$$\arg \min_w \quad w^\top \Sigma w$$ \begin{align}\text{s.t.} \quad \mathbf{1}^\top w = 1 \end{align} is the optimization problem for the minimum-variance portfolio weights, whose analytical solution, derived from the above's Lagrangean formulation, is $$w_{MV}=\frac{\Sigma^{-1}\mathbf{1}}{\mathbf{1}^T\Sigma^{-1}\mathbf{1}}$$

The max skewness portfolio, on the other hand, where $M_3$ is the coskewness matrix, has the optimization problem $$\arg \min_w \quad -w^\top M_3 (w\otimes w)$$ \begin{align}\text{s.t.} \quad \mathbf{1}^\top w = 1 \end{align}

What then is the closed-form solution of the above's Lagrangean formula (not shown here)? How can the weights be derived analytically $$w_{SK}=?$$

  • $\begingroup$ On second thought, I would like to add that the portfolio skewness is defined as third moment divided by $\sigma_P^{1.5}$, i.e. it is a normalised quantity. $\endgroup$ Dec 4, 2020 at 10:27
  • $\begingroup$ which source shows this normalization for portfolios? I have never seen it as a denominator for the portfolio skewness objective function $\endgroup$
    – develarist
    Dec 4, 2020 at 10:29
  • $\begingroup$ Simply look up the definition of skewness. Just saying... $\endgroup$ Dec 4, 2020 at 10:31
  • $\begingroup$ but the definition of non-portfolio skewness doesn't possess portfolio weight multipliers. to now say that there is a denominator for portfolio skewness renders every previous publication on portfolio skewness invalid $\endgroup$
    – develarist
    Dec 4, 2020 at 10:36
  • $\begingroup$ Hold your horses. Skew is defined as $\frac{M_3}{M_2^{1.5}}$, i.e. the normalised third central moment. During optimisation, you can of course maximize portfolio $M_3$, only. Nothing stops you from that. But if you do not normalise this expression you may arrive at strange results. You can literally see for yourself in a two-asset portfolio with weights $w,1-w$, some (co-)skewness parameters and unit variances. $\endgroup$ Dec 4, 2020 at 10:53

1 Answer 1


Unfortunately, there exist no closed form for this.

The Lagrangean reads

$$ L(w,\lambda)=w^TM_3(w\otimes w)-\lambda(w^T\mathbf{1}-1) $$

with first order conditions

$$ \begin{align} \frac{\partial L }{\partial w_i}&=3w^TM_{3,i}w-\lambda \quad \forall i \\ \frac{\partial L }{\partial \lambda}&=w^T\mathbf{1}-1 \end{align} $$

where $M_{3,i}$ is the $i$th matrix component of the $3$-dimensional skewness tensor. The derivative of $w^TM_3(w\otimes w)$ with respect to $w_i$ is easily verified algebraically, and comparison to a quadratic form.

Effectively, this is a system of quadratic forms:

$$ \begin{align} w^TM_{3,1}w&=\lambda\\ w^TM_{3,2}w&=\lambda\\ \ldots&=\lambda\\ w^TM_{3,N}w&=\lambda\\ w^T\mathbf{1}&=1 \end{align} $$ There exist no closed-form solution for this. You could try to solve this equation system using a multivariate Newton Raphson scheme and careful selection of starting values.

Answering your comment:

.... Since there is no closed-form solution for the max skewness portfolio, does that mean that we cannot derive a proof that the max skewness portfolio has higher skewness than the most skewed asset?

At least anecdotically, it is quite easy to show that for a two-asset portfolio, the boundedness of the portfolio skewness is driven by the level of the co-skewness.

Please find below two graphs for a two asset portfolio. In each case, the assets of unit variance, no covariance, and skewness of $S_{111}=0.05$, $S_{222}=-0.05$. In the first graph, the co-skews $S_{112}=S_{122}=0.0$, in the second graph they are $+0.1$ and $-0.1$, respectively. The $x$-axis shows the portfolio weight on asset 1.

No co-skewness With co-skewness

As you can see, the question whether or not portfolio skew is bounded by the asset skews is driven by *co-skewness. Again, diversification is the key.


  • $\begingroup$ In the other post, you used the closed-form solution of the minimum-variance portfolio to prove that the minimum-variance portfolio has lower variance than the lowest variance asset. Since there is no closed-form solution for the max skewness portfolio, does that mean that we cannot derive a proof that the max skewness portfolio has higher skewness than the most skewed asset? $\endgroup$
    – develarist
    Dec 4, 2020 at 0:38
  • $\begingroup$ @develarist I have updated my post on this. HTH? $\endgroup$ Dec 4, 2020 at 9:14
  • $\begingroup$ thanks for the edits. I guess the figures say that asset skewness forms bounds on portfolio skewness only if there is no coskewness, and that portfolio skewness only outperforms asset skewness if coskewness is non-zero (although getting the direction of "over-"skewness right is futile). I wonder how well these two laws both generalize to the case where there are more than 2 assets. Is there a source that similarly demonstrates the skewness-portfolio weight analysis you have shown? $\endgroup$
    – develarist
    Dec 4, 2020 at 10:34

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