{h_n} $. From here, on the event $\{X_m<0 \text{\,for\,all\,} m \geq n\}$, we get the following recursion:

(30) \begin{align} Z_{n+1}-Z_{n} & \geq-\frac{X_{n+1} }{h(n+1)} +\frac{X_{n} }{h(n)}\\ &\geq-X_n\left(\frac{1}{h(n+1)} - \frac{1}{h(n)} \right ) - \frac{ |X_n|^k }{n^\gamma h(n+1) } - \frac{ Y_{n+1} }{n^\gamma h(n+1)} \nonumber\\ &\geq X_n \frac{1-\gamma}{ k-1 } \xi_n ^{ -\frac{1-\gamma}{ 1-k }-1 }- \frac{ |X_n|^k }{n^{\gamma} h(n+1) }-\frac{ Y_{n+1} }{n^\gamma h(n+1)}, \text{where }\xi_n\in \left(n,n+1\right)\nonumber \\ &\geq \frac{ X_n }{h(n+1) n^\gamma} \left(\frac{1-\gamma}{ k-1 } \xi_n ^{ -\frac{1-\gamma}{ 1-k }-1 } h(n+1)n^\gamma-\frac{|X_n|^k }{X_n} \right) -\frac{ Y_{n+1} }{n^\gamma h(n+1)}\nonumber \end{align}

(31) \begin{align}\ge \frac{{{X_n}}}{{h(n + 1){n^\gamma }}}\left( { - {a_n}\frac{{1 - \gamma }}{{k - 1}}|h(n){|^{k - 1}} - \frac{{|{X_n}{|^k}}}{{{X_n}}}} \right) - \frac{{{Y_{1 + 1}}}}{{{n^\gamma }h(n + 1)}} \end{align}

(32) \begin{align}\ge \frac{{{X_n}}}{{h(n + 1){n^\gamma }}}\left( { - \frac{{2(1 - \gamma )}}{{k - 1}}|h(n){|^{k - 1}} - \frac{{|{X_n}{|^k}}}{{{X_n}}}} \right) - \frac{{{Y_{n + 1}}}}{{{n^\gamma }h(n + 1)}}, \end{align}

where

(33) \begin{equation} a_n = \frac{ \xi_n ^{\frac{ -(1-\gamma)}{1-k} -1 } h(n+1)n^{\gamma} }{ -|h(n) |^{k-1} } . \end{equation}

To justify the inequality (32), for large enough n, notice that $a_n \rightarrow 1$.

Define

\begin{equation*}{G'}_{s,n}=\sum_{ i=s}^{n-1} \frac{Y{i+1}}{i^{\gamma}h(i+1) } .\end{equation*}

We will see that ${G'}_{1,n}$ grows big enough so that $Z_n$ must, at certain times, get close enough to the origin so that $X_n$ surpasses a constant multiple of h(n). To this end, we have the following lemma.

We use the following theorem; for a reference see [Reference FisherFis92, Theorem 1, p. 676].

$\limsup_{n \rightarrow \infty}\sum_{i=1}^{n} \frac{X_i}{s_n\phi(s_n) }\geq \epsilon(K) $ a.s.

It is clear that ${G'_{ 1,n }}$ satisfies all the hypotheses required for the aforementioned theorem to hold.

From Lemma 6.1, it is immediate that for any random time s (not necessarily a stopping time), $\limsup_{n\rightarrow \infty} {G'}_{s,n} = \infty $ a.s.

Now, we return to the proof of Proposition 6.1. Assume that there is $n_0$ such that

\begin{equation*} X_n < -\left(\frac{3(1-\gamma)}{k-1}\right)^{\frac{1}{k-1}} n^{ \frac{1-\gamma}{1-k} } \end{equation*}

for all $n\geq n_0$. Then, since $ \frac{|x|^k}{x}$ is increasing, we get that

\begin{equation*} \frac{|X_n|^k}{X_n} <-\frac{3(1-\gamma)}{k-1} n^{ -1+\gamma } .\end{equation*}

Therefore,

\begin{equation*} -\frac{2(1-\gamma)}{ k-1 } |h(n)|^{k-1}-\frac{|X_n|^k }{X_n} > -\frac{2(1-\gamma)}{ k-1 } n^{-1+\gamma}+ \frac{3(1-\gamma)}{k-1} n^{ -1+\gamma } = \frac{(1-\gamma)}{k-1} n^{ -1+\gamma } >0 \end{equation*}

So

\begin{align*} Z_n&\geq Z_{n_0} +\sum_{i=n_0}^{n}\frac{ X_i }{h(i+1)i^\gamma} \left(\frac{2(1-\gamma)}{ k-1 } |h(i)|^{k-1}-\frac{|X_i|^k }{X_i} \right)+{G'}_{n_0,n} \\ &> Z_{n_0}+ {G'}_{n_0,n}, \end{align*}

which gives $ \limsup_{n \rightarrow \infty}Z_n = \infty $. This is a contradiction, since it would imply $X_n \geq 0$ infinitely often.

Since $ n^{\frac{1-\gamma}{1-k}} = o( n^{\frac{1-2\gamma}{2} })$, for every constant $\beta<0$ the event $\{ X_n \geq \beta n^{ \frac{1-2\gamma}{ 2} } \text{\,i.o.} \} $ holds a.s.

We define $ G_{n,u}=\sum_{ i=n}^{u-1} \frac{Y_{i+1}}{i^{\gamma} } $; this is an important quantity for the next lemma and the remainder of the section.

Proof. Define $\tau = \inf \{u\geq n| G_{n,u} \notin (-a_2n^{\frac{1-2\gamma}{2}},a_2n^{\frac{1-2\gamma}{2}} ) \} $. We calculate the stopped variance of $ G_{\tau} \,{:}\,{\raise-1.5pt{=}}\, G_{n,\tau}$. We will do so recursively; fix $m\geq n$ and calculate as follows:

\begin{align*} \mathbb{E} (( G_{ \tau \wedge {m+1} } ) ^2 |\mathcal{F}_n )-\mathbb{E} ( G_{\tau \wedge m} )^2 |\mathcal{F}_n )&= \mathbb{E} \left(1_{\tau>m} \left(2\frac{Y_{m+1}}{m^\gamma } G_{m} +\frac{ Y_{m+1}^2}{m^{2\gamma}} \right) |\mathcal{F}_n \right) \\ &=\mathbb{E} \left(1_{\tau>m} 2\frac{Y_{m+1}}{m^\gamma } G_{m}|\mathcal{F}_n \right) +\mathbb{E}\left(1_{\tau>m}\frac{ Y_{m+1}^2}{m^{2\gamma}} |\mathcal{F}_n \right) \\ &=0 + \mathbb{E}\left(1_{[\tau>m]} \mathbb{E}\left(\frac{ Y_{m+1}^2}{m^{2\gamma}} |\mathcal{F}_{m} \right) |\mathcal{F}_n \right)\\ &\geq \epsilon \frac{1}{m^{2\gamma}} \mathbb{E}(1_{[\tau>m]} |\mathcal{F}_n )\\ &\geq \epsilon \frac{1}{m^{2\gamma}} \mathbb{P}(\tau=\infty |\mathcal{F}_n ) . \end{align*}

Therefore,

(34) \begin{align} E (( G_{ \tau \wedge {m} } ) ^2 |\mathcal{F}_n )&\geq \mathbb{E} ( (G_{\tau \wedge n } )^2 |\mathcal{F}_n )+c\mathbb{P}(\tau=\infty |\mathcal{F}_n ) (n^{1-2\gamma} -(m-1)^{1-2\gamma} )\nonumber\\ &= c\mathbb{P}(\tau=\infty |\mathcal{F}_n ) (n^{1-2\gamma} -(m-1)^{1-2\gamma} ). \end{align}

Notice that since the $Y_n$ are a.s. bounded, $|G_\tau| \leq a_2 n^{\frac{1-2\gamma}{2} } +\frac{M}{n^{\gamma}}$, and since $\gamma > \gamma-1/2$, we get that $|G_\tau| \leq 2a_2n^{\frac{1-2\gamma}{2} } $ for n large enough. For m large, we can find a constant c such that $n^{1-2\gamma} -(m-1)^{1-2\gamma} \geq c'n^{1-2\gamma}$. Using (34), we obtain

\begin{equation*} \frac{ 2a_2n^{1-2\gamma}}{\epsilon c'n^{1-2\gamma}}=\frac{2a_2}{\epsilon c'}\geq \mathbb{P}(\tau=\infty |\mathcal{F}_n ). \end{equation*}

Choosing $a_2$ small enough, we may conclude $\mathbb{P} ( \tau < \infty | \mathcal{F}_n) > 1/2 $ for all n large enough.

Now we take any martingale $M_n$ starting at 0 such that it exits an interval $ (-2a,2a) $ with probability at least p, and $|M_{n+1}-M_n|< a $ a.s. Then we stop the martingale upon exiting the interval $(-2a,2a)$; that is, define $\tau_{-}$ to be the first time $M_n$ goes below $-2a$ and $\tau_{+}$ to be the first time that $M_n$ surpasses 2a, and set $\tau = \tau_{-} \wedge \tau_{+}$. Using the optimal stopping theorem for the bounded martingale $M_{\tau \wedge n}$ and taking n to infinity, we obtain

\begin{align*} 0=\mathbb{E}(M_{\tau})&\leq -2a \mathbb{P}(\tau_{-} < \tau_{+} ) +3a\mathbb{P}(\tau_{-}> \tau_{+} ) +2a\mathbb{P}(\tau =\infty)\\ &=-2ap +2a(1-p)+5a\mathbb{P}(\tau_{-}> \tau_{+} ) . \end{align*}

So $ \mathbb{P}(\tau_{-}> \tau_{+} ) \geq \dfrac{4p-2}{5} $, which implies that $\mathbb{P}(\sup_{n} M_n \geq 2a ) \geq \dfrac{4p-2}{5}$.

The previous argument applied to $G_{n,u}$, given $\mathcal{F}_n$, concludes the proof of the first part of the lemma. Indeed, since the probability p of exiting the interval is bigger than $1/2$, we may deduce that $\dfrac{4p-2}{5}>0$.

For the second part of the lemma, we use the following inequality: let $M_n$ be a martingale such that $M_0=0$ and $\mathbb{E}(M_n^2) <\infty$. Then

\begin{equation*}\mathbb{P} ( \max_{n\geq u \geq 0} M_u \geq \lambda ) \leq \frac{ \mathbb{E}( M_n^2 )}{\mathbb{E}( M_n^2 )+ \lambda^2 }\end{equation*}

(for a reference see [Reference DurrettDur13, Exercise 5.4.5, p. 213]). Let $\tau$ be the first time $G_{n,u}$ surpasses $ a_2 n^{1-2\gamma} $. Condition on $[\tau < \infty]$, and notice that $G_{n,\infty}> \frac{a_2}{2}n^{1-2\gamma}$ when $\inf_{ u\geq \tau}G_{\tau,u} > -\frac{a_2}{2}n^{1-2\gamma}$. Using the previous inequality and the fact that $\frac{x}{x+1}$ is increasing, we obtain

\begin{align*} \mathbb{P}(G_{n,\infty}\leq \frac{a_2}{2}n^{\frac{1-2\gamma}{2}} |\mathcal{F}_{\tau},[\tau <\infty] ) &\leq \mathbb{P}( \inf_{ u\geq \tau}G_{\tau,u} \leq -\frac{a_2}{2}n^{1-2\gamma} |\mathcal{F}_{\tau},[\tau <\infty] ) \\[4pt] &\leq \frac{ \mathbb{E}((G_{\tau,\infty} )^2 |\mathcal{F}_{\tau},[\tau <\infty] )}{\mathbb{E}((G_{\tau,\infty} )^2 |\mathcal{F}_{\tau},[\tau <\infty])+ \frac{a_2^2}{4}n^{1-2\gamma }} \\[4pt] &\leq\frac{ c\tau ^{1-2\gamma } }{ c\tau ^{1-2\gamma } + \frac{a_2^2}{4}n^{1-2\gamma } } \\[4pt] &\leq \frac{ c }{ c + \frac{a_2^2}{4} }. \end{align*}

Therefore,

\begin{equation*}\mathbb{P}(G_{n,\infty}\geq \frac{a_2}{2} n^{ \frac{1-2\gamma}{2}} |\mathcal{F}_n ) \geq \mathbb{P} (\tau <\infty) \frac{ \frac{a_2^2}{4} } { c + \frac{a_2^2}{4} }, \end{equation*}

which concludes the proof.

For any stopping time $\sigma$, we get the following version of the previous lemma.

Proof. For any m, n we get the lower bound $ X_m-X_n \geq G_{n,m} $. Now, we define an increasing sequence of stopping times $\tau_n$, going to infinity a.s., such that $X_{\tau_n} \geq \beta \tau_n^{\frac{1-2\gamma}{2} }$ for $|\beta|<a_1$, where $a_1$ is such that $ \mathbb{P}\left(\sup_{u\geq \sigma}G_{\tau_n,u}\geq a_1 {\tau_n}^{ \frac{1-2\gamma}{2}} |\mathcal{F}_{\tau_n} \right)> \delta_1 $, whose existence is guaranteed by Lemma 6.3. From Proposition 6.1, we can do so with all $\tau_n$ a.s. finite. Hence,

\begin{align*} \mathbb{P}\left(\sup_{\infty \geq u\geq {\tau_n}} X_u-X_{\tau_n}\geq a_1{\tau_n}^{ \frac{1-2\gamma}{2}} |\mathcal{F}_{\tau_n} \right) &\geq \mathbb{P}\left(\sup_{\infty \geq u\geq {\tau_n} } G_{\tau_n,u } \geq a_1 {\tau_n}^{ \frac{1-2\gamma}{2} }|\mathcal{F}_{\tau_n} \right)\\ &>\delta_1 >0 . \end{align*}

Therefore, by Borel–Cantelli on the event $\Big\{X_{\tau_n} \geq \beta \tau_n^{\frac{1-2\gamma}{2} } \text{\ i.o.} \Big \}$, we get $\{X_{\tau_n} \geq 0\ \text{\ i.o.} \}$. Therefore $\{X_n \geq 0\ \text{\ i.o.} \}$ holds a.s.

Proof of Theorem 6.1. Again we define an increasing sequence of stopping times $\tau_n$, going to infinity a.s., such that $X_{\tau_n} \geq 0$ this time. Since $ \mathbb{P}(G_{\tau_n,\infty}\geq a_1 \tau_n^{ \frac{1-2\gamma}{2}} |\mathcal{F}_{\tau_n} )>\delta_2$, an application of Borel–Cantelli shows that $\{X_n\geq \frac{a_1}{2}n^{ \frac{1-2\gamma}{2}} \text{\ i.o.} \}$ holds a.s. We claim that a.s. there are constants $c(\omega)>0$ and $m(\omega) $ such that $\{ X_n > c \mathrm{ for all } n\geq m \}=\{\liminf_{n\to \infty} X_n> 0\}$. Indeed, if we define $\tau_0=0$ and $\tau_{n+1} = \inf \{m>\tau_n+1| X_m \geq \frac{a_1}{2} m^{ \frac{1-2\gamma}{2}} \}$, we see that $\tau_n<\infty$ a.s. and $\tau_n \rightarrow \infty.$ This gives a corresponding filtration, namely $\mathcal{F}_n = \sigma( \tau_n)$.

To finish the claim, we show that $A=\{\liminf_{n\to \infty} X_n\leq 0\}$ has probability zero. To do so, it is sufficient to argue that there is a $\delta$ such that $ \mathbb{P}( A|\mathcal{F}_n )<1-\delta $ a.s. for all $n\geq 1$. This is immediate from the previous calculation. Indeed,

\begin{align*} \mathbb{P}( A|\mathcal{F}_n ) & \leq 1-\mathbb{P}\left(\liminf_{n } X_n \geq \frac{3a_1}{2}\tau_n^{ \frac{1-2\gamma}{2}} |\mathcal{F}_n \right)\\ &= 1-\mathbb{P} \left(\liminf_{n } X_n-\frac{a_1}{2} \tau_n^{ \frac{1-2\gamma}{2}} \geq a_1\tau_n^{ \frac{1-2\gamma}{2}} |\mathcal{F}_n \right) \\ &\leq 1-\mathbb{P} ( \liminf_{n }G_{\tau_n,n} \geq a_1\tau_n^{ \frac{1-2\gamma}{2}} |\mathcal{F}_n )\\ &<1-\delta_2. \end{align*}

The process $G_{m,\infty}$ is a.s. finite, and since the drift term $\sum_{ i\geq n } \frac{|X_i|^k}{i^{\gamma}} \to \infty$, we get that $X_n \to \infty $.

Proof of Theorem 1.4. Define $\tau=\inf \{n| X_n \in (-\epsilon, \epsilon)\}$ and $\tau ' =\inf \{n>\tau | X_n \not \in (-\epsilon ' , \epsilon ') \}$, where $\epsilon<\epsilon'$. When $\epsilon$ is small enough, we may assume that $\tau<\infty$ with positive probability; otherwise we have nothing to prove. On $\{\tau<\infty \}$, couple $X_n$ with $X'_n$ so that $\mathbb{P}(X_n=X'_n, \tau \leq n\leq \tau ' |[\tau <\infty] )=1 $, where $X_n'$ is a process that solves (29). Since $X'_n\rightarrow \infty$ a.s., we have that $\tau' <\infty$ a.s. Thus, on $\{\lim_{n \rightarrow \infty} X_n =0\} $, we have that $\{ X_n>\epsilon' \text{\ i.o.} \}$. Therefore, $\mathbb{P}(\lim_{n \rightarrow \infty} X_n =0) = 0$.

6.2. Analysis of $X_t$ when $\frac{1}{2} + \frac{1}{2k} < \gamma $, $k>1$, and $\gamma\in(1/2,1)$

Before proving the main Theorem 1.5, as described in Section 1.2 we will study a process $(X_n)_{n \geq 1}$ that satisfies

(35) \begin{equation} X_{n+1} - X_n\leq\frac{f(x)}{n^{\gamma}} + \frac{Y_{n+1}}{n^{\gamma}},\qquad \gamma \in (1/2,1),\,k\in (1,\infty), \end{equation}

where $0 < f(x) \leq |x|^k$ when $ x\in(-\epsilon,\epsilon) $, and $f(x) = |x|^k$ when $ x\in\mathbb{R}\setminus (-\epsilon,\epsilon) $.

The analysis will again rely on studying the process $Z_{n}=-\frac{X_n}{h(n)}$, where $h(t)= -t^{ \frac{1-\gamma}{1-k}}$. An important quantity related to the process $Z_n$ will be $G'_{s,n} = \sum_{i=s}^{n-1} \frac{Y_{i+1}}{i^{\gamma}h(i+1)} $.

Recall that the $Y_n$ constitute martingale differences satisfying $|Y_n|<M$ almost surely. Furthermore, we find $x_0<0$ such that $f(x)>M$ for every $x\leq x_0$. We will make use of $x_0$ in the next lemma.

Proof. We can show this by induction. Of course $X_1> -2C$. For the inductive step, we distinguish two cases. First, assume that $-2C<X_n<-C$. Then

\begin{align*} X_{n+1} &= X_n +\frac{f(X_n)}{n^\gamma} + \frac{Y_{n+1}}{n^{\gamma}} \\[4pt] &\geq -2C + \frac{f(X_n)}{n^\gamma} -\frac{M}{n^{\gamma}}\\[4pt] &> -2C. \end{align*}

Now, assume that $X_n\geq -C$. Then

\begin{align*} X_{n+1} &= X_n +\frac{|X_n|^k}{n^\gamma} + \frac{Y_n}{n^{\gamma}} \\[4pt] &\geq -C + 0 -\frac{M}{n^{\gamma}}\\[4pt] &> -2C. \end{align*}

Pick $\epsilon >0$ such that

\begin{equation*} \epsilon \leq \min \left( \frac{1}{4}, \frac{1}{2} \left(\frac{1-\gamma}{3(k-1)}\right)^{ \frac{1}{k-1} } \right) .\end{equation*}

Let $a_n$ be defined as in Section 6.1, first appearing in (31) and defined in (33).

Claim: We can find $n_0$ that satisfies the following properties:

1. 1. $a_n >1/2$ for $n\geq n_0$ a.s.

2. 2. If $-\frac{X_{n+1}}{h(n+1)}>-2\epsilon$ and $-\frac{X_{n}}{h(n)}\leq-2\epsilon$, then $-\frac{X_{n+1}}{h(n+1)} <-\epsilon$, when $n\geq n_0$.

3. 3. $\mathbb{P} ({G'}_{n_0,n} \in (\frac{-\epsilon}{2}, \frac{\epsilon}{2}) \forall n \geq n_0 |\mathcal{F}_{n_0} ) >0$.

Proof.

1. 1. This is trivial.

2. 2. Recall that $h(t)=-t^{\frac{1-\gamma}{1-k}}$. Since $|Y_n|<M$, and $X_n>-2C$ a.s., whenever $X_n<0$ we have $|X_{n+1}-X_{n}|=O(n^{-\gamma})$. Also, $ n^{-\gamma} = o(h(n) )$, since $\gamma > \frac{1-\gamma}{k-1}$. Indeed, $\gamma > \frac{1-\gamma}{k-1}$ is equivalent to $ \gamma >1/k = 1/2k+1/2k$; however, $ 1/2>1/2k$, and since $\gamma > 1/2 + 1/2k$, we conclude. Furthermore, notice that $\dfrac{h(n)}{h(n+1)}\rightarrow 1$. Calculate

   \begin{align*} -\frac{X_{n+1}}{h(n+1)} &= -\frac{X_{n+1}-X_{n} }{h(n+1)} -\frac{X_{n}}{h(n)}\cdot\frac{h(n)}{h(n+1)}\\[4pt] &\geq o(1)-2\epsilon \frac{h(n)}{h(n+1)}. \end{align*}
   Since the o(1) term and $\frac{h(n)}{h(n+1)}$ depend only on n, we conclude Part 2.

3. 3. Using the inequality (17) we find that $\frac{1-\gamma}{k-1} - \gamma <\frac{-1-\delta}{2}$ for some $\delta>0$, so

   \begin{equation*} \left(m^{\gamma} h(m+1)\right)^{-1}\sim m^{\frac{1-\gamma}{k-1} - \gamma } \leq m^{\frac{1-\gamma}{k-1} - \gamma }\leq m^{\frac{-1-\delta}{2}} .\end{equation*}
   Therefore, by Doob’s inequality we have
   \begin{align*} \mathbb{P}\left(\sup_{u\geq n_0 } ({G'}_u^{n_0} |\mathcal{F}_{n_0})^2 \geq \frac{\epsilon^2}{4}\right) &\leq \sum_{m\geq n_0 } \frac{\mathbb{E}(Y_{m+1}^2|\mathcal{F}_{n_0} ) }{m^{2\gamma} h^2(m+1) }\\[5pt] &\leq C\sum_{m\geq n_0 } \frac{1 }{m^{2\gamma} h^2(m+1) }\\[5pt] &=\sum_{m\geq n_0 } \Theta \left(m^{2\left(\frac{1-\gamma}{k-1} - \gamma \right) } \right)\\[5pt] &=\sum_{m\geq n_0 } \Theta \left(m^{-1-\delta} \right )\\[5pt] &= \Theta \left({n_0}^{-\delta} \right) \to 0. \end{align*}

Notice that the previous claim holds for any stopping time $ \tau$, in place of n. So we obtain a version of the previous lemma for stopping times.

Let

\begin{equation*} \epsilon \leq \min \left( \frac{1}{4}, \frac{1}{2} \left(\frac{1-\gamma}{3(k-1)} \right)^{ \frac{1}{k-1} } \right ) ,\end{equation*}

and define a stopping time $\tau=\inf \{n\geq n_0 | Z_n < -2\epsilon \} $.

Proof. On the event $\{X_m<0 \text{for\,all} m\geq n \}$ we use the expression for $Z_n = -\dfrac{X_n}{h(n)}$ and obtain, as in (31) (except now the inequalities are reversed),

\begin{align*} Z_{n+1}-Z_n &\leq \frac{ X_n }{h(n+1) n^\gamma} \left(-a_n \frac{1-\gamma}{ k-1 } |h(n)|^{k-1}-\frac{|X_n|^k }{X_n} \right ) -\frac{ Y_{n+1} }{n^\gamma h(n+1)}\\ & < \frac{ X_n }{h(n+1) n^\gamma} \left(- \frac{1-\gamma}{ 2(k-1) } |h(n)|^{k-1}-\frac{|X_n|^k }{X_n} \right ) -\frac{ Y_{n+1} }{n^\gamma h(n+1)}. \end{align*}

Set

\begin{equation*}D_n=\dfrac{ X_n }{h(n+1) n^\gamma} \left(- \dfrac{1-\gamma}{ 2(k-1) } |h(n)|^{k-1}-\dfrac{|X_n|^k }{X_n} \right ) .\end{equation*}

Then we have

(36) \begin{equation} Z_m - Z_{\tau}\leq \sum_{ i = \tau }^{m-1} D_i + {G'}_{\tau,m}. \end{equation}

Now we will show by contradiction that on the event $A=\{{G'}_{\tau,n} \in (-\frac{\epsilon}{2}, \frac{\epsilon}{2}) \forall n \geq \tau \}$ the process satisfies $X_n < 0$ for all $n \geq \tau$. Define $\tau_0 =\inf \{n\geq \tau| Z_n \geq 0 \}$ and $\sigma = \sup \{ \tau \leq n <\tau_0 | Z_{n-1}\leq -2\epsilon, \,Z_n>-2\epsilon \} $. Also, when $Z_n\geq -2\epsilon$ we have $X_n \geq 2\epsilon h(n)= -2\epsilon n^{ \frac{1-\gamma}{1-k}} $. So $ \frac{|X_n|^k}{ X_n } \geq -(2\epsilon)^{k-1} n^{-1+\gamma } $. Therefore, by the definition of $\epsilon$, we get

\begin{equation*}- \dfrac{1-\gamma}{ 2(k-1) } |h(n)|^{k-1}-\dfrac{|X_n|^k }{X_n} < \left(- \dfrac{1-\gamma}{ 2(k-1) }+ \dfrac{1-\gamma}{ 3(k-1) }\right)n^{-1+\gamma}=-\dfrac{1-\gamma}{ 6(k-1) } n^{-1+\gamma} <0. \end{equation*}

Hence $D_n <0$ whenever $Z_n\geq -2\epsilon$. If $\{\tau_0<\infty \}\cap A$ has positive probability, then $\{\sigma<\infty \}\cap A$ does also. Thus, on $\{\tau_0<\infty \}\cap A$,

\begin{align*} 0\leq Z_{\tau_0} &= Z_{\tau} + \sum_{ i = \tau }^{\tau_0-1} D_i + {G'}_{\tau,\tau_0} \\ &= Z_{\tau} -Z_{\sigma} +Z_{\sigma} + \sum_{ i = \tau }^{\tau_0-1} D_i + {G'}_{\tau,\tau_0} \\ &=Z_{\sigma}-{G'}_{\tau,\sigma} +{G'}_{\tau,\tau_0}+\sum_{ i = \sigma }^{\tau_0-1} D_i\\ &<-\epsilon +\frac{\epsilon}{2} +\frac{\epsilon}{2} +0=0, \end{align*}

which is a contradiction.

Now we can complete the proof of the proposition. On the event A the process satisfies $X_n<0$ for all $n>\tau$; therefore $ \limsup_{ n \rightarrow \infty} X_n \leq 0$ on A. However, by Lemma 2.4 we have $\limsup_{ n \rightarrow \infty} X_n \geq 0$ a.s. Therefore, on A, $X_n\rightarrow 0$.

Remark: On A we showed that $X_n$ converges to zero, since for all $n\geq \tau$, $X_n<0$, and the only place to converge is the origin.

Proof of Theorem 1.5. We define $\tau=\inf \{n\geq n_0| X_n \in (-\epsilon_2,-\epsilon_1 ) \}$, where $n_0$ is the same as in Lemma 6.5, and $\tau_e=\inf\{n| X_n\not \in (-3\epsilon, 3\epsilon) \}$. Let $(X'_n\,:\, n\geq \tau)$ be a process that satisfies (35). Then we couple $(X_n)$ with $(X'_n)$ on $\{ \tau <\infty \}$ so that $\mathbb{P}(X_n = X'_n, \tau \leq n \leq \tau_e | \{ \tau <\infty \} ) =1$. To show that $X'_n$ converges to zero with positive probability, first we need to verify that the conditions for Proposition 6.2, are met. The only thing we need to check is that $Z'_\tau= - \frac{X'_\tau}{h(\tau)} <-2\epsilon$. However, since $h(t)\rightarrow 0$ this is always possible by choosing $n_0$ large enough. Furthermore, by Proposition 6.2, we see that there is an event of positive probability such that $X'_n\rightarrow 0$, where $\tau_e$ is infinite conditioned on this event. Therefore, $X_n$ converges to 0 with positive probability.