5
$\begingroup$

The following question is about an Exercise 14.15 from "The Elements of Statistical Learning" by Hastie, Friedman and Tibshirani.

Generate $200$ observations of three variates $X_1, X_2 , X_3$ according to \begin{align}X_1 &= Z_1 \\ X_2 &= X_1 + 0.001 \cdot Z_2 \\ X_3 &= 10 \cdot Z_3 \end{align} where $ Z_1, Z_2, Z_3 $ are independent standard normal variables. Compute the leading principal component and factor analysis directions. Hence show that the leading principal component aligns itself in the maximal variance direction $X_3$, while the leading factor essentially ignores the uncorrelated component $X_3$, and picks up the correlated component $X_2 + X_1$ (Geoffrey Hinton, personal communication).

Why? I thought that they are both "powered by" the same matrix decomposition? What have I missed?

$\endgroup$
  • 2
    $\begingroup$ The principal reason in this particular example is (I'm sure) the PCA was based on covariances while FA was based on correlations, - the decomposed matrices were different. This looks like a bad example to really understand differences between PCA and FA. $\endgroup$ – ttnphns Nov 14 '13 at 8:52
  • 1
    $\begingroup$ @ttnphns Have you got a good example? (This difference is a tricky one for some people to understand; I have had clients who couldn't get it, and, usually, the two methods do give similar results). $\endgroup$ – Peter Flom Nov 14 '13 at 11:18
  • 1
    $\begingroup$ Not now, @Peter. But, sure, I know you can ever suggest an exercise example at least not worse than I can. The question, however, is what the OP really bothered with, what does they really want to grasp? $\endgroup$ – ttnphns Nov 14 '13 at 11:40
  • 3
    $\begingroup$ @PeterFlom this paper has an example: ophi.org.uk/wp-content/uploads/Widaman-1993.pdf $\endgroup$ – Jeremy Miles Nov 14 '13 at 17:07
  • $\begingroup$ Thanks @JeremyMiles ! I vaguely remember that paper from back when I was in grad school. I will take a look $\endgroup$ – Peter Flom Nov 14 '13 at 17:14
6
$\begingroup$

The covariance matrix in this example is given by $$\mathbf C = \left(\begin{array}{c} 1 & \sim 1 & 0 \\ \sim 1 & \sim 1 & 0 \\ 0 & 0 & 100\end{array}\right).$$

To compare PCA and FA, think about how PCA/FA loadings reconstruct the covariance matrix.

The loadings of the first principal component in PCA is a vector $\mathbf v$ that minimizes the reconstruction error $\|\mathbf C - \mathbf v \mathbf v^\top \|$. As is well-known, it is given by the leading eigenvector of $\mathbf C$ scaled by a square root of its eigenvalue, and in this case will be pointing in the $(0,0,1)$ direction (in order to reproduce the covariance of $X_3$ which would otherwise be a major source of reconstruction error).

In contrast, the loadings of the first factor in FA is a vector $\mathbf v$ that minimizes the reconstruction error $\|\mathbf C - \mathbf v \mathbf v^\top - \boldsymbol \Psi \|$, where $\boldsymbol \Psi$ is a diagonal matrix of uniquenesses. This is equivalent to saying that it minimizes the reconstruction error $\|\mathrm{offdiag}\{\mathbf C - \mathbf v \mathbf v^\top\}\|$, i.e. FA does not care about reconstructing the diagonal. Think about $\mathbf C$ with erased diagonal:$$\mathrm{offdiag}\{\mathbf C\}=\left(\begin{array}{c} & \sim 1 & 0 \\ \sim 1 & & 0 \\ 0 & 0 & \end{array}\right).$$ The goal of FA is to reconstruct this part of $\mathbf C$ and so the loadings of the first factor will be pointing in the $(1,1,0)$ direction, in order to reproduce this off-diagonal covariance between $X_1$ and $X_2$.

Note that this analysis is based on the covariance matrix. Conducting an analysis based on the correlation matrix would (in this case) lead both PCA and FA to yield similar outcomes.


My answer to the opposite question might be of interest:

For many more details about PCA vs FA issue, see my [very long] answer to this question:

$\endgroup$

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.