# Maximum skewness portfolio solution derived from its Lagrangean formulation

$$\arg \min_w \quad w^\top \Sigma w$$ \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 $$w_{SK}=?$$

• 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. – Kermittfrog Dec 4 '20 at 10:27
• which source shows this normalization for portfolios? I have never seen it as a denominator for the portfolio skewness objective function – develarist Dec 4 '20 at 10:29
• Simply look up the definition of skewness. Just saying... – Kermittfrog Dec 4 '20 at 10:31
• 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 – develarist Dec 4 '20 at 10:36
• 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. – Kermittfrog Dec 4 '20 at 10:53

## 1 Answer

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.

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.

HTH?

• 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? – develarist Dec 4 '20 at 0:38
• @develarist I have updated my post on this. HTH? – Kermittfrog Dec 4 '20 at 9:14
• 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? – develarist Dec 4 '20 at 10:34