Finding a minimum variance portfolio when using a regulariser?

Question

I am aware that the minimum variance portfolio of a market with $n$ securities can be shown to be:

\begin{equation} w^* = (1^T_n\Sigma^{-1}1_n)^{-1}\Sigma^{-1}1_n, \\ s.t. \ \ 1^T_nw = 1 \end{equation}

by using the method of Langrange multipliers or other. I am interested in demonstration of the extension: \begin{equation} w^* = \underset{w}{\mathrm{argmin}}\lbrace w^T \Sigma w + \lambda\sum_{i=1}^n\rho(w_i)\rbrace\\ s.t. \ \ 1^T_nw = 1 \end{equation}

where $\rho(.)$ is some arbitrary penalty function (e.g. $\lvert w_i\rvert$).

Perhaps you could go through the process step by step as I am getting lost when I try.

Thanks!

Answer 1

You're not going to get an analytic formula except in special cases of function $\rho(x)$. And you're probably going to want $\rho$ convex.

• If $\rho$ is convex, the problem is a convex optimization problem and can be efficiently solved numerically. If $\rho$ isn't convex, the optimization problem may be difficult to solve.

  • If $\rho(x) = |x|$ you basically have the LASSO objective which doesn't have an analytic solution (though the solution can be efficiently found numerically).

  • If $\rho(x) = x^2$, you get a clean formula.

Special case $\rho(x) = x^2$

Then $\lambda \sum_i \rho(w_i) = \lambda \mathbf{w}'I\mathbf{w}$. Your optimization problem is then:

\begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over $w_i$)} & \mathbf{w}' \left(\Sigma + \lambda I \right)\mathbf{w} \\ \mbox{subject to} & \sum w_i = 1 \end{array} \end{equation}

And it's essentially the same as your original problem. $\Sigma$ is replaced by $\Sigma + \lambda I$.

\begin{equation} w^* = \frac{\left( \Sigma + \lambda I\right)^{-1}\mathbf{1}}{\mathbf{1}'\left( \Sigma + \lambda I\right)^{-1}\mathbf{1}} \end{equation}

(Just to be explicit, I use bold letters for vectors and $I$ is the identity matrix.)

-- Update -- Motivated by the comment from @noob2, I've attached a simulated example showing how security weights (in case $n = 8$) change as $\lambda$ increases. As @noob2 pointed out, higher $\lambda$ pushes weights towards the equal weight portfolio.

(Note: I've used a random covariance matrix, not one based on actual data. So don't over generalize anything besides the long run convergence towards 1/n.)