Affine CIR Bond Pricing by Riccati Reduction

From the CIR Rate to an Affine Bond Price

A zero-coupon bond pays one unit at maturity and nothing before then. Under the risk-neutral measure, its price is the conditional expectation of the discounted payoff. For a short rate that follows the Cox–Ingersoll–Ross (CIR) model, the discount rate is random, and the pricing PDE contains both a first-derivative term and a second-derivative term. The useful feature of CIR is that an exponential-affine price form reduces this PDE to two scalar ordinary differential equations, one of which is a Riccati equation.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), with an equivalent risk-neutral measure \(\mathbb{Q}\) on the pricing horizon. Under \(\mathbb{Q}\), let \(B_t^{\mathbb{Q}}\) be Brownian motion and suppose the short rate satisfies \(dr_t=\kappa(\theta-r_t)\,dt+\sigma\sqrt{r_t}\,dB_t^{\mathbb{Q}}\), where \(\kappa,\theta,\sigma>0\) and \(r_0\ge0\). Assume the CIR solution is nonnegative and has finite first moments on finite time intervals; these properties hold for the standard CIR model. The stochastic integral is an Ito integral with progressively measurable integrand \(\sigma\sqrt{r_t}\) in \(\mathcal{L}^2\) on each finite horizon, meaning \(\mathbb{E}^{\mathbb{Q}}\int_0^T\sigma^2r_s\,ds<\infty\).

For maturity \(T\), write \(P(t,r)\) for the price at time \(t\) when the current short rate is \(r\). Risk-neutral pricing gives \(P(t,r_t)=\mathbb{E}^{\mathbb{Q}}[\exp(-\int_t^T r_s\,ds)\mid\mathcal{F}_t]\). The pricing PDE and terminal condition are \(P_t+\kappa(\theta-r)P_r+\tfrac12\sigma^2rP_{rr}-rP=0\) for \(t<T\), with \(P(T,r)=1\). Here \(\kappa(\theta-r)\) is the drift and \(\sigma\sqrt r\) is the diffusion coefficient of the short rate.

Riccati Reduction and Its Explicit Solution

Substitute the proposed form into the PDE. With \(\tau=T-t\), the derivatives are \(P_t=-P_\tau\), \(P_\tau=(A'-B'r)P\), \(P_r=-BP\), and \(P_{rr}=B^2P\). Dividing by the positive quantity \(P\) gives \(-A'+B'r-\kappa\theta B+\kappa rB+\tfrac12\sigma^2rB^2-r=0\). Equating the constant term and the coefficient of \(r\) yields \(A'=-\kappa\theta B\) and \(B'=1-\kappa B-\tfrac12\sigma^2B^2\), with \(A(0)=B(0)=0\). The nonlinear equation for \(B\) is a Riccati equation.

Proof. Let \(D(\tau)=(\gamma+\kappa)(e^{\gamma\tau}-1)+2\gamma\). Since \(\gamma>\kappa>0\), \(D(\tau)>0\) for \(\tau\ge0\). The proposed \(B\) has \(B(0)=0\) and is nonnegative. To check its differential equation, put \(x=e^{\gamma\tau}\), so \(B=2(x-1)/D\), \(D=(\gamma+\kappa)x+\gamma-\kappa\), and \(D'= \gamma(\gamma+\kappa)x\). Differentiating gives \(B'=2\gamma xD-2(x-1)D'\over D^2=4\gamma^2x/D^2\), where the numerator equality follows by expanding \(xD-(x-1)(\gamma+\kappa)x=(\gamma-\kappa)x\) and using \(2\gamma(\gamma-\kappa)=4\sigma^2\) only if? Instead, direct expansion gives \(xD-(x-1)(\gamma+\kappa)x=2\gamma x\), so the numerator is \(4\gamma^2x\). Also, direct substitution of \(B=2(x-1)/D\) into \(1-\kappa B-\tfrac12\sigma^2B^2\), using \(\gamma^2-\kappa^2=2\sigma^2\), gives \(4\gamma^2x/D^2\). Therefore \(B'=1-\kappa B-\tfrac12\sigma^2B^2\).

For the second equation, differentiate the proposed logarithm. Its derivative is \(\dfrac{\kappa+\gamma}{2}-\dfrac{\gamma(\gamma+\kappa)e^{\gamma\tau}}{D} =-\dfrac{2\kappa\gamma(e^{\gamma\tau}-1)}{D}=-\kappa B\). Multiplication by \(2\kappa\theta/\sigma^2\) is not consistent with this last equality unless? Let's calculate derivative ratio: derivative log numerator (κ+γ)/2 - γ(γ+κ)x/D. simplify factor (γ+κ)[1/2 - γx/D] = (γ+κ)(D-2γx)/(2D). D-2γx=(κ-γ)x+γ-κ=-(γ-κ)(x-1). so derivative= -(γ+κ)(γ-κ)(x-1)/(2D) = -σ²(x-1)/D = -σ² B/2. times 2κθ/σ² = -κθ B. Hence \(A'=-\kappa\theta B\). At \(\tau=0\), the logarithm has argument one, so \(A(0)=0\). Finally, the right-hand sides of both ODEs are locally Lipschitz functions of their unknowns; the standard uniqueness theorem for ordinary differential equations gives uniqueness wherever the solution exists. The displayed formulas are finite for every \(\tau\ge0\), so they provide the required solution on every finite horizon. \(\square\)

Correction to the derivative computation above: in the displayed calculation for \(B'\), expanding \(xD-(x-1)(\gamma+\kappa)x\) gives \(2\gamma x\); hence \(2\gamma\) times this expression is \(4\gamma^2x\), as stated. For \(A'\), the logarithm derivative is \(-\sigma^2B/2\), and the prefactor gives \(-\kappa\theta B\). These identities verify the two equations directly.

Worked Coefficients and Bond Prices

Why the PDE Solution Is the Bond Price

The PDE calculation produces a candidate function. To identify it with the risk-neutral expectation, apply Ito's formula to the discounted candidate along the CIR process. Define \(Y_s=\exp(-\int_t^s r_u\,du)P(s,r_s)\) for \(t\le s\le T\). The PDE makes the drift of \(Y\) zero. More explicitly, with \(P_r=-BP\), Ito's formula gives the stochastic term \(\exp(-\int_t^s r_u\,du)\sigma\sqrt{r_s}P_r(s,r_s)\,dB_s^{\mathbb{Q}}\), and the drift is the same discount factor times \(P_t+\kappa(\theta-r_s)P_r+\tfrac12\sigma^2r_sP_{rr}-r_sP\), which vanishes.

Here the stochastic integral is an Ito integral. Its integrand is progressively measurable, and it belongs to \(\mathcal{L}^2\) on a finite horizon: \(B(\tau)\) is bounded on \([0,T]\), \(0<P\le1\), and \(\mathbb{E}^{\mathbb{Q}}\int_0^T \sigma^2r_s B(T-s)^2P(s,r_s)^2\,ds \le \sigma^2\sup_{0\le u\le T}B(u)^2\int_0^T\mathbb{E}^{\mathbb{Q}}[r_s]\,ds<\infty\). The finite first-moment assumption supplies the last inequality. Therefore the Ito integral is a square-integrable martingale, and \(Y\) is a martingale.

For completeness, the bounds used here follow from the formulas: \(B(\tau)\ge0\), \(A'=-\kappa\theta B\le0\), and \(A(0)=0\), so \(0<P(t,r)\le1\) for \(r\ge0\). Nonnegative rates also imply \(0<\exp(-\int_t^s r_u\,du)\le1\). At \(s=T\), the terminal condition gives \(Y_T=\exp(-\int_t^T r_u\,du)\). Taking conditional expectations in the martingale identity \(Y_t=\mathbb{E}^{\mathbb{Q}}[Y_T\mid\mathcal{F}_t]\) proves that the PDE solution equals the risk-neutral bond price.

A Common Sign Pitfall

Changing to time to maturity reverses the time derivative: if \(\tau=T-t\), then \(P_t=-P_\tau\). Omitting this minus sign leads to the wrong Riccati equation. For example, the incorrect equation \(B'=-1-\kappa B-\tfrac12\sigma^2B^2\) would give \(B'(0)=-1\). But a short-maturity bond must initially decrease with the short rate, so its affine exponent requires \(B'(0)=1\). The correct equation \(B'=1-\kappa B-\tfrac12\sigma^2B^2\) has that initial slope and yields \(B(\tau)\ge0\), consistent with a lower bond price when the current rate is higher.

The affine form is useful because it turns a second-order pricing PDE into two scalar equations whose coefficients can be checked independently. The Riccati coefficient controls sensitivity of the price to the current short rate; the constant coefficient accounts for the mean-reverting rate dynamics and the accumulated discounting. The risk-neutral verification matters: solving the PDE is not by itself enough unless the candidate satisfies its terminal condition and represents the conditional discounted payoff.