Closed-Form Variance of the Global Minimum-Variance Portfolio
Summary
The document derives the variance of the global minimum-variance portfolio from its closed-form weights. With a symmetric covariance matrix, substituting the weights into the quadratic portfolio-variance expression simplifies the result to the reciprocal of the sum of all entries in the inverse covariance matrix. This gives a direct calculation without first evaluating the full quadratic form.
It also presents related formulas using three scalar quantities formed from the inverse covariance matrix, the vector of ones, and expected returns. These give the minimum-variance and tangency portfolio expected returns and variances, along with their covariance. The derivations assume an invertible covariance matrix and the standard unconstrained portfolio setup; they do not address constraints such as long-only weights, estimation error, or practical robustness of the inputs.
Key ideas
- The global minimum-variance portfolio variance equals the reciprocal of the all-ones quadratic form in the inverse covariance matrix.
- The simplification relies on the covariance matrix being symmetric and invertible.
- The response gives closed-form expressions for the expected return and variance of the minimum-variance portfolio.
- It also relates the tangency portfolio variance and its covariance with the minimum-variance portfolio to three inverse-covariance scalars.
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# Closed-form analytical solution for the variance of the minimum-variance portfolio?
# 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}
$$$$
How about the maximum-Sharpe ratio portfolio's variance as well?
## Answer by Kermittfrog (score 6, accepted)
https://quant.stackexchange.com/a/59149
Let
\begin{align} a&\equiv \mathbf{1}^T\mathbf{\Sigma}^{-1}\mathbf{1}\\ b&\equiv \mathbf{1}^T\mathbf{\Sigma}^{-1}\boldsymbol{\mu}\\ c&\equiv \boldsymbol{\mu}^T\mathbf{\Sigma}^{-1}\boldsymbol{\mu} \end{align}
Then \begin{align} \mathrm{E(minVarPortfolio)}& = \frac{b}{a}\\ \mathrm{V(minVarPortfolio)}& = \frac{1}{a}\\ \mathrm{E(TangencyPortfolio)}& = \frac{c}{b}\\ \mathrm{V(TangencyPortfolio)}& = \frac{c}{b^2}\\ \mathrm{Cov(MVP,Tangency)}& = \frac{1}{a}\\ \end{align}
Effectively, the covariance between any efficient portfolio and the MVP is $1/a$.
## Answer by RRL (score 9)
https://quant.stackexchange.com/a/59158
A few more steps beyond your last equation gives the answer.
With $C = \mathbf{1}^T\mathbf{\Sigma}^{-1}\mathbf{1}$, we have
$$\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}}$$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.