Calculation of break-even correlation for diversification effect in N-assets case?

Question

I'm thinking about a generalization of the following case: for 2 assets, there is a diversification effect as soon as i obtain a positive weight for the minimum-variance portfolio in the asset with the higher volatility.

If $\rho_{12}$ is the correlation coefficient and $\sigma_1 < \sigma_2$ then for $\rho_{12}<\frac{\sigma_1}{\sigma_2}$ we obtain a positive weight on $\omega_2$ in the minimum variance portfolio where $$ \omega_2 = \frac{\sigma_1^2 - \sigma_1\sigma_2\rho_{12}}{\sigma_1^2+\sigma_2^2-2\sigma_1\sigma_2\rho_{12}} $$ My question: regarding the correlation coefficient (-matrix), what is the general case for N-assets w.r.t $\rho_{i,j}$ and how to interpret this given the expression for the minimum variance portfolio weights vector as:

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

where $\boldsymbol{1}$ is the usual unity vector and $\boldsymbol{\Sigma}^{-1}$ is the inverse of the covariance matrix.

EDITED: I guess similar to the 2-asset case, i have to start with the nominator $\boldsymbol{\Sigma}^{-1} \boldsymbol{1}$?

Thank you for your help

Thomas

Answer 1

"My question: regarding the correlation coefficient (-matrix), what is the general case for N-assets w.r.t $\rho_{i,j}$ and how to interpret this given the expression for the minimum variance portfolio weights vector as:"

I don't have an answer for the general closed-form solution for individual weights in the MVP. However, answering the second question on interpretation of MVP weights vector,

I don't think $\Sigma^{-1} 1$ equals zero, actually. Because $\Sigma^{-1} 1$ is supposed to represent a vector of weights that are scaled by a scalar $1'\Sigma^{-1} 1$ that then represents the vector of minimum-variance weights. A high level mathematical proof can be understood from the perspective of the Lagrange multiplier method:

Starting with,

$min_w \sigma^2(w) = w' \Sigma w$

$\sum^n_{i=1}w_i = 1$

We can then form the Lagrangian (by adding a multiplier for the constraints):

$L(w,\lambda) = w' \Sigma w + \lambda (\sum^n_{i=1}w_i - 1)$

Taking the first derivative w.r.t. to the Lagrangian (to find the minima):

$\frac{\partial L}{\partial w_i} = 2 \Sigma w + \lambda 1 = 0$

$\Leftrightarrow w_{min} = -\frac{\lambda}{2} \Sigma^{-1} 1 = \frac{\Sigma^{-1} 1}{1' \Sigma^{-1} 1}$

Take note that $-\frac{\lambda}{2}$ is the normalization/scaling factor, which in this case is just $1'\Sigma^{-1} 1$. You can also think of the previous equation as the eigenvector of weights that correspond to the smallest eigenvalue that is the lowest portfolio volatility.