CIR Rates: Positivity, Boundary Behavior, and Transition Laws

The CIR Short-Rate Diffusion

The Cox–Ingersoll–Ross (CIR) model makes the volatility of the short rate depend on its current level. On a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$, consider the diffusion

dr_t=\kappa(\theta-r_t)\,dt+\sigma\sqrt{r_t}\,dB_t,\qquad r_0\ge0,

where $\kappa>0$ is the mean-reversion speed, $\theta>0$ is the long-run mean, and $\sigma>0$ is the volatility scale. The coefficient $\kappa(\theta-r)$ is the drift: it pulls the rate toward $\theta$. The coefficient $\sigma\sqrt r$ is the diffusion coefficient: the instantaneous random fluctuation has magnitude proportional to $\sqrt r$. In particular, the diffusion vanishes at zero.

For a pricing model, these dynamics are typically specified under the risk-neutral measure $\mathbb{Q}$, with $B_t$ then understood to be Brownian under $\mathbb{Q}$; for statistical descriptions of observed rates, one may instead specify dynamics under $\mathbb{P}$. The measure matters: the parameters used under $\mathbb{Q}$ need not be the physical-measure parameters. In what follows, write $\mathbb{R}$ for whichever measure governs the stated SDE, and take $B_t$ to be Brownian under $\mathbb{R}$. Any Ito integral displayed below has a progressively measurable integrand $H_s$ satisfying $\mathbb{E}^{\mathbb{R}}\int_0^T H_s^2\,ds<\infty$ on the interval in question.

We use the standard existence and uniqueness result for the CIR equation: for these parameters and a nonnegative initial value, it has a nonnegative continuous solution, unique up to indistinguishability. The square-root coefficient is not globally Lipschitz at zero, so the usual globally Lipschitz existence theorem does not apply directly. The boundary question is more specific: does the process started strictly above zero ever reach zero?

Feller’s Boundary Condition

Definition: Set $\nu=4\kappa\theta/\sigma^2$. The Feller condition is $2\kappa\theta\ge\sigma^2$, equivalently $\nu\ge2$. It ensures that a CIR process started at $r_0>0$ does not reach zero. If $0<\nu<2$, zero is attainable, but the nonnegative CIR solution is not absorbed there.

The factor of $2$ is important. One way to see the boundary threshold is through the scale and speed densities for a one-dimensional diffusion with drift $b(x)=\kappa(\theta-x)$ and squared diffusion coefficient $a(x)=\sigma^2x$. On $x>0$, their standard forms are

s'(x)=\exp\left(-\int^x\frac{2b(y)}{a(y)}\,dy\right) =x^{-\nu/2}e^{2\kappa x/\sigma^2}, \qquad m(x)=\frac{2}{a(x)s'(x)}.

Boundary calculation. The standard scale-speed boundary classification says that zero is inaccessible from the interior when the scale integral diverges at zero; when the scale integral is finite, accessibility is tested by the corresponding speed-integral criterion. This classification theorem is a standard result for one-dimensional diffusions; its proof is beyond this tutorial. Here, the scale integral near zero is governed by $\int_0^c x^{-\nu/2}\,dx$. It diverges exactly when $\nu/2\ge1$, or $\nu\ge2$, and is finite when $\nu<2$. For $0<\nu<2$, write $S(x)=\int^x s'(y)\,dy$, choosing any finite reference point. Then $S(x)-S(0)$ is of order $x^{1-\nu/2}$, while $m(x)$ is of order $x^{\nu/2-1}$. Their product is of order $1$, so the accessibility integral is finite. Thus zero is attainable in this regime. The CIR boundary classification also identifies this attainable boundary as instantaneously reflecting: paths remain nonnegative and do not stay at zero for a positive time interval.

Worked Example: Testing the Feller Condition

Take $\kappa=2$, $\theta=0.03$, and $\sigma=0.2$, with an initial rate $r_0>0$. Determine whether zero is attainable.

First compute $2\kappa\theta=2\cdot2\cdot0.03=0.12$. Next compute $\sigma^2=0.2^2=0.04$. Since $0.12\ge0.04$, the Feller condition holds. Equivalently, $\nu=4\cdot2\cdot0.03/0.04=0.24/0.04=6$, which is at least $2$. Therefore zero is unattainable from a strictly positive starting rate.

Check: the equivalent threshold $\sigma^2\le2\kappa\theta$ reads $0.04\le0.12$, giving the same conclusion. This says zero is not reached; it does not say that every future rate must exceed its initial value.

Worked Example: Attainable Zero Does Not Mean Absorption

Take $\kappa=1$, $\theta=\tfrac18$, and $\sigma=1$. Here $2\kappa\theta=2\cdot1\cdot\tfrac18=\tfrac14$, while $\sigma^2=1$. Thus $\tfrac14<1$, so the Feller condition fails. Also $\nu=4\cdot1\cdot\tfrac18/1=\tfrac12<2$. From a strictly positive starting value, zero is attainable. Nevertheless, the CIR solution remains nonnegative and is instantaneously reflecting at zero; it is not the process that gets stuck there.

Check: the strict inequality $0<\nu<2$ applies here because $\nu=\tfrac12$. In contrast, if $\nu=2$, the scale integral diverges at zero, and zero is unattainable from the interior.

The Conditional Transition Law

For $h>0$, define the positive scale parameter $c_h$ and the noncentrality parameter $\lambda_h$ by

c_h=\frac{\sigma^2(1-e^{-\kappa h})}{4\kappa}, \qquad \lambda_h=\frac{e^{-\kappa h}r_t}{c_h}.

A noncentral chi-square random variable with degrees of freedom $\nu>0$ and noncentrality $\lambda\ge0$, written $\chi'^2_\nu(\lambda)$, has Laplace transform $\mathbb{E}[e^{-uY}]=(1+2u)^{-\nu/2}\exp(-\lambda u/(1+2u))$ for $u\ge0$. Its conditional transition law for the CIR process is the following.

Theorem: For the CIR process with $\kappa,\theta,\sigma>0$, and $h>0$, the transition law given $\mathcal{F}_t$ is $$ r_{t+h}\mid\mathcal{F}_t\ \overset{d}{=}\ c_h\,\chi'^2_\nu(\lambda_h), \qquad \nu=\frac{4\kappa\theta}{\sigma^2}. $$

Proof. First take a deterministic starting value $r_t=x\ge0$. For $u\ge0$, set $\tau=T-s$, $q=e^{-\kappa\tau}$, and $$ D(\tau)=\frac{u e^{-\kappa\tau}}{1+\frac{u\sigma^2}{2\kappa}(1-e^{-\kappa\tau})}, \qquad A(\tau)=\left(1+\frac{u\sigma^2}{2\kappa}(1-e^{-\kappa\tau})\right)^{-2\kappa\theta/\sigma^2}. $$ Let $v(s,x)=A(T-s)e^{-D(T-s)x}$. Direct differentiation gives $v_x=-Dv$, $v_{xx}=D^2v$, and $v_s=-v_\tau$. The generator is $\mathcal{L}f(x)=\kappa(\theta-x)f'(x)+\tfrac12\sigma^2x f''(x)$, so $$ \frac{\mathcal{L}v}{v}=-\kappa\theta D+\left(\kappa D+\tfrac12\sigma^2D^2\right)x. $$ Differentiating the displayed formulas for $A,D$ gives $D'=-\kappa D-\tfrac12\sigma^2D^2$ and $A'/A=-\kappa\theta D$. Consequently $v_s+\mathcal{L}v=0$, and $v(T,x)=e^{-ux}$.

By Ito’s formula, the drift of $v(s,r_s)$ is therefore zero, and its stochastic-integral term is $\sigma\sqrt{r_s}\,v_x(s,r_s)\,dB_s$. This integrand is square-integrable on a finite interval for a deterministic starting value: $D$ is bounded there, $0\le v\le1$, and the CIR first moment is finite. Thus the Ito integral is a martingale and has conditional expectation zero. It follows that $$ \mathbb{E}^{\mathbb{R}}[e^{-u r_T}\mid\mathcal{F}_t] =\left(1+\frac{u\sigma^2}{2\kappa}(1-e^{-\kappa h})\right)^{-\nu/2} \exp\left( -\frac{u e^{-\kappa h}r_t} {1+\frac{u\sigma^2}{2\kappa}(1-e^{-\kappa h})} \right). $$ For a random finite $r_t$, the same identity follows by applying the deterministic-start calculation conditionally, or by localization and bounded convergence. Substituting $c_h$ and $\lambda_h$ into the noncentral chi-square Laplace transform gives exactly this expression: $1+2c_hu=1+\frac{u\sigma^2}{2\kappa}(1-e^{-\kappa h})$, and $\lambda_h c_hu/(1+2c_hu)=u e^{-\kappa h}r_t/(1+2c_hu)$. Laplace transforms determine probability laws on $[0,\infty)$, proving the transition formula. $\square$

The noncentral chi-square law can also be described as a Poisson mixture: if $N$ is Poisson with mean $\lambda/2$, then conditionally on $N$, the variable has a central chi-square law with $\nu+2N$ degrees of freedom. Since $\nu>0$, these component distributions have no atom at zero. Thus at any strictly positive time the transition distribution has no atom at zero, even in the parameter regime where paths can hit zero.

Worked Example: A Transition Distribution and Its Moments

Let $\kappa=1$, $\theta=2$, $\sigma=2$, and $h=\log 2$, and condition on $r_t=1$. Then $e^{-\kappa h}=e^{-\log2}=\tfrac12$, and $c_h=4(1-\tfrac12)/(4\cdot1)=\tfrac12$. The degrees of freedom are $\nu=4\cdot1\cdot2/4=2$, and $\lambda_h=(\tfrac12\cdot1)/(\tfrac12)=1$. Therefore $$ r_{t+h}\mid r_t=1\ \overset{d}{=}\ \tfrac12\chi'^2_2(1). $$ For $Y\sim\chi'^2_\nu(\lambda)$, the mean is $\nu+\lambda$ and the variance is $2(\nu+2\lambda)$, as follows from the Poisson-mixture description. Hence the conditional mean is $\tfrac12(2+1)=\tfrac32$, and the conditional variance is $(\tfrac12)^2\,2(2+2)=2$.

Check: the mean-reversion formula gives $2+(1-2)e^{-h}=2+(-1)\cdot\tfrac12=\tfrac32$. The variance formula gives $1\cdot4\cdot\tfrac12\cdot\tfrac12+2\cdot4/(2\cdot1)\cdot(\tfrac12)^2=1+1=2$, agreeing with the distribution calculation.

Conditional Moments and Interpretation

Corollary: For $h>0$, the CIR transition law has conditional mean and variance $$ \mathbb{E}^{\mathbb{R}}[r_{t+h}\mid\mathcal{F}_t] =\theta+(r_t-\theta)e^{-\kappa h}, $$ $$ \operatorname{Var}^{\mathbb{R}}(r_{t+h}\mid\mathcal{F}_t) =\frac{\sigma^2 r_t}{\kappa}e^{-\kappa h}(1-e^{-\kappa h}) +\frac{\theta\sigma^2}{2\kappa}(1-e^{-\kappa h})^2. $$

Proof. For $Y\sim\chi'^2_\nu(\lambda_h)$, the Poisson-mixture representation gives $\mathbb{E}[Y]=\nu+\lambda_h$ and $\operatorname{Var}(Y)=2(\nu+2\lambda_h)$. Since $r_{t+h}$ has the conditional law $c_hY$, its conditional mean is $c_h(\nu+\lambda_h)$. Using $c_h\nu=\theta(1-e^{-\kappa h})$ and $c_h\lambda_h=e^{-\kappa h}r_t$ gives the stated mean. Its variance is $2c_h^2(\nu+2\lambda_h)$. Substituting $c_h=\sigma^2(1-e^{-\kappa h})/(4\kappa)$, $\nu=4\kappa\theta/\sigma^2$, and $\lambda_h=e^{-\kappa h}r_t/c_h$, and then collecting the terms, gives the displayed variance. $\square$

The conditional mean moves exponentially toward $\theta$, while the conditional variance depends both on the current level and on the long-run level. The transition law is also useful computationally: sampling from a scaled noncentral chi-square distribution preserves nonnegativity, unlike a direct Gaussian approximation to a one-step rate change. These facts concern the rate dynamics and transition law; pricing a zero-coupon bond requires a separate valuation calculation.

Common Pitfall: The Factor of Two

A common error is to test $\kappa\theta\ge\sigma^2$ instead of $2\kappa\theta\ge\sigma^2$. For example, with $\kappa=1$, $\theta=\tfrac12$, and $\sigma=1$, the incorrect test gives $\kappa\theta=\tfrac12<1=\sigma^2$ and would label zero attainable. The correct calculation is $2\kappa\theta=2\cdot1\cdot\tfrac12=1=\sigma^2$, so the Feller condition holds exactly and zero is unattainable from a strictly positive initial rate. The threshold is an equality case, not a strict inequality.

Also distinguish “nonnegative” from “strictly positive.” Nonnegativity is a property of the CIR solution throughout this parameter range. Strict positivity at all times when started above zero requires the Feller condition. When that condition fails, the process can reach zero, but it is not absorbed there.

Check Your Understanding

In each question, take $\kappa,\theta,\sigma>0$ and a strictly positive initial rate unless stated otherwise.

For $\kappa=3$, $\theta=0.1$, and $\sigma=0.6$, does the Feller condition hold? Show both sides of the inequality.
Write $c_h$, $\lambda_h$, and $\nu$ for the transition law over a time interval of length $h>0$.
Explain why failure of the Feller condition permits zero to be reached but does not imply absorption.
Use the scaled noncentral chi-square representation to obtain the conditional mean of $r_{t+h}$.
For $r_t=0$ and $h>0$, what is the noncentrality parameter, and does the transition distribution have an atom at zero?