I have been generate the ARIMA using R software and getting the result, but I don't know how to write the equation.
How should I do? Thank you.
$\begingroup$
$\endgroup$
1
-
$\begingroup$ post your data ..... and I will try and help you . $\endgroup$– IrishStatCommented Apr 16, 2019 at 13:58
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
You have an ARIMA(3,2,3) model. This means that the second order differences (that is the middle "2") follow an ARMA(3,3) process.
We use the backshift operator $B$,
$$ By_t := y_t-y_{t-1}. $$
Here, you need the second order backshift,
$$ B^2y_t = B(y_t-y_{t-1}) = By_t-By_{t-1} = (y_t-y_{t-1}) - (y_{t-1}-y_{t-2}) = y_t-2y_{t-1}+y_{t-2}. $$
R's convention for ARIMA models mean that your model is
$$ B^2y_t = \phi_1B^2y_{t-1} + \phi_2B^2y_{t-2} + \phi_3B^2y_{t-3} + e_t + \theta_1e_{t-1} + \theta_2e_{t-2} + \theta_3e_{t-3}, $$
where the $\phi_i$ are your ar coefficients, $\theta_i$ are your ma coefficients, and $e_t\sim N(0,\sigma^2)$, where you can read the estimate for $\sigma^2$ off in your output.
Yes, writing all this out and collection all the $y_{t-i}$ is tedious.
Similar questions and links can be found here and here and here.
answered Apr 16, 2019 at 13:58
-
$\begingroup$ Is that normal for the sigma^2 is so big? $\endgroup$ Commented Apr 16, 2019 at 14:13
-
$\begingroup$ It depends on the scale of your time series. If you are modeling rare sales, then it is high. If you are modeling the population of a city, it makes sense. Compare it to the raw variance of your data, which should yield a larger value. $\endgroup$ Commented Apr 16, 2019 at 14:16