The CIR Transition Distribution

From CIR Dynamics to a Transition Law

The Cox–Ingersoll–Ross process has a state-dependent diffusion coefficient, so its transitions are not Gaussian in general. Its conditional law nevertheless has an explicit form: after rescaling, the process has a noncentral chi-square distribution. The key is to transform the CIR process into a squared Bessel process and then account for the change of time.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. For $\kappa>0$, $\theta\ge0$, and $\sigma>0$, suppose $X_t$ is a nonnegative CIR solution of

dX_t=\kappa(\theta-X_t)\,dt+\sigma\sqrt{X_t}\,dB_t.

Here $\kappa$ is the mean-reversion rate, $\theta$ the long-run level, and $\sigma$ the volatility parameter. The drift is $\kappa(\theta-X_t)$; the diffusion coefficient is $\sigma\sqrt{X_t}$. Assume the solution is adapted and $\mathbb{E}\int_0^T X_s\,ds<\infty$ on finite time intervals, so the Brownian integral is defined for the progressively measurable integrand $\sigma\sqrt{X_s}$ in $\mathcal{L}^2$. Fix $0\le s<t$, and write $\tau=t-s>0$.

Definition: A noncentral chi-square random variable with degrees of freedom $\delta\ge0$ and noncentrality parameter $\lambda\ge0$, written $\chi'^2_\delta(\lambda)$, has the Poisson-mixture distribution $\chi^2_{\delta+2N}$, where $N$ is Poisson with mean $\lambda/2$. For positive degrees of freedom, $\chi^2_m$ is the usual chi-square distribution with $m$ degrees of freedom. At zero degrees of freedom, define $\chi^2_0$ to be identically zero. This convention specifies the possible atom at zero when $\delta=0$.

The Squared Bessel Transform

The squared Bessel process of dimension $\delta\ge0$ satisfies $dZ_u=\delta\,du+2\sqrt{Z_u}\,dW_u$, where $W_u$ is Brownian motion in the time-changed filtration. The earlier transformation of CIR into a time-changed squared Bessel process gives the relevant change of variables. In this case, for $u=(e^{\kappa r}-e^{\kappa s})/\kappa$, set $Z_u=(4/\sigma^2)e^{\kappa r}X_r$. The starting value at $r=s$ is $y=(4/\sigma^2)e^{\kappa s}X_s$.

To verify the transformation, apply Itô’s formula to $g(r,x)=(4/\sigma^2)e^{\kappa r}x$. Its derivatives are $g_r=\kappa g$, $g_x=(4/\sigma^2)e^{\kappa r}$, $g_{xx}=0$, so the $\tfrac12 g_{xx}\sigma^2X_r\,dr$ correction is zero. Substitution into the CIR equation gives drift $(4\kappa\theta/\sigma^2)e^{\kappa r}dr$ and diffusion $2\sqrt{Z_u}\,e^{\kappa r/2}dB_r$. Since $du=e^{\kappa r}dr$, the drift becomes $\delta\,du$, where $\delta=4\kappa\theta/\sigma^2$; the time-changed Brownian integral has coefficient $2\sqrt{Z_u}$.

We derive the conditional transition law of this squared Bessel process through its Laplace transform. This also gives a direct identification of the noncentral chi-square distribution rather than relying on a density formula.

Proposition: If $Z_0=y\ge0$ solves $dZ_u=\delta\,du+2\sqrt{Z_u}\,dW_u$, then for $h>0$ and $\lambda\ge0$, $$ \mathbb{E}\!\left[e^{-\lambda Z_h}\mid\mathcal{F}_0\right] =(1+2\lambda h)^{-\delta/2} \exp\!\left(-\frac{y\lambda}{1+2\lambda h}\right). $$ The same formula holds conditionally from any starting time, with $y$ replaced by the value at that time.

Proof. The case $\lambda=0$ gives $1=1$, so take $\lambda>0$. For $0\le u\le h$, put $D=1+2\lambda(h-u)$, $a=\lambda/D$, and $v(u,z)=D^{-\delta/2}e^{-az}$. Then $v(h,z)=e^{-\lambda z}$. Differentiating gives $v_z=-av$, $v_{zz}=a^2v$, and $v_u=(\delta a-2a^2z)v$, because $D_u=-2\lambda$ and $a_u=2a^2$. For the squared Bessel generator $\delta\,\partial_z+2z\,\partial_{zz}$, this yields $v_u+\delta v_z+2zv_{zz}=0$.

Itô’s formula therefore gives zero drift for $v(u,Z_u)$; its second-order correction is $\tfrac12v_{zz}(u,Z_u)(2\sqrt{Z_u})^2du=2Z_uv_{zz}(u,Z_u)du$, precisely the second generator term. The stochastic integrand is $2\sqrt{Z_u}v_z(u,Z_u)$. It is bounded for $u\in[0,h]$ and $z\ge0$: $a$ stays between two positive finite constants, and $\sqrt{z}e^{-az}$ is bounded. Thus the integrand is progressively measurable and square-integrable in expectation on this interval. The resulting stochastic integral is a martingale, so conditional expectation gives $\mathbb{E}[v(h,Z_h)\mid\mathcal{F}_0]=v(0,y)$. Substitution of $D=1+2\lambda h$ at $u=0$ proves the formula. Starting at another time uses the same calculation on the interval after that time. $\square$

For comparison, the Poisson-mixture definition identifies the Laplace transform of $h\chi'^2_\delta(y/h)$. A central chi-square variable with $m$ degrees of freedom has Laplace transform $(1+2q)^{-m/2}$ at $q\ge0$, obtained from the gamma density (and equal to $1$ when $m=0$). Averaging over $N$ with mean $y/(2h)$ gives $(1+2\lambda h)^{-\delta/2}\exp[-y\lambda/(1+2\lambda h)]$. This is the proposition’s transform. Laplace transforms determine probability laws on $[0,\infty)$, so the two distributions agree.

The CIR Conditional Distribution

Theorem: Under the stated CIR assumptions, for $0\le s<t$, conditionally on $\mathcal{F}_s$, $$ X_t\ \overset{d}{=}\ c\,\chi'^2_{\delta}(\lambda), \qquad c=\frac{\sigma^2}{4\kappa}(1-e^{-\kappa(t-s)}),\quad \delta=\frac{4\kappa\theta}{\sigma^2},\quad \lambda=\frac{4\kappa e^{-\kappa(t-s)}X_s}{\sigma^2(1-e^{-\kappa(t-s)})}. $$ Here the equality means equality of conditional distributions; the noncentrality parameter is $\mathcal{F}_s$-measurable.

Proof. The squared Bessel time increment between $s$ and $t$ is $h=(e^{\kappa t}-e^{\kappa s})/\kappa>0$, and its starting value is $y=(4/\sigma^2)e^{\kappa s}X_s$. The proposition and distributional identification give $Z_h\mid\mathcal{F}_s\overset{d}{=}h\chi'^2_\delta(y/h)$. Since $X_t=(\sigma^2/4)e^{-\kappa t}Z_h$, its scale is $(\sigma^2/4)e^{-\kappa t}h=(\sigma^2/(4\kappa))(1-e^{-\kappa(t-s)})=c$. Its noncentrality is $$ \frac{y}{h} =\frac{(4/\sigma^2)e^{\kappa s}X_s}{(e^{\kappa t}-e^{\kappa s})/\kappa} =\frac{4\kappa e^{-\kappa(t-s)}X_s}{\sigma^2(1-e^{-\kappa(t-s)})}. $$ These are the claimed parameters. $\square$

The scale $c$ has the units needed to convert the chi-square variable back to $X_t$. The degrees of freedom depend only on the model parameters; the noncentrality also depends on the current state. When $\theta=0$, the degrees of freedom are zero, and the mixture definition must retain its zero-degree component rather than treating every component as having a density.

Worked Example: A Central Transition Probability

Take $\kappa=1$, $\theta=2$, $\sigma=2$, $X_s=0$, and $t-s=\ln 2$. Find $\mathbb{P}(X_t\le\tfrac12\mid\mathcal{F}_s)$.

Here $e^{-\kappa(t-s)}=e^{-\ln2}=\tfrac12$. The degrees of freedom are $\delta=4(1)(2)/2^2=2$, and the noncentrality is $\lambda=4(1)(\tfrac12)(0)/(2^2(1-\tfrac12))=0$. The scale is $c=(2^2/(4\cdot1))(1-\tfrac12)=\tfrac12$. Thus $X_t\mid\mathcal{F}_s\overset{d}{=}\tfrac12\chi^2_2$, and the requested event is $\{\chi^2_2\le1\}$. The chi-square distribution with two degrees of freedom has CDF $1-e^{-z/2}$ for $z\ge0$, so the probability is $1-e^{-1/2}$.

Check: The conditional mean from this distribution is $\tfrac12\cdot2=1$, which agrees with the CIR mean $\theta+(X_s-\theta)e^{-\kappa(t-s)}=2+(0-2)\tfrac12=1$.

Worked Example: A Noncentral Transition and Its Atom

Take $\kappa=1$, $\theta=0$, $\sigma=2$, $X_s=1$, and $t-s=\ln2$. Find the probability that $X_t$ is strictly positive.

Again $e^{-\kappa(t-s)}=\tfrac12$. Now $\delta=4(1)(0)/2^2=0$, $c=(2^2/4)(1-\tfrac12)=\tfrac12$, and $\lambda=4(1)(\tfrac12)(1)/(2^2(1-\tfrac12))=1$. By the mixture definition, $N$ is Poisson with mean $\lambda/2=\tfrac12$. When $N=0$, the chi-square component has zero degrees of freedom and equals zero; when $N\ge1$, it has positive degrees of freedom and is strictly positive with probability one. Therefore $$ \mathbb{P}(X_t>0\mid\mathcal{F}_s) =\mathbb{P}(N\ge1) =1-\mathbb{P}(N=0) =1-e^{-1/2}. $$ The complementary atom at zero has probability $e^{-1/2}$.

Check: The two probabilities sum to $(1-e^{-1/2})+e^{-1/2}=1$. The event is strictly $X_t>0$; including zero instead would give probability one.

Conditional Moments from the Mixture

If $Y\sim\chi'^2_\delta(\lambda)$, then $\mathbb{E}[Y]=\delta+\lambda$ and $\operatorname{Var}(Y)=2(\delta+2\lambda)$. To see both identities, condition on the Poisson variable $N$: the conditional mean of $Y$ is $\delta+2N$, and its conditional variance is $2(\delta+2N)$. Since $\mathbb{E}[N]=\operatorname{Var}(N)=\lambda/2$, the unconditional mean is $\delta+\lambda$. The law of total variance gives $2\delta+2\lambda+4(\lambda/2)=2\delta+4\lambda$, as claimed. These identities also hold for $\delta=0$, including the degenerate component.

Applying them to $X_t=cY$ gives the familiar conditional moments. The algebra is included to show how each model parameter enters:

\begin{aligned} \mathbb{E}[X_t\mid\mathcal{F}_s] &=c(\delta+\lambda) =\theta(1-e^{-\kappa\tau})+e^{-\kappa\tau}X_s,\\ \operatorname{Var}(X_t\mid\mathcal{F}_s) &=2c^2(\delta+2\lambda)\\ &=\frac{\theta\sigma^2}{2\kappa}(1-e^{-\kappa\tau})^2 +\frac{\sigma^2X_s}{\kappa}e^{-\kappa\tau}(1-e^{-\kappa\tau}). \end{aligned}

Worked Example: Checking Both Conditional Moments

Take $\kappa=2$, $\theta=3$, $\sigma=2$, $X_s=4$, and $t-s=\ln2$. Compute the scale, degrees of freedom, noncentrality, mean, and variance.

Since $e^{-2\ln2}=e^{-\ln4}=\tfrac14$, the scale is $c=(2^2/(4\cdot2))(1-\tfrac14)=\tfrac38$. The degrees of freedom are $\delta=4(2)(3)/2^2=6$, and $\lambda=4(2)(\tfrac14)(4)/(2^2(1-\tfrac14))=8/3$. Hence $\mathbb{E}[X_t\mid\mathcal{F}_s]=\tfrac38(6+\tfrac83)=\tfrac{13}{4}$. The variance is $2(\tfrac38)^2(6+2\cdot\tfrac83)=\tfrac{9}{32}\cdot\tfrac{34}{3}=\tfrac{51}{16}$.

Check: The mean formula in CIR parameters gives $3+(4-3)\tfrac14=\tfrac{13}{4}$. The variance formula gives $\frac{3\cdot4}{4}(\tfrac34)^2+\frac{4\cdot4}{2}(\tfrac14)(\tfrac34)=\tfrac{27}{16}+\tfrac{24}{16}=\tfrac{51}{16}$, agreeing with the distributional calculation.

Common Pitfall: Losing the Time-Change Factors

A frequent error is to use the squared Bessel starting value as the CIR noncentrality parameter. It is not: the noncentrality is the squared Bessel starting value divided by the elapsed squared Bessel time, $y/h$. Another error is to omit the factor $(\sigma^2/4)e^{-\kappa t}$ when converting the squared Bessel variable back to $X_t$; doing so gives the wrong scale.

For example, in the first worked example the squared Bessel clock increment is $h=(2-1)/1=1$, and its starting value is $y=(4/4)e^0\cdot0=0$. The noncentrality is $y/h=0$, and the CIR scale is $(4/4)e^{-\ln2}\cdot1=\tfrac12$. The transition is therefore $\tfrac12\chi^2_2$, not $\chi^2_2$. Its mean is $1$, whereas omitting the scale would incorrectly give mean $2$.

The transition law is useful both for exact conditional probability calculations and for checking approximations to CIR dynamics. Its three parameters have distinct roles: $c$ rescales the distribution, $\delta$ controls its degrees of freedom, and $\lambda$ records the influence of the current state. The mixture representation is especially important when $\delta=0$, because it keeps track of a possible atom at zero.

Check Your Understanding

Use the conditional law and the Poisson-mixture definition to answer the following.

For $\kappa=1$, $\theta=1$, $\sigma=2$, and $t-s=\ln2$, find $c$ and $\delta$.
For those same parameters and $X_s=2$, calculate the noncentrality parameter $\lambda$.
Explain why the CIR noncentrality is $y/h$, rather than just the transformed starting value $y$.
When $\delta=0$ and $\lambda>0$, express the mass at zero using the Poisson variable $N$.
Derive the conditional mean from $c(\delta+\lambda)$ and compare it with $\theta+(X_s-\theta)e^{-\kappa(t-s)}$.