1
$\begingroup$

In the textbook Forecasting: principles and practice by Hyndman and Athana­sopou­los, in the Classical Decomposition (Sec 6.3), in step 3 of the additive decomposition algorithm, the authors state that the seasonal indexes have to be adjusted to ensure that they add to zero. Why is that?

Any help would be appreciated.

$\endgroup$
2
$\begingroup$

Just to make coefficients meaningful. Suppose you had quarterly data and a pattern within the year. If you include a year effect the seasonal coefficients can be interpreted as average deviations from the year average, and as such should add up to zero. It is just the same reason why ANOVA effects are made to add up to zero as well.

$\endgroup$
  • $\begingroup$ For the same reason, the average of multiplicative seasonality factors has to be 1. $\endgroup$ – zbicyclist Feb 23 '15 at 4:38
  • $\begingroup$ F.Tusell, I'm new to this, but what's an effect? Could you give a small example of what you're trying to say? $\endgroup$ – An old man in the sea. Feb 23 '15 at 15:29
  • 1
    $\begingroup$ Suppose you investigate the yield of four different seeds of corn. The yield $y_{ij}$of the $j$ plot of corn sown with the $i$ kind of corn is modelled as: $y_{ij} = \alpha + \beta_i + \epsilon_{ij}$. The coefficients $\beta_i$ ($i=1,\ldots,4$) measure the effect of the different kinds of corn, and $\alpha$ the average yield. Clearly, it doesn't make sense to have all $\beta_i$ positive, for then we would subtract part of their value and include it in $\alpha$: rather, it makes sense to impose that they add up to zero. Same with the seasonal coefficients. $\endgroup$ – F. Tusell Feb 23 '15 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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