← three numbers in a yield curve certificate
One curve from short to long maturities
Exact Perron growth removes secular error; a moving invariant law has a sharp second-order cost.
A fixed-maturity expansion need not remain accurate at long maturities. In a general chain an omitted term of size $T/m^2$ is order one when $T$ is comparable to $m^2$. Matched formulas that keep the exact principal eigenvalue are uniformly accurate for every $T\geq0$, and a moving-projection formula extends the result to any finite irreducible chain.
This page states the main results. Proofs, certificates and further cases are in the full note.
The two-state matched formula
Let the regimes switch in either direction at rate $m$, and write their time-dependent pricing forcings as $g_1=\bar g+\delta$ and $g_2=\bar g-\delta$. If $a'=(Q+\operatorname{diag}g)a$ with $a(0)=\mathbf1$, put $s=\tfrac12(a_1+a_2)$.
Suppose $\bar g(t)\to\bar g_\infty$ and $\delta(t)\to\delta_\infty$. The exact principal eigenvalue of the limiting matrix is
$$\Lambda(m)=\bar g_\infty+\sqrt{m^2+\delta_\infty^2}-m.$$For a stationary initial regime, the uniformly matched log amplitude is
At fixed maturity, expanding $\Lambda$ recovers $\int\bar g+(2m)^{-1}\int\delta^2$. At long maturity, keeping $\Lambda$ exact prevents the omitted $T/m^3,T/m^5,\ldots$ eigenvalue terms from accumulating.
Assume $\delta$ is continuously differentiable, $G=\sup_t|\delta(t)|<m$,
$$V_\delta=\int_0^\infty|\delta'(t)|dt<\infty,\qquad L_\delta=\int_0^\infty|\delta(t)-\delta_\infty|dt<\infty.$$Then, simultaneously for every maturity $T\geq0$,
The right-hand side is $O(m^{-2})$ and does not grow with $T$.
The Vasicek certificate uses $\kappa=1$, levels $(8\%,2\%)$, and variances $(0.004,0.001)$. It integrates the exact Riccati system out to $\max(20,m^2)$.
| $m$ | maximum log error | proved bound | $m^2\times$error |
|---|---|---|---|
| 1 | 1.4264e-4 | 4.4103e-4 | 1.4264e-4 |
| 4 | 7.4339e-6 | 2.6939e-5 | 1.1894e-4 |
| 32 | 1.0607e-7 | 4.1815e-7 | 1.0861e-4 |
Any finite chain with fixed generator
Let $Q_0$ be any irreducible finite-state generator, possibly defective, with stationary row vector $\pi$. Let $D(t)$ be a bounded $C^1$ diagonal matrix with $D(0)=0$, $\|D'\|_\infty<\infty$, and $\int_0^\infty\|D'(t)\|dt<\infty$.
Set $A_m(t)=mQ_0+D(t)$ and let $\lambda_m(t)$ be its instantaneous Perron eigenvalue. For all sufficiently large $m$, choose positive right and left eigenvectors $r_m(t),\ell_m(t)$ by
$$A_mr_m=\lambda_mr_m,\qquad \ell_m^\top A_m=\lambda_m\ell_m^\top,\qquad \pi r_m=1,\qquad \ell_m^\top r_m=1.$$The scalar transport connection and transported exponent are
$$\gamma_m(t)=\ell_m(t)^\top r_m'(t),\qquad L_m(T)=\int_0^T\{\lambda_m(t)-\gamma_m(t)\}\,dt.$$Moving-projection theorem. Let $u_m'=A_m(t)u_m$ and $u_m(0)=\mathbf1$. There are constants $m_0,C<\infty$ such that, for every $m\ge m_0$,
$$\sup_{T\ge0}\left|\log(\pi u_m(T))-L_m(T)\right|\le \frac{C}{m^3}.$$For any fixed regime prior $\nu$, the endpoint-corrected approximation satisfies
$$\sup_{T\ge0}\left|\log(\nu u_m(T))-L_m(T)-\log(\nu r_m(T))\right| \le \frac{C_\nu}{m^2}.$$The connection term is essential for a nonreversible chain: integrating only the instantaneous eigenvalue omits transport of the moving slow eigenspace. It is the finite-dimensional spectral transport introduced in Kato's original adiabatic theorem.
A limiting-Perron corollary
The exact moving eigendata can be replaced by the simpler matched formula used on the yield-curve page. In addition to the hypotheses above, assume that $D(t)\to D_\infty$ and
$$\int_0^\infty\{\|D'(t)\|+\|D(t)-D_\infty\|\}\,dt<\infty.$$Write $g(t)=\operatorname{diag}D(t)$, $\bar g(t)=\pi g(t)$, $\widetilde g(t)=g(t)-\bar g(t)\mathbf1$, and let $Q_0^\#$ be the group inverse. Define
$$k(t)=-\pi\{\widetilde g(t)\odot Q_0^\#\widetilde g(t)\},$$and let $\Lambda_{m,\infty}$ be the exact principal eigenvalue of $mQ_0+D_\infty$. Then
For every fixed regime prior $\nu$, the same scalar composite with the instantaneous Perron endpoint factor satisfies
$$\sup_{T\geq0}|\log(\nu u_m(T))-C_m(T)-\log\{\nu r_m(T)\}| \leq {C_\nu\over m^2}.$$This is a theorem for every fixed finite irreducible generator, including defective ones; it is not merely a fixed-maturity expansion. It keeps the exact limiting Perron root, so no omitted constant-forcing eigenvalue coefficient can accumulate with maturity.
The prior-dependent statement follows by combining the same comparison of transported exponents with the moving-projection theorem above. The assumptions are important: without an integrable approach to $D_\infty$, a transient second-order coefficient can itself accumulate.
The certificate uses a defective three-state generator with $D(t)=(1-e^{-t})\operatorname{diag}(-0.6,-0.2,-0.4)$. It integrates the normalized exact system through $T=\max(20,m^2)$ and evaluates both transient integrals in $C_m$ analytically.
| $m$ | stationary error | $m^2\times$error | largest pure-start error | $m^2\times$error |
|---|---|---|---|---|
| 4 | 8.3411e-4 | 0.013346 | 7.8324e-3 | 0.125318 |
| 16 | 4.6889e-5 | 0.012003 | 6.4790e-4 | 0.165862 |
| 64 | 2.8186e-6 | 0.011545 | 4.5166e-5 | 0.184999 |
The final observed orders are $2.0198$ and $1.9472$. Checking the three pure starts controls every fixed mixture because the ratio of two positive mixtures lies between the componentwise ratios.
When the fast generator moves
The third-order stationary cancellation above uses more than a spectral gap: it uses the fixed invariant row $\pi$. Let $Q(t)$ now be a $C^1$ family of finite-state generators with a common state space, $Q(t)\mathbf1=0$, invariant rows $\pi(t)$, and uniform exponential ergodicity on the complementary subspace.
Assume $Q$ and $D$ are bounded and have finite total variation, put $A_m(t)=mQ(t)+D(t)$, take $D(0)=0$ and fix the gauge row $q=\pi(0)$. Use the same eigenvector equations with $q r_m=1$, and retain the same connection and transported exponent.
Moving-generator theorem. For every fixed regime prior $\nu$, there are $m_0,C_\nu<\infty$ such that
$$\sup_{T\geq0}\left|\log(\nu u_m(T))-L_m(T)-\log\{\nu r_m(T)\}\right| \leq {C_\nu\over m^2},\qquad m\geq m_0.$$The order is generically sharp even for the initial stationary observation $\nu=q$. Thus the fixed-$Q$ $O(m^{-3})$ statement must not be transferred to state-dependent or time-dependent switching rates.
In the two-state certificate the final measured orders are $1.9578$ for the moving generator and $2.9842$ for its fixed-generator control.
Periodic forcing
Finite total variation is sufficient for the preceding transport estimates, but it is not necessary. Return to the symmetric two-state reduction and suppose that $\bar g,\delta$ are $P$-periodic, with $\delta\in C^2$.
Write $\langle f\rangle=P^{-1}\int_0^P f(t)dt$ and $r=(a_1-a_2)/(a_1+a_2)$. For all sufficiently large $m$, the Riccati equation $r'=\delta-2mr-\delta r^2$ has a unique attracting periodic solution $r_m^*$, and set
$$\lambda_m=\left\langle\bar g+\delta r_m^*\right\rangle,$$ $$\Phi_{m,1}(T)=\int_0^T\left\{\bar g(t)-\langle\bar g\rangle+ {\delta(t)^2-\langle\delta^2\rangle\over2m}\right\}dt.$$The number $\lambda_m$ is the exact principal Floquet exponent, and the phase $\Phi_{m,1}$ is bounded and periodic.
Periodic Floquet theorem. For the stationary initial regime, $r(0)=0$, there are $m_0,C<\infty$ such that
$$\sup_{T\geq0}\left|\log s(T)-\lambda_mT-\Phi_{m,1}(T)\right| \leq {C\over m^2},\qquad m\geq m_0.$$This is genuinely uniform over arbitrarily many periods. Replacing $\lambda_m$ by only its first-order expansion generally destroys that statement.
This construction is the classical decomposition of a periodic linear system into exponential growth and a periodic factor, originating in Floquet's 1883 paper.
Any finite periodic chain: frozen and transported layers
Let $Q(t)$ and the real diagonal forcing $D(t)$ be $C^2$ and $P$-periodic, with $Q(t)$ a uniformly irreducible finite-state generator having a uniform fast spectral gap. Write $A_m(t)=mQ(t)+D(t)$ and let $U_m(t,s)$ be its evolution family.
The positive monodromy $F_m=U_m(P,0)$ has a Perron multiplier $\mu_m$, right and left vectors $v_m,w_m$, and exact Floquet exponent
$$\lambda_m=P^{-1}\log\mu_m,\qquad w_m^\top v_m=1,\qquad \pi(0)v_m=1.$$Define the periodic principal profile, its initial coefficient, and the frozen fast layer by
$$r_m(t)=e^{-\lambda_mt}U_m(t,0)v_m,\qquad c_m=w_m^\top\mathbf1,\qquad z_{m,0}=\mathbf1-c_mv_m,$$ $$b_{m,0}(t)=e^{\{A_m(0)-\lambda_mI\}t}z_{m,0}.$$Bounded-horizon frozen-layer theorem. For every fixed probability row $\nu$ and every fixed $H<\infty$,
$$\sup_{0\leq T\leq H}\left| \log\{\nu U_m(T,0)\mathbf1\}-\lambda_mT -\log\{c_m\nu r_m(T)+\nu b_{m,0}(T)\} \right|\leq {C_{\nu,H}\over m^2}.$$The exact Floquet principal term without a frozen slip is genuinely all-maturity: its error is $O(m^{-2})$ for the stationary initial row $\nu=\pi(0)$ and generically $O(m^{-1})$ for a general fixed prior.
Why the locally frozen slip is not an all-maturity object
The earlier version of this note incorrectly put a supremum over every $T\geq0$ around the frozen-slip estimates. The exact Floquet complement decays under the nonautonomous exponential dichotomy, but the locally frozen term eventually grows relative to the exact normalized price. No constant independent of maturity can then bound its log error.
In the moving three-state certificate at phase zero and $m=64$, the frozen Perron root exceeds the Floquet exponent by $0.0551616197852$. At integer periods the frozen bracket for the bond changes sign between $42P=263.893783$ and $43P=270.176968$; for $x=(1.35,0.72,1.08)$ it changes sign between $28P=175.929189$ and $29P=182.212374$. The exact prices remain strictly positive.
This counterexample does not affect the bounded-horizon theorem or the all-maturity no-slip Floquet estimates. A genuinely all-maturity matched layer must use the transported nonautonomous complementary bundle (or an explicitly stable surrogate), not the exponential of one locally frozen full matrix.
The exact transported repair uses only one-period data
For a payoff vector $x$, put $c_m(x)=w_m^\top x$ and $z_m(x)=x-c_m(x)v_m$. Let $\Pi_m=v_mw_m^\top$, and normalize the monodromy on the principal Floquet scale:
$$G_m=e^{-\lambda_mP}F_m,\qquad C_m=G_m(I-\Pi_m)=G_m-\Pi_m.$$For $T=kP+\tau$, where $k\in\mathbb N_0$ and $0\leq\tau<P$, define
$$b^{\rm tr}_m(T;x)=e^{-\lambda_m\tau}U_m(\tau,0)C_m^{k}z_m(x).$$Exact transported-layer identity. For every vector $x$ and every $T\geq0$,
$$e^{-\lambda_mT}U_m(T,0)x =c_m(x)r_m(T)+b^{\rm tr}_m(T;x).$$Consequently, for a positive payoff and probability row $\nu$,
$$\log\{\nu U_m(T,0)x\}=\lambda_mT+ \log\{c_m(x)\nu r_m(T)+\nu b^{\rm tr}_m(T;x)\}$$exactly, with a strictly positive logarithmic bracket.
This is an exact all-maturity representation, not a new asymptotic error claim. It requires the one-period monodromy and only the final within-period propagator, so it avoids both long-horizon integration and the unstable local-Perron leakage of the frozen full matrix. The locally frozen layer remains the simpler analytic approximation on fixed bounded horizons.
At the very maturities where the frozen brackets have become negative, the exact transported brackets are $0.999925971$ at $43P$ for the bond and $1.055504121$ at $29P$ for the regime-dependent payoff.
The rest of the note
The full note also covers known starting regimes and initial slips, nonzero near-end forcing, unequal and periodic switching rates, an arbitrary calendar phase, regime-dependent payoffs, an all-order recursion for the Floquet exponent, a posteriori enclosures of the exponent and price, and the critical maturity at which a truncated exponent fails.
See the full note for statements, proofs and tables, and the certificate on GitHub for the numbers.