# Hitting probability for mean-reverting stochastic process

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 $$r_m=0, which guarantee the existence of stationary distribution.
Let us first consider the variable transformation: $$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}$$

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"?

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.
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}