1
$\begingroup$

I am learning CCA and I have come across a question that I do not know how to answer. Suppose we have the following 2 sets of variables:

X as psychology traits (control, concept, motivation)
Y as academic achievements (read, write, math, science)

And suppose that all variables are real valued numbers, but use different scales:

control: [-1, 1]
concept: [0, 100]
motivation: [0, 1.0]
read: [-9, 9]
write: [-9, 9]
math: [0, 100]
science: [0,100]

If I want to do a CCA analysis on these two sets of variables, do I need to transform each variable into a uniform scale? For example, by scaling all variables into the [0,1] range, before doing the CCA?

The reason I am asking, is that CCA attempts to find the max correlation between two canonical variates CX, CY. And CX CY are each the weighted sum of their variables, and these weights are canonical weights. But if the variables are on different scales (e.g., read and math), would the canonical weights for them still be comparable (e.g., if 'read' has a weight of 0.5 and 'math' also 0.5, should these two values be interpreted the same way?)

Many thanks in advance.

$\endgroup$
  • 2
    $\begingroup$ In some sense CCA is "invariant" to rescalings: e.g. the 1st canonical correlation will stay exactly the same. But if you want to be able to attach to any meaning to the values of CCA weights then it does make sense to put all your variables on the same scale. The most common choice is to standardize the data (mean zero, variance one). $\endgroup$ – amoeba Oct 2 '18 at 14:32
  • $\begingroup$ @amoeba, Can you please explain why and how CCA is scale invariant? $\endgroup$ – Vendetta Dec 4 '18 at 8:35
  • $\begingroup$ @Vendetta The same reason that correlation coefficient is scale invariant. $\endgroup$ – amoeba Dec 4 '18 at 12:46
  • $\begingroup$ @amoeba, Please correct me if I am wrong, the correlation coefficient is invariant to scaling due to mean removal and division by variance (i.e., standardization). $\endgroup$ – Vendetta Dec 5 '18 at 5:38

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.