In the book "Portfolio Construction and Risk Budgeting" of Sherer, there is an excercise with the following prompt:

"Use matrix algebra to find the maximum Sharpe ratio portfolio. Show that the Sharpe ratio is linear homogeneous. What does this mean economically ? "

I have found this portfolio ( $ w = \frac{\Sigma^{-1}\mu}{\mu '\Sigma^{-1}\mu}$ ) and computed its sharpe ratio ( $\sqrt{\mu '\Sigma^{-1}\mu}$). So, its is homogeneous but for me it is not linear. Has someone an explanation about this linear homogeneous feature and its economic meaning ? Thank you !

  • $\begingroup$ Could you provide the reference to the exercise please? Edition and page number please $\endgroup$ Jun 27, 2019 at 17:32
  • $\begingroup$ Fourth Edition - Chapter 1: A primer on portfolio theory - Exercise 2 page 43 $\endgroup$ Jun 28, 2019 at 7:52
  • 1
    $\begingroup$ The homogeneous property is meant in the leverage/deleverage sense as in the definition of coherent risk measure. I have seen this homogeneous property in the allocation sense but can’t find the definition used in the book as it seems to come out of nowhere! $\endgroup$ Jun 28, 2019 at 23:02
  • $\begingroup$ "Linear homogeneous" must mean in terms of the allocation weights of the portfolio/overall leverage. This means that economically the Sharpe ratio is leverage-invariant (but not time invariant, which is converting Sharpe from "daily" to "annual" units causes so much trouble). $\endgroup$
    – shabbychef
    Oct 5, 2021 at 17:17

1 Answer 1


The formula you came up with doesn't appear to account for the riskless asset. isn't the maximum Sharpe ratio portfolio $\boldsymbol{\omega}= \frac{\mathbf{\Sigma}^{-1}(\boldsymbol{\mu}-r_f\cdot \boldsymbol{\iota}_N)}{{\boldsymbol{\iota}_N\mathbf{\Sigma}}^{-1}(\boldsymbol{\mu}-r_f\cdot \boldsymbol{\iota}_N)}$ because the Sharpe ratio is $\frac{\boldsymbol{\omega^{\top}\mu}-r_f}{\boldsymbol{\omega}^{\top} \mathbf{\Sigma} \boldsymbol{\omega}}$?

I think linear homogeneity has to do with the constraint that individual portfolio weights must sum to 1: $\boldsymbol\iota_N^{\top}\boldsymbol\omega=1$, or $\omega_1+\omega_2+\dots+\omega_N=1$. The constraint is made homogenous in the Lagrangean derivation of the tangency portfolio by using $\omega_1+\omega_2+\dots+\omega_N -1=0$ instead, which is linear with the rest of the Lagrangean equation in the above analytical solution's derivation due to multiplier $\lambda$ placed at the front of it (see here and search 'homog' then tangency).

Merton, Robert C. “An Analytic Derivation of the Efficient Portfolio Frontier.” The Journal of Financial and Quantitative Analysis, vol. 7, no. 4, 1972, pp. 1851–1872.

As for homogeneous expectations on the other hand, from sidebar link, investors are assumed rational and only use what data they are presented with.

Homogeneous expectations is an assumption in Harry Markowitz's Modern Portfolio Theory that all investors will have the same expectations and make the same choices given a particular set of circumstances. The assumption of homogeneous expectations states that all investors will have the same expectations regarding inputs used to develop efficient portfolios, including asset returns, variances, and covariances.

  • $\begingroup$ In the book, the maximum sharpe ratio portfolio is the portfolio minimising the variance given the condition that $w' \mu = 1$. The formula I gave is correct if I consider $\mu$ as the expected excess return, isn't it ? There is no such constraint of the sum of the weights equal to one. As long as the sum is positive, I can divide by a constant to get this condition without changing the Sharpe Ratio. That is the meaning of homogeneous for me. $\endgroup$ Jun 26, 2019 at 11:43
  • $\begingroup$ if weights summing to 1 ex-ante for me is pretty much the equivalent of you scaling weights ex-post, then how we are defining homogeneous correspond to one another (somehow). $\boldsymbol\iota_N^{\top}\boldsymbol\omega=1$, and for you its $\boldsymbol\omega=\boldsymbol\omega/\boldsymbol\iota_N^{\top}\boldsymbol\omega$. so your solution probably is just derived from a different angle, but mine and your constraints on the weights are imposed at different times during the process $\endgroup$
    – develarist
    Jun 26, 2019 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.