I read Brockwell and Davis(2016), Shumway and Stoffer(2016), and Stoica and Moses(2004). However, none of them laid out clearly the reasoning behind the presumption of stationarity when conducting spectral analysis on a time series.

I understand we typically need to detrend the time series as necessary data pre-processing for spectral analysis, because otherwise the first cosine coefficient could distort the estimates/periodogram.(frequency close to zero could have very large spectrum).

But simple detrending doesn't make the process stationary. It may well have periodicity/seasonality built in, although it may not have a unit root (unit root stationarity tests are rendered useless in this case).

Based on above, how could we carry out a spectral analysis on a periodic time series which is not stationary when spectral analysis is only defined for stationary processes?


To quote Brockwell and Davis: "[t]he summability of $|\gamma(\cdot)|$ implies that the series converges absolutely..."

When you look at the definition of the spectral density $$ f(\lambda) = \frac{1}{2\pi} \sum_{h=-\infty}^{\infty} e^{-ih\lambda} \gamma(h), \hspace{10mm} - \infty < \lambda < \infty, $$ the sum would be undefined if $\gamma(\cdot)$ was the autocovariance function of a periodic process. Intuitively periodic processes' autocovariance functions don't "die out."

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  • $\begingroup$ thanks very much for pointing it out; could you also take a look at my question in last paragraph? Shumway and Stoffer are actually using periodic process in an example to illustrate the idea of periodogram. $\endgroup$ – stucash Jun 10 '19 at 8:01
  • $\begingroup$ @stucash that isn’t an infinite sum, so there’s no risk of undefinedness. Also, it’s jusy like running a regression if you prefer to think of it like that. $\endgroup$ – Taylor Jun 10 '19 at 15:22
  • $\begingroup$ thanks a lot that’s exactly what I missed in my head, although it’s actually very obvious! $\endgroup$ – stucash Jun 10 '19 at 15:29

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