I don't understand the following definition of a moving average model from Hyndman 2021, Forecasting: principles and practice

A moving average model uses past forecast errors in a regression-like model, $$ y_t = c + \varepsilon_t + \theta_1\varepsilon_{t-1} + \theta_2\varepsilon_{t-2} + \dots + \theta_q\varepsilon_{t-q} $$ where $\varepsilon_t$ is white noise.

Wikipedia makes a similar statement

... and the $\varepsilon _{t},\varepsilon _{t-1},...,\varepsilon _{t-q}$ are white noise error terms.

But if $\varepsilon_t$ is white noise, then how can anything be predicted from it? What should come out from a linear regression model, that one tries to fit to white noise?

Brownlee 2017 on the other hand writes

The residual errors from forecasts on a time series provide another source of information that we can model. Residual errors themselves form a time series that can have temporal structure. A simple autoregression model of this structure can be used to predict the forecast error, which in turn can be used to correct forecasts.

That I can understand. But it implies, that the errors are not white noise. Otherwise there would be no structure to model. So is the definition of Hyndman and Wikipedia wrong or misleading? Is a moving average model fitted to white noise or must some structure exist in the errors so fitting the moving average model makes sense?

  • $\begingroup$ The MA model is "fitted" to $y_t$, not the $\epsilon$, as, in the MA model, $y_t$ is thought to be a linear combination of current and past $\epsilon$. That said, you are indeed right that the practical issues in fitting MA models are a bit intricate as the $\epsilon$ are not observed so that certain tricks are applied. $\endgroup$ Aug 5, 2022 at 11:16
  • $\begingroup$ Brownlee first calculates the residual errors $\varepsilon$ as the residual errors of the naive/persistance model. So I'm not sure what you mean by the $\varepsilon$ are not observed. They are observed as residual errors of the naive model. Then he fits an autoregression model using these residual errors $\varepsilon$ to compute the parameters $\theta$ of the moving average model, He is then using this moving average model (better name: autoregression of residual errors) to correct the predictions of the naive model. That seems to make sense to me. There seem to be no tricks applied here. $\endgroup$
    – asmaier
    Aug 5, 2022 at 12:07
  • $\begingroup$ Yes, you may very well not call it a trick, if you prefer. Of course, at the end, it is a well-defined algorithm. I just wasn't going to spell out details of estimating MA models in my comment. $\endgroup$ Aug 5, 2022 at 12:09
  • $\begingroup$ @asmaier: as you correctly stated, white noise cannot be predicted but that is not to say that linear combinations of past white noise ( which are really past error terms ) cannot be used to predict the current response. Essentially, the response is assumed to react to past errors. Whether it works or not in practice, is a different issue. $\endgroup$
    – mlofton
    Aug 5, 2022 at 15:02

1 Answer 1


Short answer: Yes, the model is in your described first described setup to white noise and the confusion you are facing probably originates from mixed notation across your sources.

An easy way to gain some intuition is to "split" the observed error from your average component $c$ into two different types: A (theoretically, as the occurrence lays in the past) explainable error component $e_t$ and non-explainable, white noise component $\varepsilon_t$. Note that we can rewrite your first equation as: \begin{equation} y_t=c+e_t+\varepsilon_t, \end{equation} with $e_t = \theta_1\varepsilon_{t-1}+\theta_2\varepsilon_{t-2}+...+\theta_k\varepsilon_{t-k}$. In order to estimate these kind of models in an appropriate way, one either relies on yule-walker equations or (under some circumstances feasible) reformulations into an $AR(\infty)$-representation. You may find details here. The idea of simply fitting an $AR(k)$-model to the residuals looks conceptually similar, the key difference is however that the difference equation for $e_t$ is formulated based on true, unobservable errors and not based on estimates for this error.


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