The Cox-Ingersoll-Ross SDE is
$$dr_t=a(b-r_t)dt+\sigma\cdot \sqrt{r_t}dB_t, r_0>0.$$
It can be shown that there exist a solution to this SDE. But I am wondering, how do we show that the solution is positive? I understand that since we have a square-root in the equation only a positive solution makes sense?, can we use that argument? Even though that argument might be valid I would also like another argument for positivity. So let us define a modified SDE:
$$dr_t=a(b-r_t)dt+\sigma\cdot \sqrt{|r_t|}dB_t, r_0>0.$$
Let us assume there exist a solution to this SDE, this SDE could in theory be negative some places? Is there then a way to show that this solution must be positive? My only idea is to show that for every $t$ it is positive a.s., then it is a.s. positive for the rationals, and then the result follows from continuity. But I am unsure to show positivity for a given $t$, could you please help me?
To avoid being too presumptuous about the non-negativity of the CIR process, let's instead write $$dr_t = a( b - r_t)dt + \sigma \sqrt{r_t \lor 0} dB_t \qquad (\star)$$ where $r_t \lor 0 = \max (r_t, 0)$. Let $\epsilon > 0$ and define $\tau_{-\epsilon} = \inf \{t > 0 \, : \, r_t = -\epsilon\}$; that is, the first hitting time of $r_t$ at the level $-\epsilon$. We want to show that $P(\tau_\epsilon = \infty) = 1$. Suppose not, and let $\omega \in \Omega$ be such that $\tau (\omega ) < \infty$. Then, there exists a $u > 0$ such that for all $ s\in (\tau (\omega) - u, \tau (\omega ))$ we have $r_s(\omega) <0$. On this interval the SDE $(\star)$ can be locally expressed as: $$dr_t = a(b-r_t)dt$$
But then $a(b-r_s(\omega)) > 0$ for all $s\in (\tau (\omega) - u, \tau (\omega ))$, making $r_t$ strictly increasing before hitting $-\epsilon$, a contradiction.