12.2 Determining the Global Minimum Variance Portfolio Using Matrix Algebra

The global minimum variance portfolio $\mathbf{m}=(m_{1},\ldots,m_{N})^{\prime}$ for the $N$ asset case solves the constrained minimization problem: $\begin{align} \min_{\mathbf{m}}~\sigma_{p,m}^{2}=\mathbf{m^{\prime}\Sigma m}\,s.t.\,\mathbf{m^{\prime}1}=1.\tag{12.2} \end{align}$ The Lagrangian for this problem is: $\begin{align*} L(\mathbf{m},\lambda) & =\mathbf{m^{\prime}\Sigma m}+\lambda(\mathbf{m^{\prime}1}-1), \end{align*}$ and the first order conditions (FOCs) for a minimum are: $\begin{align} \mathbf{0} & =\frac{\partial L}{\partial\mathbf{m}}=\frac{\partial\mathbf{m^{\prime}\Sigma m}}{\partial\mathbf{m}}+\frac{\partial\lambda(\mathbf{m^{\prime}1}-1)}{\partial\mathbf{m}}=2\Sigma \mathbf{m}+\lambda\mathbf{1},\tag{12.3}\\ 0 & =\frac{\partial L}{\partial\lambda}=\frac{\partial\mathbf{m^{\prime}\Sigma m}}{\partial\lambda}+\frac{\partial\lambda(\mathbf{m^{\prime}1}-1)}{\partial\lambda}=\mathbf{m^{\prime}1}-1.\tag{12.4} \end{align}$ The FOCs (12.3)-(12.4) give $N+1$ linear equations in $N+1$ unknowns which can be solved to find the global minimum variance portfolio weight vector $\mathbf{m}$ as follows. The $N+1$ linear equations describing the first order conditions have the matrix representation: $\begin{equation} \left(\begin{array}{cc} 2\Sigma & \mathbf{1}\\ \mathbf{1}^{\prime} & 0 \end{array}\right)\left(\begin{array}{c} \mathbf{m}\\ \lambda \end{array}\right)=\left(\begin{array}{c} \mathbf{0}\\ 1 \end{array}\right).\tag{12.5} \end{equation}$ The system (12.5) is of the form $\mathbf{A}_{m}\mathbf{z}_{m}=\mathbf{b}$ , where: $\mathbf{A}_{m}=\left(\begin{array}{cc} 2\Sigma & \mathbf{1}\\ \mathbf{1}^{\prime} & 0 \end{array}\right),~\mathbf{z}_{m}=\left(\begin{array}{c} \mathbf{m}\\ \lambda \end{array}\right)\textrm{ and }\mathbf{b}=\left(\begin{array}{c} \mathbf{0}\\ 1 \end{array}\right).$ Provided $\mathbf{A}_{m}$ is invertible, the solution for $\mathbf{z}_{m}$ is:⁷⁹ $\begin{equation} \mathbf{z}_{m}=\mathbf{A}_{m}^{-1}\mathbf{b}.\tag{12.6} \end{equation}$ The first $N$ elements of $\mathbf{z}_{m}$ is the portfolio weight vector $\mathbf{m}$ for the global minimum variance portfolio with expected return $\mu_{p,m}=\mathbf{m}^{\prime}\mu$ and variance $\sigma_{p,m}^{2}=\mathbf{m}^{\prime}\Sigma \mathbf{m}$ .

Example 2.6 (Global minimum variance portfolio for example data)

Using the data in Table 12.1, we can use R to compute the global minimum variance portfolio weights from (12.6) as follows:

top.mat = cbind(2*sigma.mat, rep(1, 3))
bot.vec = c(rep(1, 3), 0)
Am.mat = rbind(top.mat, bot.vec)
b.vec = c(rep(0, 3), 1)
z.m.mat = solve(Am.mat)%*%b.vec
m.vec = z.m.mat[1:3,1]
m.vec

##  MSFT  NORD  SBUX 
## 0.441 0.366 0.193

Hence, the global minimum variance portfolio has portfolio weights $m_{\textrm{msft}}=0.441$ , $m_{\textrm{nord}}=0.366$ and $m_{\textrm{sbux}}=0.193$ , and is given by the vector $\mathbf{m}=(0.441,0.366,0.193)^{\prime}$ . The expected return on this portfolio, $\mu_{p,m}=\mathbf{m}^{\prime}\mu$ , is:

mu.gmin = as.numeric(crossprod(m.vec, mu.vec))
mu.gmin

## [1] 0.0249

The portfolio variance, $\sigma_{p,m}^{2}=\mathbf{m}^{\prime}\Sigma \mathbf{m}$ , and standard deviation, $\sigma_{p,m}$ , are:

sig2.gmin = as.numeric(t(m.vec)%*%sigma.mat%*%m.vec)
sig.gmin = sqrt(sig2.gmin)
c(sig2.gmin, sig.gmin)

## [1] 0.00528 0.07268

In Figure 12.4, this portfolio is labeled “GMIN” .

Figure 12.4: Global minimum variance portfolio from example data.

$\blacksquare$

12.2.1 Alternative derivation of global minimum variance portfolio

We can use the first order conditions (12.3) - (12.4) to give an explicit solution for the global minimum variance portfolio $\mathbf{m}$ as follows. First, use (12.3) to solve for $\mathbf{m}$ as a function of $\lambda$ : $\begin{equation} \mathbf{m}=-\frac{1}{2}\cdot\lambda\Sigma^{-1}\mathbf{1}.\tag{12.7} \end{equation}$ Next, multiply both sides of (12.7) by $\mathbf{1}^{\prime}$ and use (12.4) to solve for $\lambda$ : $\begin{align*} 1 & =\mathbf{1}^{\prime}\mathbf{m}=-\frac{1}{2}\cdot\lambda\mathbf{1}^{\prime}\Sigma^{-1}\mathbf{1}\\ & \Rightarrow\lambda=-2\cdot\frac{1}{\mathbf{1}^{\prime}\Sigma^{-1}\mathbf{1}}. \end{align*}$ Finally, substitute the value for $\lambda$ back into (12.7) to solve for $\mathbf{m}$ : $\begin{equation} \mathbf{m}=-\frac{1}{2}\left(-2\right)\frac{1}{\mathbf{1}^{\prime}\Sigma^{-1}\mathbf{1}}\Sigma^{-1}\mathbf{1}=\frac{\Sigma^{-1}\mathbf{1}}{\mathbf{1}^{\prime}\Sigma^{-1}\mathbf{1}}.\tag{12.8} \end{equation}$ Notice that (12.8) shows that a solution for $\mathbf{m}$ exists as long as $\Sigma$ is invertible.

Example 2.9 (Finding global minimum variance portfolio for example data)

Using the data in Table 12.1, we can use R to compute the global minimum variance portfolio weights from (12.8) as follows:

one.vec = rep(1, 3)
sigma.inv.mat = solve(sigma.mat)
top.mat = sigma.inv.mat%*%one.vec
bot.val = as.numeric((t(one.vec)%*%sigma.inv.mat%*%one.vec))
m.mat = top.mat/bot.val
m.mat[,1]

##  MSFT  NORD  SBUX 
## 0.441 0.366 0.193

$\blacksquare$

For $\mathbf{A}_{m}$ to be invertible we need $\Sigma$ to be invertible, which requires $|\rho_{ij}|\neq1$ for $i\neq j$ and $\sigma_{i}^{2}>0$ for all $i$ .↩︎