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.

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    $\begingroup$ 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. $\endgroup$ – mark leeds Aug 27 '19 at 12:42
  • $\begingroup$ 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? $\endgroup$ – Wadstk Aug 27 '19 at 14:23
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    $\begingroup$ 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 $Var(r)-qE(r)$? $\endgroup$ – Angela Pretorius Aug 28 '19 at 17:33
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    $\begingroup$ @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. $\endgroup$ – mark leeds Aug 28 '19 at 18:24
  • $\begingroup$ Thanks, It is clear to me now $\endgroup$ – Wadstk Aug 28 '19 at 20:17

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