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Wold's representation theorem states that every covariance-stationary time series $\{Y_t\}$ can be written as the sum of two time series, one deterministic and one stochastic: $$ Y_t=\sum_{j=0}^\infty b_j\varepsilon_{t-j}+\eta_t $$ where, among other things, $\{\eta_t\}$ is a deterministic time series, such as one represented by a sine wave.

I am puzzled by the inclusion of a time-varying deterministic part. Does it not violate the definition of stationarity? Since the expected values of $\varepsilon$s are all zero, $\eta_t$ gives the expected value of $Y_t$. If $\eta_t$ varies over time (such as a sine wave does), the expected value of $Y_t$ varies over time, thus violating stationarity. What am I missing?

Update: A follow-up question.

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  • $\begingroup$ Richard, I am leaving a comment for I have not touched the topic for quite some time. But doesn't deterministic mean perfectly predictable? Mainly the term is not a time trend and that means it shouldn't violate the stationarity assumption. $\endgroup$ Commented Sep 22, 2023 at 8:42
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    $\begingroup$ @User1865345, if $\mathbb{E}(X_t)\neq\mathbb{E}(X_s)$ for some $s\neq t$, that looks like nonstationarity to me. By definition, $\mathbb{E}(X_t)=\mathbb{E}(X_s)$ for all $s,t$ if the process $\{X_t\}$ is stationary. $\endgroup$ Commented Sep 22, 2023 at 8:49
  • $\begingroup$ Check this note. Note how the concerned term is constructed. $\endgroup$ Commented Sep 22, 2023 at 9:06
  • $\begingroup$ @User1865345, thank you, this looks very relevant! And a bit challenging – no bedtime reading... $\endgroup$ Commented Sep 22, 2023 at 10:37

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The component $\eta_t$ is not "a" "deterministic" time series. It will be a time series such that it will be covariance-stationary (at least) itself.

Here the terminology "deterministic" (a terminology issue, once more), should better be "conditionally deterministic", because what it aims to convey is that $\eta_t$ is perfectly predictable given the past values of the whole process. It does not mean "deterministic" as in "a deterministic rule/function to generate it".

While a covariance stationary process could conceivably be broken into two non-stationary ones, where together the one would "cancel out" the non-stationarity of the other, it could not be decomposed into one stationary and one non-stationary component.

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  • $\begingroup$ Thank you. So $\mathbb{E}(\eta_t)$ is constant over time, right? I am still trying to get a better feeling of this "deterministic" term. I wish it were as simple as in my follow up question; is it perhaps? $\endgroup$ Commented Jan 13 at 18:18
  • $\begingroup$ @RichardHardy Yes. As is the variance. The use of the term deterministic is unfortunate here. $\endgroup$ Commented Jan 13 at 20:27
  • $\begingroup$ @RichardHardy I posted also an answer to the follow-up question. $\endgroup$ Commented Jan 13 at 21:59

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