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?

  • 1
    $\begingroup$ Both of the closed-form portfolio solutions are derived under the full investment constraint (weights have to sum to 1). So for $p=1$, the norm of the MVP and the max-Sharpe portfolio should be equal? Thus for $p=1,2$, sharp inequality ($<$) does not hold and maybe it should be changed to $\leq$? $\endgroup$
    – Pleb
    Jan 5, 2021 at 11:13
  • 2
    $\begingroup$ Your formula yields $L_1$ norm for $p=1$ but not $L_2$ norm for $p=2$. $\endgroup$ Jan 6, 2021 at 14:15


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