8
$\begingroup$

From Jon Shlens's A Tutorial on Principal Component Analysis - version 1, page 7, section 4.5, II:

The formalism of sufficient statistics captures the notion that the mean and the variance entirely describe a probability distribution. The only zero-mean probability distribution that is fully described by the variance is the Gaussian distribution. In order for this assumption to hold, the probability distribution of $x_i$ must be Gaussian.

($x_i$ denotes a random variable - the value of the $i^\text{th}$ original feature.
i.e. the quote seems to claim that for the assumption to hold, each of the original features must be normally distributed.)

Why the Gaussian assumption and why might PCA fail if the data are not Gaussian distributed?

Edit: to give more info, in the tutorial, page 12, the author gave an example of non-Gaussian distributed data that causes PCA to fail.

$\endgroup$
1
  • $\begingroup$ The link that the OP provided to Shlens' tutorial is to version 1 of the tutorial, but version 3.02 (the final version?) is now available. In version 3.02, the quoted paragraph was significantly modified. $\endgroup$ Aug 18, 2018 at 9:49

3 Answers 3

9
$\begingroup$

TL;DR

  • PCA does assume normal distribution of features See p.55 SAS book1 or Rummel, 19702 or Mardia, 19793.
  • If you expect the PCs to be independent, then PCA might fail to live to your expectations.
  • Assuming that the dataset is Gaussian distributed would guarantee that the PCs are independent.

Long Answer

PCA doesn't assume the dataset to be Gaussian distributed

Most of the sources I have found (e.g. wikipedia) don't list Gaussian distribution as a requirement of PCA.
Moreover, it seems that Shlens himself doesn't believe that anymore:
I found 2 more versions of Shlens' tutorial: version 2 and version 3.02. The latter seems to be the current version (as Shlens' web page links to it), so I will refer only to version 3.02 in my answer.

In version 3.02, the paragraph you quoted was removed from the "Summary of Assumptions" section, so that currently, the section lists only the following assumptions:

  • Linearity
  • Large variances have important structure.
  • The principal components are orthogonal.

When might PCA fail to live to our (false) expectations?

In page 10 Shlens gives an example for when one might see the result of PCA as a failure, and then explains why PCA didn't really fail:

The solution to this paradox lies in the goal we selected for the analysis. The goal of the analysis is to decorrelate the data, or said in other terms, the goal is to remove second-order dependencies in the data. In the data sets of Figure 6, higher order dependencies exist between the variables. Therefore, removing second-order dependencies is insufficient at revealing all structure in the data.

Figure 6 - examples for when one might think that PCA failed
i.e. PCs are guaranteed to be uncorrelated, so that's exactly what we should expect them to be. However, if we expected the PCs to be independent, then we would consider PCA to fail when the PCs aren't independent (e.g. the examples in figure 6).
(See this answer for another explanation of this paragraph.)

How can the assumption of Gaussian distribution help with our (false) expectations?

If we assume that the original dataset is Gaussian distributed (i.e. the features are jointly normally distributed), then by definition every linear combination of the original features is normally distributed.
Each of the PCs given by PCA is a linear combination of the original features. Thus, also every linear combination of the PCs is a linear combination of the original features, and so every linear combination of the PCs is normally distributed.
So, by definition the PCs are jointly normally distributed. PCA guarantees that the PCs are uncorrelated, and therefore they are also independent.

(Note that in the original paragraph quoted in the question, Shlens seems to claim that each of the original features should be normally distributed. However, I believe that was a mistake, and he actually meant that the original features should be jointly normally distributed (I deduced that's what he meant mainly from footnote 7 in page 10 in version 3.02). This answer explains why these conditions are not equivalent in the 2D case. Similarly, they aren't equivalent for any dimension $>1$.)

Thus, under the assumption that the original dataset is Gaussian distributed, PCA guarantees that the PCs are independent.

$\endgroup$
1
4
$\begingroup$

Someone correct me if I'm wrong, but the PCA process itself doesn't assume anything about the distribution of your data. The PCA algorithm is simple -

  1. find the direction of greatest variance in your data
  2. write down the direction of the vector pointing in that direction, and 'divide' the data along that direction by its variance in that direction, so the resulting variance in that direction is 1. This provides you with an eigenvector (direction) and associated eigenvalue (scale).
  3. repeat steps 1-2, potentially as many times as you have dimensions, but with the constraint that the next vector must be orthogonal (aka at a right angle) to all previous.

The result will be an ordered list of orthogonal vectors (eigenvectors), and scales (eigenvalues). This set of vectors/values can be viewed as a summary of your data, particularly if all you care about is your data's variance.

I think there is an implicit assumption that the orthogonality implies independence of the resulting vectors, and from what I understand that's true if the data is Gaussian but not necessarily true in general. So I suppose whether your data can be modeled as Gaussian may or may not matter, depending on your use case.

$\endgroup$
6
  • 1
    $\begingroup$ According to the tutorial (cs.princeton.edu/picasso/mats/PCA-Tutorial-Intuition_jp.pdf page 7) there are several assumptions made by PCA, including, "large variances have important dynamics" assumption and the Gaussian Distribution assumption. In page 12, the author gave an example of non-Gaussian distributed data causes PCA to fail. $\endgroup$
    – Math J
    Dec 19, 2017 at 7:02
  • $\begingroup$ Also I didn't quite understand what you meant by "orthogonality implies independence". The basis before changing to orthogonal eigenvectors are orthogonal too, but clearly don't imply independence. $\endgroup$
    – Math J
    Dec 19, 2017 at 11:27
  • 1
    $\begingroup$ My interpretation of "PCA fails" is "PCA doesn't give an intuitive interpretation". The directions of highest variance still get the principal components (i.e. eigenvectors). The author says "the largest variances do not correspond to the meaningful axes" - but meaningful is in the eyes of the analyst. The algorithm doesn't care. You're right that orthogonality does not truly imply independence, but people typically interpret each principal component as a separate source or 'factor', depending on the analysis. They are not necessarily separate...unless the underlying data is Gaussian. $\endgroup$
    – tom
    Dec 19, 2017 at 15:51
  • $\begingroup$ I think "fail" here means that it doesn't give the results we expect, and that's why Independent Component Analysis (ICA) was created (to improve it). And by "Gaussian assumption" I meant the data must be Gaussian in order for PCA to achieve what we would expect it to achieve. But what I didn't understand was why and what the math behind it is. $\endgroup$
    – Math J
    Dec 19, 2017 at 16:34
  • $\begingroup$ Regarding the "PCA fails" example on page 12, aren't the data along different principle components supposed to be un-correlated (i.e. the covariance matrix is diagonal), but it's clearly not the case as the graph shows. So perhaps that's what the author means by "fails"? $\endgroup$
    – Math J
    Dec 19, 2017 at 16:53
1
$\begingroup$

In my opinion, the following information can be considered as implicit assumptions of PCA:

  1. Data comes from a constant multivariate normal distribution.
  2. Principal components are orthogonal.
  3. More variances have an important structure.
$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.