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The information content of principal components is almost always expressed as a variance (e.g., in scree plots or in statements like "the first three PCs contain 95% of the total data variance"). The intent of this usage is to describe how much variation/information is contained in the PCs. it seems to me that variance can be a misleading measure of information contained in PCs, because it is a squared metric of variation that emphasizes large deviations from the mean over small ones. This can grossly underemphasize the importance of information contained in lower-eigenvalue PCs. The standard deviation of PCs would seem to be a much more direct, meaningful and balanced metric of the information they contain.

I am very clear on the rationale for the use of variance in statistics more generally, i.e. it is much more mathematically convenient than standard deviation. However, I'm wondering if there is a specific rationale for why variance is used a measure of variation in PCs instead of standard deviation. Are there any good references for this dilemma?

Update to clarify: I should be clear that I am not asking about why variance is used in the derivation of the principal components, but rather why it is used as a default descriptor of variation in the PCs when reporting results of the PCA. Many people seem to use "variance" and "variation" as synonymous in this context, but isn't standard deviation a measure of variation, and variance a squared measure of variation? A PC that contains 95% of the data variance might contain only 80% of the variation in the data as measured in standard deviations: isn't the latter a better descriptor?

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    $\begingroup$ Good question, +1. I think mathematical convenience is exactly the right reason. E.g. total variance (trace of covariance matrix) is preserved with orthogonal rotations, and as a result, total variance of all PCs is equal to the total variance of original variables. That's convenient. I don't think there is any other rationale. $\endgroup$ – amoeba Jan 12 '15 at 21:05
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    $\begingroup$ I wasn't clear. My point was, actually, that the immediate function is that sum. Taking root of it or inverting it or whatever would be just further "convenient" manipulations. I was with you. The "information" conveyed by sum variances or summs of squares is that they are additive across variables. $\endgroup$ – ttnphns Jan 12 '15 at 21:32
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    $\begingroup$ Loosely because variance is a thing that "components". When you look at the (orthogonal) principal directions, variances of components add ... so you can actually decompose the variation. What alternative does that? If you try doing it with some measure that doesn't decompose in that way, you can't reasonably call them 'components' in the same sense. You may be able to contrive something vaguely analogous (in that it consists of some somewhat analogous steps), but it shouldn't have the "C" in it. $\endgroup$ – Glen_b Jan 12 '15 at 22:35
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    $\begingroup$ What is variation in the [multivariate] data as measured in standard deviations? You cannot add up st. deviations to reflect the total variation, they aren't summative. $\endgroup$ – ttnphns Jan 13 '15 at 8:33
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    $\begingroup$ @whuber: Well, first, I had two days to think about it in the back of my head :) But second, there is a fine distinction here that I tried to convey in my answer. Namely, if the question is "Why do people usually use variance instead of SD to put a number on each PC?", then the answer, I believe, is largely "mathematical convenience" (plus tradition). But if the question is "Why do people usually use variance instead of SD to report a proportion of explained something by one or several PCs?", then the answer is that "using SD would not make sense". $\endgroup$ – amoeba Jan 15 '15 at 17:41
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Reporting standard deviations instead of variances

I think you are right in that standard deviation of each PC can perhaps be a more reasonable or a more intuitive (for some) measure of its "influence" than its variance. And actually it even has a clear mathematical interpretation: variances of PCs are eigenvalues of the covariance matrix, but standard deviations are singular values of the centered data matrix [only scaled by $1/\sqrt{n-1}$].

So yes, it is completely fine to report it. Moreover, e.g. R does report standard deviations of PCs rather than their variances. For example running this simple code:

irispca <- princomp(iris[-5])
summary(irispca)

results in this:

Importance of components:
                          Comp.1     Comp.2     Comp.3      Comp.4
Standard deviation     2.0494032 0.49097143 0.27872586 0.153870700
Proportion of Variance 0.9246187 0.05306648 0.01710261 0.005212184
Cumulative Proportion  0.9246187 0.97768521 0.99478782 1.000000000

There are standard deviations here, but not variances.

Explained variance

A PC that contains 95% of the data variance might contain only 80% of the variation in the data as measured in standard deviations: isn't the latter a better descriptor?

However, note that after presenting standard deviations, R does not display a "proportion of standard deviation", but instead a proportion of variance. And there is a very good reason for that.

Mathematically, total variance (being a trace of covariance matrix) is preserved under rotations. This means that the sum of variance of original variables is equal to the sum of variances of PCs. In case of the same Fisher Iris dataset, this sum is equal to $4.57$, and so we can say that PC1, having a variance of $2.05^2=4.20$ explains $92\%$ of the total variance.

But the sum of standard deviations is not preserved! The sum of standard deviations of original variables is $3.79$. The sum of standard deviations of PCs is $2.98$. They are not equal! So if you want to say that PC1 with standard deviation $2.05$ explains $x\%$ of the "total standard deviation", what would you take as this total? There is no answer, because it simply does not make sense.

The bottom line is that it is completely fine to look at the standard deviation of each PC and even compare them between each other, but if you want to talk about "explained" something, then only "explained variance" makes sense.

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  • $\begingroup$ thanks so much for taking the time to provide your outstanding answer. that question has been nagging me for a long time and now it is crystal clear. $\endgroup$ – seadollar Jan 21 '15 at 18:13

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