For the Cox-Ingersoll-Ross model $$\text{d}r_t = a(b-r_t)\text{d}t+\sigma\sqrt{r_t}\text{d}W_t$$ the condition (referred to as "Feller condition") $$2ab\geq\sigma^2$$ ensures that the solution is bounded below by zero. I see it being used all the time but I can't find a good source for citation... Can anybody help me out here?


1 Answer 1


For $x,K > 0 $, denote by $(X_t^x)_{t\geq 0}$ the unique strong solution of the CIR sde starting from $x$ at time 0.

Define the following stopping time :

$$\tau_{K}^x := \inf\left\{t\geq 0 : \quad X_t^x = K \right\} $$

Let us define furthermore the function $\psi$ defined on $\mathbb{R}_+^*$ by : $$ \forall \ x > 0 : \quad \psi(x) := \int_{1}^{x} y^{-\frac{2ab}{\sigma^2}}\exp(\frac{2ay}{\sigma^2})dy$$ For $0< \varepsilon < x < K$, define the following stopping time : $\tau_{\varepsilon, K}^x = \min(\tau_K^x, \tau_{\varepsilon}^x)$. Then, one can easily prove that $\tau_{\varepsilon, K}^x$ is a.s finite and for all $x \in (\varepsilon, K)$ we have : $$ \psi(x) = \psi(\varepsilon)\mathbb{P}\left(\tau_{\varepsilon}^x < \tau_K^x\right) + \psi(K) \mathbb{P}\left(\tau_{\varepsilon}^x > \tau_K^x\right)$$ Now, suppose that the feller condition is satisfied. Then, noticing that $\displaystyle \lim_{x\to 0^+}\psi(x) = - \infty$, one can prove that : $$\forall K> 0 : \quad \mathbb{P}\left(\tau_{0}^x < \tau_K^x\right) = 0$$ And so that : $$ \mathbb{P}\left(\tau_{0}^x < \infty\right)=0$$


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