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In the article "On the properties of equally-weighted risk contributions portfolios" of Maillard, Roncalli and Teiletche, some general results are derived.

One of them states that, if correlation $\rho$ is constant for each couple of variable, the weights $x_i$ of the ERC portfolio need to satisfy the condition $ \sigma_i(x) = \sigma_j(x)$ and with the hypothesis made just above, it is equivalent to have $x_i \sigma_i = x_j \sigma_j$. Where $ \sigma_i(x) = x_i \sigma_i( (1 - \rho)x_i \sigma_i + \rho \sum_k x_k \sigma_k ) / \sigma(x) $ and $\sigma_i$ the vol of the asset i and $\sigma(x)$ the vol of the portfolio.

I did not manage to prove the result.

I have shown that it is equivalent to have : $(x_i\sigma_i - x_j\sigma_j)( (1-\rho )(x_i\sigma_i + x_j\sigma_j) + \rho\sum_k x_k \sigma_k) = 0$

The idea would be to show that the second parenthesis is different from 0, but I did not success.

One indication of the authors is " We use the fact that constant correlation verifies $\rho \geq - \frac{1}{n-1}$".

Thank you for you help.

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Ok, I found a solution !

So, we are starting from $(x_i\sigma_i - x_j\sigma_j)((x_i\sigma_i + x_j\sigma_j)(1 - \rho) + \rho\sum_k x_k \sigma_k) = 0 $ and we will show that the elements in the second parenthesis is greater than $0$. We have:

$(x_i\sigma_i + x_j\sigma_j)(1 - \rho) + \rho\sum_k x_k \sigma_k = (x_i\sigma_i + x_j\sigma_j) + \rho(\sum_k x_k \sigma_k - x_i\sigma_i - x_j\sigma_j ) $

Note that $a = x_i\sigma_i + x_j\sigma_j$ and $ b = \sum_k x_k \sigma_k - x_i\sigma_i - x_j\sigma_j $ are positive, so the line defined by $ y = a + bx$ is increasing and crosses the horizontal axis for a negative $x$ (as $a$ is positive). We will denote this $x$ by $x^*$.

So $a + bx^* = 0 \iff x^* = \frac{-a}{b} \iff x^* = \frac{- (x_i\sigma_i + x_j\sigma_j)}{\sum_k x_k \sigma_k - x_i\sigma_i - x_j\sigma_j} \iff x^* = \frac{-1}{\frac{\sum_k x_k \sigma_k - x_i\sigma_i - x_j\sigma_j}{x_i\sigma_i + x_j\sigma_j}}$

Recall, that as we suppose constant correlation, we necessarily have $\rho \geq - \frac{1}{n-1}$

So, to get our result, we need to show that $x^* < - \frac{1}{n-1} \iff \frac{\sum_k x_k \sigma_k - x_i\sigma_i - x_j\sigma_j}{x_i\sigma_i + x_j\sigma_j} < n-1 $.

We have:

$\frac{\sum_k x_k \sigma_x - x_i\sigma_i - x_j\sigma_j}{x_i\sigma_i + x_j\sigma_j} = \frac{\sum_k x_k\sigma_k }{x_i\sigma_i + x_j\sigma_j} - 1$

Suppose that $\frac{\sum_k x_k\sigma_k}{x_i\sigma_i + x_j\sigma_j} \geq n $,

$\iff \frac{x_i\sigma_i + x_j\sigma_j} {\sum_k x_k\sigma_k} \leq \frac{1}{n} \iff \frac{x_i\sigma_i} {\sum_k x_k\sigma_k} + \frac{ x_j\sigma_j} {\sum_k x_k\sigma_k} \leq \frac{1}{n} $

As it is true for all $i$, I got that:

$\frac{x_1\sigma_1 + x_2\sigma_2 + ..+x_n\sigma_n } {\sum_k x_k\sigma_k} + \frac{ nx_j\sigma_j} {\sum_k x_k\sigma_k} \leq n\frac{1}{n} \iff 1 +\frac{ nx_j\sigma_j} {\sum_k x_k\sigma_k} \leq 1 $

It implies that $\frac{ nx_j\sigma_j} {\sum_k x_k\sigma_k} \leq 0$ which is false.

Thus, $\frac{\sum_k x_k\sigma_k }{x_i\sigma_i + x_j\sigma_j} - 1 < n - 1$

And so, the second term in the parenthesis is strictly greater than $0$ and we got the result $x_i\sigma_i = x_j \sigma_j$

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