Reading Boundary Cases Without Confusing Reachability and Absorption
For a diffusion on the nonnegative half-line, three different questions matter: can the process reach zero in finite time, what happens if it reaches zero, and is zero inaccessible from a positive starting point? These questions are related, but they are not interchangeable. In particular, a process can reach zero and then leave it, or reach zero and remain there forever. The calculations below use parameter criteria already established in the earlier material on CIR boundary behavior, Bessel processes at the origin, and the CEV-to-Bessel transformation.
All SDEs are considered on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{P})\) carrying Brownian motion \(B_t\). An SDE is understood in its integral form, with progressively measurable Ito integrands. On finite time intervals where the integrands are square-integrable, they belong to \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). At a boundary or where a coefficient is only locally controlled, the equation is interpreted up to suitable stopping times; the stopped integrands are required to be in \(\mathcal{L}^2\). The drift and diffusion coefficients are distinct: the drift multiplies \(dt\), and the diffusion coefficient multiplies \(dB_t\).
CIR: Apply the Feller Ratio Carefully
Consider the CIR equation \(dX_t=\kappa(\theta-X_t)\,dt+\sigma\sqrt{X_t}\,dB_t\), with \(X_0=x_0>0\), \(\kappa>0\), \(\theta>0\), and \(\sigma>0\). The relevant ratio is \(2\kappa\theta/\sigma^2\). The established CIR boundary classification says that zero is inaccessible when this ratio is at least \(1\), including equality, and attainable when it is less than \(1\). When it is attainable and \(\theta>0\), the standard CIR process is not absorbed at zero: its positive drift there, \(\kappa\theta\), points into the state space.
Proof. The CIR-to-squared-Bessel time-change uses dimension \(\delta=4\kappa\theta/\sigma^2\). Dividing this equality by \(2\) gives \(\delta/2=2\kappa\theta/\sigma^2\). Since division by the positive number \(2\) preserves order, \(\delta\geq2\) holds if and only if \(2\kappa\theta/\sigma^2\geq1\); likewise, \(\delta<2\) holds if and only if \(2\kappa\theta/\sigma^2<1\). The squared-Bessel origin classification therefore gives the stated equivalent boundary test. \(\square\)
Worked Example: CIR at the Critical Threshold
Take \(\kappa=2\), \(\theta=1\), \(\sigma=2\), and \(x_0=1\). Decide whether zero is attainable.
Substitute into the Feller ratio: \(2\kappa\theta/\sigma^2=2\cdot2\cdot1/2^2=4/4=1\). The ratio equals the threshold, so zero is inaccessible from \(x_0=1\). The associated dimension is \(\delta=4\kappa\theta/\sigma^2=4\cdot2\cdot1/4=2\), the same critical value.
Check: the proposition gives \(\delta/2=2/2=1\), which agrees with the direct Feller calculation. Equality belongs to the inaccessible case, not the attainable case.
Worked Example: An Accessible but Nonabsorbing CIR Boundary
Take \(\kappa=1\), \(\theta=\tfrac14\), \(\sigma=1\), and \(x_0=2\). Classify zero and state what the standard CIR process does there.
The ratio is \(2\kappa\theta/\sigma^2=2\cdot1\cdot\tfrac14/1^2=\tfrac12\), which is less than \(1\). Thus zero is attainable. The associated dimension is \(\delta=4\cdot1\cdot\tfrac14/1^2=1\), also less than \(2\). Since \(\theta>0\), the drift at zero is \(\kappa\theta=1\cdot\tfrac14=\tfrac14>0\); the standard CIR boundary is not absorbing.
Check: \(\delta/2=1/2\), matching the Feller ratio. The conclusion is “attainable, but not absorbed,” not “strictly positive at all times.”
Squared Bessel: Dimension Determines the Origin Behavior
The squared-Bessel process \(Z_t\) of dimension \(\delta\geq0\), started at \(z_0>0\), satisfies \(dZ_t=\delta\,dt+2\sqrt{Z_t}\,dB_t\). The standard origin classification is: zero is inaccessible for \(\delta\geq2\); for \(0<\delta<2\), it is attainable and instantaneously reflecting; for \(\delta=0\), it is attainable and absorbing. “Instantaneously reflecting” means that the process can hit zero but does not stay there for a positive interval as an absorbed process would.
Worked Example: Three Squared-Bessel Dimensions
For \(z_0=1\), compare dimensions \(\delta=2\), \(\delta=1\), and \(\delta=0\).
At \(\delta=2\), the equation is \(dZ_t=2\,dt+2\sqrt{Z_t}\,dB_t\); since \(2\geq2\), zero is inaccessible. At \(\delta=1\), it is \(dZ_t=dt+2\sqrt{Z_t}\,dB_t\); since \(0<1<2\), zero is attainable and instantaneously reflecting. At \(\delta=0\), it is \(dZ_t=2\sqrt{Z_t}\,dB_t\); zero is attainable and the standard squared-Bessel process is absorbed there.
Check: the drift coefficient at zero is \(\delta\). It is \(2\) at dimension \(2\), \(1\) at dimension \(1\), and \(0\) at dimension \(0\). The classifications differ at dimension \(0\): it is not enough to say only that zero is attainable.
CEV: Transform the Exponent, Then Inspect the Boundary Rule
Consider \(dS_t=\mu S_t\,dt+\sigma S_t^\beta\,dB_t\), with \(S_0=s_0>0\), \(\sigma>0\), and \(0<\beta<1\). Before zero is reached, put \(p=1-\beta>0\) and \(Y_t=S_t^p\). Ito’s formula gives a Bessel-type equation. The dimension parameter determines the local boundary classification, while the chosen boundary extension determines whether the process is absorbed or continues after a hit.
Proof. Apply Ito’s formula to \(f(s)=s^p\), with \(f'(s)=ps^{p-1}\) and \(f''(s)=p(p-1)s^{p-2}\). The correction term is \(\tfrac12 f''(S_t)\sigma^2S_t^{2\beta}\,dt\). Since \(p=1-\beta\), this gives \(dY_t=p\mu Y_t\,dt-\frac{p\beta\sigma^2}{2Y_t}\,dt+p\sigma\,dB_t\). Dividing by \(p\sigma\) and using \(Y_t=p\sigma R_t\) gives the displayed equation for \(R_t\). A Bessel drift has singular term \((\delta-1)/(2R_t)\). Equating it to \(-\beta/(2(1-\beta)R_t)\) yields \(\delta-1=-\beta/(1-\beta)\), hence \(\delta=(1-2\beta)/(1-\beta)\). Because \(1-\beta>0\), \(\delta<2\) is equivalent to \(1-2\beta<2(1-\beta)=2-2\beta\), which reduces to \(1<2\), always true. The sign of \(\delta\) is the sign of \(1-2\beta\), proving the remaining claims. \(\square\)
The linear drift \((1-\beta)\mu R_t\) does not change this local classification at zero. For \(0<\beta<\tfrac12\), the positive dimension lies below \(2\), the reflecting Bessel-type continuation reaches and leaves zero. An absorbing CEV model can instead impose absorption there; that is a boundary choice, not a conclusion from attainability. For \(\tfrac12\leq\beta<1\), the transformed dimension is nonpositive, and the usual nonnegative CEV boundary behavior is absorption. At \(\beta=1\), the equation is geometric Brownian motion, whose positive solution does not hit zero in finite time.
Worked Example: CEV with an Absorbing Boundary
Let \(\beta=\tfrac34\), \(\mu=0\), \(\sigma=1\), and \(s_0=1\), and use the usual absorbing CEV boundary convention. Decide whether zero is attainable and whether it is absorbing.
Here \(1-\beta=\tfrac14\), so the transformed dimension is \(\delta=(1-2\beta)/(1-\beta)=(1-\tfrac32)/\tfrac14=(-\tfrac12)/\tfrac14=-2\). It is below \(2\) and nonpositive. Zero is attainable in the Bessel-type classification, and under the stated CEV convention the process is absorbed on reaching zero.
Check: substituting \(\beta=\tfrac34\) in the transformed drift’s singular coefficient gives \(-\beta/(2(1-\beta))=-(\tfrac34)/(2\cdot\tfrac14)=-\tfrac32\). The Bessel expression \((\delta-1)/2\) gives \((-2-1)/2=-\tfrac32\), as required.
Worked Example: CEV Below the Half-Power Threshold
Let \(\beta=\tfrac14\), \(\mu=0\), \(\sigma=1\), and \(s_0=1\). Compare the boundary classification with the effect of choosing absorption.
The dimension is \(\delta=(1-2\beta)/(1-\beta)=(1-\tfrac12)/\tfrac34=(\tfrac12)/\tfrac34=\tfrac23\). Since \(0<\tfrac23<2\), the Bessel-type boundary is attainable and its reflecting continuation leaves zero. If the CEV model is instead defined with absorption at zero, it stays there after hitting. Thus the same exponent gives an attainable boundary in either case, but the post-hit behavior depends on the boundary rule.
Check: the singular drift coefficient is \(-\beta/(2(1-\beta))=-(\tfrac14)/(2\cdot\tfrac34)=-\tfrac16\), while \((\delta-1)/2=(\tfrac23-1)/2=-\tfrac16\). The transformation’s dimension agrees with its drift coefficient.
Common Mistake: Dropping the Factor of Two
A frequent CIR error is to test \(\kappa\theta/\sigma^2\) against \(1\), omitting the factor \(2\). For \(\kappa=1\), \(\theta=\tfrac34\), and \(\sigma=1\), the incorrect ratio is \(1\cdot\tfrac34/1^2=\tfrac34\), which might be read as evidence that zero is attainable. The correct ratio is \(2\cdot1\cdot\tfrac34/1^2=\tfrac32\), so zero is inaccessible. The dimension check agrees: \(\delta=4\cdot1\cdot\tfrac34/1^2=3\), and \(\delta\geq2\).
Always state both parts of a boundary conclusion: whether zero can be reached, and what the specified model does after reaching it. In particular, an absorbing convention is not implied just by an accessibility calculation. The distinction is important when comparing models, selecting a boundary extension, or interpreting a simulated path that approaches zero.
Check Your Understanding
For each question, distinguish the event of hitting zero from the behavior after a hit.
- For a CIR process with \(\kappa=1\), \(\theta=\tfrac12\), and \(\sigma=1\), compute the Feller ratio and classify zero.
- What is the squared-Bessel boundary classification at dimension \(\delta=2\), and why does equality matter?
- For \(\beta=\tfrac13\), compute the CEV Bessel-type dimension and decide whether the transformed origin is attainable.
- For which values of \(0<\beta<1\) is the transformed CEV dimension nonpositive?
- Explain why “zero is attainable” does not by itself imply that the process is absorbed there.