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I have a set of $N$ data series, which I put together to form a single series using a weights vector $x$. The covariance matrix is known ($\Sigma$).

We know that the volatility of the weighted series is:

$\sigma(x)=\sqrt{x^T ~ \Sigma ~ x}$

where $\sigma \in R^N$ and $\Sigma \in R^{\{N \times N\}}$.

I use Euler decomposition to do the following:

$\sigma(x) = \sum_{i=1}^N x_i \frac{\partial \sigma}{\partial x_i}$

So far, I'm happy with this.

What I want to do is to use an algorithm so that it finds me the vector $x$ such that all the series have the same contribution $x_i \partial_i \sigma(x)$ to the "global" volatility.

The optimization would be done using the following idea:

$x^* = \arg \min_x \sum_{i=1}^n \left[\frac{\sigma(x)}{n} - x_i \partial_i \sigma(x) \right]^2$

But I am struggling to compute the value of $\frac{\partial \sigma}{\partial x_i}$ to be able to run the algorithm.

I'd start by the classic step (1) :

$\frac{\partial \sigma}{\partial x_i} = \frac{\partial (\sqrt{x^T ~ \Sigma ~ x})}{\partial x_i}=-\frac{1}{2} \sigma(x)^{-\frac{3}{2}} \cdot \frac{\partial (x^T ~ \Sigma ~ x)}{\partial x_i}$

But I am not even sure this step is right, and I don't know how to carry on.

Could tell me if my initial step (1) is right, and how to proceed to get the value of $\partial_i \sigma(x)$?

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    @HenningMakholm: I'm sorry it did it too quickly. I should be ok now. I added an introduction and an extra paragraph.2011-12-19

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I found the answer I was looking for in a paper, I thought I would post it here, although the question was maybe a bit unclear.

The answer is the following:

$\partial_{x_i} \sigma(x) = \frac{(\Sigma x)_i}{\sigma(x)}$

I can hence compute the vector of the derivates for all $i$ as follows:

$\frac{\Sigma x}{\sqrt{x^T \Sigma x}}$

Back to the Euler decomposition, I can verify it by doing:

$\sum_{i=1}^N x_i \partial_{x_i} \sigma(x)=\sum_{i=1}^N x_i \frac{(\Sigma x)_i}{\sigma(x)}=\frac{1}{\sigma(x)} \sum_{i=1}^N x_i (\Sigma x)_i=\frac{1}{\sigma(x)} x^T \Sigma x=\frac{x^T \Sigma x}{\sqrt{x^T \Sigma x}}=\sqrt{x^T \Sigma x}=\sigma(x)$

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    Great, thanks. :)2011-12-22