Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a return history for a universe of risky assets and I've run a principal component algorithm and obtained a loadings matrix (num_factors by num_assets) for the first 5 factors.

I have a portfolio as well (a subset of the above universe) with weights w for each of the assets. This portfolio has a variance \sigma^2. How do I figure out the percentage of the variance in the portfolio that comes from factor 1?

share|improve this question
The algorithms usually give it to you as one of the outputs. – John May 6 '14 at 0:12
up vote 6 down vote accepted

PCA gives you a decomposition of the covariance matrix of the form $$ \Sigma = V \Lambda V^T $$ where $\Lambda$ is diagonal with the eigenvalues in the diagonal. Your portfolio variance is $$ w^T \Sigma w = (V^T w )^T \Lambda (V^T w) $$ On the other hand if you take your return matrix $R$ and define $$ F = V^T R $$ then the covariance matrix of these so called principle portfolios is $\Lambda$. You find this here by Meucci.

In fact he writes $V^{-1} R$ for the return of principle portfolios and defines the weights $w^* = V^{-1} w$ for the weight of the original portfolio on the principle portfolios.

He then defines $v_n = (w^*)^2 \lambda_n^2$ for the contribution of the n-th principle portfolio to the portfolio variance. If you relate this to the total volatility of the portfolio then you are done. Note that $V$ is orthogonal which means that $V^{-1} = V^T$.

I recommend to read the following white paper or this blog entry or this to get more details.

share|improve this answer
Thanks Richard. The links were especially helpful. Using the links I was able to determine that you can get the $v_n$ above using $v_n = (Lw)^2$ where $L$ is the loadings matrix and the results is the same as using $(w^*)^2 \lambda_n^2$. – rhaskett May 9 '14 at 18:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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