Model Choice Uses More Than a Curve Fit
Vasicek and CIR are both mean-reverting short-rate models, and both give affine zero-coupon bond prices. Their important difference for model choice is the diffusion: Vasicek has constant volatility, while CIR volatility is proportional to the square root of the current rate. Consequently, Vasicek rates can be negative, whereas CIR rates remain nonnegative when started nonnegative. CIR also has a parameter-dependent zero-boundary behavior. Fitted bond-curve residuals provide a separate diagnostic: they measure how well a model matches the observed curve, but do not by themselves determine whether its rate support or boundary assumptions are suitable.
For bond pricing, take a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{Q})\) carrying a Brownian motion \(B_t\) under the risk-neutral measure \(\mathbb{Q}\). The drift and diffusion coefficient in the dynamics below are risk-neutral coefficients. The corresponding stochastic integrals are Ito integrals. Whenever an integral is used, its integrand is progressively measurable; an integrand in \(\mathcal{L}^2\) on \([0,T]\) satisfies \(\mathbb{E}^{\mathbb{Q}}\int_0^T H_s^2\,ds<\infty\). If only local square integrability is available, the integral and SDE are understood by localization. For Vasicek, the constant diffusion integrand is in \(\mathcal{L}^2\) on every finite horizon. For a nonnegative CIR rate with finite first moments, \(\sqrt{r_s}\) is in \(\mathcal{L}^2\) on finite horizons because \(\mathbb{E}^{\mathbb{Q}}\int_0^T r_s\,ds<\infty\).
The measure matters. Bond prices are risk-neutral expectations of discounted payoffs, so the pricing dynamics are under \(\mathbb{Q}\), not necessarily the historical measure \(\mathbb{P}\). A model fitted to observed bonds is using risk-neutral parameters; historical estimates under \(\mathbb{P}\) need not coincide. Changing measure does not make the two diffusion structures interchangeable.
Support and the CIR Boundary
In Vasicek, the rate has a Gaussian conditional distribution. In CIR, the state remains nonnegative, but whether a rate starting above zero can reach zero depends on the Feller condition. The boundary classification is a standard result for the CIR diffusion; its proof uses scale and speed analysis and is not repeated here.
For Vasicek, the conditional law makes the contrast precise. The following result also explains why a Gaussian rate model assigns positive probability to negative rates, even if the current rate and long-run level are both positive.
Proof. Multiply the Vasicek equation by the deterministic integrating factor \(e^{\kappa t}\). The product rule gives \(d(e^{\kappa t}r_t)=\kappa\theta e^{\kappa t}\,dt+\sigma e^{\kappa t}\,dB_t\). Integrating and multiplying by \(e^{-\kappa t}\) yields
The deterministic integrand \(s\mapsto e^{-\kappa(t-s)}\) is progressively measurable and satisfies \(\int_0^t e^{-2\kappa(t-s)}\,ds=(1-e^{-2\kappa t})/(2\kappa)<\infty\), so it is in \(\mathcal{L}^2\). Its Ito integral is normal with mean zero and variance equal to this integral, by the distribution of Brownian integrals with deterministic integrands. This gives \(m_t\) and \(v_t\). Since \(t>0\), \(\kappa>0\), and \(\sigma>0\), \(v_t>0\). A normal distribution with positive variance has a strictly positive density on all of \(\mathbb{R}\); therefore its probability on \((-\infty,0)\) is positive. \(\square\)
Worked Example: Applying the CIR Boundary Test
Compare two CIR specifications, each with \(\kappa=1\) and \(\sigma=0.2\), but with \(\theta=0.03\) in the first and \(\theta=0.01\) in the second. Assume \(r_0>0\). Determine the boundary behavior in each case.
The common squared volatility is \(\sigma^2=(0.2)^2=0.04\). In the first specification, \(2\kappa\theta=2\cdot1\cdot0.03=0.06\), and \(0.06\geq0.04\). Zero is therefore unattainable from the interior. In the second, \(2\kappa\theta=2\cdot1\cdot0.01=0.02\), and \(0.02<0.04\). Zero is attainable, although the process remains nonnegative.
Check: the threshold value of \(\theta\) is \(\sigma^2/(2\kappa)=0.04/2=0.02\). The first long-run level, \(0.03\), is above that threshold, while the second, \(0.01\), is below it, agreeing with the two inequalities.
Bond Curves and Residuals
For maturity \(\tau=T-t>0\), let \(P(t,T)\) be the time-\(t\) price of a zero-coupon bond paying one unit at \(T\). Write its continuously compounded yield as \(y(t,\tau)=-\log P(t,t+\tau)/\tau\). Both models have affine bond prices, but their parameter dependence differs. The formulas below are the standard affine Vasicek and CIR bond-pricing results established in the earlier material on affine bond pricing.
In Vasicek, \(P(t,t+\tau)=A_V(\tau)e^{-B_V(\tau)r_t}\), where
In CIR, \(P(t,t+\tau)=A_C(\tau)e^{-B_C(\tau)r_t}\), where \(\gamma=\sqrt{\kappa^2+2\sigma^2}\) and
Residuals should be interpreted with the quote convention and objective in mind. A yield residual is not a price residual; a small yield error at one maturity need not correspond to a fixed price error at another. Weighting also matters: changing the \(w_i\) changes which maturities dominate the fit. The residual pattern across maturities can reveal more than the total error alone.
Worked Example: Comparing Two Sets of Fitted Residuals
Suppose four market yields at maturities \(0.5,1,2,5\) years give the following residuals, in basis points, after calibration: Vasicek \((4,2,-1,-5)\) and CIR \((1,-1,2,-2)\). Compute the unweighted SSE and RMSE for each fit.
For Vasicek, \(\mathrm{SSE}=4^2+2^2+(-1)^2+(-5)^2=16+4+1+25=46\) basis-points squared. Thus \(\mathrm{RMSE}=\sqrt{46/4}=\sqrt{11.5}\) basis points. For CIR, \(\mathrm{SSE}=1^2+(-1)^2+2^2+(-2)^2=1+1+4+4=10\) basis-points squared, so \(\mathrm{RMSE}=\sqrt{10/4}=\sqrt{2.5}\) basis points.
Check: the sums use four residuals in both cases, so comparing SSE also compares RMSE. Since \(10<46\), CIR has the smaller RMSE as well. The Vasicek residuals move from positive at the shorter maturities to negative at the longer ones, suggesting a maturity-dependent shape discrepancy. The CIR residuals are smaller overall but still change sign; the total error alone does not establish that CIR is preferable for the intended rate dynamics or that the residual pattern is harmless.
What Residual Orthogonality Can Tell You
Least-squares calibration has a useful local diagnostic. It does not say that the model is correct. It says that, at an interior optimum, the residual vector has no first-order component in the directions in which the fitted curve can move by changing its parameters.
Proof. Write \(J_{ij}=\partial y_i/\partial\alpha_j\). Since \(r=y^{\mathrm{mkt}}-y\), its derivative with respect to parameter \(\alpha_j\) is \(-J_{\cdot j}\), the negative of column \(j\) of \(J\). Differentiating the objective gives \(\partial L/\partial\alpha_j=-J_{\cdot j}^{\mathsf T}Wr\). At an interior local minimum of a differentiable function, each first partial derivative is zero. Therefore \(J_{\cdot j}^{\mathsf T}Wr=0\) for every parameter \(j\), which is exactly \(J^\mathsf{T}Wr=0\). \(\square\)
Worked Example: A Residual Pattern Along a Parameter Direction
Consider three maturities and one local parameter direction that changes the fitted yields in proportion to \(x=(1,2,3)^\mathsf{T}\). With unit weights, test the residual vectors \(r=(1,-2,1)^\mathsf{T}\) and \(q=(1,1,1)^\mathsf{T}\) for orthogonality to that direction. For \(q\), also find the parameter adjustment that removes its component along \(x\).
For \(r\), \(x^\mathsf{T}r=1\cdot1+2\cdot(-2)+3\cdot1=1-4+3=0\), so this residual has no first-order component along the specified parameter direction. Its squared error is \(1^2+(-2)^2+1^2=6\). For \(q\), \(x^\mathsf{T}q=1+2+3=6\), so it is not orthogonal. If a parameter adjustment \(\delta\) moves the fitted curve by \(\delta x\), the new residual is \(q-\delta x\). Minimizing \(\|q-\delta x\|^2\) gives \(\delta=(x^\mathsf{T}q)/(x^\mathsf{T}x)=6/(1+4+9)=3/7\).
The adjusted residual is \((1-3/7,1-6/7,1-9/7)=(4/7,1/7,-2/7)\). Its squared error is \(16/49+1/49+4/49=21/49=3/7\), below the original squared error \(1^2+1^2+1^2=3\). As a check, \(x^\mathsf{T}(q-\tfrac37x)=6-\tfrac37\cdot14=0\). This is a local, one-direction calculation, not a claim that a full nonlinear calibration has a unique or globally adequate fit.
Using the Diagnostics Together
Model choice should connect the data being fitted to the use of the model. A practical comparison asks at least three different questions:
- Does the support fit the application? If negative rates must be possible under the intended specification, Vasicek permits them. CIR does not. Conversely, if negative rates are ruled out by the modeling objective, Vasicek's Gaussian support may be a material mismatch.
- Does zero matter, and what does the CIR parameter condition imply? The Feller condition distinguishes an unattainable zero boundary from an attainable one for a CIR rate started above zero. Attainability is not the same as crossing into negative rates.
- What shape remains in the fitted residuals? Check errors by maturity, their sign pattern, and their scale in the units that matter. A small aggregate SSE can hide a consistent long-maturity bias or an error concentrated at a particular maturity.
Neither model is selected solely by these checks. In particular, comparing in-sample residual sums across separately calibrated models is not a complete statistical model-selection procedure: it does not account for uncertainty, parameter count, out-of-sample performance, or whether the calibration quotes are reliable. Boundary behavior is a structural feature; residual fit is an empirical feature. A useful decision considers both, along with the model's intended pricing or risk-management role.
Common Pitfall: Treating the Feller Condition as a Positivity Test
A common incorrect conclusion is: “If \(2\kappa\theta<\sigma^2\), CIR can become negative.” That does not follow. The boundary classification says zero is attainable in this case, not that the process crosses zero; CIR remains nonnegative. The opposite overstatement is also wrong: \(2\kappa\theta\geq\sigma^2\) is the condition for zero to be unattainable from a strictly positive start, not a prerequisite for the process to be nonnegative. For Vasicek, by contrast, the normal conditional law gives positive probability of negative rates at every positive horizon.
Another tempting but inadequate comparison is to select whichever model has the smaller SSE and ignore the residuals' locations and the support implications. The worked residual example shows why the sign pattern across maturities and the intended use of the rate model should accompany any summary fit statistic.
Check Your Understanding
Use the definitions and diagnostics in this tutorial to answer the following.
- For CIR parameters \(\kappa=2\), \(\theta=0.01\), and \(\sigma=0.2\), determine whether zero is attainable from \(r_0>0\). Show the two quantities being compared.
- Why does the Vasicek conditional distribution assign positive probability to a negative rate at every \(t>0\) when \(\sigma>0\)?
- Given residuals \((2,-2,1)\) basis points at three maturities, compute the unweighted SSE and RMSE.
- What does \(J^\mathsf{T}Wr=0\) imply at an interior weighted least-squares optimum, and what does it not establish about the model?
- Explain why an attainable CIR zero boundary does not imply that the CIR rate becomes negative.