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Integrated-variance cumulants

Exact fixed-order integrated-variance and independent-return cumulant rates and boundary constants, plus an explicit European option correction for a positive switched CIR variance.

The smile produced by uncorrelated stochastic volatility is often described through the variance of integrated variance. There are two points to keep separate. The Brownian return has a fourth cumulant exactly equal to three times that variance.

The risk-neutral log return does not: its drift contributes the third and fourth cumulants of integrated variance as well. With leverage, four joint cumulants replace the conditional-normal shortcut.

The exact normal-mixture identities

Let $V_T$ be any non-negative random variable and let $Z$ be a standard normal independent of it. Define

In a stochastic-volatility model, $V_T=\int_0^T v_s\,ds$, $M_T$ is the Brownian return, and $X_T$ is the risk-neutral log return when the price Brownian motion is independent of the volatility process. Conditional normality gives the cumulant-generating functions

Comparing coefficients proves the exact identities

The last two formulas are not interchangeable. The common expression $3\operatorname{Var}(V_T)$ is exact for the martingale return $M_T$ but only the leading part of the log-return cumulant when the higher cumulants of $V_T$ are negligible.

A positive switched CIR variance

Now let a stationary finite-state chain $Y_t$ switch the long-run level of a CIR variance:

The Brownian motion $B$ is independent of $Y$ and of the price Brownian motion. We keep $\kappa$ and $\xi$ fixed and assume $2\kappa\min_i\theta_i\ge\xi^2$.

Thus, from every $v_0>0$, the model stays strictly positive; equality is allowed. Write

Integrating the CIR equation and then integrating over time gives

The regime integral and the martingale are uncorrelated. Since

Itô isometry gives the exact decomposition

The first line is the integrated-variance variance of the averaged CIR model. The second is exactly the extra variance created by the hidden regimes. This separation does not require an asymptotic expansion.

The Green–Kubo correction and a uniform bound

Put $I_2(T)=\int_0^T B_\kappa(s)^2\,ds$. Replacing the narrow correlation kernel by its mass $K_m$ gives

The remainder can be bounded without fixing the maturity. Monotonicity and $0\le B_\kappa\le1/\kappa$ imply

Splitting the covariance integral at $T$ then gives

If the generator is $mQ_0$ and $|C_m(u)|\le A e^{-m\rho u}$, the right side is at most $4A/(m^2\rho^2)$. Consequently

This is a finite-rate statement, uniform over all maturities. It is stronger than merely observing a first-order slope at one fixed expiry.

When drift and vol-of-vol switch together

The correct drift coordinates are not $\kappa$ and $\theta$ separately. Write $c_i=\kappa_i\theta_i$.

If vol-of-vol also switches, put $a_i=\xi_i^2$. The regime-dependent part of the CIR variance generator is

Thus the averaged drift is $\bar c-\bar\kappa v$, whose effective mean level is $\bar c/\bar\kappa$, generally not $\bar\theta$.

The Feller test is statewise, including vol-of-vol

Suppose $Y$ is independent of $B$, every $c_i,\kappa_i,\xi_i>0$, $v_0>0$, and

Irreducibility is not needed for the exact initial-law statement. For an initial regime law $\nu$, let $\mathcal R(\nu)$ be the states reachable by a directed path of positive transition rates from $\operatorname{supp}\nu$. Then, for every $v_0>0$,

Sufficiency follows because the chain never leaves $\mathcal R(\nu)$, so the stopped-log argument below applies along the whole realized path. For necessity, choose a positive-rate path from $\operatorname{supp}\nu$ to a deficient state $j$. The event that the chain follows that path with holding times in prescribed positive intervals has positive probability.

Either zero is hit before arrival, or continuity gives $v>0$ on arrival. Conditional on the latter event, the frozen deficient CIR has positive probability of hitting zero before the next jump. If $j$ is absorbing, ordinary CIR boundary accessibility gives the same conclusion directly.

Thus an unreachable deficient regime is harmless for that initial law, but adding a single positive-rate path to it changes the answer.

The result allows both mean reversion and vol-of-vol to switch. Put $K_t=\int_0^t\kappa_{Y_u}du$, $X_t=e^{K_t}v_t$, and introduce the strictly increasing clock

Itô's formula followed by the quadratic-variation time change gives exactly

This is a squared-Bessel equation with an adapted, piecewise-constant dimension. Before a possible hit of zero,

If every $\delta_s\ge2$, the finite-variation term is nonnegative; McKean's stopped-log argument rules out $\log Z_s\to-\infty$ in finite time. Equality is included. Unlike the common-$\xi$ proof, this argument does not apply a comparison theorem across unequal diffusion coefficients.

For necessity, if some regime $i$ has $2c_i<\xi_i^2$, let $q_i=-Q_{ii}$ and let $\tau_0$ be the zero-hitting time of the CIR process frozen in that regime. Conditional on starting in $i$, the probability of reaching zero before the first switch is

where $U$ is Tricomi's confluent hypergeometric function. The formula follows by solving $\tfrac12\xi_i^2v u''+(c_i-\kappa_i v)u'=q_i u$ with $u(0)=1$ and $u(\infty)=0$.

Accessibility of $i$ then gives positive boundary-hitting probability from every communicating regime. Consequently the Feller test cannot be applied only to the averaged model.

A two-state counterexample makes the distinction concrete. Take switching rate $q=1.5$, $(c_1,c_2)=(0.04,0.16)$, $(\kappa_1,\kappa_2)=(0.8,2)$, and $(\xi_1,\xi_2)=(0.4,0.3)$. For the symmetric fast chain the correctly averaged squared vol-of-vol is $\overline{\xi^2}=0.125$, so the averaged model passes with $2\bar c/\overline{\xi^2}=1.6$.

Regime 1 nevertheless has $2c_1/\xi_1^2=0.5$. Starting from $v_0=0.04$ there, the exact probability of hitting zero before the next switch is $0.2341927683$. The boundary certificate independently evaluates the integral representation of $U$ and agrees below $9\times10^{-16}$.

It also checks a reducible four-state example: from initial state $0$, a deficient state in a disjoint closed class is irrelevant; after one edge of rate $0.5$ is added, it becomes reachable and its conditional boundary probability before exit is again $0.2341927683$.

Now let $K^{(0)}$ be the Green–Kubo matrix of the centered feature triple $(c,\kappa,a)$ for the unscaled generator $Q_0$. For a stationary-start chain with generator $mQ_0$, a regime-free polynomial payoff $f$, and fixed $T$, the semigroup expansion on every invariant finite-degree polynomial space is

where the remainder is in the finite-dimensional matrix norm and

Indeed, centering the three coefficient vectors and solving the fast Poisson equations gives $\mathcal D=\sum_{r,s\in\{c,\kappa,a\}}K^{(0)}_{rs}A_rA_s$; multiplying the differential operators yields the displayed formula. Standard finite-dimensional semigroup perturbation then gives the Duhamel term and its $O_T(m^{-2})$ remainder. In particular,

Thus reversing a nonreversible regime cycle transposes $K^{(0)}$ and flips the entire antisymmetric contribution

while preserving every symmetric term. A reversible two-state chain has no directional term.

The same theorem supplies a cumulant correction without a common Riccati coefficient. Augment the state by $z_t=\int_0^t v_sds$; the averaged generator gains $v\partial_z$, while $\mathcal D$ is unchanged because the integrated coordinate itself is not switched. Define

Then $\mathbb E[V_T^r]=a_r+b_r/m+O_T(m^{-2})$ for $r\le4$. Let $\bar\kappa_r$ be the cumulant formed from $(a_1,\ldots,a_r)$, and let $\dot\kappa_r$ be the directional derivative of that moment–cumulant polynomial in direction $(b_1,\ldots,b_r)$. The needed coefficients are

For price Brownian noise independent of the variance driver, this gives the explicit fixed-maturity corrections

Why the simultaneous-switching bound is not maturity-uniform

The dependence of the remainder on $T$ is necessary, already for the first integrated-variance moment. This is consistent with the fixed-time perturbed-semigroup framework of Kurtz (1973); the calculation below identifies the precise obstruction to strengthening the present expansion. Consider the symmetric two-state generator $Q_0=\left(\begin{smallmatrix}-1&1\\1&-1\end{smallmatrix}\right)$, stationary law $\pi=(1/2,1/2)$, and coefficient vectors $c=(0.04,0.16)$ and $\kappa=(0.8,2)$.

One may take $\xi=(0.2,0.2)$, so the statewise Feller inequalities are strict. Vol-of-vol does not enter the following mean calculation.

If $r_m$ is the stationary row vector with components $\mathbb E[v_t\mathbf1_{\{Y_t=i\}}]$, then

Consequently

With $Y_0\sim\pi$ and any fixed finite initial variance mean, the stable two-state first-moment system gives $\mathbb E[V_T]=\mu_mT+O_m(1)$ as $T\to\infty$. Thus even after the averaged and first-order long-run slopes have been matched, the residual grows linearly in maturity for every fixed $m$. An absolute $O(m^{-2})$ remainder uniform over all $T\ge0$ is therefore impossible without retaining second-order long-run information.

The obstruction also survives a joint limit. Starting from $v_0=\mu_0=1/14$ and taking $T=m^2$, the exact matrix-exponential solution satisfies

For $m=8,16,32,64,128$, the certificate obtains respectively $0.0022948980$, $0.0021929979$, $0.0021449064$, $0.0021217062$, and $0.0021103356$. This counterexample concerns simultaneous switching. It does not weaken the separate fixed-$\kappa$ covariance result above, whose special structure yields a maturity-uniform bound.

Uniformity recovered by retaining the exact slow mode

The secular obstruction can be removed at the first-moment level without returning to the full switched diffusion. Let $Q_0$ be any irreducible finite-state generator with stationary row vector $\pi$, put $K=\operatorname{diag}(\kappa)$ with $\min_i\kappa_i>0$, and assume $Y_0\sim\pi$. If

then $r_m'=r_mA_m+b$. Its stationary row vector and long-run mean are

Let $\lambda_m<0$ be the simple principal eigenvalue of the irreducible Metzler matrix $A_m$, and let $P_m$ be its Riesz projection. For $d_m=r_m(0)-r_{m,\infty}$ define the slope-and-mode composite

The exact remainder is

Analytic spectral perturbation gives $P_m=\mathbf1\pi+O(m^{-1})$, a complementary inverse of order $m^{-1}$, and a complementary semigroup bounded by $Ce^{-cmT}$. In particular $(I-P_m)\mathbf1=O(m^{-1})$, and therefore

If the initial variance is independent of the stationary regime, so $r_m(0)=v_0\pi$, the left complementary amplitude is also $O(m^{-1})$ and the rate improves to

This is the affine first-moment analogue of retaining an exact Perron exponent at the long end. The spectral step follows the isolated-eigenprojection perturbation framework of Kato (1949). It does not assert a uniform approximation for higher integrated-variance cumulants, leveraged returns, option prices, or the full transition semigroup.

For the preceding strictly positive two-state model, the certificate maximizes the exact error over $0\le T\le m^2$. At $m=8,16,32,64,128$, independent initial variance gives errors $3.834\times10^{-6}$, $5.130\times10^{-7}$, $6.635\times10^{-8}$, $8.437\times10^{-9}$, and $1.064\times10^{-9}$, with measured rate $2.987$. Allowing the conditional initial means to differ by regime gives $6.056\times10^{-5}$, $1.631\times10^{-5}$, $4.233\times10^{-6}$, $1.078\times10^{-6}$, and $2.721\times10^{-7}$, with measured rate $1.986$.

An explicit Green–Kubo slow mode

The eigenvectors in the preceding composite are not necessary at the target $O(m^{-2})$ accuracy. Let $Q_0^{\#}$ be the group inverse, normalized by $Q_0Q_0^{\#}=Q_0^{\#}Q_0=I-\mathbf1\pi$, and define

Ordinary isolated-eigenvalue perturbation gives the two explicit expansions

The second coefficient is the Poisson corrector for the switched mean-reversion rate; the first is its Green–Kubo quadratic form. Retain the exact stationary slope $\mu_m$, but replace the spectral data by

Because $(e^{\lambda T}-1)/\lambda$ and its first derivative in $\lambda$ are bounded uniformly over $T\ge0$ when $\lambda$ stays to the left of a fixed negative constant, replacing $(\lambda_m,P_m\mathbf1)$ by these first-order expressions costs only $O(m^{-2})$. Combining this with the fast-complement estimate proves

The exact slope is essential: any nonzero error in its coefficient is multiplied by $T$ and therefore cannot be bounded uniformly over all maturities. The explicit composite needs one linear solve for $\mu_m$ and one group inverse of $Q_0$, but no $m$-dependent eigenvectors.

For the two-state example, the uniform errors at $m=8,16,32,64,128$ are $1.779\times10^{-5}$, $4.888\times10^{-6}$, $1.282\times10^{-6}$, $3.283\times10^{-7}$, and $8.307\times10^{-8}$ for an independent initial variance, with rate $1.983$. For regime-dependent conditional initial means they are $8.448\times10^{-5}$, $2.200\times10^{-5}$, $5.617\times10^{-6}$, $1.419\times10^{-6}$, and $3.568\times10^{-7}$, with rate $1.992$. A nonreversible three-state check gives rates $1.954$ and $1.987$, respectively.

Removing the exact stationary solve on growing maturity windows

The exact slope is necessary on the whole half-line, but it can be replaced to any prescribed algebraic order on a correspondingly growing maturity window. Write $D_\kappa=\operatorname{diag}(\kappa)$, $D_c=\operatorname{diag}(c)$, $S=Q_0^{\#}$, and seek

All coefficients follow from one group inverse and a Poisson recursion. First set

For every $j\ge1$, continue with

The normalization makes $z_j\mathbf1=0$ and the choice of $\mu_j$ enforces the next solvability condition $\rho_j\kappa=0$. Substitution into $r_{m,\infty}(mQ_0-D_\kappa)=-\pi D_c$ proves the expansion recursively; finite-dimensional resolvent analyticity, in the sense of Kato (1949), supplies its remainder. In particular,

Replace $r_{m,\infty}$ and $\mu_m$ in the explicit Green–Kubo slow-mode composite by their order-$p$ truncations. For each fixed $p\ge1$ and any deterministic horizon $T_m$, the same fast-mode estimate gives

Consequently $T_m=O(m^{p-1})$ preserves the $O(m^{-2})$ accuracy without any $m$-dependent linear solve: one slope correction covers bounded maturities, two cover maturities of order $m$, and three cover maturities of order $m^2$. This is a joint switching-speed/maturity theorem, not an all-maturity approximation. Uniformity over every $T\ge0$ still requires the exact slope.

For the positive two-state benchmark, the recursion gives $(\mu_0,\mu_1,\mu_2,\mu_3)=(0.0714285714,-0.0036734694, 0.0020991254,-0.0011995002)$. The measured slope-remainder orders for $p=1,2,3$ are $1.994$, $2.994$, and $3.994$; the full composite orders on $T\le m^{p-1}$ are $1.990$, $1.986$, and $1.992$. A nonreversible three-state check gives slope orders $1.995$, $2.995$, and $3.997$.

Every invariant variance moment, exactly

The invariant moments used by the cumulant hierarchy themselves have an all-order closed form. For an irreducible $Q_0$, $m>0$, and $\kappa_*:=\min_i\kappa_i>0$, put $a_i=\xi_i^2$ and, under the invariant law $\Pi_m$, define the row vectors

Applying the joint generator to $v^p\mathbf1_{\{Y=i\}}$ gives the exact triangular recursion

Each $M_p$ is a nonsingular $M$-matrix. More directly, its inverse is the killed-chain resolvent

Thus every component moment is nonnegative and

This inequality also gives a moment-order-uniform conclusion that is separate from the inverse-speed expansion below. Put

A bounded-cutoff Foster–Lyapunov argument first justifies finiteness of the displayed moments. Then multiplying the exact moment equation by $\mathbf1$ gives the sharper identity

Positivity and iteration yield the rising-factorial bound. Monotone convergence then sums its exponential series to the displayed Gamma moment-generating function. In particular the invariant variance law is moment-determinate; equivalently, the bound implies Carleman's divergent series.

The conclusion is uniform in the switching speed and does not use the statewise Feller inequalities. It does not make the inverse-speed Taylor series below uniform in $p$.

The same generator inequality is a Foster–Lyapunov bound, so all positive integer invariant moments are finite. This conclusion needs mean reversion, not the statewise Feller inequality $2c_i\ge a_i$: boundary inaccessibility and moment existence are separate questions. The finite-dimensional closure is the switched-CIR instance of the polynomial-process framework of Cuchiero, Keller-Ressel and Teichmann (2012), while the displayed recursion follows here directly from the generator.

For the nonreversible three-state certificate, $(\vartheta,\beta)=(0.057142857143,3)$. Across switching speeds $m\in\{0.25,1,8,128\}$ and moments through order $80$, the largest exact-moment/majorant ratio is $0.521637381880$.

Replacing the drift levels so that every state violates the Feller inequality still gives a maximum ratio $0.468400940311$. At $\eta=14$, which is $80\%$ of the certified endpoint, the largest 80-term exact exponential partial sum is $4.813353422729$, below the rigorous full-MGF bound $125$.

An explicit uniform cumulant disk

The same majorant controls a genuinely convergent cumulant series, not just each finite list of moments. For complex $z$, let $M_m(z)=\mathbb E_{\Pi_m}e^{zv}$. Absolute convergence and the moment bound give

Consequently $M_m$ has no zero in the switching-speed-uniform disk

Indeed, $|M_m-1|\le q_r<1$ on the closed $r$-disk, so the power series for $\log(1+(M_m-1))$ gives $|K_m|\le-\log(1-q_r)=L_r$. Cauchy's estimate and summing the remaining geometric series prove the two displayed bounds. This elementary complex-analytic argument complements the classical local uniqueness and convergence theory for moment-generating functions in Curtiss (1942).

For the three-state benchmark, $R_0=3.610240795278$. Taking $r=0.8R_0$ gives $(r,q_r,L_r)=(2.888192636223,0.717917611792,1.265556093947)$.

Through cumulant order $80$ and the four switching speeds above, the largest exact cumulant/Cauchy-bound ratio is $0.204078225577$. At $z=r/2=1.444096318111$, the order-$30$ bound is $1.178640960\times10^{-9}$; an independent 80-moment bracket of the exact MGF places the largest observed log-MGF truncation error below $9.7\times10^{-45}$, with bracket width below $5.7\times10^{-85}$.

This theorem concerns the cumulant-generating series at each exact switching speed. It does not make the separate inverse-speed expansion below uniform in moment or cumulant order, and it does not interchange the cumulant-order and fast-switching limits.

Every fixed inverse-speed coefficient at every fixed moment order

The recursion in fact determines the complete formal fast-switching expansion, coefficient by coefficient. Let $S=Q_0^\#$, $\bar\kappa=\pi\kappa$, and write

Initialize $r_{0,0}=\pi$, $r_{0,n}=0$ for $n\ge1$, and $z_{p,0}=0$. For $n=0,1,\ldots$ and, within each $n$, $p=1,2,\ldots$, set

The definition of $\mu_{p,n}$ makes the source in the second line annihilate $\mathbf1$; hence multiplication by the group inverse gives its unique zero-sum solution. Substitution into the exact moment equation, followed by induction in $(n,p)$, proves that for every fixed $p$ and truncation order $N$,

Here the series is more than formal. Put $\epsilon=1/m$ and define the exact row-valued function

The apparent singularity of each factor at zero is removable:

All other possible poles of its $j$th factor are $\lambda/j$, where $\lambda$ is a nonzero eigenvalue of $D_\kappa^{-1}Q_0$. Therefore, with

The recursion above gives its Taylor coefficients, so the fixed-$(p,N)$ remainder follows directly. A fully computable Cauchy bound is also available. For $0<r<\gamma_\kappa/p$, let

Then, whenever $|\epsilon|<r$,

This is a fixed-moment convergence theorem in the sense of Kato (1949), not a moment-uniform one: the guaranteed disk shrinks like $1/p$, and neither the displayed radius nor the Cauchy constant is asserted sharp. At $n=0$ it is exactly the stationary-moment recursion of the averaged CIR model; at $n=1$ it reduces to the previously displayed first correction. If all coefficients are regime independent, every $n\ge1$ coefficient vanishes and the exact law has the stationary Gamma moments

A nonreversible three-state certificate checks moment orders one through six and inverse-speed truncations zero through three. The observed scalar remainder orders are $(0.990,1.992,2.992,3.993)$ and the regime-resolved orders are $(0.993,1.994,2.993,3.993)$, matching the predicted powers one through four. At $m=128$ the largest scalar residual after the cubic correction is $7.9\times10^{-13}$.

Independently, one block polynomial-semigroup exponential agrees with the exact recursive invariant moments within $7.3\times10^{-15}$, and the constant-coefficient Gamma reduction within $1.2\times10^{-16}$. The guaranteed Taylor radius through moment six is $0.3053256795$. At $m=8$ the ratio $\epsilon/\rho$ is $0.4093989087$, and 80-digit truncations at orders $3,6,\ldots,30$ decrease monotonically from $4.77\times10^{-8}$ to $1.55\times10^{-20}$.

When the generalized-eigenvalue disk is exact

The preceding radius is not automatically sharp: multiplication by the other moment factors can cancel a candidate resolvent pole. There is, however, a direct and checkable noncancellation criterion. Write

Suppose $\lambda_*$ is a simple nonzero generalized eigenvalue of $(Q_0,D_\kappa)$ with $|\lambda_*|=\gamma_\kappa$, put $\epsilon_*=\lambda_*/p$, and assume no earlier factor $A_j$, $j<p$, is singular at $\epsilon_*$. Choose right and left eigenvectors $x,y$ satisfying

The possible pole of $R_p$ at $\epsilon_*$ has residue

If $\mathcal R_{p,*}\ne0$, the pole is nonremovable and the row-valued Taylor series has exact radius

The scalar moment $R_p(\epsilon)\mathbf1$ has the same exact radius when $\mathcal R_{p,*}\mathbf1\ne0$. If this is also the unique pole of modulus $\rho_p$, its Taylor coefficients satisfy the sharper asymptotic identity

This follows by taking the rank-one Laurent residue of the simple matrix pencil and then expanding $(\epsilon-\epsilon_*)^{-1}$ geometrically. If the residue vanishes, or several factors resonate at the same pole, the safe disk above remains valid but the exact radius must be found after combining the principal parts; no universal sharpness claim is then made.

For the nonreversible three-state benchmark at $p=6$, the nearest candidate is the isolated real pole $\epsilon_*=-0.305325679501$. Its row-residue norm is $1.71603791912\times10^{-8}$ and its scalar residue is $8.14325368498\times10^{-9}$, so neither series cancels the pole.

The generalized-eigenvalue radius $0.305325679501$ is therefore exact in this example. The punctured-limit residue and the displayed left/right-eigenvector formula agree relatively within $3.77\times10^{-33}$; independently, the rescaled 100th Taylor coefficient recovers the residue with relative error $2.11609\times10^{-4}$.

The exact long-run second-cumulant slope and intercept

The higher-cumulant expansion above is fixed-maturity, but its first secular coefficient can still be obtained exactly when all three variance coefficients switch. Let the joint process $(Y_t,v_t)$ start in its invariant law $\Pi_m$, put $a_i=\xi_i^2$, and define the row vectors

With $A_m=mQ_0-D_\kappa$ and $B_m=mQ_0-2D_\kappa$, stationarity of the first two polynomial moments gives the finite linear systems

The centered Poisson equation for the joint generator,

has an affine solution $\phi_i(v)=\alpha_i v+\beta_i$. Its coefficients require only two more finite solves:

The second equation is solvable because $\pi D_c\alpha=-\pi D_cA_m^{-1}\mathbf1=r\mathbf1=\mu_m$. The stationary covariance identity, or direct integration of the affine conditional mean, therefore proves the exact formula

No reversibility is used. The affine conditional-mean system is exponentially stable, so the stronger fixed-rate statement is $\operatorname{Var}_{\Pi_m}(V_T)=\sigma_m^2T+O_m(1)$ as $T\to\infty$. This is the finite-dimensional switched-CIR specialization of the Poisson-equation method developed for ergodic diffusions by Pardoux and Veretennikov (2001).

The bounded term is also explicit. Put $\bar\phi_m=\mathbb E_{\Pi_m}[\phi]=r\alpha+\pi\beta$ and solve the second affine Poisson equation $-\mathcal L_m\psi=\phi-\bar\phi_m$, with $\psi_i(v)=p_iv+q_i$. Its coefficients are

Solvability follows from $\pi D_cp=-\pi D_cA_m^{-1}\alpha=r\alpha=\bar\phi_m$; the last equality uses $\pi\beta=0$. Integrating the stationary covariance twice gives the exact finite-maturity decomposition

The decay can in fact be made uniform over every half-line of switching rates. Assume $\kappa_*:=\min_i\kappa_i>0$ and fix $m_0>0$. If $(P_T\psi)(i,v)=a_m(T)_i+b_m(T)_iv$, then

The first semigroup is a finite-state chain killed at rate $\kappa_i$, so the Feynman–Kac representation gives $\lVert e^{A_mT}\rVert_\infty\le e^{-\kappa_*T}$. Irreducibility of $Q_0$ gives constants $M_Q,\gamma_Q>0$ with $\lVert e^{tQ_0}-\mathbf1\pi\rVert\le M_Qe^{-\gamma_Qt}$. Because $(r-\mu_m\pi)\mathbf1=0$, the constant term is therefore bounded by a convolution of $e^{-m\gamma_Q(T-u)}$ and $e^{-\kappa_*u}$.

Uniform boundedness of $r,s,\alpha,\beta,p,q$ for $m\ge m_0$ follows from the killed resolvents, the fixed group inverse of $Q_0$, and the stationary moment bounds $\mu_m\le\max_i c_i/\kappa_*$ and $\mathbb E_{\Pi_m}v^2\le(2\max_i c_i+\max_i a_i)\mu_m/(2\kappa_*)$. Put $\alpha_0=\min\{m_0\gamma_Q,\kappa_*\}$. At a resonant pair of rates a polynomial factor can multiply $e^{-\alpha_0T}$, but every strict subgap exponent absorbs it.

Consequently, for each fixed $0<\gamma<\alpha_0$, there is $C_\gamma<\infty$ such that

This is a joint maturity–switching-rate bound for the exact coefficients $\sigma_m^2$ and $\chi_m$; it does not make a finite $1/m$ truncation maturity-uniform, because any nonzero error in the slope is still multiplied by $T$. No reversibility is required.

For the strictly positive two-state benchmark above, with $a=(0.04,0.04)$, exact elimination gives

The same exact solves give

At $m=8,16,32,64,128$, the exact rates are $0.0015340375$, $0.0014967052$, $0.0014774262$, $0.0014676293$, and $0.0014626910$. The averaged and first-order-corrected errors have measured orders $0.996$ and $1.998$. An independent degree-two polynomial semigroup, differenced between maturities 20 and 40 to remove the bounded intercept, matches the Poisson formula within $9.1\times10^{-14}$; a nonreversible three-state check matches within $1.5\times10^{-14}$.

The exact intercepts at the same speeds are $-0.0011137851$, $-0.0010777572$, $-0.0010595646$, $-0.0010504172$, and $-0.0010458298$. Their averaged and first-order-corrected errors have measured orders $0.998$ and $1.984$. At maturity 20, the independent polynomial semigroup matches $\sigma_m^2T+\chi_m$ within $2.6\times10^{-13}$; the nonreversible three-state discrepancy is $4.4\times10^{-14}$.

At $T=0.75$, retaining the explicit remainder $\rho_m(T)$ recovers the full variance within $2.1\times10^{-17}$, including on the nonreversible example. On grids $m=1,2,\ldots,128$ and $0\le T\le12$, one common exponent $\gamma=0.4$ gives maximum rescaled envelopes $\max e^{\gamma T}|\rho_m(T)|=0.0015977624$ and $0.0004103649$ in the two-state and nonreversible three-state examples. At $T=12$ the corresponding remainder maxima are $6.13\times10^{-10}$ and $6.54\times10^{-15}$; independent degree-two semigroups verify the full decomposition within $5.1\times10^{-14}$ and $3.2\times10^{-13}$.

This theorem identifies the exact stationary long-time slope and intercept, not a maturity-uniform $O(m^{-2})$ approximation to the whole variance curve. Replacing either coefficient by a finite expansion still leaves a joint-limit error.

The exact long-run third-cumulant slope and intercept

The Poisson hierarchy closes one level further. Let $t_i=\mathbb E_{\Pi_m}[v^3\mathbf1_{\{Y=i\}}]$ and $C_m=mQ_0-3D_\kappa$. Stationarity gives the additional finite system

Put $g=v-\mu_m$ and center the first Poisson solution under the joint invariant law:

This centering is essential: the convenient normalization $\pi\beta=0$ does not imply $\mathbb E_{\Pi_m}\phi=0$. Define $c_2=\mathbb E_{\Pi_m}[g\phi^\circ]=\sigma_m^2/2$ and solve the nested Poisson equation

Its coefficients again require only finite linear solves:

Solvability of the last equation follows by taking the invariant expectation of the nested Poisson equation. Ordering the three time arguments in the stationary third-correlation integral then gives the exact, nonreversible formula

The same ordered-correlation representation also determines the bounded term exactly. Let $R_m=(-\mathcal L_m)^{-1}$ denote the centered Poisson inverse, use the joint-centered representatives $\phi^\circ=R_mg$ and $\psi^\circ=R_m(g\phi^\circ-c_2)$, and define

If $C_m(a,b)=\mathbb E_{\Pi_m}[g_0g_ag_{a+b}]$, stationarity and the six strict time orderings give

Integrating each time gap in turn yields the exact decomposition

For each fixed $m>0$, stability of the nonconstant degree-two polynomial semigroup gives constants $C_m,\gamma_m>0$ with $|\rho_{3,m}(T)|\le C_m(1+T)e^{-\gamma_mT}$. The two terms in $\chi_m$ are distinct: $\zeta$ weights the first time gap and $\omega$ weights the second.

This is the next member of the additive-functional Poisson hierarchy used above and in Pardoux and Veretennikov (2001). The broader analytic-cumulant framework via multiplicative Poisson equations is developed by Kontoyiannis and Meyn (2003); the formula here is a direct continuous-time correlation calculation using switched-CIR polynomial closure.

For the positive two-state benchmark, at $m=8,16,32,64,128$ the exact rates are $0.0000936333$, $0.0000915632$, $0.0000904367$, $0.0000898504$, and $0.0000895514$. They converge to the averaged-CIR value $0.0000892485$ at measured first order $0.991$.

An independent degree-three polynomial semigroup, differenced between maturities 20 and 40, agrees within $2.7\times10^{-13}$; a nonreversible three-state check agrees within $9.5\times10^{-14}$. Omitting the joint-invariant centering would change the $m=8$ answer by $0.0002367787$, more than twice the true rate.

At the same speeds the exact intercepts are $-0.0001359353$, $-0.0001318630$, $-0.0001297164$, $-0.0001286160$, and $-0.0001280592$. They converge to the averaged-CIR value $-0.0001274979$ at measured order $0.994$.

At maturity 20, the independent degree-three semigroup matches $\tau_mT+\chi_m$ within $3.6\times10^{-13}$; at maturity 10 the nonreversible three-state residual is $7.1\times10^{-14}$. These are exact coefficients for each fixed $m$: no switching-rate-uniform third-cumulant remainder bound or maturity-uniform finite-$1/m$ truncation is claimed.

The exact long-run fourth-cumulant slope

The hierarchy also gives the fourth coefficient, but with a lower-cumulant subtraction that is easy to miss. Let $h_\theta$ and $\lambda_m(\theta)$ be the locally analytic, normalized Feynman–Kac eigenpair

With the centered Poisson solutions above, put $c_2=\mathbb E_{\Pi_m}[g\phi^\circ]$ and $c_3=\mathbb E_{\Pi_m}[g\psi^\circ]=\tau_m/6$, and solve

Differentiating the eigenvalue equation gives $h'_0=\phi^\circ$, $h''_0=2\psi^\circ$, and $h'''_0=6\xi$. Therefore

The term $c_2\phi^\circ$ comes from the $3\lambda_m''(0)h'_0$ term in the third eigenfunction derivative; merely centering $g\psi^\circ$ is wrong. The solution $\xi$ is cubic. If $w_i=\mathbb E_{\Pi_m}[v^4\mathbf1_{\{Y=i\}}]$, its required invariant moments close with

No reversibility is used. Local eigenpair analyticity follows the isolated-eigenvalue perturbation method of Kato (1949); the multiplicative-Poisson and asymptotic-cumulant setting is treated more generally by Kontoyiannis and Meyn (2003).

For the positive two-state benchmark, at $m=8,16,32,64,128$ the exact rates are $0.0000103442$, $0.0000097177$, $0.0000094099$, $0.0000092577$, and $0.0000091822$. They converge to the averaged-CIR value $0.0000091070$ at measured order $1.003$. An independent degree-four semigroup agrees within $8.0\times10^{-13}$; a nonreversible three-state check agrees within $8.4\times10^{-15}$.

Omitting $c_2\phi^\circ$ changes the $m=8$ rate by $0.0000102515$, nearly the full correct value. The spectral-amplitude theorem below also identifies the fourth-order intercept, although not by this three-solve formula.

Every fixed long-run cumulant rate

The fourth-order calculation is one step of a recursion that closes at every fixed order. For the normalized eigenpair above, write

Normalization gives $\mathbb E_{\Pi_m}[h_k]=0$ for $k\ge1$. Differentiating the eigenvalue equation $k$ times, then taking invariant expectation, gives the exact triangular recursion

Consequently, for every fixed integer $k\ge2$ and every fixed switching rate $m>0$,

The deterministic mean term in $V_T$ does not affect orders $k\ge2$. The proof of the limit is the local spectral representation of the Feynman–Kac semigroup: its logarithm is $T\lambda_m(\theta)+O_{m,\theta}(1)$ near zero. The displayed equations then follow by Leibniz' rule.

In the switched CIR model $h_k$ is a regime-dependent polynomial of degree at most $k$, so each new step is a finite centered Poisson solve. The cases $k=2,3,4$ recover respectively $h_1=\phi^\circ$, $h_2=2\psi^\circ$, and $h_3=6\xi$, including the essential $c_2\phi^\circ$ subtraction at fourth order.

The one-state CIR model provides an independent all-order benchmark. If

the centered tilted eigenvalue follows from the stable Riccati root:

Thus, for every $k\ge2$,

The certificate evaluates the Poisson recursion through order eight. In the switched benchmark at $m=8$, the rates for orders two through eight are $1.5340375000\times10^{-3}$, $9.3633346625\times10^{-5}$, $1.0344200270\times10^{-5}$, $1.6980748064\times10^{-6}$, $3.7402863486\times10^{-7}$, $1.0398724197\times10^{-7}$, and $3.5037228249\times10^{-8}$. Orders two through four agree with the separately derived formulas within $3.7\times10^{-19}$; for the averaged one-state CIR model, orders two through eight agree with the closed formula within $8.9\times10^{-15}$ relative error.

The first inverse-speed coefficient at every fixed order

The numerical $O(m^{-1})$ convergence above has a common analytic coefficient. Let $P=\mathbf1\pi$ act on the regime coordinate, let $N=I-P$, and let $B_\theta$ denote the order-one part of the tilted polynomial generator, including multiplication by $\theta v$. On the polynomial space of any fixed target degree, eliminating the $N$ block gives the slow Schur complement

Indeed, $(Q_0|_N)^{-1}$ is $Q_0^\#$ on the fast block, and the first Schur term is $-PB_\theta Q_0^\#B_\theta P$. The regime-free pieces of $B_\theta$ preserve the $P$ and $N$ blocks, so only the centered switched coefficients survive; the result is precisely the Green–Kubo effective generator $\mathcal D$ already used in the fixed-maturity expansion.

Let $\bar\Lambda(\theta)$ be the eigenvalue continuing zero for $\bar{\mathcal L}+\theta M_v$, and choose its right and left eigenvectors $\bar h_\theta,\bar\ell_\theta$ with $\bar\ell_\theta\bar h_\theta=1$. Simple-eigenvalue perturbation then gives, locally in $\theta$,

Consequently, for every fixed $k\ge2$,

This identifies the missing finite-rate coefficient without solving the switched model at large $m$. It is the finite-dimensional slow-fast reduction of Phillips and Kokotovic (1981), followed by the isolated-eigenvalue perturbation of Kato (1949).

For the two-state benchmark, the coefficients $\eta_2,\ldots,\eta_8$ are

The uncorrected residuals at $m=128,256$ have measured orders $0.9981,0.9954,1.0018,1.0052,1.0083,1.0119,1.0163$; after subtracting $\eta_k/m$, the orders are $1.9991,1.9996,1.9897,1.9973,2.0003,2.0042,2.0652$. Independently derived rational formulas for the stationary mean and second-cumulant slope reproduce $\eta_1=-9/2450$ and $\eta_2=153/240100$ within $6.0\times10^{-18}$.

This is an all-fixed-order statement. It supplies neither constants uniform in $k$, nor convergence of an infinite cumulant series, nor an interchange of $k$, $m$, and $T$ limits. The long-maturity limit is taken first; the displayed rate expansion is not a maturity-uniform finite-$1/m$ approximation to the finite-$T$ cumulant.

The second inverse-speed coefficient

The $O_k(m^{-2})$ residual has a computable coefficient, but it is not obtained by simply taking the next fast-block product. Let $\iota$ inject a regime-free polynomial, let $\mathsf E=\pi\otimes I$ average the regime coordinate, and let $\mathsf S=Q_0^\#\otimes I$. With $A_\theta=\bar{\mathcal L}+\theta M_v$, Schur elimination of the fast block gives the eigenvalue-dependent slow pencil

The first expression for $\mathcal D$ reduces to the Green–Kubo operator above. Put $\eta(\theta)=\bar\ell_\theta\mathcal D\bar h_\theta$ and solve the reduced-resolvent equation

Ordinary simple-eigenvalue perturbation, applied after the Schur reduction, gives

The second term in $\zeta(\theta)$ is the response of the slow eigenvector to the first Green–Kubo perturbation. Omitting it generally leaves an $O(m^{-2})$ error. This is the next finite-dimensional term in the singular reduction of Phillips and Kokotovic (1981) and the isolated-eigenvalue expansion of Kato (1949).

For the two-state benchmark, $\zeta_2,\ldots,\zeta_8$ are

Between $m=12$ and $m=24$, the twice-corrected residual orders for cumulants two through eight are $3.142,3.129,3.001,3.054,3.131,2.898,3.149$. If the eigenvector-response term is dropped, the corresponding orders are only $2.012,2.041,1.953,1.974,1.997,2.025,2.061$.

Independently derived rational formulas give $\zeta_1=18/8575$ for the stationary mean and $\zeta_2=-369/1680700$ for the variance rate, agreeing within $3.2\times10^{-18}$. A nonreversible three-state certificate gives twice-corrected orders $2.955,3.030,2.917,2.967,3.048$ for cumulants two through six.

As before, the order, long-maturity limit, and coefficient data are fixed before $m\to\infty$. The $O_k(m^{-3})$ statement is local in the cumulant-generating parameter and is neither order-uniform nor a maturity-uniform approximation of the finite-$T$ cumulant.

Arbitrary-start boundary constants at every fixed order

The same spectral representation also resolves the bounded term left open by the rate recursion. Normalize the right eigenfunction by $\mathbb E_{\Pi_m}h_\theta=1$, and normalize the left eigenfunctional $\ell_\theta$ by $\ell_\theta(h_\theta)=1$. For any admissible initial law $\nu$, define the Perron amplitude

Assume the locally tilted semigroup has a simple analytic dominant eigenvalue separated by a spectral gap in a common weighted space. This is the same multiplicative-ergodic hypothesis used to justify the local cumulant-rate eigenpair; positive mean reversion and the local exponential moments of the switched CIR invariant law provide the required neighborhood of zero. Then, locally uniformly in $\theta$,

Taking a logarithm and differentiating a fixed number of times gives:

This expression is directly computable to any fixed order. Let $\ell_k=\left.\partial_\theta^k\ell_\theta\right|_{\theta=0}$, let $M_g$ denote multiplication by $g$, and retain the right derivatives $h_k$ and rates $\gamma_k$ from the preceding recursion. Differentiating the left eigen-equation and its normalization gives

Put $a_{k,\nu}=A_\nu^{(k)}(0)$, with $a_{0,\nu}=1$. Leibniz' rule and the standard moment-to-cumulant recursion then give

On the switched-CIR polynomial space these are finite bordered linear solves. For a target order $K$, the $K$-jet of the tilted semigroup involves only polynomials of degree at most $K$; no transform inversion or long-maturity extrapolation is required.

The statement holds for every fixed $k$ and includes nonstationary starts. For $k\ge2$, centering $V_T$ by its invariant mean changes neither side.

At orders two and three under a stationary start, the amplitude derivatives recover exactly the explicit Poisson intercepts above. Unlike those special formulas, the spectral form cleanly separates the initial-law contribution $\nu h_\theta$ from the terminal eigenmeasure contribution $\ell_\theta(1)$.

An exact polynomial-semigroup certificate evaluates the constants through order six for the two-state model at $m=8$. For the stationary start, orders one through six are

For the point start $Y_0=0$, $v_0=0.04$, they are

The explicit Perron recursion agrees with the independent $T=40$ polynomial semigroup within $2.7\times10^{-15}$ for the stationary start and $2.3\times10^{-15}$ for the point start. Computing $\kappa_k(V_T)-T\gamma_k$ at $T=20$ and $40$ changes the values by at most $3.2\times10^{-14}$. The stationary order-two and order-three constants match the earlier explicit Poisson formulas within $6.2\times10^{-16}$.

A finite-moment jet theorem: no MGF is needed

The analytic Perron hypothesis above is useful when one wants an actual cumulant-generating function on a neighborhood of zero, but it is stronger than necessary for any prescribed finite list of cumulants. Fix $K$.

Suppose only that the initial law has moments through degree $K$. In the truncated ring $\mathbb R[\theta]/(\theta^{K+1})$, define the formal transform jet

Polynomial closure makes this a finite-dimensional object: its coefficient vector evolves under the matrix $\mathcal L_m+\theta M_g$ on polynomials of degree at most $K$. At $\theta=0$ the constant mode is the only center direction.

The centered degree-zero block and every positive-degree block are stable, with diagonal blocks $mQ_0-jD_\kappa$ for $j\ge1$. Finite recursive Duhamel expansion, followed by the formal logarithm, therefore gives

The coefficients $\gamma_j$ and $\beta_{j,\nu}$ are exactly those produced by the right-left recursions above: those recursions are identities of finite jets, not consequences of positive exponential moments. If a local MGF does exist, this formal boundary jet agrees with the derivatives of the Perron amplitude.

Thus the fixed-order slope, boundary constant, and exponential remainder require only the corresponding finite moments. The statement is for fixed $K$; it asserts neither convergence of an infinite cumulant series nor any bound uniform in cumulant order.

A certificate makes the distinction concrete. Let $Y_0$ have its stationary two-state law and let $v_0$ be independent Pareto with shape $13/2$ and mean $0.04$. Its first six moments are finite, while its seventh moment and every positive exponential moment are infinite.

At switching speed $m=8$, the exact polynomial semigroup through degree six can still be compared with the recursively computed slope and boundary term. The maximum absolute remainder over orders one through six at maturities $T=4,6,8,10,12$ is respectively

A log-linear fit gives decay rate $1.378$. This is a finite-order numerical certificate, not evidence for an MGF: since the integrated CIR path retains a strictly positive affine loading on $v_0$ at every positive maturity, its positive transform also diverges in this example.

Every independent-return cumulant inherits the finite-jet theorem

The fourth-order log-return identity at the start of the page is one row of an exact triangular map. Fix $K$, let $Z$ be standard normal and independent of $V_T$, and put $X_T=-V_T/2+\sqrt{V_T}Z$. Conditional Gaussian moments give, in the truncated formal cumulant ring,

This is a moment identity: it requires moments only through order $K$, not an MGF. Comparing the coefficient of $s^k$ yields, for every $1\le k\le K$,

For $k=4$ the three nonzero coefficients are $3,3/2,1/16$, recovering the formula above. More importantly, the map transports the entire long-maturity decomposition. If

then exactly

where every sum runs from $\lceil k/2\rceil$ to $k$. Thus the exact $m$-dependent slopes and arbitrary-start boundary constants remain computable at every fixed return order. The same exponent in the switching-rate-uniform remainder theorem is inherited: if the variance remainders through order $K$ are bounded by $C_Ke^{-\gamma T}$, then the order-$k$ return remainder is bounded by $C_Ke^{-\gamma T}\sum_j|a_{kj}|$.

This remains fixed-order, not order-uniform. The triangular transfer itself says nothing about leverage, for which conditional normality fails; the direct tilted-generator theorem below handles that case separately.

The degree-six certificate uses the Pareto initial variance above, so the return has no two-sided MGF near zero. At $T=4,8,12$, an independent polynomial semigroup in $(v,V,M)$ agrees with the transported cumulants through order six within $8.7\times10^{-17}$, $1.3\times10^{-15}$, and $1.5\times10^{-14}$.

Transporting the exact slope, boundary, and remainder separately gives the same errors. The largest absolute return remainder across the six orders falls from $1.458\times10^{-4}$ to $1.801\times10^{-6}$ and then $1.551\times10^{-8}$.

A switching-rate-uniform remainder at every fixed order

For switched CIR, the dependence on $m$ in the preceding remainder can be removed when the cumulant order is fixed. Fix $K$, $m_0>0$, and assume $\kappa_*:=\min_i\kappa_i>0$. Let $\lVert e^{tQ_0}-\mathbf1\pi\rVert\le M_Qe^{-\gamma_Qt}$, and consider either a fixed initial law or a family of initial laws whose moments through degree $K$ are uniformly bounded.

Put $\alpha_0=\min\{m_0\gamma_Q,\kappa_*\}$. Then, for every fixed $0<\gamma<\alpha_0$, there is $C_{K,\gamma}<\infty$ such that

Here both the slope $\gamma_k(m)$ and the boundary constant $\beta_{k,\nu}(m)$ are the exact $m$-dependent finite-jet quantities above. In particular, $\beta_{k,\nu}(m)$ is defined by the formal right-left recursion when the initial law has only $K$ moments; it need not be the derivative of an actual Perron amplitude. The proof is finite-dimensional.

On polynomials of $v$ through degree $K$, the degree-$j$ diagonal block is $mQ_0-jD_\kappa$. For $j\ge1$, its Feynman–Kac semigroup is bounded by $e^{-j\kappa_*T}$; on the centered degree-zero block, finite-chain mixing gives $M_Qe^{-m\gamma_QT}$. The tilted-semigroup jet through order $K$ is obtained by finitely many Duhamel convolutions of these stable blocks.

After the exact secular term and finite boundary jet are removed, every remaining convolution contains a stable factor. An $r$-fold resonance contributes at worst $T^r e^{-\alpha_0T}$. Every strict subgap exponent absorbs this polynomial because, for integer $r\ge1$,

The killed resolvents and the stationary CIR moment recursion are uniformly bounded for $m\ge m_0$, completing the common-envelope estimate. The initial law enters only through its finite vector of component moments $\mathbb E[\mathbf1_{\{Y_0=i\}}v_0^j]$, $j\le K$; uniform boundedness of those moments gives a common bound on that vector. Hence neither the theorem nor its constant requires a positive MGF.

This is a fixed-jet refinement of the preceding multiplicative-Poisson expansion. It can be read entirely in the finite polynomial jet just described; when local exponential moments are available, it is also the finite-derivative consequence of the isolated-spectral-subspace framework of Kato (1949).

The strict inequality $\gamma<\alpha_0$ is generally necessary. Consider the symmetric chain $Q_0=\left(\begin{smallmatrix}-1&1\\1&-1\end{smallmatrix}\right)$ at $m=1$, common CIR reversion $\kappa_i=2$, drifts $c_\pm=\bar c\pm\delta$, and start in the high-drift regime at $v_0=\bar c/2$. The chain-contrast and CIR-decay rates both equal two, so direct solution of the first moment equations gives

After subtracting its exact zero slope and boundary constant $\delta/4$, the remainder is $-\delta(2T+1)e^{-2T}/4$. Hence $e^{2T}|R(T)|$ grows linearly, even though every $e^{\gamma T}|R(T)|$, $\gamma<2$, is bounded.

This resonance rules out replacing the strict-subgap statement by the endpoint exponent without an additional nonresonance assumption. The certificate takes $\bar c=0.04$, $\delta=0.02$, and $a_i=0.04$, so $2c_i\ge a_i$ in both regimes and this obstruction occurs inside the positive switched-CIR model.

For this two-state first cumulant the nonresonance condition is exact. Let the chain-contrast rate be $\lambda=2m$, let the common CIR reversion be $\kappa$, retain $c_\pm=\bar c\pm\delta$, and start in the high-drift regime at the averaged equilibrium $v_0=\bar c/\kappa$. If $\lambda\ne\kappa$, direct solution gives

Thus the endpoint exponent $\min\{\lambda,\kappa\}$ is valid for each fixed nonresonant model. As $\lambda\to\kappa$, however, the endpoint constant diverges like $|\lambda-\kappa|^{-1}$ and the continuous limit is exactly $-\delta(1+\kappa T)e^{-\kappa T}/\kappa^2$. Consequently the strict-subgap theorem is sharp uniformly over a parameter family that contains or approaches resonance; a uniform endpoint claim needs a quantitative separation $|\lambda-\kappa|\ge\eta>0$ (and the analogous separation of all stable blocks at higher cumulant orders).

The transition between the two formulas has an exact universal profile. Fix $\kappa>0$, put $\lambda_T=\kappa+z/T$, and let $R_T(z)$ be the displayed boundary remainder at that contrast rate. For $\delta\ne0$ define

Substitution in the exact nonresonant expression, followed by continuous extension at $z=0$, gives the finite-maturity identity

Hence the resonant factor $T e^{-\kappa T}$ is not an isolated pathology: it is the uniform crossover scale whenever $(\lambda-\kappa)T$ remains bounded. In particular, for $T>Z/\kappa$,

This $O(T^{-1})$ statement is uniform on every compact detuning window and does not assert an endpoint bound uniform over fixed nonzero rate separations. It instead resolves the singular boundary layer between the fixed-nonresonant and exactly resonant regimes.

The same profile is universal over arbitrary initial laws with a finite variance mean. Encode the high and low regimes by $Y=+1$ and $Y=-1$, and for an initial law $\nu$ put

No independence between $Y_0$ and $v_0$ is needed. Indeed, the two first-moment equations close as $\mathbb E Y_t=y_0e^{-\lambda t}$ and $x'(t)=-\kappa x(t)+\delta y_0e^{-\lambda t}$. Therefore, when $\lambda\ne\kappa$,

At resonance the second term has the continuous limit $\delta y_0\{1-(1+\kappa T)e^{-\kappa T}\}/\kappa^2$. For $y_0\ne0$ and $\lambda_T=\kappa+z/T$, the corresponding boundary remainder satisfies the exact identity

Consequently, for $T>Z/\kappa$,

Thus an arbitrary initial variance mean changes only the explicit $O(T^{-1})$ correction; even initial regime–variance dependence does not change the leading crossover. If $y_0=0$, the resonant forcing is absent and the remainder reduces to $-x_0e^{-\kappa T}/\kappa$, so this normalization is neither needed nor defined.

The certificate checks orders one through six for $m=1,2,\ldots,64$ and $0\le T\le10$ with one exponent $\gamma=0.35$. The largest rescaled envelopes over all six orders are $0.001598$ for the stationary two-state model, $0.03264$ for its point start, $0.02246$ for the Pareto no-MGF start above, and $0.0004104$ for the nonreversible three-state model. At $T=10$, the largest absolute remainders are respectively $7.23\times10^{-9}$, $1.94\times10^{-7}$, $9.10\times10^{-8}$, and $4.12\times10^{-13}$.

A separate exact resonant certificate agrees with the displayed first-cumulant formula to $1.69\times10^{-16}$. At $T=20$, $e^{2T}|R(T)|/T=0.01025$, approaching the exact limit $\delta/2=0.01$; for $\gamma=1.9$, the global envelope is $0.0386741023$, attained at $T=9.5$.

The new nonresonant checks put the contrast rate on both sides of the CIR rate: at $(\lambda,\kappa)=(1.4,2)$ and $(3,2)$ the closed form agrees with the independent polynomial semigroup within $2.7\times10^{-16}$. At $T=20$, the endpoint-rescaled remainders are $0.0238094214$ and $0.00999999999$, against the exact limits $1/42=0.0238095238$ and $0.01$.

The crossover certificate also takes $\lambda=2+z/T$ for seven detunings in $[-4,4]$. The exact first cumulant agrees with the polynomial semigroup within machine precision, and the finite-$T$ profile identity is verified before comparison with the displayed compact-window bound.

Three further mixed initial laws use nondegenerate regime probabilities and different conditional variance means; they independently verify the arbitrary-start identity and show numerically that only $y_0$ and $x_0$ enter. Across those laws, the largest semigroup discrepancy is $1.16\times10^{-16}$ and the largest error in the exact scaled-profile identity is $1.27\times10^{-14}$.

This is uniform in switching speed only after fixing the cumulant order and retaining the exact rate and boundary constant. It gives neither constants uniform in $k$, nor a maturity-uniform finite-$1/m$ truncation, nor a price-level theorem.

Two-term standardized cumulants with an explicit finite-time bound

The exact rate–boundary–remainder decomposition also quantifies how skewness and excess kurtosis decay with maturity. The statement is algebraic and applies without change to integrated variance, independent returns, or leveraged returns whenever the corresponding exact fixed-order decomposition above is available. Write, for $2\le j\le K$,

assume $\Gamma_2>0$, and put $Z_T=(Y_T-\mathbb EY_T)/\sqrt{\kappa_2(Y_T)}$. For fixed $j\ge3$ and $p=j/2$, direct expansion of the exact denominator gives

This is more than an asymptotic order statement. Let $\mathcal B_2=|B_2|+C_2$, $L_p=p2^{p+1}$, and $D_p=p(p+1)2^{p+1}$. Taylor's theorem for $(1+x)^{-p}$ on $|x|\le1/2$ gives, for $T\ge2\mathcal B_2/\Gamma_2$,

In particular, skewness and excess kurtosis have the explicit forms

If the preceding exact decomposition is switching-rate uniform, the same conclusion is uniform in $m\ge m_0$ provided $\inf_m\Gamma_2(m)>0$ and the displayed slopes, boundaries, and remainder constants have the corresponding uniform upper bounds. Those nondegeneracy assumptions are explicit; the normalization does not create them. The finite-jet remainder comes from the killed-semigroup argument of Kac (1949) and the isolated spectral decomposition of Kato (1949); the step on this page is the exact scalar normalization and Taylor remainder.

For the positive two-state benchmark at $m=8$, the certificate compares the formulas with the independent degree-four polynomial semigroup at $T=4,8,16,32$. For stationary start, the last-doubling residual orders improve from $1.575$ to $2.547$ for skewness and from $2.104$ to $3.057$ for excess kurtosis after the boundary term is included.

For the point start they improve from $1.765$ to $2.604$ and from $2.448$ to $3.129$. The exact normalization identity agrees to machine precision, and every tested finite-time error is below the displayed explicit Taylor bound.

Vanishing standardized cumulants at any fixed list of orders are not, by themselves, an Edgeworth theorem, a Berry–Esseen bound, or even a proof of weak convergence. No such distributional claim is made here. The result is a finite-order shape certificate with a fully stated normalization error.

Every leveraged-return cumulant has the same exact long-maturity decomposition

Conditional normality is unnecessary. Write $X_t=M_t-\tfrac12\int_0^t v_sds$ and $eta_i=\rho_i\xi_i$. Exponential conjugation of the joint generator by $e^{\theta x}$ leaves a polynomial operator on $v$ alone:

Consequently, on polynomials through degree $K$, the complete leveraged-return transform jet is finite dimensional. If $h_n$ and $\Gamma_n$ denote the $n$th derivatives at zero of the normalized right eigenvector and eigenvalue, respectively, then $h_0=1$ and

The analogous left recursion gives the arbitrary-start Perron-amplitude jet and hence its formal logarithm. It follows that, for every fixed $K$,

The reason is structural: leverage changes only $B_1$. At $\theta=0$ the centered degree-zero block still mixes at rate $m\gamma_Q$, every positive-degree block is still killed at least at rate $\kappa_*$, and the order-$K$ jet uses only finitely many convolutions with $B_1$ and $B_2$. Polynomial resonance is absorbed at every strict subgap, exactly as in the variance theorem, while its endpoint counterexample already occurs at the first unleveraged cumulant.

Thus finite initial moments suffice; no two-sided MGF is needed. This is the formal finite-jet version of the affine transform used in Heston (1993) and the general affine-process framework of Duffie, Filipović and Schachermayer (2003).

The certificate uses switching correlations $(-0.8,0.35)$ and an independent Pareto initial variance with shape $13/2$, so moments through degree six exist but no positive MGF does. A separate joint polynomial semigroup in $(M,v,z)$ agrees with the tilted-generator cumulants at $T=4,8,12$ within $5.9\times10^{-17}$, $1.8\times10^{-15}$, and $1.1\times10^{-14}$. Over $m=1,2,\ldots,64$ and $0\le T\le10$, the largest rescaled envelope among orders one through six is $0.06671$ for $\gamma=0.35$; the orderwise $T=10$ remainders are

As before, the exact $m$-dependent slope and boundary are retained. The theorem is not a maturity-uniform statement for a finite inverse-speed truncation and is not uniform in cumulant order.

Leverage can be included without conditional normality by augmenting once more with $M_t=\int_0^t\sqrt{v_s}dW_s$. Allow the instantaneous correlation itself to switch:

The two regime-dependent diffusion coefficients are $a_i=\xi_i^2$ and $\eta_i$; neither $\rho_i$ nor $\xi_i$ alone is a generator coordinate. On polynomials in $(M,v,z)$, define

The averaged model is again a valid Heston diffusion, with

There is a sharper admissibility statement. For fixed positive statewise vol-of-vol values, the exact attainable interval is

The first inequality is the triangle inequality $|\mathbb E_\pi(\rho\xi)|\le\mathbb E_\pi\xi$. It is sharp: the two endpoints are attained by $\rho_i\equiv+1$ and $\rho_i\equiv-1$, and every intermediate value is attained by a constant correlation.

Hence heterogeneous $\xi_i$ strictly contracts even the most extreme homogenized correlation away from $\pm1$. Equivalently, the determinant of the averaged instantaneous covariance matrix is nonnegative, and at either endpoint equals $v^2\operatorname{Var}_\pi(\xi)$.

The boundary parameter is compatible with the same averaging:

Thus statewise Feller admissibility implies admissibility of the averaged CIR model. The converse is false, as the explicit two-state hitting example above shows; averaging can conceal a reachable boundary in one regime. This distinction is pathwise and does not depend on the switching rate.

Let $K^{(0)}$ now be the Green–Kubo matrix of the centered feature vector $(c,\kappa,a,\eta)$. For fixed maturity on every invariant finite-degree polynomial space, the corrector is

with an $O_T(m^{-2})$ semigroup remainder. For constant $\rho$, writing $\eta_i=\rho\xi_i$ is exactly equivalent to the earlier $(\xi_i,\rho A_\eta)$ representation.

For switched $\rho$, however, separately averaging $\rho$ and $\xi$ is generally wrong; the relevant average and Poisson equation use $\eta=\rho\xi$. The new commutators are

Thus cycle reversal also flips $K^{(0),\rm anti}_{c\eta}\partial_{Mv} +\tfrac12K^{(0),\rm anti}_{a\eta}v\partial_{Mvv}$, while preserving the symmetric correction. Put $x=M-z/2$ and define

Direct differentiation of the fourth moment–cumulant polynomial gives

This is the simultaneous-$(\kappa\theta,\kappa,\xi^2,\rho\xi)$ leverage correction. It agrees with the independent-price formula above when $\eta=0$, but does not assume that reduction under leverage.

The independent certificate works on polynomials through degree five, an invariant subspace of the CIR generator. For a nonreversible three-state chain it compares the original coupled semigroup with the displayed Duhamel correction. The averaged and symmetric-only errors remain first order, whereas the full residual has rate 1.994.

Reversing the chain transposes $K^{(0)}$; after subtracting the commutator correction, the forward-minus-reverse residual has rate 1.985. The matrix identity between the displayed and abstract correctors holds to $6.9\times10^{-18}$. In the augmented degree-four system, the averaged fourth cumulants are $0.0003932318$ for $M_T$ and $0.0003946899$ for $X_T$; their $m^{-1}$ coefficients are $0.0000922515$ and $0.0000931020$.

Both corrected residuals have measured rate 1.992. At $\rho=-0.7$, the averaged $\kappa_4(X_T)$ is $0.0009887961$, its simultaneous-switching coefficient is $0.0001250489$, and the corrected residual has rate 1.990. For switched correlations $(-0.80,-0.25,0.35)$, the effective correlation is $-0.20052942$, the averaged fourth cumulant is $0.0004572129$, the first-order coefficient is $0.0003411583$, and the corrected residual has rate 1.986.

Its sharp admissible interval is $[-0.99580913,0.99580913]$, rather than merely $[-1,1]$. The certificate independently verifies the constant-$\rho$ equivalence to $2.1\times10^{-17}$, the three displayed commutators exactly on the polynomial space, and Green–Kubo transposition under cycle reversal to $1.4\times10^{-17}$. A separate 10,000-case randomized certificate checks the sharp correlation envelope and positive-semidefinite averaged covariance matrix.

The $\rho=0$ augmented calculation reproduces both independent-price numbers within numerical precision. The parameter choice satisfies the strict Feller inequality in every regime.

This extension is deliberately an operator theorem, not another closed Riccati formula. When $\kappa$ varies, the regime transform coefficients are no longer represented by one common scalar $B_s$, so the explicit one-dimensional call correction below does not carry over unchanged.

A two-state certificate

The certificate uses a symmetric chain with flip rate $m$, $\theta=(0.08,0.02)$, $\kappa=2$, $\xi=0.2$, $v_0=0.04$, and $T=1.5$. The strict Feller inequality holds in both regimes. Here $C_m(u)=0.03^2e^{-2mu}$ and $K_m=0.03^2/(2m)$.

An independent matrix exponential solves the closed ODE for every mixed moment $\mathbb E[V_T^a v_T^b\mathbf1_{\{Y_T=i\}}]$ of total degree at most four. It agrees with the covariance decomposition to better than $2\times10^{-13}$.

The scaled residual settles, confirming the $m^{-2}$ remainder. The certificate also computes $\kappa_4(X_T)$ two ways: directly from conditional raw moments and from the exact three-cumulant formula above.

What leverage changes exactly

Now allow the price and variance Brownian motions to have correlation $\rho$:

Conditional on the variance path, $M_T$ is no longer an independent centered normal with variance $V_T$. The replacement is an exact identity that uses joint cumulants.

Write $\kappa_{r,s}$ for the joint cumulant with $r$ copies of $M_T$ and $s$ copies of $V_T$. Multilinearity gives

This statement needs only fourth moments. When $\rho=0$, conditional normality reduces its five terms to

recovering the earlier formula. None of these four reductions should be assumed under leverage.

An independent affine referee

The leveraged certificate solves every polynomial moment $\mathbb E[M_T^aV_T^bv_T^c\mathbf1_{\{Y_T=i\}}]$ with $a+b+c\le4$. This system closes exactly under the switched CIR generator.

As an independent check, it also solves the affine transform. For the log-return MGF, with $B(0)=0$ and $a(0)=\mathbf1$,

Cauchy coefficient extraction from this transform agrees with the polynomial fourth cumulant below $3\times10^{-9}$.

For the certificate parameters at switching speed $m=8$, the independent shortcut is exact at $\rho=0$ but materially wrong under leverage:

The even-in-$\rho$ change in $\kappa_{4,0}$ is accompanied by odd mixed-cumulant terms, so positive and negative leverage do not give the same log-return kurtosis. At $\rho=-0.7$, the independence formula misses about 60% of the exact fourth cumulant.

An explicit leveraged option-price correction

Set $\bar\theta=(\theta_++\theta_-)/2$ and $\delta=(\theta_+-\theta_-)/2$, and let the symmetric chain flip at rate $m$. The price and variance Brownian motions may now have any fixed correlation $\rho$. For an MGF argument $s$, the Heston Riccati coefficient remains independent of the regime:

Write the two regime amplitudes as $a_\pm$, their mean as $a=(a_++a_-)/2$, their half-difference as $d=(a_+-a_-)/2$, and $r=d/a$. With $g(t)=\kappa\delta B_s(t)$, their exact equations reduce to

The stationary start and $B_s(0)=0$ remove the leading initial layer. Therefore, at fixed maturity and uniformly for $s$ in any compact subset of the MGF strip,

This has an elementary finite-rate proof. If $G=\sup|g|$, $L=\sup|g'|$, and $m\ge G$, a bootstrap in the Riccati equation gives $|r|\le G/m$. For $e=r-g/(2m)$,

so the remainder in the displayed log transform is bounded by $TG[L/(4m^2)+G^3/(2m^3)]$. This bound is fixed-maturity and compact-transform-domain; it is not the maturity-uniform cumulant bound above.

For log strike $k$, Carr–Madan damping $\alpha>0$, and $D_\alpha(u)=\alpha^2+\alpha-u^2+i(2\alpha+1)u$, the normalized call price consequently has first correction

The stated error follows directly on every truncated frequency interval. Passing to the infinite integral additionally requires the usual Heston moment-strip and Fourier-tail domination conditions. The certificate uses $k=0$, $\alpha=1$, cutoff 80, and doubles the Gauss–Legendre order from 320 to 640; both $C_{\rm av}$ and $c_1$ change by less than $2\times10^{-10}$.

This first table is the independent-price case $\rho=0$: $C_{\rm av}=0.1044747017$ and $c_1=-0.0020009463$. Its last residual rate is 2.127.

Leverage changes the Riccati solution, and hence both $C_{\rm av}$ and $c_1$, but it does not change the two-state amplitude equation or its error proof. At $\rho=-0.7$ the independent check gives:

Here $C_{\rm av}=0.1030497886$, $c_1=-0.0019979648$, and the last residual rate is 2.137. Doubling the quadrature order changes the leveraged averaged price, correction, and $m=8$ exact price by less than $2\times10^{-10}$. Thus the same full-transform option correction survives leverage; it is not an Edgeworth approximation based only on four cumulants.

The corresponding implied-volatility correction

The price result transfers directly to Black–Scholes implied volatility, but only where inversion is well conditioned. Let $C_{\rm BS}(k,\sigma)$ be the normalized zero-rate Black–Scholes call, define $\sigma_{\rm av}(k)$ by $C_{\rm BS}(k,\sigma_{\rm av}(k))=C_{\rm av}(k)$, and write $\mathcal V_{\rm av}(k)=\partial_\sigma C_{\rm BS}(k,\sigma_{\rm av}(k))$.

On a compact strike interval $K$, assume the call expansion is uniform, all prices remain strictly inside their no-arbitrage bounds, and $\mathcal V_{\rm av}\ge v_*>0$. The implicit-function theorem then gives, uniformly for $k\in K$,

The error statement has an elementary quantitative form. If $|C_m-C_{\rm av}-c_1/m|\le A/m^2$, $|c_1|\le B$, Vega is at least $v_*$ throughout the intervening volatility interval, and Black–Scholes volga is at most $M$, then

This follows by the mean-value theorem to obtain $|\sigma_m-\sigma_{\rm av}|\le(A+B)/(v_*m)$, followed by one Taylor expansion of $C_{\rm BS}$. The lower-Vega condition is essential: no strike-uniform wing claim is being made.

The table uses $\rho=-0.7$ and exact switched prices at $m=4,8,16,32$. The averaged Vegas across these strikes are $0.37391$, $0.48452$, and $0.44298$, safely away from zero. After subtracting $b_1/m$, every implied-volatility residual decays at second order.

Scope

• The chain starts in stationarity. A known or filtered starting regime adds an initial-memory term.
• The maturity-uniform Green–Kubo bound above is proved for the independent-price benchmark. The leverage section supplies an exact finite-rate identity and exact computation, not a uniform asymptotic error bound.
• The call correction assumes a stationary symmetric two-state chain, fixed correlation $\rho$, fixed maturity, and a valid damped Fourier contour. It is uniform only on compact transform contours unless separate tail bounds are supplied; no maturity-uniform leverage bound is claimed.
• The implied-volatility correction is uniform only on compact strike sets where the price expansion is uniform and Black–Scholes Vega has a positive lower bound. It deliberately excludes vanishing-Vega wings.
• The exact integrated-variance decomposition, maturity-uniform cumulant bound, and explicit Fourier call correction assume fixed $\kappa$ and switch only the CIR mean level. The simultaneous $(\kappa\theta,\kappa,\xi^2)$ results are separate fixed-maturity polynomial-generator and cumulant theorems; with leverage the feature list additionally contains $\rho\xi$ because that product is the cross-derivative coefficient. The exact two-state mean calculation above proves that their first-order remainder cannot be absolute and maturity-uniform in general. Retaining the exact stationary slope and either the exact principal first-moment mode or its explicit first-order Green–Kubo expansion restores a uniform mean approximation only. Truncating the Poisson recursion at order $p$ instead gives a joint-limit result only through $T=O(m^{p-1})$. The two affine Poisson solves identify the exact long-run second-cumulant slope and intercept under joint stationarity. For every $m\ge m_0>0$, their exact remainder has one switching-rate-uniform exponential envelope. This does not make a finite $1/m$ truncation of the slope maturity-uniform. The nested polynomial Poisson solves identify the exact third-cumulant slope and intercept for each fixed $m$, with a fixed-$m$ exponentially decaying remainder. The centered-Poisson recursion identifies every fixed-order long-run cumulant rate, the slow-fast Schur recursion identifies its first two inverse-speed coefficients with a fixed-order $O(m^{-3})$ residual, and the Perron amplitude identifies its arbitrary-start boundary constant with a fixed-$m$ exponential remainder. At each fixed order, retaining the exact $m$-dependent slope and boundary gives a switching-rate-uniform exponential envelope at every exponent strictly below the stable-block gap; the endpoint may fail by resonance. In the exact two-state first-moment subfamily, the endpoint holds for a fixed nonresonant pair but its constant is not uniform as the chain and CIR rates approach one another. The triangular finite-jet map transfers that statement to every independent-price return cumulant. These results provide no order-uniform constants or maturity-uniform finite-$1/m$ cumulant approximation. The standardized-cumulant corollary additionally requires a positive long-run variance slope; uniformity in switching speed requires a uniform positive lower bound on that slope. Its finite-order decay formulas are not an Edgeworth expansion, a Berry–Esseen estimate, or a weak-convergence theorem. No price-level extension is claimed.
• The statewise Feller theorem is pathwise and exact for every switching speed and initial regime. It is valid when vol-of-vol also switches, and is separate from the stationary-start, fixed-maturity asymptotic expansions.
• The triangular finite-jet map identifies every fixed-order independent-price return cumulant. Under leverage, conditional normality fails, but the separate tilted-generator jet above still supplies every exact fixed-order long-run slope, arbitrary-start boundary constant, and switching-rate-uniform exponential remainder. The fixed-maturity inverse-speed correction remains a distinct asymptotic statement.

The conditional integrated-variance viewpoint goes back to Hull and White (1987); the positive square-root model is from Cox, Ingersoll and Ross (1985). Boundary classification and the martingale time change use Feller (1951) and Dambis (1965). The leveraged affine referee follows the transform structure of Heston (1993); switched-Heston calibration is treated by Papanicolaou and Sircar (2014).

The finite-dimensional moment construction follows the polynomial-process framework of Cuchiero, Keller-Ressel and Teichmann (2012). The contribution here is the exact switched-CIR covariance decomposition, its finite-rate Green–Kubo bound, the explicit joint-cumulant benchmark showing precisely which terms leverage adds, full-transform European option and implied-volatility corrections, and fixed-maturity effective-generator and cumulant corrections when $\kappa\theta$, $\kappa$, and vol-of-vol switch, together with an exact long-maturity obstruction and exact finite-rate formulas for the stationary second-cumulant slope and intercept, third-cumulant slope and intercept, and every fixed-order long-run cumulant rate and arbitrary-start boundary constant in that simultaneous-switching model, together with the exact triangular transfer of those quantities and their exponential remainders to every fixed-order independent-price return cumulant, and the resulting explicit two-term standardized-cumulant expansion with a finite-time normalization bound. The damped inversion follows Carr and Madan (1999).