# Asymptotic behaviour of an integral. How should I proceed?

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 $$(4)$$?

• Is anything known about the signs of $u$ and $i$? – Iosif Pinelis Sep 22 at 13:40
• No, nothing as far as I know @IosifPinelis – Strictly_increasing Sep 22 at 13:45
• 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 $x=1$ – 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 $$a,k$$.