The conjecture on a stationary process without a good ARMA approximation

About 2 years old · traced to 2024

Let $(\epsilon_t)$ be uncorrelated white noise random variables, and define the stationary process

X_t=\epsilon_t-\frac{\epsilon_{t-1}}{2}+\frac{\epsilon_{t-2}}{3}+\cdots=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\epsilon_{t-n}.

For integers $M,N\geq 0$ , consider an ARMA model

Y_t=\sum_{n=1}^{N}q_nY_{t-n}+\sum_{n=0}^{M}p_n\epsilon_{t-n},

with $M$ moving-average terms and $N$ autoregressive terms, and the truncation

\overline{Y_t}=\sum_{n=0}^{M+N}\frac{(-1)^n}{n+1}\epsilon_{t-n}.

Conjecture on the absence of a good ARMA approximation. The process $X_t$ does not have a good ARMA model of the displayed form that is better than the simple $M+N$ -term truncation (or moving-average approximation) $\overline{Y_t}$ .

References

Primary source

Anand Ganesh, Babhrubahan Bose and Anand Rajagopalan, “On an L^2 norm for stationary ARMA processes”, arXiv:2408.10610 (2026).

Posted 1 month ago

Nourdine Dehri 1 month ago
The conjecture is underspecified because “better” is undefined. If “better” refers to the $L^2$ approximation criterion introduced in the source paper, then the conjecture is refuted by the stable AR $(1)$ model
$Y_t=\epsilon_t-\tfrac12Y_{t-1}.$

Progress summary

Refreshed 1 month ago

Claimed solved

An unverified counterexample claims that a simple stable model beats the truncation, so the conjecture may be false under mean-square error.

A 2024 paper formulates the conjecture that the logarithmic-coefficient stationary process cannot be approximated better by any finite ARMA model than by its corresponding truncation. It explicitly leaves the claim requiring further justification.

Known results

A rational-approximation theorem applies when $\limsup_{n\to\infty}|a_n|^{1/n}<1$ , but the logarithmic example fails this condition.
The same paper gives a complex-valued example where Padé approximations are not optimal ARMA approximations; the analogous real-valued question remains unproved.

Community submission (unverified; September 3, 2026)

A submitted argument chooses $M=0$ , $N=1$ , $p_0=1$ , and $q_1=-1/2$ , yielding the stable causal AR(1) process $Y_t=\epsilon_t-Y_{t-1}/2$ . It claims that, under the $L^2$ criterion, the geometric tail coefficients $2^{-n}$ give strictly smaller error than the zero tail of the $M+N=1$ truncation, and therefore refute the conjecture.

Current status (as of September 2026): The conjecture has an explicit but unverified $L^2$ counterexample claim; absent independent checking, its truth remains unsettled.

Sources

CounterexampleI found a stable autoregressive model that reduces the mean-square error by 47.9% compared with the proposed truncation. Under this criterion, it provides a counterexample to the conjecture.See full solutionHide full solution

The progress summary appears to report the status of the existing literature, but it does not address the following elementary counterexample under the $L^2$ /mean-square approximation criterion used in the source paper.

Let the white noise have variance $\sigma^2>0$ . Choose

M=0,\qquad N=1,\qquad p_0=1,\qquad q_1=-\frac12,

so that

Y_t=\epsilon_t-\frac12Y_{t-1}.

This is a stable causal AR(1) process, since $|q_1|<1$ , and it has the expansion

Y_t = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^n}\epsilon_{t-n}.

The corresponding $M+N=1$ truncation is

\overline Y_t = \epsilon_t-\frac12\epsilon_{t-1}.

Both approximations match the target coefficients at lags $0$ and $1$ . For every $n\ge2$ ,

0<\frac1{2^n}<\frac1{n+1},

and therefore

\left| \frac{(-1)^n}{n+1} - \frac{(-1)^n}{2^n} \right| < \left| \frac{(-1)^n}{n+1}-0 \right|.

Because the innovations are uncorrelated,

E|X_t-Y_t|^2 = \sigma^2 \sum_{n=2}^{\infty} \left( \frac1{n+1}-\frac1{2^n} \right)^2 < \sigma^2 \sum_{n=2}^{\infty} \frac1{(n+1)^2} = E|X_t-\overline Y_t|^2.

More explicitly,

E|X_t-Y_t|^2 = \sigma^2 \left( \frac{\pi^2}{6}+\frac43-4\log2 \right) \approx0.205679\,\sigma^2,

whereas

E|X_t-\overline Y_t|^2 = \sigma^2 \left( \frac{\pi^2}{6}-\frac54 \right) \approx0.394934\,\sigma^2.

Thus, under the $L^2$ criterion, this stable AR(1) model strictly outperforms the corresponding truncation and provides a counterexample to the conjecture.

If “better” is intended to mean something other than $L^2$ /mean-square approximation, that criterion and any additional admissibility conditions need to be stated explicitly. Otherwise, could the status be updated or could you indicate which condition excludes this AR(1) example?