# Prove that the portfolio that maximizes utility lies on the efficient frontier

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

## 1 Answer

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.

• 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? – Wadstk Oct 28 '19 at 14:05
• 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 – Attack68 Oct 28 '19 at 15:13
• But how do we know that the optimizer will find those points on the frontier? – Wadstk Oct 28 '19 at 15:30