4
$\begingroup$

I am trying to figure out whether from the following graph of the OLS residuals that the linear relationship does not hold, and that probably a cubic relationship would do better? Since both in the beginning as in the middle the variance is larger, and two bends in a regression could lower the variance for these areas. Or is that something that is not determinable from this graph?

enter image description here

[edit after comments]

The mean structure does suggest a quadratic term, and possibly a cubic term, however this last one seems small.

enter image description here

$\endgroup$
6
  • $\begingroup$ Why not just test a cubic model and see if the quadratic and cubic terms are large and/or significant? $\endgroup$
    – Peter Flom
    May 10, 2013 at 19:40
  • $\begingroup$ Thanks, I did test them, and they are significant, but not large. Just wanted to know if it is ok to justify the fact I added them in the first place, because the variance structure suggests this. $\endgroup$
    – Marloes
    May 10, 2013 at 20:00
  • $\begingroup$ I would read this plot as showing the variance of the residuals does not appreciably change over time. As a rule of thumb, variances of small- to medium-sized datasets (such as are used in the local windows in the Loess algorithm) can be expected to vary randomly by a factor of two to three up and down. Although the fall from 5 down to 2 during the first half hour looks almost significant, I suspect it is a characteristic "edge effect" from the Loess smooth coupled with sparse data during the first 10 minutes or so. This variance structure does not suggest adding more terms to the model. $\endgroup$
    – whuber
    May 10, 2013 at 20:02
  • $\begingroup$ Thanks. I do have theoretical grounds to add a quadratic term though. And this also shows up in the mean structure (see edit to question). But I probably won't be adding the cubic term. $\endgroup$
    – Marloes
    May 10, 2013 at 20:18
  • 1
    $\begingroup$ The mean structure does not suggest either a cubic or quadratic term: they could not accurately reproduce what is shown. There are several ways it could be handled: allow the slope to change discontinuously at an unknown point (that adds two parameters) or add a (decaying) exponential term, which also needs two parameters. If you want to anticipate the outcome, eyeball the break (it's around $3/4$) and introduce a variable equal to time before the break and zero after the break. Although this doesn't account for uncertainty in estimating the break, it will show you a better fit. $\endgroup$
    – whuber
    May 10, 2013 at 22:00

0

Your Answer

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

Browse other questions tagged or ask your own question.