5
$\begingroup$

Suppose we are in the mean-variance optimization setting with a vector of returns $\alpha$ and a vector of portfolio weights $\omega$.

In a robust setting, the returns are assumed to lie in some uncertainty region. I came accross a paper which lets this region, call it $U$, be given by the sphere centered at $\alpha$ with radius $\chi|\alpha|$ where $\chi$ lies between 0 and 1.

The authors then turn their attention to:

$\min_U r_{p}$

and end up with the following solution:

$\min_U r_{p}=\alpha^\intercal\omega-\chi|\alpha||\omega|$ ... ... ... (1)

They do not provide much detail as to how they arrive at this but mention the following: "this uncertainty region corresponds to a one-sigma neighborhood under a Bayesian prior of an uncertain $\alpha$ distributed normally about the estimated $\alpha$, with $\sigma=\chi|\alpha|$..."

Does anyone know how they might have arrived at equation (1)??? Thanks in advance!

$\endgroup$
3
  • 1
    $\begingroup$ Can you post the link to the paper? Reminds me of an Atillio Meucci paper $\endgroup$ Commented Jan 23, 2013 at 14:54
  • $\begingroup$ His book has appendices that show the derivation for robust Bayesian optimization (available at symmys.com) $\endgroup$
    – John
    Commented Jan 23, 2013 at 15:21
  • $\begingroup$ Hi! It is not a Meucci paper but instead a Golts and Jones (2009) working paper. It can be accessed on this link: ssrn.com/abstract=1483412. I have looked at the Meucci papers and books but they do not really help. Perhaps I missed something??? $\endgroup$ Commented Jan 23, 2013 at 16:56

1 Answer 1

3
$\begingroup$

In robust optimization, the true return is not known, we just have a prior $\alpha$ and you have to take into account a possible misestimate which can lower the true return. This is done under the assumption that the posterior return will be within the prior return $\alpha$ plus minus the error being in some $\sigma$-interval.

Now a try for a more formal answer: The posterior return vector is estimated as

$\vec{\alpha} +\vec{\chi}\cdot|\alpha|$ (1)

with $|\vec{\chi}|\leq 1$, or equivalently $\vec{\chi}^{2} \leq 1$. This exactly describes a sphere around $\vec{\alpha}$. Now the return is the product of the return vector $\vec{\alpha}+\vec{\chi}\cdot|\alpha|$ times the weight vector $\vec{\omega}$ :

$r=(\vec{\alpha}+\vec{\chi}\cdot|\alpha|)\cdot\vec{\omega}=\alpha^{T}\omega+\chi^{T}\omega|\alpha|$. (2)

Here, $\vec{\chi}$ can have any orientation. We want the minimum of the second term. $\alpha^{T}\omega$ is minimal if $\vec{\alpha}$ and $\vec{\omega}$ look in opposite direction (property of the dot product), therefore

$\min_U r_P=\alpha^{T}\omega-\chi|\alpha| |\omega|$. (3)

The first term is just the dot product of $\vec{\alpha}$ and $\vec{\omega}$, so it can be written as $|\alpha||\omega|\cos(\phi)$ where $\phi$ is the angle between the two vectors (in n dimensions). This is the next equation in the Golts and Jones working paper:

$\min_U r_P=|\alpha||\omega|(\cos(\phi)-\chi)$. (4)

$\endgroup$
6
  • $\begingroup$ @Phillipe: Can you recommend a good source or reference that will help or that you have used??? $\endgroup$ Commented Jan 24, 2013 at 10:34
  • $\begingroup$ Can you please explain, a little more clearly, how you get to the final equation from the second one? $\endgroup$ Commented Jan 24, 2013 at 20:05
  • $\begingroup$ Regarding the references, I just glimpsed over the nice working paper from Golts and Jones (2009) you cited and the pdf version of an article by Goldfarb and Iyengar (CORC Technical Report TR-2002-03 Robust portfolio selection problems). There, equations (4) and (15) seem to state the same result. Regarding the mathematical steps, I will edit my answer $\endgroup$
    – philippe
    Commented Jan 24, 2013 at 22:47
  • $\begingroup$ From eqn (2) to (3), we want to have the worst possible r. The first term on the right side of (2) is constant in our setting given our initial $\vec{\alpha}$ and $\vec{\omega}$. Now, we want to reduce that by as much as possible. In $\chi'\omega|\alpha|$ |\alpha| is again constant, so we are looking for the smallest value of $\chi'\omega$. This can be written as $|\chi||\omega|\cos(\delta)$ with $|\chi|$ and $|\omega|$ the length of the vectors and $\delta$ the angle between them. This is minimal for $\delta=\pi$ (ie both vectors have 180 degrees between them and look in opposite directions) $\endgroup$
    – philippe
    Commented Jan 24, 2013 at 23:15
  • $\begingroup$ Continuing the last comment: This gives us $-|\chi||\omega|$ (because $\cos(\delta)$ is $-1$ at that angle). If we now use the notation $\chi=|\chi|$ and $\chi$ being in the interval $[0,1]$ by assumption, we arrive at equation (3). Does that explanation help? $\endgroup$
    – philippe
    Commented Jan 24, 2013 at 23:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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