What sets the homogenization scale?

Ergodicity gives averaging. The covariance tail decides the size of the first fluctuation correction.

A fast finite-state chain has an integrable covariance, so its symmetric correction is proportional to its correlation time. That conclusion is not a consequence of stationarity and ergodicity alone.

Long memory changes the power of the small parameter. Conversely, a vanishing Green–Kubo integral can make averaging faster than the usual diffusive rate, while a periodic ergodic driver can have no convergent Green–Kubo integral at all.

The exact starting point

Let $X_t$ be a centered stationary scalar process with covariance $C(t)=\mathbb E[X_0X_t]$, and accelerate it to $X_{t/\varepsilon}$. For

stationarity alone gives the exact identity

Every result below follows from this formula or its equivalent spectral form. Ergodicity is relevant to the averaged limit, but it does not determine the rate at which the variance vanishes.

The spectral form

Write Bochner's one-sided representation as $C(t)=\int_{[0,\infty)}\cos(\omega t)\,\mu(d\omega)$. Fubini's theorem gives the exact Fejér-kernel identity

The atom is $\mu(\{0\})=\left\|P_{\rm inv}X_0\right\|_2^2$, where $P_{\rm inv}$ is projection onto the shift-invariant subspace. A centered ergodic process has $\mu(\{0\})=0$. If $\mu(\{0\})>0$, the variance converges to $T^2\mu(\{0\})$ and no fluctuation rescaling can repair the failed averaging limit.

The normalized variance $\operatorname{Var}A_\varepsilon(T)/T^2$ depends on maturity and switching speed only through $R=T/\varepsilon$, the number of fast correlation times observed:

Here $K_R(\omega)=2\{1-\cos(R\omega)\}/(R\omega)^2$ with $K_R(0)=1$. Thus $T_\varepsilon\to0$ by itself does not imply a frozen or homogenized limit.

The low-frequency theorem

Assume $\mu(\{0\})=0$, put $F(x)=\mu((0,x])$, and suppose that the cumulative spectral mass is regularly varying:

where $L$ is slowly varying. No density is required. Along every joint path $R_\varepsilon=T_\varepsilon/\varepsilon\to\infty$,

This single formula includes long memory for $0<\alpha<1$, ordinary Green–Kubo scaling at $\alpha=1$ when $L$ tends to a positive constant, and zero-Green–Kubo antipersistence for $1<\alpha<2$. For a general stationary process it is a variance theorem, not a functional limit theorem. It is the continuous-time counterpart of the spectral regular-variation criterion for partial sums proved by Deligiannidis and Utev (2013).

Covariance regimes

The same scales can be read from the covariance tail. In the rows with a regularly varying tail, $C(t)\sim c\,t^{-\alpha}L(t)$ with $L$ slowly varying (and eventually positive in the $K=0$ rows), $K=\int_0^\infty C(s)ds$ is the Green–Kubo integral, and $H(x)=\int_1^x L(s)\,ds/s$.

Under long-range dependence the Green–Kubo integral diverges and the fluctuation size is $\varepsilon^{\alpha/2}L(1/\varepsilon)^{1/2}$ rather than $\sqrt\varepsilon$. At the critical index, $C(t)\sim c/t$ gives $2cT\varepsilon\log(1/\varepsilon)$, and a logarithmic modifier changes even the critical scale.

In the antipersistent rows, nonnegativity of the variance forces $c<0$. Negative long-lag covariance cancels the short-lag contribution exactly enough to replace the usual $\varepsilon$ variance by $\varepsilon^\alpha L(1/\varepsilon)$. The fluctuation amplitude is therefore $\varepsilon^{\alpha/2}L(1/\varepsilon)^{1/2}$: smaller than $\sqrt\varepsilon$, but generally larger than $\varepsilon$.

A pure $t^{-2}$ tail produces $-2c\varepsilon^2\log(T/\varepsilon)$, not a pure $\varepsilon^2$ law. In the last row, $M<0$ for a nonzero stationary process, because

So the statement “integrable covariance implies an order-$\varepsilon$ correction” requires the qualification $K\ne0$. The regular-variation step is Karamata's integral theorem (1933).

The limit law needs more than the covariance

Let $X$ be centered, stationary and Gaussian, under the same cumulative spectral hypothesis, and set $H=1-\alpha/2\in(0,1)$. With $V(r)=\operatorname{Var}\!\left(\int_0^rX_s\,ds\right)$ and $S_R(t)=\int_0^{Rt}X_s\,ds/\sqrt{V(R)}$ on $0\le t\le1$, as $R\to\infty$,

where $B_H$ is standard fractional Brownian motion and $c_0^\gamma[0,1]$ is the separable little-Hölder space. No mixing assumption is needed beyond the spectral regular variation, but Gaussianity is essential. The long-memory half, $0<\alpha<1$, is the linear Gaussian case of Taqqu's weak-convergence theorem (1975).

Fix one $\alpha\in(0,1)$. For each integer $q\geq1$, a normalized degree-$q$ Hermite polynomial of a centered stationary unit-variance Gaussian process with covariance $(1+|t|)^{-\alpha/q}$ has covariance exactly $(1+|t|)^{-\alpha}$. After normalization by the common variance, rank one gives fractional Brownian motion, while rank $q\geq2$ gives the non-Gaussian rank-$q$ Hermite process (Taqqu (1979)).

Hence no statistic determined only by the covariance or spectral measure can identify the functional limit, even within stationary mixing transforms of Gaussian processes. The restriction $\alpha<2$ is also sharp: at the divergent critical endpoint the Gaussian finite-dimensional limit has no stochastically continuous version, and there is no càdlàg functional limit under this normalization.

Two bounded positive examples

For $0<\alpha<1$, let $G_t$ be a centered stationary Gaussian process with correlation $R(t)=(1+t^2)^{-\alpha/2}$, and define a bounded strictly positive variance by $v_t=v_0+\delta\,\operatorname{sign}(G_t)$, $0<\delta<v_0$. For integrated variance $V_\varepsilon(T)=\int_0^T v_{t/\varepsilon}dt$,

The observable is bounded and the driver is stationary and ergodic, yet the order-$\varepsilon$ Green–Kubo smile does not apply. When the Brownian return is independent of the variance driver, the convexity smile of this example is of order $\varepsilon^\alpha$, not $\varepsilon$.

Ergodic need not mean diffusive. Let $U$ be uniform on the circle, $Y_t=U+t/\varepsilon\pmod{2\pi}$ and $v(y)=v_0+\delta\cos y>0$; then exactly

The correction is order $\varepsilon^2$ with an oscillatory coefficient. The one-sided covariance integral does not converge.

Certificate numbers

The certificate checks the predicted orders on exact benchmarks. The zero-Green–Kubo benchmark has covariance $C_\alpha(t)=\operatorname{Re}(1-it)^{-\alpha}$, with $K=0$ exactly.

For the long-range covariance with a logarithmic modifier, numerical integration of the exact finite-horizon variance gives asymptotic ratios $1.02166$ for $\alpha=0.4$ and $0.94394$ at the critical $\varepsilon\log^2(1/\varepsilon)$ scale. The boundary checks obtain the exact log-corrected ratio $1.000000000000$ at $\alpha=2$.

Scope

The theorems determine the variance scale from either the covariance tail or the low-frequency spectral density, together with whether the Green–Kubo integral vanishes. They do not claim a universal option expansion for every long-memory process or nonsmooth payoff. The finite-state chain used on this site satisfies the stronger assumptions; the counterexamples delimit the generality of the claim rather than contradicting the chain calculation.

The full note also covers the vector and matrix versions with their rank caveats, the proofs, the critical Gaussian endpoint and its power-scale modulus profile, the ergodic non-Gaussian counterexample, the saturated $\varepsilon^2$ boundary with its Cesàro limits, and the remaining certificate checks.

Full note: mixing-scale-full.html. Certificate: verify_long_memory.py.