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Square-Root Diffusions and Other Models · Tutorial 251 of 600

CIR Dynamics and Parameter Calculations

Learn how the CIR parameters control mean reversion and volatility, and calculate and check its conditional moments.

Advanced 10 min read

What You'll Learn

  • Identify the drift, diffusion coefficient, and roles of the CIR parameters
  • Derive the conditional mean from the SDE
  • Derive the conditional variance using Itô's formula for the square
  • Calculate the mean-reversion half-life and finite-horizon moments
  • Check short-time and long-time limits of the conditional moments
  • Avoid confusing a finite-horizon variance with its long-run limit

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

$$ dX_t=\kappa(\theta-X_t)\,dt+\sigma\sqrt{X_t}\,dB_t, \qquad \kappa>0,\quad \theta\ge0,\quad \sigma>0. $$

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.

Definition: For a time interval of length \(h\), the CIR mean-reversion half-life is the positive \(h\) for which the initial displacement from \(\theta\) has been reduced by a factor of \(1/2\). Its value is \((\ln 2)/\kappa\), because the displacement decays by the factor \(e^{-\kappa h}\).

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 \(\mathcal{F}_s\).

Theorem: For a nonnegative CIR solution satisfying the integrability conditions above, define \(a=e^{-\kappa h}\). Then, almost surely,
$$ \mathbb{E}[X_{s+h}\mid\mathcal{F}_s] =\theta+(X_s-\theta)a, \qquad \operatorname{Var}(X_{s+h}\mid\mathcal{F}_s) =\frac{\sigma^2}{\kappa} \left(X_s a(1-a)+\frac{\theta}{2}(1-a)^2\right). $$

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

$$ d(X_t^2)=\bigl(2\kappa\theta X_t-2\kappa X_t^2+\sigma^2X_t\bigr)\,dt +2\sigma X_t^{3/2}\,dB_t. $$

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

$$ v'(u)=-2\kappa v(u)+\sigma^2m(u),\qquad v(0)=0. $$

Solving this linear equation, using the conditional mean already derived, gives

$$ v(h)=\sigma^2\int_0^h e^{-2\kappa(h-u)} \left(\theta+(X_s-\theta)e^{-\kappa u}\right)\,du. $$

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

$$ v(h)=\frac{\sigma^2}{\kappa} \left(\frac{\theta}{2}(1-a^2)+(X_s-\theta)(a-a^2)\right) =\frac{\sigma^2}{\kappa} \left(X_s a(1-a)+\frac{\theta}{2}(1-a)^2\right). $$

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

$$ 4+(1-4)\frac12=4-\frac32=\frac52. $$

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.

ParameterRole in the CIR modelEffect in the conditional moments
\(\kappa\)Mean-reversion speedSets the decay factor \(e^{-\kappa h}\)
\(\theta\)Long-run mean levelIs the limiting conditional mean
\(\sigma\)Volatility scaleEnters the conditional variance through \(\sigma^2\)

Check Your Understanding

Use the conditional moment formulas and keep the conditioning state explicit.

  1. For \(\kappa=1\), what is the mean-reversion half-life, and what is \(e^{-\kappa h}\) at that horizon?
  2. For \(X_s=x\), show from the conditional mean formula that the displacement from \(\theta\) is multiplied by \(e^{-\kappa h}\).
  3. Evaluate the conditional variance formula at \(h=0\), and find its derivative at \(h=0\).
  4. For \(\kappa=1\), \(\theta=2\), \(\sigma=2\), and \(X_s=2\), calculate the limiting conditional variance as \(h\) tends to infinity.
  5. For \(\kappa=2\), determine how many half-lives are contained in the interval \(h=\ln 2\), and justify the answer using \(e^{-\kappa h}\).