<p>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.</p>

<p>The typical mean-variance utility function is given by </p>

<p><span class="math-container">$U(x) = E(r) - \frac{\alpha}{2}Var(r)$</span> </p>

<p>where <span class="math-container">$\alpha$</span> is some risk aversion parameter. The generally accepted way to find the efficient frontier is to minimize the following equation (<a href="https://en.wikipedia.org/wiki/Modern_portfolio_theory" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Modern_portfolio_theory</a>):</p>

<p><span class="math-container">$Var(r) - q E(r)$</span></p>

<p>where, here, <span class="math-container">$q$</span> is a risk tolerance parameter - increasing the value of <span class="math-container">$q$</span> pushes us out onto the frontier. </p>

<p>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 </p>

<p><span class="math-container">$-U(x) = \frac{\alpha}{2}Var(r) - E(r)$</span> </p>

<p>I understand that the <span class="math-container">$\frac{1}{2}$</span> is simply a quadratic programming construct, but I don't know how we can change the risk aversion parameter, <span class="math-container">$\alpha$</span>, to a risk tolerance parameter, <span class="math-container">$q$</span>, and slap it onto <span class="math-container">$E(r)$</span>. 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.</p>

CC BY-SA 4.0
2019-08-27T12:42:58.343Z
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
CC BY-SA 4.0
2019-08-27T14:23:05.563Z
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
CC BY-SA 4.0
2019-08-28T17:33:44.750Z
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 Pretorius
CC BY-SA 4.0
2019-08-28T18:24:05.840Z
@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
CC BY-SA 4.0
2019-08-29T03:37:58.923Z
@Wadstk: Glad to hear it.
mark leeds
<utility><financial-economics><portfolio-theory>
Mean Variance Optimization in a Utility Maximization Framework
Wadstk
# Mean Variance Optimization in a Utility Maximization Framework

(Asked by: username_0 on 2019-08-26)

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.

**Tags:** utility, financial-economics, portfolio-theory
> **Comments:**
> **username_1**: 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.  
> **username_0**: 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?  
> **username_2**: 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 $Var(r)-qE(r)$?  
> **username_1**: @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.  
> **username_1**: @Wadstk: Glad to hear it.  

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## 0 answers

