In portfolio management one often has to solve problems of the quadratic form $$ w^T \Sigma w + w^T c \rightarrow \min_{\omega} $$ with portfolio weights $w \in \mathbb{R}^N$ a constant $c \in \mathbb{R}^N$ and a covariance matrix $\Sigma \in \mathbb{R}^{N \times N}$. Furthermore we assume real world continuous and binary (e.g. cardinality) constraints.

For estimating the covariance matrix $\Sigma$ we can e.g. use the sample covariance of the returns of all assets - let's call this an asset-by-asset model. It is known that some care has to be taken here if $N$ is big and so forth, but this is not the point of this question.

On the other hand we can define factors $(F_k)_{k=1}^K$ with $K<N$. These factors have a covariance matrix $\Sigma_F \in \mathbb{R}^{K \times K}$. Denoting by $r_i$ the return of asset $i$ with $i = 1, \ldots, N$, we can write: $$ r_i = \sum_{k=1}^K e_{i,k} F_k + \epsilon_i, $$ with the meaning that the variation of the return is described by the exposures $e_{i,k}$ to the factors and some purely idiosyncratic risk $\epsilon_i$. In this case the covariance matrix of returns is given by $$ \hat{\Sigma} = e \Sigma_F e^T + \mathop{diag}(Var[\epsilon_1],\ldots,Var[\epsilon_N]), $$ where $e \in \mathbb{R}^{N \times K}$ is the matrix of all exposures and the "diag" part adds the idiosyncratic parts of variance at the main diagonal. Note that $\hat{\Sigma} \in \mathbb{R}^{N \times N}$.

I sometimes read that a problem with covariance matrix $\hat{\Sigma}$ from the factor model is easier to solve than with $\Sigma$. Is this true? If yes, then how can we see this? My personal answer so far is: no. Because both are $N \times N$ matrices and the structure does not help in general.

Especially if the factors are not orthogonal - what do we gain? If the factors come from a PCA then we might gain something but I wonder how many PCs we would need e.g. in a global equities portfolio ...

  • $\begingroup$ In fact @DavidNehme has already answered this question when he answered this one: quant.stackexchange.com/questions/9616/… $\endgroup$
    – Richi Wa
    Dec 12, 2013 at 13:01
  • $\begingroup$ But the non-orthonogal case is still not clear to me. $\endgroup$
    – Richi Wa
    Dec 12, 2013 at 14:30
  • $\begingroup$ I'm not so sure that I've read that portfolio optimization on $\widehat{\Sigma}$ is "easier" than the optimization on $\Sigma$. You're still working with an $N\times N$ matrix, but you've ensured that it is positive definite (which is a good thing). You could always generate some random data and test the idea. It might be possible to set up the optimization in terms of principal portfolios so you're only operating on the first few $K$ principal portfolios and then translating those weights back into normal weights for constraints and TC. $\endgroup$
    – John
    Dec 12, 2013 at 21:58
  • $\begingroup$ However, if the weight constraints are too restrictive, then it might not work as you'd like. $\endgroup$
    – John
    Dec 12, 2013 at 21:59

1 Answer 1


A few points.

First, In a typical factor model, the idiosyncratic piece (what you call $Var[\epsilon_k]$) is non-negligible, which results in a $\hat\Sigma$ that is going to be well-conditioned. From a numerical point of view, this is very convenient. The more well-conditioned a covariance matrix, the less susceptible the optimization routine is to round-off errors.

Second, one can introduce variables $l_1, ..., l_K$ with constraints that $\sum_{i=1}^N e_{i,j} w_i = l_j$ and then work with $\Sigma_F$ directly in the formulation of an optimization and not use a full $N\times N$ covariance matrix.

Third, from a portfolio optimization standpoint we don't care if the factors are orthogonal or not. We can rotate them to be orthogonal. As long as $\Sigma_F$ is positive definite, we can use the Cholesky decomposition to write it as $LL^T = \Sigma_F$ and then replace the loadings matrix $e$ with a rotated one $\tilde e =eL$ so that now the factors are orthogonal. That is $$e\Sigma_Fe^T = eLL^Te^T = \tilde e \tilde e^T.$$


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