Deriving Correlation Along the Mean-Variance Efficient Frontier

Summary

The document asks how to derive the correlation between the global minimum-variance portfolio and another portfolio on the mean-variance efficient frontier, given their covariance. The answer applies the definition of correlation: divide covariance by the product of the portfolios’ standard deviations. Using the efficient-frontier variance expressions, it gives a closed-form result involving the expected-return vector and the inverse covariance matrix.

This provides a compact analytical relationship for studying how the minimum-variance portfolio co-moves with other efficient portfolios. The result depends on the stated mean-variance setup and matrix quantities, including the covariance matrix and expected returns. The source gives the formula but does not walk through its intermediate algebra or discuss assumptions such as invertibility, estimation error, or constraints on portfolio weights. It therefore serves as a formula-level derivation aid rather than a practical portfolio construction guide.

Key ideas

• Correlation is obtained by scaling covariance by the square root of the two portfolio variances.
• The efficient-frontier correlation can be expressed using expected returns and the inverse covariance matrix.
• The formula relies on the mean-variance portfolio setup and its matrix assumptions.

Tags

# Correlation between mean-variance efficient portfolios


# Correlation between mean-variance efficient portfolios












If the covariance solution between the returns series of the minimum-variance portfolio ($A$) and any other portfolio along the efficient frontier ($B$) is

$$Cov_{A, B} = \frac{1}{\mathbf{1}^T\mathbf{\Sigma}^{-1}\mathbf{1}}$$ What is the derivation of the closed-form analytical solution for the correlation between those portfolios, $\rho_{A, B}=?$

## Answer by steveo'america (score 1, accepted)

https://quant.stackexchange.com/a/59451

Just divide covariance by the square roots of the two variances. In this case you would want $$ \frac{1/a}{\sqrt{\frac{1}{a}\frac{c}{b^2}}}, $$ which takes value $$ \frac{|1^{\top}\Sigma^{-1}\mu|}{\sqrt{(1^{\top}\Sigma^{-1}1)(\mu^{\top}\Sigma^{-1}\mu)}}. $$

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.