In time series, Moving Average model $MA(q)$ is defined by $$X_t = \mu + \epsilon_t + \theta_1 \epsilon_{t-1} + \theta_2\epsilon_{t-2} + ... + \theta_q \epsilon_q$$ where $\mu$ is the mean of the process $\{X_t\}_t$ and $\epsilon_t$ is a white noise, that is, uncorrelated with zero mean and fixed variance for all t.$

For simplicity, assume that $\mu = 0.$ From equation above, $MA(q)$ model forecasts future using current and past random shocks (or stochastic terms). Does it mean that we can apply anything to forecasting in $MA(q)$ as long as it is white noise?

I get confused because it seems like the model is not forecasting future $X_t$ but rather forecasting future random shock $\epsilon_{t+1}.$


Why do you get the impression that the future random shocks $\epsilon_{t+1}$ is predicted? That would not make any sense, since in the model the series of random shocks $\dotsc,\epsilon_t, \epsilon_{t+1},\dotsc$ is supposed to represent white noise, which by definition can't be (usefully) forecast, since it is independent of the past.

So, what is forecast in this model is the future observation $X_{t+1}$. For more information about MA models see Moving-average model error terms.

  • $\begingroup$ Well, yes, that will be the best possible forecast of a white noise ... $\endgroup$ Jul 23 '19 at 12:59
  • $\begingroup$ @kjetilbhalvorsen I have such impression because MA model uses current and past random shock data. This is what gives me the impression. On the other hand, if one were to fit data using MA model, how would the calculation goes? As far as I know, we plot ACF and PACF of a time series. From there, we identify parameter $q$ in MA model. Then I do not know how to proceed. Should we calculate $\theta_1,...,\theta_q$ next? I guess the residuals $\epsilon_t, \epsilon_{t-1},...,\epsilon_{t-q}$ are computed the last. $\endgroup$
    – Idonknow
    Jul 23 '19 at 13:30
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    $\begingroup$ Sorry. I started writing an answer and thought I deleted but didn't. For MA(q) models, one assumes that the future $\epsilon_t$ are zero. So, you are predicting $X_t$ but in order to do that, you use the expectation of the future $\epsilon_t$ in your forecast. $\endgroup$
    – mlofton
    Jul 24 '19 at 16:52

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