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This question is related to recent rule changes in the Quantopian Open.

I am trying to figure out a closed form solution to a beta constrained minimum variance portfolio problem but it doesn't seem particularly tractable. Has anyone else tried this? So far, I have set up the problem

$$\begin{align} \min_w \quad& w^\prime \Sigma w \\ s.t. \quad& w^\prime\vec 1 = 1 \\ \text{and} \quad & w^\prime \beta = c \end{align}$$

where

  • $w$ is the vector of portfolio weights, our control
  • $\Sigma$ is the total covariance matrix
  • $\beta$ is the vector of CAPM-type market betas
  • $c$ a constant that the portfolio beta should be equal to

Changing constraints to Lagrange multipliers the objective becomes

$$\min_w \quad w^\prime\Sigma w - \lambda_1(w^\prime\vec 1 - 1) - \lambda_2(w^\prime\beta - c)$$

the first order conditions are

$$\begin{align} 0 &= 2w^\prime\Sigma - \lambda_1\vec 1 - \lambda_2 \beta\\ 1 &= w^\prime\vec 1\\ c &= w^\prime\beta \end{align}$$

I cannot seem to get the equations to work out nicely, perhaps no closed form solution exists but I wanted to check here and see if anyone could get something reasonable before I go to numerical optimization.

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  • $\begingroup$ The constraints can be grouped together to something like $Aw=b$, so that the lambdas are a vector. This set-up is actually pretty common. I typically refer to the derivation in edoc.hu-berlin.de/master/jiao-wei-2003-12-16/PDF/jiao.pdf $\endgroup$ – John Jun 4 '15 at 18:44
  • $\begingroup$ @John I like this reference but I would suggest pointing interested readers to page 7 (pdf page 15) when making it $\endgroup$ – user25064 Jun 5 '15 at 13:43
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Assuming that $\Sigma$ is invertible, then \begin{align} 2\omega' = \lambda_1\overrightarrow{1}'\Sigma^{-1}+\lambda_2\beta'\Sigma^{-1}. \end{align} We can then solve $\lambda_1$ and $\lambda_2$ from the system of equations \begin{align*} 2 &= \lambda_1\overrightarrow{1}'\Sigma^{-1}\overrightarrow{1}+\lambda_2\beta'\Sigma^{-1}\overrightarrow{1}\\ 2c &= \lambda_1\overrightarrow{1}'\Sigma^{-1}\beta+\lambda_2\beta'\Sigma^{-1}\beta. \end{align*} Consequently, $\omega$ can be obtained from the above equation.

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  • $\begingroup$ I was hoping for some cute $w = \ldots$ reduced form type equation but this certainly answers the question. edit - I bet mathematica could help but I don't own it $\endgroup$ – user25064 Jun 4 '15 at 16:37
  • $\begingroup$ @user25064, I believe this can be done in Excel 2010. $\endgroup$ – Gordon Jun 4 '15 at 17:24
  • $\begingroup$ I was referring to symbolic manipulation... $\endgroup$ – user25064 Jun 4 '15 at 17:25
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This is an interesting problem. I don't think the problem is set up correctly quite yet. I rewrote it slightly to correspond to how it's generally written as a quadratic program.

The optimization problem you write down fixes betas to be a certain value. That could make sense but instead I wondered if we could simply minimize beta across the portfolio while minimizing correlations. In that case, the optimization problem becomes:

\begin{align} \min_w \quad& w^\prime \Sigma w + w^\prime\lvert\beta\rvert\\ s.t. \quad& w^\prime\vec 1 = 1 \\ \text{and} \quad & w > 0 \end{align}

I don't think a closed-form solution exists to this problem. But it's quite easy (and fast) to solve this with a quadratic optimizer such as provided by cvxopt. Here is some example code:

import numpy as np import matplotlib.pyplot as plt import cvxopt as opt from cvxopt import blas, solvers import pandas as pd np.random.seed(123) # Turn off progress printing solvers.options['show_progress'] = False ## NUMBER OF ASSETS n_assets = 4 ## NUMBER OF OBSERVATIONS n_obs = 1000 # Generate returns return_vec = np.random.randn(n_assets, n_obs) betas = np.array([.1, .8, .2, .05]) def markowitz_beta(returns, betas): n = len(returns) returns = np.asmatrix(returns) # Convert to cvxopt matrices # minimize: w * mu*S * w + betas * x S = opt.matrix(np.cov(returns)) # Minimize betas, can add mean returns here if desired pbar = opt.matrix(np.abs(betas)) # Create constraint matrices # Gx < h: Every item is positive G = opt.matrix(-np.eye(n)) # negative n x n identity matrix h = opt.matrix(np.zeros(n)) # Ax = b sum of all items = 1 A = opt.matrix(1.0, (1, n)) b = opt.matrix(1.0) # CALCULATE THE OPTIMAL PORTFOLIO wt = solvers.qp(S, pbar, G, h, A, b)['x'] return np.asarray(wt).ravel() weights = markowitz_beta(return_vec, betas) print weights print betas.ravel() print np.dot(weights, betas) Output: [ 3.50677883e-01 3.34638307e-07 2.50326847e-01 3.98994936e-01] [ 0.1 0.8 0.2 0.05] 0.105083172131

As you can see, the highest betas receive very low weights.

We also wrote a blog post on this (and the code is a version from there) which you can find here: https://www.quantopian.com/posts/the-efficient-frontier-markowitz-portfolio-optimization-using-cvxopt-repost-cloning-of-nb-now-enabled

Disclaimer: I work at Quantopian.

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    $\begingroup$ You just wrote down a different problem which is totally fine. The problem I was trying to solve has a closed form solution because my constraint is an equality constraint (I happen to use $c=0$) and I allow long/short portfolios. Still interesting and illustrative, thanks $\endgroup$ – user25064 Aug 5 '15 at 12:51
  • $\begingroup$ Ah, you're probably right in that case. I wonder in which cases you would want to target a specific beta value for your portfolio though. $\endgroup$ – twiecki Aug 5 '15 at 12:54
  • $\begingroup$ Separately, I posted this NB on our forums: quantopian.com/posts/… $\endgroup$ – twiecki Aug 5 '15 at 12:54

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