3
$\begingroup$

So on wikipedia here under Examples, it is mentioned that ARIMA(0,2,2) is given by:

$$ X_t = 2X_{t-1} - X_{t-2} + (\alpha + \beta -2)\epsilon_{t-1} + (1 - \alpha)\epsilon_{t-2} + \epsilon_t \ \ \ \ \ (1) $$

My question is: how this equation has been derived?

We know that ARIMA(0,2,2) means $d=2$ (second order differencing) and $q=2$ is the MA (moving-average) order.

If $MA(q)$ is given by: $$ X_t = \mu + \epsilon_t + \theta_1\epsilon_{t-1} + ... + \theta_q\epsilon_{t-q} $$

therefore (?):

$$ X_t = \mu + \epsilon_t + \theta_1\epsilon_{t-1} + \theta_2\epsilon_{t-2} \\ X_{t-1} = \mu + \epsilon_{t-1} + \theta_1\epsilon_{t-2} + \theta_2\epsilon_{t-3} \\ X_{t-2} = \mu + \epsilon_{t-2} + \theta_1\epsilon_{t-3} + \theta_2\epsilon_{t-4} $$

After removing $\epsilon_{t-3}, \epsilon_{t-3}$, for the second order differencing we compute $X_t - 2X_{t-1} + X_{t-2}$:

$$ X_t - 2X_{t-1} + X_{t-2} = (\theta_1 - 2)\epsilon_{t-1} + (\theta_2 - 2\theta_1 + 1)\epsilon_{t-2} + \epsilon_t $$

and the coefficients of the error terms differ from those in (1).

$\endgroup$

1 Answer 1

0
$\begingroup$

Ok I realised that we just need to substitute: $$ \alpha = 2\theta_1 - \theta_2 \\ \beta = -\theta_1 + \theta_2 $$ although I do not quite understand what was the reasoning behind this particular substitution.

$\endgroup$
2
  • $\begingroup$ this is my answer for a similar question $\endgroup$
    – Taylor
    Jul 21, 2019 at 17:04
  • 1
    $\begingroup$ @Tomasz Bartkowiak: it is a weird notation but it comes from the fact that the arima(0,2,2) is equivalent to a local level model which is a structural time series model. If you look that up, you'll probably see why they wrote it in that strange way initially. They should have explained why. You can look at Andrew Harvey's blue covered text ( I forget the title ) for what the local level model is. Also, if you're not interested in structural time series, don't worry about it and don't let it confuse you. Definitely a confusing move by the author of original equation. $\endgroup$
    – mlofton
    Jul 21, 2019 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.