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Reflecting an inside-unit-circle zero across the unit circle produces an invertible moving-average model with the same spectral density of a stationary process, after rescaling the driving white noise variance. For real $|\theta |>1$, $|1-\theta e^{-i\lambda }{|}^2={\theta }^2|1-{\theta }^{-1}{e}^{-i\lambda }{|}^2$. The transformed driving sequence is a linear innovation process; without Gaussianity it need not be strong white noise.

Among the supplied candidates, choose ARMA(2,1). It has the smallest Akaike information criterion. Relative to ARMA(2,2), its AIC improvement is $1.71$; the additional second moving-average estimate is only half a standard error from zero and increases the log likelihood function by only about $0.14$. Dropping it is a reasonable parsimony choice, although that AIC gap is small. ARMA(1,1) has AIC larger by $17.55$, a much clearer loss of fit. In the chosen model the second autoregressive term is about seven standard errors from zero, so it should not be dropped merely to obtain order one. The first autoregressive term is less precisely estimated; this does not justify automatically deleting it without fitting and comparing the reduced candidate.

Using the usual positive-sign moving-average convention, the fitted autoregressive moving-average model isThese are plug-in values rounded as given, not exact population parameters. The model has zero mean. Its polynomials are $\varphi (z)=1+0.30z-0.28z^2=(1-0.4z)(1+0.7z)$ and $\theta (z)=1+0.5z$.

With angular frequency $\omega \in [-\pi ,\pi ]$, so that the autocovariance is $\gamma (h)={\int }_{-\pi }^{\pi }{e}^{ih\omega }{f}_X(\omega )d\omega$, the spectral density of a stationary process isThe frequency convention makes the normalization unambiguous.

Use cycles-per-time spectral density, with frequency $\omega \in [-1/2,1/2]$. The spectral representation theorem for a stationary time series gives the centered white noise representationwhere disjoint increments are orthogonal. Put ${\theta }_0=1$. Since the filter is finite, substitute each noise representation and interchange the finite sum with the integral:The new orthogonal increment measure is $d{Z}_X=\Theta (e^{-2\pi i\omega })d{Z}_{\epsilon }$. Its variance measure is therefore ${\sigma }^2|\Theta (e^{-2\pi i\omega }){|}^{2}d\omega$. The coefficients are real, so conjugation changes the sign of the exponent without changing the modulus. HenceWith angular frequency $\lambda =2\pi \omega$, the spectral density of a stationary process instead contains the factor $1/(2\pi )$. The convention explains its absence here. Invertibility is not needed for this finite-filter spectral calculation.

For the spectral questions take the unique weakly stationary process solving the equation, as is customary for a stable autoregressive moving-average model. The equation alone also admits nonstationary solutions differing by $C{4}^{-t}$, for which a spectral density of a stationary process need not exist. Let $B$ be the backshift operator. Since the autoregressive root is $4$, the stationary solution is causal andUsing the convention $\gamma (h)={\int }_{-\pi }^{\pi }{e}^{ih\lambda }f(\lambda )d\lambda$, the spectral density of a stationary process isThe stationary mean is zero, since $\mu -\mu /4=0$.

With angular frequency $\lambda$, the periodogram isAt Fourier frequencies $2\pi j/T$ it is the squared modulus of the normalized discrete Fourier transform. If a nonzero mean is unknown, subtract the sample mean first.

For a zero-mean stationary process, expanding the square givesUnder absolute summability of the covariances this converges to the spectral density of a stationary process. Thus the periodogram is generally biased at finite $T$, but asymptotically unbiased under this short-memory time series condition.

It is nevertheless not a pointwise estimator with statistical consistency unless it is smoothed. For Gaussian white noise, at a nonzero Fourier frequency other than the Nyquist frequency, the real and imaginary Fourier components are independent normal variables. Exactly,where $f={\sigma }^2/(2\pi )$. The variance does not decrease with $T$. Under usual short-memory time series assumptions the same exponential limit is asymptotic for general processes. Averaging nearby frequencies or using a lag-window estimator reduces variance; a frequency bandwidth tending to zero while $T$ times that bandwidth tends to infinity can give statistical consistency. The raw plot remains useful for detecting strong periodic peaks, but increasing the record length alone does not remove its pointwise noise.

For a centered record of length $T$, the periodogram is the squared modulus of its discrete Fourier transform, with normalizationUnder suitable short-memory assumptions its expectation approaches the spectral density of a stationary process, but its variance generally does not vanish. Smoothing nearby frequencies can provide an estimate with statistical consistency.

In a second-order sense a weakly stationary process has short memory when its autocovariance is absolutely summable. Then its spectral density of a stationary process is continuous and its sample-mean variance has the usual finite long-run variance of a stationary process. A central limit theorem still needs additional conditions.