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I tried my data into an $ARMA$ model which is turned out to be $ARMA(2,3)$. I want to extract the model from the parameters I got in the pic (without any transformation). Is it right if I write the model like this?

$Y_t = 0.6218 + 0.126Y_{t-1} + 0.3964Y_{t-2} + e_t + 0.1488e_{t-1} + 0.9569e_{t-2} - 0.1057e_{t-3}$

Furthermore, how do I get $e_t$ (error of prediction on period-t) since I haven't got the prediction value yet.. I mean, $Y_t$ is something I want to predict, right?

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  • $\begingroup$ I think your equation is correct. Furthermore, I think what you would do for a calculation you would fit your data to your model till the last known value and then you get the error for the forecast. See also otexts.org/fpp/8/8 . However you can use forecast() in order to come up with a forecast. $\endgroup$ – burton030 Aug 1 '18 at 7:26
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You will need to flip the signs on the MA coefficients. Here is the help page for arima():

Different definitions of ARMA models have different signs for the AR and/or MA coefficients. The definition used here has

 X[t] = a[1]X[t-1] + ... + a[p]X[t-p] + e[t] + b[1]e[t-1] + ... + b[q]e[t-q]

and so the MA coefficients differ in sign from those of S-PLUS.


In forecasting, you will plug the last few residuals into $e_{t-3}$, $e_{t-2}$ and $e_{t-1}$. Get these using residuals(). For $e_t$, as you write correctly, we don't know this yet. But we assume normally distributed errors. So in forecasting, just plug in the expectation, which is zero.

Of course, if you are not interested in point forecasting, but in predictive densities or s, it makes sense to simulate and draw $e_t~\sim N(0,\widehat{\sigma}^2)$, with $\widehat{\sigma}^2=0.003064$ as per your output.

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