# Closed-form analytical solution for the variance of the minimum-variance portfolio?

The portfolio weights vector of the minimum-variance portfolio has a closed-form analytical solution,

$$\boldsymbol{w} = \frac{\boldsymbol{\Sigma}^{-1} \boldsymbol{1} }{\boldsymbol{1}^\top \boldsymbol{\Sigma}^{-1} \boldsymbol{1}}$$

but is there a direct calculation for the same portfolio's variance $$\sigma_p^2$$?

Given that $$\sigma_p^2 = \boldsymbol{w^\top \Sigma w}$$, what is the simplification of

\begin{aligned} \sigma_p^2 & = \left( \frac{\boldsymbol{\Sigma}^{-1} \boldsymbol{1} }{\boldsymbol{1}^\top \boldsymbol{\Sigma}^{-1} \boldsymbol{1}}\right)^\top \cdot \boldsymbol{\Sigma} \cdot \frac{\boldsymbol{\Sigma}^{-1} \boldsymbol{1} }{\boldsymbol{1}^\top \boldsymbol{\Sigma}^{-1} \boldsymbol{1}} \\ & = \frac{\boldsymbol{1} ^\top(\boldsymbol{\Sigma}^\top)^{-1}}{\boldsymbol{1} ^\top\boldsymbol{\Sigma}^{-1} \boldsymbol{1} } \cdot \boldsymbol{\Sigma} \cdot \frac{\boldsymbol{\Sigma}^{-1} \boldsymbol{1} }{\boldsymbol{1}^\top \boldsymbol{\Sigma}^{-1} \boldsymbol{1}} \\ & = ? \end{aligned}



How about the maximum-Sharpe ratio portfolio's variance as well?

• Not sure I understand the question fully. Isn't any portfolio variance defined as $\sigma_P^2 = w' \Sigma w$? So once you have your weights, the variance should be easy? Nov 5, 2020 at 8:18
• I'ved edited with detail Nov 5, 2020 at 8:37

Let

\begin{align} a&\equiv \mathbf{1}^T\mathbf{\Sigma}^{-1}\mathbf{1}\\ b&\equiv \mathbf{1}^T\mathbf{\Sigma}^{-1}\boldsymbol{\mu}\\ c&\equiv \boldsymbol{\mu}^T\mathbf{\Sigma}^{-1}\boldsymbol{\mu} \end{align}

Then \begin{align} \mathrm{E(minVarPortfolio)}& = \frac{b}{a}\\ \mathrm{V(minVarPortfolio)}& = \frac{1}{a}\\ \mathrm{E(TangencyPortfolio)}& = \frac{c}{b}\\ \mathrm{V(TangencyPortfolio)}& = \frac{c}{b^2}\\ \mathrm{Cov(MVP,Tangency)}& = \frac{1}{a}\\ \end{align}

Effectively, the covariance between any efficient portfolio and the MVP is $$1/a$$.

• does the last sentence lead to analytical solutions for the many frontier portfolios that lie between the min-var and max-Sharpe, and their respective portfolio moments? Nov 5, 2020 at 11:13
• Yes, you can analytically derive the volatility of an efficient portfolio $\sigma_p$ as a function of its expected return, $\mu$, i.e. $\sigma(\mu)$. This holds only for equality constraint portfolio optimization, though. Nov 5, 2020 at 11:53
• That's basic algebra. $w_{MVP}=\frac{\Sigma^{-1}\mathbf{1}}{\mathbf{1}^T\Sigma^{-1}\mathbf{1}}$ and $w_{T}=\frac{\Sigma^{-1}\mathbf{\mu}}{\mathbf{1}^T\Sigma^{-1}\mathbf{\mu}}$. Then the covariance is $w_{MVP}^T\Sigma w_{T}$ which equals $1/a$. Nov 6, 2020 at 9:40
• @develarist, I notice that you ask many questions around (multivariate) normality, linear algebra and such. If you are such inclined, I suggest you study the materials on LinAlg in portfolio optimisation, e.g. here faculty.washington.edu/ezivot/econ424/portfolioTheoryMatrix.pdf Nov 10, 2020 at 11:43
• I think I understand what you want to say. Technically, it really doesn't matter how returns are distributed: As long as the first two moments exist, the Markowitz approach can be used to form mean/variance optimal portfolios. If the agent exhibits quadratic utility, such a portfolio would be expected utility maximizing. For general distributions and general utility functions, of course, the Markowitz approach is not optimal in a utility sense. Nov 10, 2020 at 14:15

A few more steps beyond your last equation gives the answer.

With $$C = \mathbf{1}^T\mathbf{\Sigma}^{-1}\mathbf{1}$$, we have

$$\sigma_P^2 = [C^{-1} \mathbf{\Sigma}^{-1}\mathbf{1}]^T \mathbf{\Sigma} [C^{-1}\mathbf{\Sigma}^{-1}\mathbf{1}] = C^{-2}\mathbf{1}^T(\mathbf{\Sigma}^{-1})^T\mathbf{\Sigma} \mathbf{\Sigma}^{-1}\mathbf{1}$$

Since $$[(\mathbf{\Sigma}^{-1})^T\mathbf{\Sigma}^T]^T = \mathbf{\Sigma}\mathbf{\Sigma}^{-1} = \mathbf{I} = \mathbf{I}^T$$, it follows that $$(\mathbf{\Sigma}^{-1})^T= (\mathbf{\Sigma}^T)^{-1}$$. As the covariance matrix is symmetric, this implies $$(\mathbf{\Sigma}^{-1})^T= \mathbf{\Sigma}^{-1}$$.

Thus,

$$\sigma_P^2 = C^{-2}\mathbf{1}^T\mathbf{\Sigma}^{-1}\mathbf{\Sigma} \mathbf{\Sigma}^{-1}\mathbf{1}= C^{-2}\mathbf{1}^T\ \mathbf{\Sigma}^{-1}\mathbf{1}= C^{-2}C = \frac{1}{ \mathbf{1}^T\mathbf{\Sigma}^{-1}\mathbf{1}}$$

• can $\frac{1}{ \mathbf{1}^T\mathbf{\Sigma}^{-1}\mathbf{1}}$ be reduced to a non-fraction such as a simplified version of $(\mathbf{1}^T\mathbf{\Sigma}^{-1}\mathbf{1})^{-1}$? I am not familiar with matrix fraction operations Nov 6, 2020 at 9:47
• @develarist: The term $C= \mathbf{1}^T\mathbf{\Sigma}^{-1}\mathbf{1}$ is a number -- it is the product of a $1\times n$ vector, an $n \times n$ matrix, and an $n \times 1$ vector. So $C^{-1} = 1/C$. (That is how I factored out $C^{-2}$ in the first step.)
– RRL
Nov 6, 2020 at 17:46
• wasn't my question. how can 1/$C$ be reduced to a non-fraction Nov 6, 2020 at 18:22
• The quadratic form $C= \mathbf{1}^T\mathbf{\Sigma}^{-1}\mathbf{1}$ is the sum of the elements of the inverse matrix $\mathbf{\Sigma}^{-1}$. An element $s_{ij}$ of $\mathbf{\Sigma}^{-1}$ is of the form $s_{ij} = \frac{C_{ji}}{\det \mathbf{\Sigma}}$ where $C_{ji}$ is the cofactor -- determinant of the submatrix of $\mathbf{\Sigma}$ obtained by deleting row j and column i. Hence, $\frac{1}{C} = \frac{\det\mathbf{\Sigma}}{\sum_i \sum_j C_{ij}}$
– RRL
Nov 7, 2020 at 2:40
• Does $\frac{1}{C} = \frac{\det\mathbf{\Sigma}}{\sum_i \sum_j C_{ij}} = \det \mathbf{\Sigma} (\sum_i \sum_j C_{ij})^{-1}$ qualify as a non-fraction? I can't simplify it any further.
– RRL
Nov 7, 2020 at 2:49