# Is there a considered floor for variation the 1st principal component must explain?

I am wondering if there is a considered floor to the percentage variation the 1st principal component must explain in general for PCA - ie. any lower and it is not worth doing PCA at all? Is the floor near 75%, 80% or should the 1st 3 explain a minimum of 90% or what?

As a follow on, if I have 10 X variables (index & sector returns) and only 6 are highly correlated (I take a correlation above 0.8 to be highly correlated - or is that too high?) should I just do PCA on those 6, then combine the 1st two principal components with the 4 remaining original variables and use that as my X for regression?

What I was doing was doing PCA on all 10 variables, taking the 1st 2 or 3 principal components, regressing those on Y (which is a single stock's return) then taking those PC's betas and matrix multiplying them by the eigenvectors to back out sensitivities to the original 10 factors but I am left with the situation of not knowing if all 10 factors are significant (all I know is that the 1st 2 PCs are significant)

• I would not expect any lower bound as a reasoning. In APT model you can have $n$ securities driven by $m<n$ common noise terms, entering linearly. Hence, you can expect each noise to explain equal amount of variance. By no means the first (any) of them shall explain more than half for example. – Ulysses Aug 3 '15 at 15:07
• A positive lower bound exists for the smallest eigenvalue of any symmetric positive definite matrix (see here). A positive lower bound for the variation explained by the first principal component immediately follows. I think the only way to reduce the floor to zero would be to take the extreme case where every column of your covariance matrix is linearly dependent, and for any practical work, this is obviously silly. – Colin T Bowers Aug 4 '15 at 4:22