# Mean Variance Optimization in a Utility Maximization Framework

I'm struggling to gain a broad understanding of Mean-Variance utility theory as it relates to finding the efficient frontier of a group of assets which each have some return and variance.

The typical mean-variance utility function is given by

$$U(x) = E(r) - \frac{\alpha}{2}Var(r)$$

where $$\alpha$$ is some risk aversion parameter. The generally accepted way to find the efficient frontier is to minimize the following equation (https://en.wikipedia.org/wiki/Modern_portfolio_theory):

$$Var(r) - q E(r)$$

where, here, $$q$$ is a risk tolerance parameter - increasing the value of $$q$$ pushes us out onto the frontier.

At first, I thought that the equation to be minimized was just the negative of the utility function, but the negative of the utility function is

$$-U(x) = \frac{\alpha}{2}Var(r) - E(r)$$

I understand that the $$\frac{1}{2}$$ is simply a quadratic programming construct, but I don't know how we can change the risk aversion parameter, $$\alpha$$, to a risk tolerance parameter, $$q$$, and slap it onto $$E(r)$$. This makes me think that I am missing something significant in this framework as it relates to utility, indifference curves, and the frontier. Hoping for some clarification.

• Hi: It's just a factor that is trading off variance for return so, whether you put it on Var(r) or E(r) doesn't make a difference when you actually do the optimization. Also, as you pointed out, the 1/2 is a scale factor and not necessary. – mark leeds Aug 27 '19 at 12:42
• Thank you for your comment. It seems to flip the axis in mean-standard deviation space so that E(r) is on the x axis and Var(r) is on the y axis. Is the minimization problem an inverse function? – Wadstk Aug 27 '19 at 14:23
• The value of $r$ which minimises $-U(r)$ is the same as the value of $r$ which minimises $-\frac{2}{\alpha}U(r)$. Can you write $-\frac{2}{\alpha}U(r)$ in the form $Var(r)-qE(r)$? – Angela Richardson Aug 28 '19 at 17:33
• @Wadstk: The software you are using probably treats one term as the x-axis and the other term as the y-axis but that's irrelevant also. It's still the same function being minimized. So it's not an inverse function but it just chooses the y-axis and x-axis probably based on which term includes the tradeoff parameter. – mark leeds Aug 28 '19 at 18:24
• Thanks, It is clear to me now – Wadstk Aug 28 '19 at 20:17