The CIR Model and Its Parameters
The Cox–Ingersoll–Ross (CIR) model combines a restoring drift with state-dependent noise. Its drift pulls the state toward a long-run level, while the diffusion coefficient grows with the square root of the state. The parameters therefore have distinct roles: one sets the speed of mean reversion, one sets its target, and one controls the noise scale.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider a nonnegative solution \(X_t\) of the Itô SDE
Here \(\kappa\) is the mean-reversion speed, \(\theta\) is the long-run mean level, and \(\sigma\) is the volatility scale. The drift coefficient is \(\mu(x)=\kappa(\theta-x)\); the diffusion coefficient is \(\sigma(x)=\sigma\sqrt{x}\). The drift is positive below \(\theta\), zero at \(\theta\), and negative above \(\theta\). The diffusion coefficient is state-dependent, not a constant volatility.
All stochastic integrals below are Itô integrals. On each finite time interval \([0,T]\), their integrands are progressively measurable and belong to \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_t^2\,dt<\infty\). We assume the nonnegative solution has the moment integrability needed below, in particular \(\mathbb{E}\int_0^T(X_t+X_t^3)\,dt<\infty\). These hypotheses make the stochastic integrals in the first- and second-moment calculations square-integrable martingales, so their conditional expectations of future increments are zero. The SDE is written in Itô form; it is not an ordinary chain rule or a Stratonovich equation. For \(x>0\), the equivalent Stratonovich drift would be \(\kappa(\theta-x)-\sigma^2/4\), since the correction is \(\tfrac12(\sigma\sqrt{x})(\sigma/(2\sqrt{x}))=\sigma^2/4\). The moment calculations here use the Itô form.
Conditional Mean and Variance
Fix \(s\ge0\) and \(h\ge0\), and suppose the stated integrability conditions hold on \([s,s+h]\). The conditional moments are functions of the state \(X_s\), which is known given \(\mathcal{F}_s\).
Proof. For the conditional mean, integrate the SDE from \(s\) to \(s+h\). The Itô integral has progressively measurable integrand \(\sigma\sqrt{X_t}\), and \(\mathbb{E}\int_s^{s+h}\sigma^2X_t\,dt<\infty\). It is therefore a square-integrable martingale increment with conditional expectation zero. Equivalently, applying Itô's formula to \(f(x)=x\), with \(f'(x)=1\) and \(f''(x)=0\), gives the same drift equation. If \(m(u)=\mathbb{E}[X_{s+u}\mid\mathcal{F}_s]\), conditional expectation and integration yield \(m'(u)=\kappa(\theta-m(u))\), \(m(0)=X_s\). Solving this linear equation gives \(m(u)=\theta+(X_s-\theta)e^{-\kappa u}\).
For the second moment, apply Itô's formula to \(f(x)=x^2\), whose derivatives are \(f'(x)=2x\) and \(f''(x)=2\). The Itô correction is explicitly \(\tfrac12 f''(x)\sigma^2x\,dt=\sigma^2x\,dt\). Thus
The last integrand is square-integrable on the interval by the assumed \(\mathbb{E}\int X_t^3\,dt<\infty\), so its stochastic integral is a true martingale, not merely a local martingale. Let \(q(u)=\mathbb{E}[X_{s+u}^2\mid\mathcal{F}_s]\). Taking conditional expectations gives \(q'(u)=(2\kappa\theta+\sigma^2)m(u)-2\kappa q(u)\). The conditional variance \(v(u)=q(u)-m(u)^2\) consequently satisfies
Solving this linear equation, using the conditional mean already derived, gives
Write \(a=e^{-\kappa h}\). The two integrals are \(\int_0^h e^{-2\kappa(h-u)}du=(1-a^2)/(2\kappa)\) and \(\int_0^h e^{-2\kappa(h-u)}e^{-\kappa u}du=(a-a^2)/\kappa\). Hence
The last equality follows by collecting the \(X_s\) terms and using \(1-a^2-2a+2a^2=(1-a)^2\). This proves both formulas. \(\square\)
The use of the Itô integral in this argument matters: its martingale property follows from square-integrability, whereas a general local martingale need not have zero conditional-mean increments. The stated integrability assumptions justify the conditional expectation steps without treating a local martingale as a true martingale.
Worked Parameter Calculations
Worked Example: A Half-Life Horizon
Let \(\kappa=\tfrac12\), \(\theta=4\), \(\sigma=1\), and suppose \(X_s=1\). Calculate the conditional mean and variance after one mean-reversion half-life.
The half-life is \(h=(\ln 2)/\kappa=2\ln 2\). Thus \(a=e^{-\kappa h}=e^{-\ln 2}=\tfrac12\). The conditional mean is
For the conditional variance, the bracket in the formula is \(1\cdot\tfrac12(1-\tfrac12)+\tfrac42(1-\tfrac12)^2=\tfrac14+2\cdot\tfrac14=\tfrac34\). Since \(\sigma^2/\kappa=1/(\tfrac12)=2\), the variance is \(2\cdot\tfrac34=\tfrac32\).
Check: In the integral expression for the variance, the contribution before multiplying by \(\sigma^2\) from \(\theta\) is \(4(1-(1/2)^2)/(2\cdot(1/2))=3\); the contribution from \(X_s-\theta\) is \((-3)((1/2)-(1/2)^2)/(1/2)=-\tfrac32\). Their sum is \(\tfrac32\), agreeing with the formula because \(\sigma^2=1\). The displacement of the mean from \(4\) is \(1/2\) of its initial value, as required at a half-life.
Worked Example: Calculating Parameters from a Half-Life
Suppose a CIR model has long-run mean \(\theta=3\), volatility scale \(\sigma=\tfrac12\), and mean-reversion half-life \(2\) time units. If \(X_s=1\), calculate its conditional moments after \(h=2\) time units.
The half-life relation \(e^{-\kappa h}=\tfrac12\) gives \(e^{-2\kappa}=\tfrac12\), hence \(\kappa=(\ln 2)/2\). At \(h=2\), \(a=\tfrac12\), so the mean is \(3+(1-3)\tfrac12=2\). The variance bracket is \(1\cdot\tfrac14+\tfrac32\cdot\tfrac14=\tfrac14+\tfrac38=\tfrac58\). The factor outside the bracket is \(\sigma^2/\kappa=(1/4)/(\ln 2/2)=1/(2\ln 2)\). Therefore the conditional variance is \(5/(16\ln 2)\).
Check: In the integral form, the \(\theta\)-term before multiplying by \(\sigma^2\) is \(3(1-(1/2)^2)/(2\cdot(\ln 2/2))=9/(4\ln 2)\). The \(X_s-\theta\) term is \((-2)((1/2)-(1/2)^2)/(\ln 2/2)=-1/\ln 2\). The sum is \(5/(4\ln 2)\); multiplying by \(\sigma^2=1/4\) gives \(5/(16\ln 2)\), as above.
Worked Example: Two Half-Lives and a Finite-Horizon Variance
Take \(\kappa=2\), \(\theta=3\), \(\sigma=1\), and \(X_s=1\). Find the conditional moments at \(h=\ln 2\), and compare the variance with its long-time limit.
Here \(a=e^{-2\ln 2}=\tfrac14\). Since the half-life for \(\kappa=2\) is \((\ln 2)/2\), this horizon is two half-lives, not one. The conditional mean is \(3+(1-3)/4=5/2\). For the variance, the bracket is \(1\cdot(1/4)(3/4)+(3/2)(3/4)^2=3/16+27/32=33/32\). Multiplication by \(\sigma^2/\kappa=1/2\) gives \(33/64\).
Check: The integral form gives the \(\theta\)-contribution \(3(1-(1/4)^2)/(2\cdot2)=45/64\). The \(X_s-\theta\) contribution is \((-2)((1/4)-(1/4)^2)/2=-3/16\). Their sum is \(45/64-3/16=45/64-12/64=33/64\). The long-time variance limit is \(\theta\sigma^2/(2\kappa)=3/4=48/64\), which is larger than the finite-horizon value.
Limiting Behavior and a Common Pitfall
The formulas show the transient and long-run roles of the parameters. As \(h\) decreases to zero, \(a\) tends to \(1\), so the conditional mean tends to \(X_s\) and the variance tends to zero. More precisely, the mean's initial rate of change is \(\kappa(\theta-X_s)\), while the variance's initial rate of change is \(\sigma^2X_s\), as expected from the drift and squared diffusion coefficient at the current state. As \(h\) tends to infinity, \(a\) tends to zero, so the conditional mean tends to \(\theta\) and the conditional variance tends to \(\theta\sigma^2/(2\kappa)\). These are limits of conditional moments; the formulas alone do not assert a boundary classification or a transition distribution.
Common pitfall: The long-run variance is not the variance at every positive horizon. For the parameters in the third example, using \(\theta\sigma^2/(2\kappa)=3/4\) as the variance at \(h=\ln 2\) would be incorrect: the actual value is \(33/64\). The finite-horizon formula retains both transient terms, \(X_s a(1-a)\) and \(\tfrac{\theta}{2}(1-a)^2\). A second common error is to call \(h=\ln 2\) a half-life when \(\kappa=2\); the defining factor is \(e^{-\kappa h}\), so the half-life is \((\ln 2)/2\), and \(h=\ln 2\) corresponds to two half-lives.
| Parameter | Role in the CIR model | Effect in the conditional moments |
|---|---|---|
| \(\kappa\) | Mean-reversion speed | Sets the decay factor \(e^{-\kappa h}\) |
| \(\theta\) | Long-run mean level | Is the limiting conditional mean |
| \(\sigma\) | Volatility scale | Enters the conditional variance through \(\sigma^2\) |
Check Your Understanding
Use the conditional moment formulas and keep the conditioning state explicit.
- For \(\kappa=1\), what is the mean-reversion half-life, and what is \(e^{-\kappa h}\) at that horizon?
- For \(X_s=x\), show from the conditional mean formula that the displacement from \(\theta\) is multiplied by \(e^{-\kappa h}\).
- Evaluate the conditional variance formula at \(h=0\), and find its derivative at \(h=0\).
- For \(\kappa=1\), \(\theta=2\), \(\sigma=2\), and \(X_s=2\), calculate the limiting conditional variance as \(h\) tends to infinity.
- For \(\kappa=2\), determine how many half-lives are contained in the interval \(h=\ln 2\), and justify the answer using \(e^{-\kappa h}\).