1
$\begingroup$

I came across this tutorial for Y-aware PCA using the vtreat R package. In short, Y-aware PCA is PCA on variables that have been scaled to be in y-units.

Is it valid to scale categorical independent variables in y-units?

All of my independent variables are dichotomous. It seems inconsistent to implement this y-scaling on them and then run PCA, since, typically, a method like multiple correspondence analysis would be used on categorical variables.

$\endgroup$
0
$\begingroup$

You need to run independent component analysis (ICA), which is used for performing (essentially) PCA on categorical or binary variables. One reason that PCA is unwise to use is because the covariance or correlation for binary data results in a departure from normality. That is, means and variances are biased (wrong) for binary data.

$\endgroup$
3
  • $\begingroup$ Thanks @wrtsvkrfm. Just to make sure, since scaling the binary variables to be in units of the independent variable $y$ does not make its distribution normal, ICA should be run because it can handle non-normal distributions. $\endgroup$
    – RTrain3K
    Mar 16 '17 at 12:46
  • 1
    $\begingroup$ Wouldn't say "non-normal", but rather a different probability distribution. Recall discrete (binomial, Poisson) and continuous prob dist's (normal, log-normal, Rayleigh, Cauchy, etc) prob dists. A highly skewed distribution with a long right tail is a log-normal dist - so it's non-normal. But binary is a discrete prob dist, so it's not in the same class (continuous) as the normal. ICA is for variables which are generated from discrete (categorical, binary, etc.) prob dists. $\endgroup$
    – user32398
    Mar 17 '17 at 17:07
  • $\begingroup$ JoleT, let me suggest you to edit and expand your nice answer to incorporate the comment. You may also want to say more of how ICA is used thus as you say it is. $\endgroup$
    – ttnphns
    Jul 20 '19 at 14:28

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.