The mean variance optimization to the objective: $h^T\alpha - \lambda h^T V h$ results in the solution:

$h = \frac{V^{-1} \alpha}{2 \lambda}$

Would a positive value for an asset in $\alpha$ result in a positive value in the weights vector $h$ ?

  • $\begingroup$ what is $\alpha$? $\endgroup$
    – develarist
    Nov 22, 2019 at 21:39
  • $\begingroup$ Expected returns. This is the signal generated for tomorrow's returns. $\endgroup$
    – whisperer
    Nov 25, 2019 at 19:35
  • $\begingroup$ No because in the case of two assets with very highly correlated returns and positive alphas, MVO would most likely chose to go long the one with larger alpha and short the one with smaller alpha as this would (theoretically) decrease variance by a lot. $\endgroup$ Feb 3, 2023 at 18:14

1 Answer 1


The sign of the portfolio weight $h_n$ assigned to one asset cannot be solely determined by the sign of that asset's expected return, $\alpha_n$, due to the model having to also take as an input that asset's dependence structure with the other $N-1$ assets being considered, captured within the covariance matrix $V$. Those other assets are also vying for inclusion in the portfolio.

Since the mean-variance model was recognized by Michaud to be an "error maximization" model in that it favors (overweights) assets that have high expected returns and low variance, however, even though these same assets are likely to be the most prone to misestimation due to the nature of expected return estimates being much more unreliable than asset volatility estimates, the sign of an asset's expected return does, at least, suggest that the model will "pick it" for a long position, especially if that asset has the highest expected return (plus lowest risk) within the candidate investment pool. But again, it all depends on the overall data for all assets being considered.

  • 2
    $\begingroup$ Mathematically the weights are determined by $h=\frac{1}{2\lambda} V^{−1} \alpha$. Knowing that a particular element $\alpha_i$ of the vector $\alpha$ is >0 does not guarantee that the corresponding element $h_i$ is >0. That is just not how matrix multiplication works. $h_i$ depends on the entire row $i$ from the matrix and all the elements of $\alpha$. $\endgroup$
    – nbbo2
    Apr 24, 2020 at 2:27

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