0
$\begingroup$

When maximizing mean-variance utility in a portfolio optimization framework

$max \{R - \lambda \sigma ^2\}$

where R is portfolio return, $\lambda$ is a risk aversion parameter, and $\sigma^2$ is portfolio volatility, how can I be sure that the result lies on the efficient frontier? I can show that $\lambda$ is effectively equals $\frac{R}{\sigma^2}$ but I don't quite see how this problem develops the efficient frontier

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

The efficient frontier is defined as the set of portfolios which have the highest return for a given measure of volatility, i.e. $\{S: s \in P \; s.t. \nexists \; t \in P \; \text{where} \;R(s) < R(t) \; \text{and} \; \sigma(s)=\sigma(t) \}$, where $P$ is the set of all validly constructed portfolios.

Therefore this also holds for the efficient frontier when the risk is squared, i.e. a one-to-one mapping for risk to variance.

The optimisation framework $max \{R-\lambda\sigma^2 \}$ must therefore return the efficient frontier since by definition there does not exist a valid portfolio for a given risk or variance where $R$ is greater and therefore increases the objective function above that of the efficient frontier.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks. If I understand it correctly, selecting values of $\lambda$ is equivalent to selecting values of $\frac{R}{\sigma^2}$ which is effectively a point on the efficient frontier. You are asking the optimizer to find the value of $R$ and $\sigma^2$ on the point on the frontier where the slope equals $\lambda$. But what if there are multiple points on the efficient frontier where the slope equals $\lambda$ ? In that case, there isnt a unique solution. Am i thinking about this correctly? $\endgroup$
    – Wadstk
    Oct 28, 2019 at 14:05
  • $\begingroup$ Consider two points on the eifficent frontier (R,s^2) = (3,2) and (4,3). If lambda is 1 both points are equivalent under the maximisation scheme. If lambda is <1 then return is preferred and (4,3) is optimal. If lambda is >1 then you are more risk averse and (3, 2) is optimal. Varying lambda and solving will construct the efficient frontier $\endgroup$
    – Attack68 ♦
    Oct 28, 2019 at 15:13
  • $\begingroup$ But how do we know that the optimizer will find those points on the frontier? $\endgroup$
    – Wadstk
    Oct 28, 2019 at 15:30
  • $\begingroup$ Becuase that is what optimizers do. They are algorithms designed to find the optimal points and the optimal points, by definition, lie on the efficient frontier, hence those points will be found by an optimiser. $\endgroup$
    – Attack68 ♦
    Jul 9 at 18:15

Your Answer

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