Green–Kubo geometry of joint default

Thus full rank can still be numerically useless when the design vectors nearly lie in a smaller subspace. This is the finite-dimensional frame-operator viewpoint introduced by Duffin and Schaeffer (1952).

An explicit positive design avoids an existence-only theorem. For any $\epsilon>0$, take

Here $W^\top W=\epsilon^2I+(d+2\epsilon)\mathbf1\mathbf1^\top$, so

There is also a closed-form inverse. If $y^{(r)}=Aw^{(r)}$ and $b=A\mathbf1$, then $b_r=y^{(r)}_r$ for $r $$Ae_r=\frac{y^{(r)}-b}{\epsilon}\quad(rThe spectrum also solves the optimal-design problem. For the minimally identifying case $q=d-1$, ordering the nonzero frame eigenvalues gives

Equality holds exactly when all $d-1$ nonzero frame eigenvalues agree. Equivalently, $W^\top W=cI_{d-1}$: the design vectors are orthogonal and have equal norm. Strictly positive vectors have strictly positive pairwise inner products, so every strictly positive minimal design has $\kappa(\mathcal M_W)>\sqrt2$.

The bound is nevertheless sharp. The family above satisfies

The same result has a minimax noise interpretation. If the total squared exposure is fixed at $\|W\|_F^2=B$, then

Equality has the same tight-frame condition and is therefore unattainable under strict positivity but approached by a common rescaling of the positive family as $\epsilon\to\infty$. Hence positivity creates no gap in the infimum, only nonattainment at the boundary where distinct causes become coordinate-isolated. The $1/\epsilon$ factor for small $\epsilon$ is the opposite, coalescing-design instability.

The certificate verifies the full singular spectrum for every $3\le d\le7$ and $1\le q\le d-1$, the explicit condition number and inverse for $3\le d\le8$, and both sharp optimal-design bounds.

A sharp three-cause example uses the same directed three-state cycle and cyclic hazards

Here $\|A\|_F=2.21094\times10^{-4}$ but $A\mathbf1=0$, so the equal-weight race is exactly unchanged by cycle reversal. At $w=(1,2,4)$ the predicted reversal-gap coefficients for the three causes are $(-0.0017192625,0.0051577876,-0.0034385251)$. Exact resolvents give second-order residual rates $1.998$ for both the probability vector and reversal gap.

The equal-weight design alone is the sharp $q=d-2$ counterexample. Adding just one further positive design recovers $A$ with maximum error $1.36\times10^{-19}$. The certificate also verifies the kernel-dimension formula for every $3\le d\le7$ and $1\le q\le d-1$, together with the complete frame spectrum.

Which rank bound survives?

For fixed centered feature vectors $h_1,\ldots,h_d$, the matrix $K\succeq0$ has rank at most $n-1$. In fact its rank equals the dimension spanned by those centered vectors for an irreducible finite chain. At a single maturity, the matrix of correction densities is $\operatorname{diag}(B(t))K\operatorname{diag}(B(t))$, and it has the same rank for $t>0$.

For completeness, if $f$ is centered and $w=-Q_0^\#f$, then $2\mathcal K_0(f,f)=\sum_{i,\ell\ne i}\pi_i(Q_0)_{i\ell} (w_\ell-w_i)^2$. Irreducibility makes this zero only when $w$ is constant, equivalently $f=0$. Thus the symmetric Green–Kubo form is an inner product on the $(n-1)$-dimensional centered space, which proves the claimed rank of $K$.

The integrated matrix $D(T)=K\circ J(T)$ is also positive semidefinite, because it is a time integral of positive semidefinite matrices. Its rank need not be at most $n-1$. Time-dependent diagonal rescalings rotate a low-dimensional range as $t$ varies.

In particular, a rank test applied directly to $D(T)$ is not a lower bound on the number of regimes. This distinction is absent for constant hazards: all $B_j(t)=1$ and $D(T)=TK$.

The sharp Hadamard-rank theorem

There is nevertheless an exact replacement for the false regime-count bound. Let $r=\operatorname{rank}K$ and $s=\operatorname{rank}J(T)$, where $s$ is the dimension of the span of the loading functions $B_1,\ldots,B_d$ in $L^2[0,T]$. Then

Indeed, factor $K=UU^\top$ and $J=VV^\top$, and denote their $j$th rows by $u_j$ and $v_j$. The Hadamard product has the Gram factorization

Thus the rank is exactly the dimension spanned by the $d$ rowwise tensor products $u_j\otimes v_j$ in an $rs$-dimensional space. The bound is sharp: generic factor pairs have rowwise tensor rank $\min(d,rs)$. Positivity of $D$ is the classical Schur product theorem; the tensor factorization also supplies the rank statement directly.

The corrected regime-count lower bound

If the fixed features are driven by an irreducible $n$-state chain, then $r=\operatorname{rank}K\le n-1$. Consequently the integrated rank $R=\operatorname{rank}D(T)$ and the known loading-span dimension $s=\operatorname{rank}J(T)$ imply

If only an upper bound $s\le q$ is known, the still-valid conclusion is $n\ge1+\lceil R/q\rceil$. This is the strongest possible lower bound from these ranks alone: for every $r,s\ge1$, generic Gram factors with $d\ge rs$ attain $R=rs$, and hence equality with $n=r+1$. A complete-graph $n$-state chain realizes an arbitrary rank-$r$ centered feature Gramian, while $s$ independent deterministic loading shapes realize the second factor.

The endpoints explain the apparent contradiction. With one common loading shape, $s=1$ and the familiar rule $n\ge R+1$ survives.

With $d$ independent loading shapes, a full-rank $d\times d$ integrated matrix says only $n\ge2$, and the two-regime CIR construction below attains that bound. This inference uses the rank of the loading family; it is not a claim that $D(T)$ itself is a fixed-feature Green–Kubo matrix.

Two regimes can create arbitrarily large integrated rank

The two-name example is not a dimension-two accident. For any number $d$ of names and any $T>0$, a two-regime model can have $\operatorname{rank}K=1$ but $\operatorname{rank}D(T)=d$.

Choose every centered regime feature on the single two-state contrast, with nonzero coefficient $a_j$, so $K=aa^\top$ up to scale. At zero vol-of-vol and distinct $\kappa_j>0$,

These functions are linearly independent on every positive interval: differentiating a vanishing linear combination gives a vanishing combination of distinct exponentials $e^{-\kappa_jt}$, whose Vandermonde system is nonsingular. Hence $J(T)$ is positive definite and

has full rank $d$. The conclusion persists for sufficiently small strictly positive CIR volatilities because the Riccati solutions and the Gram determinant depend continuously on $\sigma_j^2$.

Mean levels can simultaneously be chosen with a uniform strict Feller margin. Therefore no bound depending only on the number of regimes survives after maturity integration.

An exact full-rank criterion

The preceding rank-one construction is a special case of a sharp criterion that removes any numerical genericity assumption. Fix a common CIR volatility $\sigma\ge0$, pairwise distinct mean-reversion rates $\kappa_1,\ldots,\kappa_d$, and any $T>0$.

The CIR loading independence proved below makes $J(T)$ positive definite. For every positive semidefinite fixed-parameter Green–Kubo matrix $K$,

To prove sufficiency, factor $K=FF^\top$ and write $f_\ell$ for the columns of $F$. The Schur product has the exact decomposition

If a vector $x$ has zero quadratic form, positivity of $J(T)$ forces $\operatorname{diag}(f_\ell)x=0$ for every $\ell$. But $K_{ii}=\sum_\ell f_{\ell i}^2>0$, so some feature is nonzero in each coordinate and hence $x=0$. Conversely, if $K_{ii}=0$, positive semidefiniteness forces the entire $i$th row and column of $K$ to vanish, making $D(T)$ singular.

Thus heterogeneous maturity loadings create exact full rank at every positive maturity if and only if every name has nonzero fixed-parameter Green–Kubo variance. The rank of $K$ itself may be one. This does not increase fixed-feature Green–Kubo rank; it characterizes the rank created only after integrating different deterministic maturity loadings.

A five-name certificate uses the same positive CIR loading Gramian with fixed Green–Kubo ranks one, two, and three. All three integrated matrices have rank five; their smallest eigenvalues are respectively $1.57035185\times10^{-7}$, $7.36817240\times10^{-5}$, and $3.91143441\times10^{-2}$. Setting one feature row to zero makes the corresponding diagonal entry vanish and reduces the integrated rank exactly to four.

Exact rank under repeated loading shapes

The distinct-rate criterion has an exact extension when some names share a mean-reversion rate. Partition the name indices into clusters $C_1,\ldots,C_g$, where $\kappa_i=\alpha_c$ for $i\in C_c$, and assume the cluster centers $\alpha_c$ are distinct. For common CIR volatility and every $T>0$,

Indeed, factor $K=FF^\top$ and let $f_i$ be row $i$ of $F$. If $b_c(t)=B_{\alpha_c}(t)$, then $K\circ J(T)$ is the Gram matrix in $\mathbb R^r\otimes L^2[0,T]$ of the rowwise tensors

The distinct CIR loading functions $b_1,\ldots,b_g$ are linearly independent. Hence the tensor spans belonging to different clusters form a direct sum.

The span within cluster $C_c$ has dimension $\operatorname{rank}F_{C_c}=\operatorname{rank}K_{C_c,C_c}$, proving the formula. This rowwise tensor factorization is the transpose form of the product introduced by Khatri and Rao (1968).

Consequently, the integrated matrix is positive definite exactly when

The endpoints clarify the rank distinction. If every loading is distinct, each cluster is a singleton and the condition reduces to $K_{ii}>0$.

If every loading is identical, there is one cluster and integration preserves $\operatorname{rank}K$. Thus positive diagonal entries alone are not sufficient once loading shapes collide.

A six-name certificate with collision sizes $(3,2,1)$ uses fixed Green–Kubo ranks one, two, and three. The within-cluster rank sums are respectively $1+1+1$, $2+2+1$, and $3+2+1$, and the observed integrated ranks are exactly $3,5,6$. In the rank-one case every diagonal entry of $K$ is positive, yet the integrated rank is only the number of distinct loading shapes, three.

A complete heterogeneous two-name criterion

Clustering by mean reversion alone is valid only under the preceding common-volatility assumption. For two names allow separate CIR parameter pairs $(\kappa_i,\sigma_i)$, with $\kappa_i>0$ and $\sigma_i\ge0$, and let $B_i$ solve

For every $T>0$, the two-name loading Gramian $J(T)$ is positive definite exactly when

Indeed, a singular two-function Gramian makes the analytic loadings proportional on $[0,T]$. Since $B_i'(0)=1$, the proportionality constant must be one. The next two derivatives are

so equality of the loadings forces equality of both parameter pairs; the converse follows from uniqueness of the Riccati equation.

For the distinct-pair case, positive diagonal entries and the Schur decomposition above imply positive definiteness; a zero diagonal in a positive semidefinite matrix removes that coordinate. Equivalently, strict Cauchy–Schwarz for the two different loadings makes the determinant positive. This gives the complete heterogeneous-parameter answer in dimension two without assuming a common volatility.

The certificate fixes $\kappa_1=\kappa_2=1.1$ but takes $(\sigma_1,\sigma_2)=(0.18,0.46)$. The loading-Gram determinant is $9.27359551\times10^{-4}$. A rank-one positive-diagonal $K$ produces integrated eigenvalues $(1.73191413\times10^{-4},3.42690265)$, hence rank two.

When the entire parameter pair is repeated, the same $K$ remains rank one, while a positive-definite $K$ remains rank two. Equal mean reversion by itself is therefore not a loading collision.

The unrestricted heterogeneous many-name criterion

The two-name result in fact extends to every finite collection without a common-volatility or common-decay assumption. Put

so that

The coordinates $(\gamma_i,q_i)$ retain the full pair because

Loading theorem. The distinct functions $B_{\kappa_i,\sigma_i}$ are linearly independent on every interval of positive length. To prove it, expand for $t>0$:

After its constant term is removed by letting $t\to\infty$, this is an absolutely convergent generalized Dirichlet series whose rates are locally finite. Its coefficients are unique: order the rates, multiply by the exponential of the smallest remaining rate, let $t\to\infty$, and iterate.

Now suppose a nontrivial linear combination vanishes and let $\gamma_*$ be the smallest decay among its active terms. For any larger decay $\gamma_j$, a collision $n\gamma_*=m\gamma_j$ is impossible when $\gamma_j/\gamma_*$ is irrational; when that ratio is $a_j/b_j$ in lowest terms, it requires $a_j\mid n$.

Choose arbitrarily large primes $n$ exceeding every such numerator $a_j$. At rate $n\gamma_*$ only the slowest-decay group contributes, so coefficient uniqueness gives

Here the harmless nonzero factors have been absorbed into $d_i$, and large primes are odd. Within a fixed decay, distinct full parameter pairs have distinct $q_i$. Letting these primes tend to infinity and successively dividing by the largest positive $q_i^{n-1}$ kills every $q_i>0$ coefficient.

There can be at most one $q_i=0$ term; the coefficient at the collision-free first rate $\gamma_*$ kills it. Removing this decay group and repeating proves independence. A relation on a finite interval extends to all $t>0$ by analyticity, completing the argument.

The proof is the same rowwise tensor direct-sum argument as for the common-volatility cluster law, now justified without a parameter restriction. In particular, if every full pair is distinct, positive diagonal entries of $K$ are necessary and sufficient for full integrated rank, whatever the fixed rank of $K$.

The first numerical certificate deliberately creates harmonic collisions, taking $\gamma=(1,1,2,3)$ and $q=(0,0.35,0.20,0.55)$. All four full parameter pairs are distinct and the loading-Gram eigenvalues are $(5.7326\times10^{-7},6.3707\times10^{-5}, 3.6423\times10^{-2},8.8222)$, hence full rank.

A second certificate has unrestricted decay rates and full-pair cluster sizes $(3,2,1)$. Fixed Green–Kubo ranks one, two, and three give the predicted and observed integrated ranks $3,5,6$.

Short maturity has an explicit odd-power spectrum

Exact full rank at every $T>0$ is also poorly conditioned near zero maturity, because every CIR loading begins with the same term $B_i(t)=t+O(t^2)$. Write the vector loading as

and assume $V$ is nonsingular. Let $H_d$ be the shifted Hilbert matrix and $C$ its lower Cholesky factor:

If the eigenvalues of the loading Gramian $J(T)=\int_0^TB(t)B(t)^\top dt$ are ordered decreasingly, then

Consequently,

The proof rescales $t=Tx$ and factors

Exterior powers after QR and Cholesky give the individual constants. For common volatility and distinct mean-reversion rates, the CIR recurrence already displayed below makes $v_k$ a polynomial vector of degree $k-1$ with nonzero leading coefficient.

Thus $V$ is a scaled Vandermonde matrix and the hypothesis is automatic. This is the shifted version of the ordinary small-time controllability-Gramian hierarchy analyzed by Trélat and Zuazua (2026): the CIR loading vanishes at zero, shifting powers $T,T^3,\ldots$ to $T^3,T^5,\ldots$.

For rank-one fixed-feature geometry $K=aa^\top$ with every $a_i\ne0$, $D(T)=\operatorname{diag}(a)J(T)\operatorname{diag}(a)$ has the same powers after replacing $V$ by $\operatorname{diag}(a)V$. The theorem does not claim the same constants for arbitrary higher-rank $K$; that problem depends on its rowwise tensor jets. In particular, this short-maturity hierarchy is created by deterministic loading integration and does not change the fixed-parameter Green–Kubo rank.

For four positive-volatility loadings with $\kappa=(0.35,0.8,1.7,3.2)$ and common $\sigma=0.18$, the predicted constants for powers $(T^3,T^5,T^7,T^9)$ are

At $T=1/256$, the 100-digit scaled eigenvalues are

The largest relative error is $0.815\%$. The observed slopes are $(2.985538,4.972892,6.978991,8.987390)$, the normalized determinant is $0.977571$ against its $T^{24}$ limit, and the measured condition-number slope is $-6.001852$.

Higher fixed rank gives a block-jet spectrum

The rank-one hierarchy has an exact higher-rank extension. Factor the fixed-feature Gram matrix as $K=FF^\top$, with $F\in\mathbb R^{d\times r}$, and define the rowwise Taylor blocks and their cumulative ranks by

For analytic loadings, let $s_\infty$ be the eventual value of $s_q$. Then for every $T>0$,

and exactly $s_q-s_{q-1}$ positive eigenvalues of $D(T)$ have order $\Theta(T^{2q+1})$ as $T\downarrow0$. Thus the product of the positive eigenvalues has order

If $q_*=\min\{q:s_q=s_\infty\}$ and $s_1>0$, the positive-spectrum condition number is $\Theta(T^{-2(q_*-1)})$. In the full-rank case the product above is the ordinary determinant. The proof uses the nested spaces $\mathcal V_q=\operatorname{col}[A_1\ \cdots\ A_q]$: after decomposing into the new grades $\mathcal V_q\cap\mathcal V_{q-1}^{\perp}$ and rescaling the $q$th grade by $T^{q+1/2}$, the synthesis map converges to a full-row-rank polynomial map.

Its integral Gram matrix is positive definite because the grades have distinct first nonzero polynomial degrees. Exterior powers then give the stated multiplicities and powers.

Explicit block-jet determinant constant. Put $d_q=s_q-s_{q-1}$ and choose a $d\times d_q$ matrix $U_q$ with orthonormal columns spanning $\mathcal V_q\cap\mathcal V_{q-1}^{\perp}$. Define

Here $G=(G_{qp})_{q,p=1}^{q_*}$ has dimension $s_\infty$; grades with $d_q=0$ are simply omitted. Then $G$ is positive definite and

where $\operatorname{pdet}$ denotes the product of the positive eigenvalues. In particular, this is an ordinary determinant formula when $s_\infty=d$.

To see the constant directly, concatenate the $U_q$ and write $S_T=\operatorname{diag}_q(T^{q+1/2}I_{d_q})$. Taylor expansion gives

Each $C_q$ has full row rank. Hence the rows of the polynomial matrix with $q$th block $t^qC_q$ are independent, so their $L^2[0,1]$ Gram matrix $G$ is positive definite. Changing the orthonormal basis within any grade only orthogonally conjugates $G$, making $\det G$ intrinsic.

Every repeated-power constant. The same matrix gives the leading constant of each eigenvalue, not only their product. Let $G_{<q,<q}$ denote the leading block indexed by grades below $q$ and define the positive-definite Schur pivots

If $\mu_{q,1}\ge\cdots\ge\mu_{q,d_q}>0$ are the eigenvalues of $\Sigma_q$ and the positive eigenvalues of $D(T)$ are ordered decreasingly, then

For the proof, insert $\lambda=\mu T^{2q+1}$ into the characteristic determinant and eliminate the earlier grades. Their resolvent converges, after its own block scaling, to $G_{<q,<q}^{-1}$; the grade-$q$ factor tends to $\det(\Sigma_q-\mu I)$, while every later grade is of smaller order.

Induction over $q$ gives the displayed limits. The identity $\det G=\prod_q\det\Sigma_q$ recovers the pseudo-determinant constant above.

For common-volatility CIR loadings with distinct mean-reversion rates, $s_d=d$ whenever $K_{ii}>0$ for every name. Indeed, the coefficient of order $q$ is a polynomial of degree $q-1$ in $\kappa_i$ with nonzero leading coefficient. A vector annihilating the first $d$ blocks would, feature column by feature column, annihilate a Vandermonde system; hence $x_iF_{i\ell}=0$ for every $i,\ell$, and the nonzero rows implied by $K_{ii}>0$ force $x=0$.

Therefore full fixed rank gives $d$ copies of $T^3$, rank one gives $T^3,T^5,\ldots,T^{2d+1}$, and intermediate fixed ranks give repeated graded powers. This is the block/multi-input analogue of the scalar small-time controllability-Gramian hierarchy of Trélat and Zuazua (2026), shifted by the CIR zero at the origin.

A five-name 100-digit certificate with fixed rank two has cumulative ranks $(2,4,5)$ and therefore predicts powers $(3,3,5,5,7)$, determinant power $23$, and condition-number slope $-4$. Over the final dyadic step down to $T=1/512$, the measured eigenvalue slopes are $(2.986771,2.996952,4.987058,4.994626,6.988704)$, the determinant slope is $22.954111$, and the condition-number slope is $-4.001934$.

The explicit block Gram above gives $\det G=9.72067545896\times10^{-11}$; at $T=1/512$ the ratio $\det D(T)/(T^{23}\det G)$ is $0.9862517970$. The five predicted eigenvalue constants are

and the largest relative discrepancy of the five scaled eigenvalues at $T=1/512$ is $0.4003\%$. This theorem keeps the two notions of rank separate: for CIR, where $v_1$ is the all-ones vector, $s_1=\operatorname{rank}K$ is the fixed-parameter Green–Kubo rank, while later blocks describe new directions created only after integrating heterogeneous maturity loadings.

Long maturity restores effective rank one

Full algebraic rank also does not mean uniform information across maturities. Let the distinct CIR loadings have positive limits

for arbitrary fixed $\kappa_j>0$, $\sigma_j\ge0$, and pairwise distinct full pairs $(\kappa_j,\sigma_j)$. Put $r(t)=B(t)-b$. The Riccati formula gives $r(t)=O(e^{-\rho t})$ for some $\rho>0$, hence

Let $P_b$ be orthogonal projection onto $b^\perp$ and define, on that $(d-1)$-dimensional space,

The cross terms in $C$ vanish after compression because $P_bb=0$. Moreover $H$ is positive definite: if $z\perp b$ and $z^\top Hz=0$, then $z^\top r(t)=0$ for every $t$, hence $z^\top B(t)=0$; the unrestricted CIR loading independence proved above forces $z=0$.

In an orthonormal basis beginning with $b/\|b\|$, the first block of $J(T)$ is $T\|b\|^2+O(1)$, the lower block tends to $H$, and the off-diagonal block is bounded. A Schur complement gives the displayed $O(T^{-1})$ bounded-eigenvalue corrections. The same theorem holds for the rank-one Green–Kubo correction $D(T)=\operatorname{diag}(a)J(T)\operatorname{diag}(a)$ when every $a_j\ne0$, after replacing $B,b,r$ by their coordinatewise products with $a$.

The conclusion is deliberately two-sided: heterogeneous maturity loadings create exact full rank for every finite $T>0$, but at the long end only their common stationary levels accumulate linearly. The remaining $d-1$ directions contain a finite amount of transient information, so relative conditioning deteriorates like $T$.

This is a deterministic loading-Gram theorem at fixed CIR parameters. It does not make the earlier $O_L(m^{-2})$ pricing remainder uniform as $T\to\infty$ and does not interchange the switching-speed and maturity limits.

For four positive-volatility loadings with $\kappa=(0.35,0.8,1.7,3.2)$ and $\sigma=(0.12,0.31,0.18,0.44)$, the limiting bounded eigenvalues are

At $T=2048$ their measured values are $(2.66554908\times10^{-4},1.28466530\times10^{-2}, 5.82430527\times10^{-1})$. The four convergence orders, including the scaled leading eigenvalue, are $(1.005735,1.004698,1.002508,0.999960)$. The measured and predicted limits are respectively $34182.90668$ and $34225.20992$ for $\operatorname{cond}_2J(T)/T$, and $1.81725753\times10^{-5}$ and $1.82531713\times10^{-5}$ for $\det J(T)/T$.

The inverse information has a rank-one long-end correction

The spectrum identifies the scales but not the limiting covariance or the stationary–transient coupling. Write $\beta=\|b\|$, $e=b/\beta$, $P=I-ee^\top$, and let $H^+$ denote the Moore–Penrose inverse of $H$ on $\mathbb R^d$ (the inverse on $b^\perp$ and zero on $\operatorname{span}\{b\}$). With $C$ as above, define

Indeed, in an orthonormal basis $(e,U)$ with columns of $U$ spanning $b^\perp$, exponential convergence of $B(t)$ gives

The lower block is positive definite. Its block inverse is $H^+$ plus the outer product of $e-H^+g$ divided by the scalar Schur complement $\beta^2T+c_{00}-g^\top H^+g+O(e^{-\rho T})$. Expanding that reciprocal proves the operator-norm remainder.

Thus long maturity supplies unbounded precision in only one asymptotic direction. Every orthogonal exposure retains the finite transient-information floor $x^\top H^+x$; the vector $w$, rather than $e$ alone, records the finite coupling to the stationary direction.

For the four-name parameters above, the certificate evaluates the exact finite-$T$ loading Gramian by quadrature. At $T=2048$, the operator-norm error after the rank-one inverse correction is $7.98785874\times10^{-3}$, with observed order $2.006271$ between the last two doubled maturities. It also gives $T b^\top J(T)^{-1}b=1.00427059$ versus limit $1$, and transient inverse trace $3728.05152$ versus $\operatorname{tr}(H^+)=3726.17300$.

This is again a fixed-parameter deterministic loading-Gram statement. It neither changes the fixed-parameter Green–Kubo rank nor asserts a maturity-uniform pricing remainder, and it does not interchange the long-maturity and fast-switching limits.

Fixed Green–Kubo rank determines the number of long-end directions

The rank-one conclusion has a sharp finite-rank extension. Let $K$ be any fixed positive semidefinite Green–Kubo matrix of rank $r$, put $\Delta_t=\operatorname{diag}B(t)$ and $\Delta_\infty=\operatorname{diag}b$, and consider the integrated correction

This Gram representation is the continuous loading version of the Schur product theorem.

Because every $b_j>0$, the stationary matrix

also has rank $r$. Exponential loading convergence gives

Let $Q$ project onto $\operatorname{ran}A$, put $P=I-Q$, and define on $\ker A$

The unrestricted full-pair cluster law above makes the premise exact. If $C_c$ are the clusters of identical pairs $(\kappa_i,\sigma_i)$, then $D(T_0)\succ0$ for one, equivalently every, $T_0>0$ exactly when every $K_{C_c,C_c}$ is positive definite. For distinct full pairs this reduces to $K_{ii}>0$ for every name.

The finite-maturity full-rank premise supplies the missing positivity, rather than merely assuming it. For $z\in\ker A$,

If this vanishes, analyticity of the CIR loadings implies $\Delta_tz\in\ker K$ for every $t>0$, hence $D(T_0)z=0$ and therefore $z=0$. Thus $H_K$ is positive definite. The displayed spectrum and determinant then follow from the $(\operatorname{ran}A,\ker A)$ block Schur complement.

The same block calculation gives every algebraic correction to the identification volume. Write restrictions to $\operatorname{ran}A$ as $A_R=QAQ$ and define on that subspace

Let $e_j(M_K)$ be the $j$th elementary symmetric polynomial of the eigenvalues of $M_K$, with $e_0=1$. In particular,

Indeed, in the orthogonal decomposition $\operatorname{ran}A\oplus\ker A$, the lower block tends exponentially to $H_K$, while its Schur complement in the upper block is $TA_R+G_K+O(e^{-\rho T})$. Factoring $TA_R$ and using $\det(I+M_K/T)=\sum_{j=0}^re_j(M_K)T^{-j}$ proves the determinant formula.

Expanding $\operatorname{tr}\log(I+M_K/T)$ gives the displayed log-volume terms and, similarly, any fixed algebraic order. The matrix $M_K$ includes the stationary–transient coupling through $H_K$; replacing $G_K$ by $QC_KQ$ is generally wrong.

There is also a matrix-level inverse limit. Let $A^+$ and $H_K^+$ be the Moore–Penrose inverses on their complementary supports and set

This follows by inverting the same two blocks: the Schur complement on $\operatorname{ran}A$ is $TA+O(1)$, while the lower block tends to $H_K$. The correction has rank $r$.

Hence maturity integration may make $D(T)$ exactly full rank at every finite maturity, but precisely $\operatorname{rank}K$ information directions accumulate without bound at the long end. The preceding theorem is the $r=1$ specialization.

The same calculation supplies the complete matrix-valued inverse expansion, not only its first correction. Let $A^{+/2}$ denote the positive square root of $A^+$ on $\operatorname{ran}A$. For every fixed integer $p\ge0$,

Indeed, the constant proxy $TA+C_K$ has the exact block inverse

Expanding $(TI+M_K)^{-1}$ proves the formula. More quantitatively, if $T>\|M_K\|_2$, the proxy remainder after the displayed $p$th term is at most

The exponentially small difference between $D(T)$ and $TA+C_K$ adds $O(e^{-\rho T})$ by the resolvent identity, which is absorbed by every fixed algebraic remainder above. Each inverse correction has rank at most $r$. This is a fixed-parameter long-maturity expansion; it is not uniform when $K$, the CIR parameters, or the transient spectral floor vary with $T$.

The inverse expansion also gives an exact long-maturity dichotomy for equality-constrained minimum variance. For a fixed nonzero normalization vector $c$, let

This is the classical minimum-variance solution under $c^\top w=1$; the usual budget constraint takes $c=\mathbf1$. There are two cases.

Thus a normalization that sees the transient kernel has a finite long-end risk floor and its optimizer converges into that kernel. If the normalization is blind to the kernel, only the $r$ stationary Green–Kubo directions can satisfy it and the minimum variance grows linearly. Finite-maturity full rank does not remove this distinction.

The result is unconstrained apart from $c^\top w=1$; it does not claim the same closed form under long-only, leverage, or turnover constraints. The optimization criterion is the one introduced by Markowitz (1952); the two-scale limit here comes from the CIR loading expansion.

A five-name certificate takes the rank-two matrix $K=FF^\top$ with $F=(\mathbf1,x)$ and $x=(-2,-0.7,0.2,1.1,2.3)^\top$, and heterogeneous positive-volatility CIR loadings with one repeated full pair: $\kappa=(0.35,0.35,1.2,2.4,5)$ and $\sigma=(0.12,0.12,0.30,0.08,0.50)$. The repeated-pair $2\times2$ block of $K$ is positive definite, so the exact cluster criterion gives full integrated rank five. The limiting transient eigenvalues are

while the two stationary eigenvalues are $(3.11016825,45.7412552)$. At $T=2048$, all five scaled convergence orders lie between $0.999888$ and $1.004244$; $\operatorname{cond}_2D(T)/T$ is $25317.3875$ versus limit $25356.4705$, and $\det D(T)/T^2$ is $1.82305073\times10^{-2}$ versus limit $1.83449734\times10^{-2}$. Here $\vartheta_K=-12.7986046$ and $e_2(M_K)=40.5617550$.

At $T=64$, the scaled quantity $T^2\{\det D/(T^2\det_+(A)\det H_K)-1- \vartheta_K/T\}$ is $40.5617551$. After subtracting only $\vartheta_K/T$, the normalized determinant remainder at $T=2048$ is $9.67074\times10^{-6}$ with observed dyadic order $1.999992$. The inverse remainders at $T=64$ after one, two, and three terms are $8.22310\times10^{-1}$, $8.74039\times10^{-2}$, and $9.33677\times10^{-3}$, with observed dyadic orders $2.004872$, $3.040135$, and $4.084200$, respectively.

The one-term remainder at $T=2048$ is $7.20040338\times10^{-4}$. For the constant proxy at $T=64$, the three-term error is $9.33677673\times10^{-3}$, below the explicit Neumann bound $1.03296926\times10^{-2}$. For the ordinary budget vector, the predicted risk floor is $0.097070544508$ and the value at $T=2048$ is $0.097042372590$.

For a normalization vector in $\operatorname{ran}A$, the predicted limit of $v_T/T$ is $3.0766963489$, versus $3.0675084592$ at $T=2048$. The four weight and risk errors converge with observed dyadic orders between $1.000036$ and $1.004731$.

The result concerns a fixed $K$ and fixed CIR parameters. It explicitly distinguishes the $r$ stationary directions inherited from fixed Green–Kubo rank from the additional finite information created by heterogeneous maturity loadings; it asserts no switching-rate-uniform price error and no interchange of maturity and switching limits.

Full rank becomes singular at a quantified rate

Algebraic full rank does not make nearby loading parameters stably identifiable. This can be made sharp.

Fix $T>0$, a common $\sigma\ge0$, distinct real numbers $x_1,\ldots,x_d$, and put $\kappa_j=\kappa_0+\epsilon x_j$. Let $B_\kappa$ solve

and define $J_\epsilon=(\langle B_{\kappa_i},B_{\kappa_j}\rangle_{L^2[0,T]})$. If its eigenvalues are ordered $\lambda_1\ge\cdots\ge\lambda_d$, then, for fixed $d$ as $\epsilon\downarrow0$,

The leading determinant is explicit. With indices $r,s=0,\ldots,d-1$, set

Then

To prove it, Taylor-expand the $d$ vectors $B_{\kappa_0+\epsilon x_j}$ in $L^2[0,T]$ and take their exterior product. Its first nonzero term is the Vandermonde determinant times $B_{\kappa_0}\wedge\partial_\kappa B_{\kappa_0}\wedge\cdots\wedge \partial_\kappa^{d-1}B_{\kappa_0}$, with power $\epsilon^{d(d-1)/2}$; squaring its norm gives the formula. The derivative vectors are independent because the Riccati equation gives

Divided-difference column operations then give singular values of the loading synthesis map of orders $1,\epsilon,\ldots,\epsilon^{d-1}$, which proves the individual eigenvalue powers. This is the confluent counterpart of the classical Vandermonde conditioning studied by Gautschi (1978). If $K=aa^\top$ with fixed nonzero coordinates, the congruence $D_\epsilon=\operatorname{diag}(a)J_\epsilon\operatorname{diag}(a)$ has the same powers.

Thus two regimes still give full integrated rank for every $\epsilon>0$, while the weakest identified direction vanishes as $\epsilon^{2(d-1)}$. This is a fixed-$d$, fixed-$T$ statement; it is not uniform in the number of names or maturity.

The constants multiplying those powers are explicit as well. Let $W:L^2[0,T]\leftarrow\mathbb R^d$ have columns $\partial_\kappa^rB_{\kappa_0}$, $0\le r<d$, and define the scaled Vandermonde matrix

Take the Hilbert-space QR factorization $W=U_WR_W$ and the LQ factorization $V=L_VU_V$. Equivalently, $R_W$ is the upper Cholesky factor of $G^{(d)}$ and $L_V$ is the lower Cholesky factor of $VV^\top$. Then

Indeed, Taylor expansion gives the loading synthesis map as $WD_\epsilon\{V+O(\epsilon)\}$, with $D_\epsilon=\operatorname{diag}(1,\epsilon,\ldots,\epsilon^{d-1})$. After the two orthogonal changes of basis, the unique lowest-power minor in each exterior power uses the first rows and columns.

Consecutive exterior-product limits give the individual singular-value constants; squaring gives the displayed Gram-eigenvalue constants. Their product is exactly the determinant coefficient above.

The certificate uses positive volatilities $\sigma_j=0.02$, distinct $\kappa_j$ between $0.2$ and $5$, and strict Feller mean levels for $2\le d\le8$. Every integrated matrix is full rank. At $d=8$ its smallest eigenvalue is $1.4691\times10^{-8}$ and its condition number is $6.7541\times10^9$.

This is an important distinction: algebraic rank can be maximally amplified while statistical identification becomes extremely ill-conditioned. A separate 70-digit coalescence check at $d=4$ finds eigenvalue log--log slopes $(0.000316,2.000484,4.000156,5.999654)$. The explicit leading eigenvalue constants are $(3.250068477,0.02564049977,0.0007991812818,0.00001379064292)$; at $\epsilon=1/256$ their largest relative error is $1.67\times10^{-5}$.

The determinant ratio is $1.000021036$ against the displayed constant, and the condition-number slope is $-5.999901$.

Partial collisions have a cluster spectrum

The one-center hierarchy has a precise multi-center extension. Let $\alpha_1,\ldots,\alpha_g$ be distinct, let cluster $c$ contain $m_c$ distinct fixed nodes $x_{c,1},\ldots,x_{c,m_c}$, and put $\kappa_{c,j}=\alpha_c+\epsilon x_{c,j}$. Assume the finite jet family

is linearly independent in $L^2[0,T]$. If $r_1\le\cdots\le r_d$ is the sorted multiset $\biguplus_{c=1}^g\{0,1,\ldots,m_c-1\}$, then the descending eigenvalues of the loading Gram matrix satisfy, as $\epsilon\downarrow0$ with $g$, the cluster data, and $T$ fixed,

Thus exactly $g$ eigenvalues stay of constant order. If $m_{\max}=\max_c m_c$, then

The determinant coefficient is explicit. Order the jet vectors by $(c,r)$ and let $G_{\rm jet}$ be their Gram matrix. Then

For the proof, Taylor-expand separately at each $\alpha_c$ and right-multiply each cluster by its fixed scaled-Vandermonde inverse. The transformed columns are $\epsilon^r\{\partial_\kappa^rB_{\alpha_c}+O(\epsilon)\}$. Jet independence bounds the resulting synthesis map above and below by the diagonal map with entries $\epsilon^r$, proving all singular exponents; squaring gives the Gram exponents.

Taking the exterior product gives the coefficient above. This is the finite-dimensional derivative-matching structure introduced by Hermite (1878) and is consistent with the confluent-Vandermonde bounds of Gautschi (1978).

For an arbitrary analytic loading family, the theorem remains conditional on the displayed joint jet independence. For the CIR Riccati family, however, that condition is automatic. Write

Substitution in the Riccati equation gives $c_1=1$ and

Induction shows that $c_n$ is a polynomial of degree $n-1$ with leading coefficient $(-1)^{n-1}/n!$. For $d=\sum_cm_c$, form the Taylor-jet matrix

If the distinct centers are ordered increasingly, triangular change of polynomial basis and the confluent Vandermonde determinant give

Any $L^2[0,T]$ relation among the CIR jets is an analytic identity in $t$; its first $d$ Taylor coefficients are annihilated by $M$. The displayed determinant therefore forces every coefficient in the relation to vanish.

Thus, for fixed common $\sigma\ge0$ and every $T>0$, distinct CIR cluster centers alone imply joint jet independence and make $G_{\rm jet}$ positive definite. This is precisely the derivative interpolation structure in Hermite's original 1878 paper; no numerical genericity assumption is needed for CIR.

The exact rational certificate applies the recurrence to cluster sizes $(3,2,1)$ and centers $(4/5,2,4)$. The resulting six-by-six Taylor-jet determinant is $-1536/48828125=-3.145728\times10^{-5}$, exactly equal to the formula above and independent of the lower-degree volatility terms. The positive-CIR quadrature certificate then checks the asymptotic constants at 80-digit precision for the same cluster sizes, centers $(0.8,2,4)$, common volatility $0.02$, and $T=4$.

It finds eigenvalue slopes $(0.000007,0.000058,0.000344,1.999615,2.000232,3.999713)$, determinant slope $7.999970$, condition-number slope $-3.999706$, and determinant ratio $0.999999999999953$ against the explicit coefficient. Consequently, one rank-one fixed-feature Green--Kubo matrix with nonzero loadings still yields full integrated rank for every sufficiently small $\epsilon>0$; cluster multiplicities, rather than the number of regimes, determine the asymptotic conditioning.

The explicit two-name benchmark

Take a symmetric two-state generator $Q_0=\left(\begin{smallmatrix}-1&1\\1&-1\end{smallmatrix}\right)$, $\pi=(1/2,1/2)$, and two CIR names with

The smallest Feller margin $\min_{i,j}(2\kappa_j\theta_{j,i}-\sigma_j^2)$ is $0.03>0$. With $d_j=(\theta_{j,1}-\theta_{j,2})/2$, the one-dimensional centered space gives $K_{jk}=\kappa_jd_j\kappa_kd_k/(2m)$. Thus $K$ has rank one.

Yet $B_1$ and $B_2$ are linearly independent on every positive maturity interval: both have $B_j'(0)=1$, while $B_j''(0)=-\kappa_j$ differ. Their Gram matrix $J(T)$ is positive definite, and multiplication by the nonzero diagonal $(\kappa_1d_1,\kappa_2d_2)$ proves $D(T)$ has rank two.

For $T=5$ and $m=1$, the eigenvalues of $K$ are $(0,\,8.125\times10^{-4})$, whereas those of $D(5)$ are approximately $(4.143\times10^{-6},\,1.343\times10^{-3})$. The certificate checks the Feller condition and eigenvalues, then solves the exact joint-survival ODE. The residual of the pairwise formula multiplied by $m^2$ stabilizes at approximately $-8.6\times10^{-5}$ as $m$ increases from 10 to 80.

What the curves identify

Suppose $x_j,\kappa_j,\sigma_j$ are known, every name shares the same stationary-start chain, and ideal first-order marginal and pairwise survival curves are available over a maturity interval. The marginal curve for $j$, after subtracting $-B_jx_j$, has the coefficients $-\kappa_j\bar\theta_j$ on $\int B_j$ and $K_{jj}$ on $\int B_j^2$. Those two functions are independent because $B_j$ is nonconstant.

The pairwise log ratio then determines $K_{jk}$ through $2K_{jk}J_{jk}(T)$, with $J_{jk}(T)>0$ for $T>0$. Thus these curves identify $\bar\theta_j$ and the whole rate-scaled matrix $K$ at first order, although they do not identify $Q_0$, $\theta_{j,i}$, or $m$ separately.

After removing the known loadings, the correct regime-count restriction is $\operatorname{rank}K\le n-1$, or equivalently $\operatorname{rank}[\operatorname{diag}(B(t))^{-1} D'(t)\operatorname{diag}(B(t))^{-1}]\le n-1$ for $t>0$. The corresponding bound on $\operatorname{rank}D(T)$ is false. Estimating derivatives from noisy curves may be unstable; the entrywise division $K_{jk}=D_{jk}(T)/J_{jk}(T)$ uses integrated loadings instead.

Scope: fixed CIR $\kappa_j,\sigma_j$, switching mean levels, independent Brownian motions and Cox triggers, arbitrary initial regime prior independent of the specified $x_j$, and first-order fast-switching accuracy. Switching $\kappa$ or $\sigma$ changes the Riccati reduction and requires its own identification argument.