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Square-Root Diffusions and Other Models · Tutorial 253 of 600

Transforming CIR Into a Time-Changed Squared Bessel Process

Learn how the CIR mean-reversion clock and state rescaling produce a squared Bessel process, including the exact dimension and inverse construction.

Advanced 11 min read

What You'll Learn

  • Identify the time change that removes the CIR mean-reversion drift.
  • Compute the squared Bessel dimension from the CIR parameters.
  • Track the deterministic rescaling of the state and the Brownian motion.
  • Derive the CIR-to-squared-Bessel transformation using Itô’s formula.
  • Recover a CIR process from a squared Bessel process and check the coefficients.
  • Recognize common errors involving the time-change direction and exponential factors.

Why a Time Change Removes Mean Reversion

The Cox–Ingersoll–Ross (CIR) process has a drift that pulls the state toward a long-run level, while its diffusion coefficient is proportional to the square root of the state. A deterministic change of clock and a time-dependent rescaling turn these dynamics into those of a squared Bessel process. This representation is useful because it isolates the role of the CIR parameters and gives an exact connection to a standard diffusion.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Fix \(\kappa>0\), \(\theta>0\), and \(\sigma>0\). Let \(R_t\) be a nonnegative, continuous, adapted CIR solution satisfying \(dR_t=\kappa(\theta-R_t)\,dt+\sigma\sqrt{R_t}\,dB_t\), with \(R_0=r_0\ge0\). On each finite time interval, assume the integrability needed for the displayed Itô integrals; in particular, \(\mathbb{E}\int_0^T R_s\,ds<\infty\). Thus the integrand \(\sigma\sqrt{R_s}\) is progressively measurable and belongs to \(\mathcal{L}^2\) on finite intervals.

Definition: A squared Bessel process of dimension \(\delta\ge0\), driven by a Brownian motion \(W_u\), is a nonnegative continuous process \(X_u\) satisfying \(dX_u=\delta\,du+2\sqrt{X_u}\,dW_u\). The time variable is denoted by \(u\) to distinguish this clock from CIR time \(t\).

The CIR-to-squared-Bessel correspondence uses the increasing clock \(u(t)=(e^{\kappa t}-1)/\kappa\), whose inverse is \(t(u)=\log(1+\kappa u)/\kappa\). The state is multiplied by \(e^{\kappa t}\) before expressing it in the new clock. The squared Bessel dimension will be \(\delta=4\kappa\theta/\sigma^2\).

The CIR-to-Squared-Bessel Transformation

Theorem: Suppose \(R_t\) satisfies the CIR SDE above with the stated nonnegative and finite-moment assumptions. Define \(u(t)=(e^{\kappa t}-1)/\kappa\), \(t(u)=\log(1+\kappa u)/\kappa\), and \(X_u=(4/\sigma^2)e^{\kappa t(u)}R_{t(u)}\). Then, with the time-changed filtration \(\mathcal{G}_u=\mathcal{F}_{t(u)}\), there is a \(\mathcal{G}_u\)-Brownian motion \(W_u\) such that \(dX_u=\delta\,du+2\sqrt{X_u}\,dW_u\), where \(\delta=4\kappa\theta/\sigma^2\) and \(X_0=4r_0/\sigma^2\).

Proof. Write \(a=4/\sigma^2\) and first use CIR time. Set \(U_t=ae^{\kappa t}R_t\). Apply Itô’s formula to \(f(t,r)=ae^{\kappa t}r\). Its derivatives are \(f_t=\kappa ae^{\kappa t}r\), \(f_r=ae^{\kappa t}\), and \(f_{rr}=0\); hence the Itô correction \(\tfrac12 f_{rr}\sigma^2r\,dt\) is zero. Substitution of the CIR drift and diffusion gives \(dU_t=a\kappa\theta e^{\kappa t}\,dt+a\sigma e^{\kappa t}\sqrt{R_t}\,dB_t\). Since \(U_t=ae^{\kappa t}R_t\), the diffusion coefficient in this equation is \(\sigma\sqrt{a}\,e^{\kappa t/2}\sqrt{U_t}=2e^{\kappa t/2}\sqrt{U_t}\).

Now define \(W_u=\int_0^{t(u)}e^{\kappa s/2}\,dB_s\). For each finite horizon, the deterministic integrand is progressively measurable and has finite squared integral, so it belongs to \(\mathcal{L}^2\). By the Itô isometry, \(W\) is a continuous square-integrable martingale, and its quadratic variation is \([W]_u=\int_0^{t(u)}e^{\kappa s}\,ds=(e^{\kappa t(u)}-1)/\kappa=u\). The standard Lévy characterization says that a continuous local martingale starting at zero with quadratic variation \(u\) is Brownian motion in its filtration; we use that standard result here. Thus \(W_u\) is Brownian motion with respect to \(\mathcal{G}_u\).

Set \(X_u=U_{t(u)}\). Because \(du=e^{\kappa t}dt\), the drift \(a\kappa\theta e^{\kappa t}dt\) becomes \(a\kappa\theta\,du\), and the stochastic integral is expressed using the Brownian motion \(W\) just defined. Therefore \(dX_u=a\kappa\theta\,du+2\sqrt{X_u}\,dW_u\). Finally, \(a\kappa\theta=4\kappa\theta/\sigma^2=\delta\), and \(X_0=ar_0\). This is the squared Bessel SDE with the asserted dimension and initial value. \(\square\)

All stochastic integrals in this derivation have progressively measurable integrands. On finite intervals, the CIR integrability assumption gives square integrability for the transformed diffusion integrand as well; equivalently, its squared coefficient is a deterministic multiple of \(R_t\), with a bounded multiplier on each finite interval.

Reading the Factors and Clock

The theorem gives the inverse state relation \(R_t=(\sigma^2/4)e^{-\kappa t}X_{u(t)}\). The exponential in this expression decays in CIR time, while the squared Bessel clock \(u(t)\) accelerates. Keeping these two factors together is essential: changing the clock without changing the state does not remove the mean-reversion term.

Worked Example: A CIR Process of Dimension Six

Take \(\kappa=2\), \(\theta=3\), \(\sigma=2\), and \(r_0=5\). Compute the dimension, the squared Bessel clock and initial value, and the correspondence at the time \(t=(\log 2)/2\).

The dimension is \(\delta=4\kappa\theta/\sigma^2=4\cdot2\cdot3/2^2=24/4=6\). The state multiplier in the squared Bessel direction is \(4/\sigma^2=4/4=1\), so \(X_0=1\cdot5=5\). The clock is \(u(t)=(e^{2t}-1)/2\). At \(t=(\log2)/2\), \(e^{2t}=2\), and therefore \(u(t)=(2-1)/2=1/2\). The inverse relation gives \(R_{(\log2)/2}=(2^2/4)e^{-2(\log2)/2}X_{1/2}=1\cdot(1/2)X_{1/2}\).

Check: Substituting the same time into the forward relation gives \(X_{1/2}=(4/2^2)e^{2(\log2)/2}R_{(\log2)/2}=1\cdot2R_{(\log2)/2}\). This is exactly equivalent to \(R_{(\log2)/2}=X_{1/2}/2\), as required.

Worked Example: A Two-Dimensional Squared Bessel Process

Let \(\kappa=1\), \(\theta=1\), \(\sigma=\sqrt{2}\), and \(r_0=4\). Find the transformed dimension and clock, then express the CIR state at \(t=\log3\) in terms of the squared Bessel state.

Here \(\sigma^2=2\), so \(\delta=4\cdot1\cdot1/2=2\), and the state multiplier is \(4/\sigma^2=4/2=2\). Thus \(X_0=2\cdot4=8\). The clock is \(u(t)=e^t-1\). At \(t=\log3\), \(u(t)=3-1=2\). The inverse state relation yields \(R_{\log3}=(2/4)e^{-\log3}X_2=(1/2)(1/3)X_2=X_2/6\).

Check: The forward relation at this same time is \(X_2=(4/2)e^{\log3}R_{\log3}=2\cdot3R_{\log3}=6R_{\log3}\). Rearranging recovers \(R_{\log3}=X_2/6\). Both the dimension and the scale factor use \(\sigma^2=2\), not \(\sigma\).

Worked Example: A Three-Dimensional Squared Bessel Process

Take \(\kappa=3/2\), \(\theta=2\), \(\sigma=2\), and \(r_0=2\). Evaluate the clock and state conversion at the time \(t=(2/3)\log4\).

The dimension is \(\delta=4(3/2)2/2^2=12/4=3\), and \(4/\sigma^2=1\), so \(X_0=2\). Since \(\kappa t=(3/2)(2/3)\log4=\log4\), the clock is \(u(t)=(e^{\kappa t}-1)/\kappa=(4-1)/(3/2)=3\cdot(2/3)=2\). The inverse relation gives \(R_t=(2^2/4)e^{-\log4}X_2=1\cdot(1/4)X_2=X_2/4\).

Check: The forward relation gives \(X_2=(4/2^2)e^{\log4}R_t=1\cdot4R_t\), which again is equivalent to \(R_t=X_2/4\). In this example the scale multiplier equals one, but the time-dependent factor is still necessary.

Recovering CIR from a Squared Bessel Process

Proposition: Let \(X_u\) be a nonnegative squared Bessel process of dimension \(\delta=4\kappa\theta/\sigma^2\), with Brownian motion \(W_u\), and set \(u(t)=(e^{\kappa t}-1)/\kappa\). Then \(R_t=(\sigma^2/4)e^{-\kappa t}X_{u(t)}\) satisfies the CIR SDE, when driven by the Brownian motion obtained by reversing the deterministic time change.

Proof. Let \(c=\sigma^2/4\), and set \(B_t=\int_0^{u(t)}e^{-\kappa t(v)/2}\,dW_v\). The integrand is deterministic, progressively measurable, and square integrable on finite intervals. Its quadratic variation is \([B]_t=\int_0^{u(t)}e^{-\kappa t(v)}\,dv=t\), because \(dv=e^{\kappa s}ds\) when \(v=u(s)\). The same standard Lévy characterization used above shows that \(B_t\) is Brownian motion in the time-changed filtration. In differential notation, \(dW_{u(t)}=e^{\kappa t/2}dB_t\).

For the time-changed squared Bessel process, \(dX_{u(t)}=\delta e^{\kappa t}dt+2\sqrt{X_{u(t)}}e^{\kappa t/2}dB_t\). Apply Itô’s formula to \(g(t,x)=ce^{-\kappa t}x\). The derivatives are \(g_t=-\kappa ce^{-\kappa t}x\), \(g_x=ce^{-\kappa t}\), and \(g_{xx}=0\), so the Itô correction \(\tfrac12g_{xx}\,d[X_{u(\cdot)}]_t\) is zero. The drift of \(R_t=g(t,X_{u(t)})\) is \((-\kappa ce^{-\kappa t}X_{u(t)}+c\delta)dt=(-\kappa R_t+c\delta)dt\). Since \(c\delta=(\sigma^2/4)(4\kappa\theta/\sigma^2)=\kappa\theta\), this is \(\kappa(\theta-R_t)dt\). Its diffusion coefficient is \(2ce^{-\kappa t/2}\sqrt{X_{u(t)}}=2\sqrt{c}\sqrt{R_t}=\sigma\sqrt{R_t}\), since \(c=\sigma^2/4\) and \(\sigma>0\). Thus \(dR_t=\kappa(\theta-R_t)dt+\sigma\sqrt{R_t}\,dB_t\), as claimed. \(\square\)

Common Pitfall: Mixing the Clock with the State Factor

A frequent error is to use the increasing clock \(u(t)=(e^{\kappa t}-1)/\kappa\) but multiply the CIR state by \(e^{-\kappa t}\) while trying to obtain a squared Bessel process. The forward transformation uses \(e^{+\kappa t}\); the decreasing factor \(e^{-\kappa t}\) belongs to the inverse relation that constructs CIR from squared Bessel.

There is also a drift-factorization trap in the forward calculation. In Itô’s formula for \(U_t=ae^{\kappa t}R_t\), the drift contributions are \(+\kappa ae^{\kappa t}R_t\,dt\) from the time derivative and \(a\kappa e^{\kappa t}(\theta-R_t)\,dt\) from the CIR drift. They cancel their \(R_t\) terms exactly, leaving \(a\kappa\theta e^{\kappa t}dt\). No state-dependent drift remains before the clock is changed. In the inverse calculation, the drift instead becomes \(-\kappa R_t+c\delta\); its constant part is \(c\delta=\kappa\theta\). These are different calculations in opposite directions, and their exponential factors should not be interchanged.

The parameters have distinct roles: \(\kappa\) sets the rate of the clock, \(\sigma^2/4\) sets the inverse state scale, and \(4\kappa\theta/\sigma^2\) is the squared Bessel dimension. This representation identifies the diffusion dynamics, but it does not by itself settle questions about behavior at zero or provide a transition distribution.

Check Your Understanding

Use the forward and inverse relations, keeping CIR time \(t\) separate from squared Bessel time \(u\).

  1. For \(\kappa=1\), \(\theta=3\), and \(\sigma=2\), compute the squared Bessel dimension and the state multiplier \(4/\sigma^2\).
  2. Show that the inverse clock of \(u(t)=(e^{\kappa t}-1)/\kappa\) is \(t(u)=\log(1+\kappa u)/\kappa\).
  3. For \(\kappa=2\), evaluate \(u(t)\) at \(t=(\log3)/2\).
  4. In the inverse construction, compute \(c\delta\) from \(c=\sigma^2/4\) and \(\delta=4\kappa\theta/\sigma^2\).
  5. Explain why the forward state multiplier is \(e^{+\kappa t}\), while the inverse relation contains \(e^{-\kappa t}\).