A three- asset portfolio doesn't seem prone to generating corner solutions, which are very high allocations to one of the assets and $0$ to the others. Instead, when the number of assets is low, these small portfolios are fairly diversified

When the number of assets grows, however, the problem of corner solutions becomes more apparent since, even when optimizing 50 assets, it is not uncommon to find only 1 out of those 50 being given a weight of $100%$ by the mean-variance model, while the other 49 get $0$.

What explains this massive leap in the mean-variance model's pumping out of corner solutions when $N$ is very large? And am I right to view this as a bad thing since corner solutions imply overconcentration/lack of diversification


1 Answer 1


The source of the problem is twofold:

  • Dimensionality of variance directions is low (most directions have close to 0 variance)
  • Portfolio Optimization is prone to an unstable covariance matrix (which almost always is the case)

And now I will try to explain what that means in more detail and then sum it up in a simple, intuitive statement:

  • If you have a variance covariance matrix $\Sigma$, linear algebra lets us find a basis transformation to other assets that are now uncorrelated. $\Sigma = E^{'} \Lambda E$. $\Lambda$ is then a diagonal matrix with its diagonal being the variances of the uncorrelated portfolios - also called the eigenvaleus $\lambda_i$. In theory, the idea is to switch from your correlated assets into these uncorrelated portfolios, do all analysis/optimization there and then transform back. In practice, when the number of assets is higher (think of a stock index for example), only the first few variances $\lambda_i^{1/2}$ show a meaningful difference from 0. Those represent the more meaningful market factors theses stocks share. These are usually somewhere between 2 and 7 or so. These eigenvalues correspond to "dimensions of variance", meaning all other directions of variance are negligible in comparison. From this, you might already suspect that this fact could translate back into the asset space somehow.

  • The basic solution of the unconstrained portfolio optimization problem is $w^{*} \approx \Sigma^{-1} \mu$, where $\approx$ means up to a constant which I have forgotten right now. The important part is the $\Sigma^{-1}$. If we remember the eigenvalue decomposition from before, almost all $\lambda_i$ were close to 0. We know that $\Lambda_i^{-1}$ is obtained by inverting the diagonal elements, $\lambda_i^{-1}$. If the $\lambda_i$ are close to $0$, your $\Sigma^{-1}$ will change a lot if one $\lambda_i$ changes. Therefore, your optimal asset weights $w_i^{*}$ will change a lot.

So, for these two reasons, many of your assets will have similar risk and correlation profiles, making them interchangeable from an optimizers point of view. If they are interchangeable in both variance and correlation aspects, the optimization problem is easy: Put everything into the asset with the highest return. This is why you obtain lots of corner solutions.

My second point however shows, that these corner solutions are usually not stable at all. The second the $\Sigma$ changes a little bit, the corner solution can change completely.

There are plenty of means around this, but they would be beyond the scope of this answer.

  • 1
    $\begingroup$ That is not very convincing - it is true that there are usually not so many "significant" eigenvectors, but if the eigenvectors are dense, then you do NOT get corner solution. In fact, the top eigenvectors are NOT dense, but you don't try to explain that. $\endgroup$
    – Igor Rivin
    Sep 29, 2020 at 17:12
  • $\begingroup$ what else might better explain corner solutions? $\endgroup$
    – develarist
    Sep 29, 2020 at 20:49
  • $\begingroup$ @IgorRivin I agree with your statement of eigenvectors being dense. You are absolutely right there, so it does not follow straight away that the resulting portfolio will have a corner solution. However, the dimensionality argument still stands - but I should have shown it, I agree. I will edit my answer and try to reflect this later today. If you have a solution already, you are very welcome to edit my answer as well. $\endgroup$
    – vanguard2k
    Sep 30, 2020 at 7:37
  • $\begingroup$ Looking forward to it! I am very interested. $\endgroup$
    – Igor Rivin
    Sep 30, 2020 at 17:06

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