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Regarding the final output of PCA:

  1. Using rotation, I'll get a loading for my 50 components, so the equation would be $pc_1=0.05*a_1+0.02*a_2+\dots+0.04*a_{50}$. Now, can I derive the value of $pc_1$ by using my last observed return for $a_1,a_2,\dots,a_{50}$?

  2. As I looked at my data, I have found at least 10 principal components are required to explain 80% of the variance, so using same logic as above, I'll get $pc_1,pc_2,\dots,pc_{10}$. Now the regression would be $y=w_1*pc_1+w_2*pc_2+...+w_{10}*pc_{10}+e $. In each of them $w_1,w_2,\dots,w_{10}$ can be their respective eigenvalue and the remaining term is an error term?

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    $\begingroup$ How does this differ from your previous question? $\endgroup$ – user88 May 25 '12 at 11:17
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Everything sounds pretty OK except the ending of Q2. When you do regress y on the first ten PCs, w1,...,w10 are unknown weights to be estimated from your data using Ordinary Least Squares for example.

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    $\begingroup$ To add to@danas.zuokas's answer: the eigenvalues describe the variance accounted for among the 50 variables used in PCA, but this is before the dependent variable (Y) was ever entered into the analysis. The eigenvalues and loadings have nothing to do with relationships with Y. $\endgroup$ – rolando2 Jul 24 '12 at 11:33

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