Why Minimum-Variance Weights Do Not Always Have Smaller Norms

Summary

The document compares two portfolio weight formulas: the global minimum-variance portfolio and a maximum-Sharpe portfolio. It asks whether the sum of absolute weight powers, scaled by the power, must be smaller for the minimum-variance portfolio when the power is one or two. This connects portfolio choice to measures of weight concentration and exposure size.

The document does not provide a proof, assumptions, examples, or a resolution. In particular, it does not state constraints on expected returns, the risk-free rate, covariance matrix, or portfolio weights beyond the displayed formulas. Those details matter when assessing whether a general inequality can hold. The text is therefore useful as a clearly framed mathematical question, but it does not teach a completed result or establish that the proposed comparison is always true.

Key ideas

• The document gives formulas for global minimum-variance and maximum-Sharpe portfolio weights.
• It asks whether the minimum-variance weights always have a smaller scaled absolute-weight norm for powers one and two.
• It supplies no proof, counterexample, or assumptions that would settle the claim.

Tags

# Prove norm $\frac{1}{p}\sum_{i=1}^n |w_i|^p$ of min-variance portfolio $\leq$ max-Sharpe portfolio


# Prove norm $\frac{1}{p}\sum_{i=1}^n |w_i|^p$ of min-variance portfolio $\leq$ max-Sharpe portfolio












The minimum-variance portfolio weight vector is

$$\boldsymbol{w}_{MV} = \frac{\boldsymbol{\Sigma}^{-1} \boldsymbol{1} }{\boldsymbol{1}' \boldsymbol{\Sigma}^{-1} \boldsymbol{1}}$$

whereas the maximum Sharpe ratio portfolio's weights are

$$\boldsymbol{w}_{SR} = \frac{\boldsymbol{\Sigma}^{-1} \left(\boldsymbol\mu - r_f \cdot \boldsymbol{1} \right) }{\boldsymbol{1}^\top \boldsymbol{\Sigma}^{-1} \left(\boldsymbol\mu - r_f \cdot \boldsymbol{1} \right)}$$ where $\boldsymbol\mu$ and $\boldsymbol{\Sigma}$ are the asset means and covariance matrix.

How can it be shown that the $p=1,2$ norm $\frac{1}{p}\sum_{i=1}^n |w_i|^p$ of the first solution will always be smaller than or equal to the second?

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.