1
$\begingroup$

Based on a few sources I came upon, we transform the dependent variable so that the residual sum of squares (SSE) does not depend on $\lambda$ (the new parameter). I find this very confusing, because isn't the whole purpose of this procedure to find the value of $\lambda$ that minimizes the SSE? so what exactly does "depend" mean here? Sorry if my question is unclear, I feel like this is a simple problem because everywhere I've looked, it's always the same one-line explanation.

$\endgroup$
  • 2
    $\begingroup$ The point is that the SSE for the different $\lambda$ values must be made comparable. They will not be so without some action, because $y^{\lambda}$, the transformed values for different $\lambda$ values, do not have the same unit of measurement! $\endgroup$ – kjetil b halvorsen Jun 13 '17 at 21:46
  • $\begingroup$ ok that makes more sense, I guess the sources all meant the SSE's unit does not depend on $\lambda$ $\endgroup$ – lucusk Jun 13 '17 at 21:50
  • 1
    $\begingroup$ This is neither a simple issue nor can it be addressed in a single line. That's because (a) transformations are decidedly not intended to minimize sums of squared errors and (b) they have multiple purposes which need to be balanced and placed in proper perspective. I believe I have addressed your question in a reply at stats.stackexchange.com/a/60455/919 (which lists six different purposes of a Box-Cox transformation). $\endgroup$ – whuber Jun 13 '17 at 21:51
  • 1
    $\begingroup$ BTW, an example of how one can make $R^2$ arbitrarily close to $1$ (effectively reducing the SSE to an absolute minimum on a scale that permits comparison among transformed values) by means of a Box-Cox transformation of an independent variable is described at stats.stackexchange.com/a/50428/919. That example demonstrates why the intention behind a transformation cannot possibly be to minimize sums of squared errors. $\endgroup$ – whuber Jun 13 '17 at 21:54
  • $\begingroup$ @whuber, thank you for those links, they are very interesting examples, I think I understand the second link but since I'm only concerned with box-cox transformation of the dependent variable, your example of bringing R^2 arbitrarily close to 1 would not be a problem here right? $\endgroup$ – lucusk Jun 15 '17 at 14:07

Your Answer

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

Browse other questions tagged or ask your own question.