More about the deterministic part of Wold decomposition

Question

This is a follow-up on this question of mine. 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: $$ Y_t = \mu_t + \sum_{j=0}^\infty b_j\varepsilon_{t-j} $$ where $\{\mu_t\}$ is the deterministic one. According to Hansen "Econometrics" (2022) p. 472, $\mu_t$ is the projection of $Y_t$ on the history of infinite past: $\mu_t=\lim_{m\rightarrow\infty} \mathcal{P}_{t-m}(Y_t)$. For example, it could be $$ \mu_t = \left\{ \matrix{ (-1)^t \ \ \ \ \text{with probability 1/2} \\ (-1)^{t+1} \ \text{with probability 1/2} } \right\}. $$ Now if we observe only a single path of the stochastic process, we only face one of the two cases, either $\mu_t=(-1)^{t}$ or $\mu_t=(-1)^{(t+1)}$. Without additional information, we would probably just model that component as deterministic, thinking that $\{Y_t\}$ is not stationary but $\{Y_t-\mu_t\}$ is. I do not think this would cause any trouble for modeling and predicting the future elements in that path, even though it would mess up inference about the stochastic process itself and predictions of its other paths.

Thinking back about my original question (linked at the very beginning), I can say, if only all instances of Wold's decomposition were this nice! Are they, perhaps? Or does there exist a case in which $\{\mu_t\}$ is not completely deterministic within a single path?

Answer 1

$\mu_t$ is unfortunately called "deterministic", because its defining property is that it is "completely determined" by past values of the process, it is a "deterministic function" of the past values of the process:

$$\mu_t = h(Y_{t-1}, Y_{t-2},...)$$

But this means that $\mu_t$ is a stochastic process, and one that it must be covariance-stationary, otherwise it could not be a component of a stationary process (given that its other component in the Wold decomposition is stationary). It can be a degenerate process, a constant, which is trivially stationary. The mention that existed in the wikipedia article

$\mu_t$ is a deterministic time series, such as one represented by a sine wave.

was totally misleading, because it conveyed the sense that $\mu_t$ can be any process (so not necessarily covariance-stationary). But this is not correct. I write "was" because I just corrected the wikipedia entry.

The example given in Sargent (1987) Macroeconomic Theory (2nd ed) p. 289 is illuminating, because it does formulate $\mu_t$ using $\cos$ and $\sin$, but in a way that the process is covariance stationary.

Sargent models $$\mu_t = \sum_{j=1}^{m} [a_j \cos(\lambda_jt) + b_j \sin(\lambda_jt)]$$

where $\lambda_j \in [-\pi, \pi]$, while $a_j, b_j$ are random variables (that do not change with $t$), with

$$E(a_j) = E(b_j) = E(a_jb_j) = E(a_ja_k) = E(b_jb_k)= 0 ,\quad j\neq k,$$

$$E(a_j^2) = E(b_j^2) =\sigma_j^2.$$

Obviously, $E(\mu_t) = 0$. As for the covariance, we have (setting $m=1$ for simplicity, it does not affect the result),

\begin{align} {\rm Cov}(\mu_t, \mu_{t-\ell}) &= E\Big\{\big[a_1 \cos(\lambda_1t) + b_1 \sin(\lambda_1t)\big]\cdot \big[a_1 \cos[\lambda_1(t-\ell)] + b_1 \sin[\lambda_1(t-\ell)]\big]\Big\}\\ &=\sigma^2_1 \cdot\Big[\cos(\lambda_1t)\cdot \cos[\lambda_1(t-\ell)] + \sin(\lambda_1t)\cdot \sin[\lambda_1(t-\ell)]\Big] \end{align}

Then we invoke the trigonometric identity

$$\cos\gamma \cdot \cos \delta + \sin \gamma \cdot \sin \delta = \cos(\gamma-\delta)$$

to get

\begin{align} {\rm Cov}(\mu_t, \mu_{t-\ell}) &=\sigma^2_1 \cdot \Big[ \cos\big[\lambda_1t - \lambda_1(t-\ell)\big]\Big]\\ \\ &= \sigma^2_1 \cdot \cos\left[\lambda_1\ell\right] \end{align}

So the covariance of the process $\mu_t$ does not depend on the index $t$, but only on the distance $\ell$ between two instances of the process, and we have covariance stationarity of $\mu_t$.

We see that while we can use "sine and cosine" to model $\mu_t$, at the same time the whole setup must be such that it preserves covariance-stationarity of $\mu_t$: All the properties of the $a_j, b_j$ random variables are needed to arrive at the covariance-stationarity result. For example, if we had $E(a_1^2) \neq E(b_1^2)$ we would not arrive at the trigonometric identity which allowed us to eliminate $t$ from the covariance function.