CIR Behavior When Zero Is Accessible

When the CIR Boundary Can Be Reached

The Cox–Ingersoll–Ross (CIR) process has a drift that pulls the state toward its long-run level and a diffusion coefficient that vanishes at zero. When the drift’s inward push is strong enough, a process started above zero stays strictly positive. When that condition fails, zero becomes accessible: a path starting above zero can reach it in finite time. Accessibility does not mean absorption. With positive mean-reversion level, the drift at zero is positive, and the process cannot remain there for a whole interval.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Fix parameters $\kappa>0$, $\theta>0$, and $\sigma>0$. A nonnegative CIR solution with initial state $X_0=x\ge0$ satisfies

dX_t=\kappa(\theta-X_t)\,dt+\sigma\sqrt{X_t}\,dB_t.

Here $\kappa(\theta-X_t)$ is the drift and $\sigma\sqrt{X_t}$ is the diffusion coefficient. The stochastic integral is the Ito integral: on each finite time interval its integrand is progressively measurable and belongs to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T \sigma^2X_s\,ds<\infty$. We consider the standard nonnegative CIR solution, which has this finite-moment property. The boundary parameter is

\alpha=\frac{2\kappa\theta}{\sigma^2}.

The Feller strict-positivity condition is $\alpha\ge1$. For a process started from $x>0$, the earlier result on CIR nonnegativity and strict positivity gives strict positivity in this regime. The focus here is the complementary case $0<\alpha<1$: zero is accessible from the interior.

Definition: For a process started at $x>0$, zero is accessible if it has positive probability of reaching zero in finite time. Write $\tau_a=\inf\{t\ge0:X_t=a\}$, with $\inf\varnothing=\infty$. The phrase “leaves zero immediately” will mean that, after any finite time $\tau$ at which $X_\tau=0$, every interval $(\tau,\tau+\varepsilon]$, $\varepsilon>0$, contains a time at which $X_t>0$. It does not mean that the process stays strictly positive after $\tau$.

The Scale Function and Access to Zero

The generator of the CIR diffusion, acting on a twice continuously differentiable function $f$, is $\mathcal{L}f(x)=\kappa(\theta-x)f'(x)+\tfrac12\sigma^2x f''(x)$. A scale function $s$ is an increasing solution of $\mathcal{L}s=0$ on $(0,\infty)$. Choose its additive constant by setting $s(0)=0$ when the integral is finite. Solving the first-order equation for $s'$ gives the scale density

s'(x)=x^{-\alpha}\exp\left(\frac{2\kappa x}{\sigma^2}\right),\qquad s(x)=\int_0^x y^{-\alpha}\exp\left(\frac{2\kappa y}{\sigma^2}\right)\,dy.

For $0<\alpha<1$, the integrand behaves like $y^{-\alpha}$ near zero, and $\int_0^c y^{-\alpha}\,dy=c^{1-\alpha}/(1-\alpha)<\infty$. Thus $s(0)$ is finite. The standard one-dimensional diffusion boundary-classification theorem says that zero is accessible when the scale integral is finite there and the corresponding speed-integrability condition holds. For CIR, the speed density is $m(y)=2/(\sigma^2y s'(y))$, proportional near zero to $y^{\alpha-1}$; since $s(y)-s(0)$ is of order $y^{1-\alpha}$, their product is bounded near zero and integrable. This boundary-classification theorem is a standard result; its general proof is beyond this tutorial.

Theorem: Let $0<\alpha<1$, and start the CIR process at $0<x<R$. Then zero is accessible, and its probability of being hit before $R$ is $$ \mathbb{P}_x(\tau_0<\tau_R)=\frac{s(R)-s(x)}{s(R)-s(0)} =1-\frac{\int_0^x y^{-\alpha}e^{2\kappa y/\sigma^2}\,dy} {\int_0^R y^{-\alpha}e^{2\kappa y/\sigma^2}\,dy}>0. $$ Here $\mathbb{P}_x$ denotes probability for initial state $x$.

Proof. First fix $0<\varepsilon<x<R$ and put $\rho=\tau_\varepsilon\wedge\tau_R$. On $[\varepsilon,R]$, the diffusion coefficient is bounded away from zero and the coefficients are bounded, so the standard exit result for a nondegenerate diffusion on a bounded interval gives $\rho<\infty$ almost surely. The scale function is bounded on this interval. Ito’s formula, including its second-order correction, gives

ds(X_t)=s'(X_t)\kappa(\theta-X_t)\,dt +s'(X_t)\sigma\sqrt{X_t}\,dB_t +\frac12s''(X_t)\sigma^2X_t\,dt.

By $\mathcal{L}s=0$, the two $dt$ terms sum to zero. Up to time $\rho$, the stochastic integrand $s'(X_t)\sigma\sqrt{X_t}$ is bounded, progressively measurable, and square-integrable on every finite interval. The stopped stochastic integral is therefore a martingale: its conditional expectation increments are zero by the Ito integral’s martingale property, and its second moment is finite by the Ito isometry. Since $s(X_{t\wedge\rho})$ is bounded, taking expectations and then letting $t$ increase to infinity gives $\mathbb{E}_x[s(X_\rho)]=s(x)$. If $p_\varepsilon=\mathbb{P}_x(\tau_\varepsilon<\tau_R)$, continuity gives $X_\rho=\varepsilon$ on that event and $X_\rho=R$ otherwise. Therefore $s(x)=p_\varepsilon s(\varepsilon)+(1-p_\varepsilon)s(R)$, or $p_\varepsilon=(s(R)-s(x))/(s(R)-s(\varepsilon))$. As $\varepsilon\downarrow0$, this probability tends to $(s(R)-s(x))/(s(R)-s(0))>0$. The boundary-classification result stated above identifies this limiting event with hitting the accessible endpoint zero before $R$, giving the formula. $\square$

The proof uses an ordinary martingale only after stopping inside $[\varepsilon,R]$, where the integrand is bounded. It does not assert that an unstopped scale process is a true martingale on the whole state space. Local martingales need not be true martingales unless suitable integrability or uniform-integrability conditions are verified.

Worked Boundary Calculations

Worked Example: A CIR Process with Accessible Zero

Take $\kappa=2$, $\theta=1$, and $\sigma=3$, with initial state $x=1$ and upper level $R=4$. First test the boundary condition:

\alpha=\frac{2\kappa\theta}{\sigma^2} =\frac{2\cdot2\cdot1}{3^2}=\frac49<1.

Thus zero is accessible. The scale density and scale function are

s'(y)=y^{-4/9}e^{4y/9},\qquad s(z)=\int_0^z y^{-4/9}e^{4y/9}\,dy.

Substitution into the hitting formula gives the exact probability

\mathbb{P}_1(\tau_0<\tau_4) =1-\frac{\int_0^1 y^{-4/9}e^{4y/9}\,dy} {\int_0^4 y^{-4/9}e^{4y/9}\,dy}.

Both integrals are finite because the singularity at zero has exponent $-4/9>-1$; the numerator is strictly positive, and the denominator is strictly larger because its integral includes the positive contribution from $1<y<4$. Hence the displayed probability is strictly between zero and one. Check: the upper endpoint is above the initial state, so hitting zero before $4$ is a genuine competing-boundary event; the ratio is positive and less than one, as required.

Worked Example: The Critical Parameter Is Different

Set $\kappa=1$, $\theta=\tfrac12$, and $\sigma=1$. Then

\alpha=\frac{2\cdot1\cdot(1/2)}{1^2}=1.

This is the critical case, not the accessible regime $0<\alpha<1$. Its scale density is $s'(y)=y^{-1}e^{2y}$, and $\int_0^c y^{-1}e^{2y}\,dy=\infty$ for every $c>0$, because $e^{2y}\ge1$ and $\int_0^c y^{-1}\,dy=\infty$. The boundary is therefore not accessible from a positive initial state under the standard CIR boundary classification. Check: the calculation uses equality $\alpha=1$; replacing the strict condition $\alpha<1$ by $\alpha\le1$ would wrongly classify this critical case as accessible.

Why the Process Does Not Stay at Zero

At zero, the diffusion coefficient vanishes, but the drift does not: $\kappa(\theta-0)=\kappa\theta>0$. The SDE itself rules out absorption on any time interval. This conclusion is pathwise and does not require an appeal to uniqueness in law or a comparison of different solutions.

Proposition: Suppose $\kappa,\theta,\sigma>0$ and $X$ is a continuous nonnegative CIR solution. Almost surely, there is no finite stopping time $\tau$ and no $\varepsilon>0$ for which $X_t=0$ for every $t\in[\tau,\tau+\varepsilon]$. Consequently, if $X_\tau=0$, the process leaves zero immediately in the sense defined above.

Proof. The SDE’s integral form holds simultaneously for all times outside one null set. Suppose, on a path in this full-probability set, $X_t=0$ throughout $[\tau,\tau+\varepsilon]$. Subtract the integral equation at $\tau$ from the one at $\tau+u$, for any $0<u\le\varepsilon$. The drift integral becomes $\int_\tau^{\tau+u}\kappa\theta\,ds=\kappa\theta u$. The stochastic integrand is $\sigma\sqrt{X_s}=0$ throughout the interval, so the Ito integral increment is zero. Thus $X_{\tau+u}-X_\tau=\kappa\theta u$. The assumed zero path makes the left side zero, while $\kappa\theta u>0$, a contradiction. If the process did not leave zero immediately, some positive-length interval after $\tau$ would contain no positive values; nonnegativity would force it to be identically zero there, which has just been ruled out. $\square$

There is no conflict between this result and the vanishing diffusion coefficient at the boundary. At the instant $X_t=0$, the stochastic term has zero coefficient, while the positive drift prevents the state from remaining at zero. Once the process is positive, its diffusion is active again. In the accessible regime it may return to zero later; immediate departure does not mean that it cannot hit zero repeatedly, or that it remains strictly positive after its first visit.

Worked Example: Testing an Absorbed Candidate

Let $\kappa=1$, $\theta=\tfrac14$, and $\sigma=1$, and suppose a candidate path is $X_t=0$ on an interval $[\tau,\tau+u]$ with $u>0$. The parameters give $\alpha=2\cdot1\cdot\tfrac14/1^2=\tfrac12<1$, so this is an accessible-boundary regime. On the proposed zero path the diffusion increment is $\int_\tau^{\tau+u}\sqrt{X_s}\,dB_s=0$, and the drift increment is

\int_\tau^{\tau+u}\left(\frac14-X_s\right)\,ds =\int_\tau^{\tau+u}\frac14\,ds =\frac{u}{4}.

The SDE would therefore require $X_{\tau+u}-X_\tau=u/4$, but the candidate path gives $X_{\tau+u}-X_\tau=0$. Since $u/4>0$, it fails the SDE on every positive-length interval. Check: if $u=2$, the required increment is $2/4=1/2$, still not the candidate increment $0$. The contradiction comes from the drift, not from an approximation to the noise.

Interpretation and Common Pitfalls

Accessibility is a statement about a positive probability of reaching the boundary, not a claim that every path reaches it. Nor does it specify what happens after a visit; that requires examining the SDE at the boundary. For positive $\kappa\theta$, the integral equation excludes a period of absorption. The standard boundary terminology for this CIR regime is an instantaneously reflecting boundary: the process can reach zero and departs without a holding interval. The term should not be mistaken for a guarantee of permanent positivity after departure.

A common error is to infer absorption from $\sigma\sqrt{0}=0$. For example, setting $X_t=0$ after a hit and then saying “the noise is zero, so the solution stays zero” drops the drift term. The corrected substitution gives $dX_t=\kappa\theta\,dt$ while $X_t=0$, which is nonzero. Conversely, the positive drift alone does not prove that zero is inaccessible; the scale calculation shows that in the regime $\alpha<1$ it can be reached.

Parameter regime	Boundary from a positive initial state	Behavior at zero
$\alpha\ge1$	Not accessible	A path started above zero does not hit zero
$0<\alpha<1$	Accessible	Can be reached, but no positive-length interval of absorption is possible

Check Your Understanding

Use the scale density and the integral form of the SDE to justify each answer.

For $\kappa=1$, $\theta=\tfrac18$, and $\sigma=1$, compute $\alpha$ and decide whether zero is accessible.
For $0<x<R$ and $0<\alpha<1$, write the scale-function expression for $\mathbb{P}_x(\tau_0<\tau_R)$.
Why is $\int_0^c y^{-\alpha}\,dy$ finite when $0<\alpha<1$, but infinite when $\alpha=1$?
If a CIR path is zero throughout an interval of length $u>0$, what drift increment does its SDE require on that interval?
Explain why immediate departure from zero does not imply that the process stays strictly positive for all later times.

Parameter regime	Boundary from a positive initial state	Behavior at zero
\(\alpha\ge1\)	Not accessible	A path started above zero does not hit zero
\(0<\alpha<1\)	Accessible	Can be reached, but no positive-length interval of absorption is possible