# How come the deterministic part of Wold decomposition does not violate stationarity?

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.

• 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. Commented Sep 22, 2023 at 8:42
• @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. Commented Sep 22, 2023 at 8:49
• Check this note. Note how the concerned term is constructed. Commented Sep 22, 2023 at 9:06
• @User1865345, thank you, this looks very relevant! And a bit challenging – no bedtime reading... Commented Sep 22, 2023 at 10:37

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".
• 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? Commented Jan 13 at 18:18