I am not able to understand what the error/deviation/stochastic terms in moving average model stand for? What is the practical significance of the error term. Is the error term difference between the consecutive values in the series. Or is it the difference between the forecasted values and the observed values.
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$\begingroup$ Error terms in moving average model are unobservable in real life. Usually it is taken to be the difference between predicted value and true value. $\endgroup$– IdonknowCommented Dec 24, 2019 at 12:36
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1$\begingroup$ You can refer to this post for more information stats.stackexchange.com/a/74826/99818 $\endgroup$– IdonknowCommented Dec 24, 2019 at 12:38
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$\begingroup$ But we don't have any predicted value first of all. In the link: stats.stackexchange.com/a/74826/99818, I understand how mathematically it is being done. But what is the significance of doing this? And applying regression over unobservables?? $\endgroup$– FreemnCommented Dec 24, 2019 at 12:47
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$\begingroup$ Related thread: Understanding error term in AR model. $\endgroup$– Richard HardyCommented Dec 30, 2019 at 13:10
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1 Answer
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The error terms is neither the difference between the consecutive values nor the difference between the forecasted values and the observed values, though the latter is a somewhat close guess.
A moving-average model of order $q$, MA($q$), is $$ x_t=\varepsilon_t+\theta_1\varepsilon_{t-1}+\dots+\theta_q\varepsilon_{t-q}. $$ Its conditional mean, conditioning on information up to time $t-1$, $I_{t-1}$, is $$ \mathbb{E}(x_t|I_{t-1}) = \theta_1\varepsilon_{t-1}+\dots+\theta_q\varepsilon_{t-q} $$ and this is just $=x_t-\varepsilon_t$. Hence, $\varepsilon_t$ is the difference between $x_t$ and its conditional mean.
The difference between two consecutive values is \begin{aligned} x_t-x_{t-1} &= (\varepsilon_t+\theta_1\varepsilon_{t-1}+\dots+\theta_q\varepsilon_{t-q}) - (\varepsilon_{t-1}+\theta_1\varepsilon_{t-2}+\dots+\theta_q\varepsilon_{t-q-1}) \\ &= (\varepsilon_t-\varepsilon_{t-1})+\theta_1(\varepsilon_{t-1}-\varepsilon_{t-2})+\dots+\theta_q(\varepsilon_{t-q}-\varepsilon_{t-q-1}) \\ &= \varepsilon_t+(\theta_1-1)\varepsilon_{t-1}+(\theta_2-\theta_1)\varepsilon_{t-2}+\dots+(\theta_q-\theta_{q-1})\varepsilon_{t-q}-\theta_{q}\varepsilon_{t-q-1} \\ &\neq\varepsilon_t. \end{aligned}
The difference between a forecasted value and an observed value depends on the forecast. For example, the conditional mean is the optimal forecast under square loss. The conditional mean $\theta_1\varepsilon_{t-1}+\dots+\theta_q\varepsilon_{t-q}$ is unknown but can be estimated by $\hat\theta_1\hat\varepsilon_{t-1}+\dots+\hat\theta_q\hat\varepsilon_{t-q}$ where hats denote estimates of the true quantities. So when forecasting, the difference between an observed value $x_t$ and a forecasted value $\hat{x}_t$ is \begin{aligned} x_t-\hat{x}_t &= (\varepsilon_t+\theta_1\varepsilon_{t-1}+\dots+\theta_q\varepsilon_{t-q}) - (\hat\theta_1\hat\varepsilon_{t-1}+\dots+\hat\theta_q\hat\varepsilon_{t-q}) \\ &\neq\varepsilon_t. \end{aligned} If only the true values were known, the forecast error would coincide with the error term $\varepsilon_t$.
answered Dec 25, 2019 at 9:23
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$\begingroup$ In the final equation, on the l.h.s., I think it should be xt−xt^ instead of xt−xt-1. (pardon the formatting.) $\endgroup$– FreemnCommented Dec 30, 2019 at 6:44
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$\begingroup$ @Freemn, right, now corrected. $\endgroup$ Commented Dec 30, 2019 at 9:38