0
$\begingroup$

Can someone explain me where $\frac12$ came from in the expression $\frac{1}{2} \omega'\Sigma \omega$?

That is the expression to be minimized io order to get the minimal variance portfolio (with also some constraints).

$\omega$ is the weights vector and $\Sigma$ is the covariance matrix.

$\endgroup$
4
  • $\begingroup$ The normal distribution. $\endgroup$ Nov 1 '20 at 11:44
  • 7
    $\begingroup$ As it's a constant it has no effect on anything. It's only there so that when you take the derivative with respect to $\omega$ it cancels. This just makes the formulae that you get simpler to write. $\endgroup$
    – Dom
    Nov 1 '20 at 12:13
  • $\begingroup$ Please make this an answer @dom $\endgroup$
    – Bob Jansen
    Nov 1 '20 at 12:35
  • 1
    $\begingroup$ It seemed hardly worth it and now someone else has said this so I'll not duplicate. $\endgroup$
    – Dom
    Nov 1 '20 at 12:42
4
$\begingroup$

This has already been dealt with multiple times. As @Dom explains, the purpose is to simplify partial derivatives.

For exposition's sake, assume there are only two assets with weight vector $\omega=(\omega_1,\omega_2)$, then we seek to minimize a function of the form: $$f(\omega_1,\omega_2)=\frac{1}{2}\left(\omega_1^2\sigma_2^2+\omega_2^2\sigma_2^2+2\omega_1\omega_2\sigma_1\sigma_2\rho\right)-\gamma(\omega_1r_1+\omega_2r_2)$$ where $r_i$ is the expected return, $\sigma_i^2$ the return variance, $\rho$ the return's correlation and $\gamma$ some risk-aversion parameter. Then, $i\not=j$: $$\begin{align} &\frac{\partial}{\partial\omega_i}f(\omega_1,\omega_2) =\frac{1}{\color{blue}{2}}(\color{blue}{2}\omega_i\sigma_i^2+\color{blue}{2}\omega_j\sigma_i\sigma_j\rho)-\gamma r_i =\omega_i\sigma_i^2+\omega_j\sigma_i\sigma_j\rho-\gamma r_i \end{align}$$ It is just for convenience, since, as you can see, the coefficient $\color{blue}{2}$ is cancelled by the $\color{blue}{\frac{1}{2}}$

In general, quadratic optimization programs are usually reformulated with a $\frac{1}{2}$ factor. For example in Machine Learning, standard formulations of Support Vector Machines also include this factor.

$\endgroup$
2
  • 1
    $\begingroup$ is there a covariance term in the 2-asset portfolio variance formula? if so, you forgot it $\endgroup$
    – develarist
    Nov 1 '20 at 15:08
  • $\begingroup$ @develarist Of course, fixed. $\endgroup$ Nov 1 '20 at 18:44
1
$\begingroup$

A lot of things we use in economics and financial economics in particular are inconsequential, but practical. If you have a quadratic program, include this fraction conveniently gets rid of pesky constants in the first order conditions.

Specifically, $\nabla_\omega \omega' \Sigma \omega = (\Sigma + \Sigma') \omega$, using the convention that vectors are column vectors. But since $\Sigma$ is the covariance matrix, $\Sigma = \Sigma'$ and, hence, $\nabla_\omega \omega' \Sigma \omega = 2 \Sigma \omega$. It's not hard to see here why someone would put a fraction in front! Now, if you objective was, say, to minimize the variance of your portfolio $\omega' \Sigma \omega$, possibly subject to some constraint, then for any strictly monotonically increasing function $f : \mathbb{R} \rightarrow \mathbb{R}$, it is the same as minimizing $f(\omega' \Sigma \omega)$ and it so happens that multiply by a strictly positive constant is an example of that sort of function.

Another very common example is working with a Cobb-Douglas utility function. It's a bit faster to see what is going on if you solve the problem after taking the natural logarithm of the utility function.

Slightly more incidental changes can be found in the option pricing literature when GARCH dynamics and discrete time models are used. We usually include a convexity correction term in the mean equation of the underlying price so that when you take the expected value of the exponential, the values related to the error term and this addition cancel out and you're left with a simple mean value.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.