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 : $$ \mathbb{P}\left(\tau_{0}^x < \infty\right)=0$$


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.