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}

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}

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.