# Robust-Bayesian optimization in Markowitz framework

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!

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Can you post the link to the paper? Reminds me of an Atillio Meucci paper – Ram Ahluwalia Jan 23 '13 at 14:54
His book has appendices that show the derivation for robust Bayesian optimization (available at symmys.com) – John Jan 23 '13 at 15:21
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??? – Geraldine Bailey Jan 23 '13 at 16:56

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)

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@Phillipe: Can you recommend a good source or reference that will help or that you have used??? – Geraldine Bailey Jan 24 '13 at 10:34
Can you please explain, a little more clearly, how you get to the final equation from the second one? – Geraldine Bailey Jan 24 '13 at 20:05
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 – philippe Jan 24 '13 at 22:47
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) – philippe Jan 24 '13 at 23:15
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? – philippe Jan 24 '13 at 23:22