From CIR Dynamics to a Squared-Bessel Process
The square-root diffusion coefficient in the Cox–Ingersoll–Ross (CIR) equation makes the behavior at zero important: the coefficient is defined on the nonnegative state space, but the equation alone does not tell us whether a path can reach its boundary. A time change converts the CIR process into a squared-Bessel process. Its nonnegativity gives the first boundary conclusion, while a hitting-probability calculation yields the strict-positivity condition.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\geq0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Fix \(\kappa>0\), \(\sigma>0\), \(\theta\geq0\), and \(X_0=x\geq0\). A CIR solution is a continuous adapted process satisfying
The drift is \(\kappa(\theta-X_t)\), and the diffusion coefficient is \(\sigma\sqrt{X_t}\). Ito integrals below are taken with progressively measurable integrands. On each finite interval, integrands with finite expected integrated square belong to \(\mathcal{L}^2\); any locally square-integrable integral is understood by stopping and localization. We do not infer pathwise uniqueness or uniqueness in law from the calculations below.
Set \(\delta=4\kappa\theta/\sigma^2\) and define the increasing clock \(s(t)=\frac{\sigma^2}{4\kappa}(e^{\kappa t}-1)\). The representation we will use is
Here is the calculation behind the representation. Starting with a CIR solution, let \(Y_{s(t)}=e^{\kappa t}X_t\). The product rule gives \(d(e^{\kappa t}X_t)=\kappa\theta e^{\kappa t}dt+\sigma e^{\kappa t}\sqrt{X_t}\,dB_t\). Since \(ds/dt=(\sigma^2/4)e^{\kappa t}\), the deterministic time change \(dW_{s(t)}=\sqrt{ds/dt}\,dB_t\) is Brownian in the time-changed filtration. The drift becomes \(\delta\,ds\), and the noise becomes \(2\sqrt{Y_s}\,dW_s\). This is the standard deterministic Brownian time-change result; equivalently, the time-changed stochastic integral has quadratic variation \(s\), so Levy's characterization applies.
Nonnegativity from the Representation
Proof. The squared-Bessel dimension is \(\delta=4\kappa\theta/\sigma^2\geq0\), so the standard squared-Bessel process \(Y\) has nonnegative continuous paths. The clock \(s(t)\) is nonnegative for \(t\geq0\), and \(e^{-\kappa t}>0\). Therefore \(X_t=e^{-\kappa t}Y_{s(t)}\geq0\) at every time on the event where \(Y\) has nonnegative paths. That event has probability one. Conversely, starting with such a squared-Bessel process and applying the time change produces a CIR solution: the Ito calculation above, read in reverse, gives the stated drift and diffusion. \(\square\)
The argument proves nonnegativity, not strict positivity. If \(Y\) starts at zero, the representation still gives a nonnegative CIR path, but it does not assert that the path is bounded away from zero. Strict positivity requires a separate hitting argument and a positive starting value.
Worked Example: Checking the Time Change and the Drift
Take \(\kappa=2\), \(\theta=3\), \(\sigma=2\), and \(X_0=1\). Calculate the squared-Bessel dimension and time change, then verify the CIR drift and diffusion coefficients.
The dimension is \(\delta=4\kappa\theta/\sigma^2=4\cdot2\cdot3/2^2=24/4=6\). The clock is \(s(t)=\frac{2^2}{4\cdot2}(e^{2t}-1)=\frac12(e^{2t}-1)\). Thus \(X_t=e^{-2t}Y_{\frac12(e^{2t}-1)}\), with \(dY_s=6\,ds+2\sqrt{Y_s}\,dW_s\).
To check the SDE directly, use \(F(t,y)=e^{-2t}y\). Its derivatives are \(F_t=-2e^{-2t}y\), \(F_y=e^{-2t}\), and \(F_{yy}=0\); the Ito correction is \(\frac12F_{yy}\,d[Y]_t=0\). Since \(ds/dt=e^{2t}\), the drift of \(X\) is \(e^{-2t}\cdot6e^{2t}-2e^{-2t}Y_s=6-2X_t\). Its diffusion term is \(e^{-2t}2\sqrt{Y_s}\,dW_s\). In the original time, \(dW_s=e^t\,dB_t\), so this term is \(2e^{-t}\sqrt{Y_s}\,dB_t=2\sqrt{X_t}\,dB_t\). These are precisely \(2(3-X_t)\,dt\) and \(2\sqrt{X_t}\,dB_t\).
Check: The dimension contribution to the drift is \(e^{-2t}\cdot6\cdot(ds/dt)=e^{-2t}\cdot6\cdot e^{2t}=6=\kappa\theta\). Also, \(\sigma\sqrt{X_t}=2\sqrt{X_t}\), matching the diffusion just computed.
A Hitting-Probability Calculation
We now establish strict positivity in the squared-Bessel clock. Suppose \(Y_0=x>0\), let \(0<a<x<b\), and define \(\tau_a=\inf\{s\geq0:Y_s=a\}\), \(\tau_b=\inf\{s\geq0:Y_s=b\}\), and \(\tau=\tau_a\wedge\tau_b\). Continuity means that on \(\{\tau_a<\tau_b\}\), the process exits \((a,b)\) at \(a\), and on \(\{\tau_b<\tau_a\}\), it exits at \(b\).
Proof. First, the stopped process satisfies \(Y_{s\wedge\tau}=x+\delta(s\wedge\tau)+\int_0^{s\wedge\tau}2\sqrt{Y_u}\,dW_u\). The stopped integrand is bounded by \(2\sqrt b\), so on every finite interval it is in \(\mathcal{L}^2\), and its Ito integral has mean zero. Since \(Y_{s\wedge\tau}\leq b\), taking expectations gives \(\delta\mathbb{E}[s\wedge\tau]\leq b-x\). As \(\delta\geq2\), monotone convergence implies \(\mathbb{E}[\tau]\leq(b-x)/\delta<\infty\), hence \(\tau<\infty\) almost surely.
For \(\delta>2\), put \(p=1-\delta/2<0\) and \(f(y)=y^p\). Its derivatives are \(f'(y)=p y^{p-1}\) and \(f''(y)=p(p-1)y^{p-2}\). Ito's formula, up to \(\tau\), gives the drift correction explicitly:
because \(d[Y]_s=4Y_s\,ds\) and \(\delta+2(p-1)=0\). For \(\delta=2\), use \(f(y)=\log y\), with \(f'(y)=1/y\), \(f''(y)=-1/y^2\); its drift is \(2/y+\frac12(-1/y^2)4y=0\). In both cases \(f(Y_{s\wedge\tau})\) is a true martingale: the stochastic-integral integrand \(2\sqrt{Y_s}f'(Y_s)\mathbf{1}_{\{s\leq\tau\}}\) is bounded on \([a,b]\), hence square-integrable on finite time intervals. The stopped values of \(f\) are bounded, so dominated convergence as \(s\to\infty\) gives \(\mathbb{E}[f(Y_\tau)]=f(x)\).
Writing \(q=\mathbb{P}(\tau_a<\tau_b)\), the exit values give \(f(x)=qf(a)+(1-q)f(b)\). Solving yields \(q=(f(x)-f(b))/(f(a)-f(b))\). Substitution of \(f(y)=y^{1-\delta/2}\) or \(f(y)=\log y\), respectively, gives the two displayed expressions. \(\square\)
Strict Positivity Under the Feller Condition
Proof. The parameter condition is equivalent to \(\delta=4\kappa\theta/\sigma^2\geq2\). In the squared-Bessel clock, fix any upper level \(b>x=X_0\) and take a sequence \(a_n\downarrow0\) with \(a_n<x\). If \(Y\) hits zero before \(b\), continuity implies \(\tau_{a_n}<\tau_b\) for every \(n\). Therefore \(\mathbb{P}(\tau_0<\tau_b)\leq\mathbb{P}(\tau_{a_n}<\tau_b)\) for every \(n\), where \(\tau_0\) is the first hitting time of zero.
For \(\delta>2\), the proposition's numerator is fixed and its denominator \(a_n^{1-\delta/2}-b^{1-\delta/2}\) tends to infinity, so the hitting probability tends to zero. For \(\delta=2\), its numerator is fixed and \(\log b-\log a_n\) tends to infinity, with the same conclusion. Thus \(\mathbb{P}(\tau_0<\tau_b)=0\) for every \(b>x\).
If \(Y\) hits zero at a finite time, its continuous path is bounded up to that time. Hence some integer \(b>x\) is not reached before that zero hit. The event of a finite zero hit is therefore contained in the countable union, over integers \(b>x\), of \(\{\tau_0<\tau_b\}\). Each event has probability zero, so \(Y\) never hits zero at a finite time, almost surely. The time change \(s(t)\) is finite for finite \(t\), and \(X_t=e^{-\kappa t}Y_{s(t)}\); consequently \(X_t>0\) for every finite \(t\). \(\square\)
Worked Example: The Boundary Case \(\delta=2\)
Take \(\kappa=1\), \(\theta=\tfrac12\), \(\sigma=1\), and \(X_0=1\). Then \(\delta=4\cdot1\cdot\frac12/1^2=2\), so the theorem applies. In the squared-Bessel clock, calculate the probability of hitting \(a=\frac14\) before \(b=4\).
The formula for \(\delta=2\) gives \(\mathbb{P}(\tau_{1/4}<\tau_4)=(\log4-\log1)/(\log4-\log(1/4))\). Since \(\log1=0\) and \(\log(1/4)=-\log4\), this equals \(\log4/(2\log4)=\frac12\).
Check: If \(a_n=4^{-n}\), with \(n\geq1\), the same formula gives \((\log4)/( \log4+n\log4)=1/(n+1)\), which tends to zero. A positive probability of reaching a fixed positive level below the starting value is consistent with never hitting zero.
Worked Example: A Dimension Strictly Above Two
Take squared-Bessel dimension \(\delta=4\), starting value \(x=1\), and levels \(a=\frac14\), \(b=4\). Compute the probability of hitting the lower level first.
Here \(1-\delta/2=-1\), so the formula gives \(\mathbb{P}(\tau_{1/4}<\tau_4)=(1^{-1}-4^{-1})/((1/4)^{-1}-4^{-1})\). The numerator is \(1-\frac14=\frac34\); the denominator is \(4-\frac14=\frac{15}{4}\). The probability is therefore \((3/4)/(15/4)=\frac15\).
Check: For lower level \(a=\frac1{16}\), with the same \(x,b,\delta\), the numerator remains \(\frac34\), while the denominator is \(16-\frac14=\frac{63}{4}\). The probability is \((3/4)/(63/4)=\frac1{21}\), smaller than \(\frac15\) and tending to zero as \(a\downarrow0\), as required by the theorem.
A Common Pitfall: Omitting the Ito Correction
For a positive squared-Bessel process, applying the ordinary chain rule to \(\log Y_s\) would give a drift \(2/Y_s\) when \(\delta=2\), and might suggest a nonzero drift at the boundary calculation. That is wrong: the process has quadratic variation \(d[Y]_s=4Y_s\,ds\), so the second derivative contributes to the drift.
For \(\delta=2\), the two drift terms cancel. This calculation is valid only while \(Y_s>0\), which is why it is stopped inside \([a,b]\) before taking limits. The condition obtained is \(\delta\geq2\), equivalently \(2\kappa\theta\geq\sigma^2\); merely having \(\theta>0\) is not enough to invoke this strict-positivity proof.
Check Your Understanding
Use the representation and hitting calculation to answer these questions.
- For \(\kappa=1\), \(\theta=2\), and \(\sigma=2\), calculate \(\delta\). Does the strict-positivity theorem apply to a process started above zero?
- In the squared-Bessel equation, verify directly that \(y^{1-\delta/2}\) has zero generator when \(\delta>2\).
- For \(\delta=2\), \(x=2\), \(a=\frac12\), and \(b=8\), compute \(\mathbb{P}(\tau_a<\tau_b)\).
- Why is \(\mathbb{E}[\tau]<\infty\) useful in proving the two-level hitting formula?
- State the distinction between nonnegativity and strict positivity in the CIR result.