# Deriving Theoretical Result - ERC portfolio

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.

## 1 Answer

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 $$x_i\sigma_i = x_j \sigma_j$$