Why Maximum-Skewness Portfolio Weights Lack a Closed-Form Solution

# Maximum skewness portfolio solution derived from its Lagrangean formulation


# 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 $$w_{SK}=?$$

## Answer by Kermittfrog (score 5, accepted)

https://quant.stackexchange.com/a/59703

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?