Let us consider the following SDE: $$dY_t=b(Y_t)dt+\sigma(Y_t)dW_t\tag{1}$$ with $b, \sigma: (l, r)\to\mathbb{R}$, $−\infty \leq l < r \leq \infty$ bounded functions on compact intervals of $(l, r)$.
In particular, $$b(Y_t)=(u-(u+i)Y_t)$$ $$\sigma(Y_t)=o\sqrt{(Y_t)(1-Y_t)}$$ with $u$, $i$ and $o$ arbitrary parameters.
Hence, focus will be on the following SDE: $$dY_t=(u-(u+i)Y_t)dt+o\sqrt{(Y_t)(1-Y_t)}dW_t\tag{2}$$ I must check whether the process $\{X_t\}$ remains within the interval $(l,r)$ or not for each $0\leq t\leq T$.

To this, I use the Feller test for explosions. Such a test requires that the following two integrals must be defined and computed: $$p(x)=\int_c^x \exp\bigg\{-2\int_c^{\xi}\frac{b(\zeta)}{\sigma^2(\zeta)}d\zeta\bigg\}d\xi\tag{3}$$ $$v(x)=\int_c^x\frac{2(p(x)-p(y))}{p\hspace{0.1cm}'(y)\sigma^2(y)}dy\tag{4}$$ with $c\in(l,r)$.
According to Feller test, probability that the process at least touches the bounds of interval $I$ equals $1$ or is less than $1$ according to whether $v(l+)=v(r-)=\infty$ or not. Let us fix $(l,r)=(0,1)$ and $c=\frac{1}{2}$.

I would like to study the asymptotic behaviour of the integral $(4)$ with $c=\frac{1}{2}$ at bounds $l=0$ and $r=1$, but I have not any experience with analyses like that. Is there a good standard method or is it just a matter of manipulation? Could you please help me understand how could I study asymptotic behaviour of $(4)$?

  • $\begingroup$ Is anything known about the signs of $u$ and $i$? $\endgroup$ – Iosif Pinelis Sep 22 at 13:40
  • $\begingroup$ No, nothing as far as I know @IosifPinelis $\endgroup$ – Strictly_increasing Sep 22 at 13:45
  • $\begingroup$ In a certain sense, if I properly understood Feller's test spirit, it could help me find values for parameters $i$, $u$, $o$ such that bounds are NOT touched by the process. In the first instance, however, the problem would consist in understanding the asymptotic beahviour of $v(x)$ for $x=0$ and $x=1$ $\endgroup$ – Strictly_increasing Sep 22 at 13:53

Direct calculations show that for $a:=i/o^2$, $k:=u/o^2$, and $x\in(1/2,1)$ $$p(x)=\int_{1/2}^x(2-2s)^a(2s)^k\,ds,$$ whence $$p'(x)=(2-2x)^a(2x)^k\asymp(1-x)^a$$ and, for $a>-1$ and $y\in(1/2,x)$, $$p(x)-p(y)=\int_y^x(2-2s)^a(2s)^k\,ds\asymp\int_y^x(1-s)^a\,ds \asymp(1-y)^{a+1}-(1-x)^{a+1}\le(1-y)^{a+1}.$$ So, $$v(x)\ll\int_{1/2}^x\frac{dy}{(1-y)^{a+1}}\,(1-y)^{a+1}\le1.$$ So, the right endpoint, $r=1$, is not touched by the process for any $a>-1$. The consideration of the case $a\le-1$ is similar, with the same conclusion, so that the right endpoint, $r=1$, is not touched by the process for any real $a,k$.

The left endpoint, $l=0$, is considered similarly. Here we can also use the symmetry $x\leftrightarrow1-x$ and $a\leftrightarrow k$, which reduces the consideration of the left endpoint to the already completed consideration of the right endpoint. Thus, we conclude that the left endpoint, $l=0$, is not touched by the process for any real $a,k$.

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