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?

  • $\begingroup$ Hi Richard: I'm not certain about your example above but the last part of the christiano document explains why the deterministic part cannot be a time trend and why his specific example works. Also, there's a 23 page proof of the wold decomposition theorem by Herman Bierens. I read it once like 7 years ago. All I can say and remember is that it's more understandable than the informal proof that christiano ( really sargent ) provides. If you can't find it and you're interested, let me know. I know it's somewhere on the net but my eyes are closing. $\endgroup$
    – mlofton
    Commented Jan 9 at 6:38
  • $\begingroup$ @mlofton, thank you. I found a 26 page document on a site that won't let me download it without registering and earning some points on that site. I have not looked very hard elsewhere, but the document does not seem to be floating anywhere right on the surface (except for the case above). I sent prof. Bierens an e-mail, perhaps he will share the document with me... $\endgroup$ Commented Jan 9 at 8:26
  • $\begingroup$ Hi Richard: I remember it not being easy to find but that must be it. He's a very generous and kind person. I think he's Emeritus now but I bet he'll respond. If he doesn't, let me know and I'll give a search a shot. and you're welcome. It's the only proof I know that doesn't skip steps. $\endgroup$
    – mlofton
    Commented Jan 9 at 16:07
  • $\begingroup$ @mlofton, already got his answer with the document attached. Hats off to the gentleman! $\endgroup$ Commented Jan 9 at 21:03
  • 1
    $\begingroup$ glad to hear it. way back when, I asked him some questions about that document and he answered them one by one in a latex document with detailed notation-explanations etc. I was shocked. $\endgroup$
    – mlofton
    Commented Jan 9 at 21:56

1 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.

  • $\begingroup$ Thank you for your answer! It seems to address a somewhat different point than I am asking about. I wonder what you think about the more specific question I have in this post. Even so, I found your answer stimulating and I think I might have understood the gist of the matter. (I may still find out I have not, though.) Also, I think you probably need a limit in $\mu_t=h(Y_{t-1}, Y_{t-2},\dots)$ so that the past is not recent but distant. $\endgroup$ Commented Jan 14 at 8:41
  • $\begingroup$ My current understanding of $\mu_t$ is that it is predetermined within a path, for every future time point in the path. In other words, it is perfectly predictable within the path, even though each path forks out in every time instance. It stays the same in each branch stemming from the same origin in the infinite past. Thus when we observe a single path, we know the pattern of $\mu_t$ will continue deterministically in every branch of this process in the future. $\endgroup$ Commented Jan 14 at 8:45
  • $\begingroup$ @RichardHardy I got rather confused as I read presentations of Wold decomposition in various sources. So I found Wold's original book, and I will post a question on the matter, providing the source material and asking for explanations on how it evolved into the modern presentations. $\endgroup$ Commented Jan 14 at 13:45
  • $\begingroup$ @RichardHardy This is the best I can do for the topic, stats.stackexchange.com/q/636863/28746 $\endgroup$ Commented Jan 14 at 23:48

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