I quote [Delbaen and Shirakawa (2002)](https://link.springer.com/article/10.1023/A:1024125430287).



> 
> 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"?



Hitting probability for mean-reverting stochastic process
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}


