Hitting probability for mean-reverting stochastic process

Question

I quote Delbaen and Shirakawa (2002).

Starting from a stochastic differential equation of the form: $$dr_t=\alpha\left(r_{\mu}-r_t\right)dt+\beta\sqrt{\left(r_t-r_m\right)\left(r_M-r_t\right)}dW_t\tag{1}$$ with $\left\{W_t\right\}_{t\geq0}$ a standard Wiener process in the filtered probability space $\left(\Omega,\mathcal{F},\left\{\mathcal{F}_n\right\},\mathbb{P}\right)$. We assume $\alpha,\beta>0$ and $r_m=0<r_{\mu}<r_M=1$, which guarantee the existence of stationary distribution.
Let us first consider the variable transformation: $$z_t=\dfrac{r_t-r_m}{r_M-r_m}$$ whence SDE (1) can be rewritten as: $$dz_t=\alpha(\gamma-z_t)dt+\beta\sqrt{z_t(1-z_t)}dW_t\tag{2}$$ with $\gamma=\dfrac{r_{\mu}-r_m}{r_M-r_m}$.
Let us consider $r_m=0$ as a lower bound and $r_M=1$ as an upper bound. Let $\tau_y$ be the stopping time: $$\tau_y=\inf\left\{t\geq0: z_t=y\right\}, y\in\left\{r_m=0, r_M=1\right\}$$ Then, let $\rho_{x,y}$ be the probability that $z_t$ hits $y$ in finite time when it starts from $x$. Namely: $$\rho_{x,y}=\mathbb{P}\left(\tau_y<\infty|z_0=x\right)$$ Then, it holds that: $$\rho_{x,0}=\lim\limits_{y\to0, z\to0}\dfrac{B_{x,z}(p,q)}{B_{y,z}(p,q)}\tag{3}$$ $$\rho_{x,1}=\lim\limits_{y\to0, z\to0}\dfrac{B_{y,x}(p,q)}{B_{y,z}(p,q)}\tag{4}$$ where: \begin{cases} p=1-\dfrac{2\alpha\gamma}{\beta^2}\\ q=1-\dfrac{2\alpha(1-\gamma)}{\beta^2}\\ B_{x,y}(u,v)=\displaystyle{\int_x^y}z^{u-1}(1-z)^{v-1}dz \end{cases}

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What I cannot really understand is the bold part, in particular $(3)$ and $(4)$, with $B_{x,y}(u,v)$, $p$ and $q$ defined as it immediately follows below $(3)$ and $(4)$ $\bigg($For example, in $(4)$ I would expect $\lim\limits_{y\to\color{red}{1},z\to\color{red}{1}}(\cdots)\bigg)$. Why are they defined that way? Could you please give me some explanation for such "results"?

Answer 1

As the OP suggests, the confusion appears to be due to a typo in (3) and (4). Here are the corrected limits. \begin{align} \tag{$\star$} \rho_{x,0} &= \lim_{y \downarrow 0, \color{red}{z \uparrow 1}} \frac{B_{x,z}(p,q)}{B_{y,z}(p,q)} \;, \quad \rho_{x,1} = \lim_{y \downarrow 0, \color{red}{z \uparrow 1}} \frac{B_{y,x}(p,q)}{B_{y,z}(p,q)} \;. \end{align} This typo does not alter the boundary classification in the paper.

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At the outset of the corresponding section in the paper, the authors suppose $( z_t )_{t \ge 0}$ is stopped whenever it hits either boundary at $0$ or $1$. Therefore, the asymptotics of $( z_t )_{t \ge 0}$ can be determined from $$ \tag{$\dagger$} P\left[ \text{$( z_t )_{t \ge 0}$ hits $d$ before $c$} \right] = \frac{s(x)-s(c)}{s(d) - s(c)} = \frac{B_{c,x}(p,q)}{B_{c,d}(p,q)} \;, \quad \text{$0 \le c < x < d \le 1$ } \;, $$ where $B_{x,y}(u,v)$ is as defined by the OP, and $s(x)$ is a scale function of $( z_t )_{t \ge 0}$ defined as $$ s(x) := \int_{x_0}^x \exp\left( - \int_{x_0}^z \frac{2 \alpha (\gamma - y)}{\beta^2 y (1-y) } dy \right) dz = (1-x_0)^{1-q} x_0^{1-p} B_{x_0,x}(p,q) \;, $$ for some fixed $x_0 \in (0,1)$.

There are two especially interesting special cases of ($\star$) worth highlighting.

1. If $s(d) \uparrow \infty$ as $d \uparrow 1$ and $s(c) \downarrow -\infty$ as $c \downarrow 0$, then ($\dagger$) implies that: (i) both boundary points are unattainable, (ii) the process is recurrent over $(0,1)$, and (iii) ($\star$) reduces to $\rho_{x,0}=\rho_{x,1}=0$.
2. If both boundary points are attainable (i.e., $s(0)>-\infty$ and $s(1)<\infty$), then ($\dagger$) implies that: all interior states are transient, and for $x\in(0,1)$, ($\star$) becomes \begin{align} \rho_{x,0} &= \lim_{y \downarrow 0, z \uparrow 1} \frac{B_{x,z}(p,q)}{B_{y,z}(p,q)} = \frac{s(1)-s(x)}{s(1)-s(0)} >0\;, \quad \text{and} \\ \rho_{x,1} &= \lim_{y \downarrow 0, z \uparrow 1} \frac{B_{y,x}(p,q)}{B_{y,z}(p,q)} = \frac{s(x)-s(0)}{s(1)-s(0)} >0 \;. \end{align}